SOLUTIONS MANUAL for Introduction to Stochastic Finance with Market Examples By Nicolas Privault. by StudyGuide

Solutions Manual hence the risk-neutral probability measure P∗ is given by p∗ = P∗ (S1 = 5) =

2 3

and q ∗ = P∗ (S1 = 2) =

1 . 3

Exercise 1.3 a) Each of the stated conditions yields one equation, i.e.    4α + β = 1 α = 2 with solution   5α + β = 3, β = −7. Therefore, the portfolio allocation at t = 0 consists to purchase α = 2 unit of the asset S priced S0 = 4, and to borrow −β = $7 in cash. We can check that the price V0 = αS0 + β of the initial portfolio at time t = 0 is V0 = αS0 + β = 2 × 4 − 7 = $1. b) This loss is expressed as α × $2 + β = 2 × 2 − 7 = −$3. Note that the $1 received when selling the option is not counted here because it has already been fully invested into the portfolio. Exercise 1.4 a)

i) Does this model allow for arbitrage? Yes | ✓

No |

ii) If this model allows for arbitrage opportunities, how can they be realized? By shortselling | By borrowing on savings | ✓ N.A. | b)

i) Does this model allow for arbitrage? Yes |

No | ✓

ii) If this model allows for arbitrage opportunities, how can they be realized? By shortselling | By borrowing on savings | N.A. | ✓ c)

i) Does this model allow for arbitrage? Yes | ✓

No |

ii) If this model allows for arbitrage opportunities, how can they be realized? By shortselling | ✓ By borrowing on savings | N.A. | Exercise 1.5

2 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition a) We need to search for possible risk-neutral probability measure(s) P∗ such (1) (1) that IE∗ S1 = (1 + r)S0 . Letting  ! (1) (1)  S1 − S0  (1) (1) ∗ ∗ ∗  =a ,  p = P S1 = S0 (1 + a) = P  (1)  S0       !  (1) (1)  S1 − S0 (1) (1) ∗ ∗ ∗ =b , θ = P S1 = S0 (1 + b) = P (1)  S0       !  (1) (1)   S1 − S0  (1) (1) ∗ ∗ ∗  q = P S = (1 + c)S = P = c ,  1 0  (1) S0 We have   (1 + a)p∗ S0(1) + (1 + b)θ∗ S0(1) + (1 + c)q ∗ S0(1) = (1 + r)S0(1)  ∗ p + θ∗ + q ∗ = 1, from which we obtain  ∗  p a + θ∗ b + q ∗ c = r,  ∗ p + θ∗ + q ∗ = 1.

=⇒

 (1 − θ∗ )c + θ∗ b − r  ∗  ∈ (0, 1),  p = c−a  ∗ ∗    q ∗ = r − (1 − θ )a − θ b ∈ (0, 1). c−a

In order for p∗ and q ∗ to belong to the interval (0, 1) we should have   0 < (1 − θ∗ )c + θ∗ b − r < c − a,  i.e.

0 < r − (1 − θ∗ )a − θ∗ b < c − a, r−a c−r ∗    b − c < θ < c − b,

   r − c < θ∗ < r − a . b−a b−a Therefore there exists an infinity of risk-neutral probability measures depending on the value of r−a r−c c−r r−a θ∗ ∈ max , , min , , b−c b−a c−b b−a in which case the market is without arbitrage but not complete.

3 September 15, 2022

Solutions Manual b) Hedging a claim with possible payoff values Ca , Cb , Cc would require to solve  (1) (0)  (1 + a)αS0 + (1 + r)βS0 = Ca      (1) (0) (1 + b)αS0 + (1 + r)βS0 = Cb       (1) (0) (1 + c)αS0 + (1 + r)βS0 = Cc , for α and β, which is not possible in general due to the existence of three conditions with only two unknowns. Exercise 1.6 a) The risk-neutral condition IE∗ [R1 ] = 0 reads bP∗ (R1 = b) + 0 × P∗ (R1 = 0) + (−b) × (R1 = −b) = bp∗ − bq ∗ = 0, hence p∗ = q ∗ =

1 − θ∗ , 2

since p∗ + q ∗ + θ∗ = 1. b) We have " # (1) (1) ∗ S1 − S0 Var = IE∗ R12 − (IE∗ [R1 ])2 (1) S0 = IE∗ R12 = b2 P∗ (R1 = b) + 02 × P∗ (R1 = 0) + (−b)2 × (R1 = −b) = b2 (p∗ + q ∗ ) = b2 (1 − θ∗ ) = σ2 , hence θ∗ = 1 − σ 2 /b2 , and therefore p∗ = q ∗ =

σ2 1 − θ∗ = 2, 2 2b

provided that σ 2 ≤ 2b2 . Exercise 1.7 a) We denote the risk-neutral measure by p∗ = P∗ (S1 = 2), q ∗ = P∗ (S1 = 1). i) Yes |

No | ✓

Comment: No loss is possible, while a 100%

profit is possible with non-zero probability 1/3. 4

4 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition ii) Yes |

No | ✓

Comment: The (unique) risk-neutral measure

(p , q ) = (0, 1) is given by ∗

∗

$2 × p∗ + $1 × q ∗ = $1 × (1 + r) = $1 and p∗ + q ∗ = 1, and is not equivalent to P given by (p, q) = (1/3, 2/3). b) We denote the risk-neutral measure by p∗ = P∗ (S1 = 2), θ∗ = P∗ (S1 = 1), q ∗ = P∗ (S1 = 0). i) Yes | ✓

No |

Comment: The risk-neutral measure (p∗ , θ∗ , q ∗ )

is given by the equations $2 × p∗ + $1 × θ∗ + $0 × q ∗ = $1 × (1 + r) = $1 and p∗ + θ∗ + q ∗ = 1, (S.1.1) which clearly admit solutions, see (biv) below. ii) Yes | ✓

No |

Comment: Realizing arbitrage would mean

building a portfolio achieving no strictly negative return with probability one, which is impossible since the probability of 100% loss is P(S1 = 0) = 1 − 1/4 − 1/9 = 23/36 > 0. iii) Yes |

No | ✓

Comment: Examples of claims that cannot be

attained can be easily constructed in this market. For example, the claim 1{S1 >0} cannot be attained since there is no portfolio allocation (α, β) satisfying   $2 × α + β = $1 $1 × α + β = $1   $0 × α + β = $0. iv) Yes |

No | ✓

Comment: The risk-neutral measure is clearly

not unique, as for example (p∗ , θ∗ , q ∗ ) = (1/4, 1/2, 1/4) and (p∗ , θ∗ , q ∗ ) = (1/3, 1/3, 1/3) are both solutions of (S.1.1). Exercise 1.8 a) The possible values of R are a and b. b) We have IE∗ [R] = aP∗ (R = a) + bP∗ (R = b) "

5 September 15, 2022

Solutions Manual b−r r−a +b b−a b−a = r. =a

c) By Theorem 1.6, there do not exist arbitrage opportunities in this market since from Question (b) there exists a risk-neutral probability measure P∗ whenever a < r < b. d) The risk-neutral probability measure is unique hence the market model is complete by Theorem 1.13. e) Taking β−α α(1 + b) − β(1 + a) and ξ = , η= π1 (b − a) S0 (b − a) we check that

  ηπ1 + ξS0 (1 + a) = α if R = a, 

which shows that

ηπ1 + ξS0 (1 + b) = β if R = b, ηπ1 + ξS1 = C

in both cases R = a and R = b. f) We have π0 (C) = ηπ0 + ξS0 α(1 + b) − β(1 + a) β − α = + (1 + r)(b − a) b−a α(1 + b) − β(1 + a) − (1 + r)(α − β) = (1 + r)(b − a) αb − βa − r(α − β) = . (1 + r)(b − a)

(S.1.2)

g) We have IE∗ [C] = αP∗ (R = a) + βP∗ (R = b) b−r r−a =α +β . b−a b−a

(S.1.3)

h) Comparing (S.1.2) and (S.1.3) above we do obtain π0 (C) =

1 IE∗ [C] 1+r

i) The initial value π0 (C) of the portfolio is interpreted as the arbitragefree price of the option contract and it equals the expected value of the discounted payoff. j) We have 6

6 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

C = (K − S1 ) = (11 − S1 ) = +

  11 − S1 if K > S1 , 

if K ≤ S1 .

k) We have S0 = 1, a = 8, b = 11, α = 2, β = 0, hence ξ= and

0−2 2 β−α = =− , S0 (b − a) 11 − 8 3

24 α(1 + b) − β(1 + a) = . π1 (b − a) 3 × 1.05

η=

l) The arbitrage-free price π0 (C) of the contingent claim with payoff C is π0 (C) = ηπ0 + ξS0 = 6.952. Exercise 1.9 Letting R denote the price of one right, it will require 10R/3 to purchase one stock at €6.35, hence absence of arbitrage tells us that 10 R + 6.35 = 8, 3

from which it follows that R=

3 (8 − 6.35) = €0.495. 10

Note that the actual share right was quoted at €0.465 according to market data. See also Exercise 17.8 for the pricing of convertible bonds. Exercise 1.10 Let a := (152 − 180)/180 = −7/45 and b := (203 − 180)/180 = 23/180 denote the potential market returns, with r = 0.03. From the strike price K and the risk-neutral probabilities p∗r =

r−a = 0.6549 b−a

and qr∗ =

b−r = 0.3451, b−a

the price of the option at the beginning of the year is given from Proposition 1.16 as the discounted expected value 1 1 IE∗ [(K − S1 )+ ] = p∗ (K − 203)+ + qr∗ (K − 152)+ . 1+r 1+r r Equating this price with the intrinsic value (K − 180)+ of the put option yields the equation (K − 180)+ = "

1 p∗ (K − 203)+ + qr∗ (K − 152)+ 1+r r 7 September 15, 2022

Solutions Manual which requires K > 180 (the case K ≤ 152 is not considered because both the option price and option payoff vanish in this case). Hence we consider the equation K − 180 =

1 p∗ (K − 203)+ + qr∗ (K − 152)+ , 1+r r

with the following cases. i) If K ∈ [180, 203] we get (1 + r)(K − 180) = qr∗ (K − 152), hence K=

(1 + r)180 − qr∗ 152 (1 + r)180 − qr∗ 152 = = 194.11. ∗ 1 + r − qr p∗r + r

ii) If K ≥ 203 we find K=

180(1 + r) − 203p∗r − 152qr∗ < 203, r

which is out of range and leads to a contradiction. We note that the above formula K=

(1 + r)180 − qr∗ 152 28b − 180a + r(180(b − a) + 152) = p∗r + r (b + 1 − a)r − a

yields a decreasing function K(r) of r in the interval [0, 100%], although the function is not monotone over R+ . 150 140 130

K(r)

120 110 100 90 80 70 60

0.5

1.5

Fig. S.1: Strike price as a function of risk-free rate r.

8 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

Chapter 2 Exercise 2.1 Let m := $2, 550 denote the amount invested each year. a) By (2.1), the value of the plan after N = 10 years becomes m

N X

(1 + r)k = m(1 + r)

k=1

(1 + r)N − 1 , r

which in turns becomes (1 + r)N m

N X

(1 + r)k = m(1 + r)N +1

k=1

(1 + r)N − 1 , r

after N additional years without further contributions to the plan. Equating (1 + r)N − 1 A = 30835 = m(1 + r)N +1 r shows that 2N +1 N +1 A (1 + r) − (1 + r) = , r m with m = 2550, or (1 + r)21 − (1 + r)11 30835 = ≃ 12.09215, r 2550 hence r ≃ 1.23% according to Figure S.2, which is typical of an annual fixed deposit interest rate. 14 13.5 13 12.5 12 11.5 11 10.5 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 r in %

Fig. S.2: Graph of r 7−→ ((1 + r)21 − (1 + r)11 )/r. In the hypothesis r = 3.25% we would find A = m(1 + r)N +1 "

(1 + r)N − 1 = 42040.42. r 9 September 15, 2022

Solutions Manual b) Taking N = 10, m = 2, 550 and r = 0.0325, we find A2N : = m(1 + r)N

N X

(1 + r)N −k+1

k=1

= m(1 + r)N

N X

(1 + r)k

k=1

= m(1 + r)N +1

(1 + r)N − 1 r

= $42, 040.42. c) In this case, with N = 10, m = 2, 550 and r = 0.0325, we find A2N = AN = m

N X

(1 + r)N −k+1 = m(1 + r)

k=1

(1 + r)N − 1 = $30, 532.79. r

Exercise 2.2 a) Let m := $3, 581 denote the amount invested each year. After multiplying (2.1) by (1 + r)N in order to account for the compounded interest from year 11 until year 20, we get the equality A = m(1 + r)N +1

(1 + r)N − 1 r

shows that (1 + r)21 − (1 + r)11 = r

50862 ≃ 14.2033r, 3581

showing that r ≃ 2.28% according to Figure S.3. 16 15.5 15 14.5 14 13.5 13 12.5 12 11.5

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 r in %

Fig. S.3: Graph of r 7−→ ((1 + r)21 − (1 + r)11 )/r.

10 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition b) Taking N = 10, m = 3, 581 and r = 0.0325, we find A2N : = m(1 + r)N

N X

(1 + r)N −k+1

k=1

= m(1 + r)N

N X

(1 + r)k

k=1

= m(1 + r)N +1

(1 + r)N − 1 r

= $59, 037.94. c) In this case, we find A2N = m

N X

(1 + r)N −k+1 = m(1 + r)

k=1

(1 + r)N − 1 = $42, 877.61. r

Exercise 2.3 a) We find m = $10, 000. b) Taking N = 10, A = 100, 000 and r = 0.02, Proposition 2.1 shows that m=

r 0.02 × 100, 000 A= = $11, 132.65. 1 − (1 + r)−N 1 − 1.02−10

Exercise 2.4 a) The discounted value of the loan after N months is m(1 + r)−N

N −1 X

(1 + r)k = m(1 + r)−N

k=0

1 − (1 + r)N 1 − (1 + r)−N =m , 1 − (1 + r) r

which should match A = $3, 000 with m = $275 and N = 12, hence 1 − (1 + r)−12 A = = 10.909090909, r m see also Equation (3.1.2) in Proposition 3.1 of MH4514_Slides.pdf, hence r ≃ 1.49767%

per month.

b) The yearly interest rate is given by (1 + r)N − 1 = (1 + r)12 − 1 = 1.014976712 − 1 ≃ 19.529% per year. Remark: Computing the interest rate as "

11 September 15, 2022

Solutions Manual 12 × $275 − 1 = 0.1 = 10% $3000 is not correct because this implicitly means that the 12 × $275 = $3, 300 are repaid as one lump sum at the end of the 12th month, which is not the case. c) The analysis of replies to Question (c) shows that “All of the above” was the most popular answer, followed by “Block”.

Fig. S.4: Histogram of replies to Question c). Exercise 2.5 We check that for any P∗ of the form P∗ (Rt = −1) := p∗ , we have

P∗ (Rt = 0) := 1 − 2p∗ ,

P∗ (Rt = 1) := p∗ ,

IE∗ [S1 ] = S0 (2p∗ + 1 − 2p∗ ) = S0 ,

and similarly

IE∗ [S2 | S1 ] = S1 (2p∗ + (1 − 2p∗ )) = S1 ,

hence the probability measure P∗ is risk-neutral. Exercise 2.6 a) In order to check for arbitrage opportunities we look for a risk-neutral probability measure P∗ which should satisfy (1) (1) IE∗ Sk+1 | Fk = (1 + r)Sk ,

k = 0, 1, . . . , N − 1,

(1) with r = 0. Rewriting IE∗ Sk+1 | Fk as (1) IE∗ Sk+1 | Fk 12

12 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (1)

(1)

= (1 − b)Sk P∗ (Rk+1 = a | Fk ) + Sk P∗ (Rk+1 = 0 | Fk ) (1) + (1 + b)Sk P∗ (Rk+1 = b | Fk ) (1) (1) (1) = (1 − b)Sk P∗ (Rk+1 = a) + Sk P∗ (Rk+1 = 0) + (1 + b)Sk P∗ (Rk+1 = b),

k = 0, 1, . . . , N − 1, it follows that any risk-neutral probability measure P∗ should satisfy the equations   (1 + b)Sk(1) P∗ (Rk+1 = b) + Sk(1) P∗ (Rk+1 = 0) + (1 − b)Sk(1) P∗ (Rk+1 = a) = Sk(1) 

P∗ (Rk+1 = b) + P∗ (Rk+1 = 0) + P∗ (Rk+1 = −b) = 1,

k = 0, 1, . . . , N − 1, i.e.  ∗  bP (Rk = b) − bP∗ (Rk = −b) = 0, 

P∗ (Rk = b) + P∗ (Rk = −b) = 1 − P∗ (Rk = 0),

k = 1, 2, . . . , N , with solution P∗ (Rk = b) = P∗ (Rk = −b) = k = 1, 2, . . . , N . b) We have " (1) # (1) Sk+1 − Sk F Var∗ k (1) Sk   ! (1) (1) 2 S − S k+1 k Fk  − = IE∗  (1) Sk   ! (1) (1) 2 ∗  Sk+1 − Sk = IE Fk  (1) Sk

" IE∗

1 − θ∗ , 2

(1)

#!2

(1)

Sk+1 − Sk (1)

= b2 P∗σ (Rk+1 = −b | Fk ) + b2 P∗σ (Rk+1 = b | Fk ) 1 − P∗σ (Rk+1 = 0) 1 − P∗σ (Rk+1 = 0) + b2 = b2 2 2 = b2 (1 − θ∗ ) = σ2 , k = 0, 1, . . . , N − 1, hence P∗σ (Rk = 0) = θ∗ = 1 −

σ2 , b2 13 September 15, 2022

Solutions Manual and therefore P∗σ (Rk = b) = P∗σ (Rk = −b) =

1 − P∗σ (Rk = 0) σ2 = 2, 2 2b

k = 0, 1, . . . , N − 1, under the condition 0 < σ 2 < b2 . Exercise 2.7 a) The possible values of Rt are a and b. b) We have IE∗ [Rt+1 | Ft ] = aP∗ (Rt+1 = a | Ft ) + bP∗ (Rt+1 = b | Ft ) r−a b−r +b = r. =a b−a b−a c) Letting p∗ = (r − a)/(b − a) and q ∗ = (b − r)/(b − a) we have k X k (p∗ )i (q ∗ )k−i (1 + b)i (1 + a)k−i St i i=0 k X k i k−i = St (p∗ (1 + b)) (q ∗ (1 + a)) i i=0

IE∗ [St+k | Ft ] =

14 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Exercise 2.8 a) We check that IE∗ [Rt+1 | Ft ] = aP∗ (Rt+1 = a | Ft ) + bP∗ (Rt+1 = b | Ft ) r−a b−r +b = r. =a b−a b−a b) We have IE∗ Set+1 Ft =

1 E∗ [St+1 | Ft ] A0 (1 + r)t+1

1 (1 + a)St P∗ (Rt+1 = a | Ft ) + (1 + b)St P∗ (Rt+1 = b | Ft ) A0 (1 + r)t+1 1 b−r r−a (1 + a)S + (1 + b)S = t t A0 (1 + r)t+1 b−a b−a b − a + (b − a)r = Set (1 + r)(b − a) = Set , t = 0, 1, . . . , N − 1. =

c) We have  !β  N Y ∗ β IE (SN ) = S0 IE (1 + Rk )  ∗

k=1

= S0 IE∗

"N Y

# (1 + Rk )β

k=1

= S0

N Y

IE∗ (1 + Rk )β ,

k=1

after using the independence of the returns (Rk )k=1,2,...,N , with b−r r−a IE∗ (1 + Rk )β = (1 + a)β + (1 + b)β , b−a b−a

k = 0, 1, . . . , N,

hence we find N b−r r−a IE∗ (SN )β = S0β (1 + a)β + (1 + b)β . b−a b−a d) We have P∗ St ≥ αAt for some t ∈ {0, 1, . . . , N } = P∗

St ≥α t=0,1,...,n At max

15 September 15, 2022

Solutions Manual IE (MN )β αβ β N S0 βb−r βr−a = , (1 + a) + (1 + b) (1 + r)N αA0 b−a b−a ≤

since the discounted price process (Mt )t=0,1,...,N :=

St At

t=0,1,...,N

is a nonnegative martingale by part (b). e) Since (Mt )t=0,1,...,N is a nonnegative martingale, we have IE[St+1 | Ft ] = IE[Mt+1 At+1 | Ft ] = At+1 IE[Mt+1 | Ft ] = At+1 Mt ≥ At M t = St ,

t = 0, 1, . . . , N − 1,

because r ≥ 0, hence (St )t=0,1,...,N is a nonnegative submartingale. Therefore, we have IE (M )β N P∗ max St ≥ x ≤ t=0,1,...,n xβ β N S0 b−r r−a ≤ (1 + a)β + (1 + b)β . x b−a b−a

Chapter 3 Exercise 3.1 (Exercise 2.5 continued). We consider the following trinomial tree.

16 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition S2 = 4, C = 0

p∗

1 − 2p∗

S1 = 2 p∗

S2 = 0, C = 1

∗

S2 = 2, C = 0

p∗

1 − 2p∗

S0 = 1

1 − 2p∗

S1 = 1 p∗

S2 = 1, C = 0 S2 = 0, C = 1

p∗

S2 = 0, C = 1

p∗

1 − 2p∗

S1 = 0

S2 = 2, C = 0

p∗

S2 = 0, C = 1 S2 = 0, C = 1

At time t = 0, we find π0 (C) =

1 IE∗ [(K − S2 )+ ] (1 + r)2

= p∗ (p∗ + (1 − 2p∗ ) + p∗ ) + (1 − 2p∗ )p∗ + (p∗ )2 = p∗ + (1 − 2p∗ )p∗ + (p∗ )2 = 2p∗ − (p∗ )2 . At time t = 1, we find 1 IE∗ [(K − S2 )+ | S1 ] 1+r  ∗   p if S1 = 2S0 ,    = p∗ if S1 = S0 ,      1 if S1 = 0.

π1 (C) =

Exercise 3.2 We have p∗ = (r−a)/(b−a) = 1/2 and q ∗ = (b−r)/(b−a) = 1/2, and the following underlying asset price tree:

17 September 15, 2022

Solutions Manual S2 = 4, C = 3

p∗

S1 = 2 q∗

S0 = 1

S2 = 2, C = 1

p∗

q∗

S1 = 1 q∗

S2 = 1, C = 3.

We first price, and then hedge. At time t = 1, by Theorem 3.5 we have  ∗ 3p + q ∗ 4     1 + r = 3 if S1 = 2 4 p∗ + q ∗ 8 π1 (C) = V1 = and π0 (C) = V0 = = .  p∗ + 3q ∗ 3 1+r 9  4   = if S1 = 1, 1+r 3 This leads to the following option pricing tree: p∗

∗

S2 = 4, V2 = 3

V1 = 4/3 S1 = 2 q∗

V0 = 8/9 S0 = 1

S2 = 2 V2 = 1 p∗ q∗

V1 = 4/3 S1 = 1 q∗

S2 = 1 V2 = 3. Regarding hedging, if S1 = 2 the condition ξ 2 · S 2 = ξ2 S2 + η2 A2 = V2 reads   4ξ2 + η2 (1 + r)2 = 3 S1 = 2 =⇒  2ξ2 + η2 (1 + r)2 = 1, hence (ξ2 , η2 ) = (1, −4/9). On the other hand, if S1 = 1 we have

18 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

S1 = 1 =⇒

  2ξ2 + η2 (1 + r)2 = 1 

ξ2 + η2 (1 + r)2 = 3,

hence (ξ2 , η2 ) = (−2, 20/9). Finally, at time t = 0 with S0 = 1 the condition ξ 1 · S 1 = ξ1 S1 + η1 A1 = V1 yields  4   2ξ1 + η1 (1 + r) =  3    ξ + η (1 + r) = 4 , 1 1 3 hence (ξ1 , η1 ) = (0, 8/9). The results can be summarized in the following table:

S1 = 2, V1 = 4/3 S2 = 4 S0 = 1 ξ2 = 1, η2 = −4/9 V2 = 3 V0 = 8/9 S2 = 1 ξ1 = 0 V2 = 3 η1 = 8/9 S1 = 1, V1 = 4/3 S2 = 1 ξ2 = −2, η2 = 20/9 V2 = 3 Table S.1: CRR pricing and hedging table. In addition, it can be checked that the portfolio strategy (ξk , ηk )k=1,2 is selffinancing, as we have ξ1 S1 + η1 A1 =

8 3 × 9 2  4 3  2 − ×  9 2

   −2 + 20 × 3 9 2 = ξ2 S1 + η2 A1 . Exercise 3.3 a) We have IE∗ [St+1 | Ft ] = IE∗ [St+1 | St ] St ∗ = P (Rt = −0.5) + St P∗ (Rt = 0) + 2St P∗ (Rt = 1) 2 ∗ r = + q ∗ + 2p∗ St 2 = St , t = 0, 1, "

19 September 15, 2022

Solutions Manual with r = 0. b) We have the following graph:

S1 = 2

∗ =

4 p∗ = 1/ q ∗ = 1/4 r ∗ = 1/

4 1/

q ∗ = 1/4

S0 = 1

S1 = 1

4 p∗ = 1/ q ∗ = 1/4 r ∗ = 1/ 2

r∗

1/ 2

S1 = 0.5

4 p∗ = 1/ q ∗ = 1/4 r ∗ = 1/ 2

S2 = 4, C = 2.5 S2 = 2, C = 0.5 S2 = 1, C = 0 S2 = 2, C = 0 S2 = 1, C = 0 S2 = 0.5, C = 0 S2 = 1, C = 0 S2 = 0.5, C = 0 S2 = 0.25, C = 0

c) The down-an-out barrier call option is priced at time t = 0 as V0 = IE∗ [C] = 2.5 × (p∗ )2 + 0.5 × p∗ q ∗ =

3 . 16

At time t = 1 we have V1 = 2.5 × p∗ + 0.5 × q ∗ = 2.5 ×

1 3 1 + 0.5 × = 4 4 4

if S1 = 2, and V1 = 0 in both cases S1 = 1 and S1 = 0.5. d) This market is not complete, and not every contingent claim is attainable, because the risk-neutral probability measure P∗ is not unique, for example (r∗ , q ∗ , p∗ ) = (1/4, 5/8, 1/8) and (r∗ , q ∗ , p∗ ) = (1/2, 1/4, 1/4) are both risk-neutral probability measures. Exercise 3.4 The CRR model can be described by the following binomial tree.

20 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

(1 + b)2 S0

p∗

(1 + b)S0

p∗

q∗

S0 = 1

(1 + a)(1 + b)S0

p∗

q∗

(1 + a)S0 q∗

(1 + a)2 S0

a) By the formulas 1 1 IE∗ [V2 | F1 ] = IE∗ [V2 | S1 ] 1+r 1+r S0 (1 + b)2 − 8 ∗ = P (S2 = S0 (1 + b)2 | S1 ) 1+r (S0 (1 + b)2 − 8) = p∗ 1{S1 =S0 (1+b)} , 1+r

V1 =

and 1 IE∗ [V1 | F0 ] 1+r 1 (S0 (1 + b)2 − 8) = p∗ × P∗ (S1 = S0 (1 + b)) + 0 × P∗ (S1 = S0 (1 + a)) 1+r 1+r 2 (S (1 + b) − 8) 0 = (p∗ )2 , (1 + r)2

V0 =

we find the table

S1 = 3, V1 = 1/4 S0 = 1 V0 = 1/16 S1 = 1, V1 = 0

S2 = 9 V2 = 1 S2 = 3 V2 = 0 S2 = 1 V2 = 0

Table S.2: CRR pricing tree. Note that we could also directly compute V0 from "

21 September 15, 2022

Solutions Manual V0 =

1 IE∗ [V2 | F0 ]. (1 + r)2

b) When S1 = S0 (1 + b), the equation ξ2 S2 + η2 A2 = V2 reads   ξ2 S0 (1 + b)2 + η2 A0 (1 + r)2 = S0 (1 + b)2 − 8 

ξ2 S0 (1 + b)(1 + a) + η2 A0 (1 + r)2 = 0,

which yields ξ2 =

S0 (1 + b)2 − 8 S0 (b − a)(1 + b)

and η2 = −

(S0 (1 + b)2 − 8)(1 + a) . (b − a)A0 (1 + r)2

(S.3.4)

When S1 = S0 (1 + a), the equation ξ2 S2 + η2 A2 = V2 reads   ξ2 S0 (1 + a)2 + η2 A0 (1 + r)2 = 0 

ξ2 S0 (1 + b)2 + η2 A0 (1 + r)2 = 0,

which has the unique solution (ξ2 , η2 ) = (0, 0). Next, the equation ξ1 S1 + η1 A1 = V1 reads  p∗ (S0 (1 + b)2 − 8)   ξ1 S0 (1 + b) + η1 A0 (1 + r) = , 1+r   ξ1 S0 (1 + a) + η1 A0 (1 + r) = 0, which yields (1 + a)(S0 (1 + b)2 − 8) . (b − a)A0 (1 + r)2 (S.3.5) This can be summarized in the following table: ξ1 = p∗

S0 (1 + b)2 − 8 S0 (b − a)(1 + r)

and η1 = −p∗

S1 = 3, V1 = 1/4 S0 = 1 V0 = 1/16 ξ1 = 1/8 η1 = −1/16

ξ2 = 1/6, η2 = −1/8 S1 = 1, V1 = 0 ξ2 = 0, η2 = 0

S2 = 9 V2 = 1 S2 = 3 V2 = 0 S2 = 1 V2 = 0

Table S.3: CRR pricing and hedging tree. When S1 = S0 (1 + a) at time t = 1 the option price is V1 = 0 and the hedging strategy is to cut all positions: ξ2 = η2 = 0. On the other hand, 22

22 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition if S1 = S0 (1 + b) then there is a chance of being in the money at maturity and we need to increase our position in the underlying asset from ξ1 = 1/8 to ξ2 = 1/6. Note that the self-financing condition ξ1 S1 + η1 A1 = ξ2 S1 + η2 A1 ,

(S.3.6)

is verified. For example when S1 = S0 (1 + a) we have 1 1 × S1 − A1 = 0 × S1 + 0 × A1 = 0, 8 16 while when S1 = S0 (1 + b) we find 1 1 1 1 1 × S1 − A1 = × S1 − × A1 = . 8 16 6 8 4 c) We can also use the self-financing condition (S.3.6) to recover (S.3.5) by rewriting the system of equations as   ξ1 S0 (1 + b) + η1 A0 (1 + r) = ξ2 S0 (1 + b) + η2 A0 (1 + r) 

ξ1 S0 (1 + a) + η1 A0 (1 + r) = 0,

with (ξ2 , η2 ) given by (S.3.4), which recovers

V1 = ξ1 S1 + η1 A1 =

3 2 1    8 − 16 = 4 if S1 = 3,    1 − 2 = 0 if S = 1. 1 8 16

Exercise 3.5 a) We build a portfolio based at times t = 0, 1 on αt+1 units of stock and $βt+1 in cash. When S1 = 2, we should have 4α2 + β2 = 0 2α2 + β2 = 1, hence (α2 , β2 ) = (−1/2, 2). On the other hand, when S1 = 1 we should have 2α1 + β1 = 1 α1 + β1 = 0, hence (α2 , β2 ) = (1, −1). b) When S1 = 2, the price of the claim at t = 1 is "

23 September 15, 2022

Solutions Manual α2 S1 + β2 = (−1/2) × 2 + 2 × 1 = 1. When S1 = 1, the price of the claim at t = 1 is α2 S1 +β2 = 1×1−1×1 = 0. c) At time t = 1 we build a portfolio using α1 units of stock and $β1 in cash. We should have 2α1 + β1 = 1 α1 + β1 = 0, hence (α1 , β1 ) = (1, −1). d) The price of the claim C at time t = 0 is α1 S0 + β1 = 1 × 1 + (−1) × 1 = 0. e) The probabilities (p∗ , q ∗ ) = ((r − a)/(b − a), (b − r)/(b − a)) = (0, 1) are clearly risk-neutral in the sense of Definition 2.12, as they yield IE∗ [S2 | S1 ] = S1

and

IE∗ [S1 | S0 ] = S0 .

with the risk-free rate r = 0. However, this does not form a risk-neutral probability measure P∗ equivalent to P in the sense of Definition 2.14 when q = P(R1 = 0) = P(R2 = 0) > 0. In case (p, q) = (0, 1), the probabilities (p∗ , q ∗ ) = (0, 1) would yield an equivalent risk-neutral probability measure. f) According to Theorem 2.15 this model allows for arbitrage opportunities as the unique available risk-neutral probability measure P∗ are not be equivalent to the historical probability measure P when q = P(R1 = 0) = P(R2 = 0) > 0. In this case, arbitrage opportunities are easily implemented by purchasing the option at the price 0 of part (d) while receiving a strictly positive payoff at maturity. More generally, arbitrage opportunities exist when the underlying price may increase with nonzero probability, without a possibility of strict decrease. In case (p, q) = (0, 1), no arbitrage is possible as prices remain constant. Exercise 3.6 We have the model-free answer 1 IE∗ [h(SN ) | Fk ] (1 + r)N −k 1 = IE∗ [α + βSN | Fk ] (1 + r)N −k α β = + IE∗ [SN | Fk ] (1 + r)N −k (1 + r)N −k α = Ak + βSk , k = 0, 1, . . . , N. (1 + r)N

πk (C) =

The hedging portfolio strategy is to hold β units of the underlying asset priced Sk and α/(1 + r)N units of the riskless asset priced Ak = (1 + r)k at 24

24 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition time k = 0, 1, . . . , N . Note that in the particular case of the CRR model, this answer is also compatible with (3.19)-(3.20). Exercise 3.7 Call-put parity. a) The relation (x − K)+ = x − K + (K − x)+ can be verified by successively checking the cases x ≤ K and x ≥ K. b) Respectively denoting by C(k) and P (k) the call and put prices at time k = 0, 1, . . . , N , by part (a) we have C(k) = (1 + r)−(N −k) IE∗ (SN − K)+ Fk = (1 + r)−(N −k) IE∗ SN − K + (K − SN )+ Fk = (1 + r)−(N −k) IE∗ [SN | Fk ] − (1 + r)−(N −k) K +(1 + r)−(N −k) IE∗ (K − SN )+ Fk = Sk − (1 + r)−(N −k) K + (1 + r)−(N −k) IE∗ (K − SN )+ Fk = Sk − (1 + r)−(N −k) K + P (k). Exercise 3.8 a) Taking q ∗ = 1 − p∗ = 1/4, we find the binary tree

p∗ p∗

6.25 = (1 + b)2

2.5 = 1 + b q∗

S0 = 1

p∗

1.25 = (1 + a)(1 + b)

q∗

0.5 = 1 + a q∗

0.25 = (1 + a)(1 + b) b) We find the binary tree

25 September 15, 2022

Solutions Manual

p∗ p∗

S1 = 2.5 and V1 = 0 q∗

S0 = 1 and V0 = 1/64

p∗ q∗

S2 = 6.25 and V2 = 0

S2 = 1.25 and V2 = 0

S1 = 0.5 and V1 = 1/8 q∗

S2 = 0.25 and V2 = 1

and the table

S2 = 6.25 V2 = 0 S0 = 1 S2 = 1.25 V0 = 1/64 V2 = 0 S1 = 0.5, V1 = 1/8 S2 = 0.25 V2 = 1 S1 = 2.5, V1 = 0

Table S.4: CRR pricing tree. c) Here, we compute the hedging strategy from the option prices. When S1 = S0 (1 + b) we clearly have ξ2 = η2 = 0. When S1 = S0 (1 + a), the equation ξ2 S2 + η2 A2 = V2 reads   ξ2 S0 (1 + a)2 + η2 (1 + r)2 = (K − S0 (1 + a)2 ) 

ξ2 S0 (1 + b)(1 + a) + η2 (1 + r)2 = 0

hence ξ2 = −

(K − S0 (1 + a)2 ) S0 (b − a)(1 + a)

and η2 =

(K − S0 (1 + a)2 )(1 + b) . S0 (b − a)(1 + r)2

Next, at time t = 1 the equation ξ1 S1 + η1 A1 = V1 reads  q ∗ (K − (1 + a)(1 + b))   ξ1 S0 (1 + a) + η1 (1 + r) = S0 , 1+r   ξ1 S0 (1 + b) + η1 (1 + r) = 0 which yields 26

26 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition ξ1 = −

q ∗ (K − S0 (1 + a)(1 + b)) S0 (b − a)(1 + r)

and η1 =

q ∗ (K − S0 (1 + a)(1 + b))(1 + b) . S0 (b − a)(1 + r)2

This can be summarized in the following table:

S1 = 2.5, V1 = 0 S2 = 6.25 S0 = 1 V2 = 0 V0 = 1/64 ξ2 = 0, η2 = 0 S2 = 1.25 ξ1 = −1/16 S1 = 0.5, V1 = 1/8 V2 = 0 η1 = 5/64 S2 = 0.25 ξ2 = −1, η2 = 5/16 V2 = 1 Table S.5: CRR pricing and hedging tree. If S1 = S0 (1 + a) then there is a chance of being in the money at maturity and we need to short sell further by decreasing ξ1 from ξ1 = −1/16 to ξ2 = −1. Note that the self-financing condition ξ1 S1 + η1 A1 = ξ2 S1 + η2 A1 is satisfied. Exercise 3.9 a) The binary call option can be priced under the risk-neutral probability measure P∗ as 1 IE∗ [C] 1+r 1 IE∗ [1[K,∞) (SN )] = 1+r 1 = P∗ (SN ≥ K) 1+r ∗ p , = 1+r

π0 (C) =

with p∗ := P∗ (SN ≥ K). b) Investing $p∗ by purchasing one binary call option yields a potential net return of  $1 − p∗ $1   = ∗ − 1 if SN ≥ K,   p∗ p  ∗    $0 − p = −100% if SN < K. p∗ c) The corresponding expected return is

27 September 15, 2022

Solutions Manual p∗ ×

1 − 1 + (1 − p∗ ) × (−1) = 0. ∗ p

d) The corresponding expected return is p∗ × 0.86 + (1 − p∗ ) × (−1) = p∗ × 1.86 − 1, which will be negative if p∗ <

1 ≃ 0.538. 1.86

That means, the expected gain can be negative even if 0.538 > p∗ = P∗ (SN ≥ K) > 0.5. Similarly, the expected gain (1 − p∗ ) × 0.86 + p∗ × (−1) = 0.86 − p∗ × 1.86, on binary put options will be negative if 1 − p∗ > 1/1.86, i.e. if p∗ >

0.86 ≃ 0.462. 1.86

That means, the expected gain can be negative even if 1 − 0.462 > P∗ (SN < K) > 0.5. In conclusion, the average gains of both call and put options will be negative if p∗ ∈ (0.462, 0.538). Note that the average of call and put option gains will still be negative, as p∗ × 1.86 − 1 0.86 − p∗ × 1.86 0.86 − 1 + = < 0. 2 2 2 Exercise 3.10 a) Based on the price map of the put spread collar option:

28 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 130

Put spread collar price map f(S) y=S

120 110 100 90 80 70

100

110

120

130

Fig. S.5: Put spread collar price map. we deduce the following payoff function graph of the put spread collar option in the next Figure S.6. 20

Put spread collar payoff function

15 10 5 0 -5 -10 -15 -20

100 SN

110

120

130

Fig. S.6: Put spread collar payoff function. b) The payoff function can be written as −(K1 − x)+ + (K2 − x)+ − (x − K3 )+ = −(80 − x)+ + (90 − x)+ − (x − 110)+ , see also http://optioncreator.com/stp7xy2.

29 September 15, 2022

Solutions Manual 20

-(K1-x)++(K2-x)+ -(x-K3)+

15 10 5 0 -5 -10 -15 -20

K2 SN

100

110

120

130

Fig. S.7: Put spread collar payoff as a combination of call and put option payoffs.∗ Hence this collar option payoff can be realized by 1. issuing (or shorting/selling) one put option with strike price K1 = 80, and 2. purchasing and holding one put option with strike price K2 = 90, and 3. issuing (or shorting/selling) one call option with strike price K3 = 110. Exercise 3.11 a) Based on the price map of the call spread collar option: 140

Call spread collar price map f(S) y=S

130 120 110 100 90 80 70 60

100

110

120

130

Fig. S.8: Call spread collar price map. we deduce the following payoff function graph of the call spread collar option in the next Figure S.9. ∗

The animation works in Acrobat Reader on the entire pdf file.

30 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 20

Call spread collar payoff function

15 10 5 0 -5 -10 -15 -20

100

110

120

130

Fig. S.9: Call spread collar payoff function. b) The payoff function can be written as −(K1 − x)+ + (x − K2 )+ − (x − K3 )+ = −(80 − x)+ + (x − 100)+ − (x − 110)+ , see also http://optioncreator.com/st3e4cz. 20

-(K1-x)++(x-K2)+ -(x-K3)+

15 10 5 0 -5 -10 -15 -20

100 SN

110

120

130

Fig. S.10: Call spread collar payoff as a combination of call and put option payoffs.∗ Hence this collar option payoff can be realized by 1. issuing (or shorting/selling) one put option with strike price K1 = 80, and 2. purchasing and holding one call option with strike price K2 = 100, and ∗

The animation works in Acrobat Reader on the entire pdf file.

31 September 15, 2022

Solutions Manual 3. issuing (or shorting/selling) one call option with strike price K3 = 110. Exercise 3.12 We have S1 + · · · + SN ϕ(S1 ) + · · · + ϕ(SN ) IE∗ ϕ ≤ IE∗ N N IE∗ [ϕ(S1 )] + · · · + IE∗ [ϕ(SN )] = N IE∗ [ϕ(IE∗ [SN | F1 ])] + · · · + IE∗ [ϕ(IE∗ [SN | FN ])] = N IE∗ [IE∗ [ϕ(SN ) | F1 ]] + · · · + IE∗ [IE∗ [ϕ(SN ) | FN ]] ≤ N IE∗ [ϕ(SN )] + · · · + IE∗ [ϕ(SN )] = N = IE∗ [ϕ(SN )].

since ϕ is convex,

because (Sn )n∈N is a martingale, by Jensen’s inequality, by the tower property.

The above argument is implicitly using the fact that a convex function ϕ(Sn ) of a martingale (Sn )n∈N is itself a submartingale, as ϕ(Sk ) = ϕ(IE∗ [SN | Fk ]) ≤ IE∗ [ϕ(SN ) | Fk ],

k = 1, 2, . . . , N.

Exercise 3.13 (Exercise 2.7 continued). a) The condition VN = C reads   ηN πN + ξN (1 + a)SN −1 = (1 + a)SN −1 − K 

ηN πN + ξN (1 + b)SN −1 = (1 + b)SN −1 − K,

from which we deduce the (static) hedging strategy ξN = 1 and ηN = −K(1 + r)−N /π0 . b) We have   ηN −1 πN −1 + ξN −1 (1 + a)SN −2 = ηN πN −1 + ξN (1 + a)SN −2 

ηN −1 πN −1 + ξN −1 (1 + b)SN −2 = ηN πN −1 + ξN (1 + b)SN −2 ,

which yields ξN −1 = ξN = 1 and ηN −1 = ηN = −K(1 + r)−N /π0 . Similarly, solving the self-financing condition   ηt πt + ξt (1 + a)St−1 = ηt+1 πt + ξt+1 (1 + a)St−1 

ηt πt + ξt (1 + b)St−1 = ηt+1 πt + ξt+1 (1 + b)St−1

at time t yields ξt = 1 and ηt = −(1 + r)−N 32

K , π0

t = 1, 2, . . . , N. 32 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition c) We have πt (C) = Vt = ηt πt + ξt St = St − K(1 + r)−N

πt π0

= St − K(1 + r)−(N −t) . d) For all t = 0, 1, . . . , N we have (1 + r)−(N −t) IE∗ [C | Ft ] = (1 + r)−(N −t) IE∗ [SN − K | Ft ], = (1 + r)−(N −t) IE∗ [SN | Ft ] − (1 + r)−(N −t) IE∗ [K | Ft ] = (1 + r)−(N −t) (1 + r)N −t St − K(1 + r)−(N −t) = St − K(1 + r)−(N −t) = Vt = πt (C). For a future contract expiring at time N we take K = S0 (1 + r)N and the contract is usually quoted at time t using the forward price (1+r)N −t (St − K(1 + r)N −t ) = (1 + r)N −t St − K = (1 + r)N −t St − S0 (1 + r)N , or simply using (1 + r)N −t St . Future contracts are “marked to market” at each time step t = 1, 2, . . . , N via a positive or negative cash flow exchange (1 + r)N −t St − (1 + r)N −t+1 St−1 from the seller to the buyer, ensuring that the absolute difference |(1 + r)N −t St − K| has been credited to the buyer’s account if it is positive, or to the seller’s account if it is negative. Exercise 3.14 a) We write VN =

  ξN SN −1 (1 + 1/2) + ηN = (SN −1 (1 + 1/2))2 

which yields

ξN SN −1 (1 − 1/2) + ηN = (SN −1 (1 − 1/2))2 ,   ξN = 2SN −1 

ηN = −3(SN −1 )2 /4.

i) We have IE∗ [(SN )2 | FN −1 ] = p∗ (SN −1 )2 (1 + 1/2)2 + (1 − p∗ )(SN −1 )2 (1 − 1/2)2 1 = (SN −1 )2 (1 + 1/2)2 + (1 − 1/2)2 2 = 5(SN −1 )2 /4.

33 September 15, 2022

Solutions Manual ii) We have ξN −1 SN −1 + ηN −1 A0 =

  ξN −1 SN −2 (1 + 1/2) + ηN −1

ξN −1 SN −2 (1 − 1/2) + ηN −1 = VN −1 

= 5(SN −1 )2 /4   5(SN −2 (1 + 1/2))2 /4 =  5(SN −2 (1 − 1/2))2 /4, hence

  ξN −1 = 5SN −2 /2 

ηN −1 = −15(SN −2 )2 /16.

iii) We have ξN −1 SN −1 + ηN −1 A0 = 5SN −2 SN −1 /2 − 15(SN −2 )2 /16   5(SN −2 )2 (1 + 1/2)/2 − 15(SN −2 )2 /16 =  5(SN −2 )2 (1 − 1/2)/2 − 15(SN −2 )2 /16   15(SN −2 )2 /4 − 15(SN −2 )2 /16 =  5(SN −2 )2 − 15(SN −2 )2 /16   45(SN −2 )2 /16 =  5(SN −2 )2 /16, and on the other hand, ξN SN −1 + ηN A0 = 2(SN −1 )2 − 3(SN −1 )2 /4   2(SN −2 )2 (1 + 1/2)2 − 3(SN −2 )2 (1 + 1/2)2 /4 =  2(SN −2 )2 (1 − 1/2)2 − 3(SN −2 )2 (1 − 1/2)2 /4   45(SN −2 )2 /16 =  5(SN −2 )2 /16. Remark: We could also determine (ξN −1 , ηN −1 ) as in Proposition 3.12, from (ξN , ηN ) and the self-financing condition ξN −1 SN −1 + ηN −1 A0 = ξN SN −1 + ηN AN −1 , as 34

34 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

ξN −1 SN −1 + ηN −1 A0 =

  ξN −1 SN −2 (1 + 1/2) + ηN −1

ξN −1 SN −2 (1 − 1/2) + ηN −1 = ξN SN −1 + ηN A0 

= 2(SN −1 )2 − 3(SN −1 )2 /4   2(SN −2 )2 (1 + 1/2)2 − 3(SN −2 )2 (1 + 1/2)2 /4 =  2(SN −2 )2 (1 − 1/2)2 − 3(SN −2 )2 (1 − 1/2)2 /4, which recovers ξN −1 = 5SN −2 /2 and ηN −1 = −15(SN −2 )2 /16. Exercise 3.15 a) By Theorem 2.19 this model admits a unique risk-neutral probability measure P∗ because a < r < b, and from (2.16) we have P∗ (Rt = a) =

0.07 − 0.05 b−r = , b−a 0.07 − 0.02

P(Rt = b) =

r−a 0.05 − 0.02 = , b−a 0.07 − 0.02

and

t = 1, 2, . . . , N . b) There are no arbitrage opportunities in this model, due to the existence of a risk-neutral probability measure. c) This market model is complete because the risk-neutral probability measure is unique. d) We have C = (SN )2 , hence

2 e = (SN ) = h(XN ), C (1 + r)N

with

h(x) = x2 (1 + r)N .

Now we have

(S.3.7)

Vet = ṽ(t, Xt ),

where the function v(t, x) is given from Proposition 3.8 as ṽ(t, x) =

N −t X k=0

k N −t−k ! 1+a N −t 1+b ∗ k ∗ N −t−k . (p ) (q ) h x k 1+r 1+r

Using (S.3.7) and the binomial theorem, we find

35 September 15, 2022

Solutions Manual N −t X

N −t k k=0 2k 2(N −t−k) 1+b 1+a ×(p∗ )k (q ∗ )N −t−k 1+r 1+r N −t X N −t = x2 (1 + r)N k k=0 N −t−k k (b − r)(1 + a)2 (r − a)(1 + b)2 × (b − a)(1 + r)2 (b − a)(1 + r)2 N −t (r − a)(1 + b)2 (b − r)(1 + a)2 = x2 (1 + r)N + 2 2 (b − a)(1 + r) (b − a)(1 + r) N −t x2 (r − a)(1 + b)2 + (b − r)(1 + a)2 = (1 + r)N −2t (b − a)N −t N −t 2 x (r − a)(1 + 2b + b2 ) + (b − r)(1 + 2a + a2 ) = (1 + r)N −2t (b − a)N −t

ṽ(t, x) = x2 (1 + r)N

x2 r(1 + 2b + b2 ) − a(1 + 2b + b2 ) + b(1 + 2a + a2 ) − r(1 + 2a + a2 ) (1 + r)N −2t (b − a)N −t N −t (1 + r(a + b + 2) − ab) . = x2 (1 + r)N −2t

N −t

e) We have ξt1 =

1+b Xt−1 − v t, 1+a v t, 1+r 1+r Xt−1

Xt−1 (b − a)/(1 + r) 2 2 1+b − 1+a 1+r 1+r (1 + r(a + b + 2) − ab)N −t = Xt−1 (b − a)/(1 + r) (1 + r)N −2t (1 + r(a + b + 2) − ab)N −t = St−1 (a + b + 2) , t = 1, 2, . . . , N, (1 + r)N −t

representing the quantity of the risky asset to be present in the portfolio at time t. On the other hand we have Vt − ξt1 Xt Xt0 Vt − ξt1 Xt = π0

ξt0 =

= Xt (1 + r(a + b + 2) − ab)N −t

Xt − Xt−1 (a + b + 2)/(1 + r) π0 (1 + r)N −2t 36 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition St − St−1 (a + b + 2) π0 (1 + r)N (1 + a)(1 + b) , = −(St−1 )2 (1 + r(a + b + 2) − ab)N −t π0 (1 + r)N = St (1 + r(a + b + 2) − ab)N −t

t = 1, 2, . . . , N . f) Let us check that the portfolio is self-financing. We have 0 1 ξ t+1 · S t = ξt+1 St0 + ξt+1 St1

(1 + a)(1 + b) 0 S π0 (1 + r)N t N −t−1 (1 + r(a + b + 2) − ab) +(St )2 (a + b + 2) (1 + r)N −t−1 (1 + r(a + b + 2) − ab)N −t−1 = (St )2 (1 + r)N −t × ((a + b + 2)(1 + r) − (1 + a)(1 + b)) 1 = (Xt )2 (1 + r(a + b + 2) − ab)N −t (1 + r)N −3t = (1 + r)t Vt = −(St )2 (1 + r(a + b + 2) − ab)N −t−1

= ξt · St,

t = 1, 2, . . . , N.

Exercise 3.16 a) We have Vt = ξt St + ηt πt = ξt (1 + Rt )St−1 + ηt (1 + r)πt−1 . b) We have IE∗ [Rt |Ft−1 ] = aP∗ (Rt = a | Ft−1 ) + bP∗ (Rt = b | Ft−1 ) r−a b−r +b =a b−a b−a r r =b −a b−a b−a = r. c) By the result of Question (a), we have IE∗ [Vt | Ft−1 ] = IE∗ [ξt (1 + Rt )St−1 | Ft−1 ] + IE∗ [ηt (1 + r)πt−1 | Ft−1 ] = ξt St−1 IE∗ [1 + Rt | Ft−1 ] + (1 + r) IE∗ [ηt πt−1 | Ft−1 ] = (1 + r)ξt St−1 + (1 + r)ηt πt−1 = (1 + r)ξt−1 St−1 + (1 + r)ηt−1 πt−1 "

37 September 15, 2022

Solutions Manual = (1 + r)Vt−1 , where we used the self-financing condition. d) We have 1 IE∗ [Vt | Ft−1 ] 1+r 3 8 = P∗ (Rt = a | Ft−1 ) + P∗ (Rt = b | Ft−1 ) 1+r 1+r 0.25 − 0.15 0.15 − 0.05 1 = 3 +8 1 + 0.15 0.25 − 0.05 0.25 − 0.05 1 3 8 = + 1.15 2 2 = 4.78.

Vt−1 =

Problem 3.17 CRR model with transaction costs. a)

i) In the event of an increase in the stock position ξt , the corresponding cost of purchase (1 + λ)(ξt+1 − ξt )St > 0 has to be deducted from the savings account value ηt At , which becomes updated as ηt+1 At = ηt At − (1 + λ)(ξt+1 − ξt )St , hence we have ηt+1 At + (1 + λ)ξt+1 St = ηt At + (1 + λ)ξt St . ii) In the event of a decrease in the stock position ξt , the corresponding sale profit (ξt −ξt+1 )(1−λ)St > 0 has to be added to from the savings account value ηt At , which becomes updated as ηt+1 At = ηt At + (ξt − ξt+1 )(1 − λ)St , hence we have ηt+1 At + ξt+1 (1 − λ)St = ηt At + ξt (1 − λ)St .

b) We have: i) If ξt+1 (βSt−1 ) > ξt (St−1 ), (ξt (St−1 ) − ξt+1 (βSt−1 ))β ↑ Set−1 = (ηt+1 (βSt−1 ) − ηt (St−1 ))ρ. ii) If ξt+1 (βSt−1 ) < ξt (St−1 ), (ξt (St−1 ) − ξt+1 (βSt−1 )) β↓ Set−1 = (ηt+1 (βSt−1 ) − ηt (St−1 ))ρ, 38

38 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition and iii) If ξt+1 (αSt−1 ) > ξt (St−1 ), (ξt (St−1 ) − ξt+1 (αSt−1 )) α↑ Set−1 = ρηt+1 (αSt−1 ) − ρηt (St−1 ). iv) If ξt+1 (αSt−1 ) < ξt (St−1 ), (ξt (St−1 ) − ξt+1 (αSt−1 )) α↓ Set−1 = ρηt+1 (αSt−1 ) − ρηt (St−1 ). c) We find gβ (ξt (St−1 ), ξt+1 (βSt−1 )) ξt (St−1 )−ξt+1 (βSt−1 ) Set−1 = ρηt+1 (βSt−1 )−ρηt (St−1 ). and gα (ξt (St−1 ), ξt+1 (αSt−1 )) ξt (St−1 )−ξt+1 (αSt−1 ) Set−1 = ρηt+1 (αSt−1 )−ρηt (St−1 ). d) The equation is Set−1 gβ (ξt (St−1 ), ξt+1 (βSt−1 )) ξt (St−1 ) − ξt+1 (βSt−1 ) −Set−1 gα (ξt (St−1 ), ξt+1 (αSt−1 )) ξt (St−1 ) − ξt+1 (αSt−1 ) = ρηt+1 (βSt−1 ) − ρηt+1 (αSt−1 ), which can be rewritten as f (x, St−1 ) = 0

(S.3.8)

at x = ξt (St−1 ). The function x 7→ f (x, St−1 ) is continuous by construction, and its derivative is the function x 7→ gβ (x, ξt+1 (βSt−1 )) − gα (x, ξt+1 (αSt−1 )), which can only take four values β ↑ − α↑ , β ↑ − α↓ , β↓ − α↑ , β↓ − α↓ , which are all strictly positive due to the conditions  ↑  α := α(1 + λ) < β(1 − λ) =: β↓ , α↓ := α(1 − λ) < β(1 − λ) := β↓ ,   ↑ α := α(1 + λ) < β(1 + λ) =: β ↑ . Hence x 7→ f (x, St−1 ) is strictly increasing, and we have lim f (x, St−1 ) = −∞ and lim f (x, St−1 ) = ∞.

x→−∞

x→∞

39 September 15, 2022

Solutions Manual Therefore, Equation (S.3.8) admits a unique solution x = ξt (St−1 ). e) We have ξt (St−1 ) = ρ +

ηt+1 (βSt−1 ) − ηt+1 (αSt−1 ) Set−1 gβ (ξt (St−1 ), ξt+1 (βSt−1 )) − gα (ξt (St−1 ), ξt+1 (αSt−1 ))

ξt+1 (βSt−1 )gβ (ξt (St−1 ), ξt+1 (βSt−1 )) − ξt+1 (αSt−1 )gα (ξt (St−1 ), ξt+1 (αSt−1 )) , gβ (ξt (St−1 ), ξt+1 (βSt−1 )) − gα (ξt (St−1 ), ξt+1 (αSt−1 ))

and ηt (St−1 ) gα (ξt (St−1 ), ξt+1 (αSt−1 ))gβ (ξt (St−1 ), ξt+1 (βSt−1 ))ξt+1 (βSt−1 ) ρgα (ξt (St−1 ), ξt+1 (αSt−1 )) − ρgβ (ξt (St−1 ), ξt+1 (βSt−1 )) g (ξ β t (St−1 ), ξt+1 (βSt−1 ))gα (ξt (St−1 ), ξt+1 (αSt−1 ))ξt+1 (αSt−1 ) − Set−1 ρgα (ξt (St−1 ), ξt+1 (αSt−1 )) − ρgβ (ξt (St−1 ), ξt+1 (βSt−1 )) gα (ξt (St−1 ), ξt+1 (αSt−1 ))ηt+1 (βSt−1 ) − gβ (ξt (St−1 ), ξt+1 (βSt−1 ))ηt+1 (αSt−1 ) + . gα (ξt (St−1 ), ξt+1 (αSt−1 )) − gβ (ξt (St−1 ), ξt+1 (βSt−1 )) f) i) In case f (ξt+1 (αSt−1 ), St−1 ≥ 0 we have ξt (St−1 ) ≤ ξt+1 (αSt−1 ) because f is increasing, hence = Set−1

gα (ξt (St−1 ), ξt+1 (αSt−1 )) = α↑ (1 + λ)α. ii) In case f ξt+1 (αSt−1 ), St−1 < 0 we have ξt (St−1 ) > ξt+1 (αSt−1 ) because f is increasing, hence gα (ξt (St−1 ), ξt+1 (αSt−1 )) = α↓ = (1 − λ)α. Note that in case f ξt+1 (αSt−1 ), St−1 = 0 we have ξt (St−1 ) = ξt+1 (αSt−1 ) hence there is no transaction from St−1 to αSt−1 . Similarly, iii) If f ξt+1 (βSt−1 ), St−1 ≥ 0 then ξt (St−1 ) ≤ ξt+1 (βSt−1 ), hence gβ (ξt (St−1 ), ξt+1 (βSt−1 )) = β ↑ (1 + λ)β. iv) If f ξt+1 (βSt−1 ), St−1 < 0 then ξt (St−1 ) > ξt+1 (βSt−1 ), hence gβ (ξt (St−1 ), ξt+1 (βSt−1 )) = β↓ = (1 − λ)β. Note that in case f ξt+1 (βSt−1 ), St−1 = 0 we have ξt (St−1 ) = ξt+1 (βSt−1 ) hence there is no transaction from St−1 to βSt−1 . g) With the parameters N = 2, K = $2, S0 = 8, ρ = 1, α = 0.5, β = 2, and the transaction cost rate λ = 12.5%, we find the tree of asset prices

40 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

S2 = 32 S1 = 16 S0 = 8

S2 = 8 S1 = 4 S2 = 2.

Fig. S.11: Tree of market prices with N = 2. At maturity time N we use the equations (4.4.4)-(4.4.5) of Proposition 4.11, which read here h(βSN −1 ) − h(αSN −1 ) ξN SN −1 = , (β − α)SN −1

(4.4.4)

where h(t, x) = (x − K)+ , and ηN (SN −1 ) =

βh(αSN −1 ) − αh(βSN −1 ) , (β − α)AN

(4.4.5)

as the evaluation of the terminal payoff is not affected by bid/ask prices. This yields (η2 (16), ξ2 (16)) = (−2, 1) and (η2 (4), ξ2 (4)) = (−2, 1). In this case we check that f (ξ2 (16), S0 ) = f (1, 8) = 0 and f (ξ2 (4), S0 ) = f (1, 8) = 0, which yields the hedging strategy ξ1 (8) = ξ2 (16) = ξ2 (4) = 1 and η1 (8) = η1 (15) = η1 (4) = −2 as the portfolio is self-financing. This static hedging involves no transaction costs and gives the initial price V0 = 8 × 1 − 2 × 1 = $6. Due to the simplicity of the case K = $2, we now consider the case K = $4. In this case, (4.4.4) and (4.4.5) give (η2 (16), ξ2 (16)) = (−4, 1)

and (η2 (4), ξ2 (4)) = (−4/3, 2/3),

which yields f (ξ2 (16), S0 ) = f (1, 8) = gβ (1, 1)(1 − 1) − gα (1, 2/3)(1 − 2/3) − 1 1 = − α↓ + 3 3 "

−4 − (−4/3) 8

41 September 15, 2022

Solutions Manual 1 1 = − (1 − λ)α + 3 3 1 1 1 = − × 0.875 × + 3 2 3 > 0, hence

gβ (ξ1 (S0 ), ξ2 (βS0 )) = β↑ = (1 + λ)β.

We also have f (ξ2 (4), S0 ) = f (2/3, 8) = gβ (2/3, 1)(2/3 − 1) − gα (2/3, 2/3)(2/3 − 2/3) − 1 1 = − β↑ + 3 3 1 1 = − (1 + λ)β + 3 3 2 1 = − × 1.125 + 3 3 < 0, hence

−4 − (−4/3) 8

gα (ξ1 (S0 ), ξ2 (αS0 )) = α↓ = (1 − λ)α.

Therefore, we find ξ1 (S0 ) = ρ

η2 (βS0 ) − η2 (αS0 ) Se0 gβ (ξ1 (S0 ), ξ2 (βS0 )) − gα (ξ1 (S0 ), ξ2 (αS0 ))

ξ2 (βS0 )gβ (ξ1 (S0 ), ξ2 (βS0 )) − ξ2 (αS0 )gα (ξ1 (S0 ), ξ2 (αS0 )) gβ (ξ1 (S0 ), ξ2 (βS0 )) − gα (ξ1 (S0 ), ξ2 (αS0 )) η2 (βS0 ) − η2 (αS0 ) ξ2 (βS0 )(1 + λ)β − ξ2 (αS0 )(1 − λ)α =ρ + (1 + λ)β − (1 − λ)α Se0 (1 + λ)β − (1 − λ)α +

−4 − (−4/3) 1.125 × 2 − (2/3) × 0.875 × 0.5 + 2 × 1.125 − 0.5 × 0.875 8 1.125 × 2 − 0.875 × 0.5 = 0.8965,

and gα (ξ1 (S0 ), ξ2 (αS0 ))gβ (ξ1 (S0 ), ξ2 (βS0 ))ξ2 (βS0 ) ρgα (ξ1 (S0 ), ξ2 (αS0 )) − ρgβ (ξ1 (S0 ), ξ2 (βS0 )) gβ (ξ1 (S0 ), ξ1 (βS0 ))gα (ξ1 (S0 ), ξ2 (αS0 ))ξ2 (αS0 ) e −S0 ρgα (ξ1 (S0 ), ξ2 (αS0 )) − ρgβ (ξ1 (S0 ), ξ2 (βS0 )) gα (ξ1 (S0 ), ξ2 (αS0 ))η2 (βS0 ) − gβ (ξ1 (S0 ), ξ2 (βS0 ))η2 (αS0 ) + gα (ξ1 (S0 ), ξ2 (αS0 )) − gβ (ξ1 (S0 ), ξ2 (βS0 ))

η1 (S0 ) = Se0

42 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (1 − λ)α(1 + λ)βξ2 (βS0 ) e (1 + λ)β(1 − λ)αξ2 (αS0 ) − S0 ρ(1 − λ)α − ρ(1 + λ)β (1 − λ)αρ − (1 + λ)βρ (1 − λ)αη2 (βS0 ) − (1 + λ)βη2 (αS0 ) + (1 − λ)α − (1 + λ)β 0.875 × 0.5 × 1.125 × 2 − 1.125 × 2 × 0.875 × 0.5 × 2/3 =8 0.875 × 0.5 − 1.125 × 2 0.875 × 0.5 × (−4) − 1.125 × 2 × (−4/3) + 0.875 × 0.5 − 1.125 × 2 = −2.1379. = Se0

This leads to the initial option price V0 = 0.8965 × 8 − 2.1379 = 5.0345. Remark: Note that with λ = 0 and K = 4 we would find ξ1 (S0 ) =

8 = 0.88, 9

η1 (S0 ) = −

20 = −2.22, 9

and V0 =

44 = 4.88. 9

Therefore, the presence of transaction costs increases the price of the option, and requires a higher stock position and a higher level of debt. Remark: Transaction costs in the CRR model were originally introduced in Boyle and Vorst (1992). The present solution is based on the method of Mel′ nikov and Petrachenko (2005), which originally also takes into account different borrowing and lending rates ρ↑ = 1 + r↑ and ρ↓ = 1 + r↓ , which can be regarded as bid/ask prices for the riskless asset, and can also represent transaction costs. Problem 3.18 CRR model with dividends (1). a) Denoting Sb2 the asset price at time 2 before the dividend is paid at the rate α, we find that the ex-dividend asset price S2 after dividend payment is S2 = Sb2 − αSb2 , hence V2 = ξ2 S2 + η2 A2 + αξ2 Sb2 S2 = ξ2 S2 + η2 A2 + αξ2 1−α S2 = ξ2 + η2 A2 . 1−α b) Denoting Sb1 the asset price at time 1 before the dividend is paid at the rate α, we find that the ex-dividend asset price S1 after dividend payment is "

43 September 15, 2022

Solutions Manual S1 = Sb1 − αSb1 , hence V1 = ξ1 S1 + η1 A1 + αξ1 Sb1 S1 = ξ1 S1 + η1 A1 + αξ1 1−α S1 = ξ1 + η1 A1 . 1−α c) If S1 = 3 we have  9ξ2  2    1 − α + η2 2 = $1

S2 V2 = ξ2 + η 2 A2 =  1−α  3ξ2   + η2 22 = 0 1−α

if S2 = 9, if S2 = 3,

hence (ξ2 , η2 ) = ((1 − α)/6, −1/8). If S1 = 1 we have

V2 = ξ2

 3ξ2  2    1 − α + η2 2 = 0

if S2 = 3,

ξ2 + η2 22 = 0 1−α

if S2 = 1,

S2 + η 2 A2 =  1−α   

hence (ξ2 , η2 ) = (0, 0). d) We have  1−α 1 1 − 2α   V1 = ξ2 S1 + 2η2 = 3 × −2× = if S1 = 3, 6 8 4   V1 = ξ2 S1 + 2η2 = 0 × 1 + 0 × 2 = 0 if S1 = 1. e) We have

V1 = ξ1

 1 − 2α 3ξ1     1 − α + 2η1 = 4

S1 + η 1 A1 =  1−α   

ξ1 + 2η1 = 0 1−α

if S1 = 3,

if S1 = 1,

hence (ξ1 , η1 ) = ((α − 1)(2α − 1)/8, (2α − 1)/16). f) At time k = 0 we have V0 = ξ1 S0 + η1 = 44

(α − 1)(2α − 1) 2α − 1 (2α − 1)2 + = . 8 16 16 44 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition g) Multiplying the prices (Sk )k=1,2 of the original tree by

1 1−α

1 1 − 1/4

k 4 , 3

we find the prices (S k )k=1,2 = (Sk /(1 − α)k )k=1,2 as in the following tree: S 2 = 16

p∗ p∗

S1 = 4 q∗

S0 = 1

p∗

S 2 = 16/3

q∗

S 1 = 4/3 q∗

S 2 = 16/9 h) The market returns found in Question (g) are a = 1/3 and b = 3, with r = 1%. Therefore we have p∗ =

r−a 1 − 1/3 1 = = b−a 3 − 1/3 4

and q ∗ =

3−1 b−r 3 = = . 3 − 1/3 b−a 4

i) If S1 = 3 we have 1 p∗ 1 IE∗ (S2 − K)+ | S 1 = 3] = $1 × = , 1+r 2 8 which coincides with V1 = ξ2 S1 + 2η2 =

3 2 1 − = . 8 8 8

If S1 = 1 we have 1 IE∗ (S2 − K)+ | S 1 = 1] = 0, 1+r which coincides with

V1 = ξ2 S1 + 2η2 = 0.

j) At time k = 0 we have 1 (p∗ )2 1 IE∗ (S2 − K)+ ] = = , 2 (1 + r) (1 + r)2 64 "

45 September 15, 2022

Solutions Manual which coincides with V0 = ξ1 S0 + η1 =

1 1 3 − = . 64 32 64

We also have 1 1 1 1 1 p∗ IE∗ V1 ] = × = × = . 1+r 1+r 8 8 8 64 Problem 3.19 CRR model with dividends (2). a) We have (1)

  (1)   (1 + b)(1 − α)Sk−1 if Rk = b     (1 + a)(1 − α)S (1)

k−1

(1)

= (1 + Rk )(1 − α)Sk−1 , and (1)

(1)

Sk = S0

k Y

(1 + Ri ),

 if Rk = a  k = 1, 2, . . . , N,

k = 0, 1, . . . , N,

i=1

with the binary tree (1)

(1 + b)(1 − α)S0

(1)

(1 + a)(1 − α)S0 (1)

b) The asset price before dividend payment is Sk /(1−α), hence the dividend amount is (1) (1) Sk αSk (1) − Sk = , 1−α 1−α therefore, the dividend value represents a percentage α/(1 − α) of the ex(1) dividend price Sk . Moreover, the return of the risky asset satisfies the following relation " (1) # Sk+1 (1) IE∗ Fk = IE∗ (1 + Rk )Sk Fk 1−α 46

46 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition =

r−a b−r (1) (1) (1 + b)Sk + (1 + a)Sk b−a b−a (1)

= (1 + r)Sk ,

k = 0, 1, . . . , N − 1.

α (1) c) When reinvesting the dividend amount ξk Sk into the new portfolio 1−α allocation, we have (1)

(0)

Vk = ξk+1 Sk + ηk+1 Sk (1)

(0)

= ξk Sk + ηk Sk +

α (1) ξk Sk 1−α

(1)

= ξk

Sk (0) + ηk Sk , 1−α

at times k = 1, 2, . . . , N − 1. Moreover, at time N we will similarly have (1)

(1)

VN = ξN SN +

S α (1) (0) (0) ξN SN + ηN SN = ξN N + ηN SN , 1−α 1−α

therefore the self-financing condition reads (1)

V k = ξk

Sk (0) + ηk Sk , 1−α

k = 1, 2, . . . , N.

(S.3.9)

d) By the self-financing condition (S.3.9) we have (1)

(1) S S Vek − Vek−1 = ξk+1 k(0) + ηk+1 − ξk k−1 − ηk (0) Sk Sk−1 (1) (1) S ξk Sk + ηk − ξk k−1 − ηk (0) (0) Sk (1 − α) Sk−1 ! (1) (1) Sk−1 Sk = ξk − , k = 1, 2, . . . , N, (0) (0) Sk (1 − α) Sk−1

which allows us to conclude from Question (b) that IE∗ Vek | Fk−1 − Vek−1 = IE∗ Vek − Vek−1 | Fk−1 ! # " (1) (1) Sk−1 Sk ∗ − (0) = IE ξk × Fk−1 (0) Sk (1 − α) Sk−1 " # (1) (1) S Sk = ξk IE∗ − k−1 Fk−1 (0) (0) Sk (1 − α) Sk−1

47 September 15, 2022

Solutions Manual

ξk (0)

Sk = 0,

" IE∗

# ! (1) Sk (1) Fk−1 − (1 + r)Sk−1 (1 − α)

k = 1, 2, . . . , N, e therefore Vk k=0,1,...,N −1 is a martingale under P∗ . e) Assuming that the portfolio strategy attains the claim C we have C = VN e = VeN , hence by the martingale property of Vek under and C k=0,1,...,N −1 ∗ P , we find e Fk , Vek = IE∗ VeN Fk = IE∗ C k = 0, 1, . . . , N, which shows that Vk =

1 IE∗ [C | Fk ], (1 + r)N −k

k = 0, 1, . . . , N,

f) By a binomial probability computation, we have 1 IE∗ [h(SN ) | Fk ] (1 + r)N −k " ! # N Y 1 ∗ Fk = IE h x (1 + Rl ) (1 + r)N −k

Vk =

l=t+1

, (1)

x=Sk

1 (1 + r)N −k N −k X N −k (1) × (p∗ )l (q ∗ )N −k−l h Sk (1 + b)k (1 + a)N −k−l (1 − α)N −k l l=0 (1) = C0 k, Sk (1 − α)N −k , N, a, b, r .

g) We “absorb” the dividend rate α into new market returns by taking aα , bα , rα such that 1 + aα = (1 + a)(1 − α),

1 + bα = (1 + b)(1 − α),

1 + rα = (1 + r)(1 − α),

bα = −α + b(1 − α),

rα = −α + r(1 − α).

i.e. aα = −α + a(1 − α),

As a consequence, we have 1 (1 + r)N −k N −k X N −k (1) × (p∗ )l (q ∗ )N −k−l h Sk (1 + b)k (1 + a)N −k−l (1 − α)N −k l

Vk =

l=0

48 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition N −k

(1 − α)N −k X (1 + rα )N −k

N −k (1) (p∗ )k (q ∗ )N −k−l h Sk (1 + bα )k (1 + aα )N −k−l l l=0 (1) = (1 − α)N −k C0 k, Sk , N, aα , bα , rα ,

where p∗ := P∗ (Rk = b) =

r−a rα − aα = > 0, bα − aα b−a

q ∗ := P∗ (Rk = a) =

bα − rα b−r = > 0, bα − aα b−a

and k = 1, 2, . . . , N . h) We have Vek =

1 (1) Cα k, Sk , N, aα , bα , rα , (1 + r)N

k = 0, 1, . . . , N,

hence by the martingale property we have 1 (1) Cα k, Sk , N, aα , bα , rα (1 + r)k = IE∗ Vek+1 | Fk 1 (1) IE∗ Cα k + 1, Sk+1 , N, aα , bα , rα Fk = (1 + r)k+1 1 (1) = p∗ Cα k + 1, Sk (1 + bα ), N, aα , bα , rα k+1 (1 + r) (1) +q ∗ Cα k + 1, Sk (1 + aα ), N, aα , bα , rα .

