MONTE CARLO SAMPLING
MONTE CARLO SAMPLING
٢
:MONTE CARLO SAMPLING ( ﻃﺮﻳﻘﺔ ﻣﺎﻧﺘﻮ ﻛﺎرﻟﻮا١-١)
We considered before how a simple random sample
observations could be selected from a finite and existent
population using tables of random digits . We now discuss methods of generating observations from theoretical
distributions which are usually called Monte Carlo Sampling methods. We firstly present a key result from a theorem.
Theorem(1):
Let X be a continuous random variable with C.D.F. F(x). Then the random variable Z=F(x) is uniformly distribution over the 1 has the C.D.F. F(x).(The transformation x F (z) interval(0,1)
z=F(x) is called the probability integral transformation).
*proof: Since F(x) is anon decreasing function of the argument x ,the 1 defined for any value of z between 0 and inverse function F (x)
x F1 (z) satisfying F(x) z .Now : 1 as the smallest value of p(Z z) p F(X) z p X F1 (z) FF1 (z) z
For 0<z<1
Which is the C.D.F. of the uniform distribution over(0,1). Conversely, if Z is uniformly distributed over (0,1), P(X x) P F1 (Z) x P Z F(x) F(x)
MONTE CARLO SAMPLING
٣
Since Z has a c.d.f. z(0<z<1).This completes the proof.
Example (1-1-1): Consider a random variable X having an exponential distribution with c.d.f. F(x) 1 exp(x), x 0 with F(x)=0 elsewhere . Then the random variable Z F(X) 1 exp(X) Uniformly distributed over (0,1) . Also, since F1 x 1 log(1 x) ,it follows that if Z is a random variable 1 1 uniformly distributed over (0,1), then X F (Z) log(1 Z)
has the given exponential distribution.
The importance of theorem ( ) is clear. If we are able to option a stream of independent observation z1 ,z 2 ,..., on a random variable Z which is uniformly distributed over (0,1), we can generate a stream of observations x1 , x 2 ,..., from a continuous population 1 1 with a given c.d.f F(x) by setting x1 F (Z1 ), x 2 F (Z2 ),... This
method is known as the inverse transformation method. Observations for discrete random variable can also be
generated using observations on Z. Thus suppose that X is a discrete random variable which may assume the values x1 , x 2 ,..., x k with corresponding probabilities p1 , p 2 ,..., p k k
where
p i 1
i
1.
Consider the use of the transformation
MONTE CARLO SAMPLING
٤
0 Z P1
if
P1 Z P1 P2
if
K 1
P Z P P 1
2
if
1
I1
x1 X x 2 x k
We see that: i
i 1
) p(X x i ) p( p j Z p j j1
j1
i 1
i
p j p j pi j1
j1
As desired. أﻫـﻢ ﺗﻄﺒﻴـﻖ ﻟﻠﻨﻈﺮﻳـﺔ اﻟﺴـﺎﺑﻘﺔ ) (١ﻫـﻮ ﺗﻮﻟﻴـﺪ ﻣﺘﻐﻴـﺮات ﻋﺸـﻮاﺋﻴﺔ " “ pseudoﻣـﻦ ﺗﻮزﻳﻌـﺎت ﻣﻌﻴﻨـﺔ
ﺑﺎﺳﺘﺨﺪام اﻟﺤﺎﺳﺐ اﻵﻟﻲ .ﺑﻤﻌﻨﻲ آﺧـﺮ ،ﻧﺤﺼـﻞ ﻋﻠـﻰ ﻋﻴﻨـﺔ ﻋﺸـﻮاﺋﻴﺔ ﻣـﻦ ﺗﻮزﻳـﻊ ﻣﻌـﻴﻦ ﺑﺪاﻟـﺔ ﺗﻮزﻳـﻊ ﺗﺠﻤﻴﻌـﻲ ) . F(xإذا ﻟـﺪﻳﻨﺎ nأﻋـﺪاد ﻋﺸـﻮاﺋﻴﺔ ،ﻟـﺘﻜﻦ y1 , y2 ,… , ynﺗـﻢ ﺗﻮﻟﻴـﺪﻫﺎ ﻋﻠـﻰ اﻟﺤﺎﺳـﺐ اﻵﻟـﻲ
ﻣﻦ ﺗﻮزﻳﻊ ﻣﻨﺘﻈﻢ ﻓﻰ اﻟﻔﺘﺮة ) (0, 1وﺗﺒﻌﺎ ﻟﺬﻟﻚ ﻓﺈن x1 , x2 , … xnﺗﺤﺴﺐ ﻛﺎﻵﺗﻲ : i 1,2,..., n .
) x i G(y i
واﻟﺘــﻲ ﺗﻘﺎﺑــﻞ ﺗﻮﻟﻴــﺪ ﻋﻴﻨــﺔ ﻋﺸــﻮاﺋﻴﺔ ﻣــﻦ ﺗﻮزﻳــﻊ ﻟــﻪ ) . F(xﺑــﺎﻟﻄﺒﻊ ﻓــﻲ ﻛﺜﻴــﺮ ﻣــﻦ اﻷﻣﺜﻠــﺔ ﻓــﺈن ) F(xﺗﻜــﻮن 1 ﺗﻨﺎﻇﺮﻳﺔ وﻋﻠـﻰ ذﻟـﻚ ﻳﻤﻜـﻦ اﺳـﺘﺨﺪام ) . x i F ( y iأﻳﻀـﺎ ﻳﻤﻜـﻦ اﺳـﺘﺨﺪام اﻟﻤﻌﺎدﻟـﺔ ﻟﻠﻤﺘﻐﻴـﺮات
اﻟﻌﺸﻮاﺋﻴﺔ ﻣﻦ اﻟﻨﻮع اﻟﻤﺘﻘﻄﻊ .