Vek =

This yields (1)

(1 + r)Cα k, Sk , N, aα , bα , rα

(1) (1) = p∗ Cα k + 1, Sk (1 + bα ), N, aα , bα , rα + q ∗ Cα k + 1, Sk (1 + aα ), N, aα , bα , rα . i) We find the equations  (0) (1) (1)   ηk Sk + ξk (1 + aα )Sk−1 = Cα k, (1 + aα )Sk−1 , N, aα , bα , rα   η S (0) + ξ (1 + b )S (1) = C k, (1 + b )S (1) , N, a , b , r , k k k α α α α α α k−1 k−1 which imply ξk =

(1) (1) Cα k, (1 + bα )Sk−1 , N, aα , bα , rα − Cα k, (1 + aα )Sk−1 , N, aα , bα , rα (1)

(bα − aα )Sk−1 49 September 15, 2022

Solutions Manual (1)

= (1 − α)N −k

C0 k, (1 + bα )Sk−1 , N, aα , bα , rα

(1)

(bα − aα )Sk−1 (1)

− (1 − α)N −k

C0 k, (1 + aα )Sk−1 , N, aα , bα , rα (1)

(bα − aα )Sk−1

and (1)

ηk =

(1 + bα )Cα k, (1 + bα )Sk−1 , N, aα , bα , rα

(0)

(bα − aα )Sk−1

(1) (1 + aα )Cα k, (1 + aα )Sk−1 , N, aα , bα , rα − (0) (bα − aα )Sk−1 (1) (1 + bα )C0 k, (1 + bα )Sk−1 , N, aα , bα , rα N −k = (1 − α) (0) (bα − aα )Sk−1 (1) (1 + aα )C0 k, (1 + aα )Sk−1 , N, aα , bα , rα , − (1 − α)N −k (0) (bα − aα )Sk−1 k = 1, 2, . . . , N . j) A possible answer: We have ξk =

N −k X

1 (1)

N −k (p∗ )k (q ∗ )N −k−l l

(b − a)Sk−1 l=0 (1) × h (1 − α)N −k Sk (1 + b)k+1 (1 + a)N −k−l (1) −h (1 − α)N −k Sk (1 + b)k (1 + a)N −k−l+1

and ηk =

N −k X

(1) (b − a)Sk−1 l=0

N −k (p∗ )k (q ∗ )N −k−l l

(1) × (1 + b)h (1 − α)N −k Sk (1 + b)k (1 + a)N −k−l+1 (1) −(1 + a)h (1 − α)N −k Sk (1 + b)k+1 (1 + a)N −k−l , k = 1, 2, . . . , N . Differentiation with respect to α of the general term inside the above summations yields respectively (1 + a)yh′ ((1 + a)y) − (1 + b)yh′ ((1 + b)y)

(S.3.10)

for ξk , and 50

50 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (1 + b)yh′ ((1 + b)y) − (1 + a)yh′ ((1 + a)y),

(S.3.11)

(1) for ηk , with y := (1 − α)N −k Sk (1 + b)k (1 + a)N −k−l and a < b.

We note that the sign of the above quantities (S.3.10)-(S.3.11) depends on whether the function x 7−→ xh′ (x) is non-decreasing, which is the case for example for the payoff functions h(x) = (x − K)+ and h(x) = (K − x)+ of both European call and put options. In particular, when the function x 7−→ xh′ (x) is non-decreasing, the amount invested on the risky (resp. riskless) asset will be lower (resp. higher) in the presence of a higher dividend. We also note that the expected return p∗ (1 + b)(1 − α) + q ∗ (1 + a)(1 − α) = r(1 − α) and the variance p∗ (1 + b)2 (1 − α)2 + q ∗ (1 + a)2 (1 − α)2 − r2 (1 − α)2 = (1 − α)2 p∗ (1 + b)2 + q ∗ (1 + a)2 − r2 of returns are lower in the presence of dividends. Problem 3.20 a) In order to check for arbitrage opportunities we look for a risk-neutral probability measure P∗ which should satisfy (1) (1) IE∗ Sk+1 Fk = (1 + r)Sk ,

k = 0, 1, . . . , N − 1.

(1) Rewriting IE∗ Sk+1 Fk as (1) (1) (1) IE∗ Sk+1 Fk = (1 + a)Sk P∗ (Rk+1 = a | Fk ) + Sk P∗ (Rk+1 = 0 | Fk ) (1)

+(1 + b)Sk P∗ (Rk+1 = b | Fk ) (1) (1) = (1 + a)Sk P∗ (Rk+1 = a) + Sk P∗ (Rk+1 = 0) (1) ∗ +(1 + b)Sk P (Rk+1 = b),

k = 0, 1, . . . , N − 1, it follows that any risk-neutral probability measure P∗ should satisfy the equations

51 September 15, 2022

Solutions Manual  (1) (1 + r)Sk =      (1) (1) (1) (1 + b)Sk P∗ (Rk+1 = b) + Sk P∗ (Rk+1 = 0) + (1 + a)Sk P∗ (Rk+1 = a),      ∗ P (Rk+1 = b) + P∗ (Rk+1 = 0) + P∗ (Rk+1 = a) = 1, k = 0, 1, . . . , N − 1, i.e.  ∗  bP (Rk = b) + aP∗ (Rk = a) = r, 

P∗ (Rk = b) + P∗ (Rk = a) = 1 − P∗ (Rk = 0),

k = 1, 2, . . . , N , with solution P∗ (Rk = b) =

r − (1 − θ∗ )a r − (1 − P∗ (Rk = 0))a = , b−a b−a

and

(1 − P∗ (Rk = 0))b − r (1 − θ∗ )b − r = , b−a b−a k = 1, 2, . . . , N . We check that this ternary tree model is without arbitrage if and only if there exists θ∗ := P∗ (Rk = 0) ∈ (0, 1) such that P∗ (Rk = a) =

(1 − θ∗ )a < r < (1 − θ∗ )b, or 0 < θ∗ < min

r−a b−r , −a b

(S.3.12)

 r    1 − b if r ≥ 0,

   1 − r if r ≤ 0. a Condition (S.3.12) is necessary in order to have P∗ (Rk = b) > 0

and P∗ (Rk = a) > 0,

and it is sufficient because it also implies P∗ (Rk = b) = 1 − θ∗ − P∗ (Rk = a) ≤ 1 and

P∗ (Rk = a) = 1 − θ∗ − P∗ (Rk = b) ≤ 1.

b) We will show that this ternary tree model is without arbitrage if and only if a < r < b. (i) Indeed, if the condition a < r < b is satisfied there always exists θ ∈ (0, 1) such that

52 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition a < (1 − θ)a < r < (1 − θ)b < b, as can be seen by taking θ∈

r−a b−r 0, min , , −a b

hence there exists a risk-neutral probability measure P∗θ , and the market model is without arbitrage. (ii) Conversely, if this ternary tree model is without arbitrage there exists some θ = P∗ (Rt = 0) ∈ (0, 1) such that (1 − θ)a < r < (1 − θ)b. c) When r ≤ a < 0 < b the risky asset overperforms the riskless asset, therefore we can realize arbitrage by borrowing from the riskless asset to purchase the risky asset. When a < 0 < b ≤ r the riskless asset overperforms the risky asset, therefore we can realize arbitrage by shortselling the risky asset and save the profit of the short sale on the riskless asset. d) Under the absence of arbitrage condition a < r < b, every value of θ ∈ (0, 1) such that r−a b−r , 0 < θ < min −a b satisfies

(1 − θ)a < r < (1 − θ)b,

and gives rise to a different risk-neutral probability measure, hence the risk-neutral measure is not unique and by Theorem 5.11 this ternary tree model is not complete. In particular, every risk-neutral probability measure P∗θ will give rise to a different claim price 1 IE∗ C Ft , (1 + r)N −t θ

(θ)

πt (C) =

t = 0, 1, . . . , N.

e) We have " ∗

Var

(1)

∗

= IE

(1)

Sk 

(1)

Sk+1 − Sk

Sk+1 − Sk (1)

 Fk  −

" ∗

(1)

#!2

(1)

Sk+1 − Sk (1)

53 September 15, 2022

Solutions Manual  ∗

= IE

(1)

Sk+1 − Sk

 Fk  − r 2

(1)

= a2 P∗σ (Rk+1 = a | Fk ) + b2 P∗σ (Rk+1 = b | Fk ) − r2 (1 − P∗σ (Rk+1 = 0))b − r r − (1 − P∗σ (Rk+1 = 0))a = a2 + b2 − r2 b−a b−a = ab(θ − 1) + r(a + b) − r2 = σ2 , k = 0, 1, . . . , N − 1, hence P∗σ (Rk = 0) = θ = 1 +

σ 2 + r2 − r(a + b) , ab

and therefore P∗σ (Rk = b) =

r − (1 − P∗σ (Rk = 0))a σ 2 − r(a − r) = , b−a b(b − a)

P∗σ (Rk = a) =

(1 − P∗σ (Rk = 0))b − r r(b − r) − σ 2 = , b−a a(b − a)

and

k = 1, 2, . . . , N , under the condition σ 2 > max(−r(r − a), r(b − r)), in addition to the condition 0 < θ < 1, i.e. r(b − r) + rb < σ 2 < (b − r)(r − a). Finally, we find if r ∈ (a, 0], and

−r(r − a) < σ 2 < (b − r)(r − a), r(b − r) < σ 2 < (b − r)(r − a),

if r ∈ [0, b). f) In this case the ternary tree becomes a trinomial recombining tree, and the expression of the risk-neutral probability measure becomes P∗θ (Rk = b) =

r(b + 1) + (1 − θ)b , b2 + 2b

and P∗θ (Rk = a) = (b + 1)

(1 − θ)b − r , b2 + 2b

54 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition k = 1, 2, . . . , N . The market model is without arbitrage if and only if there exists θ := P∗θ (Rk = 0) ∈ (0, 1) such that −(1 − θ)

b < r < (1 − θ)b, b+1

r 0<θ <1− . b g) Using the tower property of conditional expectations, we have (1)

f k, Sk

1 IE∗ [C | Fk ] (1 + r)N −k

1 IE∗ [IE∗ [C | Fk+1 ] | Fk ] (1 + r)N −k 1 (1) IE∗ (1 + r)N −(k+1) f k + 1, Sk+1 Fk = (1 + r)N −k 1 (1) = IE∗ f k + 1, Sk+1 Fk 1+r 1 (1) (1) = f k + 1, Sk (1 + a) P∗θ (Rk = a) + f k + 1, Sk P∗θ (Rk = 0) 1+r (1) +f k + 1, Sk (1 + b) P∗θ (Rk = b) . =

h) In this case we have f (N, x) = (K − x)+ . i) See the attached code.∗† j) Taking θ = 0.5 we find the following graph: 4.0

1.0

2.0

1.0

0.5

0.0

0.74

0.24

0.0

0.82

1.0

1.32

1.5

0.25

(a) Underlying asset prices.

1.75

(b) Put option prices.

Fig. S.12: Put option prices in the trinomial model. ∗

Download the modified (trinomial) IPython notebook that can be run here. Download the corresponding (binomial) IPython notebook. The Anaconda distribution can be installed from https://www.anaconda.com/distribution/ or tried online at https://jupyter.org/try. †

55 September 15, 2022

Solutions Manual There also exists extensions of the trinomial model to five states (pentanomial model), six states (hexanomial model), etc.

Chapter 4 Exercise 4.1 If 0 ≤ s ≤ t, using the facts that IE[Bt ] = 0 and IE Bt2 = 0, t ≥ 0, we have IE[Bt Bs ] = IE[(Bt − Bs )Bs ] + IE Bs2 = IE[(Bt − Bs )] IE[Bs ] + IE Bs2 = 0+s = s, and similarly we obtain IE[Bt Bs ] = t when 0 ≤ t ≤ s, hence in general we have IE[Bt Bs ] = min(s, t), s, t ≥ 0. Exercise 4.2 We need to check whether the four properties of the definition of Brownian motion are satisfied. a) Checking Conditions 1-2-3 are easily satisfied using the time shift t 7→ c + t. As for Condition 4, Bc+t − Bc+s clearly has the centered Gaussian distribution with variance c + t − (c − s) = t − s. We conclude that (Bc+t − Bc )t∈R+ is a standard Brownian motion. b) We note that Bct2 is a centered Gaussian random variable with variance ct2 - not t, hence (Bct2 )t∈R+ is not a standard Brownian motion. c) Similarly, checking Conditions 1-2-3 does not pose any particular problem using the time change t 7→ t/c2 . As for Condition 4, Bc+t − Bc+s clearly has a centered Gaussian distribution with Var c Bt/c2 − Bs/c2 = c2 Var Bt/c2 − Bs/c2 = (t − s)c2 /c2 = t − s. As a consequence, Bt/c2 t∈R is a standard Brownian motion. + d) This process does not have independent increments, hence it cannot be a Brownian motion. For example, by (4.1) we have IE Bt + Bt/2 − Bs + Bs/2 Bs + Bs/2 = IE Bt Bs + Bt Bs/2 + Bt/2 Bs + Bt/2 Bs/2 − IE Bs Bs + Bs Bs/2 + Bs/2 Bs + Bs/2 Bs/2

56 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = s+ =

s , 2

s s s s s +s+ −s− − − 2 2 2 2 2

which differs from 0, hence the two increments are not independent. Indeed, independence of Bt + Bt/2 − Bs + Bs/2 ) and Bs + Bs/2 would yield IE Bt + Bt/2 − Bs + Bs/2 ) Bs + Bs/2 ) = IE Bt + Bt/2 − Bs + Bs/2 ) IE Bs + Bs/2 ) = 0. Exercise 4.3 By Definition 4.5, we have wT 0

2dBt = 2(BT − B0 ) = 2BT ,

which has a Gaussian distribution with mean 0 and variance 4T . On the other hand, by Definition 4.5 again, we have wT 0

(2 × 1[0,T /2] (t) + 1(T /2,T ] (t))dBt = 2(BT /2 − B0 ) + (BT − BT /2 ) = BT + BT /2 ,

which has a Gaussian distribution with mean 0 and variance Var[BT + BT /2 ] = Var[(BT − BT /2 ) + 2BT /2 ] = Var[BT − BT /2 ] + 4 Var[BT /2 ] 4T T = + 2 2 5T . = 2 Equivalently, using the Itô isometry (4.8), we have w T Var (2 × 1[0,T /2] (t) + 1(T /2,T ] (t))dBt 0 " 2 # wT = IE (2 × 1[0,T /2] (t) + 1(T /2,T ] (t))dBt 0

2 2 × 1[0,T /2] (t) + 1(T /2,T ] (t) dt w T /2 wT =4 dt + dt =

T /2

5T = . 2 "

57 September 15, 2022

Solutions Manual w 2π Exercise 4.4 By Proposition 4.12, the stochastic integral sin(t) dBt has 0 a Gaussian distribution with mean 0 and variance w 2π 1 w 2π sin2 (t)dt = (1 − cos(2t))dt = π. 0 2 0 Exercise 4.5 By the Itô formula (4.28), we have d(f (t)Bt ) = f (t)dBt + Bt df (t) + df (t) • dBt = f (t)dBt + Bt f ′ (t)dt + f ′ (t)dt • dBt = f (t)dBt + Bt f ′ (t)dt, and by integration on both sides we get wT 0

f (t)dBt +

wT 0

Bt f ′ (t)dt =

wT 0

d(f (t)Bt )

= f (T )BT − f (0)B0 = 0, since f (T ) = 0 and B0 = 0, hence the conclusion. Note that this result can also be obtained by integration by parts on the interval [0, T ], see (4.11). Exercise 4.6

r1 a) The stochastic integral 0 t2 dBt is a centered Gaussian random variable with variance " 2 # w w1 1 1 IE t2 dBt = t4 dt = . 0 0 5 r 1 −1/2 b) The stochastic integral 0 t dBt has the variance " 2 # w w1 1 1 dt = +∞. IE t−1/2 dBt = 0 t 0 r1 In fact, the stochastic integral 0 t−1/2 dBt does not exist as a random variable in L2 (Ω) because the function t 7→ t−1/2 is not in L2 ([0, 1]). Remark. Writing Relation (4.11) with f (t) = t−1/2 gives w1

BT 1 w 1 −3/2 t−1/2 dBt = √ + t Bt dt, 0 2 0 T

however this is only a formal statement as f is not in C 1 ([0, 1]). Informally, wT we can check that the term t−3/2 Bt dt has the infinite variance 0

58 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

" w

2 # w w T T Bt f ′ (t)dt = IE Bt f ′ (t)dt Bs f ′ (s)ds 0

= IE =

w w T T 0

wT wT 0

Bs Bt f ′ (s)f ′ (t)dsdt

f ′ (s)f ′ (t) IE[Bs Bt ]dsdt

0 1wT wT

s−3/2 t−3/2 min(s, t)dsdt 4 0 0 w 1 T −3/2 w t −5/2 1 w T −5/2 w T −3/2 = t t s dsdt + s dsdt 0 t 4 0 4 0 1 w T −5/2 w T −3/2 1 w T −3/2 w t −5/2 t s dsdt + t s dsdt = 0 t 4 0 4 0 = +∞,

where we used Relation (4.1) or the result of Exercise 4.1 Exercise 4.7 a) By Proposition 4.12, the probability distribution of Xn is Gaussian with mean zero and variance " 2 # w 2π Var[Xn ] = IE sin(nt)dBt 0

w 2π

sin2 (nt)dt w 1 2π 1 w 2π = cos(0)dt − cos(2nt)dt 2 0 2 0 = π, n ≥ 1. 0

b) The random variables (Xn )n≥1 have same Gaussian distribution, and they are pairwise independent as from Corollary 4.13 we have w w 2π 2π IE[Xn Xm ] = IE sin(nt)dBt sin(mt)dBt 0

w 2π 0

sin(nt) sin(mt)dt

1 w 2π 1 w 2π = cos((n − m)t)dt − cos((n + m)t)dt 2 0 2 0 =0 and the vector (Xn , Xm ) is jointly Gaussian, for n, m ≥ 1 such that n ̸= m. Note that this condition implies independence only when the random variables have a Gaussian distribution.

59 September 15, 2022

Solutions Manual Exercise 4.8 We have Xt = f (Bt ) with f (x) = sin2 x, f ′ (x) = 2 sin x cos x = sin(2x), and f ′′ (x) = 2 cos(2x), hence dXt = d sin2 (Bt ) = df (Bt )

1 = f ′ (Bt )dBt + f ′′ (Bt )dt 2 = sin(2Bt )dBt + cos(2Bt )dt. Exercise 4.9 a) Using the Itô isometry (4.16), we have w wT T IE BT3 = IE dBt T + 2 Bt dBt 0 0 w w wT T T = T IE dBt + 2 IE dBt Bt dBt 0 0 0 w T = 2 IE Bt dt 0

= 0.

IE[Bt ]dt

We also have IE

BT4

= IE

T +2

wT 0

" = IE T 2 + 4T = T 2 + 4T IE = T 2 + 4 IE = T +4 2

= T2 + 4

0 wT 0

Bt dBt

wT 0

w T

2 #

Bt dBt + 4

Bt dBt + 4 IE

2 #

Bt dBt

" wT 0

2 # Bt dBt

|Bt |2 dt

IE |Bt |2 dt tdt

T2 = T2 + 4 2 = 3T 2 .

60 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition b) If X ≃ N (0, σ 2 ), we have X ≃ BT with σ 2 = T , hence the answer to Question (a) yields IE X 3 = 0 and IE X 4 = 3σ 4 . We note that those moments can be recovered directly from the Gaussian probability density function as w∞ 2 2 1 x3 e−x /(2σ ) dx = 0 IE X 3 = √ 2 2πσ −∞ and

w∞ 2 2 1 IE X 4 = √ x4 e−x /(2σ ) dx = 3σ 4 . 2 2πσ −∞

Exercise 4.10 Taking expectation on both sides of (4.39) shows that 0 = IE (BT )3 wT = IE C + ζt,T dBt 0 w T = C + IE ζt,T dBt 0

=0 by (4.17), hence C = 0. Next, applying Itô’s formula to the function f (x) = x3 shows that (BT )3 = f (BT )

wT 1 w T ′′ f (Bt )dt = f (B0 ) + f ′ (Bt )dBt + 0 2 0 wT wT Bt2 dBt + 3 Bt dt. =3 0

By the integration by parts formula (4.11) applied to f (t) = t, we find wT 0

Bt dt = T BT −

wT 0

tdBt =

wT 0

(T − t)dBt ,

hence

(BT )3 = 3

wT tdBt Bt2 dBt + 3 T BT −

(T − t + Bt2 )dBt ,

61 September 15, 2022

Solutions Manual and we find ζt,T = 3(T − t + Bt2 ), t ∈ [0, T ]. This type of stochastic integral decomposition can be used for option hedging, cf. Section 7.5. Exercise 4.11 a) We have i h rT i h rT rt IE e 0 f (s)dBs Ft = e 0 f (s)dBs IE e t f (s)dBs Ft h rT i rt = e 0 f (s)dBs IE e t f (s)dBs w t 1wT = exp f (s)dBs + |f (s)|2 ds ,(S.4.13) 0 2 t 0 ≤ t ≤ T , where we used the Gaussian moment generating function wT 2 IE eX = eσ /2 for X ≃ N (0, σ 2 ) and the fact that f (s)dBs ≃ t w T 2 N 0, f (s)ds by Proposition 4.12. t

b) We have w t 1wt 2 IE exp f (s)dBs − f (s)ds Fu 0 2 0 w t 1wt 2 = exp − f (s)ds IE exp f (s)dBs Fu 0 2 0 w wt u 1wt 2 = exp − f (s)ds IE exp f (s)dBs + f (s)dBs Fu 0 u 2 0 w w t u 1wt 2 f (s)ds IE exp f (s)dBs Fu = exp f (s)dBs − u 0 2 0 w w u t 1wt 2 = exp f (s)dBs − f (s)ds IE exp f (s)dBs 0 u 2 0 w u 1wt 2 1wt 2 = exp f (s)dBs − f (s)ds + f (s)ds 0 2 0 2 u w u 1wu 2 f (s)dBs − = exp f (s)ds , 0 ≤ u ≤ t. 0 2 0 This result can also be obtained by directly applying (S.4.13). c) We apply the conclusion of Question (b) to the constant function f (t) := σ, t ≥ 0. Exercise 4.12 We have (1)

− St

(1)

− dSt

= dSt 62

(2)

dSt = d St

(2)

62 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (1)

(1)

(2)

= µSt dt + σ1 dWt − µSt dt + σ2 dWt (1) (2) (1) (2) = µ St − µSt dt + σ1 dWt − σ2 dWt . (1)

The process Mt := σ1 Wt

(1)

dMt • dMt = σ1 dWt

(2)

− σ2 Wt

(2)

− σ2 dWt

is a continuous martingale, with •

(1)

σ1 dWt

(2)

− σ2 dWt

(1) (1) (1) (2) (2) (2) • dW − 2σ1 σ2 dWt • dWt + σ22 dWt • dWt t 2 2 = (σ1 − 2ρσ1 σ2 + σ2 )dt.

= σ12 dWt

Therefore, letting σ :=

q σ12 − 2ρσ1 σ2 + σ22

and

Mt σ1 (1) σ2 (2) = W − Wt , σ σ t σ by the Lévy characterization theorem, see e.g. Theorem IV.3.6 in Revuz and Yor (1994), the process (Wt )t∈R+ is a standard Brownian motion with quadratic variation dWt • dWt = dt, with Wt :=

(1)

dSt = µ St

(2)

− St

(1)

dt + σ1 dWt

(2)

− σ2 dWt

= µSt dt + σdWt . Remark: Since ρ ∈ [−1, 1], we have σ12 − 2ρσ1 σ2 + σ22 ≥ σ12 − 2σ1 σ2 + σ22 = (σ1 − σ2 )2 ≥ 0. Exercise 4.13 a) Using (4.31), we have w T 2 IE exp β Bt dBt = IE eβ(BT −T )/2 0 h i 2 = e−βT /2 IE eβ(BT ) /2 e−βT /2 w ∞ βx2 /2 −x2 /(2T ) = √ e e dx 2πT −∞ e−βT /2 w ∞ (β−1/T )x2 /2 = √ e dx 2πT −∞ −βT /2 w ∞ −x2 /(2/(1/T −β)) e e p = √ dx 1 − βT −∞ 2π/(1/T − β) e−βT /2 = √ , 1 − βT "

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Solutions Manual for all β < 1/T . √ b) The function β 7→ 1/ 1 − βT can be identified as the moment generating function of the gamma distribution with shape parameter λ = 1/2, scaling parameter 1/T and probability density function y 7→

1 (yT )−1/2 e−y/T , Γ (1/2)

due to the relation 1 T −λ w ∞ βy λ−1 −y/T e y e dy = , Γ (λ) 0 (1 − βT )λ

β < 1/T,

wT therefore T + Bt dBt = BT2 has a gamma distribution with shape 0 parameter 1/2 and scaling parameter T . In other words, the square wT √ 1+ Bt dBt /T = BT2 /T of the normal random variable BT / T ≃ 0

N (0, 1) has a χ2 distribution with one degree of freedom. Exercise 4.14 a) Letting Yt = ebt Xt , we have dYt = d(ebt Xt ) = bebt Xt dt + ebt dXt = bebt Xt dt + ebt (−bXt dt + σe−bt dBt ) = σdBt , hence Yt = Y0 + and

wt 0

dYs = Y0 + σ

wt 0

dBs = Y0 + σBt ,

Xt = e−bt Yt = e−bt Y0 + σe−bt Bt = e−bt X0 + σe−bt Bt .

Alternatively, we can also search for a solution Xt of the form Xt = f (t, Bt ), with dXt = df (t, Bt ) =

∂f 1 ∂2f ∂f (t, Bt )dt + (t, Bt )dBt + (t, Bt )dt ∂t ∂x 2 ∂x2

from the Itô formula. Matching this expression to the stochastic differential equation (4.40) would yield ∂f 1 ∂2f (t, Bt )dt = −bXt = −bf (t, Bt ) (t, Bt ) + ∂t 2 ∂x2 and 64

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Introduction to Stochastic Finance with Market Examples, Second Edition ∂f (t, Bt ) = σe−bt , ∂x hence 1 ∂2f ∂f (t, x)dt = −bf (t, x) and (t, x) + ∂t 2 ∂x2

∂f (t, x) = σe−bt , ∂x

x ∈ R, which can be solved as f (t, x) = f (t, 0) + σxe−bt and ∂f (0, x) = −bf (t, 0), ∂t which gives f (t, 0) = f (0, 0)e−bt , and recovers Xt = f (t, Bt ) = X0 e−bt + σe−bt Bt ,

t ≥ 0.

b) Letting Yt = ebt Xt , we have dYt = d(ebt Xt ) = bebt Xt dt + ebt dXt = bebt Xt dt + ebt − bXt dt + σe−at dBt

= σe(b−a)t dBt , hence we can solve for Yt by integrating on both sides as wt wt Yt = Y0 + dYs = Y0 + σ e(b−a)s dBs , t ≥ 0. 0

This yields the solution Xt = e−bt Yt = e−bt X0 + σe−bt

wt 0

e(b−a)s dBs ,

t ≥ 0.

Comments: (i) This type of computation appears anywhere discounting by the factor e−bt is involved. (ii) In part (b) the solution cannot take the form Xt = f (t, Bt ) when a ̸= b. Indeed, solving ∂f (t, x) = σe−at ∂x gives f (t, x) = f (t, 0) + σxe−at , yielding ∂f 1 ∂2f ∂f (t, x) + (t, x) = (t, 0) − aσxe−at , ∂t 2 ∂x2 ∂t which cannot match −bf (t, x) unless a = b. "

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Solutions Manual wt (iii) The stochastic integral e(b−a)s dBs cannot be computed in closed 0 form. It is a centered Gaussian random variable with variance wt 0

e2(b−a)s ds =

e2(b−a)t − 1 2(b − a)

if b ̸= a, and variance t if a = b. Exercise 4.15 a) Note that the stochastic integral wT 0

1 dBs T −s

is not defined in L2 (Ω) as the function s 7→ 1/(T − s) is not in L2 ([0, T ]), and by the Itô isometry (4.8) we have " 2 # w ∞ wT 1 T 1 1 IE dBs ds = = +∞. = 2 0 T −s 0 (T − s) T −s 0 On the other hand, by (4.27) and (4.42) we have Xt dXt 1 1 d = + Xt d + dXt • d T −t T −t T −t T −t Xt dXt + dt = T − t (T − t)2 dBt =σ , T −t hence, by integration over the time interval [0, t] and using the initial condition X0 = 0, we find w t dB X0 w t Xs Xt s = + d =σ , 0 ≤ t < T. 0 0 T −s T −t T T −s b) By (4.17), we have IE[Xt ] = (T − t)σ IE

w t

0 T −s

dBs = 0,

0 ≤ t < T.

c) By the Itô isometry (4.8), we have Var[Xt ] = (T − t)2 σ 2 Var

1 dBs 0 T −s t

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Introduction to Stochastic Finance with Market Examples, Second Edition

= (T − t)2 σ 2 IE = (T − t)2 σ 2 = (T − t)2 σ 2 = σ2 t

T −t , T

1 dBs 0 T −s

2 #

ds 1 1 − T −t T

0 (T − s)2

0 ≤ t < T.

d) We have lim ∥Xt ∥L2 (Ω) = lim Var[Xt ] = σ 2 lim

t→T

t−

t2 T

= 0.

Exercise 4.16 Exponential Vašíček (1977) model (1). Applying the Itô formula (4.29) to Xt = ert = f (rt ) with f (x) = ex , we have dXt = dert

1 = f ′ (rt )drt + f ′′ (rt )drt • drt 2 1 = ert drt + ert drt • drt 2 1 = ert ((a − brt )dt + σdBt ) + ert ((a − brt )dt + σdBt )2 2 σ 2 rt = ert ((a − brt )dt + σdBt ) + e dt 2 2 σ − b log(Xt ) dt + σXt dBt = Xt a + 2 = Xt (e a − ebf (Xt ))dt + σg(Xt )dBt ,

hence

σ2 eb = b and 2 the functions f (x) and g(x) are given by f (x) = log x and g(x) = x. Note that this stochastic differential equation is that of the exponential Vasicek model. e a=a+

Exercise 4.17 Exponential Vasicek model (2). wt a) We have Zt = e−at Z0 + σ e−(t−s)a dBs . 0 wt θ b) We have Yt = e−at Y0 + 1 − e−at + σ e−(t−s)a dBs . 0 a "

67 September 15, 2022

Solutions Manual σ2 c) We have dXt = Xt θ + − a log Xt dt + σXt dBt . 2 wt θ d) We have rt = exp e−at log r0 + 1 − e−at + σ e−(t−s)a dBs . 0 a 2 e) Using the Gaussian moment generating function identity IE[eX ] = eα /2 2 for X ≃ N (0, α ) and the variance formula (4.10), we have wt θ Fu IE[rt | Fu ] = IE exp e−at log r0 + 1 − e−at + σ e−(t−s)a dBs 0 a w ru t −at −at −(t−s)a θ +σ 0 e dBs = ee log r0 + a 1−e IE exp σ e−(t−s)a dBs Fu u w r t θ e−at log r0 + a 1−e−at +σ 0u e−(t−s)a dBs IE exp σ e−(t−s)a dBs =e u wu σ 2 w t −2(t−s)a θ = exp e−at log r0 + 1 − e−at + σ e−(t−s)a dBs + e ds 0 a 2 u wu θ σ2 −at −at −(t−s)a = exp e log r0 + 1 − e +σ e dBs + 1 − e−2(t−u)a 0 a 4a wu θ = exp e−(t−u)a e−au log r0 + 1 − e−au + σ e−(u−s)a dBs 0 a σ2 θ −(t−u)a −2(t−u)a + 1−e + 1−e a 4a σ2 θ = exp e−(t−u)a log ru + 1 − e−(t−u)a + 1 − e−2(t−u)a a 4a σ2 θ e−(t−u)a −(t−u)a = ru exp 1−e + 1 − e−2(t−u)a . a 4a In particular, for u = 0 we find σ2 −at θ 1 − e−2at . IE[rt ] = r0e exp + 1 − e−at + a 4a f) Similarly, we have wt 2θ 1 − e−at + 2σ e−(t−s)a dBs Fu IE[rt2 | Fu ] = IE exp 2e−at log r0 + 0 a w r u −(t−s)a t −at 2e−at log r0 + 2θ 1−e +2σ e dB s a 0 =e IE exp 2σ e−(t−s)a dBs Fu u w r u −(t−s)a t −at −at 2θ +2σ 0 e dBs = e2e log r0 + a 1−e IE exp 2σ e−(t−s)a dBs u wu wt 2θ −at −at −(t−s)a = exp 2e log r0 + 1−e + 2σ e dBs + 2σ 2 e−2(t−s)a ds 0 u a 68

68 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition wu 2θ σ2 = exp 2e−at log r0 + 1 − e−at + 2σ 1 − e−2(t−u)a e−(t−s)a dBs + 0 a a wu 2θ −au −(u−s)a −(t−u)a −au 1−e + 2σ e dBs = exp 2e 2e log r0 + 0 a 2 2θ σ + 1 − e−(t−u)a + 1 − e−2(t−u)a a a σ2 2θ = exp 2e−(t−u)a log ru + 1 − e−(t−u)a + 1 − e−2(t−u)a a a 2 −(t−u)a σ 2θ = ru2e exp 1 − e−(t−u)a + 1 − e−2(t−u)a , a a hence Var[rt | Fu ] = IE[rt2 | Fu ] − (IE[rt | Fu ])2 σ2 −(t−u)a 2θ 1 − e−(t−u)a + 1 − e−2(t−u)a = ru2e exp a a σ2 2θ 2e−(t−u)a −(t−u)a −ru exp 1−e + 1 − e−2(t−u)a a 2a 2 −(t−u)a σ 2θ 1 − e−(t−u)a + 1 − e−2(t−u)a = ru2e exp a a σ2 −2(t−u)a × 1 − exp − 1−e . 2a σ2 θ + and g) We find lim IE[rt ] = r0 exp t→∞ a 4a 2θ σ 2 σ2 lim Var[rt ] = exp + 1 − exp − t→∞ a a 2a 2 2θ σ = exp exp −1 . a a Exercise 4.18 Cox-Ingersoll-Ross (CIR) model. a) We have rt = r0 +

wt 0

(α − βrs )ds + σ

wt√ 0

rs dBs ,

t ≥ 0.

(S.4.14)

b) Taking expectations on both sides of (S.4.14) and using the fact that the expectation of the stochastic integral with respect to Brownian motion is zero, we find u(t) = IE[rt ] "

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Solutions Manual wt wt√ rs dBs = IE r0 + (α − βrs )ds + σ 0 0 wt = IE r0 + (α − βrs )ds 0 w t = r0 + IE (α − βrs )ds 0

= r0 + = r0 +

wt 0

(α − β IE[rs ])ds (α − βu(s))ds,

which yields the differential equation u′ (t) = α − βu(t). Letting w(t) : eβt u(t), we have w′ (t) = βeβt u(t) + eβt u′ (t) = αeβt , hence IE[rt ] = u(t) = e−βt w(t) wt = e−βt w(0) + α eβs ds 0 α = e−βt u(0) + (eβt − 1) β α = e−βt r0 + 1 − e−βt , t ≥ 0. β

(S.4.15)

c) By applying Itô’s formula (4.29) to wt√ wt rs dBs , rt2 = f r0 + (α − βrs )ds + σ 0

with f (x) = x2 , we find 1 d(rt )2 = f ′ (rt )drt + f ′′ (rt )drt • drt 2 = 2rt drt + drt • drt 3/2

= rt (σ 2 + 2α − 2βrt )dt + 2σrt dBt or, in integral form, wt wt rt2 = r02 + rs (σ 2 + 2α − 2βrs )ds + 2σ rs3/2 dBs , 0

t ≥ 0.

(S.4.16)

d) Taking again the expectation on both sides of (S.4.16), we find 70

70 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition v(t) = IE[rt2 ] wt wt = IE r02 + rs (σ 2 + 2α − 2βrs )ds + 2σ rs3/2 dBs 0 0 w t 2 2 = r0 + IE rs (σ + 2α − 2βrs )ds 0

= r02 +

(σ 2 IE[rs ] + 2α IE[rs ] − 2β IE[rs2 ])ds wt = v(0) + (σ 2 u(s) + 2αu(s) − 2βv(s))ds, 0

and after differentiation with respect to t this yields the differential equation v ′ (t) = (σ 2 + 2α)u(t) − 2βv(t), t ≥ 0. By (S.4.15) we find v ′ (t) = (σ 2 + 2α)

α −βt α + r0 − e − 2βv(t), β β

t ≥ 0.

Looking for a solution of the form v(t) = c0 + c1 e−βt + c2 e−2βt ,

t ≥ 0,

we find v ′ (t) = −βc1 e−βt − 2βc2 e−2βt α α −βt = (σ 2 + 2α) + r0 − e − 2β(c0 + c1 e−βt + c2 e−2βt ) β β α α −βt = (σ 2 + 2α) + (σ 2 + 2α) r0 − e − 2βc0 − 2βc1 e−βt − 2βc2 e−βt , β β t ≥ 0, hence  α 0 = (σ 2 + 2α) − 2βc0 ,    β    α  −βc1 = (σ 2 + 2α) r0 − − 2βc1 , β and c0 = with

α (σ 2 + 2α), 2β 2

c1 =

σ 2 + 2α β

α r0 − , β

r02 = v(0) = c0 + c1 + c2 ,

which yields "

71 September 15, 2022

Solutions Manual c2 = r02 − c0 − c1 σ 2 + 2α α α r0 − = r02 − 2 (σ 2 + 2α) − 2β β β α r0 2 2 = r0 − (σ + 2α) − 2 , β 2β and IE[rt2 ] = v(t) = c0 + c1 e−βt + c2 e−2βt α σ 2 + 2α α −βt = (σ 2 + 2α) + r0 − e 2 2β β β α r0 − 2 + r02 − (σ 2 + 2α) e−2βt , β 2β

t ≥ 0.

e) We have Var[rt ] = IE[rt2 ] − (IE[rt ])2 σ 2 + 2α α α −βt 2 (σ + 2α) + r − e = 0 2β 2 β β α r0 − 2 + r02 − (σ 2 + 2α) e−2βt β 2β 2 α −βt α − + r0 − e β β α σ 2 + 2α α −βt 2 = (σ + 2α) + r − e 0 2β 2 β β r0 α − 2 e−2βt + r02 − (σ 2 + 2α) β 2β 2 ! 2 2α α −βt α α α − r0 − e − r02 − 2r0 + e−2βt − β β β β β ασ 2 σ 2 −βt ασ 2 ασ 2 −2βt e − e−2βt + − 2 e−βt + e 2 β 2β β 2β 2 ασ 2 2 σ 2 −βt = r0 e − e−2βt + 1 − e−βt , t ≥ 0. β 2β 2 = r0

Problem 4.19 a) The Itô formula cannot be applied to the function f (x) := (x − K)+ because it is not (twice) differentiable. b) The function x 7→ fε (x) can be plotted as follows with K = 1. 72

72 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

fε

K-ε

K+ε

Fig. S.13: Graph of the function x 7→ fε (x). We note that fε converges uniformly on R to the function x 7→ (x − K)+ as we have ε x ∈ R. (S.4.17) 0 ≤ fε (x) − (x − K)+ ≤ , 4 c) Applying the Itô formula to the function fε we find 1 w T ′′ f (Bt )dt 2 0 ε wT 1 = fε (B0 ) + fε′ (Bt )dBt + 1(K−ε,K+ε) (Bt )dt, 0 4ε 0

fε (BT ) = fε (B0 ) +

wT 0 wT

fε′ (Bt )dBt +

and to conclude it suffices to note that ℓ

t ∈ [0, T ] : K − ε < Bt < K + ε

wT 0

1(K−ε,K+ε) (Bt )dt.

d) The derivative fε′ (x) of fε (x) is given by  1 if x > K + ε,       1 fε′ (x) := (x − K + ε) if K − ε < x < K + ε,  2ε      0 if x < K − ε. Hence we have ∥1[K,∞) (·) − fε′ (·)∥2L2 (R+ ) = =

w∞

1[K,∞) (x) − fε′ (x) dx 2

0 w K+ε

1 + |fε′ (x)|2 dx 2 1 w K+ε ≤ 2ε + 2 x − K + ε dx 4ε K−ε 3 iK+ε 1 h = 2ε + x−K +ε 2 12ε K−ε "

K−ε

73 September 15, 2022

Solutions Manual

f'ε

K-ε

K+ε

Fig. S.14: Graph of the derivative x 7→ fε′ (x).

e) IE

i) We have w T 0

= 2ε +

2ε . 3

1[K,∞) (Bt ) − fε′ (Bt ) dt = 2

2 i

1[K,∞) (Bt ) − fε′ (Bt )

wT w∞

2 2 1 = 1[K,∞) (x) − fε′ (x) e−x /(2t) dx √ dt 0 −∞ 2πt w T w K+ε 2 2 1 ≤ 1[K,∞) (x) − fε′ (x) e−(K−ε) /(2t) dx √ dt 0 K−ε 2πt wT 2 1 ≤ ∥1[K,∞) (·) − fε′ (·)∥2L2 (R+ ) e−(K−ε) /(2t) √ dt 0 2πt w T 2 2ε 1 ≤ 2ε + e−(K−ε) /(2t) √ dt, 0 3 2πt where wT

wT 2 1 1 dt < e−K /(8t) √ dt < ∞, 0 2πt 2πt w T 2 for ε < K/2, hence lim IE 1[K,∞) (Bt ) − fε′ (Bt ) dt = 0, and 0

e−(K−ε) /(2t) √

ε→0

by the Itô isometry w 2 hw ∞ ∞ 2 i IE 1[K,∞) (Bt ) − fε (Bt ) dBt = IE 1[K,∞) (Bt ) − fε (Bt ) dt 0

we find that lim IE

ε→0

w ∞ 0

1[K,∞) (Bt )dBt −

w∞ 0

fε (Bt )dBt

= 0,

74 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition which shows that

r∞ 0

fε (Bt )dBt converges to

w∞ 0

1[K,∞) (Bt )dBt in

L2 (Ω) as ε tends to zero. ii) By (S.4.17) we have h 2 i ε IE (BT − K)+ − fε (BT ) ≤ , 4 hence fε (BT ) converges to (BT − K)+ in L2 (Ω). iii) Similarly, fε (B0 ) converges to (B0 − K)+ for any fixed value of B0 . As a consequence of (ei), (eii) and (eiii) above, the equation (4.47) shows that 1 ℓ t ∈ [0, T ] : K − ε < Bt < K + ε 2ε admits a limit in L2 (Ω) as ε tends to zero, and this limit is denoted by LK [0,T ] . The formula (4.48) is known as the Tanaka formula. Problem 4.20 a) We have 0 ≤ IE[(X − ε)+ ] w∞ 2 2 1 = √ (x − ε)e−x /(2σ ) dx 2 ε 2πσ w∞ w∞ 2 2 2 2 1 ε = √ xe−x /(2σ ) dx − √ e−x /(2σ ) dx 2 2 ε 2πσ 2πσ ε σ 2 h −x2 /(2σ2 ) i∞ = −√ e − εP(X ≥ ε) ε 2πσ 2 2 2 σ2 e−εx /(2σ ) − εP(X ≥ ε), = √ 2πσ 2 which leads to the conclusion. b) We have P(X ∈ dx and X + Y ∈ dz) P(X + Y ∈ dz) P(X ∈ dx and Y ∈ (dz) − x) = P(X + Y ∈ dz) p 2 2 2 2 2π(α2 + β 2 ) e−x /(2α )−(z−x) /(2β ) = dx 2 /(2(α2 +β 2 )) −z 2παβ e p 1/β 2 + 1/α2 −(x2 (1+β 2 /α2 )+(x2 +z2 −2xz)(1+α2 /β 2 )−z2 )/(2(α2 +β 2 )) √ = e dx 2π p 1/β 2 + 1/α2 −(x2 (2+β 2 /α2 +α2 /β 2 )+z2 α2 /β 2 −2xz(1+α2 /β 2 ))/(2(α2 +β 2 )) √ = e dx 2π

P(X ∈ dx | X + Y = z) =

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Solutions Manual 1/β 2 + 1/α2 −(x(β/α+α/β)−zα/β)2 /(2(α2 +β 2 )) √ e dx 2π p 1/β 2 + 1/α2 −(x((α2 +β 2 )/(αβ))−zα/β)2 /(2(α2 +β 2 )) √ = e dx 2π p 2 2 1/β + 1/α −(x−zα2 /(α2 +β 2 ))2 /(2/(1/α2 +1/β 2 ))) √ e dx. = 2π =

c) Given that Bu = x we decompose Bv = (Bv − B(u+v)/2 ) + (B(u+v)/2 − Bu ) + x, and apply the result of Question (b) by taking X = B(u+v)/2 − Bu

and

Y = Bv − B(u+v)/2 ,

i.e.

v−u and z = y − x, 2 which shows that the distribution of B(u+v)/2 = x + X given that Bu = x x+y v−u and Bv = y is Gaussian N , with mean 2 4 α2 = β 2 =

x+y y−x α2 z = =x+ α2 + β 2 2 2

α2 β 2 v−u . = α2 + β 2 4

and variance

d) Four linear interpolations are displayed in Figure S.15.

0.0

0.2

0.4

0.6 t

0.8

1.0

0.0

0.2

0.4

0.6

0.8

1.0

0.5

1.0

1.5

2.0

n= 3

0.0

0.5

1.0

1.5

2.0

n= 2

0.0

0.5

1.0

1.5

2.0

n= 1

0.0

0.5

1.0

1.5

2.0

n= 0

0.0

0.2

0.4

0.6

0.8

1.0

0.0

0.2

0.4

0.6

0.8

1.0

Fig. S.15: Samples of linear interpolations. (0)

e) Clearly, the statement is true for n = 0 because Z1 and B1 have the same N (0, 1) distribution. Next, assuming that it holds at the rank n, we note that the terms appearing in the sequence (n+1)

(n+1)

Z (n+1) = 0, Z1/2n+1 , Z2/2n+1 , Z3/2n+1 , Z4/2n+1 , . . . , Z1

can be written for any k = 0, 1, . . . , 2n − 1 as

76 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 

(n+1)

. . . , Z (n+1) 2k/2n+1 , 

2 (n)

(n)

= . . . , Zk/2n ,

(n)

Z2k/2n+1 + Z(2k+2)/2n+1 2



(n) = . . . , Zk/2n , N 



(n+1)

Z2k/2n+1 + Z(2k+2)/2n+1

+ N 0, 1/2

n+2

(n+1) , Z(2k+2)/2n+1 , . . .

 (n) + N 0, 1/2n+2 , Z(k+1)/2n , . . .

(n) (n) Z2k/2n+1 + Z(2k+2)/2n+1

 1  (n) ,Z

 .

(k+1)/2n , . . .

2n+2

(S.4.18) On the other hand, the result of Question (c) shows that given that B2k/2n+1 = x and B(2k+2)/2n+1 = y, the distribution of B(2k+1)/2n+1 is N

B2k/2n+1 + B(2k+2)/2n+1 (2k + 2)/2n+1 − (2k + 2)/2n+1 , 2 4 B2k/2n+1 + B(2k+2)/2n+1 1 =N , n+2 . (S.4.19) 2 2

Given that Z (n) and B (n) have same distribution, we conclude by comparing (S.4.18) and (S.4.19) that Z (n+1) and B (n+1) also have same distribution. f) We have ! (n+1)

sup |Zt

(n)

− Zt | ≥ εn

t∈[0,1]

(n) (n+1) =P max n |Z(2k+1)/2n+1 − Z(2k+1)/2n+1 | ≥ εn k=0,1,...,2 −1   n o [ (n+1) (n)  ≤P |Z(2k+1)/2n+1 − Z(2k+1)/2n+1 | ≥ εn  k=0,1,...,2n −1

≤

n 2X −1

(n+1)

(n)

P |Z(2k+1)/2n+1 − Z(2k+1)/2n+1 | ≥ εn

k=0

(n+1) (n) = 2n P |Z1/2n+1 − Z1/2n+1 | ≥ εn   (n) (n) Z n + Z1/2n 0/2 (n+1) = 2n P  Z1/2n+1 − ≥ εn  . 2 g) Since (n)

(n+1)

Z1/2n+1 = "

(n)

+ Z1/2n 2

(n)

+ N (0, 1/2n+2 ) = Z1/2n+1 + N (0, 1/2n+2 ), 77 September 15, 2022

Solutions Manual we have ! P

(n+1)

sup |Zt

(n)

− Zt | ≥ εn

t∈[0,1]

(n+1)

(n)

≤ 2n P |Z1/2n+1 − Z1/2n+1 | ≥ εn 

(n)

(n+1) = 2n P  Z1/2n+1 −

(n)

Z0/2n + Z1/2n 2

= 2n P N (0, 1/2n+2 ) ≥ εn ≤

 ≥ εn 

2n/2 −ε2n 2n+1 √ e , εn 2π

where we applied the bound of Question (a) to the Gaussian random variable (n) (n) Z0/2n + Z1/2n (n+1) ≃ N (0, 1/2n+2 ). Z1/2n+1 − 2 h) We have ! X X (n+1) (n) − Zt | ≥ εn P ∥Z (n+1) − Z (n) ∥∞ ≥ 2−n/4 = P sup |Zt n≥0

t∈[0,1]

n≥0

X 2n/2 2 n+1 √ e−εn 2 ≤ ε 2π n≥0 n 1 X 3n/4 −21+n/2 = √ 2 e < ∞, 2π n≥0 since 1+(n+1)/2

23(n+1)/4 e−2 n→∞ 23n/4 e−21+n/2 lim

= 23/4 lim e−2

1+n/2

√ ( 2−1)

n→∞

= 0.

Hence the Borel-Cantelli lemma shows that P ∥Z (n+1) − Z (n) ∥∞ ≥ 2−n/4 for infinitely many n = 0, therefore we have P ∥Z (n+1) − Z (n) ∥∞ < 2−n/4 except for finitely many n = 1. i) The result of Question (h) shows that with probability one we have lim ∥Z (p) − Z (q) ∥∞ = lim

p,q→∞

p−1 X n=q

Z (n+1) − Z (n) ∞

78 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition p−1 X

≤ lim

p,q→∞

≤ lim

p→∞

∥Z (n+1) − Z (n) ∥∞

n=q

∥Z (n+1) − Z (n) ∥∞

n≥q

= 0, hence the sequence Z is Cauchy in C0 ([0, 1]) for the ∥ · ∥∞ norm. n≥0 Since C0 ([0, 1]) is a complete space for the ∥ · ∥∞ norm, this implies that, with probability one, the sequence (Z (n) )n≥0 admits a limit in C0 ([0, 1]). (n) j) 1. By construction we have Z0 = 0 for all n ∈ N, hence Z0 = (n) limn→∞ Z0 = 0, almost surely. (n)

2. The sample trajectories t 7→ Zt are continuous, because the limit Z belongs to C0 ([0, 1]) with probability 1. 3. The result of Question (e) shows that for any fixed m ≥ 1, the sequences Zt1 − Zt0 , Zt2 − Zt1 , . . . , Ztm − Ztm−1 and Bt1 − Bt0 , Bt2 − Bt1 , . . . , Btm − Btm−1 have same distribution when the t′k s are dyadic rationals of the form tk = in /2n , k = 0, 1, . . . , n. This property extends to any sequence t0 , t1 , . . . , tm of real numbers by approximation of each tk > 0 by a sequence (in )n∈N such that tk = limn→∞ in /2n and taking the limit as n tends to infinity. 4. By a similar argument as in the above point 3, one can show that for any 0 ≤ s < t, Zt − Zs has the Gaussian distribution N (0, t − s). Problem 4.21 a) We have n (n) X IE QT = IE (BkT /n − B(k−1)T /n )2

k=1 n X

k=1

= T,

T T − (k − 1) n n

n ≥ 1.

b) We have

79 September 15, 2022

Solutions Manual  !2  n X (n) 2 2   IE (QT ) = IE (BkT /n − B(k−1)T /n ) k=1

 = IE 

n X

 (BkT /n − B(k−1)T /n )2 (BlT /n − B(l−1)T /n )2 

k,l=1

n X

IE (BkT /n − B(k−1)T /n )4

k=1

IE (BkT /n − B(k−1)T /n )2 IE (BlT /n − B(l−1)T /n )2

1≤k<l≤n

n X

(kT /n − (k − 1)T /n)2

k=1

(kT /n − (k − 1)T /n)(lT /n − (l − 1)T /n)

1≤k<l≤n

n(n − 1)T 2 T2 + n n2 2 2T = T2 + , n ≥ 1, n

hence (n) (n) 2 (n) 2 2T 2 Var QT = IE QT − IE QT = , n

n ≥ 1.

c) We have (n) (n) (n) ∥QT − T ∥2L2 (Ω) = IE (QT − IE[QT ])2 (n) = Var QT n(n + 2)T 2 − T2 n2 2T 2 = , n

hence (n)

lim ∥QT − T ∥2L2 (Ω) = lim

n→∞

showing that

2T 2 = 0, n

(n)

lim QT = T

n→∞

in L2 (Ω). d) We have

80 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition n X

(BkT /n − B(k−1)T /n )B(k−1)T /n =

1X 2 2 BkT /n − B(k−1)T /n 2 k=1

k=1 n

1X (BkT /n − B(k−1)T /n )(BkT /n − B(k−1)T /n ) − 2 k=1

1 ((BT )2 − (B0 )2 ) 2 n 1X (BkT /n − B(k−1)T /n )(BkT /n − B(k−1)T /n ) − 2 k=1

1 (n) (BT )2 − QT , 2

which converges to ((BT )2 − T )/2 in L2 (Ω) as n tends to infinity, hence wT 0

Bt dBt = lim

n X

n→∞

(BkT /n − B(k−1)T /n )B(k−1)T /n

k=1 2

(BT ) − T . 2

e) We have n (n) X e = IE (B(k−1/2)T /n − B(k−1)T /n )2 IE Q T

k=1 n X

((k − 1/2)T /n − (k − 1)T /n)

k=1

T , 2

n ≥ 1.

Next, we have  IE

e (n) 2 = IE  Q T

n X

!2  B(k−1/2)T /n − B(k−1)T /n



k=1

 = IE 

n X

 (B(k−1/2)T /n − B(k−1)T /n ) (BlT /n − B(l−1)T /n ) 2

2

k,l=1

n X

IE (B(k−1/2)T /n − B(k−1)T /n )4

k=1

IE (B(k−1/2)T /n − B(k−1)T /n )2 IE (B(l−1/2)T /n − B(l−1)T /n )2

1≤k<l≤n

81 September 15, 2022

Solutions Manual =3

n X

((k − 1/2)T /n − (k − 1)T /n)2

k=1

((k − 1/2)T /n − (k − 1)T /n)((l − 1/2)T /n − (l − 1)T /n)

1≤k<l≤n

n(n − 1)T 2 T2 + 4n 4n2 n(n + 2)T 2 = , n ≥ 1. 4n2

Finally, we find (n) e (n) 2 e (n) − T /2∥2 2 e ∥Q L (Ω) = IE (QT − IE[QT ]) T (n) e = Var Q T n(n + 2)T 2 T2 − 4n2 4 T2 = , 2n

hence

T e (n) − T /2∥2 2 lim ∥Q lim = 0, L (Ω) = n→∞ T 2n

n→∞

showing that (n)

e = lim Q T

n→∞

T 2

in L2 (Ω). f) We have n X

BkT /n − B(k−1)T /n B(k−1/2)T /n

k=1

n X

BkT /n − B(k−1/2)T /n B(k−1/2)T /n

k=1 n X

B(k−1/2)T /n − B(k−1)T /n B(k−1/2)T /n

k=1 n

1X 2 2 BkT /n − B(k−1/2)T = /n 2 k=1 n

1X BkT /n − B(k−1/2)T /n BkT /n − B(k−1/2)T /n − 2 k=1 n

1X 2 2 + B(k−1/2)T /n − B(k−1)T /n 2 k=1

82 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition n

1X B(k−1/2)T /n − B(k−1)T /n B(k−1/2)T /n − B(k−1)T /n 2 k=1

1 1X BkT /n − B(k−1/2)T /n BkT /n − B(k−1/2)T /n = (BT )2 − 2 2 k=1

n 1X + B(k−1/2)T /n − B(k−1)T /n B(k−1/2)T /n − B(k−1)T /n , 2 k=1

which converges to ((BT )2 − T + T )/2 = (BT )2 /2 in L2 (Ω) as n tends to infinity, hence wT

Bt ◦ dBt = lim

n X

n→∞

(BkT /n − B(k−1)T /n )B(k−1/2)T /n =

k=1

(BT )2 . 2

g) We have n 2 (n) X e IE B(k−α)T /n − B(k−1)T /n IE Q = T

k=1 n X

((k − α)T /n − (k − 1)T /n)

k=1

T = (1 − α) , 2

n ≥ 1.

Next, we have IE

 !2  n X (n) 2 2 e  QT = IE  (B(k−α)T /n − B(k−1)T /n ) k=1

 = IE 

n X

 (B(k−α)T /n − B(k−1)T /n ) (BlT /n − B(l−1)T /n ) 2

2

k,l=1

n X

IE (B(k−α)T /n − B(k−1)T /n )4

k=1

IE (B(k−α)T /n − B(k−1)T /n )2 IE (B(l−α)T /n − B(l−1)T /n )2

1≤k<l≤n

n X

((k − α)T /n − (k − 1)T /n)2

k=1

((k − α)T /n − (k − 1)T /n)((l − α)T /n − (l − 1)T /n)

1≤k<l≤n

= 3(1 − α)2 "

T2 n(n − 1)T 2 + (1 − α)2 n n2 83 September 15, 2022

Solutions Manual = (1 − α)2

n(n + 2)T 2 , n2

n ≥ 1.

Finally we find (n) 2 e (n) − IE Q e Q T T (n) e = Var Q T

(n)

e − (1 − α)T /2∥2 2 ∥Q L (Ω) = IE T

n(n + 2)T 2 − (1 − α)2 T 2 n2 T2 = 2(1 − α)2 , n

= (1 − α)2

hence 2

T 2 e (n) − (1 − α)T ∥2 2 = 0. lim ∥Q lim L (Ω) = (1 − α) n→∞ T n

n→∞

Next, we have n X

(BkT /n − B(k−1)T /n )B(k−α)T /n

k=1

= =

n X

(BkT /n − B(k−α)T /n )B(k−α)T /n +

k=1 n X

1 2

n X

(B(k−α)T /n − B(k−1)T /n )B(k−α)T /n

k=1 n

2 2 BkT /n − B(k−α)T /n −

k=1 n

1X (BkT /n − B(k−α)T /n )(BkT /n − B(k−α)T /n ) 2 k=1

1X 2 2 + B(k−α)T /n − B(k−1)T /n 2 k=1 n

1X + (B(k−α)T /n − B(k−1)T /n )(B(k−α)T /n − B(k−1)T /n ) 2 k=1

1X 1 (BkT /n − B(k−α)T /n )(BkT /n − B(k−α)T /n ) = (BT )2 − 2 2 k=1

1X (B(k−α)T /n − B(k−1)T /n )(B(k−α)T /n − B(k−1)T /n ), + 2 k=1

which converges to (BT )2 − αT + (1 − α)T (BT )2 + (1 − 2α)T = 2 2 in L2 (Ω) as n tends to infinity, hence

84 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition wT 0

Bt ◦ dα Bt = lim

n X

n→∞

(BkT /n − B(k−1)T /n )B(k−α)T /n

k=1 2

(BT ) + (1 − 2α)T . 2

In particular we find wT 0

Bt ◦ d0 Bt = lim

n→∞

n X

(BkT /n − B(k−1)T /n )BkT /n =

k=1

(BT )2 + T , 2

and we note that wT wT 1 wT Bt dBt + Bt ◦ d1 Bt . Bt ◦ dBt = 0 0 0 2 h) We have lim

n→∞

n X

(k − α)

k=1

T n

T T − (k − 1) n n

T X T (k − α) n→∞ n n

= lim

k=1 n

k=1

T X T T XT = lim k − α lim n→∞ n n→∞ n n n n(n + 1) T2 = T lim − α lim n→∞ n→∞ n 2n2 T2 = , 2 2

which does not depend on α ∈ [0, 1]< hence the stochastic phenomenon of the previous questions does not occur when approximating the deterrT ministic integral 0 tdt = T 2 /2 by Riemann sums. In quantitative finance we choose to use the Itô integral (which corresponds to the choice α = 1) because it is suitable for the modeling of market returns as St+∆t − St dSt ≃ = µ∆t + σ∆Bt = µ∆t + (Bt+∆t − Bt )σ St St or dSt ≃ St+∆t − St = µSt ∆t + σSt ∆Bt , = µSt ∆t + σSt (Bt+∆t − Bt ),

85 September 15, 2022

Solutions Manual based on the value St at the the left endpoint of the discretized time interval [t, t + ∆t].

Chapter 5 Exercise 5.1 For all x ∈ R, we have 2 P(ST ≤ x) = P S0 eσBT +(µ−σ /2)T ≤ x x σ2 T ≤ log = P σBT + µ − 2 S0 1 x σ2 = P BT ≤ log T − µ− σ S0 2 w (log(x/S0 )−(µ−σ2 /2)T )/σ 2 dy = e−y /(2T ) √ −∞ 2πT w (log(x/S0 )−(µ−σ2 /2)T )/(σ√T ) −z 2 /2 dz √ e = −∞ 2π 1 x σ2 √ − µ− =Φ log T , S0 2 σ T where Φ(x) :=

−∞

2 dy e−y /2 √ , 2πT

x ∈ R,

denotes the standard Gaussian cumulative distribution function. After differentiation with respect to x, we find the lognormal probability density function dP(ST ≤ x) dx ∂ w (log(x/S0 )−(µ−σ2 /2)T )/σ −y2 /(2T ) dy √ = e ∂x −∞ 2πT 2 ∂ 1 x σ √ = Φ log − µ− T ∂x S0 2 σ T 1 1 x σ2 √ φ √ = log − µ− T S0 2 xσ T σ T 2 2 2 1 √ = e−(−(µ−σ /2)T +log(x/S0 )) /(2σ T ) , xσ 2πT

f (x) =

where

2 1 φ(y) = Φ′ (y) := √ e−y /2 , 2π

x > 0,

y ∈ R,

denotes the standard Gaussian probability density function. 86

86 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Exercise 5.2 a) We have d log St =

1 1 σ2 dSt − (dSt )2 = rdt + σdBt − dt, St 2St2 2

t ≥ 0.

b) We have f (t) = f (0)ect (continuous-time interest rate compounding), and 2

St = S0 eσBt −σ t/2+rt ,

t ≥ 0,

(geometric Brownian motion). c) Those quantities can be directly computed from the expression of St as a function of the N (0, t) random variable Bt . Namely, we have 2 IE[St ] = IE S0 eσBt −σ t/2+rt 2 = S0 e−σ t/2+rt IE eσBt = S0 ert , where we used the Gaussian moment generating function (MGF) formula 2 IE eσBt = eσ t/2 for the normal random variable Bt ≃ N (0, t), t > 0. Similarly, we have 2 IE[St2 ] = IE S02 e2σBt −σ t+2rt 2 = S02 e−σ t+2rt IE e2σBt 2

= S02 eσ t+2rt ,

t ≥ 0.

d) We note that from the stochastic differential equation St = S0 + r

wt 0

Ss ds + σ

wt 0

Ss dBs ,

the function u(t) := IE[St ] satisfies the Ordinary Differential Equation (ODE) u′ (t) = ru(t) with u(0) = S0 and solution u(t) = IE[St ] = S0 ert . On the other hand, by the Itô formula we have dSt2 = 2St dSt + (dSt )2 = 2rSt2 dt + σ 2 St2 dt + 2σSt2 dBt , hence letting v(t) = IE St2 and taking expectations on both sides of St2 = S02 + 2r

wt 0

Su2 du + σ 2

wt 0

Su2 du + 2σ

wt 0

Su2 dBu ,

we find "

87 September 15, 2022

Solutions Manual v(t) = IE St2 = S02 + (2r + σ 2 ) IE = S02 + (2r + σ 2 ) = S02 + (2r + σ 2 )

w t

wt 0

w t Su2 du + 2σ IE Su2 dBu 0

IE Su2 du v(u)du,

hence v(t) := IE St2 satisfies the ordinary differential equation v ′ (t) = (σ 2 + 2r)v(t), with v(0) = S02 and solution 2 v(t) = IE St2 = S02 e(σ +2r)t , which recovers Var[St ] = IE St2 − (IE[St ])2 = v(t) − u2 (t) 2

= S02 e(σ +2r)t − S02 e2rt 2

= S02 e2rt (eσ t − 1),

t ≥ 0.

Exercise 5.3 Using the bivariate Itô formula (4.26), we find ∂f ∂f (St , Yt )dSt + (St , Yt )dYt ∂x ∂y 2 2 1∂ f 1∂ f ∂2f (St , Yt )(dSt )2 + (St , Yt )(dYt )2 + (St , Yt )dSt • dYt + 2 ∂x2 2 ∂y 2 ∂x∂y ∂f ∂f = (St , Yt )(rSt dt + σSt dBt ) + (St , Yt )(µYt dt + ηYt dWt ) ∂x ∂y σ 2 St2 ∂ 2 f η 2 Yt2 ∂ 2 f ∂2f + (St , Yt )dt + (St , Yt )dt + ρσηSt Yt (St , Yt )dt. 2 ∂x2 2 ∂y 2 ∂x∂y

df (St , Yt ) =

Exercise 5.4 Taking expectations on both sides of (5.24) shows that w T IE[ST ] = C(S0 , r, T ) + IE ζt,T dBt = C(S0 , r, T ), 0

hence C(S0 , r, T ) = IE[ST ] 2

= IE[S0 erT +σBT −σ T /2 ] 88

88 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2

= S0 erT −σ T /2 IE[eσBT ] 2

= S0 erT −σ T /2+σ T /2 = S0 e

where we used the moment generating function 2

IE[eσBT ] = eσ T /2 of the Gaussian random variable BT ≃ N (0, T ). On the other hand, the discounted asset price Xt := e−rt St satisfies dXt = σXt dBt , which shows that wT XT = X0 + σ Xt dBt . 0

Multiplying both sides by erT shows that ST = erT S0 + σ

wT 0

erT Xt dBt = erT S0 + σ

wT 0

e(T −t)r St dBt ,

which recovers the relation C(S0 , r, T ) = S0 erT , and shows that ζt,T = σe(T −t)r St , t ∈ [0, T ]. Exercise 5.5 a) We have St = f (Xt ), t ≥ 0, where f (x) = S0 ex and (Xt )t∈R+ is the Itô process given by wt wt Xt := σs dBs + us ds, t ≥ 0, 0

or in differential form dXt := σt dBt + ut dt,

t ≥ 0,

hence dSt = df (Xt )

1 = f ′ (Xt )dXt + f ′′ (Xt )(dXt )2 2

1 = ut f ′ (Xt )dt + σt f ′ (Xt )dBt + σt2 f ′′ (Xt )dt 2 1 = S0 ut eXt dt + S0 σt eXt dBt + S0 σt2 eXt dt 2 1 2 = ut St dt + σt St dBt + σt St dt. 2 b) The process (St )t∈R+ satisfies the stochastic differential equation

89 September 15, 2022

Solutions Manual dSt =

1 ut + σt2 St dt + σt St dBt . 2

Exercise 5.6 a) We have IE[St ] = 1 because the expected value of the Itô stochastic integral is zero. Regarding the variance, using the Itô isometry (4.8) we have " 2 # wt 2 Var[St ] = σ 2 IE eσBs −σ s/2 dBs 0

w t

ds 0 2 w t 2 2 =σ IE eσBs −σ s/2 ds = σ 2 IE

eσBs −σ s/2

= σ2 = σ2 = σ2 = σ2

wt 0

wt 0 wt 0 wt 0

h i 2 IE e2σBs −σ s ds 2 e−σ s IE e2σBs ds 2

e−σ s e2σ s ds 2

eσ s ds

= eσ t − 1. b) Taking f (x) = log x, we have d log(St ) = df (St )

1 = f ′ (St )dSt + f ′′ (St )(dSt )2 2 2 2 1 = σf ′ (St )eσBt −σ t/2 dBt + σ 2 f ′′ (St )e2σBt −σ t dt 2 σ σBt −σ2 t/2 σ 2 2σBt −σ2 t = e dBt − e dt. (S.5.20) St 2St2 2

c) We check that when St = eσBt −σ t/2 , t ≥ 0, we have log St = σBt − σ 2 t/2,

and d log St = σdBt −

σ2 dt. 2

On the other hand, we also find σdBt −

2 σ2 σ σ 2 2σBt −σ2 t dt = eσBt −σ t/2 dBt − e dt, 2 St 2St2

showing by (S.5.20) that the equation 90

90 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition d log St =

σ σBt −σ2 t/2 σ 2 2σBt −σ2 t e e dBt − dt St 2St2 2

is satisfied. By uniqueness of solutions, we conclude that St := eσBt −σ t/2 solves wt 2 St = 1 + σ eσBs −σ s/2 dBs , t ≥ 0. 0

Exercise 5.7 a) Leveraging with a factor β : 1 means that we invest the amount ξt St = βFt on the risky asset priced St . In this case, the fund value Ft at time t ≥ 0 decomposes into the portfolio Ft = ξt St + ηt At = β

Ft Ft St − (β − 1) At , St At

t ≥ 0,

with ξt = βFt /St and ηt = −(β − 1)Ft /At , t ≥ 0. b) We have dFt = ξt dSt + ηt dAt Ft Ft = β dSt − (β − 1) dAt St At Ft = β dSt − (β − 1)rFt dt St = βFt (rdt + σdBt ) − (β − 1)rFt dt = rFt dt + βσFt dBt ,

t ≥ 0.

The above equation shows that the volatility βσ of the fund is β times the volatility of the index. On the other hand, the risk-free rate r remains the same. c) By Proposition 5.15 we have 2

Ft = F0 eβσBt +rt−β σ t/2 2

= F0 eσBt +rt/β−βσ t/2

β 2 2 = F0 eσBt +rt−σ t/2−(1−1/β)rt−(β−1)σ t/2 2 2 β = F0 eσBt +rt−σ t/2 e−(β−1)rt−β(β−1)σ t/2 β 2 2 = S0 eσBt +rt−σ t/2 e−(β−1)rt−β(β−1)σ t/2 2

= Stβ e−(β−1)rt−β(β−1)σ t/2 ,

t ≥ 0.

Exercise 5.8 a) For t ∈ [0, T ] and i = 1, 2 we have "

91 September 15, 2022

Solutions Manual (i) (i) (i) 2 IE St = IE S0 eµt+σi Wt −σi t/2 (i) 2 (i) = S0 eµt−σi t/2 IE eσi Wt (i)

= S0 eµt−σi t/2+σi t/2 (i) = S0 eµt .

b) For all t ∈ [0, T ] and i = 1, 2, we have IE

(i) 2

(i) 2 2µt+2σi Wt(i) −σi2 t S0 e (i) (i) 2 2µt−σi2 t = S0 e IE e2σi Wt 2 2 (i) 2 2µt−σi t+2σi t = S0 e (i) 2 2µt+σi2 t = S0 e ,

= IE

hence (i) (i) 2 (i) Var St = IE St − IE St (i) 2 2µt+σi2 t (i) 2 2µt = S0 e − S0 e 2 (i) 2 2µt σi t = S0 e e − 1 , t ∈ [0, T ], i = 1, 2. c) We have (2) (1) (2) (1) (1) (2) Var St − St = Var St + Var St − 2 Cov St , St with (1) (2) (1) (2) (1) (2) 2 2 = IE S0 S0 e2µt+σ1 Wt −σ1 t/2+σ2 Wt −σ2 t/2 IE St St (1) (2) 2 2 (1) (2) = S0 S0 e2µt−σ1 t/2−σ2 t/2 IE eσ1 Wt +σ2 Wt 2 2 1 (1) (2) 2 (1) (2) IE σ1 Wt + σ2 Wt , = S0 S0 e2µt−σ1 t/2−σ2 t/2 exp 2 with (1) (2) 2 (1) 2 (1) (2) (2) 2 IE σ1 Wt + σ2 Wt = IE σ1 Wt + 2 IE σ1 Wt σ2 Wt + IE σ2 Wt = σ12 t + 2ρσ1 σ2 t + σ22 t, hence

(1) (2) (1) (2) IE St St = S0 S0 e2µt+ρσ1 σ2 t ,

and (1)

(2)

Cov St , St

(1) (2) (1) (2) (1) (2) = IE St St −IE St IE St = S0 S0 e2µt (eρσ1 σ2 t −1),

and therefore 92

92 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (2) (1) Var St − St (1) 2 2µt σ12 t (2) 2 2µt σ22 t (1) (2) = S0 e (e − 1) + S0 e (e − 1) − 2S0 S0 e2µt (eρσ1 σ2 t − 1) 2 2 (1) 2 σ1 t (2) 2 σ2 t (1) (2) (2) (1) 2 = e2µt S0 e + S0 e − 2S0 S0 eρσ1 σ2 t − S0 − S0 . 2

Exercise 5.9 Letting Xt := f (t)eσBt −σ t/2 and noting the relation 2

deσBt −σ t/2 = σf (t)eσBt −σ t/2 dBt ,

t ≥ 0,

see Proposition 5.15 and Relation (5.22) with µ = 0, we have 2

dXt = eσBt −σ t/2 f ′ (t)dt + f (t)deσBt −σ t/2 = eσBt −σ t/2 f ′ (t)dt + σf (t)eσBt −σ t/2 dBt f ′ (t) = Xt dt + σXt dBt f (t) = h(t)Xt dt + σXt dBt , hence

d f ′ (t) log f (t) = = h(t), dt f (t)

which shows that log f (t) = log f (0) +

wt 0

h(s)ds,

and 2

Xt = f (t)eσBt −σ t/2 w t σ2 = f (0) exp h(s)ds + σBt − t 0 2 w t σ2 = X0 exp h(s)ds + σBt − t , 0 2

t ≥ 0.

Exercise 5.10 a) We have St = eXt

wt 1 w t 2 Xs us eXs dBs + vs eXs ds + u e ds 0 2 0 s wt wt 2 wt σ = eX0 + σ eXs dBs + ν eXs ds + eXs ds 0 0 2 0 wt wt σ2 w t = S0 + σ Ss dBs + ν Ss ds + Ss ds. 0 0 2 0 = eX0 +

wt 0

93 September 15, 2022

Solutions Manual b) Let r > 0. The process (St )t∈R+ satisfies the stochastic differential equation dSt = rSt dt + σSt dBt when r = ν + σ 2 /2. c) We have Var[Xt ] = Var[(BT − Bt )σ] = σ 2 Var[BT − Bt ] = (T − t)σ 2 ,

t ∈ [0, T ].

d) Let the process (St )t∈R+ be defined by St = S0 eσBt +νt , t ≥ 0. Using the time splitting decomposition ST = St

ST = St e(BT −Bt )σ+ντ , St

we have P(ST > K | St = x) = P(St e(BT −Bt )σ+(T −t)ν > K | St = x) = P(xe(BT −Bt )σ+(T −t)ν > K) = P(e(BT −Bt )σ > Ke−(T −t)ν /x) 1 BT − Bt > √ log Ke−(T −t)ν /x =P √ T −t σ T −t ! log Ke−(T −t)ν /x √ = 1−Φ σ τ ! −(T −t)ν log Ke /x √ =Φ − σ τ log(x/K) + ντ √ =Φ , σ τ where τ = T − t. Problem 5.11 (Exercise 4.19 continued). a) The option payoff is (BT − K)+ at maturity. b) We can ignore what happens between two crossings as every crossing resets the portfolio to its state right before the previous crossing. Based on this, It is clear that every of the four possible scenarios will lead to a portfolio value (BT − K)+ at maturity: i) If B0 < 1 and BT < 1 we issue the option for free and finish with an empty portfolio and zero payoff. ii) If B0 < 1 and BT > 1 we issue the option for free and finish with one AUD and one SGD to refund, which yields the payoff BT − 1 = (BT − 1)+ .

94 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition iii) If B0 > 1 and BT < 1 we purchase one AUD and borrow one SGD at the start, however the AUD will be sold and the SGD refunded before maturity, resulting into an empty portfolio and zero payoff. iv) If B0 > 1 and BT > 1 we purchase one AUD and borrow one SGD right before maturity, which yields the payoff BT − 1 = (BT − 1)+ . Therefore we are hedging the option in all cases. Note that P(BT = K) = 0 so the case BT = 1 can be ignored with probability one. c) Since the portfolio strategy is to hold AU$1 when Bt > K and to and borrow SG$1 when Bt > K, we let ξt := 1(K,∞) (Bt )

and ηt := −1(K,∞) (Bt ),

t ∈ [0, T ],

which is called a stop-loss/start-gain strategy. wt d) Noting that ηs dAs = 0 because At = A0 is constant, t ∈ [0, T ], we find 0 by the Itô-Tanaka formula (4.48) that wt 0

ηs dAs +

wt 0

ξs dBs =

wT 0

1[K,∞) (Bt )dBt

1 . = (BT − K)+ − (B0 − K)+ − LK 2 [0,T ]

e) Question (d) shows that (BT − K)+ = (B0 − K)+ +

wt 0

ηs dAs +

wt 0

1 , ξs dBs + LK 2 [0,T ]

i.e. the initial premium (B0 − K)+ plus the sum of portfolio profits and losses is not sufficient to cover the terminal payoff (BT − K)+ , and that we fall short of this by the positive amount 12 LK [0,T ] > 0. Therefore the portfolio allocation (ξt , ηt )t∈[0,T ] is not self-financing. Additional comments: The stop-loss/start-gain strategy described here is difficult to implement in practice because it would require infinitely many transactions when Brownian motion crosses the level K, as illustrated in Figure S.16.

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Solutions Manual

-1

-0.5

0.5

Fig. S.16: Brownian crossings of level 1.∗ The arbitrage-free price of the option can in fact be computed as the expected discounted option payoff πt = e−(T −t)r IE∗ [(BT − K)+ | Ft ] = e−(T −t)r IE∗ [(BT − Bt + x − K)+ | Ft ]x=Bt = e−(T −t)r IE∗ [(BT − Bt + x − K)+ ]x=Bt w∞ 2 dy = e−(T −t)r (y + Bt − K)+ e−y /(2(T −t)) p −∞ 2π(T − t) w∞ 2 dy = e−(T −t)r (y + Bt − K)e−y /(2(T −t)) p K−Bt 2π(T − t) w∞ 2 dy = e−(T −t)r ye−y /(2(T −t)) p K−Bt 2π(T − t) w∞ 2 dy +(Bt − K)e−(T −t)r e−y /(2(T −t)) p K−Bt 2π(T − t) w∞ 2 dy = e−(T −t)r ye−y /2 √ √ (K−Bt )/ T −t 2π w∞ 2 dy −(T −t)r +(Bt − K)e e−y /2 √ √ (K−Bt )/ T −t 2π e−(T −t)r h −y2 /2 i∞ = √ −e √ (K−Bt )/ T −t 2π Bt − K +(Bt − K)e−(T −t)r Φ √ T −t e−(T −t)r −(K−Bt )2 /(2(T −t)) = √ e 2π 96

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Introduction to Stochastic Finance with Market Examples, Second Edition +(Bt − K)e−(T −t)r Φ

Bt − K √ T −t

=: g(t, Bt ), where the function g(t, x) :=

e−(T −t)r −(K−x)2 /(2(T −t)) √ e +(x−K)e−(T −t)r Φ 2π

x−K √ T −t

t ∈ [0, T ),

solves the Black-Scholes heat equation ∂g 1 ∂2g ∂g (t, x) + r (t, x) + (t, x) = 0 ∂t ∂x 2 ∂2x with terminal condition g(T, x) = (x − K)+ . The Delta gives the amount to be invested in AUD at time t and is given by ξt =

∂g (t, Bt ) ∂x

e−(T −t)r −(K−Bt )2 /(2(T −t)) = (K − Bt ) √ e 2π(T − t) e−(T −t)r −(Bt −K)2 /(2(T −t)) + (Bt − K) p e + e−(T −t)r Φ 2π(T − t) ! Bt − K −(T −t)r =e Φ p (T − t)

Bt − K √ T −t

=: h(t, Bt ), with h(t, x) := e−(T −t)r Φ

x−K √ T −t

0.5

0.6

t ∈ [0, T ),

and h(T, x) = 1[K,∞) (x). Bt

0.1

0.2

0.3

0.4

0.7

0.8

0.9

Fig. S.17: Brownian path started at B0 > 1.

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Solutions Manual (Bt-K)+ g(t,Bt)

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Fig. S.18: Risk-neutral pricing of the FX option by πt (Bt ) = g(t, Bt ) vs stop-loss/startgain pricing.

1[K,∞ )(Bt) h(t,Bt) 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Fig. S.19: Delta hedging of the FX option by ξt = h(t, Bt ) vs the stop-loss/start-gain strategy.

The “one or nothing” stop-loss/start-gain strategy is not self-financing because in practice there is an impossibility to buy/sell the AUD at exactly SGD1.00 to the existence of an order book that generates a gap between bid/ask prices as in the sample of Figure S.20 with 383.16964 < 384.07141.

Fig. S.20: Bitcoin XBT/USD order book. 98

98 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition The existence of the order book will force buying and selling within a certain range [K − ε, K + ε], typically resulting into selling lower than K = 1.00 and buying higher than K = 1.00. This potentially results into a trading loss that can be proportional to the time ℓ t ∈ [0, T ] : K − ε < Bt < K + ε spent by the exchange rate (Bt )t∈[0,T ] within the range [K − ε, K + ε]. The Itô-Tanaka formula (4.48) (BT − K)+ = (B0 − K)+ +

wT 0

1 2

1[K,∞) (Bt )dBt + LK [0,T ] ,

precisely shows that the trading loss equals half the local time LK [0,T ] spent by (Bt )t∈[0,T ] at the level K. When ε is small we have 1 K 1 L ≃ ℓ({t ∈ [0, T ] : K − ε < Bt < K + ε}), 2 [0,T ] 4ε therefore the trading loss is proportional to the time spent by Brownian motion (Bt )t∈R+ within the interval (K − ε, K + ε), with proportionality coefficient 1/(4ε). Bt

K+ε 1 K-ε

0.05

0.1

0.15

0.2

Fig. S.21: Time spent by Brownian motion within the range (K − ε, K + ε). More generally, we could show that there is no self-financing (buy and hold) portfolio that can remain constant over time intervals, and that the selffinancing portfolio has to be constantly re-adjusted in time as illustrated in Figure S.19. This invalidates the stop-loss/start-gain strategy as a selffinancing portfolio strategy.

Chapter 6 Exercise 6.1 "

99 September 15, 2022

Solutions Manual a) By the Itô formula, we have dVt = dg(t, Bt ) =

∂g ∂g 1 ∂2g (t, Bt )dt. (S.6.21) (t, Bt )dt + (t, Bt )dBt + ∂t ∂x 2 ∂x2

Consider a hedging portfolio with value Vt = ηt At + ξt Bt , satisfying the self-financing condition dVt = ηt dAt + ξt dBt = ξt dBt ,

t ≥ 0.

(S.6.22)

By respective identification of the terms in dBt and dt in (S.6.21) and (S.6.22) we get  ∂g 1 ∂2g     0 = ∂t (t, Bt )dt + 2 ∂x2 (t, Bt )dt,     ξt dBt = ∂g (t, Bt )dBt , ∂x hence

 ∂g 1 ∂2g     0 = ∂t (t, Bt ) + 2 ∂x2 (t, Bt ),     ξt = ∂g (t, Bt ), ∂x

and

 ∂g 1 ∂2g     0 = ∂t (t, Bt ) + 2 ∂x2 (t, Bt ),

    ξt = ∂g (t, Bt ), ∂x hence the function g(t, x) satisfies the heat equation 0=

∂g 1 ∂2g (t, x) + (t, x), ∂t 2 ∂x2

x > 0,

(S.6.23)

with terminal condition g(T, x) = x2 , and ξt is given by the partial derivative ∂g (t, Bt ), t ≥ 0. ξt = ∂x b) In order to solve (S.6.23) we substitute a solution of the form g(t, x) = x2 + f (t) in to the partial differential equation, which yields 1 + f ′ (t) = 0 with the terminal condition f (T ) = 0. Therefore we have f (T − t) = T − t, and g(t, x) = x2 + f (t) = x2 + T − t, 0 ≤ t ≤ T. c) By (6.3), we have

100

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Introduction to Stochastic Finance with Market Examples, Second Edition ξt = ξt (Bt ) =

∂g (t, Bt ) = 2Bt , ∂x

0 ≤ t ≤ T,

which recovers the value of ξt found page 163 in the power option example. We also have ηt At = ηt A0 = g(t, Bt ) − ξt Bt = T − t − Bt2 ,

0 ≤ t ≤ T.

Exercise 6.2 By the Itô formula, we have dVt = dg(t, St )

(S.6.24)

p ∂g ∂g 1 ∂2g ∂g (t, St )dt + β(α − St ) (t, St )dt + σ 2 St 2 (t, St )dt + σ St (t, St )dBt . = ∂t ∂x 2 ∂x ∂x By respective identification of the terms in dBt and dt in (6.36) and (S.6.24) we get  rg(t, St )dt + β(α − St )ξt dt − rξt St dt     ∂g ∂g 1 ∂2g   = (t, St )dt + β(α − St ) (t, St )dt + σ 2 St 2 (t, St )dt, ∂t ∂x 2 ∂x    p p  ∂g   σξt St dBt = σ St (t, St )dBt , ∂x hence  2   rg(t, St ) + β(α − St )ξt − rξt St = ∂g (t, St ) + β(α − St ) ∂g (t, St ) + 1 σ 2 St ∂ g (t, St ),   ∂t ∂x 2 ∂x2     ξt = ∂g (t, St ), ∂x and

 ∂g 1 ∂2g   (t, St ) + σ 2 St 2 (t, St ),  rg(t, St ) − rξt St =  ∂t 2 ∂x

    ξt = ∂g (t, St ), ∂x hence the function g(t, x) satisfies the PDE ∂g ∂g 1 ∂2g (t, x) + rx (t, x) + σ 2 x 2 (t, x), ∂t ∂x 2 ∂x and ξt is given by the partial derivative rg(t, x) =

ξt =

∂g (t, St ), ∂x

x > 0,

t ≥ 0.

Exercise 6.3 "

101 September 15, 2022

Solutions Manual a) Let Vt := ξt St + ηt At denote the hedging portfolio value at time t ∈ [0, T ]. Since the dividend yield δSt per share is continuously reinvested in the portfolio, the portfolio change dVt decomposes as dVt =

ηt dAt + ξt dSt {z } |

trading profit and loss

δξ S dt | t{zt }

dividend payout

= rηt At dt + ξt ((µ − δ)St dt + σSt dBt ) + δξt St dt = rηt At dt + ξt (µSt dt + σSt dBt ) = rVt dt + (µ − r)ξt St dt + σξt St dBt ,

t ≥ 0.

b) By Itô’s formula we have dg(t, St ) =

∂g ∂g (t, St )dt + (µ − δ)St (t, St )dt ∂t ∂x 1 ∂2g ∂g + σ 2 St2 2 (t, St )dt + σSt (t, St )dBt , 2 ∂x ∂x

hence by identification of the terms in dBt and dt in the expressions of dVt and dg(t, St ), we get ∂g ξt = (t, St ), ∂x and we derive the Black-Scholes PDE with dividend rg(t, x) =

∂g ∂g 1 ∂2g (t, x) + (r − δ)x (t, x) + σ 2 x2 2 (t, x). ∂t ∂x 2 ∂x

(S.6.25)

c) In order to solve (S.6.25) we note that, letting f (t, x) := e(T −t)δ g(t, x) and substituting g(t, x) = e−(T −t)δ f (t, x) into the PDE (S.6.25), we have rf (t, x) = δf (t, x) +

∂f ∂f 1 ∂2f (t, x) + (r − δ)x (t, x) + σ 2 x2 2 (t, x), ∂t ∂x 2 ∂x

hence f (t, x) := e(T −t)δ g(t, x), satisfies the standard Black-Scholes PDE with interest rate r − δ, i.e. we have (r − δ)f (t, x) =

∂f 1 ∂2f ∂f (t, x) + (r − δ)x (t, x) + σ 2 x2 2 (t, x), ∂t ∂x 2 ∂x

with same terminal condition f (T, x) = g(T, x) = (x − K)+ , hence we have f (t, x) = Bl(x, K, σ, r − δ, T − t) = xΦ dδ+ (T − t) − Ke−(r−δ)(T −t) Φ dδ− (T − t) , where dδ± (T − t) := 102

log(x/K) + (r − δ ± σ 2 /2)(T − t) √ . σ T −t 102 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Consequently, the pricing function of the European call option with dividend rate δ is g(t, x) = e−(T −t)δ f (t, x) = e−(T −t)δ Bl(x, K, σ, r − δ, T − t) = xe−(T −t)δ Φ dδ+ (T − t) − Ke−(T −t)r Φ dδ− (T − t) ,

0 ≤ t ≤ T.

We also have g(t, x) = Bl xe−(T −t)δ , K, σ, r, T − t

= e−(T −t)δ Bl x, Ke(T −t)δ , σ, r, T − t ,

0 ≤ t ≤ T.

d) As in Proposition 6.4, we have ξt =

∂g (t, St ) = e−(T −t)δ Φ dδ+ (T − t) , ∂x

0 ≤ t < T.

Exercise 6.4 a) We check that gc (t, 0) = 0, as when x = 0 we have d+ (T −t) = d− (T −t) = −∞ for all t ∈ [0, T ). On the other hand, we have  +∞, x > K,      x = K, lim d+ (T − t) = lim d− (T − t) = 0,  t↗T t↗T     −∞, x < K, which allows us to recover the boundary condition gc (T, x) = lim gc (t, x) t↗T    xΦ(+∞) − KΦ(+∞) = x − K,     x K = − = 0,  2 2      xΦ(−∞) − KΦ(−∞) = 0,

 x>K      x=K

x<K

= (x − K)+

     

at t = T . Regarding the Delta of the European call option, we find      Φ(+∞) = 1, x > K          1 lim Φ(d+ (T − t)) = Φ(0) = , x=K   t↗T 2           Φ(−∞) = 0, x < K "

103 September 15, 2022

Solutions Manual see Figure 6.5. Similarly, we can check that     +∞,        lim d− (T − t) = 0, T →∞           −∞,

σ2 , 2

and limT →∞ d+ (T − t) = +∞, hence lim Bl(x, K, σ, r, T − t)

T →∞

= x lim Φ(d+ (T − t)) − lim T →∞

= x,

T →∞

e−(T −t)r Φ(d− (T − t))

t ≥ 0.

b) We check that gp (t, 0) = Ke−(T −t)r and gp (t, ∞) = 0 as when x = 0 we have d+ (T − t) = d− (T − t) = −∞ and as x tends to infinity we have d+ (T − t) = d− (T − t) = +∞ for all t ∈ [0, T ). On the other hand, we have   KΦ(+∞) − xΦ(+∞) = K − x, x < K             K x gp (T, x) = − = 0, x = K = (K − x)+   2 2           KΦ(−∞) − xΦ(−∞) = 0, x>K at t = T . Regarding the Delta of the European put option, we find   Φ(−∞) = 0, x>K            1 − lim Φ(−d+ (T − t)) = −Φ(0) = − , x = K   t↗T 2           −Φ(+∞) = −1, x < K see Figure 6.11. Similarly, we can check that

104

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Introduction to Stochastic Finance with Market Examples, Second Edition     +∞,        lim d− (T − t) = 0, T →∞           −∞,

σ2 , 2

and limT →∞ d+ (T − t) = +∞, hence lim Blp (x, K, σ, r, T − t) = 0,

T →∞

t ≥ 0.

Exercise 6.5 (Exercise 3.14 continued). a) Substituting g(x, t) = x2 f (t) in (6.37), we find f ′ (t) = −(r + σ 2 )f (t), hence 2 2 f (t) = f (0)e−(r+σ )t = f (T )e(r+σ )(T −t) , 2

hence g(x, t) = f (T )x2 e(r+σ )(T −t) = x2 e(r+σ )(T −t) due to the terminal condition g(x, T ) = x2 . 2 ∂g g(St , t) = 2St e(r+σ )(T −t) , and b) We have ξt = ∂x 1 (g(St , t) − ξt St ) At 2 2 1 S 2 e(r+σ )(T −t) − 2St2 e(r+σ )(T −t) = A0 ert t 2 S2 = − t e(T −2t)r+(T −t)σ , t ∈ [0, T ]. A0

ηt =

Exercise 6.6 a) Counting approximately 46 days to maturity, we have (r − σ 2 /2)(T − t) + log(St /K) √ σ T −t (0.04377 − (0.9)2 /2)(46/365) + log(17.2/36.08) p = 0.9 46/365 = −2.461179058,

d− (T − t) =

and

p d+ (T − t) = d− (T − t) + 0.9 46/365 = −2.14167602.

From the standard Gaussian cumulative distribution table we get Φ(d+ (T − t)) = Φ(−2.14) = 0.0161098 "

105 September 15, 2022

Solutions Manual and

Φ(d− (T − t)) = Φ(−2.46) = 0.00692406,

hence f (t, St ) = St Φ(d+ (T − t)) − Ke−(T −t)r Φ(d− (T − t)) = 17.2 × 0.0161098 − 36.08 × e−0.04377×46/365 × 0.00692406 = HK$ 0.028642744. For comparison, running the corresponding Black-Scholes R script of Figure 6.20 yields BSCall(17.2, 36.08, 0.04377, 46/365, 0.9) = 0.02864235. b) We have ηt =

∂f (t, St ) = Φ(d+ (T − t)) = Φ(−2.14) = 0.0161098, ∂x

(S.6.26)

hence one should only hold a fractional quantity equal to 16.10 units in the risky asset in order to hedge 1000 such call options when σ = 0.90. c) From the curve it turns out that when f (t, St ) = 10 × 0.023 = HK$ 0.23, the volatility σ is approximately equal to σ = 122%. This approximate value of implied volatility can be found under the colDerivative Warrant Search umn “Implied Volatility (IV.)” on this set of market data from the Hong Kong Stock Exchange:

http://www.hkex.com.hk/dwrc/searc

Updated: 6 November 2008 Basic Data

Strike Entitlement Ratio^

DW Issuer Code

Call /Put

DW Type

Listing (D-M-Y)

Maturity Strike Entitle(D-M-Y) ment Ratio^ Link to Relevant Exchange Traded Options

01897

00066 Call Standard 18-12-2007 23-12-2008

04348

36.08

Total Issue Size

O/S (%)

Del (%

10 138,000,000 16.43

0.7

00066 Call Standard 18-12-2007 23-02-2009 38.88 10 300,000,000 0.25 0.7 Market Data 04984 AA 00066 Call Standard 02-06-2005 22-12-2008 12.88 10 300,000,000 0.36 8.0 Total O/S Delta IV. Day Day Closing T/O UL Base Listing Supplem Issue (%) (%) (%) High Low Price # ('000) Price Document & 05931 SB 00066 Call Standard 27-03-2008 29-12-2008 27.868 10 200,000,000 0.04 2.2 Size ($) ($) ($) ($) Announc

36.08

09133 CT 16.43 000660.780 Call125.375 Standard 31-01-2008 08-12-2008 10 200,000,000 10 138,000,000 0.000 0.000 0.023 36.88 0 17.200

0.15

0.4

38.88

13436 SG 0.25 000660.767 Call 88.656 Standard 14-05-2008 30-04-2009 32 0 17.200 10 200,000,000 10 300,000,000 0.000 0.000 0.024 Corporation. Fig. S.22: Market data for the warrant #01897 on the MTR

0.10

1.0

12.88

13562 BP 0.36 000668.075 Call128.202 Standard 26-05-2008 08-12-2008 10 300,000,000 0.000 0.000 0.540

30 0 17.200 10 150,000,000

0.00

0.9

8 27.868

13688 RB 0.04 000662.239 Call126.132 Standard 04-06-2008 20-02-2009 10 200,000,000 10 200,000,000 0.000 0.000 0.086 26.6 0 17.200

7.17

0.7

106 13764

28 0 17.200 10 106 300,000,000 0.31

0.5

36.88

SG 0.15 000660.416 Call133.443 Standard 13-06-2008 26-02-2009 10 200,000,000 0.000 0.000 0.010

September 15, 13785 ML 0.10 000661.059 Call 61.785 Standard 17-06-2008 19-01-2009 102022 100,000,000 10 200,000,000 0.000 0.000 0.031 27.38 0 17.200

13821 JP 0.00 000660.987 Call127.080 Standard 18-06-2008 18-12-2008 10 100,000,000 10 150,000,000 0.000 0.000 0.026 28.8 0 17.200

0.50

0.7

0.81

0.6

Introduction to Stochastic Finance with Market Examples, Second Edition Remark: a typical value for the volatility in standard market conditions would be around 20%. The observed volatility value σ = 1.22 per year is actually quite high. Exercise 6.7 a) We find h(x) = x − K. b) Letting g(t, x), the PDE rewrites as (x − α(t))r = −α′ (t) + rx, hence α(t) = α(0)ert and g(t, x) = x − α(0)ert . The final condition g(T, x) = h(x) = x − K yields α(0) = Ke−rT and g(t, x) = x − Ke−(T −t)r . c) We have ∂g (t, St ) = 1, ξt = ∂x hence ηt =

Vt − ξt St g(t, St ) − St St − Ke−(T −t)r − St = = = −Ke−rT . At At At

Note that we could also have directly used the identification Vt = g(St , t) = St − Ke−(T −t)r = St − Ke−rT At = ξt St + ηt At , which immediately yields ξt = 1 and ηt = −Ke−rT . d) It suffices to take K = 0, which shows that g(t, x) = x, ξt = 1 and ηt = 0. Exercise 6.8 a) We develop two approaches. (i) By financial intuition. We need to replicate a fixed amount of $1 at maturity T , without risk. For this there is no need to invest in the stock. Simply invest g(t, St ) := e−(T −t)r at time t ∈ [0, T ] and at maturity T you will have g(T, ST ) = e(T −t)r g(t, St ) = $1. (ii) By analysis and the Black-Scholes PDE. Given the hint, we try plugging a solution of the form g(t, x) = f (t), not depending on the variable x, into the Black-Scholes PDE (6.38). Given that here we have ∂g (t, x) = 0, ∂x

∂2g (t, x) = 0, ∂x2

and

∂g (t, x) = f ′ (t), ∂t

we find that the Black-Scholes PDE reduces to rf (t) = f ′ (t) with the terminal condition f (T ) = g(T, x) = 1. This equation has "

107 September 15, 2022

Solutions Manual for solution f (t) = e−(T −t)r and this is also the unique solution g(t, x) = f (t) = e−(T −t)r of the Black-Scholes PDE (6.38) with terminal condition g(T, x) = 1. b) We develop two approaches. (i) By financial intuition. Since the terminal payoff $1 is risk-free we do not need to invest in the risky asset, hence we should keep ξt = 0. Our portfolio value at time t becomes Vt = g(t, St ) = e−(T −t)r = ξt St + ηt At = ηt At with At = ert , so that we find ηt = e−rT , t ∈ [0, T ]. This portfolio strategy remains constant over time, hence it is clearly self-financing. (ii) By analysis. The Black-Scholes theory of Proposition 6.1 tells us that ξt = and ηt =

∂g (t, x) = 0, ∂x

Vt − ξt St Vt e−(T −t)r = = = e−rT . At At ert

Exercise 6.9 Log contracts. a) Substituting the function g(x, t) := f (t) + log x in the PDE (6.39), we have σ2 , 0 = f ′ (t) + r − 2 hence σ2 f (t) = f (0) − r − t, 2 σ2 T in order to match the terminal condition with f (0) = r − 2 g(x, T ) := log x, hence we have σ2 g(x, t) = r − (T − t) + log x, x > 0. 2 b) Substituting the function h(x, t) := u(t)g(x, t) = u(t)

r−

σ2 2

(T − t) + log x

in the PDE (6.39), we find u′ (t) = ru(t), hence u(t) = u(0)ert = e−(T −t)r , with u(T ) = 1, and we conclude to 108

108 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

h(x, t) = u(t)g(x, t) = e−(T −t)r x > 0, t ∈ [0, T ]. c) We have ξt =

r−

e−(T −t)r ∂h (t, St ) = , ∂x St

σ2 2

(T − t) + log x ,

0 ≤ t ≤ T,

and 1 h(t, St ) − ξt St At e−rT σ2 (T − t) + log x − 1 , = r− A0 2

ηt =

0 ≤ t ≤ T. Exercise 6.10 Binary options. a) From Proposition 6.1, the function Cd (t, x) solves the Black-Scholes PDE  ∂Cd ∂Cd 1 ∂ 2 Cd   rCd (t, x) = (t, x) + rx (t, x) + σ 2 x2 (t, x), ∂t ∂x 2 ∂x2   Cd (T, x) = 1[K,∞) (x). b) We can check by direct differentiation that the Black-Scholes PDE is satisfied by the function Cd (t, x), together with the terminal condition Cd (T, x) = 1[K,∞) (x) as t tends to T . Exercise 6.11 a) By (4.35) we have St = S0 eαt + σ

wt 0

e(t−s)α dBs .

b) By the self-financing condition (5.8) we have dVt = ηt dAt + ξt dSt = rηt At dt + αξt St dt + σξt dBt = rVt dt + (α − r)ξt St dt + σξt dBt ,

(S.6.27)

t ≥ 0. Rewriting (6.41) under the form of an Itô process St = S0 +

wt 0

vs ds +

wt 0

us dBs ,

t ≥ 0,

with "

109 September 15, 2022

Solutions Manual ut = σ,

and vt = αSt ,

t ≥ 0,

the application of Itô’s formula Theorem 4.24 to Vt = C(t, St ) shows that ∂C ∂C (t, St )dt + ut (t, St )dBt ∂x ∂x 2 ∂C 1 ∂ C + (t, St )dt + |ut |2 2 (t, St )dt ∂t 2 ∂x 1 ∂2C ∂C ∂C ∂C (t, St )dt + αSt (t, St )dt + σ 2 2 (t, St )dt + σ (t, St )dBt . = ∂t ∂x 2 ∂x ∂x (S.6.28)

dC(t, St ) = vt

Identifying the terms in dBt and dt in (S.6.27) and (S.6.28) above, we get  ∂C ∂C σ2 ∂ 2 C   (t, St ) + rSt (t, St ) + (t, St ),  rC(t, St ) = ∂t ∂x 2 ∂x2  ∂C   ξt = (t, St ), ∂x hence the function C(t, x) satisfies the usual Black-Scholes PDE rC(t, x) =

∂C ∂C 1 ∂2C (t, x) + rx (t, x) + σ 2 2 (t, x), ∂t ∂x 2 ∂x

x > 0,

0 ≤ t ≤ T,

with the terminal condition C(T, x) = ex , x ≥ 0. c) Based on (6.42), we compute  ∂C σ2   (t, x) = r + xh′ (t) + h(t)h′ (t) C(t, x),    ∂t 2r    ∂C (t, x) = h(t)C(t, x)   ∂x    2    ∂ C (t, x) = (h(t))2 C(t, x), ∂x2

(S.6.29)

hence the substitution of (6.42) into the Black-Scholes PDE (S.6.29) yields the ordinary differential equation xh′ (t) +

σ2 σ2 ′ h (t)h(t) + rxh(t) + (h(t))2 = 0, 2r 2

x > 0,

0 ≤ t ≤ T,

which reduces to the ordinary differential equation h′ (t) + rh(t) = 0 with terminal condition h(T ) = 1 and solution h(t) = e(T −t)r , t ∈ [0, T ], which yields σ 2 2(T −t)r C(t, x) = exp −(T − t)r + xe(T −t)r + (e − 1) . 4r 110

110 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition d) We have ξt =

∂C σ 2 2(T −t)r (t, St ) = exp St e(T −t)r + (e − 1) . ∂x 4r

Exercise 6.12 2

a) Noting that φ(x) = Φ′ (x) = (2π)−1/2 e−x /2 , we have the √ ∂h (S, d) = Sφ d + σ T − Ke−rT φ(d) ∂d √ 2 2 S K = √ e− d+σ T /2 − √ e−rT e−d /2 2π 2π √ 2 2 2 S K = √ e−d /2−σ T d−σ T /2 − √ e−rT e−d /2 , 2π 2π hence the vanishing of

∂h (S, d∗ (S)) at d = d∗ (S) yields ∂d

√ 2 2 2 S K √ e−d∗ (S)/2−σ T d∗ (S)−σ T /2 − √ e−rT e−d∗ (S)/2 = 0, 2π 2π

log(S/K) + rT − σ 2 T /2 √ . We can also check that σ T ∂2h ∂ S − d∗ (S)+σ√T 2 /2 K −rT −d2∗ (S)/2 √ √ (S, d (S)) = e − e e ∗ ∂d2 ∂d 2π 2π √ S − d (S)+σ√T 2 /2 2 K ∗ = − d∗ (S) + σ T √ e + √ d∗ (S)e−rT e−d∗ (S)/2 2π 2π √ K 2 2 K = − d∗ (S) + σ T √ e−rT e−d∗ (S)/2 + √ d∗ (S)e−rT e−d∗ (S)/2 2π 2π √ K 2 = −σ T √ e−rT e−d∗ (S)/2 < 0, 2π √ hence the function d 7−→ h(S, d) := SΦ d + σ T − Ke−rT Φ(d) admits a maximum at d = d∗ (S), and √ h(S, d∗ (S)) = SΦ d∗ (S) + σ T − Ke−rT Φ(d∗ (S)) log(S/K) + (r + σ 2 /2)T log(S/K) + (r − σ 2 /2)T √ √ = SΦ − Ke−rT Φ σ T σ T i.e. d∗ (S) =

is the Black-Scholes call option price. b) Since ∂h ∂d (S, d∗ (S)) = 0, we find ∆= "

∂h ∂h d h(S, d∗ (S)) = (S, d∗ (S)) + d′∗ (S) (S, d∗ (S)) dS ∂S ∂d 111 September 15, 2022

Solutions Manual √ = Φ d∗ (S) + σ T = Φ

log(S/K) + rT + σ 2 T /2 √ σ T

Exercise 6.13 a) When σ > 0 we have ∂ ∂ ∂gc = xΦ′ d+ (T − t) d+ (T − t) − Ke−(T −t)r Φ′ d− (T − t) d− (T − t) ∂σ ∂σ ∂σ ∂ = xΦ′ d+ (T − t) d+ (T − t) ∂σ ∂ −Ke−(T −t)r Φ′ d+ (T − t) e(T −t)r+log(x/K) d− (T − t) ∂σ ∂ = xΦ′ d+ (T − t) d+ (T − t) − d− (T − t) ∂σ √ ∂ = xΦ′ d+ (T − t) σ T −t ∂σ √ = x T − tΦ′ d+ (T − t) , where we used the fact that 2 1 Φ′ d− (T − t) = √ e−(d− (T −t)) /2 2π 2 1 = √ e−(d− (T −t)) /2+(T −t)r+log(x/K) 2π = Φ′ d+ (T − t) e(T −t)r+log(x/K) . Relation (6.43) can be obtained from the equalities 2 2 d+ (T − t) − d− (T − t) = d+ (T − t) + d− (T − t) d+ (T − t) − d− (T − t) x = 2r(T − t) + 2 log . K Due to the call-put parity relation (6.23), the Black-Scholes call and put Vega are identical, i.e. √ ∂gp ∂gc = = x T − tΦ′ d+ (T − t) . ∂σ ∂σ The Black-Scholes European call and put prices are increasing functions of the volatility parameter σ > 0. b) We have ∂gc ∂ ∂ = xΦ′ (d+ (T − t)) d+ (T − t) − Ke−(T −t)r Φ′ (d− (T − t)) d− (T − t) ∂r ∂r ∂r 112

112 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition +(T − t)Ke−(T −t)r Φ(d− (T − t)) ∂ = xΦ′ (d+ (T − t)) d+ (T − t) ∂r −Ke−(T −t)r Φ′ (d+ (T − t))e(T −t)r+log(x/K)

∂ d− (T − t) ∂r

+(T − t)Ke−(T −t)r Φ(d− (T − t)) ∂ = xΦ′ (d+ (T − t)) d+ (T − t) − d− (T − t) + (T − t)Ke−(T −t)r Φ(d− (T − t)) ∂r ∂ √ σ T − t + (T − t)Ke−(T −t)r Φ(d− (T − t)) = xΦ′ (d+ (T − t)) ∂r = (T − t)Ke−(T −t)r Φ(d− (T − t)), where we used the fact that 2 1 Φ′ (d− (T − t)) = √ e−(d− (T −t)) /2 2π 2 1 = √ e−(d− (T −t)) /2+(T −t)r+log(x/K) 2π = Φ′ (d+ (T − t))e(T −t)r+log(x/K) .

The same relationship is used to simplify the formulas of the Black-Scholes Delta and Vega. We note that the Black-Scholes European call price is an increasing function of the interest rate parameter r. Regarding put option prices gp (t, x), the call-put parity relation (6.23) yields ∂ ∂gp = gc − (x − Ke−r(T −t) ) ∂r ∂r = (T − t)Ke−(T −t)r Φ(d− (T − t)) − (T − t)Ke−r(T −t) = (T − t)Ke−(T −t)r (Φ(d− (T − t)) − 1) = −(T − t)Ke−(T −t)r Φ(−d− (T − t)), therefore the Black-Scholes European call price is a decreasing function of the interest rate parameter r. Exercise 6.14 a) Given that p∗ =

rN − aN 1 = bN − aN 2

and q ∗ =

bN − rN 1 = , bN − a N 2

Relation (3.14) reads "

113 September 15, 2022

Solutions Manual ve(t, x) =

p 1 ve t + T /N, x(1 + rT /N )(1 − σ T /N ) 2 p 1 + ve t + T /N, x(1 + rT /N )(1 + σ T /N ) . 2

After letting ∆T := T /N and applying Taylor’s formula at the second order we obtain √ 1 ve t + ∆T, x 1 + r∆T − σ ∆T − ve(t, x) 2 √ 1 + ve t + ∆T, x 1 + r∆T + σ ∆T − ve(t, x) + o ∆T 2 √ ∂e 1 ∂e v v (t, x) ∆T (t, x) + x r∆T − σ ∆T = 2 ∂t ∂x √ 2 ∂ 2 ve x2 + (t, x) + o(∆T ) r∆T − σ ∆T 2 ∂x2 √ ∂e 1 ∂e v v + ∆T (t, x) + x r∆T + σ ∆T (t, x) 2 ∂t ∂x √ 2 ∂ 2 ve x2 + r∆T + σ ∆T (t, x) + o(∆T ) + o ∆T 2 ∂x2 2 ∂ 2 ve ∂e v ∂e v x2 √ = ∆T (t, x) + rx∆T (t, x) + σ ∆T (t, x) + o ∆T , ∂t ∂x 2 ∂x2

which shows that o ∆T ∂e v ∂e v σ 2 ∂ 2 ve (t, x) + rx (t, x) + x2 (t, x) = − , ∂t ∂x 2 ∂x2 ∆T hence as N tends to infinity (or as ∆T tends to 0) we find∗ 0=

∂e v ∂e v σ 2 2 ∂ 2 ve (t, x) + rx (t, x) + x (t, x), ∂t ∂x 2 ∂x2

showing that the function v(t, x) := e(T −t)r ve(t, x) solves the classical Black-Scholes PDE rv(t, x) =

∂v ∂v σ2 2 ∂ 2 v (t, x) + rx (t, x) + x (t, x). ∂t ∂x 2 ∂x2

b) Similarly, we have (1)

ξt (x) =

v (t, (1 + bN )x) − v (t, (1 + aN )x) x(bN − aN )

∗ The notation o(∆T ) is used to represent any function of ∆T such that lim∆T →0 o(∆T )/∆T = 0.

114

114 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition p p v t, (1 + r/N )(1 + σ T /N )x − v t, (1 + r/N )(1 − σ T /N )x p = 2x(1 + r/N )σ T /N ∂v → (t, x), ∂x as N tends to infinity. Problem 6.15 a) When the risk-free rate is r = 0 the two possible returns are (5 − 4)/4 = 25% and (2 − 4)/4 = −50%. Under the risk-neutral probability measure given by P∗ (S1 = 5) = (4−2)/(5−2) = 2/3 and P∗ (S1 = 2) = (5−4)/(5− 2) = 1/3 the expected return is 2 × 25%/3 − 50%/3 = 0%. In general the expected return can be shown to be equal to the risk-free rate r. b) The two possible returns become (3×5−4−2×4)/4 = 75% and (3×2−4− 2 × 4)/4 = −150%. Under the risk-neutral probability measure given by P∗ (S1 = 5) = (4−2)/(5−2) = 2/3 and P∗ (S1 = 2) = (5−4)/(5−2) = 1/3 the expected return is 2×75%/3−150%/3 = 0%. Similarly to Question (a), the expected return can be shown to be equal to the risk-free rate r when r ̸= 0. c) We decompose the amount Ft invested in one unit of the fund as Ft =

βF | {z t}

Purchased/sold

− (β − 1)Ft , | {z }

Borrowed/saved

meaning that we invest the amount βFt in the risky asset St , and borrow/save the amount −(β − 1)Ft from/on the saving account. d) We have Ft = ξt St + ηt At = β

Ft Ft St − (β − 1) At , St At

t ≥ 0,

with ξt = βFt /St and ηt = −(β − 1)Ft /At , t ≥ 0. e) We have dFt = ξt dSt + ηt dAt Ft Ft = β dSt − (β − 1) dAt St At Ft = β dSt − (β − 1)rFt dt St = βFt (rdt + σdBt ) − (β − 1)rFt dt = rFt dt + βσFt dBt ,

(S.6.30)

t ≥ 0.

By (S.6.30), the return of the fund Ft is β times the return of the risky asset St , up to the cost of borrowing (β − 1)r per unit of time. "

115 September 15, 2022

Solutions Manual f) The discounted fund value (e−rt Ft )t∈R+ is a martingale under the riskneutral probability measure P∗ as we have d(e−rt Ft ) = βσe−rt Ft dBt , g) We have

t ≥ 0. 2

Ft = F0 eβσBt +rt−β σ t/2 and hence

β 2 2 Stβ = S0 eσBt +rt−σ t/2 = F0 eβσBt +βrt−βσ t/2 , 2

Ft = Stβ e−(β−1)rt−β(β−1)σ t/2 ,

t ≥ 0.

Note that when β = 0 we have Ft = ert , i.e. in this case the fund Ft coincides with the money market account. h) We have e−(T −t)r IE∗ (FT − K)+ | Ft log(Ft /K) + (r + β 2 σ 2 /2)(T − t) √ = Ft Φ |β|σ T − t 2 2 log(F t /K) + (r − β σ /2)(T − t) √ − Ke−(T −t)r Φ , |β|σ T − t t ∈ [0, T ). i) We have log(Ft /K) + (r + β 2 σ 2 /2)(T − t) √ Φ |β|σ T − t 2

=Φ

log(Stβ e−(β−1)rt−β(β−1)σ t/2 /K) + (r + β 2 σ 2 /2)(T − t) √ |β|σ T − t

=Φ

log(Stβ /K) − (β − 1)rt − β(β − 1)σ 2 t/2 + (r + β 2 σ 2 /2)(T − t) √ |β|σ T − t 2

log(Stβ /(Ke(β−1)rT −(T /2−t)(β−1)βσ )) + (T − t)βr + (T − t)βσ 2 /2 √ |β|σ T − t log(St /Kβ (t)) + (r + σ 2 /2)(T − t) √ =Φ , 0 ≤ t < T, σ T −t

=Φ

if β > 0, with Kβ (t) := K 1/β e(β−1)(rT /β−(T /2−t)σ ) . j) When β < 0 we find that the Delta of the call option on FT with strike price K is

116

116 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Φ

log(Ft /K) + (r + β 2 σ 2 /2)(T − t) √ |β|σ T − t

log(Stβ /Kβ ) + (T − t)βr + (T − t)βσ 2 /2 √ =Φ |β|σ T − t log(St /Kβ (t)) + (r + σ 2 /2)(T − t) √ , =Φ − σ T −t

0 ≤ t < T,

which coincides, up to a negative sign, with the Delta of the put option 2 on ST with strike price Kβ (t) := K 1/β e(β−1)(rT /β−(T /2−t)σ ) .

Chapter 7 Exercise 7.1 (Exercise 6.1 continued). Since r = 0 we have P = P∗ and g(t, Bt ) = IE∗ BT2 | Ft = IE∗ (BT − Bt + Bt )2 | Ft = IE∗ (BT − Bt + x)2 x=B t = IE∗ (BT − Bt )2 + 2x(BT − Bt ) + x2 x=B t = IE∗ (BT − Bt )2 + 2x IE∗ [BT − Bt ] + Bt2 = Bt2 + T − t,

0 ≤ t ≤ T,

hence ξt is given by the partial derivative ξt =

∂g (t, Bt ) = 2Bt , ∂x

0 ≤ t ≤ T,

with g(t, Bt ) − ξt Bt A0 B 2 + (T − t) − 2Bt2 = t A0 (T − t) − Bt2 , 0 ≤ t ≤ T. = A0

ηt =

Exercise 7.2 Since BT ≃ N (0, T ), we have 2 IE[ϕ(ST )] = IE ϕ S0 eσBT +(r−σ /2)T 2 2 1 w∞ = √ ϕ(S0 eσy+(r−σ /2)T )e−y /(2T ) dy 2πT −∞ "

117 September 15, 2022

Solutions Manual w∞ 2 2 2 1 dx ϕ(x)e−((σ /2−r)T +log x) /(2σ T ) = √ 2 x 2πσ T −∞ w∞ = ϕ(x)g(x)dx, −∞

under the change of variable 2

x = S0 eσy+(r−σ /2)T , i.e. y= where

with dx = σS0 eσy+(r−σ /2)T dy = σxdy,

(σ 2 /2 − r)T + log(x/S0 ) σ

and dy =

dx , σx

2 2 2 1 g(x) := √ e−((σ /2−r)T +log(x/S0 )) /(2σ T ) x 2πσ 2 T

is the lognormal probability density function with location parameter (r − √ σ 2 /2)T + log S0 and scale parameter σ T . Exercise 7.3 a) By the Itô formula, we have p(p − 1) p−2 St dSt • dSt 2 p(p − 1) p−2 = pStp−1 (rSt dt + σSt dBt ) + St (rSt dt + σSt dBt ) • (rSt dt + σSt dBt ) 2 p(p − 1) p = prStp dt + σpStp dBt + σ 2 St dt 2 p(p − 1) = pr + σ 2 Stp dt + σpStp dBt . 2

dStp = pStp−1 dSt +

b) By the Girsanov Theorem 7.3, letting 1 p(p − 1) ν := (p − 1)r + σ 2 , pσ 2 the drifted process bt := Bt + νt, B

0 ≤ t ≤ T,

is a standard (centered) Brownian motion under the probability measure Q defined by ν2 dQ(ω) = exp −νBT − T dP(ω). 2 Therefore, the differential of (Stp )t∈R+ can be written as 118

118 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition p(p − 1) Stp dt + σpStp dBt pr + σ 2 2 = (r + pσν) Stp dt + σpStp dBt

dStp =

= rStp dt + σpStp (dBt + νdt) bt , = rStp dt + σpStp dB bt , and hence the discounted process Set := e−rt Stp satisfies dSet = σpSet dB (Set )t∈R+ is a martingale under the probability measure Q. Exercise 7.4 We have IE∗ [ϕ(pST1 + qST2 )] ≤ IE∗ [pϕ(ST1 ) + qϕ(ST2 )]

since ϕ is convex,

= p IE∗ [ϕ(ST1 )] + q IE∗ [ϕ(ST2 )] = p IE∗ [ϕ(IE∗ [ST2 | FT1 ])] + q IE∗ [ϕ(ST2 )]

because (St )t∈R+ is a martingale,

≤ p IE∗ [IE∗ [ϕ(ST2 ) | FT1 ]] + q IE∗ [ϕ(ST2 )]

by Jensen’s inequality,

= p IE∗ [ϕ(ST2 )] + q IE∗ [ϕ(ST2 )]

by the tower property,

= IE∗ [ϕ(ST2 )],

because p + q = 1,

see Exercise 13.7 for an extension to arbitrary summations. Remark: This kind of technique can provide an upper price estimate from Black-Scholes when the actual option price is difficult to compute: here the closed-form computation would involve a double integration of the form h i 2 2 IE∗ [ϕ(pST1 + qST2 )] = IE∗ ϕ pS0 eσBT1 −σ T1 /2 + qS0 eσBT2 −σ T2 /2 h i 2 2 = IE∗ ϕ S0 eσBT1 −σ T1 /2 p + qe(BT2 −BT1 )σ−(T2 −T1 )σ /2 2 2 1 w∞ w∞ ϕ S0 eσx−σ T1 /2 p + qeσy−(T2 −T1 )σ /2 = 2π −∞ −∞ 2 2 dxdy ×e−x /(2T1 )−y (2(T2 −T1 )) p T1 (T2 − T1 ) + 2 2 1 w∞ w∞ S0 eσx−σ T1 /2 p + qeσy−(T2 −T1 )σ /2 − K = 2π −∞ −∞ 2 2 dxdy ×e−x /(2T1 )−y (2(T2 −T1 )) p T1 (T2 − T1 ) 1 w = 2π {(x,y)∈R2 : S0 eσx (p+qeσy−(T2 −T1 )σ2 /2 )≥Keσ2 T1 /2 } 2

(S0 eσx−σ T1 /2 (p + qeσy−(T2 −T1 )σ /2 ) − K) 2

×e−x /(2T1 )−y (2(T2 −T1 )) p = ···

dxdy T1 (T2 − T1 )

119 September 15, 2022

Solutions Manual Exercise 7.5 a) The European call option price C(K) := e−rT IE∗ [(ST − K)+ ] decreases with the strike price K, because the option payoff (ST − K)+ decreases and the expectation operator preserves the ordering of random variables. b) The European put option price C(K) := e−rT IE∗ [(K − ST )+ ] increases with the strike price K, because the option payoff (K − ST )+ increases and the expectation operator preserves the ordering of random variables. Exercise 7.6 a) Using Jensen’s inequality and the martingale property of the discounted asset price process (e−rt St )t∈R+ under the risk-neutral probability measure P∗ , we have + e−(T −t)r IE∗ [(ST − K)+ | Ft ] ≥ e−(T −t)r IE∗ [ST − K | Ft ] + = e−(T −t)r e(T −t)r St − K −(T −t)r + = St − Ke , 0 ≤ t ≤ T. Black-Scholes European call price Lower bound 80 70 60 50 40 30 20 10 0 120 100 Underlying (HK$)

15 Time to maturity T-t

Fig. S.23: Lower bound vs Black-Scholes call price. b) Similarly, by Jensen’s inequality and the martingale property, we find + e−(T −t)r IE∗ [(K − ST )+ | Ft ] ≥ e−(T −t)r IE∗ [K − ST | Ft ] + = e−(T −t)r K − e(T −t)r St + = Ke−(T −t)r − St , 0 ≤ t ≤ T.

120

120 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Black-Scholes European put price Lower bound 30 25 20 15 10 5 0 120

100 Underlying (HK$) 80 20

5 10 Time to maturity T-t

Fig. S.24: Lower bound vs Black-Scholes put option price. We may also use the fact that a convex function of the martingale (ert St )t∈R+ under the risk-neutral probability measure P∗ is a submartingale, showing that ert IE∗ [(e−rT K − e−rT ST )+ | Ft ] ≥ ert (e−rT K − e−rt St )+ + = Ke−(T −t)r − St , 0 ≤ t ≤ T. Exercise 7.7 a)

(i) The bull spread option can be realized by purchasing one European call option with strike price K1 and by short selling (or issuing) one European call option with strike price K2 , because the bull spread payoff function can be written as x 7−→ (x − K1 )+ − (x − K2 )+ . see http://optioncreator.com/st3ce7z. 150

(x-K1)+ -(x-K2)+

100

-50

100 ST

150

200

Fig. S.25: Bull spread option as a combination of call and put options.∗ "

121 September 15, 2022

Solutions Manual (ii) The bear spread option can be realized by purchasing one European put option with strike price K2 and by short selling (or issuing) one European put option with strike price K1 , because the bear spread payoff function can be written as x 7−→ −(K1 − x)+ + (K2 − x)+ , see http://optioncreator.com/stmomsb. 150

-(K1-x)+ (K2-x)+

100

-50

100 ST

150

200

Fig. S.26: Bear spread option as a combination of call and put options.∗ b)

(i) The bull spread option can be priced at time t ∈ [0, T ) using the Black-Scholes formula as Bl(St , K1 , σ, r, T − t) − Bl(St , K2 , σ, r, T − t). (ii) The bear spread option can be priced at time t ∈ [0, T ) using the Black-Scholes formula as Bl(St , K2 , σ, r, T − t) − Bl(St , K1 , σ, r, T − t).

Exercise 7.8 a) Using (a), the payoff of the long box spread option is given in terms of K1 and K2 as (x−K1 )+ −(K1 −x)+ −(x−K2 )+ +(K2 −x)+ = x−K1 −(x−K2 ) = K2 −K1 . b) By standard absence of arbitrage, the long box spread option payoff is priced (K2 − K1 )/(1 + r)N −k at times k = 0, 1, . . . , N . ∗ ∗

The animation works in Acrobat Reader on the entire pdf file. The animation works in Acrobat Reader on the entire pdf file.

122

122 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition c) From Table S.6 below we check that the strike prices suitable for a long box spread option on the Hang Seng Index (HSI) are K1 = 25, 000 and K2 = 25, 200.

Table S.6: Call and put options on the Hang Seng Index (HSI). d) Based on the data provided, we note that the long box spread can be realized in two ways. i) Using the put option issued by BI (BOCI Asia Ltd.) at 0.044. In this case, the box spread option represents a short position priced ×8, 000 −0.370 ×11, 000 +0.044 ×10, 000 = −92 |0.540 {z } ×7, 500 −0.064 | {z } | {z } | {z }

Long call

Short put

Short call

Long put

index points, or −92 × $50 = −$4, 600 on 02 March 2021. Note that according to Table 2 in the test paper, option prices are quoted in index points (to be multiplied by the relevant option/warrant entitlement ratio), and every index point is worth $50. ii) Using the put option issued by HT (Haitong Securities) at 0.061. In this case, the box spread option represents a long position priced ×8, 000 −0.370 ×11, 000 +0.061 ×10, 000 = +78 0.540 | {z } ×7, 500 −0.044 | {z } | {z } | {z }

Long call

Short put

Short call

Long put

index points, or 78 × $50 = $3, 900 on 02 March 2021. e) As the option built in di) represents a short position paying $4, 600 today with an additional $50 × (K2 − K1 ) = 200 = $10, 000 payoff at maturity on June 29, I would definitely enter this position. As for the option built in dii) it is less profitable because it costs $3, 900, however it is still profitable taking into account the $10, 000 payoff at maturity on June 29. "

123 September 15, 2022

Solutions Manual By the way, the put option at 0.061 has zero turnover (T/O). In the early 2019 Robinhood incident, a member of the Reddit community /r/WallStreetBets realized a loss of more than $57,000 on $5,000 principal by attempting a box spread. This was due to the use of American call and put options that may be exercised by their holders at any time, instead of European options with fixed maturity time N . Robinhood subsequently announced that investors on the platform would no longer be able to open box spreads, a policy that remains in place as of early 2021 Remark. Searching for arbitrage opportunities via the existence of profitable long box spreads is a way to test the efficiency of the market. The data used for this test in Table 7.1 was in fact modified market data. The original 02 March 2021 data is displayed in Table S.7, and shows that the call option with strike price K2 = 25, 200 was actually not available for trading (N/A) at that time, with 0% outstanding quantity and zero turnover (T/O).

Table S.7: Original call/put options on the Hang Seng Index (HSI) as of 02/03/2021.

Exercise 7.9 a) The payoff function can be written as (x − K1 )+ + (x − K2 )+ − 2(x − (K1 + K2 )/2)+ = (x − 50)+ + (x − 150)+ − 2(x − 100)+ ,

(S.7.31)

see also https://optioncreator.com/stnurzg.

124

124 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 150

(x-K1)++(x-K2)+ -2(x-K1/2-K2/2)+

100

-50

100 ST

150

200

Fig. S.27: Butterfly option as a combination of call options.∗ Hence the butterfly option can be realized by: 1. purchasing one call option with strike price K1 = 50, and 2. purchasing one call option with strike price K2 = 150, and 3. issuing (or selling) two call options with strike price (K1 +K2 )/2 = 100. b) From (S.7.31), the long call butterfly option can be priced at time t ∈ [0, T ) using the Black-Scholes call formula as Bl(St , K1 , σ, r, T −t)+Bl(St , K2 , σ, r, T −t)−2Bl(St , (K1 +K2 )/2, σ, r, T −t). c) For example, in the discrete-time Cox-Ross-Rubinstein (Cox et al. (1979)) model, denoting by ϕ(x) the payoff function, the self-financing replicating portfolio strategy (ξt (St−1 ))t=1,2,...,N hedging the contingent claim with payoff C = ϕ(SN ) is given as in Proposition 3.10 by i h QN QN IE∗ ϕ x(1 + b) j=t+1 (1 + Rj ) − ϕ x(1 + a) j=t+1 (1 + Rj ) ξt (x) = (b − a)(1 + r)N −t St−1 with x = St−1 . Therefore, ξt (x) will be positive (holding) when x = St−1 is sufficiently below (K1 + K2 )/2, and ξt (x) will be negative (short selling) when x = St−1 is sufficiently above (K1 + K2 )/2. ∗

The animation works in Acrobat Reader.

125 September 15, 2022

Solutions Manual

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1

200 150

10 100

Time to maturity T-t 5

50 0

0 Underlying price

Fig. S.28: Delta of a butterfly option with strike prices K1 = 50 and K2 = 150. Exercise 7.10 a) We have Ct = e−(T −t)r IE∗ [ST − K | Ft ] = e−(T −t)r IE∗ [ST | Ft ] − Ke−(T −t)r = ert IE∗ e−rT ST Ft − Ke−(T −t)r = ert e−rt St − Ke−(T −t)r = St − Ke−(T −t)r . We can check that the function g(x, t) = x−Ke−(T −t)r satisfies the BlackScholes PDE rg(x, t) =

∂g σ2 2 ∂ 2 g ∂g (x, t) + rx (x, t) + x (x, t) ∂t ∂x 2 ∂x2

with terminal condition g(x, T ) = x−K, since ∂g(x, t)/∂t = −rKe−(T −t)r and ∂g(x, t)/∂x = 1. b) We simply take ξt = 1 and ηt = −Ke−rT in order to have Ct = ξt St + ηt ert = St − Ke−(T −t)r ,

0 ≤ t ≤ T.

Note again that this hedging strategy is constant over time, and the relation ξt = ∂g(St , t)/∂x for the option Delta, cf. (S.6.26), is satisfied. Exercise 7.11 Option pricing with dividends (Exercise 6.3 continued). bt )t∈R a) Let P∗ denote the probability measure under which the process (B + defined by bt = µ − r dt + dBt dB σ is a standard Brownian motion. Under absence of arbitrage the asset price process (St )t∈R+ has the dynamics 126

126 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition dSt = (µ − δ)St dt + σSt dBt bt , = (r − δ)St dt + σSt dB and the discounted asset price process (Set )t∈R+ = (e−rt St )t∈R+ satisfies bt . dSet = −δ Set dt + σ Set dB Assuming that the dividend yield δSt per share is continuously reinvested in the portfolio, the self-financing portfolio condition dVt =

ηt dAt + ξt dSt {z } |

Trading profit and loss

δξt St dt | {z }

Dividend payout

bt ) + δξt St dt = rηt At dt + ξt ((r − δ)St dt + σSt dB bt ) = rηt At dt + ξt (rSt dt + σSt dB bt , = rVt dt + σξt St dB

t ≥ 0.

In other words, no arbitrage is induced by the dividend payout. This yields dVet = d e−rt Vt = −re−rt Vt dt + e−rt dVt bt = σξt e−rt St dB bt = σξt Set dB = ξt (dSet + δ Set dt),

t ≥ 0.

Therefore, we have Vet − Ve0 =

dVeu wt bu = σ ξu Seu dB 0 wt wt ξu dSeu + δ Seu du, = 0

t ≥ 0.

Here, the asset price process (eδt St )t∈R+ with added dividend yield satisfies the equation bt , d eδt St = reδt St dt + σeδt St dB and after discount, the process (e−rt eδt St )t∈R+ = (e−(r−δ)t St )t∈R+ is a martingale under P∗ . b) We have wt bu , Vet = Ve0 + σ ξu Seu dB t ≥ 0, 0

which is a martingale under P∗ from Proposition 7.1, hence "

127 September 15, 2022

Solutions Manual Vet = IE∗ VeT

= e−rT IE∗ [VT | Ft ] = e−rT IE∗ [C | Ft ], which implies Vt = ert Vet = e−(T −t)r IE∗ [C | Ft ],

0 ≤ t ≤ T.

c) After discounting the payoff (ST − K)+ at the continuously compounded interest rate r, we obtain Vt = e−(T −t)r IE∗ (ST − K)+ | Ft + 2 = e−(T −t)r IE∗ S eσBbT +(r−δ−σ /2)T − K | F t

−(T −t)δ

−(T −t)(r−δ)

∗

bT +(r−δ−σ2 /2)T σB

S0 e

−K

| Ft

= e−(T −t)δ Bl(x, K, σ, r − δ, T − t) = e−(T −t)δ St Φ dδ+ (T − t) − Ke−(T −t)(r−δ) Φ dδ− (T − t) = e−(T −t)δ St Φ dδ+ (T − t) − Ke−(T −t)r Φ dδ− (T − t) , 0 ≤ t < T, where dδ+ (T − t) :=

log(St /K) + (r − δ + σ 2 /2)(T − t) √ |σ| T − t

dδ− (T − t) :=

log(St /K) + (r − δ − σ 2 /2)(T − t) √ . |σ| T − t

and

We also have g(t, x) = Bl xe−(T −t)δ , K, σ, r, T − t

= e−(T −t)δ Bl x, Ke(T −t)δ , σ, r, T − t ,

0 ≤ t ≤ T.

d) In view of the pricing formula Vt = e−(T −t)δ St Φ dδ+ (T − t) − Ke−(T −t)r Φ dδ− (T − t) , the Delta of the option is identified as ξt = e−(T −t)δ Φ dδ+ (T − t) ,

0 ≤ t < T,

which recovers the result of Exercise 6.3-(d). Exercise 7.12 We start by pricing the “inner” at-the-money option with payoff (ST2 − ST1 )+ and strike price K = ST1 at time T1 as 128

128 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition e−(T2 −T1 )r IE∗ (ST2 − ST1 )+ | FT1 (r + σ 2 /2)(T2 − T1 ) + log(ST1 /ST1 ) √ = ST1 Φ σ T2 − T1 (r − σ 2 /2)(T2 − T1 ) + log(ST1 /ST1 ) √ −ST1 e−(T2 −T1 )r Φ σ T2 − T1 r + σ 2 /2 p r − σ 2 /2 p = ST1 Φ T2 − T1 − ST1 e−(T2 −T1 )r Φ T2 − T1 , σ σ where we applied (7.24) with T = T2 , t = T1 , and K = ST1 . As a consequence, the forward start option can be priced as e−(T1 −t)r IE∗ e−(T2 −T1 )r IE∗ (ST2 − ST1 )+ | FT1 Ft = e−(T1 −t)r r − σ 2 /2 p r + σ 2 /2 p T2 − T1 − ST1 e−(T2 −T1 )r Φ T2 − T1 × IE∗ ST1 Φ σ σ

= e−(T1 −t)r r − σ 2 /2 p r + σ 2 /2 p T2 − T1 − e−(T2 −T1 )r Φ T2 − T1 IE∗ [ST1 | Ft ] × Φ σ σ r − σ 2 /2 p r + σ 2 /2 p T2 − T1 − e−(T2 −T1 )r Φ T2 − T1 = St Φ , σ σ 0 ≤ t ≤ T1 . Exercise 7.13 From the Black-Scholes formula we have # ! " + (r + σ 2 /2)(Tk − Tk−1 ) − log K STk p −K FTk−1 = Φ e−(Tk −Tk−1 )r IE∗ STk−1 σ Tk − Tk−1 ! 2 (r − σ /2)(Tk − Tk−1 ) − log K −(Tk −Tk−1 )r p −Ke Φ . σ Tk − Tk−1 As a consequence, for t = T0 = 0 we have " # " " ## + + STk STk IE∗ −K −K Ft = IE∗ IE∗ FTk−1 STk−1 STk−1 " ! 2 (r + σ /2)(Tk − Tk−1 ) − log K p = IE∗ e(Tk −Tk−1 )r Φ σ Tk − Tk−1 !# (r − σ 2 /2)(Tk − Tk−1 ) − log K p −KΦ σ Tk − Tk−1

129 September 15, 2022

Solutions Manual (r + σ 2 /2)(Tk − Tk−1 ) − log K p σ Tk − Tk−1 ! (r − σ 2 /2)(Tk − Tk−1 ) − log K p . σ Tk − Tk−1

= e(Tk −Tk−1 )r Φ − KΦ

Hence, the cliquet option can be priced at time t = T0 = 0 as " + # n X STk ∗ −rTk e IE −K STk−1 k=1

n X

! (r + σ 2 /2)(Tk − Tk−1 ) − log K p σ Tk − Tk−1 !! 2 (r − σ /2)(Tk − Tk−1 ) − log K p . σ Tk − Tk−1

e−rTk−1 Φ

k=1

−Ke−rTk Φ

Exercise 7.14 (Exercise 6.9 continued). We have C(t, St ) = e−(T −t)r IE∗ log ST | Ft 2 bT − B bt )σ + r − σ (T − t) | Ft = e−(T −t)r IE∗ log St + (B 2 2 σ (T − t), 0 ≤ t ≤ T. = e−(T −t)r log St + e−(T −t)r r − 2 Exercise 7.15 (Exercise 6.5 continued). a) By (5.20), for all t ∈ [0, T ], we have C(t, St ) = e−(T −t)r IE[ST2 | Ft ] S2 = e−(T −t)r IE St2 T2 Ft S 2t S = e−(T −t)r St2 IE T2 Ft S 2t ST −(T −t)r 2 =e St IE St2 2(BT −Bt )σ−(T −t)σ2 +2(T −t)r −(T −t)r 2 =e St IE e 2 = e−(T −t)r St2 e−(T −t)σ +2(T −t)r IE e2(BT −Bt )σ 2

= St2 e(r+σ )(T −t) , where we used the Gaussian moment generating function (MGF) formula 130

130 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2 IE e2(BT −Bt )σ = e2(T −t)σ for the normal random variable BT − Bt ≃ N (0, T − t), 0 ≤ t < T . b) For all t ∈ [0, T ], we have ξt =

2 ∂C (t, x)|x=St = 2St e(r+σ )(T −t) , ∂x

i.e.

ξt St = 2St2 e(r+σ )(T −t) = 2C(t, St ), and ηt =

2 C(t, St ) − ξt St e−rt 2 (r+σ2 )(T −t) St e − 2St2 e(r+σ )(T −t) = At A0 S2 2 = − t eσ (T −t)+(T −2t)r , A0

i.e. ηt At = −St2

2 At σ2 (T −t)+(T −2t)r e = −St2 eσ (T −t)+(T −t)r = −C(t, St ). A0

As for the self-financing condition, we have 2

dC(t, St ) = d St2 e(r+σ )(T −t)

= −(r + σ 2 )e(r+σ )(T −t) St2 dt + e(r+σ )(T −t) d St2

= −(r + σ 2 )e(r+σ )(T −t) St2 dt + e(r+σ )(T −t) 2St dSt + σ 2 St2 dt 2

= −re(r+σ )(T −t) St2 dt + 2St e(r+σ )(T −t) dSt , and 2

ξt dSt + ηt dAt = 2St e(r+σ )(T −t) dSt − r

St2 σ2 (T −t)+(T −2t)r e At dt A0

= 2St e(r+σ )(T −t) dSt − rSt2 eσ (T −t)+(T −t)r dt, which recovers dC(t, St ) = ξt dSt + ηt dAt , i.e. the portfolio strategy is self-financing. Exercise 7.16 (Exercise 6.11 continued). a) The discounted process Xt := e−rt St satisfies dXt = (α − r)Xt dt + σe−rs dBs , which is a martingale when α = r by Proposition 7.1, as in this case it becomes a stochastic integral with respect to a standard Brownian motion. "

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Solutions Manual This fact can be recovered by directly computing the conditional expectation IE[Xt | Fs ] and showing it is equal to Xs . By (4.35), see Exercise 6.11, we have wt St = S0 eαt + σ e(t−s)α dBs , 0

hence Xt = S0 + σ

wt 0

e−rs dBs ,

t ≥ 0,

and wt IE[Xt | Fs ] = IE S0 + σ e−ru dBu Fs 0 w t = IE[S0 ] + σ IE e−ru dBu Fs 0 w w s t −ru = S0 + σ IE e dBu Fs + σ IE e−ru dBu 0 s w ws t = S0 + σ e−ru dBu + σ IE e−ru dBu 0 s ws −ru = S0 + σ e dBu

= Xs ,

0 ≤ s ≤ t.

b) We rewrite the stochastic differential equation satisfied by (St )t∈R+ as bt , dSt = αSt dt + σdBt = rSt dt + σdB where

α−r St dt + dBt , σ which allows us to rewrite (4.35), by taking α := −r therein, as bt := dB

wt wt bs = S0 ert + σ e(t−s)r dB bs . St = ert S0 + σ e−rs dB 0

Taking ψt :=

α−r St , σ

(S.7.32)

0 ≤ t ≤ T,

bt )t∈R is a standard Brownian in the Girsanov Theorem 7.3, the process (B + motion under the probability measure Pα defined by w T dPα 1wT 2 := exp − ψt dBt − ψt dt 0 dP 2 0 ! 2 w T 1 α−r wT 2 α−r = exp − St dBt − St dt , 0 0 σ 2 σ

132

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Introduction to Stochastic Finance with Market Examples, Second Edition and (Xt )t∈R+ is a martingale under Pα . c) Using (S.7.32) under the risk-neutral probability measure P∗ , we have C(t, St ) = e−(T −t)r IEα [exp(ST ) | Ft ] wT bu Ft = e−(T −t)r IEα exp erT S0 + σ e(T −u)r dB 0 wt wT −(T −t)r rT bu + σ bu Ft =e IEα exp e S0 + σ e(T −u)r dB e(T −u)r dB 0 t wT (T −t)r (T −u)r b = exp − (T − t)r + e St IEα exp σ e dBu Ft t

= exp − (T − t)r + e(T −t)r St

wT bu IEα exp σ e(T −u)r dB t

σ2 w T

e2(T −u)r du = exp − (T − t)r + e St exp 2 t σ 2 2(T −t)r (e − 1) , 0 ≤ t ≤ T. = exp − (T − t)r + e(T −t)r St + 4r

(T −t)r

d) We have ξt =

∂C σ 2 2(T −t)r (t, St ) = exp St e(T −t)r + (e − 1) ∂x 4r

and C(t, St ) − ξt St At e−(T −t)r σ 2 2(T −t)r = exp St e(T −t)r + (e − 1) At 4r σ 2 2(T −t)r St − exp St e(T −t)r + (e − 1) . At 4r

ηt =

e) We have σ 2 2(T −t)r dC(t, St ) = re−(T −t)r exp St e(T −t)r + (e − 1) dt 4r σ 2 2(T −t)r −rSt exp St e(T −t)r + (e − 1) dt 4r σ 2 2(T −t)r σ2 (e − 1) dt − e(T −t)r exp St e(T −t)r + 2 4r σ 2 2(T −t)r + exp St e(T −t)r + (e − 1) dSt 4r 1 σ 2 2(T −t)r + e(T −t)r exp St e(T −t)r + (e − 1) σ 2 dt 2 4r "

133 September 15, 2022

Solutions Manual σ 2 2(T −t)r = re−(T −t)r exp St e(T −t)r + (e − 1) dt 4r σ 2 2(T −t)r (T −t)r (e − 1) dt + ξt dSt . −rSt exp St e + 4r On the other hand, we have ξt dSt + ηt dAt = ξt dSt σ 2 2(T −t)r (e − 1) dt +re−(T −t)r exp St e(T −t)r + 4r σ 2 2(T −t)r −rSt exp St e(T −t)r + (e − 1) dt, 4r showing that

dC(t, St ) = ξt dSt + ηt dAt ,

and confirming that the strategy (ξt , ηt )t∈R+ is self-financing. Exercise 7.17 a) Using (S.7.32) under the risk-neutral probability measure P∗ , we have C(t, St ) = e−(T −t)r IEα [ST2 | Ft ] " 2 wT −(T −t)r bu =e IEα erT S0 + σ e(T −u)r dB

# Ft

= e−(T −t)r IEα = e−(T −t)r IEα

" "

erT S0 + σ

bu + σ e(T −u)r dB bu e(T −u)r dB

wT t

bu e(T −u)r dB

2 Ft

w T bu IEα bu +2σe−(T −t)r erT S0 + σ e(T −u)r dB e(T −u)r dB 0 t " # 2 wT bu +σ 2 e−(T −t)r IEα e(T −u)r dB Ft

= e−(T −t)r erT S0 + σ

wt 0

bu e(T −u)r dB

+ σ 2 e−(T −t)r IEα

" w

bu e(T −u)r dB

2 wt wT bu + σ 2 e−(T −t)r = e−(T −t)r erT S0 + σ e(T −u)r dB e2(T −u)r du 0

σ 2 (T −t)r e − e−(T −t)r = e(T −t)r St2 + 2r sinh((T − t)r) = e(T −t)r St2 + σ 2 , 0 ≤ t ≤ T. r 134

134 September 15, 2022

2 #

Introduction to Stochastic Finance with Market Examples, Second Edition b) We find ξt =

∂C (t, St ) = 2e(T −t)r St , ∂x

0 ≤ t ≤ T.

Exercise 7.18 (Exercise 5.8 continued, see Proposition 4.1 in Carmona and (2) (1) Durrleman (2003)). Letting α := IE∗ [ST ] = ert S0 − S0 and (2) (1) η 2 := Var∗ St − St (1) 2 σ12 t (2) 2 σ22 t (1) (2) (2) (1) 2 = e2rt S0 e + S0 e − 2S0 S0 eρσ1 σ2 t − S0 − S0 , we approximate 2 2 e−rt w ∞ (x − K)+ e−(x−α) /(2η ) dx e−rt IE∗ [(ST − K)+ ] ≃ p 2πη 2 −∞ 2 2 e−rt w ∞ = p (x − K)e−(x−α) /(2η ) dx 2πη 2 K Ke−rt w ∞ −(x−α)2 /(2η2 ) e−rt w ∞ −(x−α)2 /(2η2 ) = p xe dx − p e dx 2πη 2 K 2πη 2 K w −rt w ∞ −rt ∞ 2 2 ηe Ke = √ (x + α)e−x /2 dx − √ e−x /2 dx 2π (K−α)/η 2π (K−α)/η ηe−rt h −x2 /2 i∞ K −α =−√ e − (K − α)e−rt Φ − η (K−α)/η 2π ηe−rt −(K−α)2 /(2η2 ) K − α = √ e − (K − α)e−rt Φ − . η 2π

Remark: We note that the expected value IE∗ [ϕ(ST − K)] can be exactly computed from w∞w∞ (2) (1) ϕ x − y φ1 (x)φ2 (y)dxdy, IE∗ [ϕ(ST )] = IE∗ ϕ ST − ST = 0

where φi (x) =

1 (−(r − σi2 /2)T + log(x/S0 ))2 √ exp − 2 2σi T xσi 2πT (i)

is the lognormal probability density function of ST , i = 1, 2. In particular, we have w∞ (2) (1) IE∗ [ϕ(ST )] = IE∗ ϕ ST − ST = ϕ z φ(z)dz, 0

where "

135 September 15, 2022

Solutions Manual w∞ φ(z) = φ1 (z + y)φ2 (y)dydy w∞ 0 −(−(r−σ12 /2)T +log((z+y)/S0 ))2 /(2σ12 T )−(−(r−σ22 /2)T +log(y/S0 ))2 /(2σ22 T ) = e 0

dy 2πT σ1 σ2 (z + y)y

is the probability density function of ST . 1.6 1.4 1.2 1 0.8 0.6 0.4 Gaussian PDF Integral formula

0.2 0 -1

-0.5

0 z

0.5

Fig. S.29: Gaussian approximation of spread probability density function. 0.12

Integral evaluation Monte Carlo estimation Gaussian approximation

0.1 0.08 0.06 0.04 0.02 0

0.2

0.4

0.6

0.8

Fig. S.30: Gaussian approximation of spread option prices.