ﻣﺜﺎل):(٢-١-١ إذا ﻛﺎن Xﻣﺘﻐﻴﺮاً ﻋﺸﻮاﺋﻴﺎً ﺑﺤﻴـﺚ أن ) X ~ BIN(1,1/2ﻓـﺈن داﻟـﺔ اﻟﺘﻮزﻳـﻊ اﻟﺘﺠﻤﻴﻌـﻲ ﺗﻜـﻮن ﻋﻠـﻰ اﻟﺸﻜﻞ :
MONTE CARLO SAMPLING
٥
x0 0 x 1
F( x ) 0 1/ 2 1
1 x
واﻟﺪاﻟﺔ ) G(yﺳﻮف ﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ : G ( y) 0
0 y 1/ 2 1 y 1. 2
1
ﻋﻤﻮﻣﺎ ﻟﺘﻮﻟﻴﺪ ﻣﺸﺎﻫﺪات ﻷي ﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ Xﻣﻦ اﻟﻨﻮع اﻟﻤﺘﻘﻄﻊ و ﻟﻴﻜﻦ ﻓﻀﺎء اﻟﻤﺘﻐﻴﺮ Xﻫﻮ { = A }… . b1 , b2 , b3 ,ﻓﻰ ﻫﺬﻩ اﻟﻤﻨﺎﻗﺸﺔ ﺳﻮف ﻧﻔﺘﺮض وﺟﻮد ﺳﺘﺔ ﻗﻴﻢ ﻓﻘﻂ ﻓﻰ اﻟﻔﻀﺎء Rوأن 0 . < b1 < b2 < …. < b6ﻟـﻴﻜﻦ ) pi = P(X = bi) = f(biﺣﻴـﺚ . i 1,2,3,...,6ﻫـﺬﻩ اﻻﺣﺘﻤــﺎﻻت ﻣﻮﺿــﺤﺔ ﻣــﻊ ) F(xﻓــﻰ ﺷــﻜﻞ ) .( ٤-٨ﻟﺘﻮﻟﻴــﺪ أي ﻣﺸــﺎﻫﺪة ﻣــﻦ ، Xﻧﻘــﻮم ﺑﺘﻮﻟﻴــﺪ ﻋــﺪد ﻋﺸﻮاﺋﻲ Y = yﺣﻴﺚ Yﻳﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻟﻤﻨﺘﻈﻢ ﻓﻰ اﻟﻔﺘﺮة ) . (0, 1إذا ﻛﺎﻧﺖ y p1ﻓـﺈن X = b1 وإذا ﻛﺎﻧـ ـ ـ ــﺖ p1 y p1 p 2ﻓـ ـ ـ ــﺈن y p1 .... p 6 1
b2
=
Xوﻫﻜـ ـ ـ ــﺬا ﺣﺘـ ـ ـ ــﻰ اﻟﻮﺻـ ـ ـ ــﻮل
p1 ... p 5ﻓﺈن . X = b6ﻓﻌﻠﻰ ﻋﻠﻰ ﺳﺒﻴﻞ اﻟﻤﺜﺎل :
P[p1 Y p1 p 2 ] p1 p 2 p1 p 2
وﻫﻮ اﻻﺣﺘﻤﺎل اﻟﺼﺤﻴﺢ ﻋﻨﺪﻣﺎ . X = b2ﺗﺒﻌﺎ ﻟﻬﺬﻩ اﻟﻄﺮﻳﻘﺔ ﻳﻤﻜﻨﻨﺎ اﻟﺤﺼﻮل ﻋﻠـﻰ اﻻﺣﺘﻤـﺎﻻت اﻟﻤﻄﻠﻮﺑـﺔ ﻋﻨﺪﻣﺎ . X = b6 , … , X = b2 , X = b1
MONTE CARLO SAMPLING
٦
ﻣﺜﺎل ):(٣-١-١ ﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﻟﻪ داﻟﺔ اﻟﻜﺜﺎﻓﺔ اﻹﺣﺘﻤﺎﻟﻴﺔ اﻟﺘﺎﻟﻴﺔX:إذا ﻛﺎن 6
5
4
3
2
1
X
1/6
1/6
1/6
1/6
1/6
1/6
)P(X=x
اﻟﻤﻄﻠﻮب ﺗﻮﻟﻴﺪ ﺑﻴﺎﻧﺎت ﺗﺘﺒﻊ ﻫﺬا اﻟﺘﻮزﻳﻊ ﺑﺤﻴﺚ:
, x=1,2,3,4,5,6
f(x)=1/6 =0 e.w
اﻟﺤﻞ: 1 X 1 6 1 1 1 Z X 2 6 6 6 2 3 Z X 3 6 6 3 4 Z X 4 6 6 4 5 Z X 5 6 6 5 6 Z X 6 6 6
0 Z
MONTE CARLO SAMPLING
ﺣﻴﺚ Zﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻟﻤﻨﺘﻈﻢ
٧
)U(0,1
ﻣﺜﺎل ):(٤-١-١ إذا ﻛﺎن ) Y ~ UNIF (0, 1أﺷﺮح ﻛﻴﻒ ﻳﻤﻜﻦ ﺗﻮﻟﻴﺪ ﻣﺸﺎﻫﺪة ﻟﻬﺎ داﻟﻪ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل اﻟﺘﺎﻟﻴﺔ : 1
1
1 ) f ( x ) ( x 2 (1 x ) 2 4
0 x 1
= 0 elsewhere .
اﻟﺤﻞ :
ﻳﻤﻜﻦ ﻛﺘﺎﺑﺔ اﻟﺪاﻟﺔ ﻋﻠﻰ اﻟﺸﻜﻞ :
ﺣﻴﺚ :
f ( x ) a f1 ( x ) (1 a ) f 2 ( x ) , 0 a 1
f1 ( x ) ( 1 - a) f 2 ( x ) dx 1 .
a
f ( x ) dx -
ﺣﻴﺚ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f1 ( xﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ : 1
0 x 1
2 x
f1 (x )
ﺑﺪاﻟﺔ ﺗﻮزﻳﻊ ﺗﺠﻤﻴﻌﻲ ﻋﻠﻰ اﻟﺸﻜﻞ : F1 ( x ) x 1
0 x 1 x 1.
1 2 ﺑﻮﺿﻊ Y F1 (X) Xﻓﺈن . X F1 (Y ) Yأﻳﻀﺎ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f 2 ( xﺗﻜﻮن
ﻋﻠﻰ اﻟﺸﻜﻞ :
1
0 x 1
ﺑﺪاﻟﺔ ﺗﻮزﻳﻊ ﺗﺠﻤﻴﻌﻲ ﻋﻠﻰ اﻟﺸﻜﻞ :
1 f 2 ( x ) (1 x ) 2 2 0 elsewhere .
MONTE CARLO SAMPLING
0 x 1
٨ 1 x) 2
F2 ( x ) (1
x0
0
x 1
1
ﺑﻮﺿﻊ : 1 Y F2 (X ) (1 X ) 2
ﻓـﺈن :
أي أن اﻟﺪاﻟﺔ ﻳﻤﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻋﻠﻰ اﻟﺸﻜﻞ :
X F21 (X ) 1 Y 2 1
1
1 1 2 1 f ( x ) ( x ) ((1 x ) 2 ). 2 2 2
اﻵن ﻧﻠﻘﻰ ﻋﻤﻠﺔ وﻧﻜﺘﺐ X = Y2إذا ﻛﺎن اﻟﻨﺎﺗﺞ وﺟﻪ ﺑﻴﻨﻤﺎ ) X = ( 1-Y2إذا ﻛﺎن اﻟﻨﺎﺗﺞ ﻛﺘﺎﺑﺔ.
ﻣﺜﺎل):(٥-١-١
إذا ﻛﺎن Yﻣﺘﻐﻴﺮاً ﻋﺸﻮاﺋﻴﺎً ﺣﻴﺚ ) Y ~ UNIF (0,1أﺷﺮح ﻛﻴﻒ ﻳﻤﻜﻦ اﺳﺘﺨﺪام Yﻟﻠﺤﺼﻮل ﻋﻠﻰ
ﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎل : 0 x 1
1 1 2 1 f ( x ) 3 (( x ) 1 2x 2 ) , 2 8
= 0 elsewhere .
اﻟﺪاﻟﺔ ﻳﻤﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻋﻠﻰ اﻟﺸﻜﻞ : 1
1
12 1 3 3 f ( x ) ( x ) 2 (1 2 x ) 2 ( 2x 1) 2 4 2 8 8 2 3 3 f1 ( x ) f 2 ( x ) f 3 ( x ). 8 8 8
ﺣﻴﺚ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f1 ( xﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ :
MONTE CARLO SAMPLING
٩ 1 f1 ( x ) 12 ( x ) 2 2
0 x 1
0 elswhere.
ﺑﺪاﻟﺔ ﺗﻮزﻳﻊ ﺗﺠﻤﻴﻌﻲ ﻋﻠﻰ اﻟﺸﻜﻞ : 1 1 F1 ( x ) 4( x ) 3 2 2 0 x 0 1 x 1.
0 x 1
ﺑﻮﺿـﻊ : Y 1 1/ 3 1 1 1 ) Y F1 (X ) 4( X ) 3 4 8 ﻓـﺈن 2 2 2
( X F 1 ( y)
أﻳﻀﺎ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f 2 ( xﺗﻜﻮن ﻋﻠﻰ اﻟﺸﻜﻞ : 1 2
0x
1 2
| | 1 2x
1 2
) f 2 ( x ) (1 2x
0 elsewhere . 1 ) (1 Y 2 2 ﻟﻬﺬﻩ اﻟﺪاﻟﺔ ﻧﺄﺧﺬ
X
وذﻟﻚ ﻷن :
ﺣﻴﺚ Xﻫﻨﺎ ﻣﺘﻐﻴﺮا ﻋﺸﻮاﺋﻴﺎ ﻟﻪ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f 2 ( x
1 ) P (X x ) P( (1 Y 2 ) x 2 1 1 (1 2x ) 2 .
ﺗﻤﺜﻞ داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﻤﻘﺎﺑﻠﺔ ﻟﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) . f 2 ( x وأﺧﻴﺮا داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ) f 3 ( xﻋﻠﻰ اﻟﺸﻜﻞ : 1 x 1 2
1
1- 2 x 2
1 2
)f 3 ( x ) ( 2x 1
0 elsewhere .
MONTE CARLO SAMPLING
١٠
F3 ( x )
1 ( 2x 1) 2
1 (1 Y 2 ) 2 وذﻟﻚ ﻷﻧﻪ ﻳﻤﻜﻦ إﺛﺒﺎت أن داﻟﺔ اﻟﺘﻮزﻳﻊ ﻟﻬﺬﻩ اﻟﺪاﻟﺔ ﻧﺄﺧﺬ . f 3 ( x ) ﻫﻲ ﻟﻤﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﻟﻪ اﻟﺪاﻟﺔ X
A ( ﻧﻠﻘﻰ ﻋﻤﻠﺔ ﺛﻼث ﻣﺮات وﺑﻔﺮض أن٨-٨ ) وﻋﻠﻰ ذﻟﻚ ﻟﺘﻮﻟﻴﺪ ﻣﺸﺎﻫﺪة ﺗﺘﺒﻊ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻓﻰ
اﻟﺤﺎدﺛﺔ اﻟﺤﺼﻮل ﻋﻠﻰ وﺟﻬﻴﻦ وﻛﺘﺎﺑﺔ ﻓﻰBاﻟﺤﺎدﺛﺔ ﻇﻬﻮر ﺛﻼﺛﺔ وﺟﻮﻩ أو ﺛﻼﺛﺔ ﻛﺘﺎﺑﺔ ) ﻓﻰ أي ﺗﺮﺗﻴﺐ ( و اﻟﺤﺎدﺛﺔ اﻟﺤﺼﻮل ﻋﻠﻰ أﺛﻨﻴﻦ ﻛﺘﺎﺑﺔ ووﺟﻪC أى ﺗﺮﺗﻴﺐ و
: ﻛﺎﻟﺘﺎﻟﻲX ) ﻓﻰ أي ﺗﺮﺗﻴﺐ ( وﻋﻠﻰ ذﻟﻚ ﻧﻌﺮف اﻟﻤﺘﻐﻴﺮ اﻟﻌﺸﻮاﺋﻲ 1
Y 13 1 X 2 4 8
A إذا وﻗﻌﺖ
X
1 (1 Y 2 ) 2
B إذا وﻗﻌﺖ
X
1 (1 Y 2 ) 2
C إذا وﻗﻌﺖ
Example(1-1-6): Suppose that a random sample of 10 observation is required from the standard exponential distribution with p.d.f.