Exercise 7.19 (Exercise 6.2 continued). If C is a contingent claim payoff of the form C = ϕ(ST ) such that (ξt , ηt )t∈[0,T ] hedges the claim payoff C, the arbitrage-free price of the claim payoff C at time t ∈ [0, T ] is given by πt (X) = Vt = e−r(T −t) IE∗ [ϕ(ST ) | Ft ], 136

0 ≤ t ≤ T, 136 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition where IE∗ denotes expectation under the risk-neutral measure P∗ . Hence, from the noncentral Chi square probability density function fT −t (x) =

2β x + rt e−β(T −t) 2β exp − 2 −β(T −t) σ 1−e σ 2 1 − e−β(T −t) ! p 4β rt xe−β(T −t) , × I2αβ/σ2 −1 σ 2 1 − e−β(T −t)

αβ/σ2 −1/2

rt e−β(T −t)

of ST given St , x > 0, we find g(t, St ) = e−r(T −t) IE∗ [ϕ(ST ) | Ft ] αβ/σ2 −1/2 w∞ β(x+St e−β(T −t) ) 2βe−r(T −t) x −2 = 2 ϕ(x) e σ2 (1−e−β(T −t) ) St e−β(T −t) σ 1 − e−β(T −t) 0 ! p 4β xSt e−β(T −t) dx × I2αβ/σ2 −1 σ 2 1 − e−β(T −t) 0 ≤ t ≤ T , under the Feller condition 2αβ ≥ σ 2 . Exercise 7.20 a) We have ∂f (t, x) = (r − σ 2 /2)f (t, x), ∂t and

∂f (t, x) = σf (t, x), ∂x

∂2f (t, x) = σ 2 f (t, x), ∂x2

hence dSt = df (t, Bt ) ∂f ∂f 1 ∂2f = (t, Bt )dt + (t, Bt )dBt + (t, Bt )dt ∂t ∂x 2 ∂x2 1 1 2 = r − σ f (t, Bt )dt + σf (t, Bt )dBt + σ 2 f (t, Bt )dt 2 2 = rf (t, Bt )dt + σf (t, Bt )dBt = rSt dt + σSt dBt . b) We have IE eσBT Ft = IE e(BT −Bt +Bt )σ Ft "

137 September 15, 2022

Solutions Manual = eσBt IE e(BT −Bt )σ Ft = eσBt IE e(BT −Bt )σ 2

= eσBt +σ (T −t)/2 . c) We have 2 IE[ST | Ft ] = IE eσBT +rT −σ T /2 Ft 2 = erT −σ T /2 IE eσBT Ft 2

= erT −σ T /2 eσBt +σ (T −t)/2 rT +σBt −σ 2 t/2

= e(T −t)r+σBt +rt−σ t/2 =e

(T −t)r

St ,

0 ≤ t ≤ T.

d) We have Vt = e−(T −t)r IE[C | Ft ] = e−(T −t)r IE[ST − K | Ft ] = e−(T −t)r IE[ST | Ft ] − e−(T −t)r IE[K | Ft ] = St − e−(T −t)r K,

0 ≤ t ≤ T.

e) We take ξt = 1 and ηt = −Ke−rT /A0 , t ∈ [0, T ]. f) We find VT = IE[C | FT ] = C. Exercise 7.21 Binary options. (Exercise 6.10 continued). a) By definition of the indicator (or step) functions 1[K,∞) and 1[0,K] we have    1 if x ≥ K,  1 if x ≤ K, 1[K,∞) (x) = resp. 1[0,K] (x) =   0 if x < K, 0 if x > K, which shows the claimed result by the definition of Cb and Pb . b) We have πt (Cb ) = e−(T −t)r IE[Cb | Ft ] = e−(T −t)r IE 1[K,∞) (ST ) St = e−(T −t)r P(ST ≥ K | St ) = Cb (t, St ). c) We have πt (Cb ) = Cb (t, St ), where 138

138 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Cb (t, x) = e−(T −t)r P(ST > K | St = x) (r − σ 2 /2)(T − t) + log(St /K) √ = e−(T −t)r Φ σ T −t = e−(T −t)r Φ (d− (T − t)) , with d− (T − t) =

(r − σ 2 /2)(T − t) + log(St /K) √ . σ T −t

d) The price of this modified contract with payoff Cα = 1[K,∞) (ST ) + α1[0,K) (ST ) is given by πt (Cα ) = e−(T −t)r IE 1[K,∞) (ST ) + α1[0,K) (ST ) St = e−(T −t)r P(ST ≥ K | St ) + αe−(T −t)r P(ST ≤ K | St ) = e−(T −t)r P(ST ≥ K | St ) + αe−(T −t)r (1 − P(ST ≥ K | St )) = αe−(T −t)r e−(T −t)r + (1 − α)P(ST ≥ K | St ) (T − t)r − (T − t)σ 2 /2 + log(St /K) √ = αe−(T −t)r + (1 − α)e−(T −t)r Φ . σ T −t

1.2 1 0.8 0.6 0.4 0.2 0

15 200

10 150

Time to maturity T-t

100 0

50 0

Underlying (HK$)

Fig. S.31: Price of a binary call option. e) We note that

1[K,∞) (ST ) + 1[0,K] (ST ) = 1[0,∞) (ST ), almost surely since P(ST = K) = 0, hence πt (Cb ) + πt (Pb ) = e−(T −t)r IE[Cb | Ft ] + e−(T −t)r IE[Pb | Ft ] = e−(T −t)r IE[Cb + Pb | Ft ] "

139 September 15, 2022

Solutions Manual = e−(T −t)r IE 1[K,∞) (ST ) + 1[0,K] (ST ) Ft = e−(T −t)r IE 1[0,∞) (ST ) Ft = e−(T −t)r IE[1 | Ft ] = e−(T −t)r ,

0 ≤ t ≤ T.

f) We have πt (Pb ) = e−(T −t)r − πt (Cb ) = e−(T −t)r − e−(T −t)r Φ

(T − t)r − (T − t)σ 2 /2 + log(x/K) √ σ T −t

= e−(T −t)r (1 − Φ(d− (T − t))) = e−(T −t)r Φ(−d− (T − t)). g) We have ∂Cb (t, St ) ∂x ∂ (T − t)r − (T − t)σ 2 /2 + log(x/K) √ = e−(T −t)r Φ ∂x σ T −t x=St 1 −(d− (T −t))2 /2 −(T −t)r p e =e σSt 2(T − t)π > 0.

ξt =

The Black-Scholes hedging strategy of such a call option does not involve short selling because ξt > 0 at all times t, cf. Figure S.32 which represents the risky investment in the hedging portfolio of a binary call option.

Fig. S.32: Risky hedging portfolio value for a binary call option. Figure S.33 presents the risk-free hedging portfolio value for a binary call option. 140

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Introduction to Stochastic Finance with Market Examples, Second Edition

Fig. S.33: Risk-free hedging portfolio value for a binary call option. h) Here, we have ∂Pb (t, St ) ∂x (T − t)r − (T − t)σ 2 /2 + log(x/K) ∂ √ = e−(T −t)r Φ − ∂x σ T −t x=St 1 −(T −t)r −(d− (T −t))2 /2 p = −e e σ 2(T − t)πSt < 0.

ξt =

The Black-Scholes hedging strategy of such a put option does involve short selling because ξt < 0 for all t. Exercise 7.22 Applying Itô’s formula to e−rt ϕ(St ))t∈R+ and using the fact that the expectation of the stochastic integral with respect to (Bt )t∈R+ is zero, cf. Relation (4.17), we have C(x, T ) = IE e−rT ϕ(ST ) S0 = x (S.7.33) wT wT −rt −rt ′ = IE ϕ(x) − r e ϕ(St )dt + r e St ϕ (St )dt 0 0 w wT 1 T −rt ′′ +σ e−rt St ϕ′ (St )dBt + e ϕ (St )σ 2 (St )dt S0 = x 0 2 0 w w T T −rs e ϕ(St )dt S0 = x + r IE e−rt St ϕ′ (St )dt S0 = x = ϕ(x) − r IE 0 0 w T 1 −rt ′′ 2 + IE e ϕ (St )σ (St )dt S0 = x 0 2 wT wT = ϕ(x) − re−rt IE ϕ(St ) S0 = x dt + r e−rt IE St ϕ′ (St ) S0 = x dt 0

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Solutions Manual +

1 w T −rt ′′ e IE ϕ (St )σ 2 (St ) S0 = x dt, 2 0

hence, by differentiation with respect to T we find ∂ −rT e IE[ϕ(ST ) | S0 = x] ∂T = −re−rT IE ϕ(ST ) S0 = x + re−rT IE ST ϕ′ (ST ) S0 = x 1 + e−rT IE ϕ′′ (ST )σ 2 (ST ) S0 = x . 2

ThetaT =

Problem 7.23 Chooser options. a) We take conditional expectations in the equality (ST − K)+ − (K − ST )+ = ST − K to find C(t, St , K, T ) − P (t, St , K, T ) = e−(T −t)r IE∗ [(ST − K)+ | Ft ] − e−(T −t)r IE∗ [(K − ST )+ | Ft ] = e−(T −t)r IE∗ [ST − K | Ft ] = e−(T −t)r IE∗ [ST | Ft ] − Ke−(T −t)r = St − Ke−(T −t)r ,

0 ≤ t ≤ T.

b) The price this contract at time t ∈ [0, T ] can be written as e−(T −t)r IE∗ [P (T, ST , K, U ) | Ft ] h i = e−(T −t)r IE∗ e−(U −T )r IE∗ (K − SU )+ | FT | Ft = e−(U −t)r IE∗ (K − SU )+ | Ft = P (t, St , K, U ). c) From the call-put parity (7.47) the payoff of this contract can be written as max(P (T, ST , K, U ), C(T, ST , K, U )) = max(P (T, ST , K, U ), P (T, ST , K, U ) + ST − Ke−(U −T )r ) = P (T, ST , K, U ) + max(ST − Ke−(U −T )r , 0). d) The contract of Question (c) is priced at any time t ∈ [0, T ] as e−(T −t)r IE∗ [max(P (T, ST , K, U ), C(T, ST , K, U )) | Ft ] = e−(T −t)r IE∗ [P (T, ST , K, U ) | Ft ] h i +e−(T −t)r IE∗ max(ST − Ke−(U −T )r , 0) | Ft 142

142 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = e−(T −t)r IE∗ [e−(U −T )r IE∗ [(K − SU )+ | FT ] | Ft ] +e−(T −t)r IE∗ max ST − Ke−(U −T )r , 0 Ft = e−(U −t)r IE∗ [(K − SU )+ | Ft ] +e−(T −t)r IE∗ max ST − Ke−(U −T )r , 0 Ft = P (t, St , K, U ) + C t, St , Ke−(U −T )r , T .

(S.7.34)

100 90 80 70 60 50 40 30 20 10 8

Time to maturity T-t

100

120

140

Underlying

Fig. S.34: Black-Scholes price of the maximum chooser option. e) By (S.7.34) and Relation (6.3) in Proposition 6.1 we have ∂P ∂C t, St , Ke−(U −T )r , T + (t, St , K, U ) ∂x ∂x (U −T )r 2 log(e St /K) + (r + σ /2)(T − t) √ =Φ σ T −t log(St /K) + (r + σ 2 /2)(U − t) √ −Φ − σ U −t log(St /K) + (U − t)r + (T − t)σ 2 /2 √ =Φ σ T −t log(St /K) + (r + σ 2 /2)(U − t) √ −Φ − . σ U −t

ξt =

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Solutions Manual

1 0.5 0 -0.5 -1 2.5 2 1.5 Time to maturity T-t

1 0.5 0

140

120

100 Underlying

Fig. S.35: Delta of the maximum chooser option. f) From the call-put parity (7.47) the payoff of this contract can be written as min(P (T, ST , K, U ), C(T, ST , K, U )) = min C(T, ST , K, U ) − ST + Ke−(U −T )r , C(T, ST , K, U )

= C(T, ST , K, U ) + min(−ST + Ke−(U −T )r , 0) = C(T, ST , K, U ) − max(ST − Ke−(U −T )r , 0). g) The contract of Question (f) is priced at any time t ∈ [0, T ] as e−(T −t)r IE∗ min P (T, ST , K, U ), C(T, ST , K, U ) Ft = e−(T −t)r IE∗ [C(T, ST , K, U ) | Ft ] −e−(T −t)r IE∗ max ST − Ke−(U −T )r , 0 Ft = e−(T −t)r IE∗ [e−(U −T )r IE∗ [(SU − K)+ | FT ] | Ft ] −e−(T −t)r IE∗ max ST − Ke−(U −T )r , 0 Ft = e−(U −t)r IE∗ (SU − K)+ | Ft −e−(T −t)r IE∗ max ST − Ke−(U −T )r , 0 Ft = C(t, St , K, U ) − C t, St , Ke−(U −T )r , T .

144

(S.7.35)

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Introduction to Stochastic Finance with Market Examples, Second Edition

8 7 6 5 4 3 2 1 0 8 6 4

Time to maturity T-t

2 0

100

120

140

Underlying

Fig. S.36: Black-Scholes price of the minimum chooser option. h) By (S.7.35) and Relation (6.3) in Proposition 6.1 we have ∂C ∂C (t, St , K, U ) − t, St , Ke−(U −T )r , T ∂x ∂x log(St /K) + (r + σ 2 /2)(U − t) √ =Φ σ U −t log(e(U −T )r St /K) + (r + σ 2 /2)(T − t) √ −Φ σ T −t log(St /K) + (r + σ 2 /2)(U − t) √ =Φ σ U −t log(St /K) + (U − t)r + (T − t)σ 2 /2 √ −Φ . σ T −t

ξt =

0.6 0.4 0.2 0 -0.2 -0.4 8 6 Time to maturity T-t

4 2 0

100

120

140

Underlying

Fig. S.37: Delta of the minimum chooser option. i) Such a contract is priced as the sum of a European call and a European put option with maturity U , and is priced at time t ∈ [0, T ] as P (t, St , K, U ) + C(t, St , K, U ). Its hedging strategy is the sum of the hedging strategies of "

145 September 15, 2022

Solutions Manual Questions (e) and (h), i.e. log(St /K) + (r + σ 2 /2)(U − t) √ ξt = Φ σ T −t log(St /K) + (r + σ 2 /2)(U − t) √ −Φ − σ T −t log(St /K) + (r + σ 2 /2)(U − t) √ = 2Φ − 1. σ T −t j) When U = T , the contracts of Questions (c), (f) and (i) have the respective payoffs • max((ST − K)+ , (K − ST )+ ) = |ST − K|, • min((ST − K)+ , (K − ST )+ ) = 0, and • (ST − K)+ + (K − ST )+ = |ST − K|, where |ST − K| is known as the payoff of a straddle option. Problem 7.24 a) The self-financing condition reads dVt = ηt dAt + ξt dSt = rηt At dt + µξt St dt + σξt St dBt = rVt dt + (µ − r)ξt St dt + σξt St dBt , hence VT = V0 + = Vt +

wT 0

wT t

(rVt + (µ − r)ξt St )dt + σ

(rVs + (µ − r)ξs Ss )ds + σ

wT t

ξt St dBt ξs Ss dBs .

b) The portfolio value Vt rewrites as wT wT µ−r Vt = VT − rVs + πs ds − πs dBs t t σ wT wT bs . πs dB = VT − r Vs ds − t

c) We have Vt = VT − r

wT t

Vs ds −

wT t

bs , πs dB

hence 146

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Introduction to Stochastic Finance with Market Examples, Second Edition bt , dVt = rVt dt + πt dB and after discounting we find dVet = −re−rt Vt dt + e−rt dVt bt ) = −re−rt Vt dt + e−rt (rVt dt + πt dB bt , = e−rt πt dB which shows that VeT = V0 +

wT 0

bt , e−rt πt dB

after integration in t ∈ [0, T ]. d) We have dVt = du(t, St ) ∂u ∂u ∂u (t, St )dt + µSt (t, St )dt + σSt (t, St )dBt = ∂t ∂x ∂x 1 2 2 ∂2u + σ St 2 (t, St )dt. (S.7.36) 2 ∂x e) By matching the Itô formula (S.7.36) term by term to the BSDE (7.50) we find that Vt = u(t, St ) satisfies the PDE ∂u ∂u 1 ∂2u ∂u (t, x) + µ (t, x) + σ 2 x2 2 (t, x) + f t, x, u(t, x), σx (t, x) = 0. ∂t ∂x 2 ∂x ∂x f) In this case we have ∂u 1 ∂u ∂u ∂2u (t, x) + µx (t, x) + σ 2 x2 2 (t, x) − ru(t, x) − (µ − r)x (t, x) = 0, ∂t ∂x 2 ∂x ∂x which recovers the Black-Scholes PDE ru(t, x) =

∂u ∂u 1 ∂2u (t, x) + rx (t, x) + σ 2 x2 2 (t, x). ∂t ∂x 2 ∂x

g) In the Black-Scholes model the Delta of the European call option is given by (r + σ 2 /2)(T − t) + log(St /K) √ ξt = Φ , σ T −t hence πt = σξt St = σSt Φ

(r + σ 2 /2)(T − t) + log(St /K) √ σ T −t

0 ≤ t ≤ T.

h) Replacing the self-financing condition with "

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Solutions Manual dVt = ηt dAt + ξt dSt − γSt (ξt )− dt = rηt At dt + µξt St dt + σξt St dBt − γSt (ξt )− dt = rVt dt + (µ − r)ξt St dt − γSt (ξt )− dt + σξt St dBt , we get the BSDE Vt = VT −

wT t

wT rVs + (µ − r)πs + γ(πs )− ds − πs dBs . t

i) In this case we have f (t, x, u, z) = −ru −

µ−r z − γz − σ

and the BSDE reads dVt = ru(t, St )dt + (µ − r)ξt St dt − γSt (ξt )− dt + σξt St dBt . j) We find the nonlinear PDE ∂u ∂u 1 ∂2u (t, x) + rx (t, x) + σ 2 x2 2 (t, x) − γσx ∂t ∂x 2 ∂x

∂u (t, x) ∂x

−

= ru(t, x), (S.7.37)

with the terminal condition u(T, x) = g(x). k) The self-financing condition reads dVt = r1{ηt >0} At ηt dt + R1{ηt <0} At ηt dt + ξt dSt = rAt ηt dt + (R − r)1{ηt <0} At ηt dt + ξt dSt

= rVt dt − rSt ξt dt − (R − r)(ηt At )− dt + ξt dSt = rVt dt + (µ − r)St ξt dt − (R − r)(Vt − ξt St )− dt + σξt St dBt , which yields the BSDE Vt = VT −

wT t

wT rVs + (µ − r)πs − (R − r)(Vs − ξs Ss )− ds − πs dBs , t

hence we have f (t, x, u, z) = −ru −

z − (µ − r) z + (R − r) u − σ σ

and the nonlinear PDE ∂u ∂u 1 ∂2u (t, x) + µ (t, x) + σ 2 x2 2 (t, x) + f ∂t ∂x 2 ∂x

t, x, u(t, x), σx

∂u (t, x) ∂x

rewrites as

148

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Introduction to Stochastic Finance with Market Examples, Second Edition − ∂u ∂u ∂u 1 ∂2u . (t, x)+r (t, x)+ σ 2 x2 2 (t, x) = ru(t, x)+(r−R) u(t, x) − x (t, x) ∂t ∂x 2 ∂x ∂x l) The sum of profits and losses of the portfolio (ξt , ηt )t∈R+ is V0 +

wT 0

ηt dAt +

wT 0

ξt dSt = V0 +

wT 0

dVt +

= VT + UT − U0

wT 0

dUt

> VT = C, hence the corresponding portfolio strategy superhedging the claim payoff VT = C. Exercise 7.25 Girsanov Theorem. For all n ≥ 1, let (n)

ψt

:= 1{ψt ∈[−n,n]} ψt ,

0 ≤ t ≤ T.

(n)

Since (ψt )t∈[0,T ] is a bounded process it satisfies the Novikov integrability condition (7.11), hence for all n ≥ 1 and random variable F ∈ L1 (Ω) we have w w· T 1 w T (n) 2 IE[F ] = IE F B· + ψs(n) ds exp − ψs(n) dBs − (ψs ) ds , 0 0 2 0 which yields w w· T 1 w T (n) 2 (ψs ) ds IE[F ] = lim IE F B· + ψs(n) ds exp − ψs(n) dBs − n→∞ 0 0 2 0 w w· T 1 w T (n) 2 ≥ IE lim inf F B· + ψs(n) ds exp − ψs(n) dBs − (ψs ) ds n→∞ 0 0 2 0 w w· T 1wT 2 ψs dBs − = IE F B· + ψs ds exp − (ψs ) ds , 0 0 2 0 where we applied Fatou’s Lemma.∗ Problem 7.26 a) We have Cov(dSt /St , dMt /Mt ) Var[dMt /Mt ] Cov((r + α)dt + β(dMt /Mt − rdt) + σS dWt , µdt + σM dBt ) = Var[µdt + σM dBt ] ∗

IE[limn→∞ Fn ] ≤ limn→∞ IE[Fn ] for any sequence (Fn )n∈N of nonnegative random variables, provided that the limits exist.

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Solutions Manual Cov (r + α)dt + β(µdt + σM dBt − rdt) + σS dWt , µdt + σM dBt Var[µdt + σM dBt ] Cov βσM dBt + σS dWt , σM dBt = Var[σM dBt ] Cov βσM dBt , σM dBt + Cov σS dWt , σM dBt = Var[σM dBt ] Cov βσM dBt , σM dBt = Var[σM dBt ] Cov σM dBt , σM dBt =β Var[σM dBt ] = β. =

b) We have

dMt − r St dt + σS St dWt Mt = (r + α)St dt + βSt (µdt + σM dBt − rdt) + σS St dWt = (r + α + β(µ − r))St dt + St βσM dBt + σS dWt q M dBt + σS dWt 2 + σ 2 βσp . = (r + α + β(µ − r))St dt + St β 2 σM S 2 + σ2 β 2 σM S

dSt = (r + α)St dt + β

Now, we have βσM dBt + σS dWt p 2 + σ2 β 2 σM S

!2 = =

βσM dBt

+ βσM σS dBt • dWt + σS dWt 2 + σ2 β 2 σM S

2 β 2 σM (dBt )2 + σS2 (dWt )2 2 + σ2 β 2 σM S

2 β 2 σM dt + σS2 dt 2 + σ2 β 2 σM S = dt.

By the characterization of Brownian motion as the only continuous martingale whose quadratic variation is dt, it follows that the process (Zt )t∈R+ defined by βσM dBt + σS dWt dZt = p 2 + σ2 β 2 σM S is a standard Brownian motion, see e.g. Theorem 7.36 page 203 of Klebaner (2005). Hence, we have dSt = (r + α + β(µ − r))St dt + St βσM dBt + σS dWt

150

150 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = (r + α + β(µ − r))St dt + St

2 + σ 2 dZ . β 2 σM t S

In what follows we assume that β is allowed to depend locally on the state of the benchmark market index on Mt , as β(Mt ), t ∈ R+ . c) We take µ−r dBt∗ = dBt + dt (S.7.38) σM and

dWt∗ = dWt +

α dt σS

(S.7.39)

in order to have  dM t  = µdt + σM dBt = rdt + σM dBt∗ ,    Mt     dSt dMt = (r + α)dt + β(M ) × − rdt + σS dWt  t  St Mt    ∗  = (r + α)dt + σM β(Mt )dBt + σS dWt   = rdt + σM β(Mt )dBt∗ + σS dWt∗ .

(S.7.40)

d) By the Girsanov theorem, (Bt∗ )t∈[0,T ] is a standard Brownian motion under the probability measure P∗B defined by its Radon-Nikodym density dP∗B µ−r (µ − r)2 = exp − T , BT − 2 dP σM 2σM and (Wt∗ )t∈[0,T ] is a standard Brownian motion under the probability measure P∗W defined by its Radon-Nikodym density dP∗W α α2 = exp − WT − 2 T . dP σS 2σS We conclude that (Bt∗ )t∈[0,T ] and (Wt∗ )t∈[0,T ] are independent standard Brownian motions under the probability measure P∗ defined by its RadonNikodym density dP∗ dP∗B dP∗W µ−r α (µ − r)2 α2 = × = exp − BT − WT − T − T . 2 dP dP dP σM σS 2σM 2σS2 Indeed, for any sequence t0 = 0 < t1 < · · · < tn−1 < tn = T we have ∗ dP IE∗ f Bt∗1 − Bt∗0 , . . . , Bt∗n − Bt∗n−1 = IE f Bt∗1 − Bt∗0 , . . . , Bt∗n − Bt∗n−1 dP h ∗ ∗ ∗ ∗ = IE f Bt1 − Bt0 , . . . , Btn − Btn−1

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Solutions Manual α µ−r (µ − r)2 α2 T− × exp − BT − WT − 2 T 2 σM 2σM σS 2σS µ−r (µ − r)2 ∗ ∗ ∗ ∗ = IE f Bt1 − Bt0 , . . . , Btn − Btn−1 exp − BT − T 2 σM 2σM α α2 × IE exp − WT − 2 T σS 2σS (µ − r)2 µ−r = IE f Bt∗1 − Bt∗0 , . . . , Bt∗n − Bt∗n−1 exp − BT − T 2 σM 2σM = IE f Bt1 − Bt0 , . . . , Btn − Btn−1 , and similarly for (Wt∗ )t∈[0,T ] . e) By (S.7.40), the discounted price processes (Set )t∈R+ := (e−rt St )t∈R+ satisfy

and

ft )t∈R := (e−rt Mt )t∈R (M + +

 ft = σM M ft dBt∗ ,  dM  e dSt = σM β(Mt )Set dBt∗ + σS Set dWt∗ ,

hence by the Girsanov theorem of Question (d) the discounted two ft dimensional process Set , M is a martingale under the probability t∈R+ measure P∗ , showing that P∗ is a risk-neutral probability measure. Therefore, by Theorem 6.8 the market made of St and Mt is without arbitrage opportunities due to the existence of a risk-neutral probability measure P∗ . f) The self-financing condition for the portfolio strategy (ξt , ζt , ηt )t∈[0,T ] reads ηt+dt At+dt + ξt+dt St+dt + ζt+dt Mt+dt , = ηt At+dt + ξt St+dt + ζt Mt+dt which yields

At+dt dηt + St+dt dξt + Mt+dt ζt = 0,

i.e. dAt • dηt + dSt • dξt + dMt • dζt + At dηt + St dξt + Mt dζt = 0, hence dVt = ηt dAt + ξt dSt + ζt dMt +At dηt + dηt • dAt + St dξt + dξt • dSt + Mt dζt + dζt • dMt = ηt dAt + ξt dSt + ζt dMt = 0. 152

152 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition g) By the self-financing condition we have dVt = df (t, St , Mt ) = ξt dSt + ζt dMt + ηt dAt = ξt (rSt dt + σM β(Mt )St dBt∗ + σS St dWt∗ ) + ζt (rMt dt + σM Mt dBt∗ ) + rηt At dt = rξt St dt + σM β(Mt )ξt St dBt∗ + σS ξt St dWt∗ + rζt Mt dt + σM ζt Mt dBt∗ + rηt At dt = rξt St dt + rηt At dt + σS ξt St dWt∗ + rζt Mt dt + (σM β(Mt )ξt St + σM ζt Mt )dBt∗ = rVt dt + σS ξt St dWt∗ + (σM β(Mt )ξt St + σM ζt Mt )dBt∗ .

(S.7.41)

On the other hand, by the Itô formula for two state variables, we have ∂f ∂f 1 ∂2f (t, St , Mt )dt + (t, St , Mt )dSt + (t, St , Mt )(dSt )2 ∂t ∂x 2 ∂x2 ∂f 1 ∂2f ∂2f + (t, St , Mt )dMt + (t, St , Mt )dSt • dMt (t, St , Mt )(dMt )2 + ∂y 2 ∂y 2 ∂x∂y ∂f ∂f (t, St , Mt )dt + (t, St , Mt )(rSt dt + σM β(Mt )St dBt∗ + σS St dWt∗ ) = ∂t ∂x 2 1∂ f 2 + (t, St , Mt )(σM β 2 (Mt )St2 + σS2 St2 )dt 2 ∂x2 ∂f 1 ∂2f 2 + (t, St , Mt )(rMt dt + σM Mt dBt∗ ) + (t, St , Mt )σM Mt2 dt ∂y 2 ∂y 2 ∂2f 2 + σM St Mt β(Mt ) (t, St , Mt )dt ∂x∂y ∂f ∂f ∂f = (t, St , Mt )dt + rSt (t, St , Mt )dt + σM β(Mt )St (t, St , Mt )dBt∗ ∂t ∂x ∂x 1 ∂2f ∂f ∗ 2 2 (t, St , Mt )(σM β (Mt )St2 dt + σS2 St2 dt) + σS St (t, St , Mt )dWt + ∂x 2 ∂x2 ∂f ∂f + rMt (t, St , Mt )dt + σM Mt (t, St , Mt )dBt∗ ∂y ∂y 1 ∂2f ∂2f 2 2 2 + (t, St , Mt )dt (t, St , Mt )σM Mt dt + σM St Mt β(Mt ) 2 ∂y 2 ∂x∂y ∂f ∂f ∂f = (t, St , Mt )dt + rMt (t, St , Mt )dt + rSt (t, St , Mt )dt ∂t ∂y ∂x 2 2 1 2 1 2 2 2∂ f 2∂ f + σ M Mt (t, St , Mt )dt + σM β (Mt )St 2 (t, St , Mt )dt 2 ∂y 2 2 ∂x 1 ∂2f ∂2f 2 + σS2 St2 2 (t, St , Mt )dt) + σM St Mt β(Mt ) (t, St , Mt )dt 2 ∂x ∂x∂y ∂f ∂f + σM β(Mt )St (t, St , Mt ) + σM Mt (t, St , Mt ) dBt∗ (S.7.42) ∂x ∂y ∂f + σS St (t, St , Mt )dWt∗ . ∂x

df (t, St , Mt ) =

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Solutions Manual By identification of the terms in dt in (S.7.41) and (S.7.42), we find ∂f ∂f (t, St , Mt ) + rSt (t, St , Mt ) ∂t ∂x 2 1 2 2 2 2∂ f + (σS + σM β (Mt ))St 2 (t, St , Mt ) 2 ∂x ∂f ∂2f 1 2 ∂2f 2 + rMt (t, St , Mt ) + σM Mt2 2 (t, St , Mt ) + σM St Mt β(Mt ) (t, St , Mt )dt, ∂y 2 ∂y ∂x∂y

rf (t, St , Mt ) =

which yields the PDE rf (t, x, y) (S.7.43) 2 ∂f 1 2 2 ∂ f ∂f 2 (t, x, y) + rx (t, x, y) + x (σS + σM β 2 (y)) 2 (t, x, y) = ∂t ∂x 2 ∂x 1 2 2 ∂2f ∂2f ∂f 2 (t, x, y) + σM xyβ(y) (t, x, y), +ry (t, x, y) + σM y ∂y 2 ∂y 2 ∂x∂y with the terminal condition f (T, x, y) = h(x, y),

x, y > 0.

h) By identification of terms in dBt∗ and dWt∗ in (S.7.41) and (S.7.42), we find ∂f ξt = (t, St , Mt ) ∂x and σM β(Mt )St

∂f ∂f (t, St , Mt )+σM Mt (t, St , Mt ) = σM β(Mt )ξt St +σM ζt Mt , ∂x ∂y

hence ζt =

∂f (t, St , Mt ), ∂y

and by the relation Vt = ξt St + ζt Mt + ηt At we find ηt =

Vt − ξt St − ζt Mt At f (t, St , Mt ) − St

∂f ∂f (t, St , Mt ) − Mt (t, St , Mt ) ∂x ∂y , A0 ert

0 ≤ t ≤ T.

i) When the option payoff depends only on ST we can look for a solution of (S.7.43) of the form f (t, x), in which case (S.7.43) simplifies to rf (t, x, y) =

154

∂f ∂f 1 ∂2f 2 (t, x, y) + rx (t, x, y) + x2 (σS2 + σM β 2 (y)) 2 (t, x, y), ∂t ∂x 2 ∂x (S.7.44) 154 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition When β(Mt ) = β is a constant, (S.7.44) becomes the Black-Scholes PDE with squared volatility parameter 2 σ 2 := σS2 + σM β2.

When the option is the European call option with strike price K on ST , its solution is given by the Black-Scholes function

f (t, x) = Bl(x, K, σ, r, T − t) = xΦ d+ (T − t) − Ke−(T −t)r Φ d− (T − t) , with d+ (T − t) :=

2 β 2 )/2)(T − t) log(x/K) + (r + (σS2 + σM √ , |σ| T − t

d− (T − t) :=

2 β 2 )/2)(T − t) log(x/K) + (r − (σS2 + σM √ , |σ| T − t

and ξt =

∂f (t, St , Mt ) = Φ d+ (T − t) , ∂x

with ηt = −

K −(T −t)r K e Φ d− (T − t) = − e−T r Φ d− (T − t) , At A0

0 ≤ t < T.

j) Similarly to Question (i), when the option is the European put option with strike price K on ST , its solution is given by the Black-Scholes put price function f (t, x) = Ke−(T −t)r Φ − d− (T − t) − xΦ − d+ (T − t) , with ζt =

∂f (t, St , Mt ) = −Φ − d+ (T − t) , ∂y

and ηt =

K −T r e Φ − d− (T − t) , A0

0 ≤ t < T. 0 ≤ t < T.

Remark. By the answer to Question (b) we have q 2 + σ 2 dZ dSt = (r + α + β(µ − r))St dt + St β 2 σM t S where (Zt )t∈R+ is a standard Brownian motion, hence the answers to Questions (i) and j can be recovered from the pricing relation "

155 September 15, 2022

Solutions Manual f (t, St ) = e−(T −t)r IE∗ ϕ(ST ) | Ft ],

0 ≤ t ≤ T.

Problem 7.27 1) a) If suffices to let τn := T , n ≥ 1. Then, the sequence (τn )n≥1 clearly satisfies Conditions (v)−(vi), and the process (Mτn ∧t )t∈[0,T ] = (Mt )t∈[0,T ] is a (true) martingale under P. b) Applying a) the local martingale property to a suitable sequence (τn )n≥1 of stopping times, and b) Fatou’s lemma to the non-negative sequence (Mτn ∧t )n≥1 , we have IE[Mt | Fs ] = IE lim Mτn ∧t Fs n→∞

≤ lim inf IE[Mτn ∧t | Fs ] n→∞

= lim inf Mτn ∧s n→∞

= lim Mτn ∧s n→∞

= Ms ,

0 ≤ s ≤ t ≤ T,

which shows that (Mt )t∈[0,T ] is a supermartingale. c) Since (Mt )t∈[0,T ] is a supermartingale by Question (1b), for any t ∈ [0, T ] we have IE[MT | Ft ] − Mt ≤ 0 a.s., and there exists t ∈ [0, T ] such that IE[IE[MT | Ft ] − Mt ] < 0, otherwise we would have IE[MT | Ft ]−Mt = 0 a.s. for all t ∈ [0, T ], and (Mt )t∈[0,T ] would be a martingale by the tower property.∗ Therefore, using again the tower property, we find IE[MT − M0 ] = IE[IE[MT | Ft ] − M0 ] ≤ IE[IE[MT | Ft ] − Mt ] < 0. d) We have C(0, M0 ) − P (0, M0 ) = IE[(MT − K)+ − (K − MT )+ ] = IE[MT − K] < IE[M0 ] − K, showing that call-put parity C(0, M0 ) − P (0, M0 ) = IE[M0 ] − K is not satisfied. 2) a) The stochastic differential equation can be rewritten as dSt = σ(t, St )dBt √ where σ(t, x) = x/ T − t, t ∈ [0, T − ε], satisfies the global Lipschitz condition ∗

156

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Introduction to Stochastic Finance with Market Examples, Second Edition |x − y| x−y ≤ , |σ(t, x) − σ(t, y)| = √ T −ε T −t

x, y ∈ R.

Hence by e.g. Theorem V-7 in Protter (2004) this stochastic differential equation admits unique (strong) solution such that St = S0 +

0 T −s

dBs ,

0 ≤ t ≤ T − ε.

Next, we have 1 w t ds dB √ s − 0 T −s 2 0 T −s w t dBs 1 T √ = S0 exp − log 0 T −t T −s 2 r w t t dBs √ = S0 1 − exp , 0 ≤ t ≤ T − ε. 0 T T −s

St = S0 exp

b) We have ST = 0, as can be checked from the graphs of Question (d) below. c) Consider the stopping times 1 τn := 1− T ∧ inf{t ∈ [0, T ] : |St | ≥ n}, n ≥ 1. n for all n ≥ 1 the stopped process (Sτn ∧t )t∈[0,T ] is given by w τn ∧t

wt Su S dBu = S0 + 1[0,τn ] (u) √ u dBu , 0 0 T −u T −u √ 0 ≤ t ≤ T , and the process (1[0,τn ] (u)Su / T − u)0≤u≤τn ∧t is square integrable as w w T (1−1/n)T S2 n2 IE 1[0,τn ] (u) u du ≤ IE 1[0,τn ] (u) du , 0 0 T −u T −u Sτn ∧t = S0 +

√

hence by Proposition 8.1 the stopped process (Sτn ∧t )t∈[0,T ] is a (true) martingale under P for all n ≥ 1, and therefore (St )t∈[0,T ] is a local martingale on [0, T ]. Finally, we note that since 0 = IE[ST ] ̸= S0 , the process (St )t∈[0,T ] is not a martingale. d) The following code solves the stochastic differential equation dSt = √ St dBt / 1 − t by the Euler scheme. N=10000; t <− 0:N; dt <− 1.0/N; dB <− rnorm(N,mean=0,sd=sqrt(dt));S <− rep(0,N+1);S[1]=1.0 for (k in 2:(N−1)){S[k]=S[k−1]+S[k−1]∗dB[k]/sqrt(1−k∗dt)} plot(t∗dt, S, xlab = "t", ylab = "", type = "l", ylim = c(0,1.05∗max(S)), col = " blue", xaxs = "i", yaxs = "i",cex.axis=1.6,cex.lab=1.8)

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Solutions Manual

0.0

0.2

0.4

0.6

0.8

1.0

√ Fig. S.38: Sample path of dSt = St dBt / 1 − t.

3) a) The following code solves the stochastic differential equation dSt = St2 dBt by the Euler scheme.

N=10000; t <− 0:N; dt <− 1.0/N; dB <− rnorm(N+1,mean=0,sd=sqrt(dt));S <− rep(0,N+1);S[1]=1.0 for (k in 2:(N+1)){S[k]=S[k−1]+S[k−1]^2∗dB[k]} plot(t∗dt, S, xlab = "t", ylab = "", type = "l", ylim = c(1.05∗min(S),1.05∗max(S)), col = "blue", xaxs = "i", yaxs = "i",cex.axis=1.6,cex.lab=1.8)

0.0

0.2

0.4

0.6

0.8

1.0

Fig. S.39: Sample path of dSt = St2 dBt .

158

158 September 15, 2022

1.0

1.5

2.0

2.5

3.0

3.5

Introduction to Stochastic Finance with Market Examples, Second Edition

0.0

0.2

0.4

0.6

0.8

1.0

Fig. S.40: Sample path of dSt = St2 dBt . b) With the change of variable y = 1/x and dy = −dx/x2 , we have w∞ IE[ST ] = xφT (x)dx 0 w∞ S0 (1/x − 1/S0 )2 (1/x + 1/S0 )2 √ = exp − − exp − dx 2 0 x 2T 2T 2πT w∞ S w∞ S (y − 1/S0 )2 (y + 1/S0 )2 √ 0 exp − √ 0 exp − = dy − dy 0 0 2T 2T 2πT 2πT 2 w∞ w 2 ∞ y y S S √ 0 exp − √ 0 exp − = dy − dy −1/S0 1/S0 2t 2T 2πT 2πT w∞ w∞ y2 y2 S S √ 0 exp − √ 0 exp − = dy − dy √ √ −1/(S0 T ) 1/(S0 T ) 2 2 2π 2π √ √ = S0 Φ(1/(S0 T )) − S0 Φ(−1/(S0 T )) √ = S0 (1 − 2Φ(−1/(S0 T ))). c) We have 1 1 √ IE[ST ] = 2S0 Φ − 2 S0 T √ w w0 1/(S0 T ) 2 dx −x2 /2 dx √ − e e−x /2 √ = 2S0 −∞ −∞ 2π 2π w 1/(S0 √T ) 2 dx = 2S0 e−x /2 √ 0 2π √ w 1/ T −(y/S0 )2 /2 dy √ , =2 e 0 2π hence by the dominated convergence theorem we have "

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Solutions Manual

S0 →∞

w 1/√T

2 dy e−(y/S0 ) /2 √ 2π −(y/S0 )2 /2 dy √ =2 lim e 0 S0 →∞ 2π w 1/√T dy =2 1√ 0 2π r 2 = . πT

lim IE[ST ] = lim 2

0 S0 →∞ w 1/√T

d) With the change of variable y = 1/x and dy = −dx/x2 , we have w∞ IE[(ST − K)+ ] = (x − K)+ φT (x)dx 0 w∞ x−K (1/x − 1/S0 )2 (1/x + 1/S0 )2 √ = S0 exp − − exp − dx 3 K x 2T 2T 2πT w 1/K S (y − 1/S0 )2 (y + 1/S0 )2 √ 0 = exp − − exp − dy 0 2T 2T 2πT w 1/K y 2 (y − 1/S0 ) (y + 1/S0 )2 √ −KS0 exp − − exp − dy 0 2T 2T 2πT w 1/K−1/S0 S w 1/K+1/S0 S y2 y2 √ 0 exp − √ 0 exp − = dy − dy −1/S0 1/S 2T 2T 0 2πT 2πT w 1/K−1/S0 y + 1/S 2 y 0 √ −KS0 exp − dy −1/S0 2T 2πT w 1/K+1/S0 y − 1/S y2 0 √ +KS0 exp − dy 1/S0 2T 2πT 1 1 1 √ − √ = S0 Φ − S0 Φ − √ K T S0 T S T 0 1 1 1 √ + √ √ −S0 Φ + S0 Φ K T S0 T S0 T √ √ 1 1 1 1 √ − √ √ + √ +KS0 T φ − KS0 T φ K T S0 T K T S0 T 1 1 1 √ − √ −KΦ + KΦ − √ K T S0 T S0 T 1 1 1 √ + √ √ −KΦ + KΦ K T S0 T S T 0 1 1 1 √ − √ = S0 Φ − S0 Φ − √ K T S0 T S T 0 1 1 1 √ + √ √ −S0 Φ + S0 Φ K T S0 T S0 T

160

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Introduction to Stochastic Finance with Market Examples, Second Edition √ √ 1 1 1 1 √ − √ √ + √ − KS0 T φ +KS0 T φ K T S0 T K T S0 T 1 1 1 1 √ − √ √ + √ −KΦ − KΦ +1 K T S0 T K T S T 0 1 1 1 √ − √ = S0 Φ − S0 Φ − √ K T S0 T S0 T 1 1 1 √ + √ √ + S0 Φ −S0 Φ K T S0 T S0 T √ √ 1 1 1 1 √ − √ √ + √ +KS0 T φ − KS0 T φ K T S0 T K T S0 T 1 1 1 1 √ + √ √ − √ − KΦ , +KΦ S0 T K T K T S0 T 2 √ where φ(z) := e−z / 2π is the standard normal probability density function, see Relation (2.1.2) in Jacquier (2017). e) We have IE[(ST − K)+ ] ≤ IE[ST ] w 1/√T 2 dy =2 e−(y/S0 ) /2 √ 0 2π w 1/√T dy √ ≤2 0 2π r 2 ≤ . πT

Fig. S.41: “Infogrames” stock price curve.

Problem 7.28 "

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Solutions Manual a) Relation (7.54) can be checked to hold first on the event Aα , and then on its complement Acα . Taking the Q-expectation on both sides of (7.54) yields dP dP − α (21Aα − 1) ≥ IEQ − α (21A − 1) , IEQ dQ dQ i.e. IEQ

dP dP (21Aα − 1) −α IEQ [21Aα −1] ≥ IEQ (21A − 1) −α IEQ [21A −1], dQ dQ

i.e. 2P(Aα ) − 1 − α(2Q(Aα ) − 1) ≥ 2P(A) − 1 − α(2Q(A) − 1), which shows that P(Aα ) − P(A) ≥ α(Q(Aα ) − Q(A)), allowing us to conclude to P(Aα ) − P(A) ≥ 0 since α ≥ 0. b) We check that dQ∗ /dP∗ ≥ 0 since C ≥ 0, and w Q∗ (Ω) = dQ∗ Ω w dQ∗ = dP∗ Ω dP∗ w C = dP∗ Ω IEP∗ [C] C = IEP∗ IEP∗ [C] IEP∗ [C] = IEP∗ [C] = 1. In the next questions we consider a nonnegative contingent claim payoff C ≥ 0 with maturity T > 0, priced e−rT IEP∗ [C] at time 0 under the riskneutral measure P∗ . Budget constraint. We assume that no more than a certain fraction β ∈ (0, 1] of the claim price e−rT IEP∗ [C] is available to construct the initial hedging portfolio V0 at time 0. Since a self-financing portfolio process (Vt )t∈R+ started at V0 = βe−rT IEP∗ [C] may not able to hedge the claim C when β < 1, we will attempt to maximize the probability P(VT ≥ C) of successful hedging. 162

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Introduction to Stochastic Finance with Market Examples, Second Edition For this, given A an event we consider the portfolio process (VtA )t∈[0,T ] hedging the claim C 1A , priced V0A = e−rT IEP∗ [C 1A ] at time 0, and such that VTA = C 1A at maturity T . c) Using the probability measure Q∗ , we rewrite the condition (7.56) as IEP∗ [C 1A ] dQ∗ ≤ β, Q∗ (A) = IEQ∗ [1A ] = IEP∗ 1A ∗ = dP IEP∗ [C] i.e. Q∗ (A) ≤ Q∗ (Aα ) = β. By the Neyman-Pearson Lemma, for any event A, the inequality Q∗ (A) ≤ Q∗ (Aα ) = β implies P(A) ≤ P(Aα ), which shows that the event A = Aα realizes the maximum under the required condition. d) The obvious inequality is P(Aα ) ≤ P(C 1Aα ≥ C) = P VTAα ≥ C . In other direction, we note that the event Bα := {C 1Aα ≥ C} = the VTAα ≥ C satisfies (7.56), as e−rT IEP∗ VTBα = e−rT IEP∗ C 1Bα = e−rT IEP∗ C 1{C 1Aα ≥C} = e−rT IEP∗ C 1Aα = βe−rT IEP∗ [C], where the last equality IEP∗ C 1Aα = β IEP∗ [C] follows from Q∗ (Aα ) = β ∗ and the definition (7.55) of Q . Therefore, by the result of Question (c) we have P(C 1Aα ≥ C) = P VTAα ≥ C = P(Bα ) ≤ P(Aα ). e) This hedging strategy starts from the initial amount V0Aα = e−rT IEP∗ [C 1Aα ] = βe−rT IEP∗ [C], and it satisfies P VTAα ≥ C = P(C 1Aα ≥ C) = P(Aα ) which is the maximum hedging probability under the constraint (7.56). f) We have 2

St = S0 eσBt +rt−σ t/2 = S0 eσBt = S0 eBt ,

t ≥ 0.

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Solutions Manual g) We have P VTAα ≥ C = P(Aα ) dP > α =P dQ∗ dP dQ∗ =P >α ∗ ∗ dP dP dP C =P > α dP∗ IEP∗ [C] = P (αC < IEP∗ [C]) = P (ST − K)+ < IEP∗ [C]/α = P ST − K < IEP∗ [C]/α = P ST < K + IEP∗ [C]/α

= P S0 eBT < K + IEP∗ [C]/α

= P (BT < log(K + IEP∗ [C]/α)) log(K + IEP∗ [C]/α) √ . =Φ T h) We have α=

exp

√

IEP∗ [C] . T Φ−1 P VTAα ≥ C −K

i) We have IEP∗ [C 1Aα ] = IEP∗ [(ST − K)+ 1Aα ] h i = IEP∗ (ST − K)+ 1{BT <log(K+IEP∗ [C]/α)} h i + = IEP∗ eBT − K 1{BT <log(K+IEP∗ [C]/α)} w log(K+IEP∗ [C]/α)/√T 2 dx = ex − K e−x /2 √ √ (log K)/ T 2π w log(K+IEP∗ [C]/α)/√T x −x2 /2 dx √ = e e √ (log K)/ T 2π √ w log(K+IEP∗ [C]/α)/ T −x2 /2 dx √ e −K √ (log K)/ T 2π w log(K+IEP∗ [C]/α)/√T 1/2−(x−1)2 /2 dx √ e = √ (log K)/ T 2π −K Φ(log(K + IEP∗ [C]/α)) − Φ(log K) √ √ w −1/ T +log(K+IEP∗ [C]/α)/ T 2 dx = e1/2−x /2 √ √ (−1+log K)/ T 2π −K Φ(log(K + IEP∗ [C]/α)) − Φ(log K) 164

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Introduction to Stochastic Finance with Market Examples, Second Edition =

−1 + log(K + IEP∗ [C]/α) −1 + log K √ √ e Φ −Φ T T log(K + IEP∗ [C]/α) log K √ −K Φ −Φ √ , T T √

hence e−rT IEP∗ [C 1Aα ] −1 + log K −1 + log(K + IEP∗ [C]/α) √ √ −Φ = e−rT +1/2 Φ T T log(K + IEP∗ [C]/α) log K −rT √ −Ke Φ −Φ √ . T T j)

i) We have d+ (T ) = 1, d− (T ) = 0, hence by the Black-Scholes formula we find √ IEP∗ [(ST − K)+ ] = eS0 Φ(d+ (T )) − KΦ(d− (T )) = 1.64872 × 0.84134 − 1/2 = 0.88713, and e−rT IEP∗ [(ST − K)+ ] = S0 Φ(d+ (T )) − Ke−T r Φ(d− (T )) = 0.84134 − 0.60653/2 = 0.53807. ii) By the result of Question (h), we have IEP∗ [C] exp Φ−1 P VTAα ≥ C −K 0.88714 = exp (Φ−1 (0.9)) − 1 0.88714 = 1.28 = 0.34165. e −1

α=

iii) By the result of Question (i), we find e−rT IEP∗ [C 1Aα ] = Φ(−1 + log(K + IEP∗ [C]/α)) − Φ(−1 + log K) −Ke−1/2 Φ(log(K + IEP∗ [C]/α)) − Φ(log K) = Φ(−1 + log(1 + IEP∗ [C]/α)) − Φ(−1) −e−1/2 Φ(log(1 + IEP∗ [C]/α)) − Φ(0) = Φ(−1 + log(1 + 0.88713/0.34165)) − Φ(−1) −e−1/2 Φ(log(3.596604712)) − Φ(0) "

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Solutions Manual = Φ(0.27999) − Φ(−1)

−0.60653 × Φ(1.279990265) − Φ(0) = (0.61026 − 0.158655) − 0.60653 × (0.899726 − 0.5) = 0.20915. In addition, we find β=

e−rT IEP∗ [C 1Aα ] 0.20915 = = 37.53%. e−rT IEP∗ [C] 0.53807

Chapter 8 Exercise 8.1 We need to compute the average w T 1 1 wT 1 wT IE vt dt = IE[vt ]dt = u(t)dt, 0 T T 0 T 0 where u(t) := IE[vt ]. Taking expectation on both sides of the equation vt = v0 − λ

wt 0

(vs − m)ds + η

wt√ 0

vs dBs ,

we find u(t) = IE[vt ] wt wt√ = IE v0 − λ (vs − m)ds + η vs dBs 0 0 w t (vs − m)ds = v0 − λ IE 0

= v0 − λ = v0 − λ

wt 0

(IE[vs ] − m)ds (u(s) − m)ds,

t ≥ 0,

hence by differentiation with respect to t ∈ R we find the ordinary differential equation u′ (t) = λm − λu(t), cf. e.g. Exercise 4.18-(b). This equation can be rewritten as (eλt u(t))′ = λeλt u(t) + eλt u′ (t) = λmeλt , which can be integrated as

166

166 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition eλt u(t) =

u(0) + λm

wt 0

eλs ds

= IE[v0 ] + m(eλt − 1) = meλt + IE[v0 ] − m

t ∈ R+ ,

from which we conclude that u(t) = m + (IE[v0 ] − m)e−λt ,

t ∈ R+ ,

and w T 1 wT 1 IE vt dt = u(t)dt 0 T T 0 1 wT = m + (IE[v0 ] − m)e−λt dt T 0 IE[v0 ] − m w T −λt = m+ e dt 0 T 1 − e−λT = m + IE[v0 ] − m . λT Exercise 8.2 a) By e.g. Exercise 4.18-(b), we have IE[vt ] = IE[v0 ]e−λt + m(1 − e−λt ),

t ∈ R+ ,

hence w T 1 1 IE (dSt )2 2 0 St T w 2 p T 1 1 (1) = IE β + v dB (r − αv )S dt + S t t t t t 0 St2 T w T 1 = IE (β + vt )dt 0 T 1 wT =β+ IE[vt ]dt, T 0

VST =

which yields 1 wT (IE[v0 ]e−λt + m(1 − e−λt ))dt T 0 w 1 T =β+ (IE[v0 ]e−λt + m(1 − e−λt ))dt T 0 wT 1 = β + m + (IE[v0 ] − m) e−λt dt 0 T

VST = β +

167 September 15, 2022

Solutions Manual = β + m + (IE[v0 ] − m)

eλT − 1 . λT

Note that if the process (vt )t∈R+ is started in the gamma stationary distribution then we have IE[v0 ] = IE[vt ] = m, t ∈ R+ , and the variance swap rate VST = β + m becomes independent of the time T . (2) b) The stochastic differential equation dσt = ασt dBt is solved as (2)

σt = σ0 eαBt −α t/2 ,

t ∈ R+ ,

hence we have w T 1 1 2 IE (dS ) t 0 St2 T w T 1 1 (1) 2 = IE σt St dBt 2 0 St T w T 1 = IE σt2 dt 0 T σ02 w T h 2αBt(2) −α2 t i IE e dt = T 0 h i 2 wT (2) 2 σ = 0 e−α t IE e2αBt dt T 0 σ 2 w T −α2 t+2α2 t e dt = 0 T 0 2 wT 2 σ eα t dt = 0 T 0 2 σ2 = 20 (eα T − 1). α T

VST =

Exercise 8.3 2 2 a) Taking x = R0,T and x0 = IE R0,T , we have R0,T ≈

2 2 2 2 2 2 R0,T R0,T − IE R0,T − IE R0,T q IE R0,T + 2 3/2 , (S.8.45) 2 − 8 IE R0,T 2 IE R0,T

2 2 provided that R0,T is sufficiently close to IE R0,T . b) Taking expectations on both sides of (S.8.45), we find IE∗ [R0,T ] ≈

168

2 2 2 2 2 2 IE R0,T − IE R0,T IE R0,T − IE[R0,T ] q + − IE R0,T 2 3/2 2 8 IE R0,T 2 IE R0,T

168 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2 2 2 R0,T − IE R0,T 2 3/2 8 IE R0,T 2 q Var R0,T 2 = IE R0,T − 2 3/2 , 8 IE R0,T =

2 IE IE R0,T −

2 2 provided that R0,T is sufficiently close to IE R0,T . Exercise 8.4 We have " # "N 2 # 2 T wT X St STk = IE log dNt IE log 0 STk−1 Stn=1 w T 2 = IE ZNt- dNt 0

=λ =λ

0 wT 0 2

ZNt-

2 dt

(η 2 + δ 2 )dt

= λ(η + δ 2 )T. Exercise 8.5 2

a) We have St = S0 eσBt −σ t/2+rt , t ∈ R+ . 2 b) Letting Set := e−rt St , t ∈ R+ , we have SeT = S0 eσBT −σ T /2 and dSet = e σ St dBt , hence wT SeT = S0 + σ Set dBt , 0

and " # −rT e e ST e−rT ST SeT ∗ ST 2 IE log = 2 IE log S0 S0 S0 S0 " ! # w T Se 2 σ T t = 2 IE∗ 1+σ dBt σBT − 0 S0 2 " # " # w T Se w T Se σ2 T t t ∗ ∗ ∗ 2 2 + 2σ IE BT dBt − σ T IE dBt = 2 IE σBT − 0 S0 0 S0 2 " # wT w T Se t dBt = −σ 2 T + 2σ 2 IE∗ dBt 0 0 S0 " # w T Se t = −σ 2 T + 2σ 2 IE∗ dt 0 S0 ∗

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Solutions Manual

= −σ 2 T + 2σ 2

IE∗

= −σ 2 T + 2σ 2

# Set dt S0

= −σ T + 2σ T 2

= σ 2 T. Alternatively, we could also write " # −rT e e ST e−rT ST SeT ∗ ST 2 IE log = 2 IE log S0 S0 S0 S0 h i ∗ σBT −σ 2 T /2 σBT −σ 2 T /2 = 2 IE e log e 2 σ2 T = 2 IE∗ eσBT −σ T /2 σBT − 2 2 ∗ −σ 2 T /2 σBT 2 = 2σe IE BT e − σ T IE∗ eσBT −σ T /2 2 ∂ ∗ σBT = 2σe−σ T /2 IE e − σ2 T ∂σ 2 ∂ 2 = 2σe−σ T /2 eσ T /2 − σ 2 T ∂σ 2 2 = 2σ 2 T e−σ T /2 eσ T /2 − σ 2 T ∗

= σ 2 T. Exercise 8.6 a) By the Itô formula, we have log

w T dS ST 1 w T σt2 t = log ST − log S0 = − dt. 0 St S0 2 0 St2

b) By (8.47) we have w w T T dSt ST IE∗ σt2 dt Ft = 2 IE∗ Ft − 2 IE∗ log Ft 0 0 St S0 w t dS ST u + 2r(T − t) − 2 IE∗ log Ft . =2 0 Su S0 c) At time t ∈ [0, T ] we check that w t dSu 2 Lt + e−(T −t)r St + 2e−rT + (T − t)r − 1 At 0 Su St w t dS u −(T −t)r = Lt + 2r(T − t)e + 2e−(T −t)r 0 Su 170

170 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = Vt . d) By (8.48) we have w t dS u dVt = d Lt + 2r(T − t)e−(T −t)r + 2e−(T −t)r 0 Su = dLt − 2re−(T −t)r dt + 2r2 (T − t)e−(T −t)r dt w t dS dSt u dt + 2e−(T −t)r +2re−(T −t)r 0 Su St w t dSu −(T −t)r 2 −rT = dLt + e dSt + 2e + (T − t)r − 1 dAt , 0 Su St with dAt = rert dt, hence the portfolio is self-financing.

Exercise 8.7 By second differentiation of the moment generating function (8.9), we find the two expressions # " " 2 2 # 4 2 ST ST ST = 4 IE∗ log IE∗ R0,T = 4 IE∗ log + 2 log −4 IE∗ R0,T , F0 F0 F0 and " !# 2 4 ST ST ST IE∗ R0,T = 4 IE∗ log − 2 log F0 F0 F0 " 2 # ST ST 2 = 4 IE∗ log − 4 IE∗ R0,T . F0 F0

Chapter 9 Exercise 9.1 ∂f x ∂C (T − t, x, K) = T − t, and ∂x ∂z K ∂C ∂ x (T − t, x, K) = Kf T − t, ∂K ∂K K x x ∂f x − T − t, = f T − t, K K ∂z K 1 x ∂C = C(T − t, x, K) − (T − t, x, K), K K ∂x

a) We have

hence "

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Solutions Manual ∂C 1 K ∂C (T − t, x, K) = C(T − t, x, K) − (T − t, x, K). ∂x x x ∂K 1 ∂2f x ∂2C (T − t, x, K) = T − t, and b) We have ∂x2 K ∂z 2 K ∂2C (T − t, x, K) ∂K 2 x ∂f x x ∂f x x2 ∂f x =− 2 T − t, + 2 T − t, + 3 T − t, K ∂z K K ∂z K K ∂z K x2 ∂ 2 f x = 3 2 T − t, K ∂z K x2 ∂ 2 C = 2 (T − t, x, K), K ∂x2 hence K 2 ∂2C ∂2C (T − t, x, K) = 2 (T − t, x, K). ∂x2 x ∂K 2 c) Noting that ∂C ∂C (T − t, x, K) = − (T − t, x, K), ∂t ∂T we can rewrite the Black-Scholes PDE as ∂C (T − t, x, K) ∂T 1 K ∂C +rx C(T − t, x, K) − (T − t, x, K) x x ∂K σ 2 x2 K 2 ∂ 2 C + (T − t, x, K), 2 x2 ∂K 2

rC(T − t, x, K) = −

i.e. ∂C ∂C σ 2 x2 K 2 ∂ 2 C (T − t, x, K) = −rK (T − t, x, K) + (T − t, x, K). ∂T ∂K 2 x2 ∂K 2 Remarks: 1. Using the Black-Scholes Greek Gamma expression ∂2C 1 √ (T − t, x, K) = Φ′ (d+ (T − t)) ∂x2 σx T − t 2 1 p = e−(d+ (T −t)) /2 , σx 2π(T − t)

172

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Introduction to Stochastic Finance with Market Examples, Second Edition we can recover the lognormal probability density function φT (y) of geometric Brownian motion ST as follows: ∂2C (T − t, x, K) ∂K 2 2 x ∂2C (T − t, x, K) = e(T −t)r 2 K ∂x2 (T −t)r 2 e x p = e−(d+ (T −t)) /2 σK 2 2π(T − t) 2 1 p = e−(d− (T −t)) /2 σK 2π(T − t)

φT (K) = e(T −t)r

1 σK

2π(T − t)

exp −

(r − σ 2 /2)(T − t) + log(x/K) 2(T − t)σ 2

2 ! ,

knowing that −

2 1 1 d− (T − t) = − 2 2

log(x/K) + (r − σ 2 /2)(T − t) √ |σ| T − t

2 1 log(x/K) + (r + σ 2 /2)(T − t) x √ + (T − t)r + log 2 K |σ| T − t 2 1 x = − d+ (T − t) + (T − t)r + log , 2 K =−

which can be obtained from the relation 2 2 d+ (T − t) − d− (T − t) = d+ (T − t) + d− (T − t) d+ (T − t) − d− (T − t) x = 2r(T − t) + 2 log . K 2. Using the expressions of the Black-Scholes Greeks Delta and Theta we can also recover ∂C 2 ∂T

(T − t, x, K) + rK K2

− =2

∂C (T − t, x, K) ∂K

∂2C (T − t, x, K) ∂K 2

1 x ∂C C(T − t, x, K) − (T − t, x, K) K K ∂x 2 ∂ C x2 2 (T − t, x, K) ∂x

∂C (T − t, x, K) + rK ∂t

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Solutions Manual √ xσΦ′ (d+ (T − t))/(2 T − t) + rKe−(T −t)r Φ(d− (T − t)) √ x2 Φ′ (d+ (T − t))/(xσ T − t) rC(T − t, x, K) − rxΦ(d+ (T − t)) √ +2 x2 Φ′ (d+ (T − t))/(xσ T − t)

= σ2 . Exercise 9.2 We have 2

∂C e−(K−S0 ) /(2T ) √ (S0 , K, T ) = −(K − S0 ) ∂K 2πT K − S0 K − S0 K − S0 + √ φ − √ −Φ − √ T T T K − S0 = −Φ − √ , T and 2 1 ∂C 2 e−(K−S0 ) /(2T ) , (S0 , K, T ) = √ ∂K 2 2πT

which is the Gaussian probability density function of ST = S0 + BT . We also have r 2 ∂C 1 (K − S0 )2 T −(K−S0 )2 /(2T ) (S0 , K, T ) = √ e−(K−S0 ) /(2T ) − e ∂T 2T 2 2π 2 2πT 2 (K − S0 ) K − S0 . + φ − √ 2T 3/2 T 2 1 e−(K−S0 ) /(2T ) = √ 2 2πT 1 ∂C 2 = (S0 , K, T ), 2 ∂K 2 hence v v u ∂C M u ∂C M u ∂C M r (t, y) + 2ry (t, y) u u2 u 2 (t, y) 1 1 u ∂t ∂y u ∂t , |σ(t, y)| = u = = = u 2 M t t 2 ∂2C M y2 |y| 2∂ C y (t, y) y (t, y) ∂y 2 ∂y 2 and the equation satisfied by (St )t∈R+ is dSt = St σ(t, St )dBt =

174

St dBt = sign (St )dBt = dWt , |St | 174 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition where dWt := sign (St )dBt is also a standard Brownian motion by the Lévy characterization theorem, σ(t, y) = 1/y, and St = S0 + Bt . Indeed, as in Quiz 2 of FE8815, the price of the call option in the Bachelier model is given by C(S0 , K, T ) = IE[(ST − K)+ ] = IE[(S0 + BT − K)+ ] w∞ 2 dx = (x + S0 − K)e−x /(2T ) √ K−S0 2πT w∞ w∞ 2 dx −x2 /(2T ) dx √ − (K − S0 ) e−x /(2T ) √ = xe K−S0 K−S0 2πT 2πT r w∞ 2 T h −x2 /(2T ) i∞ −y /2 dy √ = −e − (K − S0 ) √ e (K−S0 )/ T 2π K−S0 2π r K − S0 T −(K−S0 )2 /(2T ) e − (K − S0 )Φ − √ = . 2π T Exercise 9.3 a) We have ∂C ∂C ∂MC ′ (K, S, r, τ ) = (K, S, σimp (K), r, τ ) + σimp (K) (K, S, σimp (K), r, τ ). ∂K ∂K ∂σ b) We have ∂C ∂C ′ (K, S, σimp (K), r, τ ) + σimp (K) (K, S, σimp (K), r, τ ) ≤ 0, ∂K ∂σ which shows that ∂C (K, S, σimp (K), r, τ ) ′ σimp (K) ≤ − ∂K ∂C (K, S, σimp (K), r, τ ) ∂σ c) We have ∂P ∂P ′ (K, S, σimp (K), r, τ ) + σimp (K) (K, S, σimp (K), r, τ ) ≥ 0, ∂K ∂σ which shows that ∂P ′ σimp (K) ≥ − ∂K ∂P

∂σ

(K, S, σimp (K), r, τ ) (K, S, σimp (K), r, τ )

Exercise 9.4 "

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Solutions Manual a) We have K +S 2 2 K +S − S0 = σ0 + β 2 β = σ0 + (K − (2S0 − S))2 . 4

σ in %

σimp (K, S) ≃ σloc

5 σimp in % σloc in % 0

100

150

200

σimp in % σloc in % 0

100

150

200

(a) At the money K = S0 .

(b) Out of the money K > S0 .

Fig. S.42: Implied vs local volatility. b) We find ∂ ∂S

S, K, T, σimp (K, S), r

∂Bl x, K, T, σimp (K, S), r x=S ∂x ∂σimp ∂Bl x, K, T, σ, r σ=σimp (K,S) + ∂S ∂σ β = ∆ + ν (K − (2S0 − S)), 2 =

where

∂Bl x, K, T, σimp (K, S), r x=S ∂x is the Black-Scholes Delta and ∆=

ν=

∂Bl S, K, T, σ, r σ=σimp (K,S) ∂σ

is the Black-Scholes Vega, cf. §2.2 of Hagan et al. (2002). Exercise 9.5 We take t = 0 for simplicity. We start by showing that for every λ > 0, we have 176

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Introduction to Stochastic Finance with Market Examples, Second Edition wτ i 2erτ w F0 w∞ dK dK 1 h C(τ, K) 2−p P (τ, K) 2−p + . IE exp λ σt2 dt − 1 = pλ λ λ 0 F0 0 λ S0 K K By Lemma 8.2, we have wτ i 1 S pλ 1 h τ IE exp λ σt2 dt − 1 = IE −1 , 0 λ λ F0

λ > 0.