f(x)=exp(-x),x>0. using the first five columns of digits we obtain the observations shown below which serve as a sample of independent observations from the uniform distribution
over(0,1).The associated observations from the exponential
distribution can be generated using the transformation X=log(1-Z) ,or since 1-Z is itself uniformly distributed , by using X=- log Z
MONTE CARLO SAMPLING
١١ z1 0.55463
x 1 log(0.55463) 0.589
z 2 0.15389
x 2 log(0.15389) 1.872
z 3 0.85941
x 3 log(0.85941) 0.151
z 4 0.61149
x 4 log(0.61149) 0.492
z 5 0.05219
x 5 log(0.05219) 2.953
z 6 0.41417
x 6 log(0.41417) 0.881
z 7 0.283 57
x 7 log(0.28357) 1.260
z 8 0.17783
x 8 log(0.17783) 1.727
z 9 0.40950
x 9 log(0.40950) 0.893
z10 0.82995
x 10 log(0.82995) 0.186
(1-2)GENERATION OF OBSERVATIONS FROM SOME STANDARD DISTRIBUTION In this section we discuss how the Monte Carlo sampling
method may be used to draw random samples of observation from some of the more important statistical distributions. In
some cases we shall see that the inverse transformation method form being available for the which relies on a simple analytical
c.d.f. of the underlying distribution is not applicable ,so other
techniques have to be used. Throughout the discussion we shall
let Z denote a random variable having the uniform distribution over(0,1).
(1-2-1)Exponential Distribution
MONTE CARLO SAMPLING
١٢
Let X be arandom variable having the exponential
distribution with p.d.f.
x 1 exp( ), f (x) e.w o
0x
The c.d.f. is F(x) 1 exp( x / ), 0 x and is zero elsewhere . 1 Sitting Z F(X) 1 exp(x / ) ,we option X log(1 Z) F (Z) .
Thus using the inverse transformation method ,values for X
may be generated from values for the uniform random variable Z using X log(1 Z) ,or since 1-Z is it self uniformly distributed, from
X log(Z)
(1-2-2)Weibull Distribution: Let X be random variable having the Weibull distribution
with p.d.f.
1 x exp (x / ) ,0 x f (x) o elsewhere
The c.d.f. is
F(x) 1 exp ( x / ) ,0 x
elsewhere. Setting
Z F(X) 1 exp ( X / ) ,
and is zero
we option or since 1
(1-Z) has the same distribution as Z we use X (log Z) 1
X log(1 Z)
*Proof:
MONTE CARLO SAMPLING
١٣
X F(X) 1 exp ,
0X
X Z 1 exp X 1 Z exp X ln 1 Z
X ln 1 Z 1 ln 1 Z X 1
ln 1 Z X 1
( ln Z) X
(1-2-3)Gamma Distribution with Integer Shape Parameter Let X be random variable having the gamma distribution exp x / x 1 ,0 x () f (x) o ,x 0
Where the parameter is a positive integer. Instead of using the inverse transformation method which requires the inverse function F ,we use the result that the random variables X may 1
be thought of as the sum of exponentially distributed random x 1 exp( ) .Thus if Z1 , Z2 ,..., Z variables each with p.d.f.
represent a set of independent random variables uniformly
MONTE CARLO SAMPLING
١٤
distributed over ( 0,1 ) , then we may set
X log Zi log Zi i 1 i1 We require observations from the uniform distribution over
( 0,1 ) to generate a single observation form this special gamma distribution .
Proof: f (x) 0
1 x x 1 exp( ), 0 x () elsewhere
ﻳﻤﻜﻦ اﻟﺘﻌﺒﻴﺮ ﻋﻨﻪX ﺑﺪﻻ ﻣﻦ اﺳﺘﺨﺪام اﻟﻨﻈﺮﻳﺔ ﻓﺈﻧﻨﺎ ﺳﻮف ﻧﺴﺘﺨﺪم اﻟﻨﺘﻴﺠﺔ أن. ﻗﻴﻤﺔ ﺻﺤﻴﺤﺔ ﺣﻴﺚ :ﻛﻤﺠﻤﻮع ﻣﺘﻐﻴﺮات ﻋﺸﻮاﺋﻴﺔ ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ اﻷﺳﻲ وﻛﻞ ﻣﺘﻐﻴﺮ ﻟﻪ ﺗﻮزﻳﻊ أﺳﻲ ﻋﻠﻰ اﻟﺸﻜﻞ 1 x f (x) exp( ) X (t) (1 t)1
(t) (1 t)1 i 1 xi i 1
(1 t)
.واﻟﺘﻲ ﺗﻤﺜﻞ اﻟﺪاﻟﺔ اﻟﻤﻮﻟﺪة ﻟﺘﻮزﻳﻊ ﺟﺎﻣﺎ
(1-2-4) Beta Distribution:
:ﺗﻮزﻳﻊ ﺑﻴﺘﺎ
: ﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻴﺘﺎ ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎﻟﻴﺔX ﻟﻴﻜﻦ
MONTE CARLO SAMPLING
١٥
( )x 1 (1 x)1 ()() f (x) 0 elsewhere Z1 Uniform
,0 x 1
Z2 Uniform 1
Y1 Z1 ,Y2 Z2
1
Y1 Z1 ,Y2 Z2 f (y1 , y 2 ) f (z1 ,z 2 ) | J |
| J |
z1 y1
z1 y 2
z 2 y1
z 2 y 2
y11
0
0
y 21
y11y 21 0 y1 1, 0 y 2 1
:ﺳﻮف ﻧﺄﺧﺬ اﻟﺘﺤﻮﻳﻠﺔ اﻟﺘﺎﻟﻴﺔ
MONTE CARLO SAMPLING
١٦ X
Y1 , W Y1 Y2 Y1 Y2
y1 xw y 2 w y1 w wx w(1 x) dy1 dy1 dx dx J dy 2 dy 2 dx dx w x w 1 x w(1 x) xw w wx wx w h(x | 0 W 1) 1
g(x, w)dw
1
0 1
(1 1)
g(x, w)dxdw 0
0
sin ce : 1
1
g(x, w)dw x 0
1
(1 x)1 w 1dw
0
1
x
1
(1 x)
1
w 0
x 1 (1 x)1 sin ce : 1
1
(1 2)
1
x 1 (1 x)1 0 g(x, w)dxdw 0
(, ) ( )() ( ) ( )
0
(1 3)
MONTE CARLO SAMPLING
١٧
:(1-1) ( واﻟﺘﻌﻮﻳﺾ ﻓﻲ1-3) ( و1-2) ﻣﻦ x 1 (1 x)1 h(x | 0 w 1) ()()
1 x 1 (1 x)1 ,0 x 1 ,
Example (1-2-1): Suppose that we require a sample of four independent
observations from the beta distribution with parameters 2 and
14
Use the following stream of uniform observations on z : 0.706, 0.392, 0.020, 0.882, 0.670, 0.922, 0.441, 0.717, 0.577, 0.799, 0.055, 0.628 These are arranged in pairs and the transformations applied in the above steps. Z1
0.706
0.020
0.670
0.441
0.577
0.055
Z2
0.392 0.840
0.882 0.141
0.922 0.819
0.717 0.664
0.799 0.760
0.628 0.235
0.024 0.864
0.605 0.746
0.722 1.541
0.264 0.928
0.408 1.168
0.156 0.391
0.972
0.189
Re ject
0.716
Re ject
0.601
1
Y 1 Z1 2 Y 2 Z 24 Y 1 Y 2 X
Y1 Y 1 Y 2
(1-2-5) normal Distribution:
:اﻟﺘﻮزﻳﻊ اﻟﻄﺒﻴﻌﻲ
: (0,1) ﻣﺘﻐﻴﺮﻳﻦ ﻋﺸﻮاﺋﻴﻴﻦ ﻣﺴﺘﻘﻠﻴﻦ ﻳﺘﺒﻌﺎن ﺗﻮزﻳﻊ ﻣﻨﺘﻈﻢ ﻓﻲ اﻟﻔﺘﺮةX1 ,X 2 إذا ﻛﺎن ﻟﺪﻳﻨﺎ
MONTE CARLO SAMPLING
١٨ X1
2 ln Z 1 cos(2 Z 2 )
X2
2 ln Z 1 sin(2 Z 2 )