Using Relation (9.18), i.e. M

φτ (K) = erτ

∂2C ∂2P (τ, y) = erτ 2 (τ, y), ∂y 2 ∂y

we have i 1 S pλ wτ 1 h τ IE exp λ σt2 dt − 1 = IE −1 0 λ λ F0 p 1 pλ λ = IE Sτ − F0 λF0pλ 1 w ∞ pλ K φτ (K)dK − F0pλ = pλ 0 λF0 w w∞ F0 1 K pλ φτ (K)dK + K pλ φτ (K)dK − F0pλ = pλ 0 F0 λF0 w∞ w F0 2 2 1 pλ rτ rτ pλ ∂ P pλ ∂ C (τ, K)dK + e (τ, K)dK − S K . = e K 0 0 F0 λS0pλ ∂K 2 ∂K 2 Next, integrating by parts over the intervals [0, F0 ] and [F0 , ∞) and using the boundary conditions P (τ, 0) = C(τ, ∞) = 0,

∂P ∂C (τ, 0) = (τ, ∞) = 0, ∂K ∂K

with the relation ∂P ∂C (τ, K) − (τ, K) − e−rτ = 0 ∂K ∂K and the call-put parity P (τ, F0 ) − C(τ, F0 ) = S0 − F0 erτ = 0 as boundary conditions, we find wτ i 1 h IE exp λ σt2 dt − 1 0 λ w F0 ∂P 1 ∂P erτ S0pλ (τ, F0 ) − pλ erτ K pλ −1 (τ, K)dK = 0 λS0pλ ∂K ∂K "

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Solutions Manual w∞ ∂C ∂C K pλ −1 (τ, F0 ) − pλ erτ (τ, K)dK − S0pλ F0 ∂K ∂K w w∞ F0 erτ ∂C pλ −1 ∂P = −pλ pλ K K pλ −1 (τ, K)dK + (τ, K)dK 0 F0 λS0 ∂K ∂K w F0 pλ erτ pλ −1 S0 P (τ, F0 ) + (pλ − 1) K pλ −2 P (τ, K)dK = 0 λS0pλ w∞ K pλ −2 C(τ, K)dK −S0pλ −1 C(τ, F0 ) + (pλ − 1) F0 w∞ pλ (pλ − 1) rτ w F0 pλ −2 = e K P (τ, K)dK + K pλ −2 C(τ, K)dK pλ 0 F λS0 0 w∞ dK dK 2erτ w F0 P (τ, K) 2−p + C(τ, K) 2−p = pλ λ λ 0 F0 S0 K K w∞ 2erτ w F0 dK dK = pλ P (τ, K) 2−p + C(τ, K) 2−p . λ λ 0 F0 S0 K K −erτ S0pλ

Finally, taking

pλ := p− λ = 1/2 −

p 1/4 + 2λ

and letting λ tend to zero, we find hw τ i wτ i 1 h IE σt2 dt = lim IE exp λ σt2 dt − 1 0 0 λ→0 λ w∞ 2erτ w F0 dK dK = lim pλ P (τ, K) 2−p + C(τ, K) 2−p λ λ 0 F0 λ→0 S0 K K w F0 dK w ∞ dK rτ = 2e P (τ, K) 2 + C(τ, K) 2 . 0 F0 K K Exercise 9.6 (Exercise 8.7 continued). Taking ϕ(x) = (log(x/F0 ))2 with y = F0 , we have x 2 x 2 and ϕ′′ (x) = 2 1 − log , ϕ′ (x) = log x F0 x F0 hence 2 ST log = ϕ(F0 ) + (ST − F0 )ϕ′ (F0 ) F0 w F0 w∞ (ST − z)+ ϕ′′ (z)dz + (z − ST )+ ϕ′′ (z)dz + F0 0 w F0 K dK (K − ST )+ 1 − log =2 0 F0 K 2 w∞ K dK +2 (ST − K)+ 1 − log . F0 F0 K 2 178

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Introduction to Stochastic Finance with Market Examples, Second Edition Therefore, we have " 2 # w∞ w F0 dK dK ST ∗ IE∗ [(ST − K)+ ] 2 IE =2 IE∗ [(K − ST )+ ] 2 + 2 log F0 0 F0 K K w F0 w∞ K K dK dK IE∗ [(ST − K)+ ] log −2 IE∗ [(K − ST )+ ] log −2 0 F0 F0 K 2 F0 K 2 w w∞ F0 K dK K dK 2 = IE∗ R0,T − 2erT P (T, K) log − 2erT C(T, K) log , F0 0 F0 K 2 F0 K 2 and w F0 w∞ 4 F0 dK K dK IE∗ R0,T = 8erT −8erT . P (T, K) log C(T, K) log 2 0 F0 K K F0 K 2 (S.9.46) Alternatively, taking ϕ(x) = (x/F0 )(log(x/F0 ))2 with y = F0 , we have ϕ′ (x) = and ϕ′′ (x) =

1 F0

log

x F0

2 x log F0 F0

2 x 2 2 log + = xF0 F0 xF0 xF0

1 + log

x F0

hence

log

ST F0

w F0

2 K 1 + log dK 0 KF0 F0 w∞ 2 K + (ST − K)+ 1 + log dK. F0 KF0 F0 (K − ST )+

Therefore, we have " 2 # w F0 2 K ST = IE[(K − ST )+ ] 1 + log dK IE∗ log 0 F0 KF0 F0 w∞ 2 K + IE[(ST − K)+ ] 1 + log dK., F0 KF0 F0 and 4 8 rT w F0 K dK IE∗ R0,T = e P (T, K) 1 + log 0 F0 F0 K 2 8 rT w ∞ K dK . + e P (T, K) 1 + log − 4 IE∗ R0,T F0 F0 F0 K Exercise 9.7 "

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Solutions Manual a) We have w ∞ e−νx − e−µx 0

xρ+1

dx =

w ∞ e−νx − e−µx

dx xρ+1 −νx −µx ∞ 1 e 1 w ∞ −νe−νx + µe−µx −e =− + dx ρ ρ x ρ 0 xρ 0 w w ∞ ∞ ν µ =− e−νx x−ρ dx + e−µx x−ρ dx ρ 0 ρ 0 µρ − ν ρ Γ (1 − ρ). = ρ 0

b) Taking ν = 0 and µ = Rt,T , we find w ∞ dλ 2 ρ IE∗ 1 − e−λRt,T ρ+1 0 Γ (1 − ρ) λ w∞ −λR2 dλ ρ ∗ t,T = 1 − IE e , Γ (1 − ρ) 0 λρ+1

IE∗ [Rt,T ] =

see § 3.1 in Friz and p Gatheral (2005) with ρ = 1/2. c) Letting p± 1/4 − 2λ, we have λ := 1/2 ± w∞ dλ 2 ρ 1 − IE∗ e−λRt,T Γ (1 − ρ) 0 λρ+1 " ± #! p w∞ ± ρ ST λ dλ = 1 − e−rpλ T IE∗ Γ (1 − ρ) 0 S0 λρ+1 " # p± w∞ ρ ST λ dλ = IE∗ 1 − Γ (1 − ρ) 0 F0 λρ+1 " # p± w 1/8 ρ ST λ dλ IE∗ 1 − = Γ (1 − ρ) 0 F0 λρ+1 " r # w∞ ρ ST i√ ST dλ + IE∗ 1 − exp ± 8λ − 1 log Γ (1 − ρ) 1/8 F0 2 F0 λρ+1 # " p± w 1/8 dλ ρ ST λ 8ρ ∗ = + IE 1 − ρ + 1 Γ (1 − ρ) 0 F0 λρ+1 "r # w∞ ρ ST 1√ ST dλ IE∗ cos 8λ − 1 log − Γ (1 − ρ) 1/8 F0 2 F0 λρ+1 w 1/8 dλ ρ ρ = IE∗ [ϕλ (ST )] ρ+1 + IE∗ [ψ(ST )] , Γ (1 − ρ) 0 λ Γ (1 − ρ)

IE∗ [Rt,T ] =

where 180

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Introduction to Stochastic Finance with Market Examples, Second Edition

ϕλ (x) = 1 −

x F0

p± λ ,

we have ±

ϕ′λ (x) = −p± λ

xpλ −1 p± F0 λ

± and ϕ′′λ (x) = −p± λ (pλ − 1)

xpλ −2 p± F0 λ

= 2λ

xpλ −2 p±

F0 λ

hence with y := F0 we have ϕλ (y) = 0 and IE∗ [ϕλ (ST )] 

 w F0 w∞ −2 −2 p± p± λ λ p± +K +K λ (ST − F0 ) − 2λ (K − ST ) dK − 2λ (ST − K) dK  p± p± 0 F0 F0 F λ F λ

∗

= IE

= 2λ

w F0 0

+ K

∗

IE [(K − ST ) ]

p± −2 λ p±

dK +

F0 λ

w∞ F0

∗

+ K

IE [(ST − K) ]

−2 p± λ p±

F0 λ w∞ ± ± erT w F0 C(T, K)K pλ −2 dK . = 2λ ± P (T, K)K pλ −2 dK + p F0 0 F0 λ Taking now ψ(x) :=

w∞ 1/8

1−

x cos F0

x 1√ 8λ − 1 log 2 F0

dλ , λρ+1

we have w∞ 1 x 1√ dλ ψ ′ (x) = − √ 8λ − 1 log cos 2 F0 λρ+1 2 F0 x 1/8 w∞ 1√ x √ dλ 1 + √ sin 8λ − 1 log 8λ − 1 ρ+1 , 2 F0 λ 2 F0 x 1/8 which converges provided that ρ > 1/2, while ψ ′′ (x) cannot be written as a converging integral but can be estimated numerically from ψ ′ (x). Hence, we have IE∗ [Rt,T ] ρerT Γ (1 − ρ)  w w∞ w 1/8 F0 ± ± dλ 2 √ P (T, K)K pλ −2 dK + C(T, K)K pλ −2 dK × 0 F0 0 ± 1/4−2λ λp F0 w F0 w∞ + P (T, K)ψ ′′ (K)dK + C(T, K)ψ ′′ (K)dK . =

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Solutions Manual library(quantmod) today <− as.Date(Sys.Date(), format="%Y−%m−%d"); getSymbols("^SPX", src = " yahoo") lastBusDay=last(row.names(as.data.frame(Ad(SPX)))) S0 = as.vector(tail(Ad(SPX),1)); T = 30/365;r=0.02;F0 = S0∗exp(r∗T) maturity<− as.Date("2021−07−07", format="%Y−%m−%d") # Choose a maturity in 30 days SPX.OPTS <− getOptionChain("^SPX", maturity) Call <− as.data.frame(SPX.OPTS$calls); Put <− as.data.frame(SPX.OPTS$puts) Call_OTM <− Call[Call$Strike>F0,];Call_OTM$dif = c(min(Call_OTM$Strike)−F0, diff(Call_OTM$Strike)) Put_OTM <−Put[Put$Strike<F0,];Put_OTM$dif = c(diff(Put_OTM$Strike), F0− max(Put_OTM$Strike))

pl <− function(lambda){return( 1/2+sqrt(1/4−2∗lambda ))}; rho=0.9 g1 <− function(x){ f1 <− function(lambda){ − cos ( 0.5∗sqrt(lambda∗8−1)∗log (x/ F0))/lambda^(rho+1)/sqrt(x∗F0)/2}; return(f1)} g2 <− function(x){ f2 <− function(lambda){ sin ( 0.5∗sqrt(lambda∗8−1)∗log (x/F0)) /lambda^(rho+1)/sqrt(lambda∗8−1)/sqrt(x∗F0)/2}; return(f2)} g3 <− function (x) { integrate(g1(x), lower=0.125, upper=Inf,stop.on.error = FALSE) $value} g4 <− function (x) { if (x>F0) {integrate(g2(x), lower=0.125, upper=1000000,stop.on. error = FALSE)$value} else {integrate(g2(x), lower=0.125, upper=100000,stop. on.error = FALSE)$value}} eps=1;psi2nd <− function(x){(g3(x+eps)+g4(x+eps)−g3(x)−g4(x))/eps} f <− function(lambda){ return (2∗(sum(Put_OTM$Last∗Put_OTM$Strike∗∗(pl( lambda)−2) ∗Put_OTM$dif)) +sum(Call_OTM$Last ∗Call_OTM$Strike∗∗(pl( lambda)−2) ∗Call_OTM$dif)/F0∗∗pl(lambda)/lambda∗∗rho)} (sum(Put_OTM$Last∗as.numeric(lapply(Put_OTM$Strike,psi2nd)) ∗Put_OTM$dif) +sum(Call_OTM$Last∗as.numeric(lapply(Call_OTM$Strike,psi2nd)) ∗Call_ OTM$dif)+integrate(Vectorize(f), lower=0, upper=0.125)$value)∗rho∗exp(r∗T)/ gamma(1−rho)

Chapter 10 Exercise 10.1 a) By differentiating (10.2) with respect to T , we find ∂ P(τa < T ) ∂T ∂ =2 P(WT > a) ∂T 2 ∂ w ∞ −x2 /(2T ) = √ e dx 2πT ∂T a w ∞ 2 ∂ −y 2 /2 = √ dy √ e 2π ∂T a/ T 2 a = √ e−a /(2T ) , T > 0. 2πT 3

φτa (T ) =

182

(S.10.47) 182

September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition b) By differentiating (10.13) with respect to T , we find ∂ P(τ̃a < T ) ∂T ∂ =− P(τ̃a ≥ T ) ∂T ∂ bT ≤ a =− P X ∂T 0 ∂ a − µT ∂ −a − µT √ √ =− Φ + e2µa Φ ∂T ∂T T T a µ −(a−µT )2 /(2T ) √ +√ = e 2πT 2 2πT 3 2 µ a √ e2µa−(a+µT ) /(2T ) −√ + 3 2πT 2 2πT 2 a = √ e−(a−µT ) /(2T ) , T > 0. 2 2πT 3

φτ̃a (T ) =

c) By differentiating (10.15) with respect to T , for x > S0 we find ∂ P(τ̂a < T ) ∂T ∂ =− P(τ̂x ≥ T ) ∂T ∂ =− P MT ≤ x ∂T 0 −(r − σ 2 /2)T + log(x/S0 ) ∂ √ =− Φ ∂T σ T 1−2r/σ2 ∂ −(r − σ 2 /2)T − log(x/S0 ) S0 √ Φ + x ∂T σ T log(x/S0 ) 1 2 = √ exp − 2 ((r − σ /2)T − log(x/S0 ))2 , 2σ T σ 2πT 3

φτ̂x (T ) =

T > 0,

which can also be recovered from (S.10.47) by taking a := log(S0 /x)/σ and µ := r/σ − σ/2. Similarly, when 0 < x < S0 we can differentiate (10.18) in Corollary 10.8 to find ∂ P(τ̂x < T ) ∂T ∂ = P mT ≤ x ∂T 0 ∂ −(r − σ 2 /2)T + log(x/S0 ) √ = Φ ∂T σ T

φτ̂x (T ) =

183 September 15, 2022

Solutions Manual 1−2r/σ2 S0 ∂ (r − σ 2 /2)T + log(x/S0 ) √ Φ x ∂T σ T log(S0 /x) 1 2 = √ exp − 2 ((r − σ /2)T − log(x/S0 ))2 , 2σ T σ 2πT 3 +

T > 0,

which yields φτ̂x (T ) =

| log(S0 /x)| 1 √ exp − 2 ((r − σ 2 /2)T − log(x/S0 ))2 , 3 2σ T σ 2πT

T > 0,

for all x > 0. Exercise 10.2 a) We use Relation (10.14) and the integration by parts identity w∞ w∞ v ′ (z)u(z)dz = u(+∞)v(+∞) − u(0)v(0) − v(z)u′ (z)dz 0

with u(y) = Φ

−y − µT /σ √ T

and v ′ (y) =

2µ 2µy/σ ye σ

which satisfy u′ (y) = − √

2 1 e−(y+µT /σ) /(2T ) 2πT

and v(y) = ye2µy/σ −

σ 2µy/σ e , 2µ

we have ft = σ IE max (Wt + µt/σ) IE max W t∈[0,T ] t∈[0,T ] r w∞ −y − µT /σ 2 w ∞ −(y−µT /σ)2 /(2T ) √ =σ ye dy − 2µ ye2µy/σ Φ dy 0 πT 0 T r w∞ 2 w ∞ −(y−µT /σ)2 /(2T ) =σ ye dy − σ v ′ (y)u(y)dy 0 πT 0 r w∞ 2 w ∞ −(y−µT /σ)2 /(2T ) =σ u′ (y)v(y)dy ye dy − σu(+∞)v(+∞) + σu(0)v(0) + σ 0 πT 0 ! r √ σ2 2 w ∞ −(y−µT /σ)2 /(2T ) µ T =σ ye dy − Φ − 0 πT 2µ σ w∞ 2 σ w ∞ 2µy/σ−(y+µT /σ)2 /(2T ) σ2 −√ ye dy + √ e2µy/σ−(y+µT /σ) /(2T ) dy 2πT 0 2µ 2πT 0 r 2 w ∞ −(y−µT /σ)2 /(2T ) σ w ∞ −(y−µT /σ)2 /(2T ) =σ ye dy − √ ye dy πT 0 2πT 0 184

184 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition √ ! w∞ 2 µ T σ2 σ2 Φ − + √ e−(y−µT /σ) /(2T ) dy 2µ σ 2µ 2πT 0 √ ! σ2 σ w ∞ −(y−µT /σ)2 /(2T ) µ T ye dy − = √ Φ − 2µ σ 2πT 0 w∞ 2 σ2 + √ e−(y−µT /σ) /(2T ) dy 2µ 2πT 0 √ ! 2 σ2 µ T µT σ w∞ e−y /(2T ) dy − Φ − y+ = √ σ 2µ σ 2πT −µT /σ w 2 ∞ 2 σ + √ e−y /(2T ) dy 2µ 2πT −µT /σ √ ! 2 2 µ T σ w∞ µT + σ 2 /(2µ) w ∞ σ2 √ Φ − = √ ye−y /(2T ) dy + e−y /(2T ) dy − −µT /σ 2µ σ 2πT −µT /σ 2πT ! √ √ ! √ ! i∞ 2 2 2 σ h T T T σ µ σ µ µ = √ −T e−y /(2T ) + µT Φ + Φ − Φ − σ 2µ σ 2µ σ −µT /σ 2πT ! ! r √ √ 2 2 σ µ T σ µ T T −µ2 T /2 e + µT + Φ − Φ − . =σ 2π 2µ σ 2µ σ −

As σ tends to zero, we find   µT Φ(+∞) = µT if µ ≥ 0, f IE max Wt =  t∈[0,T ] µT Φ(−∞) = 0 if µ ≤ 0.

We also have r ft = σ T e−µ2 T /2 + µT Φ IE max W 2π t∈[0,T ] r T −µ2 T /2 =σ e + µT Φ 2π

√ ! √ ! √ !! µ T σ2 µ T µ T + Φ −Φ − σ 2µ σ σ √ ! √ w 2 µ T /σ 2 µ T σ + √ e−y /2 dy. √ σ 2µ 2π −µ T /σ

Hence, as µ tends to zero we find r ft = σ T e−µ2 T /2 + µT Φ IE max W 2π t∈[0,T ]

r √ ! µ T T +σ + o(µ), σ 2π

[µ → 0],

and for µ = 0 and σ = 1 we recover the average maximum of standard Brownian motion r 2T , IE max Wt = π t∈[0,T ] "

185 September 15, 2022

Solutions Manual which represents two times the expected maximum 2 1 w∞ max(y, 0)e−y /(2T ) dy 2πT −∞ 1 w ∞ −y2 /(2T ) ye dy = √ 2πT 0 i h ∞ 2 1 = √ −T e−y /(2T ) 0 2πT r T = . 2π

IE[max(WT , 0)] = √

b) By part (a) the identity in distribution (−Wt )t∈R+ ≈ (Wt )t∈R+ , we have ft = σ IE min (Wt + µt/σ) IE min W t∈[0,T ] t∈[0,T ] = −σ IE max (−Wt − µt/σ) t∈[0,T ] = −σ IE max (Wt − µt/σ) t∈[0,T ] r √ ! √ ! T −µ2 T /2 σ2 −µ T σ2 µ T = −σ e + µT + Φ − Φ . 2π 2µ σ 2µ σ In particular, as σ tends to zero, we find   µT Φ(+∞) = µT if µ ≤ 0, f IE min Wt =  t∈[0,T ] µT Φ(−∞) = 0 if µ ≥ 0. Exercise 10.3 a) We have St = S0 eσWt , t ∈ R+ . b) We have 2 IE[ST ] = S0 IE eσWT = S0 eσ T /2 . c) We have P and

w∞ 2 dx , e−x /(2T ) √ max Wt ≥ a = 2 a t∈[0,T ] 2πT

a ≥ 0,

wa 2 dx max Wt ≤ a = 2 e−x /(2T ) √ , 0 t∈[0,T ] 2πT

a ≥ 0,

hence the probability density function φ of max Wt is given by t∈[0,T ]

186

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Introduction to Stochastic Finance with Market Examples, Second Edition r φ(a) =

2 −a2 /(2T ) e 1[0,∞) (a), πT

a ∈ R.

d) We have w∞ IE M0T = S0 IE exp σ max Wt = S0 eσx φ(x)dx 0

t∈[0,T ]

2S0 w ∞ −(x−σT )2 /(2T )+σ2 T /2 2S0 w ∞ σx−x2 /(2T ) e dx e dx = √ = √ 2πT 0 2πσ 2 T 0 w w ∞ ∞ 2 2S0 σ2 T /2 2S0 2 −x2 /2 = √ e e−x /(2T ) dx = √ eσ T /2 dx √ e −σT −σ T 2π 2πT √ w σ T 2 2 = 2S0 eσ T /2 e−x /2 dx −∞ √ 2 = 2S0 eσ T /2 Φ σ T √ = 2 IE[ST ]Φ σ T . Remarks: (i) From the inequality 0 ≤ IE[(WT − σT )+ ] 2 1 w∞ = √ (x − σT )+ e−x /(2T ) dx 2πT −∞ 2 1 w∞ (x − σT )e−x /(2T ) dx = −√ 2πT σT 1 w ∞ −x2 /(2T ) σT w ∞ −x2 /(2T ) = √ xe dx − √ e dx σT 2πT 2πT σT r σT w ∞ −x2 /2 T w∞ −x2 /2 dx − √ dx = √ xe √ e 2π σ T 2π σ T r √ T h −x2 /2 i∞ = e √ − σT Φ − σ T 2π σ T r √ T −σ2 T /2 = e − σT 1 − Φ σ T , 2π we get

2 √ e−σ T /2 Φ σ T ≥1− √ , σ 2πT

hence √ 2 IE M0T = 2S0 eσ T /2 Φ σ T 2

σ 2 T /2

≥ 2S0 e "

e−σ T /2 1− √ σ 2πT

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Solutions Manual ! 2 e−σ T /2 = 2 IE[ST ] 1 − √ σ 2πT 1 σ 2 T /2 = 2S0 e . − √ σ 2πT (ii) We observe that the ratio between the expected gains by at √ selling the maximum and selling at time T is given by 2Φ σ T , which cannot be greater than 2. 2

2 Φ(σT1/2)

ratio

1.5

0.5

1.5

2.5

3.5

time T

Fig. S.43: Average return by selling at the maximum vs selling at maturity. e) By a symmetry argument, we have P min Wt ≤ a = P − max (−Wt ) ≤ a t∈[0,T ] t∈[0,T ] = P − max Wt ≤ a t∈[0,T ] = P max Wt ≥ −a t∈[0,T ]

w∞

−a

2 dx e−x /(2T ) √ , 2πT

a < 0,

i.e. the probability density function φ of min Wt is given by t∈[0,T ]

r φ(a) =

2 −a2 /(2T ) e 1(−∞,0] (a), πT

a ∈ R.

f) We have IE mT0 = S0 IE exp σ min Wt = S0 188

−∞

t∈[0,T ]

e φ(x)dx σx

188 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2S0 w 0 σx−x2 /(2T ) e dx = √ 2πT −∞ w 0 2 2 2S0 = √ e−(x−σT ) /(2T )+σ T /2 dx 2πT −∞ 2S0 σ2 T /2 w −σT −x2 /(2T ) e = √ e dx −∞ 2πT √ w −σ T 2 2S0 2 = √ eσ T /2 e−x /2 dx −∞ 2π √ 2 = 2S0 eσ T /2 Φ − σ T √ = 2 IE[ST ]Φ − σ T . Remarks: (i) From the inequality 0 ≤ IE[(−σT − WT )+ ] 2 1 w∞ = √ (−σT − x)+ e−x /(2T ) dx 2πT −∞ 2 1 w −σT = −√ (σT + x)e−x /(2T ) dx 2πT −∞ σT w −σT −x2 /(2T ) 1 w −σT −x2 /(2T ) = −√ e dx − √ xe dx 2πT −∞ 2πT −∞ r √ √ T w −σ T −x2 /2 σT w −σ T −x2 /2 e dx − xe dx = −√ −∞ 2π −∞ 2π r √ √ T h −x2 /2 i−σ T = −σT Φ − σ T + e 2π −∞ r √ T −σ2 T /2 = −σT Φ − σ T + e , 2π we get √ 2 eσ T /2 Φ − σ T ≤

1 √ , σ 2πT

hence

IE mT0 ≤

2S √ 0 . σ 2πT

(ii) The ratio between the expected √ gains by maturity T vs selling at the minimum is given by 2Φ − σ T , which is at most 1 and tends to 0 as σ and T tend to infinity.

189 September 15, 2022

Solutions Manual 2

2 Φ(-σT1/2) 2 Φ(σT1/2)

ratio

1.5

0.5

1.5

2.5

3.5

time T

Fig. S.44: Average returns by selling at the minimum vs selling at maturity. √ (iii) Given that IE M0T = 2 IE[ST ]Φ σ T , we find the bound √ √ 2 IE[ST ]Φ − σ T ≤ IE[ST ] ≤ 2 IE[ST ]Φ σ T , with equality if σ = 0 or T = 0. We also have √ 2 2 IE[ST ] − IE M0T = 2eσ T /2 1 − Φ σ T √ 2 = 2eσ T /2 Φ − σ T T = IE m0 , hence we have IE mT0 +IE M0T = 2 IE[ST ], and

IE[ST ]−IE mT0 = IE M0T −IE[ST ],

2S0 ≤ IE M0T ≤ 2 IE[ST ]. 2 IE[ST ] − √ σ 2πT

Exercise 10.4 (Exercise 10.3 continued). a) Regarding call option prices we have, assuming K ≥ S0 , " + # + IE M0T − K = S0 IE exp σ max Wt − K t∈[0,T ]

w∞

(S0 eσx − K)+ φ(x)dx + 2 2 w∞ = √ S0 eσx − K e−x /(2T ) dx 0 2πT 2 2 w∞ = √ S0 eσx − K e−x /(2T ) dx 2πT σ−1 log(K/S0 ) 2 2S0 w ∞ = √ eσx−x /(2T ) dx 2πT σ−1 log(K/S0 ) 190

190 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2 2K w ∞ e−x /(2T ) dx −√ 2πT σ−1 log(K/S0 ) 2 2 2S0 w ∞ = √ e−(x−σT ) /(2T )+σ T /2 dx 2πT σ−1 log(K/S0 ) 2 2K w ∞ e−x /(2T ) dx −√ 2πT σ−1 log(K/S0 ) 2 2S0 σ2 T /2 w ∞ = √ e e−x /(2T ) dx −σT +σ −1 log(K/S0 ) 2πT 2 2K w ∞ −√ e−x /(2T ) dx 2πT σ−1 log(K/S0 ) √ √ 2 = 2S0 eσ T /2 Φ σ T + σ −1 log(S0 /K)/ T √ −2KΦ σ −1 log(S0 /K)/ T .

When K ≤ S0 , by “completion of the square” and use of the Gaussian cumulative distribution function Φ(·), we find IE

max St − K

t∈[0,T ]

+ i

i max St − K h i = IE max St − IE[K] t∈[0,T ] h i = IE max St − K t∈[0,T ] = S0 IE exp σ max Wt −K t∈[0,T ] w∞ = S0 eσx φ(x)dx − K 0 2S0 w ∞ σx−x2 /(2T ) e dx = √ 2πT 0 w ∞ 2 2 2S0 = √ e−(x−σT ) /(2T )+σ T /2 dx − K 2πT 0 2S0 σ2 T /2 w ∞ −x2 /(2T ) = √ e e dx − K −σT 2πT √ 2 = 2S0 eσ T /2 Φ σ T − K √ 2 = 2S0 eσ T /2 1 − Φ − σ T − K √ = 2 IE[ST ]Φ σ T − K, = IE

t∈[0,T ]

hence 2

e−σ T /2 IE

M0T − K

√ 2 = 2S0 Φ σ T − Ke−σ T /2 .

191 September 15, 2022

Solutions Manual Recall that when r = σ 2 /2 the price of the finite expiration American call option price is the Black-Scholes price with maturity T , with BlCall (S0 , K, σ, r, T ) √ √ √ 2 = S0 Φ σ T + σ −1 log(S0 /K)/ T − Ke−σ T /2 Φ σ −1 log(S0 /K)/ T

≤

√ √ √  −1 −σ 2 T /2 Φ σ −1 log(S0 /K)/ T   2S0 Φ σ T + σ log(S0 /K)/ T − 2Ke   if K ≥ S0 ,  √  2   2S Φ σ T − Ke−σ T /2   0 if K ≤ S0 .   2 × BlCall (S0 , K, σ, r, T ) if K ≥ S0 ,

√ 2 2S0 Φ σ T − Ke−σ T /2 if K ≤ S0 , √ 2 = max 2 × BlCall (S0 , K, σ, r, T ), 2S0 Φ σ T − Ke−σ T /2 .

2.0



1.0 0.0

0.5

price

1.5

Upper bound Black−Scholes call price

0.0

0.5

1.0

1.5

2.0

Fig. S.45: Black-Scholes call price upper bound with S0 = 1. b) Regarding put option prices we have, assuming S0 ≥ K, " + # + IE K − mT0 = S0 IE K − exp σ min Wt t∈[0,T ]

w∞ 0

(K − S0 e ) φ(x)dx σx +

+ 2 2 w0 = √ K − S0 eσx e−x /(2T ) dx 2πT −∞ 2 2 w σ−1 log(K/S0 ) = √ K − S0 eσx e−x /(2T ) dx 2πT −∞ 192

192 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 2K w σ−1 log(K/S0 ) −x2 /(2T ) = √ e dx 2πT −∞ −1 w σ log(K/S ) 0 2 2S0 −√ eσx−x /(2T ) dx 2πT −∞ 2K w σ−1 log(K/S0 ) −x2 /(2T ) e = √ dx 2πT −∞ 2S0 w σ−1 log(K/S0 ) −(x−σT )2 /(2T )+σ2 T /2 e dx −√ 2πT −∞ −1 w σ log(K/S0 ) 2 2K = √ e−x /(2T ) dx 2πT −∞ 2S0 σ2 T /2 w −σT +σ−1 log(K/S0 ) −x2 /(2T ) −√ e e dx −∞ 2πT √ −1 = 2KΦ(−σ log S0 /K)/ T √ √ 2 −2S0 eσ T /2 Φ − σ T − σ −1 log(S0 /K)/ T , with

e−σ T /2 IE

K − mT0

√ 2 = Ke−σ T /2 − 2S0 Φ − σ T

if S0 ≤ K. Therefore we deduce the bounds BlPut (S0 , K, σ, r, T )

√ √ √ 2 = Ke−σ T /2 Φ − σ −1 log(S0 /K)/ T − S0 Φ − σ T − σ −1 log(S0 /K)/ T ≤ American put option price √ √ √  2 2Ke−σ T /2 Φ − σ −1 log(S0 /K)/ T − 2S0 Φ − σ T − σ −1 log(S0 /K)/ T    if S0 ≥ K, ≤   √  2 Ke−σ T /2 − 2S0 Φ − σ T if S0 ≤ K,  if S0 ≥ K,  2 × BlPut (S0 , K, σ, r, T ) = √ 2  Ke−σ T /2 − 2S0 Φ − σ T if S0 ≤ K, √ 2 = max 2 × BlPut (S0 , K, σ, r, T ), Ke−σ T /2 − 2S0 Φ − σ T

for the finite expiration American put option price when r = σ 2 /2.

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0.0

0.2

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price

0.6

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Upper bound Black−Scholes put price

0.0

0.5

1.0

1.5

2.0

Fig. S.46: Black-Scholes put price upper bound with S0 = 1.

Exercise 10.5 (Exercise 10.4 continued). a) Using the expression r b x + µT 2 −(x−µT )2 /(2T ) √ φ T (x) = e + 2µe2µx Φ , X 0 πT T

x ≤ 0.

of the probability density function of the minimum b T ft = min (Wt + µt) X 0 := min W t∈[0,T ]

t∈[0,T ]

ft = Wt + µt over t ∈ [0, T ] given in Propoof drifted Brownian motion W sition 10.7, we find h i w0 b IE min St = S0 eσx φ T (x)dx X 0 −∞ t∈[0,T ] r w0 2 −(x−µT )2 /(2T ) = S0 eσx e dx −∞ πT w0 x + µT √ +2µS0 eσx e2µx Φ dx −∞ T √ √ 2 2µS0 = 2S0 eσ T /2−µσT Φ (µ − σ) T + Φ −µ T 2µ − σ √ 2µS0 σ2 T /2−µσT − e Φ (µ − σ) T , 2µ − σ with µ := r/σ − σ/2, which yields

194

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Introduction to Stochastic Finance with Market Examples, Second Edition i σ2 r − σ 2 /2 √ T min St = S0 1 − Φ 2r σ t∈[0,T ] σ2 r + σ 2 /2 √ +S0 1 + T . Φ − 2r σ h i See Exercise 12.1-(b) for the computation of IE mint∈[0,1] St when r = 0. IE

b) When S0 ≤ K, we have IE

h + i h i K − min St = IE K − min St t∈[0,T ] t∈[0,T ] h i = K − IE min St t∈[0,T ] σ2 r − σ 2 /2 √ = K − S0 1 − T Φ 2r σ 2 σ r + σ 2 /2 √ −S0 erT 1 + Φ − T . 2r σ

Next, when S0 ≥ K we have, using the probability density function b φ T (x), X 0

b h + i h T + i K − min St = IE K − S0 min eσX 0 =

t∈[0,T ]

−∞

t∈[0,T ]

σx +

K − S0 e

X 0

(x)dx

+ 2 2 w0 = S0 K − S0 eσx e−(x−µT ) /(2T ) dx πT −∞ w0 + x + µT √ +2µS0 K − S0 eσx e2µx Φ dx −∞ T (r − σ 2 /2)T + log(S0 /K) √ = KΦ − σ T 1−2r/σ2 S0 (r − σ 2 /2)T + log(K/S0 ) √ +K Φ K σ T −2r/σ2 σ2 S0 (r − σ 2 /2)T + log(K/S0 ) √ −S0 1 − Φ 2r K σ T σ2 (r + σ 2 /2)T + log(S0 /K) rT √ −S0 e 1+ Φ − . 2r σ T r

In Figure S.47, using a finite expiration American put option pricer from the R fOptions package, we plot the graph of American put option price vs (S.10.48)-(S.10.49), together with the European put option price, according to the following R code. "

195 September 15, 2022

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Price

0.6

0.8

Upper bound American put price Black−Scholes put price

0.0 0.0

0.5 0.5

1.0 1.0

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2.0

Fig. S.47: “Optimal exercise” put price upper bound with S0 = 1. d1 <− function(S,K,r,T,sig) {return((log(S/K)+(r+sig^2/2)∗T)/(sig∗sqrt(T)))} d2 <− function(S,K,r,T,sig) {return(d1(S, K, r, T, sig) − sig ∗ sqrt(T))} BSPut <− function(S, K, r, T, sig){return(K∗exp(−r∗T) ∗ pnorm(−d2(S, K, r, T, sig)) − S∗pnorm(−d1(S, K, r, T, sig)))} Optimal_Put_Option <− function(S,K,r,T,sig){ if (r==0) {if (S>=K) {return(K∗pnorm(d1(K,S,0,T,sig))−S∗(1+sig∗sig∗T/2 +log(S/K))∗ pnorm(−d1(S,K,0,T,sig)) +S∗sig∗sqrt(T/(2∗pi))∗exp(−d1(S,K,0,T,sig)∗d1(S,K,0,T, sig)/(2∗sig∗sig∗T)))} else {return(K−2∗S∗(1+sig∗sig∗T/4)∗pnorm(−sig∗sqrt(T)/2) +S∗sig∗sqrt(T/(2∗pi))∗exp (−sig∗sig∗T/8))}} else {if (S>=K) {return(K∗pnorm(−d2(S,K,r,T,sig)) +K∗(S/K)∗∗(1−2∗r/sig/sig)∗pnorm( d2(K,S,r,T,sig)) −2∗S∗exp(r∗T)∗pnorm(−d1(S,K,r,T,sig)) −S∗(1−sig∗sig/2/r)∗(S/K )∗∗(−2∗r/sig/sig)∗pnorm(d1(K,S,r,T,sig)) +S∗exp(r∗T)∗(1−sig∗sig/2/r)∗pnorm(−d1 (S,K,r,T,sig)))} else {return(K−S∗(1−sig∗sig/2/r)∗pnorm((r−sig∗sig/2)∗sqrt(T)/sig) −S∗(1+sig∗sig/2/r)∗ pnorm(−(r+sig∗sig/2)∗sqrt(T)/sig))}}} r=0.5;sig=1;S=1;T=1 library(fOptions) curve(BAWAmericanApproxOption("p",S,x,T,r,b=0,sig,title = NULL, description = NULL )@price, from=0.01, to=2 , xlab="K", lwd = 3, ylim=c(0,1),ylab="",col="orange") par(new=TRUE) curve(BSPut(S,x,r,T,sig), from=0, to=2 , xlab="K", lwd = 3, ylim=c(0,1), ylab="Price", col="blue") par(new=TRUE) curve(Optimal_Put_Option(S,x,r,T,sig), from=0, to=2 , xlab="K", lwd = 3, ylim=c(0,1), ylab="",col="red") grid (lty = 5) legend(.0,1.0,legend=c("Upper bound","American put price","Black−Scholes put price"),col =c("red","orange", "blue"), lty=1:1, cex=1.)

c) When r = 0 and S0 ≤ K, we find r √ ! h + i σ2 T σ T T −σ2 T /8 K− min St = K−2S0 1 + Φ − +σS0 e . 4 2 2π t∈[0,T ] (S.10.48) Next, when r = 0 and S0 ≥ K, we find 2 h + i σ T /2 + log(K/S0 ) √ IE K − min St = KΦ (S.10.49) t∈[0,T ] σ T

196

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Introduction to Stochastic Finance with Market Examples, Second Edition S0 σ2 T σ 2 T /2 + log(S0 /K) √ −S0 1 + log + Φ − K 2 σ T r T −(σ2 T /2+log(S0 /K))2 /(2σ2 T ) e . +S0 σ 2π From the above R code we can check that when r ≃ 0 the price of the finite expiration American put option coincides with the price of the standard European put option, as noted in Proposition 15.9. Exercise 10.6 a) We have P (τa ≥ t) = P (Xt > a) w∞ = φXt (x)dx ra w 2 ∞ −x2 /(2t) = e dx, πt y

y > 0.

b) We have d P (τa ≤ t) dt d w∞ = φXt (x)dx a dt r r 1 2 −3/2 w ∞ x2 −x2 /(2t) 1 2 −3/2 w ∞ −x2 /(2t) t e dx + t e dx =− a a 2 π 2 π t r w w ∞ ∞ 2 2 2 1 2 −3/2 = t − e−x /(2t) dx + ae−a /(2t) + e−x /(2t) dx a a 2 π 2 a e−a /(2t) , t > 0. = √ 2πt3

φτa (t) =

c) We have w ∞ −2 IE (τa )−2 = t φτa (t)dt 0 a w ∞ −7/2 −a2 /(2t) t e dt = √ 2π 0 w ∞ 2 2 2a = √ x4 e−a x /2 dx 2π 0 3 = 4, a by the change of variable x = t−1/2 , i.e. x2 = 1/t, t = x−2 , and dt = −2x−3 dx. "

197 September 15, 2022

Solutions Manual Remark: We have a w ∞ −1/2 −a2 /(2t) IE[τa ] = √ t e dt = +∞. 2π 0 Exercise 10.7 Starting from the probability density function r 2 2 1 2 φX (a, b) = 1 (2a − b)eµb−(2a−b) /(2T )−µ T /2 {a≥max(b,0)} T b0 ,W eT T πT fT := WT + µT and its maximum X bT = of the drifted Brownian motion W 0 ft , we take µ := r/σ − σ/2 and let the functions f and g be defined max W t∈[0,T ] as 1 x y 1 f (x, y) := log and g(x, y) := log , σ S0 σ S0 with the Jacobian ∂f ∂x |J(x, y)| = ∂g ∂x

∂f 1 ∂y σx = ∂g 0 ∂y

0 1 1 = σ 2 xy , σy

x, y > 0, which yields the joint density φM0T ,ST (x, y) = |J(x, y)|φX b0T ,W e T (f (x, y), g(x, y)) ! r 2 2 x2 µ y 1 x2 T 1 1{x≥max(y,S0 )} log exp log − 2 log − µ2 = 3 σ T xy πT S0 y σ S0 2σ T S0 y 2 of ST and its maximum M0T = maxt∈[0,T ] St over t ∈ [0, T ], x, y > 0.

Chapter 11 Exercise 11.1 Barrier options. a) By (11.26) and (12.15) we find ∂g St St T −t T −t ξt = (t, St ) = Φ δ+ − Φ δ+ ∂y K B −2r/σ2 2 K 2r St B B T −t T −t + e−(T −t)r 1 − 2 Φ δ− − Φ δ− B σ B KSt St 198

198 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition −1−2r/σ2 2 B B T −t T −t Φ δ+ − Φ δ+ KSt St 2 ! 1 2 K St T −t exp − δ+ − p 1− , B 2 B σ 2π(T − t) +

2r σ2

St B

0 < St ≤ B, 0 ≤ t ≤ T , cf. also Exercise 7.1-(ix) of Shreve (2004) and Figure 11.11 above. At maturity for t = T we find ξT = 1[K,B] (ST ). b) We find P(YT ≤ a and WT ≥ b) = P(WT ≤ 2a − b),

a < b < 0,

hence fYT ,WT (a, b) =

dP(YT ≤ a and WT ≥ b) dP(YT ≤ a and WT ≤ b) =− , dadb dadb

a, b ∈ R, satisfies r fYT ,WT (a, b) =

2 (b − 2a) −(2a−b)2 /(2T ) 1(−∞,min(0,b)] (a) e πT T

r  2 (b − 2a) −(2a−b)2 /(2T )   e , πT T =    0,

a < min(0, b), a > min(0, b).

c) We find f

b Y

b (a, b) = 1(−∞,min(0,b)] (a)

T ,WT

1 T

2 2 2 (b − 2a)e−µ T /2+µb−(2a−b) /(2T ) πT

 r  2 2 2 1  (b − 2a)e−µ T /2+µb−(2a−b) /(2T ) , T πT =    0,

a < min(0, b), a > min(0, b).

d) The function g(t, x) is given by the Relations (11.10) and (11.11) above. Exercise 11.2 a) By Corollary 10.8, the probability density function of the minimum "

199 September 15, 2022

Solutions Manual m∆τ 0 =

min Sτ +s

s∈[0,∆τ ]

with Sτ = B is given by (−(r − σ 2 /2)∆τ + log(x/B))2 1 √ exp − 2 2σ ∆τ σx 2π∆τ 1−2r/σ2 B 1 ((r − σ 2 /2)∆τ + log(x/B))2 + √ exp − 2 2σ ∆τ σx 2π∆τ x 1−2r/σ2 B 1 2r (r − σ 2 /2)∆τ + log(x/B) √ , + −1 Φ 2 x σ x σ ∆τ

φm∆τ (x) = 0

0 < x ≤ B, see also Proposition 10.7 for the probability density function of ft = Wt +µt over t ∈ [0, T ]. the minimum of the drifted Brownian motion W Hence, we have + i wB min Sτ +s − K Fτ = (x − K)+ φm∆τ (x)dx 0 0 2 σ r σ √ log(B/K) + (r − σ 2 /2)∆τ √ = B 1+ er∆τ Φ − + ∆τ + KΦ − 2r σ 2 σ ∆τ σ2 log(B/K) + (r + σ 2 /2)∆τ r∆τ √ −B 1 + e Φ − 2r σ ∆τ 2 σ r σ √ +B 1 − Φ − ∆τ 2r σ 2 2r/σ2 σ2 K log(B/K) − (r − σ 2 /2)∆τ √ +B Φ − − K, 2r B σ ∆τ

s∈[0,∆τ ]

with r > 0. b) When r = 0, we find i σ√ σ2 ∆τ Φ − ∆τ Fτ = B 2 + 2 2 s∈[0,∆τ ] ! ! 2 2 log(B/K) + σ2 ∆τ log(B/K) − σ2 ∆τ σ2 B √ √ + KΦ − . −B 1 + ∆τ + log Φ − 2 K σ ∆τ σ ∆τ √ 2 2 1 − √ σB ∆τ e−σ ∆τ /8 − e−d+ /2 − K. 2π

min Sτ +s − K

c) By the solution of Exercise 10.1-(c), the probability density function of τB is given by | log(S0 /B)| 1 √ φτB (x) = exp − 2 ((r − σ 2 /2)x − log(B/S0 ))2 , x > 0. 2σ x σ 2πx3 200

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Introduction to Stochastic Finance with Market Examples, Second Edition d) We have h e−∆τ IE e−τ 1[0,T ] (τ )

min

t∈[τ,τ +∆τ ]

h h = e−∆τ IE e−rτ 1[0,T ] (τ ) IE

where IE and

mint∈[τ,τ +∆τ ] St − K

+ i +

min

St − K

min

St − K

t∈[τ,τ +∆τ ]

h = e−∆τ IE e−rτ 1[0,T ] (τ ) IE h

St − K

t∈[τ,τ +∆τ ]

Fτ

i Fτ ,

i Fτ is given by Questions (a)-(b),

w T −rx IE e−rτ 1[0,T ] (τ ) = e φτB (x)dx 0 w S0 T −rxτ 1 1 √ = log e exp − 2 ((r − σ 2 /2)x − log(B/S0 ))2 dx 3 0 B 2σ x σ 2πx p ! log(B/S0 ) r − σ 2 /2 − (r − σ 2 /2)2 + 2rσ 2 S0 = log exp B σ2 wT p 2 1 1 √ × exp − 2 x (r − σ 2 /2)2 + 2rσ 2 − log(B/S0 ) dx 3 0 σ 2πx 2σ x ! p log(B/S0 ) r − σ 2 /2 − (r − σ 2 /2)2 + 2rσ 2 = exp σ2 wT p 2 1 1 √ exp − 2 x (r − σ 2 /2)2 + 2rσ 2 − log(B/S0 ) × dx 3 0 σ 2πx 2σ x √ ! p r−σ2 /2− (r−σ2 /2)2 +2rσ2 /σ2 log(B/S0 ) − T (r − σ 2 /2)2 + 2rσ 2 B √ = Φ S0 σ T ! p r−σ2 /2+√(r−σ2 /2)2 +2rσ2 /σ2 log(B/S0 ) + T (r − σ 2 /2)2 + 2rσ 2 B √ + Φ , S0 σ T where the last identity follows from Proposition 10.4 and the relation a − µT −a − µT b0T ≤ a √ √ Φ − e2µa Φ =P X T T = P(τ̃a ≤ T ) wT = φτ̃a (x)dx 0 w 2 a 1 T √ = e−(a−µx) /(2x) dx, T > 0. 2 0 2πx3

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Solutions Manual with a := log(B/S0 )/σ, for a Brownian motion with drift µ = see Exercise 10.1-(b).

p (r − σ 2 /2)2 + 2rσ 2 /σ,

Exercise 11.3 Barrier forward contracts. a) Up-and-in barrier long forward contract. We have 



    −(T −t)r −(T −t)r   e IE[C | Ft ] = e IE  (ST − K) 1       max Su > B     0≤u≤T = 1  

 (St − Ke−(T −t)r ) + 1    

max S > B     0≤u≤t u

    Ft     

 ϕ(t, St ),  

max S ≤ B     0≤u≤t u

(S.11.50) where the function T −t T −t ϕ(t, x) := xΦ δ+ (x/B) − Ke−(T −t)r Φ δ− (x/B) 2 T −t +B(B/x)2r/σ Φ −δ+ (B/x) 2 T −t −Ke−(T −t)r (B/x)−1+2r/σ Φ −δ− (B/x) solves the Black-Scholes PDE with the terminal condition 2r/σ2 ! B K 1[B,∞) (x), B−x ϕ(T, x) = x − K + x B as in the proof of Proposition 11.3. Note that only the values of ϕ(t, x) with x ∈ [0, B] are used for pricing.

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20 15 10 5 0 75 70 Underlying

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Time in days

Fig. S.48: Price of the up-and-in long forward contract with K = 60 < B = 80. As for the hedging strategy, we find T −t 2 ∂ϕ 1 T −t (t, St ) = Φ δ+ (x/B) + √ e−(δ+ (x/B)) /2 ∂x 2π T −t 2 2 1 2r T −t − √ Ke−(T −t)r−(δ− (x/B)) /2 − 2 (B/x)1+2r/σ Φ −δ+ (B/x) σ x 2π T −t 2 2 1 + √ (B/x)1+2r/σ e−(δ+ (B/x)) /2 2π 2 K(1 − 2r/σ 2 ) −(T −t)r T −t − e (B/x)2r/σ Φ −δ− (B/x) B T −t 2 2 K − √ (B/x)2r/σ e−(T −t)r−(δ− (B/x)) /2 B 2π 2r 2 T −t T −t = Φ δ+ (x/B) − 2 (B/x)1+2r/σ Φ −δ+ (B/x) σ T −t T −t 2 2 1 B + √ (1 − K/B) e−(δ+ (x/B)) /2 + e−(T −t)r−(δ− (x/B)) /2 x 2π K T −t 2 −(T −t)r 2r/σ 2 − (1 − 2r/σ )e (B/x) Φ −δ− (B/x) , B

ξt =

since by (12.22) we have T −t

e−(δ− and

T −t

e−(δ−

T −t

(B/x))2 /2

= er(T −t) (x/B)2r/σ e−(δ+

(x/B))2 /2

= er(T −t) (B/x)2r/σ e−(δ+

(x/B))2 /2

(B/x))2 /2

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0.3 0.25 0.2 0.15 0.1 0.05 0 75 70 Underlying

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Fig. S.49: Delta of the up-and-in long forward contract with K = 60 < B = 80. b) Up-and-out barrier long forward contract. We have 



      e−(T −t)r IE[C | Ft ] = e−(T −t)r IE  (ST − K) 1       max S < B u    0≤u≤T  = 1  

 ϕ(t, St ),  

    Ft     

(S.11.51)

max S ≤ B     0≤u≤t u

where the function T −t T −t ϕ(t, x) := xΦ −δ+ (x/B) − Ke−(T −t)r Φ −δ− (x/B) 2 T −t −B(B/x)2r/σ Φ −δ+ (B/x) T −t −(T −t)r −1+2r/σ 2 +Ke (B/x) Φ −δ− (B/x) solves the Black-Scholes PDE with the terminal condition ϕ(T, x) = (x − K)1[0,B] (x) −

B x

2r/σ2 B−x

K B

1[B,∞) (x).

Note that only the values of ϕ(t, x) with x ∈ [B, ∞) are used for pricing.

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150 100

75 70 Underlying

Fig. S.50: Price of the up-and-out long forward contract with K = 60 < B = 80. As for the hedging strategy, we find T −t 2 ∂ϕ 1 T −t (t, St ) = Φ −δ+ (x/B) − √ e−(δ+ (x/B)) /2 ∂x 2π T −t 2 2 1 2r T −t + √ Ke−(T −t)r−(δ− (x/B)) /2 + 2 (B/x)1+2r/σ Φ −δ+ (B/x) σ x 2π T −t 2 2 1 − √ (B/x)1+2r/σ e−(δ+ (B/x)) /2 2π 2 K(1 − 2r/σ 2 ) −(T −t)r T −t + e (B/x)2r/σ Φ −δ− (B/x) B T −t 2 2 K + √ (B/x)2r/σ e−(T −t)r−(δ− (B/x)) /2 B 2π 2r 2 T −t T −t (B/x) = Φ −δ+ (x/B) + 2 (B/x)1+2r/σ Φ −δ+ σ T −t T −t 2 1 1 B −(T −t)r−(δ− (x/B))2 /2 − √ e−(δ+ (x/B)) /2 − √ e 2π 2π x T −t T −t 2 K 1 K −(T −t)r−(δ− (x/B))2 /2 e + √ e−(δ+ (x/B)) /2 + √ B 2π 2π x 2 K T −t + (1 − 2r/σ 2 )e−(T −t)r (B/x)2r/σ Φ −δ− (B/x) B 2r 2 T −t T −t = Φ −δ+ (x/B) + 2 (B/x)1+2r/σ Φ −δ+ (B/x) σ T −t T −t 2 2 1 B − √ (1 − K/B) e−(δ+ (x/B)) /2 + e−(T −t)r−(δ− (x/B)) /2 x 2π 2r/σ2 K 2r B B T −t + 1 − 2 e−(T −t)r Φ −δ− , B σ x x

ξt =

by (12.22).

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Underlying

80 100

200 Time in days

150

Fig. S.51: Delta of the up-and-out long forward contract price with K = 60 < B = 80. c) Down-and-in barrier long forward contract. We have 

     Ft     

      e−(T −t)r IE[C | Ft ] = e−(T −t)r IE  (ST − K) 1       min S < B u    0≤u≤T  = 1  

 (St − Ke−(T −t)r ) + 1    

min S < B     0≤u≤t u

 ϕ(t, St )  

min S ≥ B    0≤u≤t u 

(S.11.52) where the function T −t T −t ϕ(t, x) := xΦ −δ+ (x/B) − Ke−(T −t)r Φ −δ− (x/B) 2 T −t +B(B/x)2r/σ Φ δ+ (B/x) 2 T −t −Ke−(T −t)r (B/x)−1+2r/σ Φ δ− (B/x) solves the Black-Scholes PDE with the terminal condition 2r/σ2 ! B K ϕ(T, x) = x − K + B−x 1[0,B] (x). x B

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18 16 14 12 10 8 6 4 2 0 120

140

160 180 Time in days

200

220

100

80 85 Underlying

Fig. S.52: Price of the down-and-in long forward contract with K = 60 < B = 80. As for the hedging strategy, we find ξt =

∂ϕ (t, St ) ∂x

2r 2 T −t T −t (x/B) + 2 (B/x)1+2r/σ Φ δ+ = Φ −δ+ (B/x) σ T −t T −t 2 2 1 B − √ (1 − K/B) e−(δ+ (x/B)) /2 + e−(T −t)r−(δ− (x/B)) /2 x 2π 2 K T −t (B/x) . + (1 − 2r/σ 2 )e−(T −t)r (B/x)2r/σ Φ δ− B

0.6 0.5 0.4 0.3 0.2 0.1 0 120

140

160 180 Time in days

200

220

100

80 85 Underlying

Fig. S.53: Delta of the down-and-in long forward contract with K = 60 < B = 80. d) Down-and-out barrier long forward contract. We have 



    −(T −t)r −(T −t)r   e IE[C | Ft ] = e IE  (ST − K) 1       min Su > B    0≤u≤T 

    Ft     

207 September 15, 2022

Solutions Manual = 1  

 ϕ(t, St )  

(S.11.53)

min S ≥ B    0≤u≤t u 

where the function x x T −t T −t − Ke−(T −t)r Φ δ− ϕ(t, x) := xΦ δ+ B B 2 B T −t −B(B/x)2r/σ Φ δ+ x ! 2 B T −t +Ke−(T −t)r (B/x)−1+2r/σ Φ δ− x solves the Black-Scholes PDE with the terminal condition 2r/σ2 K B ϕ(T, x) = (x − K)1[B,∞) (x) − B − x 1[0,B] (x). B x Note that ϕ(t, x) above coincides with the price of (11.11) of the standard down-and-out barrier call option in the case K < B, cf. Exercise 11.1-(d).

40 35 30 25 20 15 10 5 0

220

200

180 160 Time in days

140

120

100 80

100 95 Underlying

Fig. S.54: Price of the down-and-out long forward contract with K = 60 < B = 80. As for the hedging strategy, we find ∂ϕ (t, St ) ∂x x 2r 2 T −t T −t = Φ δ+ − 2 (B/x)1+2r/σ Φ δ+ (B/x) B σ T −t T −t 2 2 1 K B +√ 1− e−(δ+ (x/B)) /2 + e−(T −t)r−(δ− (x/B)) /2 B x 2π 2r/σ2 K B 2r B T −t − 1 − 2 e−(T −t)r Φ δ− . B σ x x

ξt =

208

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220

200

180

160 Time in days

140

120

100 80

100 95 Underlying

Fig. S.55: Delta of the down-and-out long forward contract with K = 60 < B = 80. e) Up-and-in barrier short forward contract. The price of the up-and-in barrier short forward contract is identical to (S.11.50) with a negative sign. f) Up-and-out barrier short forward contract. The price of the up-and-out barrier short forward contract is identical to (S.11.51) with a negative sign. Note that ϕ(t, x) coincides with the price of (11.8) of the standard up-and-out barrier put option in the case B < K. g) Down-and-in barrier short forward contract. The price of the down-and-in barrier short forward contract is identical to (S.11.52) with a negative sign. h) Down-and-out barrier short forward contract. The price of the down-andout barrier short forward contract is identical to (S.11.53) with a negative sign. Exercise 11.4 When B < K, we find Vegadown-and-out-call r T − t −(δ+T −t (St /K))2 /2 = St e 2π 1−2r/σ2 2 2 2 4r St B B B St T −t T −t − 3 Φ δ+ − Ke−(T −t)r Φ δ− log σ B St KSt KSt B r 1−2r/σ2 2 T −t 2 2 T −tB St − e−(δ+ (B /K/St )) /2 . 2π St B When B > K, we find Vegadown-and-out-call T −t 2 St = √ e−(δ+ (St /K)) /2 2π

K −1 B

! ! T −t √ δ− (St /B) √ + T −t + T −t σ 209 September 15, 2022

Solutions Manual 4r σ3

1−2r/σ2

B2 St B B T −t T −t Φ δ+ − Ke−(T −t)r Φ δ− log St St St B ! ! T −t √ δ− (B/St ) √ 1 B 2 −(δ+T −t (St /B))2 /2 K e −1 + T −t + T −t . −√ B σ 2π St −

St B

The corresponding formulas for the down-and-in qcall option can be obtained from the parity relation (11.2) and the value St Black-Scholes Vega, see Table 6.1.

T −t T −t −(δ+ (St /K))2 /2 of the 2π e

Exercise 11.5 We have h i IE∗ [C] = IE∗ 1{ST ≥K} 1{M0T ≤B} = IE∗ 1 1 T σ X b σ W b T ≥K} {S e {S0 e 0 ≤B} 0 w∞ w∞ cT ≤ y) σy = 1{S0 e ≥K} 1{S0 eσx ≤B} dP(Xb0T ≤ x, W −∞ y∨0 w∞ w∞ = 1{S0 eσy ≥K} 1{S0 eσx ≤B} fXbT ,Wb T (x, y)dxdy −∞ y∨0 r w 2 σ−1 log(B/S0 ) 1 = T πT −∞ w∞ 2 2 1{S0 eσy ≥K} 1{S0 eσx ≤B} (2x − y)e−µ T /2+µy−(2x−y) /(2T ) dxdy y∨0 r 2 2 1 2 w σ−1 log(B/S0 ) w σ−1 log(B/S0 ) = (2x − y)e−µ T /2+µy−(2x−y) /(2T ) dxdy, T πT σ−1 log(K/S0 ) y∨0 if B ≥ S0 (otherwise the option price is 0), with µ = r/σ − σ/2 and y ∨ 0 = max(y, 0). Next, letting a = y ∨ 0 and b = σ −1 log(B/S0 ), we have wb a

(2x − y)e2x(y−x)/T dx =

T (1 − e2b(y−b)/T ), 2

hence, letting c = σ −1 log(K/S0 ), we have 2

1 w b µy−y2 /(2T ) e (1 − e2b(y−b)/T )dy 2πT c wb 2 eµy−y /(2T ) dy

IE∗ [C] = e−µ T /2 √ 2

= e−µ T /2 √

1 2πT

−µ2 T /2−2b2 /T

−e

√

1 w b y(µ+2b/T )−y2 /(2T ) e dy. 2πT c

Using the relation √ 210

2 1 w b γy−y2 /(2T ) −c + γT −b + γT √ √ e dy = eγ T /2 Φ −Φ , 2πT c T T 210 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition we find h i + IE∗ [C] = IE∗ (ST − K) 1{M0T ≤B} −b + µT −c + µT √ √ −Φ =Φ T T −c + (µ + 2b/T )T −b + (µ + 2b/T )T −µ2 T /2−2b2 /T +(µ+2b/T )2 T /2 √ √ −e Φ −Φ T T S0 S0 T T = Φ δ− − Φ δ− K B 2 B B −µ2 T /2−2b2 T +(µ+2b/T )2 T /2 −e Φ δ− − Φ δ− , KS0 S0 0 ≤ x ≤ B. Given the relation −

µ2 T b2 T −2 + 2 T 2

2 2r 2b B µ+ = −1 + 2 log , T σ S0

we get h i e−rT IE∗ [C] = e−rT IE∗ 1{ST ≥K} 1{M0T ≤B} S0 S0 T T − Φ δ− = e−rT Φ δ− K B 1−2r/σ2 2 ! S0 B B T T − Φ δ− − Φ δ− . B KS0 S0

Exercise 11.6 a) For x = B and t ∈ [0, T ] we check that B T −t T −t − Φ δ+ g(t, B) = B Φ δ+ (1) K B T −t T −t −(T −t)r −e K Φ δ− − Φ δ− (1) K B T −t T −t −B Φ δ+ − Φ δ+ (1) K B T −t T −t −(T −t)r +e − Φ δ− (1) K Φ δ− K = 0, and the function g(t, x) is extended to x > B by letting "

211 September 15, 2022

Solutions Manual g(t, x) = 0,

x > B.

b) For x = K and t = T , we find    +∞ if s > 1,    0 if s = 1, δ± (s) = −∞ × 1{s<1} + ∞ × 1{s>1} = 0      −∞ if s < 1, hence when x < K < B we have g(T, x) = x (Φ (−∞) − Φ (−∞)) −K (Φ (−∞) − Φ (−∞)) 2r/σ2 B −B (Φ (+∞) − Φ (+∞)) x 2r/σ2 B +K (Φ (+∞) − Φ (+∞)) K = 0, c) when K < x < B, we get g(T, x) = x (Φ (+∞) − Φ (−∞)) −K (Φ (+∞) − Φ (−∞)) 2r/σ2 B (Φ (+∞) − Φ (+∞)) −B x 2r/σ2 B +K (Φ (+∞) − Φ (+∞)) K = x − K. Finally, for x > B we obtain g(T, K) = x (Φ (+∞) − Φ (+∞)) −K (Φ (+∞) − Φ (+∞)) 2r/σ2 B (Φ (−∞) − Φ (−∞)) −B x 2r/σ2 B +K (Φ (−∞) − Φ (−∞)) K = 0.

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Introduction to Stochastic Finance with Market Examples, Second Edition Exercise 11.7 a) The price at time t ∈ [0, T ] of the European knock-out call option is given by EKOCt = e−(T −t)r IE∗ (ST − K)+ 1{ST ≤B} Ft , 0 ≤ t ≤ T. 100 90 80 70 60 50 40 30 20 10 0

100

150

200

Fig. S.56: Payoff function of the European knock-out call option. Using the relation 2

ST = St e(T −t)r+(BbT −Bbt )σ−(T −t)σ /2 ,

0 ≤ t ≤ T,

we have e−(T −t)r IE∗ (ST − K)+ 1{ST ≤B} Ft 2 b b = e−(T −t)r IE∗ (xe(T −t)r+ BT −Bt σ−(T −t)σ /2 − K)+ i ×1 (T −t)r+(Bb −Bb )σ−(T −t)σ2 /2 t T {xe ≤B} x=St = e−(T −t)r IE∗ (em(x)+X − K)+ 1{em(x)+X ≤B} x=S , t

where m(x) := (T − t)r − and

0 ≤ t ≤ T,

σ2 (T − t) + log x 2

bT − B bt σ ≃ N (0, (T − t)σ 2 ) X := B

under P∗ . Next, as in Lemma 7.7 we note that if X is a centered Gaussian random variable with variance v 2 > 0 and B ≥ K, for any m ∈ R we have IE (em+X − K)+ 1{em+X ≤B} w ∞ 2 2 1 = √ (em+x − K)+ 1{em+x ≤B} e−x /(2v ) dx 2πv 2 −∞ 2 2 1 w −m+log B m+x (e − K)e−x /(2v ) dx = √ 2πv 2 −m+log K "

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Solutions Manual = √

em 2πv 2

w −m+log B −m+log K

ex−x /(2v ) dx − √

K 2πv 2

w −m+log B −m+log K

e−x /(2v ) dx

em+v /2 w −m+log B −(v2 −x)2 /(2v2 ) K w (−m+log B)/v −x2 /2 = √ e dx − √ e dx 2 −m+log K 2π (−m+log K)/v 2πv 2 em+v /2 w −v2 −m+log B −y2 /(2v2 ) e dy = √ 2πv 2 −v2 −m+log K −K Φ((m − log K)/v) − Φ((m − log B)/v) 2 = em+v /2 Φ(v + (m − log K)/v) − Φ(v + (m − log B)/v) −K Φ((m − log K)/v) − Φ((m − log B)/v) . 2

Hence, the price of the European knock-out call option is given, if B ≥ K, by EKOCt = e−(T −t)r IE∗ (ST − K)+ 1{ST ≤B} Ft 2 m(St ) − log K m(St ) − log K = e−(T −t)r em(St )+σ (T −t)/2 Φ v + −Φ v+ v v m(S ) − log K m(S ) − log B t t − Ke−(T −t)r Φ −Φ v v (T − t)r + (T − t)σ 2 /2 + log(St /K) √ = St Φ σ T −t (T − t)r + (T − t)σ 2 /2 + log(St /B) √ − St Φ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /K) −(T −t)r √ − Ke Φ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /B) √ + Ke−(T −t)r Φ , σ T −t 0 ≤ t ≤ T , with EKOCt = 0 if B ≤ K.

16 14 12 10 8 6 4 2 0 120

80 Time to maturity (days)

80 70

40 0

Underlying

Fig. S.57: Price map of the European knock-out call option.

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Introduction to Stochastic Finance with Market Examples, Second Edition b) By computations similar to part (a) we find that, if B ≤ K, (T − t)r − (T − t)σ 2 /2 + log(St /B) √ EKIPt = Ke−(T −t)r Φ − σ T −t (T − t)r + (T − t)σ 2 /2 + log(St /B) √ −St Φ − . σ T −t 160 140 120 100 80 60 40 20 0

100

150

200

Fig. S.58: Payoff function of the European knock-in put option. When B ≥ K, we find the Black-Scholes put option price EKIPt = e−(T −t)r IE∗ (K − ST )+ 1{ST ≤B} Ft = e−(T −t)r IE∗ (K − ST )+ Ft (T − t)r − (T − t)σ 2 /2 + log(St /K) √ = Ke−(T −t)r Φ − σ T −t (T − t)r + (T − t)σ 2 /2 + log(St /K) √ − St Φ − , σ T −t 0 ≤ t ≤ T. 35 30 25 20 15 10 5 500

Underlying

90 0

120

Time to maturity (days)

Fig. S.59: Price map of the European knock-in put option. c) Using the in-out parity relation EKOCt + EKICt = e−(T −t)r IE∗ [(ST − K)+ | Ft ] "

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Solutions Manual (T − t)r + (T − t)σ 2 /2 + log(St /K) √ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /K) −(T −t)r √ , −Ke Φ σ T −t

= St Φ

which is the price of the European call put option with strike price K, the price at time t ∈ [0, T ] of the European knock-in call option is given, if B ≥ K, as EKICt = e−(T −t)r IE∗ (ST − K)+ 1{ST ≥B} Ft (T − t)r + (T − t)σ 2 /2 + log(St /B) √ = St Φ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /B) −(T −t)r √ −Ke Φ , σ T −t 0 ≤ t ≤ T. 160 140 120 100 80 60 40 20 0

100 x

150

200

Fig. S.60: Payoff function of the European knock-in call option. When B ≤ K, we find the Black-Scholes call option price EKICt = e−(T −t)r IE∗ (ST − K)+ 1{ST ≥B} Ft = e−(T −t)r IE∗ (ST − K)+ Ft (T − t)r + (T − t)σ 2 /2 + log(St /K) √ = St Φ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /K) −(T −t)r √ −Ke Φ . σ T −t

216

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30 25 20 15 10 90

5 80

0 120

Time to maturity (days)

Underlying

Fig. S.61: Price map of the European knock-in call option. d) Using the in-out parity relation EKOPt + EKIPt = e−(T −t)r IE∗ [(K − ST )+ | Ft ], which is the price of the European put option with strike price K, we find that the price at time t ∈ [0, T ] of the European knock-in put option is given, if B ≤ K, as EKOPt = e−(T −t)r IE∗ (K − ST )+ 1{ST ≥B} Ft = e−(T −t)r IE∗ [(K − ST )+ | Ft ] − EKIPt (T − t)r − (T − t)σ 2 /2 + log(St /K) √ = Ke−(T −t)r Φ − σ T −t (T − t)r + (T − t)σ 2 /2 + log(St /K) √ −St Φ − σ T −t (T − t)r + (T − t)σ 2 /2 + log(St /B) √ +St Φ − σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /B) √ −Ke−(T −t)r Φ − . σ T −t 100 90 80 70 60 50 40 30 20 10 0

100

150

200

Fig. S.62: Payoff function of the European knock-out put option. When B ≥ K, we have "

217 September 15, 2022

Solutions Manual EKIPt = e−(T −t)r IE∗ (K − ST )+ 1{ST ≥B} Ft = 0.

14 12 10 8 6 4 2 0 120

90 80 70 80 Time to maturity (days)

Underlying

40 0

Fig. S.63: Price map of the European knock-out put option. In addition, by the results of Questions (d) and (c) we can verify the call-put parity relation EKICt − EKIPt = e−(T −t)r IE∗ (ST − K)1{ST ≥B} Ft (T − t)r + (T − t)σ 2 /2 + log(St /B) √ = St Φ σ T −t (T − t)r − (T − t)σ 2 /2 + log(St /B) √ −Ke−(T −t)r Φ . σ T −t

Chapter 12 Exercise 12.1 a) This probability density function is given by r 2 −(x−σT /2)2 /(2T ) −x − σT /2 √ e − σeσx Φ , x 7−→ πT T

x ≥ 0.

b) We have 2 IE min St = S0 IE min eσBt −σ t/2 t∈[0,T ] t∈[0,T ] h i −σ maxt∈[0,T ] (Bt +σt/2) = S0 IE e r ! w∞ 2 −(x−σT /2)2 /(2T ) −x − σT /2 √ = S0 e−σx e − σeσx Φ dx 0 πT T 218

218 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition w ∞ −x − σT /2 2S0 w ∞ −(x+σT /2)2 /(2T ) √ e dx − S0 σ Φ dx = √ 0 2πT 0 T 2S0 w ∞ −x2 /(2T ) σS0 w ∞ −(x+σT /2)2 /(2T ) = √ e dx − √ xe dx 2πT σT /2 2πT 0 w w ∞ ∞ 2 2 2S0 σS0 = √ e−x /(2T ) dx − √ (x − σT /2)e−x /(2T ) dx 2πT σT /2 2πT σT /2 r √ T −σ2 T /8 = 2S0 (1 + σ 2 T /4)Φ(−σ T /2) − S0 σ e . (S.12.54) 2π 1 0.9 0.8 0.7

price

0.6 0.5 0.4 0.3 0.2 0.1 0

time T

Fig. S.64: Expected minimum of geometric Brownian motion over [0, T ]. c) We have IE

K − min St

+ #

t∈[0,T ]

! r √ ! σ2 T σ T T −σ2 T /8 Φ − −σ e . 2 1+ 4 2 2π

2.0

= K − S0

= IE K − min St

0.0

0.5

Price 1.0

1.5

Upper bound Black−Scholes put price

1.0

1.2

1.4

1.6

1.8

2.0

Fig. S.65: Black-Scholes put price upper bound with S0 = 1. "

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Solutions Manual The derivative with respect to time is given by √ 2 ∂ σ2 σ2 T σ √ e−σ T /8 IE min St = S0 Φ − σ T /2 − 2S0 1 + ∂T 2 4 t∈[0,T ] 4 2πT r σS0 −σ2 T /8 S0 σ 3 T −σ2 T /8 −√ e e + 8 2π 8πT ! √ S0 σ 2 T S0 σ −σ2 T /8 3σ 2 T −√ = Φ −σ e 1+ . 2 2 4 2πT 0

derivative

-0.2

-0.4

-0.6

-0.8

-1

time T

Fig. S.66: Time derivative of the expected minimum of geometric Brownian motion. On the other hand, when r > 0 we have 2r/σ2 t St m0 σ 2 mt0 T −t T −t IE∗ mT0 | Ft = mt0 Φ δ− − St Φ δ− t m0 2r St St σ2 St T −t (T −t)r . +St e 1+ Φ −δ+ 2r mt0 When r tends to 0, this minimum tends to log(St /mt0 ) − σ 2 T /2 log(St /mt0 ) + σ 2 T /2 √ √ mt0 Φ + St Φ − σ T σ T t 2r/σ2 t ! 1 St m0 m0 T −t T −t 2 (T −t)r +σ St lim e Φ −δ+ − Φ δ , − r→0 2r mt0 St St where t 2r/σ2 t ! St m0 m0 T −t e Φ − Φ δ− mt0 St St t 2 1 log(St /m0 ) + σ T /2 + rT √ (1 + (T − t)r)Φ − = lim r→0 2r σ T

1 lim r→0 2r

220

(T −t)r

T −t −δ+

220 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition mt 2r log(mt0 /St ) − σ 2 T /2 + rT √ − 1 + 2 log 0 Φ σ St σ T mt0 1 2 log(St /mt0 ) + σ 2 T /2 √ = T − t + 2 log Φ − 2 σ St σ T √ w −(log(St /mt0 )+σ 2 T /2+rT )/(σ T ) 2 1 + lim √ e−y /2 dy r→0 r 8π −∞ w (− log(St /mt0 )−σ2 T /2+rT )/(σ√T ) 2 e−y /2 dy − −∞ mt 1 2 log(St /mt0 ) + σ 2 T /2 √ = T − t + 2 log 0 Φ − 2 σ St σ T √ 1 w (− log(St /mt0 )−σ2 T /2+rT )/(σ T ) −y2 /2 − lim √ e dy √ r→0 r 8π (− log(St /mt0 )−σ 2 T /2−rT )/(σ T ) 1 2 mt log(St /mt0 ) + σ 2 T /2 √ = T − t + 2 log 0 Φ − 2 σ St σ T √ T −((log(St /mt0 )+σ2 T /2)/(σ√T ))2 /2 − √ e , σ 2π hence log(St /mt0 ) + σ 2 T /2 log(St /mt0 ) − σ 2 T /2 √ √ + St Φ − σ T σ T St mt0 log(St /mt0 ) + σ 2 T /2 2 √ + (T − t)σ + 2 log Φ − 2 St σ T r T −((log(St /mt0 )+σ2 T /2)/(σ√T ))2 /2 −σSt e . 2π

IE∗ mT0 | Ft = mt0 Φ

In particular, when T tends to infinity we find that IE∗ mT0 | Ft lim = 0, r ≥ 0. ∗ T →∞ IE [ST | Ft ] When t = 0 we have S0 = m00 , and we recover r √ ! σ2 T T T −σ2 T /8 IE∗ mT0 = 2 S0 + Φ −σ − σS0 e . 4 2 2π Exercise 12.2 a) By (S.12.54), we have h i IE max St = IE eσ maxt∈[0,T ] (Bt −σt/2) t∈[0,T ]

221 September 15, 2022

Solutions Manual i h = S0 IE e−(−σ) maxt∈[0,T ] (Bt −(−σ)t/2) r √ T −σ2 T /8 = 2S0 (1 + σ 2 T /4)Φ(σ T /2) + S0 σ e . 2π 1.9 1.8 1.7 1.6

price

1.5 1.4 1.3 1.2 1.1 1

0.1

0.2

0.3

0.4

0.5

time T

Fig. S.67: Expected maximum of geometric Brownian motion over [0, T ]. b) We have IE

S0 max eσBt −σ t/2 − K t∈[0,T ]

σ2 T 4

2 = IE S0 max eσBt −σ t/2 − K t∈[0,T ]

r √ ! T T −σ2 T /8 Φ σ + S0 σ e − K. 2 2π

2.0

= 2S0 1 +

+ #

0.0

0.5

Price 1.0

1.5

Upper bound Black−Scholes call price

0.0

0.2

0.4

0.6

0.8

1.0

Fig. S.68: Black-Scholes call price upper bound with S0 = 1. The derivative with respect to time is given by √ ! ∂ S0 σ 2 T S0 σ −σ2 T /8 3σ 2 T IE max St = Φ σ +√ e 1+ . ∂T 2 2 4 t∈[0,T ] 2πT 222

222 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 10

derivative

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

time T

Fig. S.69: Time derivative of the expected maximum of geometric Brownian motion. Note that when r > 0 we have St St σ2 T −t T −t (T −t)r Φ δ IE∗ M0T | Ft = M0t Φ −δ− + S e 1 + t + M0t 2r M0t t 2r/σ2 t 2 M0 σ M0 T −t −St Φ −δ− . 2r St St When r tends to 0, this maximum tends to log(St /M0t ) + σ 2 T /2 log(St /M0t ) − σ 2 T /2 √ √ + St Φ M0t Φ − σ T σ T t 2r/σ2 t ! 1 St M0 M0 T −t T −t +σ 2 St lim e(T −t)r Φ δ+ − Φ −δ , − r→0 2r M0t St St where 2r/σ2 t ! M0t M0 T −t e Φ − Φ −δ− St St    St σ2 log M t + 2 T + rT 1  0  √ = lim (1 + (T − t)r)Φ  r→0 2r σ T 2r log(St /M0t ) + σ 2 T /2 − rT Mt √ − 1 + 2 log 0 Φ σ St σ T 1 2 log(St /M0t ) + σ 2 T /2 M0t √ = T − t + 2 log Φ 2 σ St σ T √ w (log(St /M0t )+σ 2 T /2+rT )/(σ T ) 2 1 + lim √ e−y /2 dy r→0 r 8π −∞ w (log(St /M0t )+σ2 T /2−rT )/(σ√T ) 2 − e−y /2 dy

1 lim r→0 2r

(T −t)r

T −t δ+

St M0t

−∞

223 September 15, 2022

Solutions Manual Mt 2 log(St /M0t ) + σ 2 T /2 √ log 0 Φ 2 σ St σ T √ 1 w (log(St /M0t )+σ2 T /2+rT )/(σ T ) −y2 /2 + lim √ e dy √ r→0 r 8π (log(St /M0t )+σ 2 T /2−rT )/(σ T ) Mt 1 2 log(St /M0t ) + σ 2 T /2 √ = T − t + 2 log 0 Φ 2 σ St σ T √ T −((log(St /M0t )+σ2 T /2)/(σ√T ))2 /2 , + √ e σ 2π =

1 2

T −t+

hence log(St /M0t ) − σ 2 T /2 log(St /M0t ) + σ 2 T /2 √ √ IE∗ M0T | Ft = M0t Φ − + St Φ σ T σ T St M0t log(St /M0t ) + σ 2 T /2 2 √ + (T − t)σ + 2 log Φ 2 St σ T r T −((log(St /M0t )+σ2 T /2)/(σ√T ))2 /2 e . +σSt 2π In particular, when T tends to infinity we find that  σ2  1 + if r > 0, IE∗ M0T | Ft 2r lim = ∗ T →∞ IE [ST | Ft ]   +∞ if r = 0. When t = 0 we have S0 = M00 , and we recover r √ ! T T −σ2 T /8 σ2 T Φ σ + σS0 e . IE∗ M0T = 2 S0 + 4 2 2π Exercise 12.3 a) We have P

wa 2 dx , min Bt ≤ a = 2 e−x /(2T ) √ −∞ t∈[0,T ] 2πT

a < 0,

i.e. the probability density function φ of sup Bt is given by t∈[0,T ]

r φ(a) =

2 −a2 /(2T ) e 1(−∞,0] (a), πT

a ∈ R.

b) We have 224

224 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition IE

min St = S0 IE exp σ min Bt

t∈[0,T ]

2S0 w 0 σx−x2 /(2T ) 2S0 w 0 −(x−σT )2 /(2T )+σ2 T /2 e dx = √ e dx = √ 2πT −∞ 2πσ 2 T −∞ √ w w −σ T 2 2S0 σ2 T /2 −σT −x2 /(2T ) 2S0 2 = √ e e dx = √ eσ T /2 e−x /2 dx −∞ −∞ 2π 2πT √ √ 2 = 2S0 eσ T /2 Φ − σ T = 2 IE[ST ] 1 − Φ σ T , hence √ IE ST − min St = IE[ST ] − IE min St = IE[ST ] − 2 IE[ST ] 1 − Φ σ T t∈[0,T ]

t∈[0,T ]

√ √ 1 2 = IE[ST ] 2Φ σ T − 1 = 2S0 eσ T /2 Φ σ T − , 2

and √ √ 2 e−σ T /2 IE ST − min St = S0 2Φ σ T −1 = S0 1−2Φ −σ T . t∈[0,T ]

Remark: We note that the price of the lookback option converges to S0 as T goes to infinity. 1

2 (Φ(σT1/2)-1)

0.8

Price

0.6

0.4

0.2

Time T

Fig. S.70: Lookback call option price as a function of T with S0 = 1. Exercise 12.4 We have IE∗ e−rτ 1{τ ≤T } 1{M0τ −Sτ ≥K} wT w∞w∞ = e−rt f(τ,Sτ ,Mτ ) (t, x, y)dydxdt 1 0 K+x w T w ∞ w y−K = e−rt f(τ,Sτ ,Mτ ) (t, x, y)dxdydt 1

225 September 15, 2022

Solutions Manual for T ≥ 1, and IE∗ e−rτ 1{τ ≤T } 1{M0τ −Sτ ≥K} = 0 if T ∈ [0, 1]. Exercise 12.5 a)

i) The boundary condition (12.3a) is explained by the fact that f (t, 0, y) = e−(T −t)r IE∗ M0T − ST St = 0, M0t = y = e−(T −t)r IE∗ M0t − ST St = 0, M0t = y = e−(T −t)r IE∗ M0t M0t = y − e−(T −t)r IE∗ [ST | St = 0] = ye−(T −t)r , since IE∗ [ST | St = 0] = 0 as St = 0 implies ST = 0 from the relation 2

ST = St eσ(BT −Bt )+(µ−σ /2)(T −t) ,

0 ≤ t ≤ T.

ii) The boundary condition (12.3b), i.e. ∂f (t, x, y)y=x = 0, ∂y

0 ≤ x ≤ y,

is illustrated in the following Figure S.71, see also Figure 12.3.