x 1 2 2 ln z 1 cos 2 (2 z 2 ) x 2 2 2 ln z 1 sin 2 (2 z 2 ) x 1 2 x 2 2 2 ln z 1 cos 2 (2 z 2 ) sin 2 (2 z 2 ) 2 ln z 1 x 12 x 2 2 ln z 1 2 x 12 x 2 2 exp z1 2 x2 sin 2 z 2 tan 2 z 2 x 1 cos 2 z 2
tan 1
x2 2 z 2 x1
x tan 1 2 x1 z2 2
J
z1 x1 z 2 x1
z1 x 2
x1e
q 2
x 2 1 1 2 z 2 2 x 2 x1 1 x 2 x1
x 2e 1 2
x12 x 2 2 q
x1e
q 2
x2 1 2 2 x1 x 2 2
x 2e
q 2
x1 1 2 2 x1 x 2 2
1 x12 x 2 2 q 1 q exp( ) exp ( ) 2 x12 x 2 2 2 2 2
q 2
1 x 2 x1 1 x1 1
2
MONTE CARLO SAMPLING
١٩
w1 x1 (1 2 )x 2 w 2 x1 1
0
1 0 1 2 0 1 1 0
2 1
1 1 var(w1 ) 1 var(w 2 ) 1 w1 ,w 2
:(٢-٢-١) ﻣﺜﺎل إذا ﻛﺎن ﻟﺪﻳﻨﺎ ﻋﻴﻨﻪ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ ﺧﻤﺲ ﻣﺸﺎﻫﺪات ﻣﻦ ﺗﻮزﻳﻊ ﻃﺒﻴﻌﻲ E(X1 ) 10
,E(X 2 ) 20
Z: .186 .345
, var(X1 ) 4
.976
, .7
:اﻟﻤﻄﻠﻮب ﺗﻮﻟﻴﺪ ﺑﻴﺎﻧﺎت ﺣﻴﺚ
.577
.927 .455
, var(X 2 ) 9
.305 .279 .779 .231
:اﻟﺤﻞ 1 2
X1 (2log 0.186) cos 2(0.927) 1.390 1 2
X 2 (2log 0.186) sin 2(0.927) 1.197 1 2
W1 (0.7)(1.390) (1 .49) (1.197) 1.828 W2 X1 1.390 V1 10 4(1.828) 13.656 V2 20 9(1.390) 24.170
:وﺑﺎﻟﻤﺜﻞ ﻟﺒﺎﻗﻲ اﻟﻘﻴﻢ ﺗﺼﺒﺢ ﺑﺎﻟﺼﻮرة
MONTE CARLO SAMPLING
٢٠
V1 13.656 12.733 10.574 6.920 12.478 V 24.170 21.854 18.497 12.067 24.773 2
(1-2-6) Binomial Distribution:
ﺗﻮزﻳﻊ ذي اﻟﺤﺪﻳﻦ:
ﻟﻴﻜﻦ Xﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ذي اﻟﺤﺪﻳﻦ n P(X x) p x (1 p) n x , x 0,1,2,...,n x n
and X i 0
where X i 1 with probablity p
X Xi i 1
with probablity (1 p),i 1,...,n we set X i 1 if 0 Z p and X i 0 if p Z 1
ﻋﻤﻠﻴﻪ ﺑﻮاﺳﻮن
(1-2-7) Poisson Distribution:
ﻳﻮﺟﺪ ﻋﺪد ﻛﺒﻴﺮ ﻣﻦ اﻟﻀﻮاﻫﺮ اﻟﺘﻲ ﺗﺘﻐﻴﺮ ﺧﻼل اﻟﺰﻣﻦ . tﻓﺈذا ﻛﺎن ﻋﺪد اﻟﺘﻐﻴﺮات ﺧﻼل اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ )
(0,tﻫﻮ ) X (tﻓﻴﻜﻮن ) X ( tﻋﺒﺎرﻩ ﻋﻦ داﻟﻪ ﻓﻲ اﻟﺰﻣﻦ وﻳﺘﺒﻊ ﺗﻮزﻳﻊ اﺣﺘﻤﺎل ﻣﻌﻴﻦ .ﻓﻲ ﻫﺬﻩ اﻟﺤﺎﻟﻪ
ﻻﻧﻜﻮن ﺑﺼﺪد ﻣﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ واﺣﺪ Xﺑﻞ ﻟﻜﻞ t>0ﻳﻜﻮن ﻟﺪﻳﻨﺎ ﻣﺘﻐﻴﺮ ) X( tوﻫﺬﻩ اﻟﻤﺠﻤﻮﻋﻪ ﻣـﻦ اﻟﻤﺘﻐﻴﺮات اﻟﻌﺸﻮاﺋﻴﻪ ﺗﺪﻋﻰ ﺑﺎﻟﻌﻤﻠﻴﺔ اﻟﻌﺸﻮاﺋﻴﺔ.
ﻓﺈذا ﻛﺎﻧﺖ tﺗﺄﺧﺬ ﻗﻴﻤﺎ ﻣﺘﺼﻠﻪ ﻓﺄن اﻟﻌﻤﻠﻴﻪ ﺗﺴﻤﻰ ﻋﻤﻠﻴﻪ ﺗﺼﺎدﻓﻴﻪ ﺑﺰﻣﻦ ﻣﺘﺼﻞ وﻳﺮﻣﺰ ﻟﻬﺎ
ﺑﺎﻟﺮﻣﺰ )(X (t ), t 0