Lookback put price

St= 60

80 y=Mt0

Fig. S.71: Graph of the lookback put option price (2D) with St = 60. iii) Condition (12.3c) follows from the fact that f (T, x, y) = IE∗ M0T − ST ST = x, M0T = y = y − x. b)

i) The boundary condition (12.14a) is explained by the fact that f (t, x, 0) = e−(T −t)r IE∗ ST − mT0 St = x, mt0 = 0 = e−(T −t)r IE∗ ST St = x, mt0 = 0 = e−(T −t)r IE∗ [ST | St = x]

226

226 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = e−(T −t)r x,

x > 0.

ii) Condition (12.14b) follows from the fact that f (T, x, y) = IE∗ ST − mT0 ST = x, mT0 = y = x − y. We have

f (t, x, x) = xC(T − t),

with τ C(τ ) = 1 − e−rτ Φ δ− (1) σ2 σ2 τ τ (1) , Φ − δ+ (1) + e−rτ Φ δ− − 1+ 2r 2r

τ > 0,

hence

∂f (t, x, x) = C(T − t), ∂x while we also have ∂f (t, x, y)y=x = 0, ∂y

0 ≤ t ≤ T,

0 ≤ x ≤ y,

see also Figure 12.8.

Chapter 13 Exercise 13.1 We have w w T T IE St dt = IE[St ]dt τ

= S0 = S0 = S0 = S0

τ wT

τ rT

IE[eσBt +rt−σ t/2 ]dt 2

ert−σ t/2 IE[eσBt ]dt ert dt − erτ , r

0 ≤ τ ≤ T,

and " 2 # w wT wT T IE St dt = IE St dt Su du τ

227 September 15, 2022

Solutions Manual = IE =2

w w T T τ

Su St dtdu

wT wu

IE[Su St ]dtdu τ wT wu 2 2 IE[eσBu +ru−σ u/2 eσBt +rt−σ t/2 ]dtdu τ τ wT wu 2 2 = 2S02 eru−σ u/2+rt−σ t/2 IE[eσBu +σBt ]dtdu τ τ wT wu 2 2 = 2S02 e(r−σ /2)u e(r−σ /2)t IE[eσBu +σBt ]dtdu τ τ wT wu 2 2 = 2S02 e(r−σ /2)u e(r−σ /2)t IE[e2σBt +σ(Bu −Bt ) ]dtdu τ τ wT w u 2 2 = 2S02 e(r−σ /2)u e(r−σ /2)t IE[e2σBt ] IE[eσ(Bu −Bt ) ]dtdu τ τ wT wu 2 2 2 2 = 2S02 e(r−σ /2)u e(r−σ /2)t e2σ t eσ (u−t)/2 dtdu τ τ wT wu 2 = 2S02 eru ert+σ t dtdu τ τ 2 2S 2 w T (2r+σ2 )u e − eru e(r+σ )τ du = 2 0 σ +r τ 2 2 2 re(σ +2r)T − (σ 2 + 2r)erT +(σ +r)τ + (σ 2 + r)e(σ +2r)τ = 2S02 , (σ 2 + r)(σ 2 + 2r)r τ

= 2S02

Exercise 13.2 a) The integral

0 ≤ τ ≤ T.

rs ds has a centered Gaussian distribution with variance " 2 # w w wT T T IE rs ds = σ 2 IE Bs Bt dsdt 0

=σ

wT wT 0

IE[Bs Bt ]dsdt

wT wT

min(s, t)dsdt 0 0 wT wt = 2σ sdsdt 0 0 wT t2 dt = σ2 = σ2

=T b) Since the integral

rT 0

rs ds is a random variable with probability density φ(x) = p

228

0 2

3σ

1 2πT 3 /3

e−3x /(2πT ) , 228 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition we have e−rT IE

" w

+ # ru du − κ

= e−rT

w∞

−∞

(x − κ)+ φ(x)dx

w∞ 2 2 3 e−rT (x − κ)e−3x /(2σ T ) dx = p 2 3 2πσ T /3 κ p 2 e−rT w ∞ √ 2 3 (x σ 2 T 3 /3 − κ)e−x /2 dx = √ κ/ σ T /3 2π p 2 2 e−rT σ 2 T 3 /3 w ∞ e−rT w ∞ √ 2 3 xe−x /2 dx − κ √ √ √ = e−x /2 dx σ T /3 κ/ 2π 2π κ/ σ2 T 3 /3 p p e−rT σ 2 T 3 /3 h −x2 /2 i∞ e−rT √ =− e − κ √ (1 − Φ(κ/ σ 2 T 3 /3)) √ κ/ σ 2 T 3 /3 2π 2π p p e−rT σ 2 T 3 /3 −3κ2 /(2σ2 T 3 ) e−rT √ = e − κ √ (1 − Φ(κ/ σ 2 T 3 /3)) 2π 2π ! r r σ 2 T 3 −3κ2 /(2σ2 T 3 ) 3 e−rT −rT =e e − κ √ Φ −κ . 6π σ2 T 3 2π Exercise 13.3 We have " # + w 1 T 1 wT −(T −t)r e IE Su du − κ Ft = e−(T −t)r IE Su du − κ Ft T 0 T 0 w T 1 = e−(T −t)r IE Su du Ft − κe−(T −t)r T 0 w w t T 1 1 Su du Ft + e−(T −t)r IE Su du Ft − κe−(T −t)r = e−(T −t)r IE 0 t T T w wt T 1 1 Su du + e−(T −t)r IE Su du Ft − κe−(T −t)r = e−(T −t)r t T 0 T wt wT −(T −t)r 1 −(T −t)r 1 =e Su du + e IE[Su | Ft ]du − κe−(T −t)r T 0 T t w w 1 t 1 T = e−(T −t)r Su du + e−(T −t)r St e(u−t)r du − κe−(T −t)r T 0 T t w w t T −t 1 St = e−(T −t)r Su du + e−(T −t)r eru du − κe−(T −t)r T 0 T 0 1 wt St (T −t)r = e−(T −t)r Su du + e−(T −t)r (e − 1) − κe−(T −t)r T 0 rT 1 wt 1 − e−(T −t)r = e−(T −t)r Su du + St − κe−(T −t)r , T 0 rT t ∈ [0, T ], cf. Geman and Yor (1993) page 361. We check that the function f (t, x, y) = e−(T −t)r (y/T − κ) + x(1 − e−(T −t)r )/(rT ) satisfies the PDE "

229 September 15, 2022

Solutions Manual rf (t, x, y) =

∂f ∂f ∂f 1 ∂2f (t, x, y) + x (t, x, y) + rx (t, x, y) + x2 σ 2 2 (t, x, y), ∂t ∂y ∂x 2 ∂x

t, x > 0, and the boundary conditions f (t, 0, y) = e−(T −t)r (y/T − κ), 0 ≤ t ≤ T , y ∈ R+ , and f (T, x, y) = y/T − κ, x, y ∈ R+ . However, the condition limy→−∞ f (t, x, y) = 0 is not satisfied because we need to take y > 0 in the above calculation. Exercise 13.4 a) We have 1 wT Su du − K Ft T 0 w 1 T = e−(T −t)r IE∗ Su du Ft − Ke−(T −t)r T 0 e−(T −t)r ∗ w t e−(T −t)r ∗ w T = IE IE Su du Ft + Su du Ft − Ke−(T −t)r 0 t T T −(T −t)r w T −(T −t)r w t e e Su du + IE∗ [Su | Ft ]du − Ke−(T −t)r = 0 t T T e−(T −t)r w t e−(T −t)r w T (u−t)r = Su du + e St du − Ke−(T −t)r 0 t T T e−(T −t)r w t e−rT w T ru = Su du + St e du − Ke−(T −t)r 0 t T T e−(T −t)r w t St = Su du + 1 − e−(T −t)r − Ke−(T −t)r . 0 T rT

e−(T −t)r IE∗

b) Using the relation (x − K)+ − (K − x)+ = x − K,

K, x ∈ R,

we have C(t, K) − P (t, K) = e−(T −t)r IE∗

1 wT Su du − K T 0 #

+ Ft

1 wT Su du Ft T 0 " # + + 1 wT 1 wT ∗ −(T −t)r =e IE Su du − K − K− Su du Ft T 0 T 0 w 1 T = e−(T −t)r IE∗ Su du − K Ft T 0 −e−(T −t)r IE∗

230

K−

230 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition =

e−(T −t)r w t St 1 − e−(T −t)r − Ke−(T −t)r . Su du + 0 T rT

(S.13.55)

c) Any self-financing portfolio strategy (ξt )t∈R+ with price process (Vt )t∈R+ has to satisfy the equation dVt = ηt dAt + ξt dSt = rηt At dt + µξt St dt + σξt St dBt = rVt dt + (µ − r)ξt St dt + σξt St dBt ,

t ∈ R+ .

On the other hand, by part (a) we have −(T −t)r w t St e Ss ds + (1 − e−(T −t)r ) − Ke−(T −t)r 0 T rT r −(T −t)r w t e−(T −t)r St = e Ss dsdt + St dt − e−(T −t)r dt 0 T T T 1 − e−(T −t)r −(T −t)r + dSt − rKe dt rT w −(T −t)r t r 1−e = e−(T −t)r Ss dsdt + dSt − rKe−(T −t)r dt 0 T rT 1 − e−(T −t)r = rVt dt + dSt − St (1 − e−(T −t)r )dt rT 1 − e−(T −t)r = rVt dt + ((µ − r)St dt + σSt dBt ), rT

dVt = d

hence

1 − e−(T −t)r , t ∈ [0, T ], rT which can be recovered by differentiating the pricing function ξt =

e−(T −t)r x y+ 1 − e−(T −t)r − Ke−(T −t)r T rT wt in (S.13.55) with respect to x = St , with y = Su du. We also have 0

Vt =

−(T −t)r w t

Ss ds + ξt St − Ke−(T −t)r ,

t ∈ [0, T ].

d) The following code yields $7.906436 for the price of the long forward contract. T=1;t=63/252;r=0.0209;K=80;dt=1/252;S=as.numeric(last(futures)); exp(−(T−t)∗r)∗sum(futures)∗dt/T+S∗(1−exp(−(T−t)∗r))/(r∗T)−K∗exp(−(T−t)∗r)

231 September 15, 2022

Solutions Manual Exercise 13.5 The geometric mean price G satisfies w w 1 T 1 t 1 wT G = exp log Su du = exp log Su du + log Su du T 0 T 0 T t w 1 t T −t 1 wT Su = exp log Su du + log St + log du T 0 T T t St w 1 t T −t = exp log Su du + log St T 0 T w 1 T + (r(u − t) + (Bu − Bt )σ − (u − t)σ 2 /2)du T t w 1 t T −t = exp log Su du + log St T 0 T w 1 T −t σ wT + (ru − σ 2 u/2)du + (Bu − Bt )du T 0 T t w t σ wT 1 (T − t)2 log Su du + (r − σ 2 /2) + (Bu − Bt )du = (St )(T −t)/T exp T 0 2T T t where

wT t

Bu du is centered Gaussian with conditional variance wT

2 Bu du

= IE

" w

= IE

" w

# Ft = IE

wT t

2 (Bu − Bt )du

# Ft

2 # T (Bu − Bt )du

T −t

w T −t w u 0

2 # w T −t w T −t (Bu − Bt )du = IE [Bs Bu ] dsdu

sdsdu =

w T −t 0

u2 du =

(T − t)3 . 3

Hence, letting m :=

1 wt T −t (T − t)2 log Su du + log St + (r − σ 2 /2), T 0 T 2T

X :=

σ wT Bu du, T t

and v 2 = (T − t)σ 2 /3, we find " # w + 1 T ∗ −(T −t)r e IE exp log Su du − K Ft T 0 w (T − t)2 σ2 1 t = (St )(T −t)/T e−(T −t)r exp log Su du + (2r − σ 2 ) + (T − t) T 0 4T 6

232

232 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition  ×Φ 

(T − t)σ 2 /3 + T1 

−Ke−(T −t)r Φ  T

rt 0

(T −t)/T

log Su du + log t K p σ (T − t)/3

−t) + (T2T (r − σ 2 /2)

 

 (T −t)/T S −t)2 log Su du + log t K + (T2T (r − σ 2 /2) , p σ (T − t)/3

0 ≤ t ≤ T . In case t = 0, we get " w + # 1 T log Su du − K e−rT IE∗ exp T 0 2

log(S0 /K) + T (r + σ 2 /6)/2 p σ T /3 ! log(S0 /K) + T (r − σ 2 /2)/2 p . σ T /3

= S0 e−T (r+σ /6)/2 Φ −Ke−rT Φ

Exercise 13.6 Under the above condition we have, taking t ∈ [τ, T ], " # + 1 wT e−(T −t)r IE∗ Ft rs ds − K T −τ τ # " + 1 wT ∗ −(T −t)r =e IE Λt + rs ds − K Ft T −τ t 1 wT = e−(T −t)r IE∗ Λt + rs ds − K Ft T −τ t w −(T −t)r T e = e−(T −t)r (Λt − K) + IE∗ rs ds Ft t T −τ e−(T −t)r w T ∗ −(T −t)r IE [rs | Ft ]ds, t ∈ [τ, T ], =e (Λt − K) + t T −τ where IE∗ [rs | Ft ] = vt e−(s−t)λ + m(1 − e−(s−t)λ ), hence " IE∗

1 wT rs ds − K T −τ τ

# Ft = IE∗

Λt +

0 ≤ s ≤ t,

1 wT rs ds − K T −τ t

+ Ft

1 wT ∗ IE [rs | Ft ]ds T −τ t 1 wT = Λt − K + rt e−(s−t)λ + m(1 − e−(s−t)λ ) ds T −τ t = Λt − K +

233 September 15, 2022

Solutions Manual w T −t 1 e−(T −t)r (rt − m) e−λs ds + m(T − t) 0 T −τ T −τ 1 w T −t −λs T −t = Λt − K + (rt − m) e ds + m T −τ 0 T −τ −(T −t)λ 1−e T −t = Λt − K + (rt − m) + m . (T − τ )λ T −τ = Λt − K +

Exercise 13.7 This question extends Exercise 7.4 to n ≥ 3. If (St )t∈R+ is a martingale then for any convex payoff function ϕ we can write h i h i ST1 + · · · + STn ϕ(ST1 ) + · · · + ϕ(STn ) ≤ IE∗ n n IE∗ [ϕ(ST1 )] + · · · + IE∗ [ϕ(STn )] = n IE∗ [ϕ(IE∗ [STn | FT1 ])] + · · · + IE∗ [ϕ(IE∗ [STn | FTn ])] = n IE∗ [IE∗ [ϕ(STn ) | FT1 ]] + · · · + IE∗ [IE∗ [ϕ(STn ) | FTn ]] ≤ n IE∗ [ϕ(STn )] + · · · + IE∗ [ϕ(STn )] = n = IE∗ [ϕ(STn )].

IE∗ ϕ

since ϕ is convex,

because (St )t∈R+ is a martingale, by Jensen’s inequality, by the tower property,

On the other hand, if (St )t∈R+ is only a submartingale then the above argument still applies to a convex non-decreasing payoff function ϕ such as ϕ(x) = (x − K)+ . Exercise 13.8 Taking t ∈ [τ, T ], under the condition Λt :=

1 wt Ss ds ≥ K, T −τ τ

we have # + 1 wT e IE Ss ds − K Ft T −τ τ " # + 1 wT = e−(T −t)r IE∗ Λt + Ss ds − K Ft T −τ t 1 wT Ss ds − K Ft = e−(T −t)r IE∗ Λt + T −τ t e−(T −t)r ∗ w T −(T −t)r =e (Λt − K) + IE Ss ds Ft t T −τ −(T −t)r w T e = e−(T −t)r (Λt − K) + IE∗ [Ss | Ft ]ds t T −τ −(T −t)r

234

∗

234 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition e−(T −t)r w T (s−t)r e ds t T −τ −(T −t)r w T −t e = e−(T −t)r (Λt − K) + St ers ds 0 T −τ e−(T −t)r (T −t)r (e − 1) = e−(T −t)r (Λt − K) + St (T − τ )r = e−(T −t)r (Λt − K) + St

= e−(T −t)r (Λt − K) + St

1 − e−(T −t)r , (T − τ )r

t ∈ [τ, T ].

Exercise 13.9 The Asian option price can be written as # " + 1 wT ∗ −r(T −t) Ft = St Ib E (UT )+ | Ut e IE Su du − K T 0 = St h(t, Ut ) = St g(t, Zt ), which shows that

g(t, Zt ) = h(t, Ut ),

and it remains to use the relation Ut =

1 − e−(T −t)r + e−(T −t)r Zt , rT

t ∈ [0, T ].

Exercise 13.10 i) By change of variable. We note that Zet = e−(T −t)r Zt , where 1 1 wt 1 Λt Zt := Su du − K = −K , 0 ≤ t ≤ T, St T 0 St T and the pricing function g(t, Zt ) satisfies the Rogers-Shi PDE ∂g 1 ∂g 1 ∂2g (t, z) + − rz (t, z) + σ 2 z 2 2 (t, z) = 0. ∂t T ∂z 2 ∂z −(T −t)r (T −t)r Letting ze := e z and ge(t, ze) := g t, e ze = g(t, z) = ge t, e−(T −t)r z , we note that

235 September 15, 2022

Solutions Manual  ∂ ∂g   (t, z) = ge t, e−(T −t)r z   ∂t ∂t       ∂e g ∂e g    = t, e−(T −t)r z + re−(T −t)r z t, e−(T −t)r z   ∂t ∂x      ∂e g ∂e g (t, ze) + re z (t, ze), =  ∂t ∂x       ∂g ∂e g ∂e g   (t, z) = e−(T −t)r t, e−(T −t)r z = e−(T −t)r (t, ze),    ∂z ∂e z ∂e z      2 2 2    ∂ g (t, z) = e−2(T −t)r ∂ ge t, e−(T −t)r z = e−2(T −t)r ∂ ge (t, ze), ∂z 2 ∂e z2 ∂e z2 hence ∂g 1 ∂g 1 ∂2g (t, z) + − rz (t, z) + σ 2 z 2 2 (t, z) ∂t T ∂z 2 ∂z 1 ∂e g ∂e g ∂e g = (t, ze) + re z (t, ze) + − rz e−(T −t)r (t, ze) ∂t ∂x T ∂e z 1 2 2 −2(T −t)r ∂ 2 ge + σ z e (t, ze) 2 ∂e z2 ∂e g 1 1 ∂e g ∂ 2 ge = (t, ze) + e−(T −t)r (t, ze) + σ 2 ze2 2 (t, ze), ∂t T ∂e z 2 ∂e z

and the (simpler) PDE ∂e g 1 ∂e g 1 ∂ 2 ge (t, ze) + e−(T −t)r (t, ze) + σ 2 ze2 2 (t, ze) = 0. ∂t T ∂e z 2 ∂e z ii) Using the Itô formula. Given that dZet = d(e−(T −t)r Zt ) = re−(T −t)r Zt dt + e−(T −t)r dZt = rZet dt + e−(T −t)r dZt , and

dSt = rSt dt + σSt dBt ,

under the risk-neutral probability measure P∗ , an application of Itô’s formula to the discounted portfolio price leads to et d e−rt St ge t, Z et dSt + St de = e−rt − re g t, Zet g t, Zet dt + ge t, Z g t, Zet + dSt • de g g et dSt + St ∂e et dt + St ∂e et t, Z t, Zet dZ = e−rt −rSt ge t, Zet dt + ge t, Z ∂t ∂z 236

236 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition 1 ∂ 2 ge et 2 + dSt • de g t, Zet + e−rt St 2 t, Zet dZ 2 ∂z g −rt et dSt + St ∂e et dt =e −rSt ge t, Zet dt + ge t, Z t, Z ∂t ∂e g ∂e g t, Zet dt + St e−(T −t)r t, Zet dZt +rZet St ∂z ∂z 1 −rt ∂ 2 ge e e + e g t, Zet St 2 t, Zt dZt + dSt • de 2 ∂z g et dt + σSt ge t, Zet dBt + rZet St ∂e = e−rt −rSt ge t, Zet dt + rSt ge t, Z t, Zet dt ∂z g −rt −(T −t)r 2 ∂e e +e e St Zt −r + σ t, Zt dt ∂z 1 ∂e g g et dt − σe−(T −t)r St Zt ∂e + e−(T −t)r St t, Z t, Zet dBt T ∂z ∂z 1 2 e2 ∂ 2 ge ∂e g −rt 2 e e +e σ Zt St 2 t, Zt dt − σ St Zt t, Zet dt 2 ∂z ∂z 1 ∂e g 1 e2 ∂ 2 ge ∂e g et + e−(T −t)r = e−rt St t, Z t, Zet + σ 2 Z t, Zet dt t 2 ∂t T ∂z 2 ∂z ∂e g −rt e e e t, Zt dBt . +St e σe g t, Zt − σ Zt ∂z Since the discounted portfolio price process is a martingale under the risk-neutral probability measure P∗ , the sum of components in dt should vanish in the above expression, which yields 2 1 ∂e g ∂e g et + 1 σ 2 Zet2 ∂ ge t, Z et = 0, t, Zet + e−(T −t)r t, Z ∂t T ∂z 2 ∂z 2

and the PDE ∂e g 1 ∂e g 1 ∂ 2 ge (t, ze) + e−(T −t)r (t, ze) + σ 2 ze2 2 (t, ze) = 0, ∂t T ∂e z 2 ∂e z under the terminal condition ge(T, ze) = ze+ , ze ∈ R. Exercise 13.11 a) When Λt /T ≥ K we have f (t, St , Λt ) = e−(T −t)r

Λt −K T

+ St

1 − e−(T −t)r , rT

see Exercise 13.8. b) When Λt /T ≥ K we have "

237 September 15, 2022

Solutions Manual ξt =

1 − e−(T −t)r rT

and

ηt At = e(T −t)r

Λt −K , T

0 ≤ t ≤ T.

c) At maturity we have f (T, ST , ΛT ) = (ΛT /T − K)+ , hence ξT = 0 and e−rT A0

ηT AT = AT

ΛT −K T

1{ΛT >KT } =

ΛT −K T

+ .

d) By Proposition 13.12 we have ∂g 1 Λt 1 Λt ξt = f (t, St , Λt ) − −K t, −K St T ∂z St T where the function g(t, z) satisfies f (t, x, y) = xg(t, (y/T − K)/x)) and g(t, z) = ze−(T −t)r +

1 − e−(T −t)r , rT

z > 0,

and solves the PDE ∂g (t, z) + ∂t

1 − rz T

∂g 1 ∂2g (t, z) + σ 2 z 2 2 (t, z) = 0, ∂z 2 ∂z

under the terminal condition g(T, z) = z + , hence letting h(t, z) := e(T −t)r

∂g (t, z), ∂z

we have e(T −t)r

∂g (t, z) + e(T −t)r ∂t

1 − rz T

∂g 1 ∂2g (t, z) + σ 2 z 2 2 (t, z) = 0, ∂z 2 ∂z

with h(t, z) = 1, z > 0, hence 2 ∂2g ∂g 1 ∂ g (t, z) − re(T −t)r (t, z) + e(T −t)r − rz (t, z) ∂t∂z ∂z T ∂z 2 ∂2g 1 ∂3g +σ 2 ze(T −t)r 2 (t, z) + e(T −t)r σ 2 z 2 3 (t, z) = 0, ∂z 2 ∂z

e(T −t)r

∂h (t, z) + ∂t

1 + (σ 2 − r)z T

∂h 1 ∂2h (t, z) + σ 2 z 2 2 (t, z) = 0, ∂z 2 ∂z

with the terminal condition h(T, z) = 1{z>0} . On the other hand, we have ηt =

238

1 (f (t, St , Λt ) − ξt St ) At

238 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Λt ∂g 1 Λt −K t, −K T ∂z St T e−(T −t)r Λt 1 Λt = − K h t, −K . At T St T =

1 At

Exercise 13.12 Asian options with dividends. When reinvesting dividends, the portfolio self-financing condition reads dVt =

ηt dAt + ξt dSt {z } |

Trading profit and loss

δξ S dt | t{zt }

Dividend payout

= rηt At dt + ξt ((µ − δ)St dt + σSt dBt ) + δξt St dt = rηt At dt + ξt (µSt dt + σSt dBt ) = rVt dt + (µ − r)ξt St dt + σξt St dBt ,

t ∈ R+ .

On the other hand, by Itô’s formula we have dgδ (t, St , Λt ) ∂gδ ∂gδ ∂gδ = (t, St , Λt )dt + (t, St , Λt )dΛt + (µ − δ)St (t, St , Λt )dt ∂t ∂y ∂x 1 ∂ 2 gδ ∂gδ (t, St , Λt )dBt + σ 2 St2 (t, St , Λt )dt + σSt 2 ∂x2 ∂x ∂gδ ∂gδ ∂gδ = (t, St , Λt )dt + St (t, St , Λt )dt + (µ − δ)St (t, St , Λt )dt ∂t ∂y ∂x 1 ∂ 2 gδ ∂gδ + σ 2 St2 (t, St , Λt )dt + σSt (t, St , Λt )dBt , 2 ∂x2 ∂x hence by identification of the terms in dBt and dt in the expressions of dVt and dgδ (t, St ), we get ∂gδ (t, St , Λt ), ξt = ∂x and we derive the Black-Scholes PDE with dividend rgδ (t, x, y) =

∂gδ ∂gδ (t, x, y) + y (t, x, y) (S.13.56) ∂t ∂y 1 ∂ 2 gδ ∂gδ +(r − δ)x (t, x, y) + σ 2 x2 (t, x, y). ∂x 2 ∂x2

Defining f (t, x, y) := e(T −t)δ gδ (t, x, y) and substituting gδ (t, x, y) = e(T −t)δ f (t, x, y) in (S.13.56) yields the equation

239 September 15, 2022

Solutions Manual ∂f ∂f (t, x, y) + (t, x, y) ∂y ∂t ∂f 1 ∂2f +(r − δ)x (t, x, y) + σ 2 x2 2 (t, x, y), ∂x 2 ∂x

rf (t, x, y) = δf (t, x, y) + y

i.e. (r − δ)f (t, x, y) =

∂f ∂f (t, x, y) + y (t, x, y) ∂t ∂y ∂f 1 ∂2f +(r − δ)x (t, x, y) + σ 2 x2 2 (t, x, y), ∂x 2 ∂x

whose solution f (t, x, y) is the Asian option pricing function with modified interest rate r − δ and no dividends, under the terminal condition f (T, x, y) = gδ (T, x, y) =

y T

−K

Therefore the Asian option price gδ (t, St , Λt ) with dividend rate δ can be recovered from the relation gδ (t, x, y) = e(T −t)δ f (t, x, y),

t ∈ [0, T ], x, y > 0.

Note that we can also define h(t, x, y) := gδ t, xe−δ(T −t) , y and substituting

gδ (t, x, y) = h t, xeδ(T −t) , y

in (S.13.56) yields the equation ∂h ∂h (t, x, y) + (t, x, y) ∂y ∂t ∂h 1 ∂2h +rx (t, x, y) + σ 2 x2 2 (t, x, y), ∂x 2 ∂x

rh(t, x, y) = y

whose solution h(t, x, y) is the Asian option pricing function with interest rate r and no dividends, under the terminal condition h(T, x, y) = gδ (T, x, y) =

y T

−K

Finally, the Asian option price gδ (t, St , Λt ) with dividend rate δ can be also recovered from the relation gδ (t, x, y) = h t, xe−(T −t)δ , y , t ∈ [0, T ], x, y > 0.

240

240 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

Chapter 14 Exercise 14.1 a) The process ((2 − Bt )+ )t∈R+ is a convex function x 7−→ (2 − x)+ of the Brownian martingale (Bt )t∈R+ , hence it is a submartingale by Proposition 14.4-(a). b) Taking σ := 1 and µ := σ 2 /2 > 0, the process eBt can be written as 2

eBt = eσBt −σ t/2+µt = eµt eσBt −σ t/2 ,

t ∈ R+ ,

hence it is a submartingale as the driftless geometric Brownian motion 2 eσBt −σ t/2 is a martingale. c) When t > 0, the question “is ν > t?” cannot be answered at time t without waiting to know the value of B2t at time 2t > t. Therefore ν is not a stopping time. d) For any t ∈ R+ , the question “is τ > t?” can be answered based on the observation of the paths of (Bs )0≤s≤t and of the (deterministic) curve (es/2 + αses/2 )0≤s≤t up to the time t. Therefore τ is a stopping time. e) Since τ is a stopping time and eBt −t/2 t∈R+ is a martingale, the Stopping Time Theorem 14.7 shows that eBt∧τ −(t∧τ )/2 t∈R+ is also a martingale ∗ and, in particular, its expected value IE eBt∧τ −(t∧τ )/2 = IE eB0∧τ −(0∧τ )/2 = IE eB0 −0/2 = 1 is constantly equal to 1 for all t ≥ 0. This shows that h i IE eBτ −τ /2 = IE lim eBt∧τ −(t∧τ )/2 t→∞ = lim IE eBt∧τ −(t∧τ )/2 t→∞

= 1. Next, we note that eBτ = (α + βτ )eτ /2 at time τ , hence α + βτ = eBτ −τ /2 and α + β IE[τ ] = IE[α + βτ ] = IE eBτ −τ /2 = 1, i.e. IE[τ ] = (1 − α)/β. Remark: This argument also recovers IE[τ ] = 0 when α = 1, however it fails when (α > 1 and β > 0) and when (α < 1 and β < 0), because τ is not a.s. finite (P(τ < ∞) < 1) in those cases. Exercise 14.2 Stopping times. ∗

We let t ∧ τ := min(t, τ ).

241 September 15, 2022

Solutions Manual a) When 0 ≤ t < 1 the question “is ν > t?” cannot be answered at time t without waiting to know the value of B1 at time 1. Therefore ν is not a stopping time. b) For any t ∈ R+ , the question “is τ > t?” can be answered based on the observation of the paths of (Bs )0≤s≤t and of the (deterministic) curve αe−s/2 0≤s≤t up to the time t. Therefore τ is a stopping time. Since τ is a stopping time and (Bt )t∈R+ is a martingale, the Stopping Time Theorem 14.7 shows that eBt∧τ −(t∧τ )/2 t∈R+ is also a martingale and in particular its expected value IE eBt∧τ −(t∧τ )/2 = IE eB0∧τ −(0∧τ )/2 = IE eB0 −0/2 = 1 is constantly equal to 1 for all t. This shows that h i IE eBτ −τ /2 = IE lim eBτ ∧t −(τ ∧t)/2 = lim IE eBτ ∧t −(τ ∧t)/2 = 1. t→∞

t→∞

Next, we note that we have e

Bτ

= αe

−τ /2

α IE e−τ = IE eBτ −τ /2 = 1,

at time τ , hence

i.e.

1 IE e−τ = ≤ 1. α

Remark: note that this argument fails when α < 1 because in that case τ is not a.s. finite. c) For any t ∈ R+ , the question “is τ > t?” can be answered based on the observation of the paths of (Bs )0≤s≤t and of the (deterministic) curve (1 + αs)0≤s≤t up to the time t. Therefore τ is a stopping time. Since τ is a stopping time and (Bt )t∈R+ is a martingale, the Stopping 2 Time Theorem 14.7 shows that (Bt∧τ − (t ∧ τ ))t∈R+ is also a martingale and in particular its expected value 2 2 IE[Bt∧τ − (t ∧ τ )] = IE[B0∧τ − (0 ∧ τ )] = IE[B02 − 0] = 0

is constantly equal to 0 for all t. This shows that h i 2 2 IE[Bτ2 − τ ] = IE lim (Bt∧τ − (t ∧ τ )) = lim IE (Bt∧τ − (t ∧ τ )) = 0. t→∞

t→∞

Next, we note that Bτ2 = 1 + ατ at time τ , hence 1 + α IE[τ ] = IE[1 + ατ ] = IE[Bτ2 ] − IE[τ ] = 0, i.e.

1 . 1−α Remark: Note that this argument is valid whenever α ≤ 1 and yields IE[τ ] = +∞ when α = 1, however it fails when α > 1 because in that case IE[τ ] =

242

242 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition τ is not a.s. finite. Exercise 14.3 a) By the Stopping Time Theorem 14.7, for all n ≥ 0 we have i h √ 1 = IE e 2rBτL ∧n −r(τL ∧n) h √ i h √ i = IE e 2rBτL ∧n −r(τL ∧n) 1{τL <n} + IE e 2rBτL ∧n −r(τL ∧n) 1{τL ≥n} h √ i i h √ = IE e 2rBτL −rτL 1{τL <n} + IE e 2rBn −rn 1{τL ≥n} h √ i √ = eL 2r IE e−rτL 1{τL <n} + IE e 2rBn −rn 1{τL ≥n} . The first term above converges to √ √ eL 2r IE e−rτL 1{τL <∞} = eL 2r IE e−rτL

as n tends to infinity, by dominated or monotone convergence and the fact that r > 0. The second term can be bounded as i h √ i h √ √ 0 ≤ IE e 2rBn −rn 1{τL ≥n} ≤ e−rn IE eL 2r 1{τL ≥n} ≤ e−rn eL 2r , which tends to 0 as n tends to infinity because r > 0. Therefore we have i h √ √ 1 = lim IE e 2rBτL ∧n −r(τL ∧n) = eL 2r IE e−rτL , n→∞

√ −L 2r L which yields IE e−rτ for any r ≥ 0. When r < 0 we could in e −rτ = fact show that IE e L = +∞. b) In order to maximize the quantity IE e−rτL BτL = IE e−rτL BτL 1{τL <∞} −rτL = L IE e 1{τL <∞} = L IE e−rτL √

= Le−L 2r , we differentiate √ √ √ √ ∂ (Le−L 2r ) = e−L 2r − L 2re−L 2r = 0, ∂L √ which yields the optimal level L∗ = 1/ 2r.

This shows that when the value of r is “large” the better strategy is to √ opt for a “small gain” at the level L∗ = 1/ 2r rather than to wait for a "

243 September 15, 2022

Solutions Manual longer time. Exercise 14.4 See e.g. Theorem 6.16 page 161 of Klebaner (2005). By the Itô formula, we have 1 w t ′′ f (Bs )ds 2 0 wt 1 w t ′′ 1 w t ′′ f (Bs )ds − f (Bs )ds = f (B0 ) + f ′ (Bs )dBs + 0 2 0 2 0 wt = f (B0 ) + f ′ (Bs )dBs ,

Xt = f (Bt ) −

hence the process (Xt )t∈R+ is a martingale. By the Stopping Time Theorem 14.7 we have f (x) = IE[X0 | B0 = x] = IE[Xτ ∧t | B0 = x]

w τ ∧t 1 IE f ′′ (Bs )ds | B0 = x 0 2 = IE[f (Bτ ∧t ) | B0 = x] + IE[τ ∧ t | B0 = x],

= IE[f (Bτ ∧t ) | B0 = x] −

since f ′′ (y) = −2 for all y ∈ R. We note that, by dominated convergence, h i IE τ | B0 = x = IE lim (τ ∧ t) | B0 = x t→∞

= lim IE[τ ∧ t | B0 = x] t→∞

≤ |f (x)| + max |f (y)| y∈[a,b]

< ∞, hence IE[τ ] < ∞ and therefore P(τ < ∞) = 1, allowing us to write limt→∞ (τ ∧ t) = τ < ∞ with probability P(τ < ∞) = 1. Next, we have f (x) = lim IE[f (Bτ ∧t ) | B0 = x] + lim IE[τ ∧ t | B0 = x] t→∞ t→∞ h i h i = IE lim f (Bτ ∧t ) | B0 = x + IE lim (τ ∧ t) | B0 = x t→∞

t→∞

= IE[f (Bτ ) | B0 = x] + IE[τ | B0 = x] = IE[τ | B0 = x], since f ′′ (x) = −2 and f (a) = f (b) = 0 with Bτ ∈ {a, b}. Remarks. i) The above exchanges between limt→∞ and the expectation operator IE[ · | B0 = x] is justified by the dominated convergence theorem, since 244

244 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition |f (Bτ ∧t )| ≤ max |f (y)|, y∈[a,b]

t ∈ R+ .

ii) The function f (x) can be determined by searching for a quadratic solution of the form f (x) = α+βx+γx2 , which shows that f ′′ (x) = 2γ = −2 hence γ = −1, and  2  f (a) = α + βa − a = 0,  

f (b) = α + βb − b2 = 0,

hence α = −ab and β = a + b. Therefore, we have IE[τ | B0 = x] = f (x) = −ab + (a + b)x − x2 = (x − a)(b − x). Exercise 14.5 We use the Stopping Time Theorem 14.7 and the fact that 2 eσBt −σ t/2 t∈R+ is a martingale for all σ ∈ R. By the stopping time theorem, for all n ≥ 0 we have 2 1 = IE eσBτ ∧n −σ (τ ∧n)/2 σBτ ∧n −σ2 (τ ∧n)/2 2 = IE e 1{τ <n} + IE eσBτ ∧n −σ (τ ∧n)/2 1{τ ≥n} 2 2 = IE eσBτ −σ τ /2 1{τ <n} + IE eσBn −σ n/2 1{τ ≥n} 2 2 (S.14.57) = eσα IE eσβτ −σ τ /2 1{τ <n} + IE eσBn −σ n/2 1{τ ≥n} . 2

Under the condition σ 2 ≥ 2σβ we have 0 ≤ eσβτ −σ τ /2 ≤ 1, hence by dominated or monotone convergence we find 2 2 2 lim IE eσβτ −σ τ /2 1{τ <n} = IE eσβτ −σ τ /2 lim 1{τ <n} = IE eσβτ −σ τ /2 1{τ <∞} ,

n→∞

and the first term in (S.14.57) converges to 2 2 eσα IE e−(σ /2−σβ)τ 1{τ <∞} = eσα IE e−(σ /2−σβ)τ as n tends to infinity. In what follows we use the solutions σ± = β ± of the equation r = σ 2 /2 − σβ, and we distinguish two cases.

β 2 + 2r

a) If α ≥ 0, we have Bn ≤ α + βn, n ≤ τ , hence the second term above can be bounded as 2 0 ≤ IE eσ+ Bn −nσ+ /2 1{τ ≥n} 2 ≤ e−nσ+ /2 IE eασ+ +βσ+ n 1{τ ≥n} 2

≤ eασ+ −nσ+ /2+βσ+ n =e

ασ+ −rn

, 245 September 15, 2022

Solutions Manual which tends to 0 as n tends to infinity. Therefore, we have 2 2 1 = lim IE eσ+ Bτ ∧n −σ+ (τ ∧n)/2 = eασ+ IE e−(σ+ /2−βσ+ )τ , n→∞

which yields 2 IE e−rτ = IE e−(σ+ /2−βσ+ )τ = e−ασ+ √

= e−αβ−α √β +2r

β 2 +2r

−αβ−|α|

with P(τ < +∞) = IE[1{τ <∞} ] = lim IE 1{τ <∞} e−rτ = r→0

−1

−2

−2αβ e if β ≥ 0, 1 if β ≤ 0.

−2 0

0.1

0.2

0.3

τ 0.4

0.5

0.6

(a) α = 1, β = 0.5

0.7

0.8

0.9

0.1

0.2

τ 0.3

0.4

0.5

0.6

0.7

0.8

0.9

(b) α = 1, β = −0.5

Fig. S.72: Hitting times of a straight line started at α < 0. b) If α ≤ 0, we have Bn ≥ α + βn, n ≤ τ , hence the second term above can be bounded as 2 0 ≤ IE eσ− Bn −nσ− /2 1{τ ≥n} 2 2 ≤ e−nσ− /2 IE eασ− +βnσ− 1{τ ≥n} ≤ eασ− −nσ− /2+βnσ− = eασ− −rn , which tends to 0 as n tends to infinity. Therefore, we have 2 2 1 = lim IE eσ− Bτ ∧n −σ− (τ ∧n)/2 = eασ− IE e−(σ− /2−βσ− )τ , n→∞

which yields 2 IE e−rτ = IE e−(σ− /2−βσ− )τ = e−ασ− √ 2 = e−αβ+α β +2r 246

246 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition √ 2 = e−αβ−|α| β +2r , with P(τ < +∞) = IE[1{τ <∞} ] = lim IE 1{τ <∞} e−rτ = r→0

−2

1 if β ≥ 0, e−2αβ if β ≤ 0.

−2 0

0.1 τ

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

(a) α = −1, β = 0.5

0.2

0.3

τ 0.4

0.5

0.6

0.7

0.8

0.9

(b) α = −1, β = −0.5

Fig. S.73: Hitting times of a straight line started at α < 0. Exercise 14.6 a) Letting A0 := 0, An+1 := An + IE[Mn+1 − Mn | Fn ], and

Nn := Mn − An ,

n ≥ 0, (S.14.58)

n ∈ N,

247 September 15, 2022

Solutions Manual = IE[Mn+1 | Fn ] − Mn ≥ 0,

n ∈ N,

since (Mn )n∈N is a submartingale. (iii) By induction we have An = An−1 + IE[Mn − Mn−1 | Fn−1 ],

n ≥ 1,

which is Fn−1 -measurable provided that An is Fn−1 -measurable, n ≥ 1. (iv) This property is obtained by construction in (S.14.58). b) For all bounded stopping times σ and τ such that σ ≤ τ a.s., we have IE[Mσ ] = IE[Nσ ] + IE[Aσ ] ≤ IE[Nσ ] + IE[Aτ ] = IE[Nτ ] + IE[Aτ ] = IE[Mτ ], by (14.7), since (Mn )n∈N is a martingale and (An )n∈N is non-decreasing.

Chapter 15 Exercise 15.1 The option payoffs at immediate exercise are given as follows: 2/3 ∗ p = 2/3 ∗ p =

(K − S0 )+ = 0.3 q∗

= 1/ 3

(K − S2 )+ = 0

(K − S1 )+ = 0.05 q∗ =

1/3

2/3 ∗ p =

(K − S2 )+ = 0.17

(K − S1 )+ = 0.35 q∗ =

1/3

(K − S2 )+ = 0.44

On the other hand, the expected payoffs are given by:

248

248 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

(K − S2 )+ = 0 IE∗ [(K − S2 )+ | S1 = 1.2] = 0.17/3 (K − S2 )+ = 0.17 IE∗ [(K − S2 )+ | S1 = 0.9] = 0.26 (K − S2 )+ = 0.44 Consequently, at time t = 1 we would exercise immediately if S1 = 0.9, and wait if S1 = 1.2. At time t = 0 with S0 = 1 the initial value of the option is (0.34/3 + 0.35)/3 = 1.39/9 ≃ 0.154 < 0.25 so we would exercise immediately as well. Exercise 15.2 2

a) Taking f (x) := Cx−2r/σ , we have 2 2 1 2r2 2r rxf ′ (x) + σ 2 x2 f ′′ (x) = −C 2 x−2r/σ + Cr 1 + 2 x−2r/σ 2 σ σ = Crx−2r/σ

= rf (x), and the condition limx→∞ f (x) = 0 is satisfied since r > 0. b) The conditions f (L∗ ) = K − L∗ and f ′ (L∗ ) = −1 read  ∗ −2r/σ 2 = K − L∗ ,   C(L )   − 2r C(L∗ )−1−2r/σ2 = −1, σ2 i.e.

 ∗ −2r/σ 2 = K − L∗   C(L )   2r (K − L∗ ) = L∗ , σ2

hence  2rK   L∗ =   2r + σ 2      C = "

Kσ 2 2r + σ 2

2rK 2r + σ 2

2r/σ2

σ2 2r

2rK 2r + σ 2

1+2r/σ2 . 249 September 15, 2022

Solutions Manual Exercise 15.3 a) This an American put option with strike price considered at S0 ≥ L∗ , hence by Propositions 15.2 and 15.4 the price of this option is (K − L∗ ) IE∗ e−τL∗ = (K − L∗ )

S0 L∗

−2r/σ2 .

b) This an American put option with strike price K and immediately exercised at S0 ≤ L∗ , hence by Propositions 15.2 and 15.4 the price of this option is K − S0 . b − K exercised at the c) This is an American call option with strike price 2K ∗ b optimal level L = K, hence by Equation (15.23) the price of this option is b − K)) IE∗ e−τL∗ = (K b − (2K b − K)) IE∗ e−τL∗ = (K − K) b S0 . (L∗ − (2K b K In conclusion, the pricing function is obtained by pasting together the pricing functions of American call and put options. 60

Price function Payoff function

50 40 30 20 10 0

^-K 2K

^ K

200

Fig. S.74: American butterfly payoff and price functions.

Exercise 15.4 a) Given the value ∂ BSp (x, T ) = −Φ(−d+ (x, T )) ∂x of the Delta of the Black-Scholes put option see Proposition 6.7, the smooth fit condition states that at x = S ∗ , the left derivative of (15.31), which is 2

2 ∂ ∂ (S ∗ )2r/σ BSp (x, T ) + α (x/S ∗ )−2r/σ = −Φ(−d+ (x, T )) + α 1+2r/σ2 , ∂x ∂x x

x > S ∗ , should match the right derivative of (15.32), which is −1, hence −1 = −Φ(−d+ (S ∗ , T )) − 250

2rα ∗ −1 (S ) , σ2 250 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition which yields α∗ =

σ2 S ∗ σ2 S ∗ (1 − Φ(−d+ (S ∗ , T ))) = Φ(d+ (S ∗ , T )), 2r 2r

and  2  σ 2 (S ∗ )1+2r/σ   BSp (x, T ) + Φ(d+ (S ∗ , T )), 2 2r/σ 2rx f (x, T ) ≃    K − x,

x > S∗, x ≤ S∗.

Note that at maturity (T = 0 here) we have d+ (S ∗ , 0) = −∞ since S ∗ < K, hence Φ(d+ (S ∗ , 0)) = 0 and f (x, 0) = K − x as expected. b) Equating (15.31) to (15.32) at x = S ∗ yields the equation K − S ∗ = BSp (x, T ) + α∗ , i.e. 1 = e−rT Φ(−d− (S ∗ , T )) +

S∗ K

σ2 2r

Φ(d+ (S ∗ , T )),

which can be used to determine the value of S ∗ , and then the corresponding value of α. The proposed strategy is to exercise the put option as soon as the underlying asset price reaches the critical level S ∗ . 16

S*=87.8 L*=85.71 (K-x)+

Option price

12 10 8 6 4 2 0

85 L* = 85.7

90 S* = 87.3

100

105

Underlying x

Fig. S.75: Perpetual vs finite expiration American put option price. The plot in Figure S.75 yields a finite expiration critical price S ∗ = 87.3 which is expectedly higher than the perpetual critical price L∗ = 85.71, with K = 100, σ = 10%, and r = 3%. The perpetual price, however, appears higher than the finite expiration price. In Figure S.76 we plot the graph of a Barone-Adesi and Whaley (1987) approximation, together with the European put option price, using the R fOptions package. Note that this approximation is valid only for certain parameter ranges. "

251 September 15, 2022

Solutions Manual

r=0.1;sig=0.15;T=0.5;K=100;library(ragtop) library(fOptions);payoff <− function(x){return(max(K−x,0))};vpayoff <− Vectorize( payoff) par(new=TRUE) curve(vpayoff, from=85, to=120, xlab="", lwd = 3, ylim=c(0,10),ylab="",col="red") par(new=TRUE) curve(blackscholes(callput=−1, x, K, r, T, sig, 0)$Price, from=85, to=120, xlab="", lwd = 3, ylim=c(0,10),ylab="",col="orange") par(new=TRUE) curve(BAWAmericanApproxOption("p",x,K,T,r,b=0,sig,title = NULL, description = NULL)@price, from=85, to=120 , xlab="Underlying asset price", lwd = 3,ylim=c (0,10),ylab="",col="blue") grid (lty = 5);legend(105,9.5,legend=c("Approximation","European payoff","Black−Scholes put"),col=c("blue","red","orange"),lty=1:1, cex=1.)

Approximation

European payoff Black−Scholes put

100

105

110

115

120

Underlying asset price

Fig. S.76: American put price approximation.

Exercise 15.5 a) We have τϵ =

 ϵ 

if Z = 1,

+∞ if Z = 0.

b) First, we note that Ft =

  {∅, Ω} 

Next, we have hence

if t = 0,

{∅, Ω, {Z = 0}, {Z = 1}} if t > 0. {τϵ > 0} = {Z = 0}, {τϵ > 0} ∈ / F0 = {∅, Ω},

and therefore τ0 is not an (Ft )t∈R+ -stopping time. 252

252 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition c)

i) For t = 0 we have {τϵ > 0} = {Z = 0} ∪ {Z = 1} = Ω, hence {τϵ > 0} ∈ F0 = {∅, Ω}. ii) For 0 < t < ϵ we have {τϵ > t} = Ω, hence {τϵ > t} ∈ Ft = {∅, Ω, {Z = 0}, {Z = 1}}. iii) For t > ϵ we have {τϵ > t} = {Z = 0}, hence {τϵ > t} ∈ Ft = {∅, Ω, {Z = 0}, {Z = 1}}. Therefore τϵ is an (Ft )t∈R+ -stopping time when ϵ > 0. Note that here the filtration (Ft )t∈R+ is not right-continuous, as {∅, Ω} = F0 ̸= F0+ :=

Ft = {∅, Ω, {Z = 0}, {Z = 1}}.

t>0

Exercise 15.6 a) This intrinsic payoff is κ − S0 . b) We note that the process (Zt )t∈R+ defined as Zt :=

St S0

e−(r−δ)λt+λσ t/2−λ σ t/2

λ 2 2 2 2 = e(r−δ)t+σBbt −σ t/2 e−(r−δ)λt+λσ t/2−λ σ t/2 2 2 = eλσBbt −λ σ t/2 , t ≥ 0, is a geometric Brownian motion without drift, hence a martingale, under the risk-neutral probability measure P∗ . c) The parameter λ should satisfy the equation r = (r − δ)λ −

σ2 λ(1 − λ), 2

i.e. λ2 σ 2 /2 + λ(r − δ − σ 2 /2) − r = 0. This equation admits two solutions p −(r − δ − σ 2 /2) ± (r − δ − σ 2 /2)2 + 4rσ 2 /2 λ± = , σ2 d) We have

253 September 15, 2022

Solutions Manual (λ )

0 ≤ Zt − =

St S0

λ−

St S0

λ−

L S0

λ−

≤ ≤

e−rt

0 ≤ t < τL ,

since λ− < 0 and St > L for t ∈ [0, τL ). e) By the Stopping Time Theorem 14.7 we have IE∗ [ZτL ] = IE∗ [Z0 ] = 1, which rewrites as " ∗

SτL S0

# −((r−δ)λ−λσ 2 /2+λ2 σ 2 /2)τL

= 1,

or, given the relation SτL = L,

L S0

h i 2 2 2 IE∗ e−((r−δ)λ−λσ /2+λ σ /2)τL = 1,

i.e. IE∗ e−rτL =

S0 L

λ ,

provided that we choose λ such that −((r − δ)λ − λσ 2 /2 + λ2 σ 2 /2) = −r, i.e. λ=

−(r − δ − σ 2 /2) ±

(S.15.59)

(r − δ − σ 2 /2)2 + 4rσ 2 /2 , σ2

and we choose the negative solution p −(r − δ − σ 2 /2) − (r − δ − σ 2 /2)2 + 4rσ 2 /2 λ := σ2 since S0 /L = x/L > 1 and the expectation IE∗ e−rτL < 1 is lower than 1 as r ≥ 0. f) This follows from (15.9) and the fact that r > 0. Using the fact that SτL = L < K when τL < ∞, we find h i h i IE∗ e−rτL (K − SτL )+ S0 = x = IE∗ e−rτL (K − SτL )+ 1{τL <x} S0 = x h i = IE∗ e−rτL (K − L)1{τL <x} S0 = x 254

254 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition i h = (K − L) IE∗ e−rτL 1{τL <x} S0 = x i h = (K − L) IE∗ e−rτL S0 = x . Next, noting that τL = 0 if S0 ≤ L, for all L ∈ (0, K) we have IE∗ e−rτL (K − SτL )+ S0 = x  0 < x ≤ L,  K − x, =  −rτL IE e (K − L)+ S0 = x , x ≥ L.  0 < x ≤ L,  K − x, =  (K − L) IE e−rτL S0 = x , x ≥ L.  0 < x ≤ L,   K − x,  √ = 2 /2)2 +4rσ 2 /2 −(r−a−σ2 /2)− (r−a−σ   σ2  (K − L) x , x ≥ L. L g) In order to compute L∗ we observe that, geometrically, the slope of x 7−→ fL (x) = (K − L)(x/L)λ− at x = L∗ is equal to −1, i.e. fL′ ∗ (L∗ ) = λ− (K − L∗ )

(L∗ )λ− −1 = −1, (L∗ )λ−

hence λ− (K − L∗ ) = L∗ ,

or L∗ =

λ− K < K. λ− − 1

Equivalently we may recover the value of L∗ from the optimality condition x λ − x λ− +1 ∂fL (x) =− − λ− x(K − L) = 0, ∂L L L at L = L∗ , hence −

x λ− L

hence L∗ =

− λ− (K − L)xλ− L−λ− −1 = 0, λ− 1 K= K, 1 − λ− 1 − 1/λ−

and sup IE∗ e−rτL (K − SτL )+ L∈(0,K)

1 S0 = x = − λ−

K 1 − 1/λ−

1−λ−

xλ− .

h) For x ≥ L we have "

255 September 15, 2022

Solutions Manual fL∗ (x) = (K − L∗ )

x λ− L∗

!λ− λ− x K λ− λ− − 1 λ− −1 K λ K x(λ− − 1) − = − λ− − 1 λ− K λ− λ K x λ− − 1 − = − λ− − 1 −λ− −K λ− −1 λ− x λ− − 1 = −λ− −K x λ− λ − 1 λ− K − = . K λ− 1 − λ− = K−

(S.15.60)

i) Let us check that the relation fL∗ (x) ≥ (K − x)+

(S.15.61)

holds. For all x ≤ K we have x λ− λ − 1 λ− K − +x−K K λ− 1 − λ− ! x λ− λ − 1 λ− 1 x − =K + −1 . K λ− 1 − λ− K

fL∗ (x) − (K − x) =

Hence it suffices to take K = 1 and to show that for all L∗ =

λ− ≤x≤1 λ− − 1

we have fL∗ (x) − (1 − x) =

xλ− 1 − λ−

λ− − 1 λ−

λ−

+ x − 1 ≥ 0.

Equality to 0 holds for x = λ− /(λ− − 1). By differentiation of this relation we get λ λ− − 1 − 1 +1 λ− 1 − λ− λ −1 λ− − 1 − = xλ− −1 +1 λ− ≥ 0,

fL′ ∗ (x) − (1 − x)′ = λ− xλ− −1

256

256 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition hence the function fL∗ (x) − (1 − x) is non-decreasing and the inequality holds throughout the interval [λ− /(λ− − 1), K]. On the other hand, using (S.15.59) it can be checked by hand that fL∗ given by (S.15.60) satisfies the equality 1 (r − δ)xfL′ ∗ (x) + σ 2 x2 fL′′∗ (x) = rfL∗ (x) 2 for x ≥ L∗ =

(S.15.62)

λ− K. In case λ− − 1 0 ≤ x ≤ L∗ =

we have

λ− K < K, λ− − 1

fL∗ (x) = K − x = (K − x)+ ,

hence the relation 1 rfL∗ (x) − (r − δ)xfL′ ∗ (x) − σ 2 x2 fL′′∗ (x) (fL∗ (x) − (K − x)+ ) = 0 2 always holds. On the other hand, in that case we also have 1 (r − δ)xfL′ ∗ (x) + σ 2 x2 fL′′∗ (x) = −(r − δ)x, 2 and to conclude we need to show that 1 (r − δ)xfL′ ∗ (x) + σ 2 x2 fL′′∗ (x) ≤ rfL∗ (x) = r(K − x), 2

(S.15.63)

which is true if δx ≤ rK. Indeed, by (S.15.59) we have (r − δ)λ− = r + λ− (λ− − 1)σ 2 /2 ≥ r, hence δ

λ− ≤ r, λ− − 1

since λ− < 0, which yields δx ≤ δL∗ ≤ δ

λ− K ≤ rK. λ− − 1

j) By Itô’s formula and the relation "

257 September 15, 2022

Solutions Manual bt dSt = (r − δ)St dt + σSt dB we have d feL∗ (St ) = −re−rt fL∗ (St )dt + e−rt dfL∗ (St ) σ 2 −rt 2 ′′ e St fL∗ (St ) = −re−rt fL∗ (St )dt + e−rt fL′ ∗ (St )dSt + 2 2 σ = e−rt −rfL∗ (St ) + (r − δ)St fL′ ∗ (St ) + St2 fL′′∗ (St ) dt 2 bt , +e−rt σSt fL′ ∗ (St )dB and from Equations (S.15.62) and (S.15.63) we have 1 (r − δ)xfL′ ∗ (x) + σ 2 x2 fL′′∗ (x) ≤ rfL∗ (x), 2 hence

t 7−→ e−rt fL∗ (St )

is a supermartingale. k) By the supermartingale property of t 7−→ e−rt fL∗ (St ), for all stopping times τ we have fL∗ (S0 ) ≥ IE∗ e−rτ fL∗ (Sτ ) S0 ≥ IE∗ e−rτ (K − Sτ )+

S0 ,

by (S.15.61), hence fL∗ (S0 ) ≥

sup

τ stopping time

IE∗ e−rτ (K − Sτ )+

S0 .

l) The stopped process t 7−→ e−rt∧τL∗ fL∗ (St∧τL∗ ) is a martingale since it has vanishing drift up to time τL∗ by (S.15.62), and it is constant after time τL∗ , hence by the Stopping Time Theorem 14.7 we find h i fL∗ (S0 ) = IE∗ e−rτ fL∗ (SτL∗ ) S0 h i = IE∗ e−rτ fL∗ (L∗ ) S0 h i = IE∗ e−rτ (K − SτL∗ )+ S0 h i ≤ sup IE∗ e−rτ (K − Sτ )+ S0 . τ stopping time

258

258 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition m) By combining the above results and conditioning at time t instead of time 0 we deduce that i h fL∗ (St ) = IE∗ e−r(τL∗ −t) (K − SτL∗ )+ St  λ−   K − St , 0 < St ≤ K,   λ− − 1  = λ −1 λ   λ− −St − λ− − 1 −    , St ≥ K, −K λ− λ− − 1 for all t ∈ R+ , where τL∗ = inf{u ≥ t : Su ≤ L}. We note that the perpetual put option price does not depend on the value of t ≥ 0. Exercise 15.7 a) We have (λ)

= (St )λ e−t((r−δ)λ−λ(1−λ)σ /2) = (S0 )λ eλσBbt −λ σ t/2 ,

which is a driftless geometric Brownian motion, and therefore a martingale under P∗ . b) The condition is r = (r − δ)λ − λ(1 − λ)σ 2 /2, with solutions p δ − r + σ 2 /2 − (δ − r + σ 2 /2)2 + 2rσ 2 λ− = ≤ 0, σ2 p δ − r + σ 2 /2 + (δ − r + σ 2 /2)2 + 2rσ 2 λ+ = ≥ 1. σ2 c) Due to the inequality (λ )

0 ≤ Zt + = (St )λ+ e−rt ≤ Lλ+ , which holds because λ+ > 0 and St ≤ L, 0 ≤ t < τL , we note that since (λ ) limt→∞ Zt + = 0, we have h i h i λ+ −rτL Lλ+ IE∗ e−rτL = IE∗ SτL e 1{τL <∞} = IE∗ Zτ(λL+ ) 1{τL <∞} h i (λ ) (λ ) = IE∗ lim ZτL+∧t = lim IE∗ ZτL+∧t t→∞ t→∞ (λ ) = IE∗ Z0 + = (S0 )λ+ . Therefore, we have "

259 September 15, 2022

Solutions Manual S0 = x = IE∗ (SτL − K)e−rτL 1{τL <∞} S0 = x = (L − K) IE∗ e−rτL 1{τL <∞} S0 = x = (L − K) IE∗ e−rτL S0 = x x λ+ = (L − K) , L

IE∗ e−rτL (SτL − K)+

when S0 = x > L. In order to maximize sup IE∗ e−rτL (K − SτL )+

S0 = x ,

L∈(0,K) λ

we differentiate L 7−→ (L − K) (x/L) + with respect to L, to find x λ+ L

− λ+ (L − K)xλ+ L−λ+ −1 = 0,

hence L∗δ =

K λ+ K= , λ+ − 1 1 − 1/λ+

and sup IE∗ e−rτL (K − SτL )+ L∈(0,K)

1 S0 = x = λ+

K 1 − 1/λ+

1−λ+

xλ+ .

We note that as δ ↘ 0 we have λ+ ↘ 1 and L∗δ ↗ ∞, and since

K 1 − 1/λ+

1−λ+

λ+ − 1 → 1, = exp (λ+ − 1) log λ+ K

we find that the perpetual American call option price without dividend (δ = 0) is S0 = x. Exercise 15.8 a) By the definition (15.36) of S1 (t) and S2 (t) we have α S1 (t) Zt = e−rt S2 (t) S2 (t) = e−rt S1 (t)α S2 (t)1−α 2

= S1 (0)α S2 (0)1−α e(ασ1 +(1−α)σ2 )Wt −σ2 t/2 , which is a martingale when σ22 = (ασ1 + (1 − α)σ2 )2 , i.e. 260

260 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition ασ1 + (1 − α)σ2 = ±σ2 , which yields either α = 0 or α=

2σ2 σ1 + σ2 =1+ > 1, σ2 − σ1 σ2 − σ1

since 0 ≤ σ1 < σ2 . b) We have IE e−rτL (S1 (τL ) − S2 (τL ))+ = IE e−rτL (LS2 (τL ) − S2 (τL ))+ = (L − 1)+ IE e−rτL S2 (τL ) . (S.15.64) c) Since τL ∧ t is a bounded stopping time we can write α α S1 (τL ∧ t) S1 (0) = IE e−r(τL ∧t) S2 (τL ∧ t) (S.15.65) S2 (0) S2 (0) S2 (τL ∧ t) α α S (t) S1 (τL ) 1{τL <t} + IE e−rt S2 (t) 1 1{τL >t} = IE e−rτL S2 (τL ) S2 (τL ) S2 (t) We have e−rt S2 (t)

S1 (t) S2 (t)

1{τL >t} ≤ e−rt S2 (t)Lα 1{τL >t} ≤ e−rt S2 (t)Lα ,

hence by a uniform integrability argument, α S1 (t) 1 lim IE e−rt S2 (t) {τL >t} = 0, t→∞ S2 (t) and letting t go to infinity in (S.15.65) shows that α α S1 (0) S1 (τL ) S2 (0) = IE e−rτL S2 (τL ) = Lα IE e−rτL S2 (τL ) , S2 (0) S2 (τL ) since S1 (τL )/S2 (τL ) = L/L = 1. The conclusion IE e−rτL (S1 (τL ) − S2 (τL ))+ = (L − 1)+ L−α S2 (0)

S1 (0) S2 (0)

(S.15.66)

then follows by an application of (S.15.64). d) In order to maximize (S.15.66) as a function of L we consider the derivative L−1 1 ∂ = α − α(L − 1)L−α−1 = 0, α ∂L L L which vanishes for

L∗ =

α , α−1 261 September 15, 2022

Solutions Manual and we substitute L in (S.15.66) with the value of L∗ . e) In addition to r = σ22 /2 it is sufficient to let S1 (0) = κ and σ1 = 0 which yields α = 2, L∗ = 2, and we find sup

τ stopping time

IE e−rτ (κ − S2 (τ ))+ =

1 κ 2 , S2 (0) 2

which coincides with the result of Proposition 15.4. Exercise 15.9 a) It suffices to check the sign of the quantity (λ − 1)(λ + 2r/σ 2 ),

(S.15.67)

in (15.38), which is positive when λ ∈ (−∞, −2r/σ ]∪[1, ∞), and negative when −2r/σ 2 ≤ λ ≤ 1. b) The sign of (S.15.67) is positive when λ ∈ (−∞, 1] ∪ [−2r/σ 2 , ∞), and negative when 1 ≤ λ ≤ −2r/σ 2 . c) By the Stopping Time Theorem 14.7, for any n ≥ 0 we have h i (λ) xλ = IE∗ e−r(τL ∧n) ZτL ∧n | S0 = x h i h i = IE∗ Zτ(λ) 1{τL <n} | S0 = x + e−rn IE∗ Zn(λ) 1{τL >n} | S0 = x L ≥ IE∗ e−rτL (SτL )λ 1{τL <n} | S0 = x = Lλ IE∗ e−rτL 1{τL <n} | S0 = x . 2

(λ) By the results of Questions (a)-(b), the process Zt t∈R is a martin+ 2 gale when λ ∈ {1, −2rσ /2}. Next, letting n to infinity, by monotone convergence we find  max(1,−2r/σ 2 )   x , x ≥ L,  L x λ  ≤ IE∗ e−rτL 1{τL <∞} | S0 = x ≤  L min(1,−2r/σ2 )    x , 0 < x ≤ L. L

d) We note that P∗ (τL < ∞) = 1 by (14.15), hence if −σ 2 /2 ≤ r < 0 we have IE∗ e−rτL (K − SτL )+ 1{τL <∞} | S0 = x  x −2r/σ2   (K − L) , x ≥ L,  L = (K − L) IE∗ e−rτL | S0 = x ≤    (K − L) x , 0 < x ≤ L. L 262

262 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Similarly, if r ≤ −σ 2 /2 we have IE∗ e−rτL (K − SτL )+ 1{τL <∞} | S0 = x = (K − L) IE∗ e−rτL | S0 = x  x  x ≥ L, (K − L) ,   L ≤ 2    (K − L) x −2r/σ , 0 < x ≤ L. L e) This follows by noting that (K − L)(x/L) = (K/L − 1)x increases to ∞ when L tends to zero. f) If −σ 2 /2 ≤ r < 0 we have IE∗ e−rτL (SτL − K)+ 1{τL <∞} | S0 = x = (L − K) IE∗ e−rτL 1{τL <∞} | S0 = x  x −2r/σ2   , x ≥ L,  (L − K) L ≤    (L − K) x , 0 < x ≤ L. L If r ≤ −σ 2 /2 we have IE∗ e−rτL (SτL − K)+ 1{τL <∞} | S0 = x = (L − K) IE∗ e−rτL 1{τL <∞} | S0 = x  x  (L − K) , x ≥ L,   L ≤ 2    (L − K) x −2r/σ , 0 < x ≤ L. L g) This follows by noting that for fixed x > 0, the quantity (L − K)x/L = (1 − K/L)x increases to x when L tends to infinity. Exercise 15.10 Perpetual American binary options. a) Similarly, for x ≥ K, immediate exercise is the optimal strategy and we have CbAm (t, x) = 1. When x < K the optimal exercise level of the perpetual American binary call option is L∗ = K with the optimal exercise time τK , and by e.g. (4.4.22) page 135 we have IE∗ e−(τ −t)r 1{Sτ ≥K} St = x CbAm (t, x) = sup τ ≥t τ stopping time

= IE∗ e−(τK −t)r St = x x = , x < K. K "

263 September 15, 2022

Solutions Manual

American binary put price

1.4 1.2 1 0.8 0.6 0.4 0.2 0

100

150

200

250

Underlying x

Fig. S.77: Perpetual American binary put price map with K = 100. b) For x ≤ K, immediate exercise is the optimal strategy and we have PbAm (t, x) = 1. When x > K the optimal exercise level of the perpetual American binary put option is L∗ = K with the optimal exercise time τK , and by e.g. (4.4.11) page 125 we have PbAm (t, x) = sup IE∗ e−(τ −t)r 1{Sτ ≤K} St = x τ ≥t τ stopping time

= IE∗ e−(τK −t)r St = x x −2r/σ2 , x > K. = K

American binary call price

1.4 1.2 1 0.8 0.6 0.4 0.2 0

100

150

200

Underlying x

Fig. S.78: Perpetual American binary call price map with K = 100.

Exercise 15.11 Finite expiration American binary options. a) The optimal strategy is as follows: (i) if St ≥ K, then exercise immediately. (ii) if St < K, then wait. b) The optimal strategy is as follows: 264

264 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (i) if St > K, then wait. (ii) if St ≤ K, exercise immediately. c) Based on the answers to Question (a) we set

and

CdAm (t, T, K) = 1,

0 ≤ t < T,

CdAm (T, T, x) = 0,

0 ≤ x < K.

d) Based on the answers to Question (b), we set PdAm (t, T, K) = 1, and

PdAm (T, T, x) = 0,

0 ≤ t < T, x > K.

e) Starting from St ≤ K, the maximum possible payoff is clearly reached as soon as St hits the level K before the expiration date T , hence the discounted optimal payoff of the option is e−r(τK −t) 1{τK <T } . f) From Relation (10.13), we find that the first hitting time τa of the level a by a µ-drifted Brownian motion (Wu + µu)u∈R+ satisfies −a − µu a − µu √ √ − e2µa Φ , u > 0, P(τa ≤ u) = Φ u u and by differentiation with respect to u this yields the probability density function fτa (u) =

2 a ∂ P(τa ≤ u) = √ e−(a−µu) /(2u) 1[0,∞) (u) ∂u 2πu3

of the first hitting time of level a by Brownian motion with drift µ. Given the relation 2

Su = St eσWu−t −(u−t)σ /2+(u−t)r = St e(Wu−t +(u−t)µ)σ ,

u ≥ t,

with µ = r/σ − σ/2, we find that (Su )u∈[t,∞) hits the level K at a time τK = t + τa , such that 2

SτK = St eσWτa −σ τa /2+rτa = St e(Wτa +µτa )σ = K, i.e. a = Wτa + µτa =

K 1 log . σ St

Therefore, the probability density function of the first hitting time τK of level K after time t by (Su )u∈[t,∞) is given by

265 September 15, 2022

Solutions Manual s 7−→ p

2π(s − t)3

with µ :=

1 σ

e−(a−(s−t)µ) /(2(s−t)) ,

σ2 r− 2

and a :=

s > t,

1 K log , σ x

given that St = x. Hence, for x ∈ (0, K) we have CdAm (t, T, x) = IE e−r(τK −t) 1{τK <T } | St = x wT 2 a = e−(s−t)r p e−(a−(s−t)µ) /(2(s−t)) ds t 2π(s − t)3 w T −t 2 a = e−rs √ e−(a−µs) /(2s) ds 0 2πs3 2 ! w T −t log(K/x) 1 σ2 K √ = exp −rs − 2 − r− s + log ds 0 2σ s 2 x σ 2πs3 (r/σ2 −1/2)±(r/σ2 +1/2) K = x 2 ! w T −t log(K/x) 1 σ2 K √ × exp − 2 ± r+ s + log ds 0 2σ s 2 x σ 2πs3 2r/σ2 w ∞ 2 1 K 1 x w ∞ −y2 /2 e dy + √ = √ e−y /2 dy y+ 2π K y− 2π x x (r + σ 2 /2)(T − t) + log(x/K) √ = Φ K σ T −t x −2r/σ2 −(r + σ 2 /2)(T − t) + log(x/K) √ + Φ , 0 < x < K, K σ T −t where y± =

1 σ T −t √

σ2 K ± r+ (T − t) + log , 2 x

and we used the decomposition K 1 σ2 K 1 σ2 K log = r+ s + log + − r+ s + log . x 2 2 x 2 2 x We check that

CdAm (T, T, K) = Φ(0) + Φ(0) = 1,

and 2

CdAm (T, T, x) =

266

x −2r/σ x Φ (−∞) + Φ (−∞) = 0, K K

x < K,

266 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition since t = T , which is consistent with the answers to Question (c). In addition, as T tends to infinity we have (r + σ 2 /2)(T − t) + log(x/K) x √ lim CdAm (t, T, x) = lim Φ T →∞ K T →∞ σ T −t x −2r/σ2 −(r + σ 2 /2)(T − t) + log(x/K) √ + lim Φ T →∞ K σ T −t x = , 0 < x < K, K which is consistent with the answer to Question (a) of Exercise 15.10. T=5 T=20

American binary call price

1.4 1.2 1 0.8 0.6 0.4 0.2 0

100

150

200

Underlying x

Fig. S.79: Finite expiration American binary call price map with K = 100. g) Starting from St ≥ K, the maximum possible payoff is clearly reached as soon as St hits the level K before the expiration date T , hence the discounted optimal payoff of the option is e−r(τK −t) 1{τK <T } . h) Using the notation and answer to Question (f), for x > K we find PdAm (t, T, x) = IE e−r(τK −t) 1{τK <T } | St = x w T −t 2 a = e−rs √ e−(a−µs) /2s) ds 0 2πs3 2 ! w T −t log(x/K) 1 σ2 x √ = exp −rs − 2 r− s + log ds 0 2σ s 2 K σ 2πs3 ( r2 − 12 )±( r2 + 12 ) σ K σ = x 2 ! w T −t log(x/K) 1 σ2 x √ × exp − 2 ∓ r+ s + log ds 0 2σ s 2 K σ 2πs3 2 1 x w ∞ −y2 /2 1 x 2r/σ w ∞ −y2 /2 = √ e dy + √ e dy y+ 2π K y− 2π K "

267 September 15, 2022

Solutions Manual =

x −(r + σ 2 /2)(T − t) − log(x/K) √ Φ K σ T −t x −2r/σ2 (r + σ 2 /2)(T − t) − log(x/K) √ Φ + , K σ T −t

with y± = We check that

1 σ T −t √

x > K,

σ2 x ∓ r+ (T − t) + log . 2 K

PdAm (T, T, K) = Φ(0) + Φ(0) = 1,

and 2

PdAm (T, T, x) =

x −2r/σ x Φ (−∞) + Φ (−∞) = 0, K K

0 < x < K,

since t = T , which is consistent with the answers to Question (c). In addition, as T tends to infinity we have x −(r + σ 2 /2)(T − t) − log(x/K) √ lim PdAm (t, T, x) = lim Φ T →∞ K T →∞ σ T −t x −2r/σ2 (r + σ 2 /2)(T − t) − log(x/K) √ lim Φ + T →∞ K σ T −t x −2r/σ2 = , x > K, K which is consistent with the answer to Question (b) of Exercise 15.10 T=5 T=20

American binary put price

1.4 1.2 1 0.8 0.6 0.4 0.2 0

100

150

200

250

Underlying x

Fig. S.80: Finite expiration American binary put price map with K = 100. i) The call-put parity does not hold for American binary options since for x ∈ (0, K) we have

268

268 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (r + σ 2 /2)(T − t) + log(x/K) x √ CdAm (t, T, x) + PdAm (t, T, x) = 1 + Φ K σ T −t x −2r/σ2 −(r + σ 2 /2)(T − t) + log(x/K) √ Φ + , K σ T −t while for x > K we find x −(r + σ 2 /2)(T − t) − log(x/K) √ CdAm (t, T, x) + PdAm (t, T, x) = 1 + Φ K σ T −t x −2r/σ2 (r + σ 2 /2)(T − t) − log(x/K) √ + Φ . K σ T −t Exercise 15.12 American forward Contracts. a) For all (bounded) stopping times τ ∈ [t, T ], since the discounted asset price process Seu u∈[t,∞) := e−(u−t)r Su )u∈[t,∞) is a martingale and the stopped process Seτ ∧u = e−(τ ∧u−t)r Sτ ∧u )u∈[t,∞) is also a maru∈[t,∞)

tingale by the Stopping Time Theorem 14.7, we have h i IE∗ e−r(τ −t) (K − Sτ ) Ft = K IE∗ e−r(τ −t) Ft − IE∗ e−r(τ −t) Sτ Ft = K IE∗ e−r(τ −t) Ft − IE∗ Seτ ∧T Ft = K IE∗ e−r(τ −t) Ft − Seτ ∧t = K IE∗ e−r(τ −t) Ft − Set = K IE∗ e−r(τ −t) Ft − St , and the above quantity is clearly maximized by taking τ = t. Hence we have f (t, St ) = sup IE∗ e−r(τ −t) (K − Sτ ) Ft = K − St , t≤τ ≤T τ stopping time

and the optimal strategy is to exercise immediately (or to avoid purchasing the option) at time t and price K due to the effect of time value of money when r > 0. b) Similarly to the above, we have IE∗ e−r(τ −t) (Sτ − K) Ft = IE∗ e−r(τ −t) Sτ Ft − K IE∗ e−r(τ −t) Ft = St − K IE∗ e−r(τ −t) Ft , since τ ∈ [t, T ] is bounded and (e−rt St )t∈R+ is a martingale. As the above quantity is clearly maximized by taking τ = T , we have

269 September 15, 2022

Solutions Manual f (t, St ) =

sup

t≤τ ≤T τ stopping time

IE∗ e−r(τ −t) (Sτ − K)

Ft = St − e−(T −t)r K,

and the optimal strategy is to wait until the maturity time T in order to exercise at price K, due to the effect of time value of money when r > 0. c) Regarding the perpetual American long forward contract, since the dis counted asset price process Seu u∈[t,∞) := e−(u−t)r Su )u∈[t,∞) is a martingale, by the Stopping Time Theorem 14.7, for all stopping times τ ≥ t we have∗ , IE∗ e−r(τ −t) (Sτ − K) Ft = IE∗ e−r(τ −t) Sτ Ft − K IE∗ e−r(τ −t) Ft = IE∗ Seτ Ft − K IE∗ e−r(τ −t) Ft = Set − K IE∗ e−r(τ −t) Ft = St − K IE∗ e−r(τ −t) Ft ≤ St ,

t ≥ 0.

On the other hand, for all fixed T > 0 we have IE∗ e−r(T −t) (ST − K) Ft = e−r(T −t) IE∗ [ST | Ft ] − e−r(T −t) IE∗ [K | Ft ] = St − e−r(T −t) K,

t ∈ [0, T ],

hence (St − e−r(T −t) K) ≤

sup

t≤τ ≤T τ stopping time

IE∗ e−r(τ −t) (Sτ − K) Ft ≤ St ,

T ≥ t,

and letting T → ∞ we get St = lim (St − e−r(T −t) K) T →∞ ≤ sup IE∗ e−r(τ −t) (Sτ − K) Ft τ ≥t τ stopping time

≤ St , hence we have f (t, St ) =

sup

τ ≥t τ stopping time

IE∗ e−r(τ −t) (Sτ − K) Ft = St ,

and the optimal strategy τ ∗ = +∞ is to wait indefinitely. Regarding the perpetual American short forward contract, we have ∗

using Fatou’s Lemma.

270

270 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition f (t, St ) =

sup

IE∗ e−r(τ −t) (K − Sτ ) Ft

sup

IE∗ e−r(τ −t) (K − Sτ )+

τ ≥t τ stopping time

≤

τ ≥t τ stopping time

= fL∗ (St ), with L∗ =

(S.15.68) 2r K<K 2r + σ 2

as defined in (15.12). On the other hand, for τ = τL∗ we have (K − SτL∗ ) = (K − L∗ ) = (K − L∗ )∗ since 0 < L∗ = 2Kr/(2r + σ 2 ) < K, hence fL∗ (St ) = IE∗ e−r(τL∗ −t) (K − SτL∗ )+ Ft = IE∗ e−r(τL∗ −t) (K − SτL∗ ) Ft ≤ sup IE∗ e−r(τ −t) (K − Sτ ) Ft τ ≥t τ stopping time

= f (t, St ), which, together with (S.15.68), shows that f (t, St ) = fL∗ (St ), i.e. the perpetual American short forward contract has same price and exercise strategy as the perpetual American put option. Exercise 15.13 a) We have 2

Yt = e−rt (S0 ert+σBbt −σ t/2 )−2r/σ −2r/σ 2 −rt−2r 2 t/σ 2 +2r B bt /σ+rt =S e 0

−2r/σ 2 2r B bt /σ−(2r/σ)2 t/2

= S0 and

Zt = e−rt St = S0 eσBbt −σ t/2 ,

t ≥ 0, t ≥ 0,

which are both martingales under P because they are standard geometric Brownian motions with respective volatilities σ and 2r/σ. b) Since (Yt )t∈R+ and (Zt )t∈R+ are both martingales and τL is a stopping time, we have ∗

271 September 15, 2022

Solutions Manual −2r/σ 2

= IE∗ [Y0 ] = IE∗ [YτL ] 2 = IE∗ e−rτL Sτ−2r/σ L 2 = IE∗ e−rτL L−2r/σ 2 = L−2r/σ IE∗ e−rτL ,

hence

x −2r/σ2 IE∗ e−rτL = L 2

if S0 = x ≥ L (note that in this case YτL ∧t remains bounded by L−2r/σ ), and S0 = IE∗ [Z0 ] = IE∗ [ZτL ] = IE∗ e−rτL SτL = IE∗ e−rτL L = L IE∗ e−rτL , hence

x IE∗ e−rτL = L if S0 = x ≤ L. Note that in this case ZτL ∧t remains bounded by L. c) We find IE e−rτL (K − SτL ) S0 = x = (K − L) IE∗ e−rτL S0 = x  K −L   0 < x ≤ L,  x L , (S.15.69) =  x −2r/σ2    (K − L) , x ≥ L. L d) By differentiating ∂ IE e−rτL (K − SτL ) S0 = x ∂L  −2r/σ2 2r K x   − 1 − 1 ,   L σ2 L =     − Kx , L2

0 < L < x, L > x,

and check that the minimum occurs for L∗ = x. e) The value L∗ = x shows that the optimal strategy for the American finite expiration short forward contract is to exercise immediately starting from S0 = x, which is consistent with the result of Exercise 15.12-(a), since given any stopping time τ upper bounded by T we have IE e−rτ (K−Sτ ) = K IE e−rτ −IE e−rτ Sτ = K IE e−rτ −S0 ≤ K−S0 . 272

272 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

Exercise 15.14 a) The option payoff equals (κ − St )p if St ≤ L. b) We have i h fL (St ) = IE∗ e−r(τL −t) ((κ − SτL )+ )p Ft h i = IE∗ e−r(τL −t) ((κ − L)+ )p Ft h i = (κ − L)p IE∗ e−r(τL −t) Ft . c) We have h i fL (x) = IE∗ e−r(τL −t) (κ − SτL )+ Ft = x  (κ − x)p , 0 < x ≤ L,    2 2r/σ =  L   (κ − L)p , x ≥ L. x d) By the differentiation

d (κ − x)p = −p(κ − x)p−1 we find dx

∂fL (x) 2r = 2 (κ − L)p ∂L σ L hence the condition

(S.15.70)

2r/σ2 2r/σ2 L L − p(κ − L)p−1 , x x

∂fL′ ∗ (x) = 0 reads ∂L |x=L∗

2r (κ − L∗ ) − p = 0, σ 2 L∗

or L∗ =

2r κ < κ. 2r + pσ 2

e) By (S.15.70) the price can be computed as  (κ − St )p , 0 < St ≤ L∗ ,    p −2r/σ2 f (t, St ) = fL∗ (St ) =  pσ 2 κ 2r + pσ 2 St   , St ≥ L∗ , 2r + pσ 2 2r κ using (14.13) as in the proof of Proposition 15.4, since the process u 7−→ e−ru fL∗ (Su ),

u ≥ t,

is a nonnegative supermartingale.

273 September 15, 2022

Solutions Manual Exercise 15.15 a) The option payoff is κ − (St )p . b) We have h i fL (St ) = E∗ e−r(τL −t) (κ − (SτL )p ) Ft h i = E∗ e−r(τL −t) (κ − Lp ) Ft h i = (κ − Lp )E∗ e−r(τL −t) Ft . c) We have h i fL (x) = E∗ e−r(τL −t) (κ − (SτL )p ) St = x

0 < x ≤ L,

 κ − xp ,   2

  (κ − Lp ) x −2r/σ , x ≥ L. L

d) We have 2

fL′ ∗ (L∗ ) = − i.e. or

2r (L∗ )−2r/σ −1 (κ − (L∗ )p ) = −p(L∗ )p−1 , 2 σ (L∗ )−2r/σ2 2r (κ − (L∗ )p ) = p(L∗ )p , σ2

L∗ =

2rκ 2r + pσ 2

1/p

< (κ)1/p .

(S.15.71)

Remark: We may also compute L by maximizing L 7−→ fL (x) for all fixed x. The derivative ∂fL (x)/∂L can be computed as 2r/σ2 ! ∂fL (x) ∂ L p = (κ − L ) ∂L ∂L x 2r/σ2 2r/σ2 2r L L = −pLp−1 + 2 L−1 (κ − Lp ) , x σ x ∗

and equating ∂fL (x)/∂L to 0 at L = L∗ yields −p(L∗ )p−1 +

2r ∗ −1 (L ) (κ − (L∗ )p ) = 0, σ2

which recovers (S.15.71). e) We have 274

274 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition

fL∗ (St ) =

 κ − (St )p ,   

0 < St ≤ L∗ , 2

(S )−2r/σ    (κ − (L∗ )p ) ∗t −2r/σ2 , St ≥ L∗ (L )  κ − (St )p , 0 < St ≤ L∗ ,  

2   σ p(S )−2r/σ2 (L∗ )p+2r/σ2 , S ≥ L∗ , t t 2r  p κ − (S ) , t    −2r/(pσ2 ) =  pσ 2 κ 2r + pσ 2 Stp   < κ, 2r + pσ 2 2r κ

0 < St ≤ L∗ , St ≥ L∗ ,

however we cannot conclude as in Exercise 15.14-(e) since the process u 7−→ e−ru fL∗ (Su ),

u ≥ t,

does not remain nonnegative when p > 1, so that (14.13) cannot be applied as in the proof of Proposition 15.4.

Chapter 16 Exercise 16.1 a) We have Xt Nt X0 (σ−η)Bt −(σ2 −η2 )t/2 = d e N0 2 2 X0 = (σ − η)e(σ−η)Bt −(σ −η )t/2 dBt N0 2 2 X0 + (σ − η)2 e(σ−η)Bt −(σ −η )t/2 dt 2N0 2 2 X0 2 (σ − η 2 )e(σ−η)Bt −(σ −η )t/2 dt − 2N0 Xt 2 Xt Xt =− (σ − η 2 )dt + (σ − η)dBt + (σ − η)2 dt 2Nt Nt 2Nt Xt Xt (σ − η)dBt = − η(σ − η)dt + Nt Nt Xt = (σ − η)(dBt − ηdt) Nt

bt = d dX

275 September 15, 2022

Solutions Manual = (σ − η)

Xt b b t dB bt , dBt = (σ − η)X Nt

b bt = dBt − ηdt is a standard Brownian motion under P. where dB b) By change of numéraire, we have b N0 (XT − λNT )+ = N0 Ib bT − λ)+ . IE[(XT − λNT )+ ] = IE E (X NT bt is a driftless geometric Brownian Next, by the result of Question (a), X b hence we have motion with volatility σ − η under P, √ ! √ ! b /λ) σ b /λ) σ b T log(X b T bT −λ)+ ] = X b0 Φ log(X √0 √0 + −λΦ − , Ib E[(X 2 2 σ b T σ b T by the Black-Scholes formula with zero interest rate and volatility paramb0 we eter σ b = σ − η. By multiplication by N0 and the relation X0 = N0 X conclude to (16.36), i.e. b (X bT − λ)+ IE (XT − λNT )+ = N0 IE b0 Φ(d+ ) − λN0 Φ(d− ) = N0 X = X0 Φ(d+ ) − λN0 Φ(d− ). c) We have σ b = σ − η. Exercise 16.2 a) By the Girsanov Theorem 16.7, the processes (1)

bt dB

(1)

= dBt −

1 1 (1) (1) (2) (1) (1) dNt • dBt = dBt − (2) dSt • dBt = dBt −ηρdt, Nt St

and (2)

bt dB

(2)

= dBt −

1 1 (2) (2) (2) (2) (2) dNt • dBt = dBt − (2) dSt • dBt = dBt − ηdt Nt St

b2 . are both standard Brownian motions (and martingales) under P b) We have ! (1) St (1) b dSt = d (2) St (1)

= 276

(2)

(1)

(2)

d eσBt −ηBt −(σ −η )t/2

276 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition (1)

(1)

(2)

eσBt −ηBt −(σ −η )t/2 (2) S0 σ2 η2 σ2 − η2 (1) (2) × σdBt + dt − ηdBt + dt − dt − σηρdt 2 2 2 (1) (1) (2) b b b = St σdBt − ηdBt . =

(1) c) We note that the driftless geometric Brownian motion (Sbt )t∈R can be written as (1) (1) c dSbt = σ bSbt dW t,

b2 . In order to dect )t∈R is a standard Brownian motion under P where (W termine σ b we note that ct • dW ct σ b2 dt = σ b2 dW (1) (1) b b dSt dSt • = (1) (1) b b S St t (1) bt(2) · σdBt(1) − ηdB bt(2) = σdBt − ηdB = (σ 2 + η 2 − 2σηρ)dt, hence σ b2 = σ 2 + η 2 − 2σηρ. We conclude by applying the change of numéraire formula + (1) (2) + (2) b(1) e−rT IE∗ ST − λST = S0 Ib E ST − λ and the Black-Scholes formula to the the driftless geometric Brownian (1) motion (Sbt )t∈R . Exercise 16.3 We have Nt = P (t, T ) and from (17.25) and the relations P (t, T ) = F (t, rt ) and P (t, S) = G(t, rt ) we find    dP (t, S) = rt dt + σ(t, rt ) ∂ log G(t, rt )dWt ,    P (t, S) ∂x   dP (t, T ) ∂ dNt   = = rt dt + σ(t, rt ) log F (t, rt )dWt .  Nt P (t, T ) ∂x By the Girsanov Theorem (16.12) we also have ct = dWt − dW

dNt ∂ • dW = dW − σ(t, r ) log F (t, rt )dt, t t t Nt ∂x

hence

277 September 15, 2022

Solutions Manual dP (t, S) ∂ ∂ ∂ ct . = rt dt+σ 2 (t, rt ) log F (t, rt ) log G(t, rt )dt+σ(t, rt ) log G(t, rt )dW P (t, S) ∂x ∂x ∂x Using the relation P (t, S) = G(t, rt ) we can also write dP (t, S) = rt P (t, S)dt+σ 2 (t, rt )

∂ ∂ ∂ ct . log F (t, rt ) G(t, rt )dt+σ(t, rt ) G(t, rt )dW ∂x ∂x ∂x

Exercise 16.4 Forward contract. Taking Nt := P (t, T ), t ∈ [0, T ], we have w T (P (T, S) − K) IE∗ exp − Ft rs ds (P (T, S) − K) Ft = Nt Ib E t NT (P (T, S) − K) b = P (t, T )IE Ft P (T, T ) = P (t, T )Ib E[P (T, S) − K | Ft ] = P (t, T )Ib E[P (T, S) | Ft ] − KP (t, T ) P (T, S) = P (t, T )Ib E | Ft − KP (t, T ) P (T, T ) P (t, S) = P (t, T ) − KP (t, T ) P (t, T ) = P (t, S) − KP (t, T ), since t 7−→

P (t, T ) P (t, S) = Nt P (t, T )

b The corresponding (static) is a martingale under the forward measure P. hedging strategy is given by buying one bond with maturity S and by short selling K units of the bond with maturity T . Remark: The above result can also be obtained by a direct argument using the tower property of conditional expectations: w T IE∗ exp − rs ds (P (T, S) − K) Ft t w w S T ∗ rs ds = IE exp − rs ds IE∗ exp − FT − K Ft t T w w T S = IE∗ exp − rs ds IE∗ exp − rs ds − K FT Ft t T w w S T ∗ ∗ = IE IE exp − rs ds − K exp − rs ds FT Ft t t w w S T = IE∗ exp − rs ds − K exp − rs ds Ft t

278

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Introduction to Stochastic Finance with Market Examples, Second Edition = P (t, S) − KP (t, T ),

t ∈ [0, T ].

Exercise 16.5 a) We choose Nt := St as numéraire because this allows us to write the option payoff as (ST (ST −K))+ = NT (ST −K)+ . In this case, the forward b satisfies measure P b NT dP ST = e−rT = e−rT , dP N0 S0 or b|F dP NT ST t = e−(T −t)r = e−(T −t)r , dP|Ft Nt St

0 ≤ t ≤ T.

b) By the change of numéraire formula of Proposition 16.5, the option price becomes e−(T −t)r IE∗ (ST (ST − K))+ Ft = IE∗ e−(T −t)r NT (ST − K)+ Ft = Nt Ib E (ST − K)+ Ft = St Ib E (ST − K)+ Ft . (S.16.72) c) In order to compute (S.16.72) it remains to determine the dynamics of b Since St = S0 eσBt +rt−σ2 t/2 , we have (St )t∈R+ under P. b 2 dP ST = e−rT = eσBT −σ T /2 , dP S0 bt := Bt −σt is a standard Brownian hence by the Girsanov Theorem 7.3, B b with motion under P, 2

ST = S0 erT +σBT −σ T /2 2 2 = S e(r+σ )T +σBbT −σ T /2 0

= St e(r+σ )(T −t)+(BbT −Bbt )σ−(T −t)σ /2 ,

0 ≤ t ≤ T.

d) According to the above, (S.16.72) becomes + b ST − K + Ft ST (ST − K) Ft = St IE + 2 2 = St Ib E S0 erT +σ T +σBbT −σ T /2 − K Ft 2 2 b S e(r+σ )(T −t)+(BbT −Bbt )σ−(T −t)σ /2 − K + F = S IE

e−(T −t)r IE∗

= St e(T −t)(r+σ ) Bl(St , K, r + σ 2 , σ, T − t),

0 ≤ t ≤ T,

since the Black-Scholes formula with interest rate r + σ 2 reads "

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Solutions Manual + 2 2 2 Ft E St e(r+σ )(T −t)+(BbT −Bbt )σ−(T −t)σ /2 − K e−(T −t)(r+σ ) Ib = Bl(St , K, r + σ 2 , σ, T − t),

0 ≤ t ≤ T.

Remarks: i) The option price can be rewritten using other Black-Scholes parametrizations, such as for example 2 St Bl St e(T −t)(r+σ ) , K, 0, σ, T − t , or

2 2 St e(T −t)(r+σ ) Bl St , Ke−(T −t)(r+σ ) , 0, σ, T − t ,

however we prefer to choose the simplest possibility. ii) Deflated (or forward) processes such as St /N1 = 1 or e−(T −t)r ∗ IE (ST (ST − K))+ Ft = Ib E (ST − K)+ Ft , Nt

0 ≤ t ≤ T,

b are martingales under the forward measure P. iii) This option can also be priced via an integral calculation instead of using change of numéraire, as follows: e−(T −t)r IE∗ [ST (ST − K)+ | Ft ] 2 = e−(T −t)r IE∗ St e(T −t)r+(BT −Bt )σ−(T −t)σ /2 2 ×(St e(T −t)r+(BT −Bt )σ−(T −t)σ /2 − K)+ Ft 2 2 = St e−(T −t)σ /2 IE∗ (St e(T −t)r+2(BT −Bt )σ−(T −t)σ /2 − Ke(BT −Bt )σ )+ Ft 2 2 = St e−(T −t)σ /2 IE∗ (xe(T −t)r+2(BT −Bt )σ−(T −t)σ /2 − Ke(BT −Bt )σ )+ x=St 2 = St e−(T −t)σ /2 IE∗ (em(x)+2X − KeX )+ x=St , 0 ≤ t ≤ T, where X ≃ N (0, v 2 ) with v 2 = (T − t)σ 2 and m(x) = (T − t)r − (T − t)σ 2 /2 + log x. Next, we note that IE

em+2X − KeX

=√ =√

1 2πv 2 em

2πv 2

=√

1 2πv 2

w∞

−∞

em+2x − Kex

w∞

(em+2x − Kex )e−x /(2v ) dx

w∞

e2x−x /(2v ) dx − √

−m+log K

+ −x2 /(2v2 ) e dx

K 2πv 2

w∞

−m+log K

ex−x /(2v ) dx

2 2 2 2 2 2 em+2v w ∞ Kev /2 w ∞ = √ e−(2v −x) /(2v ) dx − √ e−(v −x) /(2v ) dx 2 −m+log K 2π −m+log K 2πv 2 2 2 2 em+2v w ∞ Kev /2 w ∞ −x2 /(2v 2 ) √ = √ e dx − e−x /(2v ) dx 2πv 2 −2v2 −m+log K 2πv 2 −v2 −m+log K 2

280

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Introduction to Stochastic Finance with Market Examples, Second Edition 2 2 2 em+2v w ∞ Kev /2 w ∞ −x2 /2 √ e e−x /2 dx = √ dx − 2π (−2v2 −m+log K)/v 2π (−v2 −m+log K)/v 2 2 m − log K m − log K = em+2v Φ 2v + − Kev /2 Φ v + , v v

hence + e−(T −t)r IE∗ ST ST − K Ft + 2 = St e−(T −t)σ /2 IE∗ em(x)+2X − KeX x=St (T − t)(r + σ 2 ) + (T − t)σ 2 /2 + log(St /K) 2 (T −t)(r+σ 2 ) √ = St e Φ σ T −t (T − t)(r + σ 2 ) − (T − t)σ 2 /2 + log(St /K) √ −KSt Φ σ T −t 2

= St e(T −t)(r+σ ) Bl(St , K, r + σ 2 , σ, T − t),

0 ≤ t ≤ T.

Exercise 16.6 2

a) Knowing that St = S0 eσWt +rt−σ t/2 , we check that the discounted numéraire process 2

e−rt Nt := Stn e−(n−1)nσ t/2−nrt 2

= S0n enσWt +nrt−nσ t/2 e−(n−1)nσ t/2−nrt 2

= S0 enσWt −(nσ) t/2 ,

0 ≤ t ≤ T,

is a martingale under P , and ∗

b 2 2 dP NT Sn = e−rT = Tn e−nσ T −nrT = enσWT −(nσ) T /2 . dP∗ N0 S0

(S.16.73)

ct := b) From Equation (S.16.73) and the Girsanov theorem, the process W b and we have Wt − σnt is a standard Brownian motion under P, 2

St = S0 ert+σWt −σ t/2 b t +nσt)−σ2 t/2 = S ert+σ(W 0

b t +σ t/2 , = S0 e(r+nσ )t+σW

t ≥ 0.

c) We have e−(T −t)r IE∗ STn 1{ST ≥K} Ft = IE∗ e−(T −t)r STn 1{ST ≥K} Ft 2 = e(n−1)nσ T /2+(n−1)rT IE∗ NT 1{ST ≥K} Ft "

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Solutions Manual 2 b 1{S ≥K} Ft = e(n−1)nσ T /2+(n−1)rT Nt IE T 2

b T ≥ K | Ft ) = e(n−1)nσ T /2+(n−1)rT Nt P(S 2 log(St /K) + (r + (n − 1/2)σ 2 )(T − t) √ = e(n−1)nσ T /2+(n−1)rT Nt Φ σ T −t log(St /K) + (r + (n − 1/2)σ 2 )(T − t) n (n−1)nσ 2 (T −t)/2+(n−1)r(T −t) √ = St e Φ , σ T −t 0 ≤ t ≤ T , n ≥ 0. d) The power call option with payoff (STn − K n )+ is priced at time t ∈ [0, T ] as + e−(T −t)r IE∗ STn − K n Ft 2 log(St /K) + (r + (n − 1/2)σ 2 )(T − t) √ = Stn e(n−1)nσ (T −t)/2+(n−1)r(T −t) Φ σ T −t log(St /K) + (r − σ 2 /2)(T − t) n −(T −t)r √ −K e Φ , n ≥ 1. σ T −t Exercise 16.7 Bond options. a) Itô’s formula yields P (t, S) P (t, S) S ζ (t) − ζ T (t) (dWt − ζ T (t)dt) d = P (t, T ) P (t, T ) P (t, S) S ct , = ζ (t) − ζ T (t) dW (S.16.74) P (t, T ) b by the Girsanov ct )t∈R is a standard Brownian motion under P where (W + Theorem 7.3. b) From (S.16.74) or (19.7) we have w wt t P (0, S) P (t, S) 2 cs − 1 ζ S (s) − ζ T (s) dW = exp ζ S (s) − ζ T (s) ds , 0 P (t, T ) P (0, T ) 2 0 hence w wu u P (u, S) P (t, S) 2 cs − 1 = exp ζ S (s) − ζ T (s) dW ζ S (s) − ζ T (s) ds , t P (u, T ) P (t, T ) 2 t t ∈ [0, u], and for u = T this yields w wT T P (t, S) 2 cs − 1 exp ζ S (s) − ζ T (s) dW ζ S (s) − ζ T (s) ds , P (T, S) = t P (t, T ) 2 t b denote the forward measure associated to the since P (T, T ) = 1. Let P numéraire 282

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Introduction to Stochastic Finance with Market Examples, Second Edition Nt := P (t, T ),

0 ≤ t ≤ T.

c) For all S ≥ T > 0 we have i h rT IE e− t rs ds (P (T, S) − K)+ Ft # " + P (t, S) 1wT S 2 Ft ζ (s) − ζ T (s) ds − K = P (t, T )Ib E exp X − P (t, T ) 2 t X+m(t,T,S) + b Ft , = P (t, T )IE e −K where X is a centered Gaussian random variable with variance wT 2 v 2 (t, T, S) = ζ S (s) − ζ T (s) ds t

given Ft , and 1 P (t, S) m(t, T, S) = − v 2 (t, T, S) + log . 2 P (t, T ) Recall that when X is a centered Gaussian random variable with variance v 2 , the expectation of (em+X − K)+ is given, as in the standard BlackScholes formula, by 2

IE[(em+X − K)+ ] = em+v /2 Φ(v + (m − log K)/v) − KΦ((m − log K)/v), where Φ(z) =

−∞

2 dy e−y /2 √ , 2π

z ∈ R,

denotes the Gaussian cumulative distribution function and for simplicity of notation we dropped the indices t, T, S in m(t, T, S) and v 2 (t, T, S). Consequently we have h rT i IE e− t rs ds (P (T, S) − K)+ Ft v 1 P (t, S) v 1 P (t, S) + log − KP (t, T )Φ − + log . = P (t, S)Φ 2 v KP (t, T ) 2 v KP (t, T ) d) The self-financing hedging strategy that hedges the bond option is obtained by holding a (possibly fractional) quantity v 1 P (t, S) Φ + log 2 v KP (t, T ) of the bond with maturity S, and by shorting a quantity

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Solutions Manual 1 P (t, S) v KΦ − + log 2 v KP (t, T ) of the bond with maturity T . Exercise 16.8 a) The process

e−rt S2 (t) = S2 (0)eσ2 Wt +(µ−r)t

is a martingale if r−µ=

1 2 σ . 2 2

b) We note that 2

e−rt Xt = e−rt e(r−µ)t−σ1 t/2 S1 (t) 2

= e−rt e(σ2 −σ1 )t/2 S1 (t) −µt−σ12 t/2

S1 (t)

µt−σ12 t/2 σ1 Wt +µt

= S1 (0)e

= S1 (0)eσ1 Wt −σ1 t/2 is a martingale, where 2

Xt = e(r−µ)t−σ1 t/2 S1 (t) = e(σ2 −σ1 )t/2 S1 (t). c) By (16.38) we have Xt b X(t) = Nt 2

= e(σ2 −σ1 )t/2

S1 (t) S2 (t)

S1 (0) (σ22 −σ12 )t/2+(σ1 −σ2 )Wt e S2 (0) S1 (0) (σ22 −σ12 )t/2+(σ1 −σ2 )W b t +σ2 (σ1 −σ2 )t = e S2 (0) S1 (0) (σ1 −σ2 )W b t +σ2 σ1 t−(σ22 +σ12 )t/2 = e S2 (0) S1 (0) (σ1 −σ2 )W b t −(σ1 −σ2 )2 t/2 , = e S2 (0) =

where

ct := Wt − σ2 t W

b defined by is a standard Brownian motion under the forward measure P 284

284 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition rT b NT dP = e− 0 rs ds dP N0 −rT S2 (T ) =e S2 (0)

= e−rT eσ2 WT +µT = eσ2 WT +(µ−r)T 2

= eσ2 WT −σ2 t/2 . 2 2 b d) Given that Xt = e(σ2 −σ1 )t/2 S1 (t) and X(t) = Xt /Nt = Xt /S2 (t), we have 2

e−rT IE[(S1 (T ) − κS2 (T ))+ ] = e−rT IE[(e−(σ2 −σ1 )T /2 XT − κS2 (T ))+ ] −(σ22 −σ12 )T /2

= e−rT e

(σ22 −σ12 )T /2

IE[(XT − κe S2 (T ))+ ] (σ22 −σ12 )T /2 + b IE[(XT − κe ) ]

−(σ22 −σ12 )T /2 b

= S2 (0)e

b T −(σ1 −σ2 ) T /2 − κe(σ2 −σ1 )T /2 )+ ] b X b0 e(σ1 −σ2 )W = S2 (0)e−(σ2 −σ1 )T /2 IE[( 2 2 b0 Φ0 (T, X b0 ) − κe(σ22 −σ12 )T /2 Φ0 (T, X b0 ) = S2 (0)e−(σ2 −σ1 )T /2 X −

2 2 b0 Φ0 (T, X b0 ) = S2 (0)e−(σ2 −σ1 )T /2 X + 2

b0 ) −κS2 (0)e−(σ2 −σ1 )T /2 e(σ2 −σ1 )T /2 Φ0− (T, X −(σ22 −σ12 )T /2

= X0 e

b0 ) − κS2 (0)Φ0− (T, X b0 ) Φ0+ (T, X

−(σ22 −σ12 )T /2

= S1 (0)e

b0 ) − κS2 (0)Φ0− (T, X b0 ), Φ0+ (T, X

where (σ − σ2 )2 − (σ22 − σ12 ) √ log(x/κ) √ + 1 T 2|σ1 − σ2 | |σ1 − σ2 | T  √ log(x/κ)   √ + σ1 T , σ1 > σ2 , Φ    |σ1 − σ2 | T =  √  log(x/κ)   √ − σ1 T , σ1 < σ2 , Φ |σ1 − σ2 | T

log(x/κ) (σ − σ2 )2 + (σ22 − σ12 ) √ √ − 1 T 2|σ1 − σ2 | |σ1 − σ2 | T  √ log(x/κ)   √ + σ2 T , σ1 > σ2 , Φ    |σ1 − σ2 | T =  √  log(x/κ)   √ − σ2 T , σ1 < σ2 , Φ |σ1 − σ2 | T

Φ0+ (T, x) = Φ

and Φ0− (T, x) = Φ

285 September 15, 2022

Solutions Manual if σ1 ̸= σ2 . In case σ1 = σ2 , we find e−rT IE[(S1 (T ) − κS2 (T ))+ ] = e−rT IE[S1 (T )(1 − κS2 (0)/S1 (0))+ ] = (1 − κS2 (0)/S1 (0))+ e−rT IE[S1 (T )] = (S1 (0) − κS2 (0))1{S1 (0)>κS2 (0)} . Exercise 16.9 We have e−(T −t)r IE∗ 1{RT ≥κ} Rt = e−(T −t)r P∗ RT ≥ κ Rt f 2 = e−(T −t)r P∗ Rt e(WT −Wt )σ+(r−r )(T −t)−(T −t)σ /2 ≥ κ Rt f 2 = e−(T −t)r P∗ xe(WT −Wt )σ+(r−r )(T −t)−(T −t)σ /2 ≥ κ x=Rt (r − rf )(T − t) − (T − t)σ 2 /2 − log(κ/Rt ) √ = e−(T −t)r Φ , σ T −t after applying the hint provided, with η 2 := (T − t)σ 2

and µ := (r − rf )(T − t) − (T − t)σ 2 /2.

Remark: Binary options are often proposed at the money, i.e. κ = Rt , with a short time to maturity, for example the small value T − t ≃ 30 seconds = 0.000000951 = 9.51 × 10−7 year−1 , in which case we have e−(T −t)r IE∗ 1{RT ≥κ} Rt = e−(T −t)r Φ

r − rf σ − σ 2

√

T −t

≃ Φ(0) 1 = . 2 Taking for example r − rf = 0.02 = 2% and σ = 0.3 = 30%, we have r − rf σ √ 0.02 0.3 p − T −t= − 9.51 × 10−7 = −0.000081279 σ 2 0.3 2 and e−(T −t)r IE∗ 1{RT ≥κ} Rt = e−(T −t)r Φ(−0.000081279) = e−r×0.000000951 × 0.499968 = 0.49996801 1 ≃ , 2 286

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Introduction to Stochastic Finance with Market Examples, Second Edition with r = 0.02 = 2%. Exercise 16.10 a) It suffices to check that the definition of (WtN )t∈R+ implies the correlation identity dWtS • dWtN = ρdt by Itô’s calculus. b) We let q σ bt =

and dWtX =

(σtS )2 − 2ρσtR σtS + (σtR )2

p σN σtS − ρσtN dWtS − 1 − ρ2 t dWt , σ bt σ bt

t ≥ 0,

which defines a standard Brownian motion under P due to the definition of σ bt . ∗

Exercise 16.11 p a) We have σ b = (σ S )2 − 2ρσ R σ S + (σ R )2 . et = e−rt Xt = e(a−r)t St /Rt , t ≥ 0, we have b) Letting X " # + h i + ST ∗ IE −κ Ft = e−aT IE∗ XT − eaT κ Ft RT + eT − e(a−r)T κ = e−(a−r)T IE∗ X Ft b2 /2)(T − t) 1 St et Φ (r − a +√σ = e−(a−r)T X + √ log κRt σ b T −t σ b T −t 2 1 S (r − a − σ b /2)(T − t) t √ + √ log −κe(a−r)T Φ κRt σ b T −t σ b T −t St (r−a)(T −t) (r − a + σ b2 /2)(T − t) 1 St √ = e Φ + √ log Rt κRt σ b T −t σ b T −t (r − a − σ b2 /2)(T − t) 1 St √ −κΦ + √ log , κRt σ b T −t σ b T −t hence the price of the quanto option is " # + ST e−(T −t)r IE∗ −κ Ft RT St −a(T −t) (r − a + σ b2 /2)(T − t) 1 St √ + √ log = e Φ Rt κRt σ b T −t σ b T −t 2 (r − a − σ b /2)(T − t) 1 St √ −κe−r(T −t) Φ + √ log . κRt σ b T −t σ b T −t "

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Solutions Manual

Chapter 17 Exercise 17.1 a) We have wt a drt = r0 de−bt + d 1 − e−bt + σd e−bt ebs dBs 0 b wt wt −bt −bt −bt bs = −br0 e dt + ae dt + σe d e dBs + σ ebs dBs de−bt 0 0 wt −bt −bt −bt bt bs = −br0 e dt + ae dt + σe e dBt − σb e dBs e−bt dt 0 wt = −br0 e−bt dt + ae−bt dt + σdBt − σb ebs dBs e−bt dt 0 a = −br0 e−bt dt + ae−bt dt + σdBt − b rt − r0 e−bt − 1 − e−bt dt b = (a − brt )dt + σdBt , which shows that rt solves (17.46). b) We note that wt a 1 − e−bt + σ e−(t−u)b dBu 0 b a = r0 e−bs e−(t−s)b + e−(t−s)b 1 − e−bs b ws wt a + 1 − e−(t−s)b + σe−(t−s)b e−(s−u)b dBu + σ e−(t−u)b dBu 0 s b wt a = rs e−(t−s)b + 1 − e−(t−s)b + σ e−(t−u)b dBu , 0 ≤ s ≤ t. s b

rt = r0 e−bt +

Hence, assuming that rs has the N (a/b, σ 2 /(2b)) distribution, the distribution of rt is Gaussian with mean a 1 − e(t−s)b b a a = e−(t−s)b + 1 − e(t−s)b b b a = , b

IE[rt ] = e−(t−s)b IE[rs ] +

and variance wt a Var[rt ] = Var rs e−(t−s)b + 1 − e(t−s)b + σ e−(t−u)b dBu s b wt = Var rs e−(t−s)b + σ e−(t−u)b dBu s w t −(t−s)b = Var rs e + Var σ e−(t−u)b dBu s

288

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Introduction to Stochastic Finance with Market Examples, Second Edition w t = e−2(t−s)b Var rs + σ 2 Var e−(t−u)b dBu s

−2(t−s)b σ

+σ e du s 2b w 2 t−s σ = e−2(t−s)b + σ2 e−2bu du 0 2b σ2 , t ≥ 0. = 2b =e

−2(t−u)b

Exercise 17.2 a) The zero-coupon bond price P (t, T ) in the Vasicek model is given by log P (t, T ) = A(T − t) + rt C(T − t), where C(T − t) := −

0 ≤ t ≤ T,

1 1 − e−(T −t)b , b

and A(T − t) :=

4ab − 3σ 2 σ 2 − 2ab + (T − t) 4b3 2b2 2 σ − ab −(T −t)b σ2 + e − 3 e−2(T −t)b b3 4b

0 ≤ t ≤ T. (S.17.75)

Since limT →∞ C(T − t)/(T − t) = 0 and lim A(T − t)/(T − t) = (σ 2 − 2ab)/(2b2 ),

T →∞

we find r∞ = − lim

T →∞

σ 2 − 2ab a σ2 log P (t, T ) =− = − 2. 2 T −t 2b b 2b

b) We have log

P (t, T ) = log P (t, T ) − log P (0, T ) P (0, T ) = A(T − t) − A(T ) + rt C(T − t) − r0 C(T ) 2 σ 2 − 2ab σ − ab σ 2 −(T −t)b −(T −t)b = −t − e + e 2b2 b3 4b3 2 2 r0 σ σ − ab r t +e−bT − − 1 − e−(T −t)b + 1 − e−bT , 4b3 b3 b b

hence "

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Solutions Manual lim log

T →∞

P (t, T ) = P (0, T )

a σ2 − 2 b 2b

t−

rt − r0 rt − r0 =− + r∞ t, b b

and∗ lim log

T →∞

2 2 P (t, T ) = e−(rt −r0 )/b+t(a/b−σ /(2b )) = e−(rt −r0 )/b+r∞ t , P (0, T )

which shows that the yield of the long-bond return is the asymptotic bond yield r∞ . Remark: By Relations (17.32)-(17.34), the Vasicek bond price P (t, T ) can be rewritten in terms of the asymptotic bond yield r∞ as 2

P (t, T ) = e−(T −t)r∞ +(rt −r∞ )C(T −t)−σ C (T −t)/(4b) ,

0 ≤ t ≤ T,

see e.g. Relation (3.12) in Brody et al. (2018). Exercise 17.3 An estimator of σ can be obtained from the orthogonality relation n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

n−1 X − σ 2 ∆t(r̃tl )2γ = σ 2 (r̃tl )2γ (Zl )2 − ∆t

l=0

≃ 0, which is due to the independence of ttl and Zl , l = 0, . . . , n − 1, and yields n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

σ b2 = l=0 ∆t

n−1 X

2 .

(r̃tl )2γ

l=0

Regarding the estimation of γ, we can combine the above relation with the second orthogonality relation n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

− σ 2 ∆t(r̃tl )2γ r̃tl

l=0

= σ2

n−1 X

(r̃tl )2γ+1 (Zl )2 − ∆t

l=0

≃ 0. One may also attempt to minimize the residual ∗

The log function is continuous on (0, ∞).

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Introduction to Stochastic Finance with Market Examples, Second Edition n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

− σ 2 ∆t(r̃tl )2γ

l=0

by equating the following derivatives to zero, as n−1

2 2 ∂ X r̃tl+1 − a∆t − (1 − b∆t)r̃tl − σ 2 ∆t(r̃tl )2γ ∂σ l=0

= −4σ

n−1 X

(r̃tl )2γ

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

− σ 2 ∆t(r̃tl )2γ

l=0

= 0, hence n−1 X

(r̃tl )2γ r̃tl+1 − a∆t − (1 − b∆t)r̃tl

− σ 2 ∆t

l=0

n−1 X

(r̃tl )4γ = 0,

l=0

which yields n−1 X

(r̃tl )2γ r̃tl+1 − a∆t − (1 − b∆t)r̃tl

σ b2 = l=0 ∆t

n−1 X

2 (S.17.76)

. (r̃tl )4γ

l=0

We also have n−1

2 2 ∂ X r̃tl+1 − a∆t − (1 − b∆t)r̃tl − σ 2 ∆t(r̃tl )2γ ∂γ l=0

= −4σ 2 ∆t

n−1 X

(r̃tl )2γ

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

− σ 2 ∆t(r̃tl )2γ log r̃tl

l=0

= 0, which yields n−1 X

σ b = 2

(r̃tl )2γ r̃tl+1 − a∆t − (1 − b∆t)r̃tl

l=0

∆t

n−1 X

log r̃tl ,

(S.17.77)

(r̃tl )4γ log r̃tl

l=0

and shows that γ can be estimated by matching Relations (S.17.76) and (S.17.77), i.e. "

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Solutions Manual n−1 X

(r̃tl )2γ r̃tl+1 − a∆t − (1 − b∆t)r̃tl

l=0 n−1 X

(r̃tl )4γ

l=0 n−1 X

(r̃tl )2γ r̃tl+1 − a∆t − (1 − b∆t)r̃tl

= l=0

n−1 X

log r̃tl .

(r̃tl )4γ log r̃tl

l=0

Remarks. i) Estimators of a and b can be obtained by minimizing the residual n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl

l=0

as in the Vašíček (1977) model, i.e. from the equations n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl = 0

l=0

and

n−1 X

r̃tl+1 − a∆t − (1 − b∆t)r̃tl r̃tl = 0.

l=0

ii) Instead of using the (generalised) method of moments, parameter estimation for stochastic differential equations can be achieved by maximum likelihood estimation. Exercise 17.4 a) We have rt = r0 + at + σBt , and F (t, rt ) = F (t, r0 + at + σBt ), hence by Proposition 17.2 the PDE satisfied by F (t, x) is −xF (t, x) +

∂F 1 ∂2F ∂F (t, x) + a (t, x) + σ 2 2 (t, x) = 0, ∂t ∂x 2 ∂x

(S.17.78)

with terminal condition F (T, x) = 1. b) Using the relation rt = r0 + at + σBt and the fact that the stochastic wT integral (T − s)dBs is independent of Ft , we have t

292

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Introduction to Stochastic Finance with Market Examples, Second Edition w T Ft F (t, rt ) = IE∗ exp − rs ds t wT wT ∗ Ft = IE exp −r0 (T − t) − a sds − σ Bs ds t t wT 2 2 Ft = IE∗ e−r0 (T −t)−a(T −t )/2 exp −(T − t)σBt − σ (T − s)dBs t wT ∗ −r0 (T −t)−a(T 2 −t2 )/2−(T −t)σBt (T − s)dBs Ft IE exp −σ =e t wT ∗ −r0 (T −t)−a(T −t)(T +t)/2−(T −t)σBt =e IE exp −σ (T − s)dBs t σ2 w T = exp −(T − t)rt − a(T − t)2 /2 + (T − s)2 ds 2 t = exp −(T − t)rt − a(T − t)2 /2 + (T − t)3 σ 2 /6 , hence F (t, x) = exp −(T − t)x − a(T − t)2 /2 + (T − t)3 σ 2 /6 . Note that the PDE (S.17.78) can also be solved by looking for a solution of the form F (t, x) = eA(T −t)+xC(T −t) , in which case one would find A(s) = −as2 /2 + σ 2 s3 /6 and C(s) = −s. c) We check that the function F (t, x) of Question (b) satisfies the PDE (S.17.78) of Question (a), since F (T, x) = 1 and σ2 −xF (t, x) + x + a(T − t) − (T − t)2 F (t, x) − a(T − t)F (t, x) 2 σ2 + (T − t)2 F (t, x) = 0. 2 Exercise 17.5 We check from (17.51) and the differentiation rule d f (t)dt that w t 1 drt = αβd St du + r0 dSt 0 Su wt 1 wt 1 = αβSt d du + αβ dudSt + r0 dSt 0 Su 0 Su w t St St dSt = αβ dt + αβ du + r0 dSt 0 Su St St dSt = αβdt + (rt − r0 St ) + r0 dSt St dSt = αβdt + rt St = αβdt + rt (−βdt + σdBt ) = β(α − rt )dt + σdBt , "

wt 0

f (u)du =

t ≥ 0. 293 September 15, 2022

Solutions Manual Exercise 17.6 a) Applying the Itô formula 1 df (rt ) = f ′ (rt )drt + f ′′ (rt )(drt )2 2 to the function f (x) = x2−γ with f ′ (x) = (2 − γ)x1−γ and f ′′ (x) = (2 − γ)(1 − γ)x−γ , we have dRt = drt2−γ = df (rt )

1 = f ′ (rt )drt + f ′′ (rt )(drt )2 2 γ/2 = f ′ (rt ) (βrtγ−1 + αrt )dt + σrt dBt drt + 2 1 ′′ γ/2 f (rt ) (βrtγ−1 + αrt )dt + σrt dBt 2 σ 2 ′′ γ/2 = f ′ (rt ) (βrtγ−1 + αrt )dt + σrt dBt drt + f (rt )rtγ dt 2 σ2 γ/2 = (2 − γ)rt1−γ (βrtγ−1 + αrt )dt + σrt dBt + (2 − γ)(1 − γ)rtγ rt−γ dt 2 σ2 1−γ/2 = (2 − γ)(β + αrt2−γ )dt + (2 − γ)(1 − γ)dt + σ(2 − γ)rt dBt 2 2 p σ = (2 − γ) β + (1 − γ) + αRt dt + (2 − γ)σ Rt dBt . 2 We conclude that the process Rt = rt2−γ follows the CIR equation p dRt = b(a − Rt )dt + η Rt dBt with initial condition R0 = r02−γ and coefficients 1 σ2 β + (1 − γ) , and η = (2 − γ)σ. b = (2 − γ)α, a = α 2 b) By Itô’s formula and the relation P (t, T ) = F (t, rt ), t ∈ [0, T ], we have rt rt rt d e− 0 rs ds P (t, T ) = −rt e− 0 rs ds P (t, T )dt + e− 0 rs ds dP (t, T ) rt

= −rt e− 0 rs ds F (t, rt )dt + e− 0 rs ds dF (t, rt ) rt rt rt ∂F ∂F = −rt e− 0 rs ds F (t, rt )dt + e− 0 rs ds (t, rt )dt + e− 0 rs ds (t, rt )drt ∂t ∂x 294

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Introduction to Stochastic Finance with Market Examples, Second Edition rt

+ e− 0 rs ds

1 ∂2F (t, rt )(drt )2 2 ∂x2

rt rt ∂F −(1−γ) γ/2 = −rt e− 0 rs ds F (t, rt )dt + e− 0 rs ds (t, rt )((βrt + αrt )dt + σrt dBt ) ∂x rt ∂F 1 ∂2F (t, rt ) + σ 2 rtγ 2 (t, rt ) dt + e− 0 rs ds ∂t 2 ∂x r γ/2 ∂F − 0t rs ds σrt =e (t, rt )dBt ∂x rt ∂F −(1−γ) (t, rt ) + αrt ) + e− 0 rs ds −rt F (t, rt ) + (βrt ∂x 2 1 ∂ F ∂F + σ 2 rtγ 2 (t, rt ) + (t, rt ) dt. (S.17.79) 2 ∂x ∂t rt

Given that t 7−→ e− 0 rs ds P (t, T ) is a martingale, the above expression (S.17.79) should only contain terms in dBt and all terms in dt should vanish inside (S.17.79). This leads to the identities  2   r F (t, r ) = (βr−(1−γ) + αr ) ∂F (t, r ) + 1 σ 2 rγ ∂ F (t, r ) + ∂F (t, r )   t t t t t t t t  ∂x 2 ∂x2 ∂t    rt rt  ∂F    d e− 0 rs ds P (t, T ) = σe− 0 rs ds rtγ/2 (t, rt )dBt , ∂x and to the PDE xF (t, x) =

∂F ∂F σ2 γ ∂ 2 F (t, x) + (βx−(1−γ) + αx) (t, x) + x (t, x). ∂t ∂x 2 ∂x2

Exercise 17.7

a) The discounted bond prices process e− 0 rs ds F (t, rt ) is a martingale, and we have rt d e− 0 rs ds F (t, rt ) rt ∂F ∂F σ2 ∂ 2 F 2 = e− 0 rs ds −rt F (t, rt )dt + (t, rt )dt + (t, rt )drt + (t, r )(dr ) t t ∂t ∂x 2 ∂x2 r √ ∂F ∂F − 0t rs ds (t, rt )dt + (t, rt )(−art dt + σ rt dBt ) =e −rt F (t, rt )dt + ∂t ∂x 2 2 r − 0t rs ds σ ∂ F +rt e (t, rt )dt, 2 ∂x2 hence F (t, x) satisfies the affine PDE

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Solutions Manual −xF (t, x) +

∂F ∂F σ2 ∂ 2 F (t, x) = 0. (t, x) − ax (t, x) + x ∂t ∂x 2 ∂x2

(S.17.81)

b) Plugging F (t, x) = eA(T −t)+xC(T −t) into the PDE (S.17.81) shows that 2 eA(T −t)+xC(T −t)

−x − A′ (T − t) − xC ′ (T − t) − axC(T − t) +

σ x 2 C (T − t) 2

= 0.

Taking successively x = 0 and x = 1 in the above relation then yields the two equations  ′ A (T − t) = 0,   2   −1 − C ′ (T − t) − aC(T − t) + σ C 2 (T − t) = 0. 2

Remark: The initial condition A(0) = 0 shows that A(s) = 1, and it can be shown from the condition C(0) = 0 that C(T − t) = with γ = (2006).

√

2(1 − eγ(T −t) ) , 2γ + (a + γ)(eγ(T −t) − 1)

t ∈ [0, T ],

a2 + 2σ 2 , see e.g. Eq. (3.25) page 66 of Brigo and Mercurio

Exercise 17.8 a) The payoff of the convertible bond is given by max(αSτ , P (τ, T )). b) We have max(αSτ , P (τ, T )) = P (τ, T )1{αSτ ≤P (τ,T ))} + αSτ 1{αSτ >P (τ,T ))} = P (τ, T ) + (αSτ − P (τ, T ))1{αSτ >P (τ,T ))}

= P (τ, T ) + (αSτ − P (τ, T ))+ = P (τ, T ) + α(Sτ − P (τ, T )/α)+ , where the latter European call option payoff has the strike price K := P (τ, T )/α. c) From the Markov property applied at time t ∈ [0, τ ], we will write the corporate bond price as a function C(t, St , rt ) of the underlying asset price and interest rate, hence we have i h rτ C(t, St , rt ) = IE∗ e− t rs ds max(αSτ , P (τ, T )) Ft . The martingale property follows from the equalities 296

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Introduction to Stochastic Finance with Market Examples, Second Edition i h rτ rt rt e− 0 rs ds C(t, St , rt ) = e− 0 rs ds IE∗ e− t rs ds max(αSτ , P (τ, T )) Ft i h rτ = IE∗ e− 0 rs ds max(αSτ , P (τ, T )) Ft . d) We have rt d e− 0 rs ds C(t, St , rt ) rt

= −rt e− 0 rs ds C(t, St , rt )dt + e− 0 rs ds

∂C (t, St , rt )dt ∂t

∂C (1) (t, St , rt )(rSt dt + σSt dBt ) ∂x rt ∂C (2) +e− 0 rs ds (t, St , rt )(γ(t, rt )dt + η(t, St )dBt ) ∂y rt rt 1 ∂2C σ2 ∂ 2 C (t, St , rt )dt +e− 0 rs ds St2 2 (t, St , rt )dt + e− 0 rs ds η 2 (t, rt ) 2 ∂x 2 ∂y 2 2 rt ∂ C +ρσSt η(t, rt )e− 0 rs ds (t, St , rt )dt. (S.17.82) ∂x∂y rt The martingale property of e− 0 rs ds C(t, St , rt ) t∈R+ shows that the sum of terms in factor of dt vanishes in the above Relation (S.17.82), which yields the PDE +e− 0 rs ds

∂C ∂C ∂C (t, x, y)dt + ry (t, x, y) + γ(t, y) (t, x, y) ∂t ∂x ∂y 2 2 2 σ 2∂ C 1∂ C ∂2C + x (t, x, y) + η 2 (t, y) (t, x, y) + ρσxη(t, y) (t, x, y), 2 2 2 ∂x 2 ∂y ∂x∂y

0 = − yC(t, x, y) +

with the terminal condition C(τ, x, y) = max(αx, F (τ, y)),

where F (τ, rτ ) = P (τ, T )

is the bond pricing function. e) The convertible bond can be priced as h rτ i IE∗ e− t rs ds max(αSτ , P (τ, T )) Ft h rτ i h rτ i = IE∗ e− t rs ds P (τ, T ) Ft + α IE∗ e− t rs ds (Sτ − P (τ, T )/α)+ Ft b (Sτ /P (τ, T ) − 1/α)+ Ft . = P (t, T ) + αP (t, T )IE (S.17.83) f) By Proposition 16.8 we find ct , dZt = (σ − σB (t))Zt dW

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Solutions Manual ct )t∈R is a standard Brownian motion under the forward measure where (W + Pb. g) By modeling (Zt )t∈R+ as the geometric Brownian motion ct , dZt = σ(t)Zt dW Relation (S.17.83) shows that the convertible bond is priced as P (t, T ) + αSt Φ(d+ ) − P (t, T )Φ(d− ), where d+ =

1 v(t, T )

d− =

1 v(t, T )

log log

v 2 (t, τ ) St + P (t, T ) 2

v 2 (t, τ ) St − P (t, T ) 2

rT and v 2 (t, T ) = t σ 2 (s, T )ds, 0 < t < T . Exercise 17.9 We have 1 c 1 + 1− n n (1 + r) r (1 + r) n c 1 nc =− − , 1 − + (1 + r)n+1 r2 (1 + r)n r(1 + r)n+1

∂ ∂ Pc (0, n) = ∂r ∂r

hence 1+r ∂ Pc (0, n) Pc (0, n) ∂r n (1 + r)c 1 nc − − 1− + n 2 n (1 + r) r (1 + r) r(1 + r)n =− 1 c 1 + 1− n n (1 + r) r (1 + r) 1+r n (c((1 + r) − 1)) + nc −nr − r =− r + c ((1 + r)n − 1) 1+r 1 + r − nr − (r + c((1 + r)n − 1)) + nc r =− r + c ((1 + r)n − 1) 1+r 1 + r + n(c − r) = − r r + c ((1 + r)n − 1) (1 − c/r)n 1+r c ((1 + r)n − 1) = + , 1 + c ((1 + r)n − 1) /r r r + c ((1 + r)n − 1)

Dc (0, n) = −

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Introduction to Stochastic Finance with Market Examples, Second Edition with D0 (0, n) = n. We note that 1+r 1 + r + n(c − r) 1 lim Dc (0, n) = lim − =1+ . n→∞ n→∞ r r + c ((1 + r)n − 1) r When n becomes large, the duration (or relative sensitivity) of the bond price converges to 1+1/r whenever the (nonnegative) coupon amount c is nonzero, otherwise the bond duration of Pc (0, n) is n. In particular, the presence of a nonzero coupon makes the duration (or relative sensitivity) of the bond price bounded as n increases, whereas the duration n of P0 (0, n) goes to infinity as n increases. As a consequence, the presence of the coupon tends to put an upper limit the risk and sensitivity of bond prices with respect to the market interest rate r as n becomes large, which can be used for bond immunization. Note that the duration Dc (0, n) can also be written as the relative average ! n X n k 1 + c Dc (0, n) = Pc (0, n) (1 + r)n (1 + r)k k=1

of zero coupon bond durations, weighted by their respective zero-coupon prices. Exercise 17.10 a) We have P (t, T ) = P (s, T ) exp

ru du +

wt s

σuT dBu −

1wt T 2 |σu | du , 2 s

0 ≤ s ≤ t ≤ T. b) We have

rt rt d e− 0 rs ds P (t, T ) = e− 0 rs ds σtT P (t, T )dBt ,

which gives a martingale after integration, from the properties of the Itô integral. c) By the martingale property of the previous question we have h rT i h i rT IE e− 0 rs ds Ft = IE P (T, T )e− 0 rs ds Ft rt

= P (t, T )e− 0 rs ds ,

0 ≤ t ≤ T.

d) By the previous question we have h rT i rt P (t, T ) = e 0 rs ds IE e− 0 rs ds Ft h rt i rT = IE e 0 rs ds e− 0 rs ds Ft "

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Solutions Manual i h rT = IE e− t rs ds Ft ,

0 ≤ t ≤ T,

since e− 0 rs ds is an Ft -measurable random variable. e) We have w t P (t, S) P (s, S) 1wt S 2 = exp (σuS − σuT )dBu − (|σu | − |σuT |2 )du s P (t, T ) P (s, T ) 2 s w t 1wt S P (s, S) exp (σuS − σuT )dBuT − (σu − σuT )2 du , = s P (s, T ) 2 s 0 ≤ t ≤ T , hence letting s = t and t = T in the above expression, we find w T 2 P (t, S) 1wT S P (T, S) = exp σsS − σsT dBsT − σs − σsT ds . t P (t, T ) 2 t f) We have P (t, T )Ib E (P (T, S) − κ)+ " + # P (t, S) r T (σsS −σsT )dBsT −r T (σsS −σsT )2 ds/2 t et −κ = P (t, T )Ib E P (t, T ) X + ∗ Ft = P (t, T ) IE e −κ 2 1 v(t) + (m(t) + v 2 (t)/2 − log κ) = P (t, T )em(t)+v (t)/2 Φ 2 v(t) v(t) 1 −κP (t, T )Φ − + (m(t) + v 2 (t)/2 − log κ) , 2 v(t) where m(t) := log

2 P (t, S) 1 w T S − σs − σsT ds, P (t, T ) 2 t

and X is a centered Gaussian random variable with variance wT 2 v 2 (t) := σsS − σsT ds, t

given Ft . This yields P (t, T )Ib E (P (T, S) − κ)+ v(t) 1 P (t, S) v(t) 1 P (t, S) = P (t, S)Φ + log − κP (t, T )Φ − + log . 2 v(t) κP (t, T ) 2 v(t) κP (t, T ) Exercise 17.11 (Exercise 4.18 continued). From Proposition 17.2, the bond pricing PDE is

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Introduction to Stochastic Finance with Market Examples, Second Edition  ∂F 1 ∂F ∂2F   (t, x) = xF (t, x) − (α − βx) (t, x) − σ 2 x2 2 (t, x) ∂t ∂x 2 ∂x   F (T, x) = 1. We search for a solution of the form F (t, x) = eA(T −t)−xB(T −t) , with A(0) = B(0) = 0, which implies  ′   A (s) = 0   B ′ (s) + βB(s) + 1 σ 2 B 2 (s) = 1, 2 hence in particular A(s) = 0, s ∈ R, and B(s) solves a Riccati equation, whose solution can be checked to be B(s) = with γ =

2(eγs − 1) , 2γ + (β + γ)(eγs − 1)

β 2 + 2σ 2 .