وﺗﺴﻤﻰ اﻟﻌﻤﻠﻴﻪ اﻟﺘﺼﺎدﻓﻴﻪ ﺑﻌﻤﻠﻴﻪ ﺑﻮاﺳﻮن أذا ﻛﺎن ﻗﺎﻧﻮن اﺣﺘﻤﺎﻟﻬﺎ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮﺳﻮان . أي ان اﻟﻤﺘﻐﻴﺮ اﻟﻌﺸﻮاﺋﻲ ) X ( tﻳﻤﺜﻞ ﻋﺪد ﺣﺎﻻت اﻟﻨﺠﺎح ) اﻟﺘﻐﻴﺮات ( ﻓﻲ اﻟﻔﺘﺮﻩ ) ( 0,tﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮاﺳﻮن .
ﻫﻨﺎك ﻇﻮاﻫﺮ ﻛﺜﻴﺮﻩ ﻳﻜﻮن ﻋﺪد اﻟﺘﻐﻴﺮات ﻓﻴﻬﺎ ) X ( tﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮاﺳﻮن .ﻓﺈذا ﻛﻨﺎ ﻧﺮاﻗﺐ او ﻧﻼﺣﻆ
وﻗﻮع اﺣﺪاث ﻳﻤﻜﻦ ان ﺗﻘﻊ ﻓﻲ أي وﻗﺖ ﻣﻦ اﻟﺰﻣﻦ ﻣﺜﻼ اذا ﻛﻨﺎ ﻧﺮاﻗﺐ -:
#ﻋﺪد اﻟﻄﺎﺋﺮات ) X ( tاﻟﺘﻲ ﺗﺤﻴﻂ ﻓﻲ اﻟﻄﺎﺋﺮﻩ ﻣﻦ ﺧﻼل اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ ) . (0,t
#ﻋﺪد اﻟﺴﻴﺎرات ) X (tاﻟﺘﻲ ﺗﺼﻞ اﻟﻰ ﻣﺤﻄﺔ ﺑﻨﺰﻳﻦ ﻣﻦ ﺧﻼل اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ ) . (0,t #ﻋﺪد اﻟﺒﻮاﺧﺮ ) X( tاﻟﺘﻲ ﺗﺼﻞ اﻟﻰ ﻣﻴﻨﺎء ﻣﻦ ﺧﻼل اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ ) . ( 0,t
MONTE CARLO SAMPLING
٢١
#ﻋﺪد ﺣﻮادث اﻟﺴﻴﺎرات ) X( tاﻟﺘﻲ ﺗﻘﻊ ﻋﻠﻰ اﻟﻄﺮﻳﻖ ﺧﻼل اﻟﻔﺘﺮة اﻟﺰﻣﻨﻴﺔ ) . ( 0,t
#ﻋﺪد اﻟﻌﻤﻼء ) X ( tاﻟﺬﻳﻦ ﻳﺼﻠﻮن اﻟﻰ ﺷﺒﺎﺑﻴﻚ اﺣﺪ اﻟﺒﻨﻮك ﺧﻼل اﻟﻔﺘﺮة اﻟﺰﻣﻨﻴﺔ ) . ( 0,t #ﻋﺪد اﻟﻤﻜﺎﻟﻤﺎت اﻟﺘﻠﻔﻮﻧﻴﻪ ) X ( tاﻟﺘﻲ ﺗﺼﻞ اﻟﻰ ﺳﻨﺘﺮال ﻣﻦ ﺧﻼل اﻟﻔﺘﺮة اﻟﺰﻣﻨﻴﺔ ) . ( 0,t #ﻋﺪد اﻟﺠﺰﻳﺌﺎت اﻟﺘﻲ ﺗﺼﺪر ﻣﻦ ﻣﻌﺪن ﻣﺸﺒﻊ ﺧﻼل اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ ) . ( 0,t
اﻟﺪاﻟﻪ اﻻ ﺣﺘﻤﺎﻟﻴﻪ ﻟﻌﺪد اﻟﺤﻮادث ) x ( tاﻟﺘﻲ ﺗﻘﻊ ﻓﻲ اﻟﻔﺘﺮﻩ اﻟﺰﻣﻨﻴﻪ ) ( 0,tﻫﻲ
ﻣﺜﺎل ):(٣-٢-١
exp(t)(t) x , x 0,1,... !x
f (t)
إذا ﻛﺎﻧﺖ اﻟﺒﻮاﺧﺮ ﺗﺼﻞ اﻟﻰ ﻣﻴﻨﺎء ﻣﺎ ﺑﻤﻌﺪل ٣ﺑﻮاﺧﺮ ﻓﻲ اﻟﺴﺎﻋﻪ ﻓﻤﺎ اﺣﺘﻤﺎل أن ﺗﺼﻞ ﺑﺎﺧﺮﺗﺎن ﺧﻼل ٤٠ دﻗﻴﻘﻪ ؟ اﻟﺤﻞ: 3 ﻣﻌﺪل وﺻﻮل اﻟﺒﻮاﺧﺮ ﻓﻲ اﻟﺪﻗﻴﻘﻪ 60
اﻟﻔﺘﺮة اﻟﺰﻣﻨﻴﺔ t=40دﻗﻴﻘﺔ
ﻓﺈذا ﻛﺎن Xﻫﻲ ﻋﺪد اﻟﺒﻮاﺧﺮ ﻓﺈن Xﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮاﺳﻮن ﺑﻤﻌﻠﻤﺔ 3 )2 60 e 2 2 x f (x) , x 0,1, 2,3... !x e 2 2 2 p(X 2) .27 !2 (t 40
ﺗﻮزﻳﻊ اﻟﻔﺘﺮات ﺑﻴﻦ اﻟﺤﻮادث:
إذا ﻛﺎن ﻋﺪد اﻟﺤﻮادث ) X(tاﻟﺘﻲ ﺗﻘﻊ ﻓﻲ اﻟﻔﺘﺮة ) ( 0,tﻣﺘﻐﻴﺮا ﻋﺸﻮاﺋﻴﺎ ﻣﺘﻘﻄﻌﺎ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺑﻮاﺳﻮن
ﺑﻤﻌﻠﻤﺔ ) tﻣﺘﻮﺳﻂ ﻋﺪد اﻟﺤﻮادث ﻓﻲ وﺣﺪة اﻟﺰﻣﻦ( ﻓﺈن أﻃﻮال اﻟﻔﺘﺮات اﻟﺰﻣﻨﻴﺔ اﻟﺘﻲ ﺗﻔﺼﻞ ﺑﻴﻦ ﻟﺤﻈﺎت
وﻗﻮع اﻟﺤﻮادث ﺗﻤﺜﻞ ﻣﺘﻐﻴﺮا ﻋﺸﻮاﺋﻴﺎ ﻣﺘﺼﻼ ﻟﻪ ﺗﻮزﻳﻊ اﺳﻲ .ﺑﻔﺮض أن Tﻃﻮل اﻟﻔﺘﺮة ﻣﻦ ﺑﺪاﻳﺔ اﻟﺰﻣﻦ إﻟﻰ
اﻟﺤﺎدﺛﺔ اﻷوﻟﻰ ﻳﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﺗﻮزﻳﻊ Tﻣﻦ ﺗﻮزﻳﻊ ).X(t
إذا ﻟﻢ ﺗﻘﻊ ﺣﺎدﺛﺔ ﻓﻲ اﻟﻔﺘﺮة ) (0,tﻓﻬﺬا ﻳﻌﻨﻲ أن T>tأي أن :
MONTE CARLO SAMPLING
٢٢