Exercise 17.12 a) We have

 1    y0,1 = − T log P (0, T1 ) = 9.53%,   1  1 y0,2 = − log P (0, T2 ) = 9.1%, T2    P (0, T2 ) 1   log = 8.6%,  y1,2 = − T2 − T1 P (T1 , T2 )

with T1 = 1 and T2 = 2. b) We have Pc (1, 2) = ($1 + $0.1) × P0 (1, 2) = ($1 + $0.1) × e−(T2 −T2 )y1,2 = $1.00914, and Pc (0, 2) = ($1 + $0.1) × P0 (0, 2) + $0.1 × P0 (0, 1) = ($1 + $0.1) × e−(T2 −T2 )y0,2 + $0.1 × e−(T2 −T2 )y0,1 = $1.00831. Exercise 17.13

√ a) The discretization rtk+1 := rtk + (a − brtk )∆t ± σ ∆t does not lead to a binomial tree because rt2 can be obtained in four different ways from rt0 , "

301 September 15, 2022

Solutions Manual as √ rt2 = rt1 (1 − b∆t) + a∆t ± σ ∆t  √ √  (rt0 (1 − b∆t) + a∆t + σ ∆t)(1 − b∆t) + a∆t + σ ∆t      √ √     (rt0 (1 − b∆t) + a∆t + σ ∆t)(1 − b∆t) + a∆t − σ ∆t = √ √    (rt0 (1 − b∆t) + a∆t − σ ∆t)(1 − b∆t) + a∆t + σ ∆t      √ √   (rt0 (1 − b∆t) + a∆t − σ ∆t)(1 − b∆t) + a∆t − σ ∆t. b) By the Girsanov Theorem, the process (rt /σ)t∈[0,T ] with drt a − brt = dt + dBt σ σ is a standard Brownian motion under the probability measure Q with Radon-Nikodym density dQ 1 wT 1 wT = exp − (a − brt )dBt − 2 (a − brt )2 dt dP σ 0 2σ 0 1 wT 1 wT ≃ exp − 2 (a − brt )(drt − (a − brt )dt) − 2 (a − brt )2 dt σ 0 2σ 0 1 wT 1 wT = exp − 2 (a − brt )drt + 2 (a − brt )2 dt . σ 0 2σ 0 In other words, if we generate (rt /σ)t∈[0,T ] and the increments σ −1 drt ≃ √ ± ∆t as a standard Brownian motion under Q, then, under the probability measure P such that w dP 1 T 1 wT 2 = exp (a − br )dr − (a − br ) dt , t t t dQ σ2 0 2σ 2 0 the process

drt a − brt − dt σ σ will be a standard Brownian motion under P, and the samples dBt =

drt = (a − brt )dt + σdBt of (rt )t∈[0,T ] will be distributed as a Vasicek process under P. c) Approximating the standard Brownian increment σ −1 drt under Q by √ ± ∆t, we have

302

302 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Y Y 1 a − brt √ a − brt √ ± ∆t = 1± ∆t 2 2σ σ 0<t<T 0<t<T ! Y a − brt √ = exp log 1± ∆t σ 0<t<T ! X a − brt √ = exp log 1 ± ∆t σ 0<t<T ! X a − brt √ 1 X (a − brt )2 ≃ exp ± ∆t − ∆t σ 2 σ2 0<t<T 0<t<T w w T T 1 1 2 (a − br )dr − (a − br ) dt ≃ exp t t t σ2 0 2σ 2 0 dP = . dQ

2T /∆T

d) We check that IE[∆rt1 | rt0 ] = (a − brt0 )∆t √ √ = p(rt0 )σ ∆t − (1 − p(rt0 ))σ ∆t √ √ = σp(rt0 ) ∆t − σq(rt0 ) ∆t, with p(rt0 ) =

1 a − brt0 √ + ∆t 2 2σ

and q(rt0 ) =

1 a − brt0 √ − ∆t. 2 2σ

Similarly, we have IE[∆rt2 | rt1 ] = (a − brt1 )∆t √ √ = p(rt1 )σ ∆t − (1 − p(rt1 ))σ ∆t √ √ = σp(rt1 ) ∆t − σq(rt1 ) ∆t, with p(rt1 ) =

1 a − brt1 √ + ∆t, 2 2σ

q(rt1 ) =

1 a − brt1 √ − ∆t. 2 2σ

Exercise 17.14 a) We have P (1, 2) = IE∗

100 100 100 = + . 1 + r1 2(1 + r1u ) 2(1 + r1d )

b) We have "

303 September 15, 2022

Solutions Manual P (0, 2) =

100 100 + . 2(1 + r0 )(1 + r1u ) 2(1 + r0 )(1 + r1d )

c) We have P (0, 1) = 91.74 = 100/(1 + r0 ), hence r0 =

100 − P (0, 1) = 100/91.74 − 1 = 0.090037061 ≃ 9%. P (0, 1)

d) We have 83.40 = P (0, 2) =

P (0, 1) P (0, 1) + 2(1 + r1u ) 2(1 + r1d )

√

and r1u /r1d = e2σ ∆t , hence 83.40 = P (0, 2) =

P (0, 1)

√

2(1 + r1d e2σ ∆t )

P (0, 1) 2(1 + r1d )

or √ √ √ P (0, 1) P (0, 1) e2σ ∆t (r1d )2 + 2r1d eσ ∆t cosh σ ∆t 1 − +1− = 0, 2P (0, 2) P (0, 2) and √ √ P (0, 1) r1d = e−σ ∆t cosh σ ∆t −1 2P (0, 2)  s 2 √ P (0, 1) P (0, 1) 2 ± − 1 cosh σ ∆t + − 1 2P (0, 2) P (0, 2) = 0.078684844 ≃ 7.87%, and √

r1u = r1d e2σ ∆t √ √ P (0, 1) = eσ ∆t cosh σ ∆t −1 2P (0, 2)  s 2 √ P (0, 1) P (0, 1) 2 ± − 1 cosh σ ∆t + − 1 2P (0, 2) P (0, 2) = 0.122174525 ≃ 12.22%. We also find √ √ 1 rd 1 ru µ= σ ∆t + log 1 = −σ ∆t + log 1 = 0.085229181 ≃ 8.52%. ∆t r0 ∆t r0

304

304 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Exercise 17.15 a) When n = 1 the relation (17.55) shows that fe(t, t, T1 ) = f (t, t, T1 ) with F (t, x) = c1 e−(T1 −1)x and P (t, T1 ) = c1 ef (t,t,T1 ) , hence D(t, T1 ) := −

1 ∂F (t, f (t, t, T1 )) = T1 − t, P (t, T1 ) ∂x

0 ≤ t ≤ T1 .

b) In general, we have 1 ∂F t, fe(t, t, Tn ) P (t, Tn ) ∂x n X 1 ˜ (Tk − t)ck e−(Tk −t)f (t,t,Tn ) = P (t, Tn )

D(t, Tn ) = −

k=1

n X

(Tk − t)wk ,

k=1

where wk :=

ck ˜ e−(Tk −t)f (t,t,Tn ) P (t, Tn ) ˜

ck e−(Tk −t)f (t,t,Tn ) n X cl e−(Tl −t)f (t,t,Tl ) l=1

˜ ck e−(Tk −t)f (t,t,Tn ) , n X ˜ cl e−(Tl −t)f (t,t,Tn )

k = 1, 2, . . . , n,

l=1

and the weights w1 , w2 , . . . , wn satisfy n X

wk = 1.

k=1

c) We have C(t, Tn ) = = =

n X k=1 n X k=1

∂2F 1 t, fe(t, t, Tn ) P (t, Tn ) ∂x2

(Tk − t)2 wk (Tk − t − D(t, Tn ))2 wk + 2D(t, Tn )

n X

(Tk − t)wk − (D(t, Tn ))2

k=1

305 September 15, 2022

Solutions Manual = (D(t, Tn ))2 +

n X

(Tk − t − D(t, Tn ))2 wk

k=1

= (D(t, Tn ))2 + (S(t, Tn ))2 , where (S(t, Tn ))2 :=

n X

(Tk − t − D(t, Tn ))2 wk .

k=1

d) We have n

D(t, Tn ) =

X 1 ck B(Tk − t)eA(Tk −t)+B(Tk −t)fα (t,t,Tn ) P (t, Tn ) k=1

= =

n X

1 eA(Tn −t)+B(Tn −t)fα (t,t,Tn ) n X

ck B(Tk − t)eA(Tk −t)+B(Tk −t)fα (t,t,Tn )

k=1

A(Tk −t)−A(Tn −t)+(B(Tk −t)−B(Tn −t))fα (t,t,Tn )

ck B(Tk − t)e

k=1

e) We have D(t, Tn ) =

n −(Tn −t)b 1X −e−(Tk −t)b fα (t,t,Tn )/b 1 − e−(Tk −t)b ck eA(Tk −t)−A(Tn −t)+ e b k=1

(e−(Tn −t)b −e−(Tk −t)b ) 1X (Tn −t)αb = 1 − e−(Tk −t)b ck eA(Tk −t)−A(Tn −t) (P (t, t + α(Tn − t)) . b

k=1

Chapter 18 Exercise 18.1 b we a) By partial differentiation with respect to T under the expectation IE, have h rT i ∂P ∂ (t, T ) = IE∗ e− t rs ds Ft ∂T ∂T h i rT = IE∗ − rT e− t rs ds Ft b T | Ft ]. = −P (t, T )IE[r b) As a consequence of Question (a), we find f (t, T ) = − 306

∂P 1 b T | Ft ], (t, T ) = IE[r P (t, T ) ∂T

0 ≤ t ≤ T,

(S.18.84) 306

September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition see Relation (22) page 10 of Mamon (2004). c) The martingale property of (f (t, T ))t∈[0,T ] under the forward measure Ib E follows from Relation (S.18.84) and the tower property of conditional expectations as in e.g. (7.1) or (7.42). Remark. In the Vasicek model, by (17.2) and (19.9) we have wT a rT = r0 e−bT + (1 − e−bT ) + σ e−(T −s)b dWs 0 b w T a cs e−(T −s)b dW = r0 e−bT + (1 − e−bT ) + σ 0 b w σ 2 T −(T −s)b − e (1 − e−(T −s)b )ds b 0 wT a cs = r0 e−bT + (1 − e−bT ) + σ e−(T −s)b dW 0 b 2 wT 2 wT σ σ − e−(T −s)b ds + e−2(T −s)b ds, b 0 b 0 hence wt a cs Ib E[rT | Ft ] = r0 e−bT + 1 − e−bT + σ e−(T −s)b dW 0 b σ 2 w T −(T −s)b σ 2 w T −2(T −s)b − e ds + e ds b 0 b 0 wt a −bT −bT −(T −s)b = r0 e + 1−e +σ e dWs 0 b 2 wt σ + e−(T −s)b 1 − e−(T −s)b ds b 0 σ 2 w T −2(T −s)b σ 2 w T −(T −s)b e ds + e ds − 0 b b 0 a a = r0 e−bT + 1 − e−bT + e−(T −t)b rt − r0 e−bt − 1 − e−bt b b σ 2 w t −2(T −s)b σ 2 w t −(T −s)b e ds − e ds + b 0 b 0 σ 2 w T −(T −s)b σ 2 w T −2(T −s)b − e ds + e ds b 0 b 0 σ2 w T σ 2 w T −2(T −s)b a a − e−(T −s)b ds + e ds = + e−(T −t)b rt − b b b t b t σ 2 w T −t 2 w T −t a a σ = + e−(T −t)b rt − − e−bs ds + e−2bs ds b b b 0 b 0 σ2 a a σ2 = + e−(T −t)b rt − − 2 1 − e−(T −t)b + 2 1 − e−2(T −t)b b b b b a σ2 a σ2 σ2 = − 2 + e−(T −t)b rt − + 2 − 2 e−2(T −t)b , b 2b b b b "

307 September 15, 2022

Solutions Manual which recovers (18.31). Exercise 18.2 We have w T2 P (0, T2 ) = exp − f (t, s)ds = e−r1 T1 −r2 (T2 −T1 ) , 0

t ∈ [0, T2 ],

and w T2 P (T1 , T2 ) = exp − f (t, s)ds = e−r2 (T2 −T1 ) , T1

t ∈ [0, T2 ],

from which we deduce r2 = −

1 log P (T1 , T2 ), T2 − T1

and 1 T2 − T1 − log P (0, T2 ) T1 T1 1 1 = log P (T1 , T2 ) − log P (0, T2 ) T1 T1 P (0, T2 ) 1 . = − log T1 P (T1 , T2 )

r1 = −r2

Exercise 18.3 (Exercise 4.15 continued). T

a) We check that P (T, T ) = eXT = 1. b) We have 1 XtS − XtT − µ(S − T ) S−T wt 1 wt 1 1 = µ−σ (S − t) dBs − (T − t) dBs 0 S−s 0 T −s S−T 1 wt S−t T −t = µ−σ − dBs S−T 0 S−s T −s w t (T − s)(S − t) − (T − t)(S − s) 1 = µ−σ dBs S−T 0 (S − s)(T − s) w t σ (s − t)(S − T ) dBs . = µ+ S − T 0 (S − s)(T − s)

f (t, T, S) = −

c) We have f (t, T ) = µ − σ 308

t−s

0 (T − s)2

dBs . 308 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition d) We note that lim f (t, T ) = µ − σ

T ↘t

0 t−s

dBs

does not exist in L2 (Ω). e) By Itô’s calculus we have 1 XtT dP (t, T ) = σdBt + σ 2 dt + µdt − dt P (t, T ) 2 T −t log P (t, T ) 1 dt, = σdBt + σ 2 dt − 2 T −t

t ∈ [0, T ].

f) Letting 1 XtT rtT := µ + σ 2 − 2 T −t w t dB 1 s = µ + σ2 − σ , 0 T −s 2 by Question (e) we find that dP (t, T ) = σdBt + rtT dt, P (t, T )

0 ≤ t ≤ T.

g) The equation of Question (f) can be solved as σ2 t w t T P (t, T ) = P (0, T ) exp σBt − + rs ds , 0 2

0 ≤ t ≤ T,

hence the process w t σ2 t , P (t, T ) exp − rsT ds = P (0, T ) exp σBt − 0 2

0 ≤ t ≤ T,

is a martingale under P∗ , with the relation w w t T P (t, T ) exp − rsT ds = IE∗ P (T, T ) exp − rsT ds Ft 0 0 w T = IE∗ exp − rsT ds Ft , 0 ≤ t ≤ T, 0

showing that w w T t rsT ds IE∗ exp − rsT ds Ft P (t, T ) = exp 0 0 w w t T rsT ds exp − rsT ds Ft = IE∗ exp 0

309 September 15, 2022

Solutions Manual w T = IE∗ exp − rsT ds Ft , t

0 ≤ t ≤ T.

h) By Question (g) we have 2 P (t, T ) − r t rsT ds dPT Ft = IE = eσBt −σ t/2 , e 0 dP P (0, T )

0 ≤ t ≤ T.

bt := Bt − σt is a standard Browi) By the Girsanov Theorem, the process B nian motion under PT . j) We have wT S−T log P (T, S) = −µ(S − T ) + σ dBs 0 S−s wt S−T wT S−T = −µ(S − T ) + σ dBs + σ dBs 0 S−s t S−s w T S−T S−T = log P (t, S) + σ dBs t S−s S−t w wT S−T T S−T S−T b log P (t, S) + σ dBs + σ 2 ds = t S−s t S−s S−t w T S−T S−T b S−t = log P (t, S) + σ dBs + (S − T )σ 2 log , t S−s S−t S−T 0 < T < S. k) We have P (t, T ) IET (P (T, S) − K)+ Ft X + = P (t, T ) IE e − K | Ft 2 vt 1 = P (t, T )emt +vt /2 Φ + (mt + vt2 /2 − log κ) 2 vt vt 1 −κP (t, T )Φ − + (mt + vt2 /2 − log κ) 2 vt 1 1 mt +vt2 /2 = P (t, T )e Φ vt + (mt − log κ) − κP (t, T )Φ (mt − log κ) , vt vt with mt =

S−T S−t log P (t, S) + (S − T )σ 2 log S−t S−T

and vt2 = σ 2

w T (S − T )2

ds (S − s)2 1 1 = (S − T )2 σ 2 − S−T S−t

310

310 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = (S − T )σ 2

(T − t) , (S − t)

hence P (t, T ) IET (P (T, S) − K)+ Ft (S−T )σ2 2 S−t = P (t, T )(P (t, S))(S−T )(S−t) evt /2 S−T (S−T )σ2 !! 1 (P (t, S))(S−T )(S−t) S − t ×Φ vt + log vt κ S−T (S−T )σ2 !! (S−T )(S−t) 1 (P (t, S)) S−t −κP (t, T )Φ log . vt κ S−T Exercise 18.4 a) In the Vašíček (1977) model, by (17.32) we have w T P (t, T ) = IE exp − rs ds t w wT T = IE exp − h(s)ds − Xs ds t t w w T T = exp − h(s)ds IE exp − Xs ds t t w T = exp − h(s)ds + A(T − t) + Xt C(T − t) , t

0 ≤ t ≤ T,

w T hence, since X0 = 0 we find P (0, T ) = exp − h(s)ds + A(T ) . 0

b) By the identification w T P (t, T ) = exp − h(s)ds + A(T − t) + Xt C(T − t) t w T f (t, s)ds , = exp − t

we find wT t

h(s)ds =

wT t

f (t, s)ds + A(T − t) + Xt C(T − t),

and by differentiation with respect ot T this yields h(T ) = f (t, T ) + A′ (T − t) + Xt C ′ (T − t), "

0 ≤ t ≤ T, 311 September 15, 2022

Solutions Manual where A(T −t) =

4ab − 3σ 2 σ 2 − 2ab σ 2 − ab −b(T −t) σ 2 −2b(T −t) + (T −t)+ e − 3e . 4b3 2b2 b3 4b

Given an initial market data curve f M (0, T ), the matching f M (0, T ) = f (0, T ) can be achieved at time t = 0 by letting h(T ) := f M (0, T )+A′ (T ) = f M (0, T )+

σ 2 − 2ab σ 2 − ab −bT σ 2 −2bT − e + 2e , 2b2 b2 2b

T > 0. Note however that in general, at time t ∈ (0, T ] we will have h(T ) = f (t, T ) + A′ (T − t) + Xt C ′ (T − t) = f M (0, T ) + A′ (T ), and the relation f (t, T ) = f M (0, T ) + A′ (T ) − A′ (T − t) − Xt C ′ (T − t),

t ∈ [0, T ],

will allow us to match market data at time t = 0 only, i.e. for the initial curve. In any case, model calibration is to be done at time t = 0. Exercise 18.5 (Exercise 4.12 continued, p see also Proposition 4.1 in Carmona and Durrleman (2003)). Letting σ := σ12 − 2ρσ1 σ2 + σ22 , we have dSt = rSt dt + σdWt , where (Wt )t∈[0,T ] , is a standard Brownian motions under P∗ , hence St = S0 ert + σ

wt 0

e(t−s)r dWs ,

t ≥ 0,

The spread ST has a Gaussian distribution with mean α := IE∗ [ST ] = S0 erT and variance η 2 := Var∗ [ST ] w T = Var σ e(T −s)r dBs 0

= σ =

wT 0

2 e(T −s)r ds

σ 2 2rT e −1 , 2r

and probability density function 2 ! S0 erT − x φ(x) = exp − 2 2rT , σ e − 1 /r σ e2rT − 1 p

√

312

r/π

x ∈ R. 312 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition Hence, we have 2 2 e−rt w ∞ (x − K)+ e−(x−α) /(2η ) dx e−rt IE∗ [(ST − K)+ ] = p 2πη 2 −∞ 2 2 e−rt w ∞ = p (x − K)e−(x−α) /(2η ) dx 2πη 2 K e−rt w ∞ −(x−α)2 /(2η2 ) Ke−rt w ∞ −(x−α)2 /(2η2 ) xe e = p dx − p dx 2πη 2 K 2πη 2 K w −rt −rt w ∞ ∞ 2 2 Ke ηe (x + α)e−x /2 dx − √ e−x /2 dx = √ 2π (K−α)/η 2π (K−α)/η ηe−rt h −x2 /2 i∞ K −α =−√ e − (K − α)e−rt Φ − η (K−α)/η 2π ηe−rt −(K−α)2 /(2η2 ) K − α = √ e − (K − α)e−rt Φ − . η 2π

Exercise 18.6 From the definition L(t, t, T ) =

1 T −t

P (t, T ) =

1 , 1 + (T − t)L(t, t, T )

P (t, S) =

1 . 1 + (S − t)L(t, t, S)

we have and similarly

1 −1 , P (t, T )

Hence we get 1 P (t, T ) −1 S − T P (t, S) 1 1 + (S − t)L(t, t, S) = −1 S − T 1 + (T − t)L(t, t, T ) 1 (S − t)L(t, t, S) − (T − t)L(t, t, T ) = . S−T 1 + (T − t)L(t, t, T )

L(t, T, S) =

When T = one year and L(0, 0, T ) = 2%, L(0, 0, 2T ) = 2.5% we find 1 2T L(0, 0, 2T ) − T L(0, 0, T ) 2 × 0.025 − 0.02 L(t, T, S) = = = 2.94%, T 1 + T L(0, 0, T ) 1 + 0.02 so we would not prefer a spot rate at L(T, T, 2T ) = 2% over a forward contract with rate L(0, T, 2T ) = 2.94%. "

313 September 15, 2022

Solutions Manual Exercise 18.7 (Exercise 17.4 continued). a) By Definition 18.1, we have 1 f (t, T, S) = (log P (t, T ) − log P (t, S)) S−T 1 σ2 σ2 = −(T − t)rt + (T − t)3 − −(S − t)rt + (S − t)3 S−T 6 6 1 σ2 3 3 (T − t) − (S − t) . = rt + S−T 6 b) We have f (t, T ) = lim f (t, T, S) S↘T

∂ log P (t, T ) ∂T ∂ σ2 =− −(T − t)rt + (T − t)3 ∂T 6 σ2 2 = rt − (T − t) . 2 =−

c) We have

dt f (t, T ) = (T − t)σ 2 dt + σdBt .

d) The HJM condition (18.28) is satisfied since the drift of dt f (t, T ) equals rT σ t σds. Exercise 18.8 a) We have Xt = X0 e−bt + σ and

wt 0

e−(t−s)b dBs(1)

e−(t−s)b dBs(2) ,

t ∈ R+ ,

Var[Xt ] = Var[Yt ] =

σ2 (1 − e−2bt ), 2b

t ∈ R+ ,

Yt = Y0 e−bt + σ

see (17.2). b) We have

see page 480, and therefore wt wt Cov(Xt , Yt ) = Cov X0 e−bt + σ e−(t−s)b dBs(1) , Y0 e−bt + σ e−(t−s)b dBs(2) 0

314

314 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition w wt t = σ 2 Cov e−(t−s)b dBs(1) , e−(t−s)b dBs(2) 0 0 w wt t 2 −(t−s)b (1) −(t−s)b = σ IE e dBs e dBs(2) 0

= ρσ =ρ

wt 0

σ 2b

−2(t−s)b

1 − e−2bt ,

t ∈ R+ .

c) We have Cov(log P (t, T1 ), log P (t, T2 )) = Cov log F1 (t, Xt , T1 )F2 (t, Yt , T2 )eρU (t,T1 ) , log F1 (t, Xt , T1 )F2 (t, Yt , T2 )eρU (t,T2 ) = Cov C1T1 + Xt AT1 1 + C2T1 + Yt AT2 1 , C1T2 + Xt AT1 2 + C2T2 + Yt AT2 2 = Cov Xt AT1 1 + Yt AT2 1 , Xt AT1 2 + Yt AT2 2 = AT1 1 AT1 2 Var[Xt ] + AT2 1 AT2 2 Var[Yt ] + AT1 1 AT2 2 + AT1 2 AT2 1 Cov(Xt , Yt ) = AT1 1 AT1 2 + AT2 1 AT2 2 + ρ AT1 1 AT2 2 + AT1 2 AT2 1 Var[Xt ]. When Cov(Xt , Yt ) = ρ Var[Xt ] = ρ Var[Yt ], we find the correlation Cov(log P (t, T1 ), log P (t, T2 )) = p

Cov(log P (t, T1 ), log P (t, T2 ))

Var[log P (t, T1 )] Var[log P (t, T2 )] AT1 1 AT1 2 + AT2 1 AT2 2 + ρ AT1 1 AT2 2 + AT1 2 AT2 1 = q 2 2 AT1 1 + AT2 1 + ρ AT1 1 AT2 1 + AT1 1 AT2 1 ×q

1 AT1 2

+ AT2 2

+ ρ A1T2 AT2 2 + AT1 2 AT2 2

When ρ = 1, we find Cov(log P (t, T1 ), log P (t, T2 )) AT1 1 AT1 2 + AT2 1 AT2 2 + AT1 1 AT2 2 + AT1 2 AT2 1 q 2 2 A1 + A2 + AT1 1 AT2 1 + AT1 1 AT2 1 AT1 2 + AT2 2 + AT1 2 AT2 2 + AT1 2 AT2 2 AT1 1 + AT2 1 AT1 2 + AT2 2

= q =

T1 2

AT1 1 + AT2 1 AT1 2 + AT2 2 = ±1.

For example, if AT1 1 = 4, AT1 2 = 1, AT2 1 = 1 and AT2 2 = 4, we find Cov(log P (t, T1 ), log P (t, T2 )) = "

8 + 17ρ . 17 + 8ρ 315 September 15, 2022

Solutions Manual 1

Correlation

0.5

-0.5

-1

-0.5

0.5

Fig. S.81: Log bond prices correlation graph in the two-factor model.

Exercise 18.9 a) We have f (t, x) = f (0, x) + α b) We have

wt 0

s2 ds + σ

wt 0

ds B(s, x) = r + αtx2 + σB(t, x).

rt = f (t, 0) = r + B(t, 0) = r.

c) We have w T P (t, T ) = exp − f (t, s)ds t w T −t w T −t = exp −(T − t)r − αt s2 ds − σ B(t, x)dx 0 0 w T −t α = exp −(T − t)r − t(T − t)3 − σ B(t, x)dx , 0 3

t ∈ [0, T ].

d) Using (18.43), we have " 2 # w w T −t T −t w T −t IE B(t, x)dx = IE[B(t, x)B(t, y)]dxdy 0

w T −t w T −t

= 2t

w T −t w y 0

min(x, y)dxdy

xdxdy =

1 t(T − t)3 . 3

e) By Question (d) we have w T −t α IE[P (t, T )] = IE exp −(T − t)r − t(T − t)3 − σ B(t, x)dx 0 3 316

316 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition w T −t α B(t, x)dx = exp −(T − t)r − t(T − t)3 IE exp −σ 0 3 2 w T −t α σ 3 = exp −(T − t)r − t(T − t) exp Var B(t, x)dx 0 3 2 2 α σ = exp −(T − t)r − t(T − t)3 + t(T − t)3 , 0 ≤ t ≤ T. 3 6 f) By Question (e) we check that the required relation is satisfied if α σ2 − t(T − t)3 + t(T − t)3 = 0, 3 6 i.e. α = σ 2 /2. Remark: In order to derive an analog of the HJM absence of arbitrage condition in this stochastic string model, one would have to check whether the discounted bond price e−rt P (t, T ) can be a martingale by doing stochastic calculus with respect to the Brownian sheet B(t, x). g) We have w T IE exp − rs ds (P (T, S) − K)+ 0 " + # w S−T α = e−rT IE exp −(S − T )r − T (S − T )3 + σ B(T, x)dx − K 0 3 m+X −rT + =e IE (xe − K) , where x = e−(S−T )r , m = −αT (S − T )3 /3, and X=σ

w S−T 0

B(T, x)dx ≃ N (0, σ 2 t(T − t)3 /3).

Given the relation α = σ 2 /2, this yields w T IE exp − rs ds (P (T, S) − K)+ 0

= e−rS Φ σ

T (S − T )3 /12 +

log(e−(S−T )r /K) p σ T (S − T )3 /3

log(e−(S−T )r /K) p −Ke Φ −σ σ T (S − T )3 /3 ! p log(e−(S−T )r /K) = P (0, S)Φ σ T (S − T )3 /12 + p σ T (S − T )3 /3 −rT

T (S − T )3 /12 +

317 September 15, 2022

Solutions Manual

−KP (0, T )Φ −σ

T (S − T )3 /12 +

log(e−(S−T )r /K) p σ T (S − T )3 /3

! .

Chapter 19 Exercise 19.1 a) We price the floorlet at t = 0, with T1 = 9 months, T2 = 1 year, κ = 4.5%. The LIBOR rate (L(t, T1 , T2 ))t∈[0,T1 ] is modeled as a driftless geometric Brownian motion with volatility coefficient σ b = σ1,2 (t) = 0.1 under the forward measure P2 . The discount factors are given by P (0, T1 ) = e−9r/12 ≃ 0.970809519 and

P (0, T2 ) = e−r ≃ 0.961269954,

with r = 3.95%. b) By (19.21), the price of the floorlet is h r T2 i IE∗ e− 0 rs ds (κ − L(T1 , T1 , T2 ))+ = P (0, T2 ) κΦ − d− (T1 ) − L(0, T1 , T2 )Φ − d+ (T1 ) , (S.19.85) where d+ (T1 ) =

log(L(0, T1 , T2 )/κ) + σ 2 T1 /2 √ , σ1 T1

d− (T1 ) =

log(L(0, T1 , T2 )/κ) − σ 2 T1 /2 √ , σ T1

and

are given in Proposition 19.5, and the LIBOR rate L(0, T1 , T2 ) is given by L(0, T1 , T2 ) =

P (0, T1 ) − P (0, T2 ) (T2 − T1 )P (0, T2 )

e−3r/4 − e−r 0.25e−r = 4(er/4 − 1) =

≃ 3.9695675%. Hence, we have d+ (T1 ) =

318

log(0.039695675/0.045) + (0.1)2 × 0.75/2 √ ≃ −1.404927033, 0.1 × 0.75 318 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition and d− (T1 ) =

log(0.039695675/0.045) − (0.1)2 × 0.75/2 √ ≃ −1.491529573, 0.1 × 0.75

hence i h r T2 IE∗ e− 0 rs ds (κ − L(T1 , T1 , T2 ))+ = 0.961269954 × (κΦ(1.491529573) − L(0, T1 , T2 ) × Φ(1.404927033)) = 0.961269954 × (0.045 × 0.932089 − 0.039695675 × 0.919979) ≃ 0.52147141%. Finally, we need to multiply (S.19.85) by the notional principal amount of $1 million per interest rate percentage point, i.e. $10, 000 per percentage point or $100 per basis point, which yields $5214.71. Exercise 19.2 a) We price the swaption at t = 0, with T1 = 4 years, T2 = 5 years, T3 = 6 years, T4 = 7 years, κ = 5%, and the swap rate (S(t, T1 , T4 ))t∈[0,T1 ] is modeled as a driftless geometric Brownian motion with volatility coefficient σ b = σ1,4 (t) = 0.2 under the forward swap measure P1.4 . The discount factors are given by P (0, T1 ) = e−4r , P (0, T2 ) = e−5r , P (0, T3 ) = e−6r , P (0, T4 ) = e−7r , where r = 5%. b) By Proposition 19.17 the price of the swaption is (P (0, T1 ) − P (0, T4 ))Φ(d+ (T1 − t)) −κΦ(d− (T1 ))(P (0, T2 ) + P (0, T3 ) + P (0, T4 )), where d+ (T1 ) and d− (T1 ) are given in Proposition 19.17, and the LIBOR swap rate S(0, T1 , T4 ) is given by P (0, T1 ) − P (0, T4 ) P (0, T1 , T4 ) P (0, T1 ) − P (0, T4 ) = P (0, T2 ) + P (0, T3 ) + P (0, T4 ) e−4r − e−7r = −5r e + e−6r + e−7r e3r − 1 = 2r e + er + 1 e0.15 − 1 = 0.1 e + e0.05 + 1 = 0.051271096.

S(0, T1 , T4 ) =

By Proposition 19.17 we also have "

319 September 15, 2022

Solutions Manual d+ (T1 ) =

log(0.051271096/0.05) + (0.2)2 × 4/2 √ = 0.526161682, 0.2 4

d− (T1 ) =

log(0.051271096/0.05) − (0.2)2 2 × 4/2 √ = 0.005214714, 0.2 4

and

Hence, the price of the swaption is given by (e−4r − e−7r )Φ(0.526161682)

(S.19.86)

−κΦ(0.005214714)(e−5r + e−6r + e−7r ) = (0.818730753 − 0.70468809) × 0.700612 −0.05 × 0.50208 × (0.818730753 + 0.740818221 + 0.70468809) = 2.3058251%. Finally, we need to multiply (S.19.86) by the notional principal amount of $10 million, i.e. $100, 000 by interest percentage point, or $1, 000 by basis point, which yields $230, 582.51. Exercise 19.3 Taking t = 0, we have T1 = 3, T2 = 4 and T3 = 5. The LIBOR swap rate S(t, T1 , T3 ) is modeled as a driftless geometric Brownian motion with volatility σ = 0.1 under the forward swap measure Pbi,j . The receiver swaption is priced using the Black-Scholes formula as h r T1 i + IE∗ e− t rs ds P (T1 , T1 , T3 ) (κ − S(T1 , T1 , T3 )) Ft = κΦ(−d− (T1 − t))

2 X

(Tk+1 − Tk )P (t, Tk+1 )

k=1

−(P (t, T1 ) − P (t, T3 ))Φ(−d+ (T1 − t)), where κ = 5%, r = 2% and P (t, T1 ) = e−3r = 0.9417, P (t, T2 ) = e−4r = 0.9231, P (t, T3 ) = e−5r = 0.9048. Hence, P (t, T1 , T3 ) = P (t, T2 ) + P (t, T3 ) = 0.92311 + 0.90483 = 1.82794 and S(t, T1 , T3 ) =

0.9417 − 0.9048 P (t, T1 ) − P (t, T3 ) = = 0.02018. P (t, T1 , T3 ) 1.82794

We also have d+ (T1 − t) =

320

log(S(t, T1 , T3 )/κ) + σ 2 (T1 − t)/2 √ σ T1 − t 320 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = and

log(2.018/5) + 0.12 × 3/2 √ = −5.1518 0.1 3

d− (T1 − t) = d+ (T1 − t) − σ

T1 − t = −5.3250,

hence h r T1 i + IE∗ e− t rs ds P (T1 , T1 , T3 ) (κ − S(T1 , T1 , T3 )) Ft = 0.05 × 1.82794 × Φ(5.3250) − (0.9417 − 0.9048) × Φ(5.1518) = 0.05 × 1.82794 × 0.9999999 − (0.9417 − 0.9048) × 0.9999999 = 0.054496 = 5.4496% = 544.96 bp, which yields $54, 496 after multiplication by the $10, 000 notional principal. Exercise 19.4 a) We have 1 dP (t, T2 ) 1 P (t, T2 ) + P (t, T2 )d = + dP (t, T2 ) • d d P (t, T1 ) P (t, T1 ) P (t, T1 ) P (t, T1 ) dP (t, T2 ) dP (t, T1 ) dP (t, T1 ) • dP (t, T1 ) = + P (t, T2 ) − + P (t, T1 ) (P (t, T1 ))2 (P (t, T1 ))3 • dP (t, T1 ) dP (t, T2 ) − (P (t, T1 ))2 1 = rt P (t, T2 )dt + ζ2 (t)P (t, T2 )dWt P (t, T1 ) P (t, T2 ) − rt P (t, T1 )dt + ζ1 (t)P (t, T1 )dWt (P (t, T1 ))2 P (t, T2 ) + (rt P (t, T1 )dt + ζ1 (t)P (t, T1 )dWt )2 (P (t, T1 ))3 1 (rt P (t, T1 )dt + ζ1 (t)P (t, T1 )dWt ) • (rt P (t, T2 )dt + ζ2 (t)P (t, T2 )dWt ) − (P (t, T1 ))2 P (t, T2 ) P (t, T2 ) = ζ2 (t) dWt − ζ1 (t) dWt P (t, T1 ) P (t, T1 ) P (t, T2 ) P (t, T2 ) + (ζ1 (t))2 dt − ζ1 (t)ζ2 (t) dt P (t, T1 ) P (t, T1 ) P (t, T2 ) P (t, T2 ) =− ζ1 (t)(ζ2 (t) − ζ1 (t))dt + (ζ2 (t) − ζ1 (t))dWt P (t, T1 ) P (t, T1 )

321 September 15, 2022

Solutions Manual P (t, T2 ) (ζ2 (t) − ζ1 (t))(dWt − ζ1 (t)dt) P (t, T1 ) P (t, T2 ) c P (t, T2 ) c = (ζ2 (t) − ζ1 (t)) dWt = (ζ2 (t) − ζ1 (t)) d Wt , P (t, T1 ) P (t, T1 ) =

ct = dWt − ζ1 (t)dt is a standard Brownian motion under the where dW b T1 -forward measure P. b) From Question (a) or (19.7) we have P (T1 , T2 ) P (T1 , T1 ) w w T1 T1 P (t, T2 ) cs − 1 = (ζ2 (s) − ζ1 (s))dW (ζ2 (s) − ζ1 (s))2 ds exp t P (t, T1 ) 2 t P (t, T2 ) X−v2 /2 = e , P (t, T1 )

P (T1 , T2 ) =

where X is a centered Gaussian random variable with variance v 2 = w T1 b Hence by the hint (ζ2 (s) − ζ1 (s))2 ds, independent of Ft under P. t (19.43) with x := P (t, T2 )/P (t, T1 ) and κ := K/x, we find h r T1 i IE∗ e− 0 rs ds (K − P (T1 , T2 ))+ Ft = P (t, T1 )Ib E (K − P (T1 , T2 ))+ | Ft 1 K P (t, T2 ) v 1 K v + log − Φ − + log = P (t, T1 ) KΦ 2 v x P (t, T1 ) 2 v x v 1 K v 1 K = KP (t, T1 )Φ + log − P (t, T2 )Φ − + log . 2 v x 2 v x Exercise 19.5 bS is defined from the numéraire Nt := P (t, S) and a) The forward measure P this gives Ft = P (t, S)Ib E[(κ − L(T, T, S))+ | Ft ]. b) The LIBOR rate L(t, T, S) is a driftless geometric Brownian motion with bS . Indeed, the LIBOR rate volatility σ under the forward measure P bt = Xt /Nt L(t, T, S) can be written as the forward price L(t, T, S) = X where Xt = (P (t, T ) − Pr (t, S))/(S − T ) and rNt = P (t, S). Since both dist t counted bond prices e− 0 rs ds P (t, T ) and e− 0 rs ds P (t, S) are martingales ∗ under P , the same is true of Xt . Hence L(t, T, S) = Xt /Nt becomes a bS by Proposition 16.4, and commartingale under the forward measure P bS amounts to removing any “dt” term in puting its dynamics under P (19.44) after rewriting the equation in terms of the standard Brownian bS , i.e. we have ct )t∈R under P motion (W + 322

322 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition ct , dL(t, T, S) = σL(t, T, S)dW 2

b t −σ t/2 , 0 ≤ t ≤ T . which is solved as L(t, T, S) = L(0, T, S)eσW c) We find Ft = P (t, S)Ib E[(κ − L(T, T, S))+ | Ft ] 2 b T −W b t )σ )+ | F ] = P (t, S)Ib E[(κ − L(t, T, S)e−(T −t)σ /2+(W t bt Φ(−d+ (T − t))) = P (t, S)(κΦ(−d− (T − t)) − X

= κP (t, S)Φ(−d− (T − t)) − P (t, S)L(t, T, S)Φ(−d+ (T − t)) = κP (t, S)Φ(−d− (T − t)) − (P (t, T ) − P (t, S))Φ(−d+ (T − t))/(S − T ), 2

where em = L(t, T, S)e−(T −t)σ /2 , v 2 = (T − t)σ 2 , and √ log(L(t, T, S)/κ) σ T − t √ + , d+ (T − t) = 2 σ T −t and d− (T − t) =

√ log(L(t, T, S)/κ) σ T − t √ − , 2 σ T −t

because L(t, T, S) is a driftless geometric Brownian motion with volatility bS . σ under the forward measure P Exercise 19.6 a) We have P (Ti , Ti , Tj ) =

j−1 X

cl+1 P (Ti , Tl+1 ).

l=i

b) It suffices to let c̃l = 1, l = i + 1, . . . , j − 1. and c̃j = cj + 1/κ. c) The swaption can be priced as h r Ti i + IE∗ e− t rs ds P (Ti , Ti ) − P (Ti , Tj ) − κP (Ti , Ti , Tj ) Ft   !+ j−1 rT X ∗  − t i rs ds = IE e 1−κ cl+1 P (Ti , Tl+1 ) Ft  l=i



rT ∗  − t i rs ds

= κ IE

j−1 X l=i



r Ti

= κ IE∗ e− t

rs ds

j−1 X

cl+1 Fl+1 (Ti , γκ ) −

j−1 X

!+ cl+1 Fl+1 (Ti , rTi )

 Ft 

l=i

!+ cl+1 (Fl+1 (Ti , γκ ) − Fl+1 (Ti , rTi ))

 Ft 

l=i

323 September 15, 2022

Solutions Manual "

r Ti

= κ IE∗ e− t

rs ds

1{rTi ≤γκ }

j−1 X

# cl+1 (Fl+1 (Ti , γκ ) − Fl+1 (Ti , rTi ))

l=i

" ∗

= κ IE

j−1 rT X − t i rs ds

# cl+1 (Fl+1 (Ti , γκ ) − Fl+1 (Ti , rTi ))

l=i

=κ

j−1 X

∗

cl+1 IE

rT − t i rs ds

(Fl+1 (Ti , γκ ) − P (Ti , Tl+1 ))

l=i

=κ

j−1 X

b i (Fl+1 (Ti , γκ ) − P (Ti , Tl+1 ))+ Ft , cl+1 P (t, Ti )IE

l=i

which is a weighted sum of bond put option prices with strike prices Fl+1 (Ti , γκ ), l = i, i + 1, . . . , j − 1. Exercise 19.7 a) We have

dP (t, Ti ) = rt dt + ζ (i) (t)dBt , P (t, Ti )

i = 1, 2,

and P (T, Ti ) = P (t, Ti ) exp

rs ds +

wT t

ζ (i) (s)dBs −

1 w T (i) |ζ (s)|2 ds , 2 t

0 ≤ t ≤ T ≤ Ti , i = 1, 2, hence log P (T, Ti ) = log P (t, Ti ) +

wT t

rs ds +

wT t

ζ (i) (s)dBs −

1 w T (i) |ζ (s)|2 ds, 2 t

0 ≤ t ≤ T ≤ Ti , i = 1, 2, and 1 d log P (t, Ti ) = rt dt + ζ (i) (t)dBt − |ζ (i) (t)|2 dt, 2

i = 1, 2.

In the present model, we have drt = σdBt , where (Bt )t∈R+ is a standard Brownian motion under P, by the solution of Exercise 17.4 and (17.25) we have ζ (i) (t) = −(Ti − t)σ, Letting

0 ≤ t ≤ Ti ,

i = 1, 2.

(i)

dBt = dBt − ζ (i) (t)dt, defines a standard Brownian motion under Pi , i = 1, 2, and we have 324

324 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition w T P (t, T1 ) 1 w T (1) P (T, T1 ) = exp (ζ (1) (s) − ζ (2) (s))dBs − (|ζ (s)|2 − |ζ (2) (s)|2 )ds t P (T, T2 ) P (t, T2 ) 2 t w T P (t, T1 ) 1 w T (1) (1) (2) (2) = exp (ζ (s) − ζ (s))dBs − (ζ (s) − ζ (2) (s))2 ds , t P (t, T2 ) 2 t which is an Ft -martingale under P2 and under P1,2 , and w T P (t, T2 ) 1 w T (1) P (T, T2 ) = exp − (ζ (1) (s) − ζ (2) (s))dBs(1) − (ζ (s) − ζ (2) (s))2 ds , t P (T, T1 ) P (t, T1 ) 2 t which is an Ft -martingale under P1 . b) We have 1 (log P (t, T2 ) − log P (t, T1 )) T2 − T1 1 σ2 ((T1 − t)3 − (T2 − t)3 ). = rt + T2 − T1 6

f (t, T1 , T2 ) = −

c) We have P (t, T2 ) 1 d log df (t, T1 , T2 ) = − T2 − T1 P (t, T1 ) 1 1 (2) (1) =− (ζ (t) − ζ (t))dBt − (|ζ (2) (t)|2 − |ζ (1) (t)|2 )dt T2 − T1 2 1 1 (2) =− (ζ (2) (t) − ζ (1) (t))(dBt + ζ (2) (t)dt) − (|ζ (2) (t)|2 − |ζ (1) (t)|2 )dt T2 − T1 2 1 1 (2) (2) (1) (2) (1) =− (ζ (t) − ζ (t))dBt − (ζ (t) − ζ (t))2 dt . T2 − T1 2 d) We have 1 P (T, T2 ) log T2 − T1 P (T, T1 ) w T 1 1 = f (t, T1 , T2 ) − (ζ (2) (s) − ζ (1) (s))dBs − (|ζ (2) (s)|2 − |ζ (1) (s)|2 )ds t T2 − T1 2 w T 1 1 w T (2) = f (t, T1 , T2 ) − (ζ (2) (s) − ζ (1) (s))dBs(2) − (ζ (s) − ζ (1) (s))2 ds t T2 − T1 2 t w T 1 1 w T (2) (2) (1) (1) = f (t, T1 , T2 ) − (ζ (s) − ζ (s))dBs + (ζ (s) − ζ (1) (s))2 ds . t T2 − T1 2 t

f (T, T1 , T2 ) = −

Hence f (T, T1 , T2 ) has a Gaussian distribution given Ft with conditional mean 1 w T (2) m1 := f (t, T1 , T2 ) − (ζ (s) − ζ (1) (s))2 ds 2 t "

325 September 15, 2022

Solutions Manual under P1 , resp. m2 := f (t, T1 , T2 ) +

1 w T (2) (ζ (s) − ζ (1) (s))2 ds 2 t

under P2 , and variance v2 =

wT 1 (ζ (2) (s) − ζ (1) (s))2 ds. 2 (T2 − T1 ) t

Hence we have h r T2 i (T2 − T1 ) IE∗ e− t rs ds (f (T1 , T1 , T2 ) − κ)+ Ft = (T2 − T1 )P (t, T2 ) IE2 (f (T1 , T1 , T2 ) − κ)+ Ft = (T2 − T1 )P (t, T2 ) IE2 (m2 + X − κ)+ Ft 2 2 v = (T2 − T1 )P (t, T2 ) √ e−(κ−m2 ) /(2v ) + (m2 − κ)Φ((m2 − κ)/v) . 2π e) We have L(T, T1 , T2 ) = S(T, T1 , T2 ) 1 P (T, T1 ) −1 = T2 − T1 P (T, T2 ) 1 = T2 − T1 w T P (t, T1 ) 1 w T (1) × exp (ζ (1) (s) − ζ (2) (s))dBs − (|ζ (s)|2 − |ζ (2) (s)|2 )ds − 1 t P (t, T2 ) 2 t 1 = T2 − T1 w T P (t, T1 ) 1 w T (1) (ζ (1) (s) − ζ (2) (s))dBs(2) − (ζ (s) − ζ (2) (s))2 ds − 1 × exp t P (t, T2 ) 2 t 1 = T2 − T1 w T 1 w T (1) P (t, T1 ) × exp (ζ (1) (s) − ζ (2) (s))dBs(1) + (ζ (s) − ζ (2) (s))2 ds − 1 , t P (t, T2 ) 2 t and, by Itô calculus, 1 P (t, T1 ) d T2 − T1 P (t, T2 ) 1 P (t, T1 ) 1 (1) = (ζ (t) − ζ (2) (t))dBt + (ζ (1) (t) − ζ (2) (t))2 dt T2 − T1 P (t, T2 ) 2 1 (1) 2 (2) 2 − (|ζ (t)| − |ζ (t)| )dt 2

dS(t, T1 , T2 ) =

326

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Introduction to Stochastic Finance with Market Examples, Second Edition 1 + S(t, T1 , T2 ) (ζ (1) (t) − ζ (2) (t))dBt + ζ (2) (t)(ζ (2) (t) − ζ (1) (t))dt)dt T2 − T1 1 (1) = + S(t, T1 , T2 ) (ζ (1) (t) − ζ (2) (t))dBt + (|ζ (2) (t)|2 − |ζ (1) (t)|2 )dt T2 − T1 1 (2) = + S(t, T1 , T2 ) (ζ (1) (t) − ζ (2) (t))dBt , t ∈ [0, T1 ], T2 − T1

1 + S(t, T1 , T2 ) is a geometric Brownian motion, with hence t 7→ T2 −T 1

1 + S(T, T1 , T2 ) T2 − T1 1 = + S(t, T1 , T2 ) T2 − T1 w T 1 w T (1) × exp (ζ (1) (s) − ζ (2) (s))dBs(2) − (ζ (s) − ζ (2) (s))2 ds , t 2 t 0 ≤ t ≤ T ≤ T1 . f) We have h r T2 i (T2 − T1 ) IE∗ e− t rs ds (L(T1 , T1 , T2 ) − κ)+ Ft h r T1 i = (T2 − T1 ) IE∗ e− t rs ds P (T1 , T2 )(L(T1 , T1 , T2 ) − κ)+ Ft = P (t, T1 , T2 ) IE1,2 (S(T1 , T1 , T2 ) − κ)+ Ft . The forward measure P2 is defined by P (t, T2 ) − r t rs ds dP2 , e 0 Ft = IE∗ dP P (0, T2 ) and the forward swap measure is defined by dP1,2 P (t, T2 ) − r t rs ds IE∗ Ft = e 0 , dP P (0, T2 )

0 ≤ t ≤ T2 ,

0 ≤ t ≤ T1 ,

(2) hence P2 and P1,2 coincide up to time T1 and Bt t∈[0,T1 ] is a standard Brownian motion until time T1 under P2 and under P1,2 , consequently under P1,2 we have

L(T, T1 , T2 ) = S(T, T1 , T2 ) r r T (1) (2) (2) (1) (2) 2 1 T 1 1 =− + + S(t, T1 , T2 ) e t (ζ (s)−ζ (s))dBs − 2 t (ζ (s)−ζ (s)) ds , T2 − T1 T2 − T1 has same distribution as

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Solutions Manual 1 T2 − T1

P (t, T1 ) X−Var[X]/2 e −1 , P (t, T2 )

where X is a centered Gaussian random variable with variance w T1 (ζ (1) (s) − ζ (2) (s))2 ds t

given Ft . Hence, we have h r T2 i (T2 − T1 ) IE∗ e− t rs ds (L(T1 , T1 , T2 ) − κ)+ Ft = P (t, T1 , T2 ) r T1 (1) (ζ (s) − ζ (2) (s))2 ds 1 1 ×Bl + S(t, T1 , T2 ), t ,κ + , T1 − t . T2 − T1 T1 − t T2 − T1 Exercise 19.8 a) The LIBOR rate L(t, T, S) is a driftless geometric Brownian motion with bS . deterministic volatility function σ(t) under the forward measure P Explanation: The LIBOR rate L(t, T, S) can be written as the forward bt = Xt /Nt where Xt = (P (t, T ) − P (t, S))/(S − T ) price L(t, T, S) = X rt andr Nt = P (t, S). Since both discounted bond prices e− 0 rs ds P (t, T ) and t e− 0 rs ds P (t, S) are martingales under P∗ , the same is true of Xt . Hence bS L(t, T, S) = Xt /Nt becomes a martingale under the forward measure P b by Proposition 16.4, and computing its dynamics under PS amounts to removing any “dt” term in the original SDE defining L(t, T, S), i.e. we find ct , dL(t, T, S) = σ(t)L(t, T, S)dW 0 ≤ t ≤ T, hence L(t, T, S) = L(0, T, S) exp

cs − σ(s)dW

wt 0

σ 2 (s)ds/2 ,

bS . ct )t∈R is a standard Brownian motion under P where (W + b) Choosing the annuity numéraire Nt = P (t, S), we have h rS i h rS i IE∗ e− t rs ds ϕ(L(T, T, S)) Ft = IE∗ e− t rs ds NS ϕ(L(T, T, S)) Ft = Nt Ib E ϕ(L(T, T, S)) Ft = P (t, S)Ib E[ϕ(L(T, T, S)) | Ft ]. c) Given the solution

328

328 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition w T

σ 2 (s)ds/2 0 0 w wT T c = L(t, T, S) exp σ(s)dWs − σ 2 (s)ds/2 ,

L(T, T, S) = L(0, T, S) exp

cs − σ(s)dW

we find P (t, S)Ib E [ϕ(L(T, T, S)) | Ft ] h i r r b s − tT σ2 (s)ds/2 F b ϕ L(t, T, S)e tT σ(s)dW = P (t, S)IE t w∞ 2 2 2 dx , = P (t, S) ϕ L(t, T, S)ex−η /2 e−x /(2η ) p −∞ 2πη 2 wT cs is a centered Gaussian variable with variance η 2 := because σ(s)dW t wT b σ 2 (s)ds, independent of Ft under the forward measure P. t

Chapter 20 Exercise 20.1 For any x ∈ [0, T ], we have P(T − TNT > x and NT ≥ 1) P(NT ≥ 1) P(NT − NT −x = 0 and NT ≥ 1) = P(NT ≥ 1) P(NT − NT −x = 0 and NT −x ≥ 1) = P(NT ≥ 1) P(NT − NT −x = 0)P(NT −x ≥ 1) = P(NT ≥ 1)

P(T − TNT > x | NT ≥ 1) =

e−(T −(T −x))λ (1 − e−(T −x)λ ) 1 − e−λT e−(T −(T −x))λ − e−λT = 1 − e−λT e−λx − e−λT = , 0 ≤ t ≤ T. 1 − e−λT =

329 September 15, 2022

Solutions Manual Nt 6 5 4 3 2 1 0

TNT

T −x

We note that P(T − TNT > 0 | NT ≥ 1) = 1 and P(T − TNT > T | NT ≥ 1) = 0. Exercise 20.2 a) When t ∈ [0, T1 ), the equation reads dSt = −ηλSt- dt = −ηλSt dt, which is solved as St = S0 e−ηλt , 0 ≤ t < T1 . Next, at the first jump time t = T1 we have ∆St := St − St- = ηSt- dNt = ηSt- , which yields St = (1 + η)St- , hence ST1 = (1 + η)ST1- = S0 (1 + η)e−ηλT1 . Repeating this procedure over the Nt jump times contained in the interval [0, t] we get St = S0 (1 + η)Nt e−ληt , t ≥ 0. b) When t ∈ [0, T1 ) the equation reads dSt = −ηλSt- dt = −ηλSt dt, which is solved as St = S0 e−ηλt , 0 ≤ t < T1 . Next, at the first jump time t = T1 we have dSt = St − St- = dNt = 1, which yields St = 1 + St- , hence ST1 = 1 + ST1- = 1 + S0 e−ηλT1 , and for t ∈ [T1 , T2 ) we will find St = (1 + S0 e−ηλT1 )e−(t−T1 )ηλ ,

T1 ≤ t < T2 .

More generally, the equation can be solved by letting Yt := eηλt St and noting that (Yt )t∈R+ satisfies dYt = eληt dNt , which has the solution Yt = Y0 + 330

wt 0

eηλs dNs ,

t ≥ 0, 330 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition hence in general we have St = e−ηλt S0 +

wt 0

e−(t−s)ηλ dNs ,

t ≥ 0,

Exercise 20.3 a) Taking expectations on both sides of (20.40), we have u(t) = IE[St ] wt wt wt = IE S0 + µ Ss ds + σ Ss dBs + η Ss- dYs 0 0 0 w w w t t t = IE[S0 ] + IE µ Ss ds + IE σ Ss dBs + IE η Ss- dYs 0

= S0 + µ = S0 + µ

wt 0

IE[Ss ]ds + 0 + ηλ IE[Z] IE[Ss- ]ds 0 wt u(s)ds + ηλ IE[Z] u(s)ds, , t ≥ 0. 0

b) The above equation can be rewritten in differential form as u′ (t) = µu(t) + ηλ IE[Z]u(t) with u(0) = S0 , which admits the solution IE[St ] = u(t) = u(0)e(µ+λ IE[Z])t = S0 e(µ+λ IE[Z])t ,

t ≥ 0.

Exercise 20.4 a) We have

Xt =

 X0 eαt , 0 ≤ t < T1 ,         X0 eαT1 + σ e(t−T1 )α = X0 eαt + σe(t−T1 )α , T1 ≤ t < T2 ,      X0 eαT1 + σ e(T2 −T1 )α + σ e(t−T2 )α    = X0 eαt + σe(t−T1 )α + σe(t−T2 )α , T2 ≤ t < T3 ,

and more generally the solution (Xt )t∈R+ can be written as Xt = X0 eαt + σ

Nt X k=1

e(t−Tk )α = X0 eαt + σ

wt 0

e(t−s)α dNs ,

t ≥ 0.

(S.20.87) b) Letting f (t) := IE[Xt ] and taking expectation on both sides of the stochastic differential equation dXt = αXt dt + σdNt we find "

331 September 15, 2022

Solutions Manual df (t) = αf (t)dt + σλdt, or

f ′ (t) = αf (t) + σλ.

Letting g(t) = f (t)e−αt , we check that g ′ (t) = σλe−αt , hence g(t) = g(0) +

wt 0

g ′ (s)ds = g(0) + σλ

wt 0

λ e−αs ds = f (0) + σ (1 − e−αt ), α

and f (t) = IE[Xt ] = g(t)eαt λ = f (0)eαt + σ (eαt − 1) α λ = X0 eαt + σ (eαt − 1), α

t ≥ 0.

We could also take the expectation on both sides of (S.20.87) and directly find wt λ f (t) = IE[Xt ] = X0 eαt +σλ e(t−s)α ds = X0 eαt +σ (eαt −1), t ≥ 0. 0 α Exercise 20.5 a) We have Xt = X0

Nt Y

(1 + σ) = X0 (1 + σ)Nt = (1 + σ)Nt , t ∈ R+ .

k=1

b) By stochastic calculus and using the relation dXt = σXt- dNt , we have w wt t dSt = d S0 Xt + rXt Xs−1 ds = S0 dXt + rd Xt Xs−1 ds 0 0 w w w t t t = S0 dXt + rXt d Xs−1 ds + r Xs−1 ds dXt + rdXt • d Xs−1 ds 0 0 0 w t −1 −1 −1 = S0 dXt + rXt Xt dt + r Xs ds dXt + rdXt • (Xt dt) 0 w wt t = S0 dXt + rdt + r Xs−1 ds dXt = rdt + S0 + r Xs−1 ds dXt 0 0 wt −1 = rdt + σ S0 Xt- + rXtXs ds dNt = rdt + σSt- dNt . 0

c) We have 332

332 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition IE[Xt /Xs ] = IE[(1 + σ)Nt −Ns ] X = (1 + σ)k P(Nt − Ns = k) k≥0

= e−(t−s)λ

(1 + σ)k

k≥0

X ((t − s)(1 + σ)λ)k ((t − s)λ)k = e−(t−s)λ k! k! k≥0

= e−(t−s)λ e(t−s)(1+σ)λ = e(t−s)λσ ,

0 ≤ s ≤ t.

Remarks: We could also let f (t) = IE[Xt ] and take expectation in the equation dXt = σXt- dNt to get f ′ (t) = σλf (t)dt and f (t) = IE[Xt ] = f (0)eλσt = eλσt . Note that the relation IE[Xt /Xs ] = IE[Xt ]/ IE[Xs ], which happens to be true here, is wrong in general. d) We have wt wt IE[St ] = IE S0 Xt + rXt Xs−1 ds = S0 IE[Xt ] + r IE[Xt /Xs ]ds 0

= S0 e

λσt

= S0 eλσt +

(t−s)λσ

0 λσt

ds = S0 e

− 1)r , λσ

λσt

wt 0

λσs

t ≥ 0.

Exercise 20.6 a) Since IE[Nt ] = λt, the expectation IE[Nt − 2λt] = −λt is a decreasing function of t ∈ R+ , and (Nt − 2λt)t∈R+ is a supermartingale. b) We have St = S0 ert−λσt (1 + σ)Nt , t ≥ 0. c) The stochastic differential equation dSt = rSt dt + σSt- (dNt − λdt) contains a martingale component (dNt − λdt) and a positive drift rSt dt, therefore (St )t∈R+ is a submartingale. d) Given that σ > 0 we have ((1 + σ)k − 1)+ = (1 + σ)k − 1, hence e−rT IE∗ [(ST − K)+ ] = e−rT IE∗ [(S0 e(r−σλ)T (1 + σ)NT − K)+ ] = e−rT IE∗ [(S0 e(r−σλ)T (1 + σ)NT − S0 e(r−λσ)T )+ ] = S0 e−σλT IE∗ [((1 + σ)NT − 1)+ ] X = S0 e−σλT ((1 + σ)k − 1)+ P(NT = k) k≥0

= S0 e−σλT

((1 + σ)k − 1)P(NT = k)

k≥0

333 September 15, 2022

Solutions Manual = S0 e−σλT

(1 + σ)k P(NT = k) − S0 e−σλT

k≥0

= S0 e−σλT −λT

P(NT = k)

k≥0

X (T (1 + σ)λ)k k!

k≥0

− S0 e−σλT

= S0 (1 − e−σλT ), where we applied the exponential identity ex =

X xk k≥0

to x := T (1 + σ)λ. Exercise 20.7 a) For all k = 1, 2, . . . , Nt we have XTk − XTk- = a + σXTk- , hence

XTk = a + (1 + σ)XTk- ,

and continuing by induction, we obtain XTk = a + (1 + σ)a + · · · + (1 + σ)k−1 a + X0 (1 + σ)k =a

(1 + σ)k − 1 + X0 (1 + σ)k , σ

which shows that Xt = XTNt (1 + σ)Nt − 1 = X0 (1 + σ)Nt + a σ a a = (1 + σ)Nt X0 + − , t ≥ 0. σ σ This result can also be obtained by noting that a a XTk + = (1 + σ) XTk- + , k = 1, 2, . . . , Nt . σ σ b) We have IE[(1 + σ)Nt ] = e−λt

X n≥0

(1 + σ)k

(λt)k = eσλt , k!

t ≥ 0,

hence 334

334 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition IE[Xt ] = X0 eλσt + a

a a eλσt − 1 − , = eλσt X0 + σ σ σ

Exercise 20.8 We have St = S0 ert

Nt Y

t ≥ 0.

(1 + ηZk ), t ∈ R+ .

k=1

Exercise 20.9 We have  !2  NT X Var[YT ] = E  Zk − E[YT ]  k=1

 =

E

n≥0

NT X

 NT = k  P(NT = k)

Zk − λtE[Z]

k=1

−λt

X λn tn n≥0

= e−λt

X λ n tn n≥0

= e−λt

 E

n X

Zk − λtE[Z]



k=1

 E

n X

!2 − 2λtE[Z]

k=1

n X

 Zk + λ2 t2 (E[Z])2 

k=1

X λ n tn n≥0

 X

×E 2

Zk Zl +

n X

|Zk |2 − 2λtE[Z]

k=1

1≤k<l≤n

= e−λt

!2 

n X

 Zk + λ2 t2 (E[Z])2 

k=1

X λ n tn n≥0

×(n(n − 1)(E[Z])2 + nE |Z|2 − 2nλt(E[Z])2 + λ2 t2 (E[Z])2 ) X λn tn X λn tn = e−λt (E[Z])2 + e−λt E |Z|2 (n − 2)! (n − 1)! n≥2

−2e−λt λt(E[Z])2

n≥1

X λn tn + λ2 t2 (E[Z])2 ) (n − 1)!

n≥1

= λtE |Z|2 , or, using the moment generating function of Proposition 20.6, Var[YT ] = E |YT |2 − (E[YT ])2 ∂2 E[eαYT ]|α=0 − λ2 t2 (E[Z])2 ∂αw2 ∞ = λt |y|2 µ(dy) = λtE |Z|2 . =

−∞

335 September 15, 2022

Solutions Manual Exercise 20.10 a) Applying the Itô formula (20.24) to the function f (x) = ex and to the process Xt = µt + σWt + Yt , we find 1 dSt = µ + σ 2 St dt + σSt dWt + (St − St- )dNt 2 1 2 = µ + σ St dt + σSt dWt + (S0 eµt+σWt +Yt − S0 eµt+σWt +Yt− )dNt 2 1 = µ + σ 2 St dt + σSt dWt + (S0 eµt+σWt +Yt- +ZNt − eµt+σWt +Yt− )dNt 2 1 2 = µ + σ St dt + σSt dWt + St- (eZNt − 1)dNt , 2 hence the jumps of St are given by the sequence (eZk − 1)k≥1 . b) The discounted process e−rt St satisfies 1 d(e−rt St ) = e−rt µ − r + σ 2 St dt+σe−rt St dWt +e−rt St- (eZNt −1)dNt . 2 Hence by the Girsanov Theorem 20.20, choosing u, λ̃, ν̃ such that 1 µ − r + σ 2 = σu − λ̃ IEν̃ [eZ − 1], 2 shows that d(e−rt St ) = σe−rt St (dWt +udt)+e−rt St-

eZNt −1 dNt − λ̃ IEν̃ eZ −1 dt

is a martingale under (Pu,λ̃,ν̃ ). Exercise 20.11 a) We have St = S0 eµt

Nt Y

(1 + Zk ) = S0 exp µt +

k=1

Nt X

! Xk

t ≥ 0.

k=1

b) We have the discounted asset price process Set := e−rt St = S0 exp (µ − r)t +

Nt X

! Xk

t ≥ 0,

k=1

satisfies the stochastic differential equation dSet = (µ − r)Set dt + XNt Set- dNt 336

336 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition = (µ − r + λ IE[Z])Set dt + XNt − λ IE[Z] Set- dNt

t ≥ 0,

hence it is a martingale if 2

0 = µ − r + λ IE[Z] = µ − r + λ IE[eXk − 1] = µ − r + (eσ /2 − 1)λ. c) We have e−(T −t)r IE[(ST − κ)+ | St ]  =e

−(T −t)r

IE  S0 exp µT +

NT X

! Xk



−κ

St 

k=1

 =e

−(T −t)r

IE  St exp (T − t)µ +

NT X



−κ

St 

k=Nt +1

 =e

−(T −t)r

!+ 

NT X

IE  x exp (T − t)µ +

−κ



k=Nt +1

 = e−(T −t)r IE  x exp (T − t)µ +

NT −Nt X

x=St

!+ 

! −κ

k=1

−(T −t)r

(T −t)µ+

Pn k=1

x=St

+ −κ

P(NT − Nt = n)

x=St

n≥0

= e−(r+λ)(T −t)

((T − t)λ)n n! x=St

Pn + xe(T −t)µ+ k=1 Xk − κ

n≥0

= e−(T −t)λ

X ((T − t)λ)n n!

n≥0

−(T −t)λ

X n≥0



Bl(St e(µ−r)(T −t)+nσ /2 , r, nσ 2 /(T − t), κ, T − t)

((T − t)λ)n 2 , St e(µ−r)(T −t)+nσ /2 Φ(d+ ) − κe−(T −t)r Φ(d− ) n!

with 2

log(St e(µ−r)(T −t)+nσ /2 /κ) + (T − t)r + nσ 2 /2 √ σ n log(St /κ) + (T − t)µ + nσ 2 √ = , σ n

d+ =

d− =

log(St e(µ−r)(T −t)+nσ /2 /κ) + (T − t)r − nσ 2 /2 √ σ n

337 September 15, 2022

Solutions Manual =

log(St /κ) + (T − t)µ √ , σ n 2

and µ = r + (1 − eσ /2 )λ. Exercise 20.12 a) We have

d(eαt St ) = σeαt (dNt − βdt),

hence eαt St = S0 + σ

wt 0

eαs (dNs − βd),

and St = S0 e−αt + σ

wt 0

e−(t−s)α (dNs − βds),

t ≥ 0.

(S.20.88)

b) We have f (t) = IE[St ]

w t 0

wt e−(t−s)α dNs − βσ e−(t−s)α ds

= S0 e−αt + σ IE = S0 e

−αt

+ λσ

−(t−s)α

ds − βσ

−(t−s)α

1 − e−αt = S0 e−αt + (λ − β)σ α λ−β β − λ −αt =σ + S0 + σ e , α α

t ≥ 0.

c) By rewriting (S.20.88) as wt wt e−(t−s)α ds + σ e−(t−s)α (dNs − βds) St = S0 − αS0 0 0 wt −(t−s)α = S0 + σ e (dNs − (β + αS0 /σ)ds) 0 wt wt −(t−s)α = S0 + σ e (λ − β − αS0 /σ)ds) + σ e−(t−s)α (dNs − λds), 0

t ∈ R+ , we check that the process (St )t∈R+ is a submartingale, provided that λ − β − αS0 /σ ≥ 0, i.e. S0 + (β − λ)σ/α ≤ 0. We also check that this condition makes the expectation f (t) = IE[St ] decreasing in Question (b). d) Since, given that NT = n the jump times (T1 , T2 , . . . , Tn ) of the Poisson process (Nt )t∈R+ are independent uniformly distributed random variables over [0, T ], hence we can write wT IE[ϕ(ST )] = IE ϕ S0 e−αT + σ e−(T −s)α (dNs − βds), 0

338

338 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition X

P(NT = n)

n≥0

wT NT = n × IE ϕ S0 e−αT + σ e−(T −s)α (dNs − βds) 0

= e−λT

X (λT )n n!

n≥0

" × IE ϕ S0 e−αT + σ

n X

e−(T −Tk )α − σβ

k=1

= e−λT ×

! e−(T −s)α ds

# NT = n

X λn n≥0

···

wT 0

ϕ S0 e−αT + σ

n X

e−(T −sk )α − σβ

k=1

1 − e−αT α

! ds1 · · · dsn ,

T ≥ 0. Exercise 20.13 a) From the decomposition Yt − λt(t + IE[Z]) = Yt − λ IE[Z]t − λt2 as the sum of a martingale and a decreasing function, we conclude that t 7→ Yt − λt(t + IE[Z]) is a supermartingale. b) Writing dSt = µSt dt + σSt- dYt r−µ = rSt dt + σSt- dYt − dt σ = rSt dt + σSt- dYt − λ̃ IE[Z]dt ,

0 ≤ t ≤ T,

we conclude that (St )t∈[0,T ] is a martingale under Pλ̃ provided that µ−r = −λ̃ IE[Z]dt, σ i.e. λ̃ =

r−µ . σ IE[Z]

We note that λ̃ < 0 if µ < r, hence in this case there is no risk-neutral probability measure and the market admits arbitrage opportunities as the risky asset always overperforms the risk-free interest rate r. c) We have e−(T −t)r IEλ̃ [ST − κ | Ft ] = ert IEλ̃ [e−rT ST | Ft ] − Ke−(T −t)r = St − Ke−(T −t)r , "

339 September 15, 2022

Solutions Manual since (St )t∈[0,T ] is a martingale under Pλ̃ . Exercise 20.14 a) We have St = S0 eµt

Nt Y

(1 + Zk ),

t ≥ 0.

k=1

b) Letting Xk = log(1 + Zk ), k ≥ 1, we find that e

−rt

St = S0 exp (µ − r)t +

Nt X

! Xk

t ≥ 0,

k=1

and (20.43) can be rewritten for the discounted price process Set := e−rt St , as

t ≥ 0,

dSet = (µ − r + λ IE[Z])Set dt + Set- (dYt + λ IE[Z]),

which becomes a martingale if 0 = µ − r + λ IE[Z] = µ − r + λ

w∞

zν(dz).



−∞

c) We have e−(T −t)r IE[(ST − κ)+ | St ]   !+ NT Y µT −(T −t)r St  Zk − κ =e IE  S0 e k=1

 =e

−(T −t)r

NT Y

IE  St e(T −t)µ

n≥0

 = e−(r+λ)(T −t)

IE  St e(T −t)µ

n≥0

−(r+λ)(T −t)

w∞

−∞

···

NT Y

!+ Zk − κ

k=Nt +1

 St 

((T − t)λ)n n!

X ((T − t)λ)n n≥0

St  P(NT − Nt = n)

Zk − κ

k=Nt +1

w∞

−∞

n! St e(T −t)µ

n Y

!+ zk − κ

ν(dz1 ) · · · ν(dzn ).

k=1

Exercise 20.15 340

340 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition a) The discounted price process (e−rt St )t∈[0,T ] is a martingale, hence it is both a submartingale and a supermartingale. b) The discounted price process (e−rt St )t∈[0,T ] is a supermartingale. c) The discounted price process (e−rt St )t∈[0,T ] is a submartingale. e , the discounted price process (e−rt St )t∈[0,T ] d) Under the probability measure P λ̃ is a martingale, hence it is both a submartingale and a supermartingale.

Chapter 21 Exercise 21.1 a) We have IE[Nt − αt] = IE[Nt ] − αt = λt − αt, hence Nt − αt is a martingale if and only if α = λ. Given that d(e−rt St ) = ηe−rt St- (dNt − αdt), we conclude that the discounted price process e−rt St is a martingale if and only if α = λ. b) Since we are pricing under the risk-neutral probability measure we take α = λ. Next, we note that ST = S0 e(r−ηλ)T (1 + η)NT = St e(r−ηλ)(T −t) (1 + η)NT −Nt ,

0 ≤ t ≤ T,

hence the price at time t of the option is e−(T −t)r IE[|ST |2 | Ft ] = e−(T −t)r IE[|St |2 e2(r−ηλ)(T −t) (1 + η)2(NT −Nt ) | Ft ] = |St |2 e(r−2ηλ)(T −t) IE[(1 + η)2(NT −Nt ) | Ft ] = |St |2 e(r−2ηλ)(T −t) IE[(1 + η)2(NT −Nt ) ] X = |St |2 e(r−2ηλ)(T −t) (1 + η)2n P(NT − Nt = n) n≥0

= |St | e

2 (r−2ηλ−λ)(T −t)

(1 + η)2n

n≥0

(λ(T − t))n n!

2 (r−2ηλ−λ)(T −t)+(1+η)2 λ(T −t)

= |St | e

= |St |2 e(r+η λ)(T −t) ,

0 ≤ t ≤ T.

Exercise 21.2 a) Regardless of the choice of a particular risk-neutral probability measure Pu,λ̃,ν̃ , we have e−(T −t)r IEu,λ̃,ν̃ [ST − K | Ft ] = ert IEu,λ̃,ν̃ [e−rT ST | Ft ] − Ke−(T −t)r "

341 September 15, 2022

Solutions Manual = ert e−rt St − Ke−(T −t)r = St − Ke−(T −t)r = f (t, St ), for

f (t, x) = x − Ke−(T −t)r ,

t, x > 0.

b) Clearly, holding one unit of the risky asset and shorting a (possibly fractional) quantity Ke−rT of the riskless asset will hedge the payoff ST − K, and this (static) hedging strategy is self-financing because it is constant in time. ∂f (t, x) = 1 we have c) Since ∂x e ∂f aλ (t, St- ) + (f (t, St- (1 + a)) − f (t, St- )) ∂x Stξt = e σ 2 + a2 λ e a λ (St- (1 + a) − St- ) σ2 + St= e σ 2 + a2 λ = 1, 0 ≤ t ≤ T, σ2

which coincides with the result of Question (b). Exercise 21.3 a) We have

1 St = S0 exp µt + σBt − σ 2 t (1 + η)Nt . 2

b) We have 1 Set = S0 exp (µ − r)t + σBt − σ 2 t (1 + η)Nt , 2 and

dSet = (µ − r + λη)Set dt + η Set- (dNt − λdt) + σ Set dWt ,

hence we need to take

µ − r + λη = 0,

since the compensated Poisson process (Nt − λt)t∈R+ is a martingale. c) We have e−r(T −t) IE∗ [(ST − κ)+ | St ] " # + 1 2 ∗ −r(T −t) NT St =e IE S0 exp µT + σBT − σ T (1 + η) − κ 2 342

342 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition + 2 St = e−r(T −t) IE∗ St eµ(T −t)+(BT −Bt )σ−(T −t)σ /2 (1 + η)NT −Nt − κ X P(NT − Nt = n) = e−r(T −t) n≥0

× IE∗

+ 2 St St eµ(T −t)+(BT −Bt )σ−(T −t)σ /2 (1 + η)n − κ

= e−(r+λ)(T −t)

X (λ(T − t))n n≥0

× IE∗

+ 2 St St e(r−λη)(T −t)+(BT −Bt )σ−(T −t)σ /2 (1 + η)n − κ

(λ(T − t))n n! n≥0 (λ(T − t))n X = e−λ(T −t) , St e−λη(T −t) (1 + η)n Φ(d+ ) − κe−r(T −t) Φ(d− ) n!

= e−λ(T −t)

Bl(St e−λη(T −t) (1 + η)n , r, σ 2 , T − t, κ)

n≥0

with log(St e−λη(T −t) (1 + η)n /κ) + (r + σ 2 /2)(T − t) √ σ T −t log(St (1 + η)n /κ) + (r − λη + σ 2 /2)(T − t) √ = , σ T −t

d+ =

and log(St e−λη(T −t) (1 + η)n /κ) + (r − σ 2 /2)(T − t) √ σ T −t log(St (1 + η)n /κ) + (r − λη − σ 2 /2)(T − t) √ = . σ T −t

d− =

Exercise 21.4 a) The discounted process Set = e−rt St satisfies the equation dSet = YNt Set- dNt , , and it is a martingale since the compound Poisson process YNt dNt is centered with independent increments as IE[Y1 ] = 0. b) We have NT Y ST = S0 erT (1 + Yk ), k=1

hence "

343 September 15, 2022

Solutions Manual  !+  NT Y + −rT rT  (1 + Yk ) − κ  IE (ST − κ) = e IE S0 e

−rT

k=1

 = e−rT

IE  S0 erT

n≥0

NT Y

IE  S0 e

k≥0

−rT −λT

 NT = n P(NT = n)

(1 + Yk ) − κ

k=1

 −rT −λT

!+ 

 (λT ) n!

(1 + Yk ) − κ

k=1

X (λT )n w 1 k≥0

n Y

2n n!

−1

···

−1

S0 e

n Y

!+ (1 + yk ) − κ

dy1 · · · dyn .

k=1

Exercise 21.5 a) We find α = λ where λ is the intensity of the Poisson process (Nt )t∈R+ . b) We have e−(T −t)r IE[ST − κ | Ft ] = ert IE[e−rT ST | Ft ] − e−(T −t)r IE[κ | Ft ] = ert IE[e−rt St | Ft ] − e−(T −t)r κ = St − e−(T −t)r κ, since the process (e−rt St )t∈R+ is a martingale. Exercise 21.6 a) We have

St = S0 e(r−λα)t (1 + α)Nt ,

t ≥ 0.

b) We have ST e−(T −t)r IE∗ [ϕ(ST ) | Ft ] = e−(T −t)r IE∗ ϕ x St |x=St ∗ −(T −t)r (r−λα)(T −t) NT −Nt =e IE ϕ(xe (1 + α) ) |x=S t

−(r+λ)(T −t)

∞ X ((T − t)λ)k k=0

ϕ St e

(r−λα)(T −t)

(1 + α)k ,

0 ≤ t ≤ T.

c) We have dVt = rηt ert dt + ξt dSt = rηt ert dt + ξt (rSt dt + αSt- (dNt − λdt)) = rVt dt + αξt St- (dNt − λdt) = rf (t, St )dt + αξt St- (dNt − λdt). 344

(S.21.89) 344 September 15, 2022

Introduction to Stochastic Finance with Market Examples, Second Edition d) By the Itô formula with jumps we have d(e−rt f (t, St )) = −re−rt f (t, St )dt + e−rt df (t, St ) = −re−rt f (t, St )dt ∂f ∂f ∂f +e−rt (t, St )dt + rSt (t, St )dt − λSt (t, St )dt ∂t ∂x ∂x + f (t, (1 + α)St- ) − f (t, St- ) dNt = −re−rt f (t, St )dt ∂f ∂f ∂f (t, St ) + rSt (t, St ) − λSt (t, St ) +e−rt ∂t ∂x ∂x −rt +λe (f (t, (1 + α)St ) − f (t, St )) dt +e−rt f (t, (1 + α)St- ) − f (t, St- ) dNt −λe−rt IE[f (t, (1 + α)x) − f (t, x)]|x=St- dt. Since the discounted price process (e−rt f (t, St ))t∈R+ is a martingale under the risk-neutral measure, d(e−rt f (t, St )) must reduce to its martingale component, i.e. the sum of “dt” terms vanishes, and we get d(e−rt f (t, St )) = e−rt f (t, (1 + α)St- ) − f (t, St- ) dNt − λe−rt IE[f (t, (1 + α)x) − f (t, x)]|x=St- dt = e−rt f (t, (1 + α)St- ) − f (t, St- ) dNt − λe−rt f (t, (1 + α)St- ) − f (t, St- ) dt, or equivalently df (t, St ) = rf (t, St )dt+ f (t, (1+α)St- )−f (t, St- ) (dNt −λ)dt. (S.21.90) Finally, by identification of the terms in the above formula (S.21.90) with those appearing in (S.21.89), we obtain αξt St- (dNt − λdt) = f (t, (1 + α)St- ) − f (t, St- ) (dNt − λdt), which yields the hedging strategy ξt =

1 f (t, (1 + α)St- ) − f (t, St- ) , αSt-

0 < t ≤ T.

Exercise 21.7 a) We have IE[Nt | Fs ] = eθYs −sm(θ) IE e(Yt −Ys )θ−(t−s)m(θ) Fs = eθYs −sm(θ) IE e(Yt −Ys )θ−(t−s)m(θ) = Ns , "

0 ≤ s ≤ t. 345

September 15, 2022

Solutions Manual b) We have NT IEθ e−rt St | Fs = IE eYt Fs Ns N t Fs = IE eYt Ns = eYs IE eYt −Ys e(Yt −Ys )θ−(t−s)m(θ) | Fs = eYs e−(t−s)m(θ) IE e(1+θ)(Yt −Ys ) = eYs e−(t−s)m(θ) e(t−s)m(θ+1) , hence we should have m(θ) = m(θ + 1). For example, when (Yt )t∈R+ = (Nt − t)t∈R+ is a compensated Poisson process we have m(θ) = eθ − θ − 1 and the condition reads eθ + 1 = eθ+1 , i.e. θ = − log(e − 1). c) We have NT e−(T −t)r IEθ [(ST − K)+ | Ft ] = e−(T −t)r IE (ST − K)+ Ft Nt # " θ ST −(T −t)((1+r)θ+m(θ)) + Ft . =e IE (ST − K) St

Chapter 22 Exercise 22.1 For all j = 1, 2, . . . , M − 1 we have Bj,j−1 + Bj,j + Bj,j+1 = 1 + r∆t, hence when the terminal condition is a constant ϕ(T, x) = c > 0 we get −(N −i) T , ϕ(ti , x) = c(1 + r∆t)−(N −i) = c 1 + r N

i = 0, . . . , N.

In particular, when the number N of discretization steps tends to infinity, denoting by [x] the integer part of x ∈ R we find ϕ(s, x) = lim ϕ t[N s/T ] , x N →∞ −(N −[N s/T ]) T = c lim 1 + r N →∞ N −[N (T −s)/T ] T = c lim 1 + r N →∞ N

346

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Introduction to Stochastic Finance with Market Examples, Second Edition −(T −s)/T T 1+r N →∞ N

= c lim

= ce−r(T −s) , for all s ∈ [0, T ], as expected. Exercise 22.2 a) We have bN = X btN + rX btN (tk+1 − tk ) + σ X btN (Wt X − Wtk ), tk+1 k+1 k k k which yields btN = X btN X 0 k

k Y

1 + r(ti − ti−1 ) + (Wti − Wti−1 )σ ,

k = 0, 1, . . . , N.

i=1

b) We have btN = X btN + (r − σ 2 /2)X btN (tk+1 − tk ) + σ X btN (Wt X − W tk ) k+1 k+1 k k k 1 2 bN + σ Xtk (Wtk+1 − Wtk )2 , 2 which yields btN = X btN X 0 k

k Y i=1

1 1 + (r − σ 2 /2)(ti − ti−1 ) + (Wti − Wti−1 )σ + (Wti − Wti−1 )2 σ 2 . 2

347 September 15, 2022