)P(T>t)=P(X(t)=0 =e - t f (T) e - t
ﻓﺈذا ﻛﺎﻧﺖ T1ﺗﻤﺜﻞ ﻃﻮل اﻟﻔﺘﺮة ﻣﻦ اﻟﺒﺪاﻳﻪ اﻟﻰ اﻟﺤﺎدﺛﺔ اﻷوﻟﻰ
T2ﺗﻤﺜﻞ ﻃﻮل اﻟﻔﺘﺮة ﻣﻦ اﻟﺤﺎدﺛﺔ اﻷوﻟﻰ اﻟﻰ اﻟﺤﺎدﺛﺔ اﻟﺜﺎﻧﻴﺔ T3ﺗﻤﺜﻞ ﻃﻮل اﻟﻔﺘﺮة ﻣﻦ اﻟﺤﺎدﺛﺔ اﻟﺜﺎﻧﻴﺔ اﻟﻰ اﻟﺤﺎدﺛﺔ اﻟﺜﺎﻟﺜﺔ
ﻓﺈن اﻟﻤﺘﻐﻴﺮات … T1,T2,ﻫﻲ ﻣﺘﻐﻴﺮات ﻋﺸﻮاﺋﻴﺔ ﻣﺴﺘﻘﻠﺔ ﻳﺘﺒﻊ ﻛﻞ ﻣﻨﻬﺎ ﺗﻮزﻳﻊ اﺳﻲ ﺑﺪﻻﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎل f (t) et
اذا ﻛﺎن T1 T2 ,... Tnاﻟﺰﻣﻦ ﻣﻦ اﻟﺒﺪاﻳﺔ اﻟﻰ ﻇﻬﻮر اﻟﺤﺎدﺛﺔ nأي أﻧﻪ ﻳﻤﺜﻞ ﻣﺠﻤﻮع nﻣﻦ اﻟﻤﺘﻐﻴﺮات اﻟﻌﺸﻮاﺋﻴﺔ اﻟﻤﺴﺘﻘﻠﺔ ﻛﻞ ﻣﻨﻬﺎ ﻳﺘﺒﻊ ﺗﻮزﻳﻊ اﺳﻲ ﺑﻤﻌﻠﻤﺔ وﻋﻠﻰ ذﻟﻚ T1 T2 ,... Tnﻳﺘﺒﻊ ﺗﻮزﻳﻊ ﺟﺎﻣﺎ ﺑﻤﻌﺎﻟﻢ و. n
(1-2-8)Geometric Distribution:
إذا ﻛﺎن:
, x 1, 2,.... w 1
p
x 1
f x 1 p
x
p w 1 p w 1
Fx
w x
w
x 1
p 1 p w 0
ﻟﺤﺴﺎب ﻫﺬا اﻟﻤﺠﻤﻮع ﺳﻮف ﻧﺴﺘﻔﻴﺪ ﻣﻦ اﻟﻌﻼﻗﺎت اﻟﺘﺎﻟﻴﺔ: a 1 p k x 1
1 a k 1 1 a
k
aw
w 0
x 11
1 1 p Fx p 1 1 p x
1 1 p x p 1 1 p , x 1 11p
p Z 1 p x
p Z 1 1 p , Z F x x
Z
ﻣﻨﺘﻈﻢ
MONTE CARLO SAMPLING
٢٣
ﻣﻨﺘﻈﻢ1 Z
P X x P Z 1 1 p
x
P 1 Z 1 p P Z 1 p P Z 1 1 p
x
x
x
P X x P X x P X x 1
P 1 p
p Z 1 p Z 1 p
p Z 1 p
1 p e
x
x ln 1 p
x
x
x 1
x 1
Z Z
x ln 1 p ln z
x
ln Z ln 1 p
x s
. s أﺻﻐﺮ رﻗﻢ ﺻﺤﻴﺢ ﻳﺴﺎوي أو أﻛﺒﺮ ﻣﻦs ﺣﻴﺚ
let Z 0.2 1 p 2 ln Z S ln 1 p
S
s أﻗﻞ رﻗﻢ ﺻﺤﻴﺢ أﻛﺒﺮ أو ﻳﺴﺎوي
1.609 1.609 2.32 0.693 0.693 s أﻗﻞ رﻗﻢ ﺻﺤﻴﺢ أﻛﺒﺮ ﻣﻦ أو ﻳﺴﺎوي
X3
MONTE CARLO SAMPLING
٢٤
ﻣﺜﺎل ):(٤-٢-١ إذا ﻛﺎﻧﺖ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻤﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ xﻋﻠﻰ اﻟﺸﻜﻞ: x ,0 x 2 2 0 , e.w .
f x
اﻟﺤﻞ: x
t t2 x x2 F x dt | ,0 x c 2 2 4 0 4 0 0, x 0 1, x 2 let x2 4 4Z x 2 x 2 Z F x Z
ﻧﻘﻮم ﺑﺘﻮﻟﻴﺪ ﺑﻴﺎﻧﺎت Z1 , Z2 ,....., Zn
ﻧﺘﺒﻊ اﻟﺘﻮزﻳﻊ ﻣﻨﺘﻈﻢ u 0,1وﺑﺎﻟﺘﻌﻮﻳﺾ x 2 Zﻧﺤﺼﻞ ﻋﻠﻰ اﻟﻘﻴﻤﺔ اﻟﻌﺸﻮاﺋﻴﺔ اﻟﺘﻲ ﻣﺸﺎﻫﺪﺗﻬﺎ ﺗﺘﺒﻊ x f x ,0 x 2 2 اﻟﺘﻮزﻳﻊ
ﻣﺜﺎل ):(٥-٢-١ إذا ﻛﺎن Xﻣﺘﻐﻴﺮاً ﻋﺸﻮاﺋﻴﺎً ﺑﺪاﻟﺔ ﻛﺜﺎﻓﺔ اﺣﺘﻤﺎل )ﺗﻮزﻳﻊ ﻟﻮﺟﺴﺘﻲ( ﻋﻠﻰ اﻟﺸﻜﻞ: , x
ex 2
1 e x
f x
0 elsewhere
اﻟﺤﻞ:
داﻟﺔ اﻟﺘﻮزﻳﻊ ﻟﻠﻤﺘﻐﻴﺮ xﻋﻠﻰ اﻟﺸﻜﻞ:
MONTE CARLO SAMPLING
٢٥ x
F x
e t
1 e t
dt
2
x
2
e 1 e t
t
dt
x
1 e t 1
F x
1 , x 1 e x
x F1 z Z F x
1 1 e x
1 z 1 1 1 e x ln 1 x z z 1 x ln 1 z Z1 1 e x
:(٦-٢-١) ﻣﺜﺎل
ﺗﺘﺒﻊ ﺗﻮزﻳﻊ ﻛﻮﺷﻲ؟n=10 اﻟﻤﻄﻠﻮب ﺗﻮﻟﻴﺪ )ﻣﺤﺎﻛﺎة( ﻋﻴﻨﺔ ﻋﺸﻮاﺋﻴﺔ ﻣﻦ اﻟﺤﺠﻢ :اﻟﺤﻞ
: داﻟﺔ اﻟﺘﻮزﻳﻊ اﻟﺘﺠﻤﻴﻌﻲ ﻟﺘﻮزﻳﻊ ﻛﻮﺷﻲ ﻫﻲ F(x)
1 1 tan x
2
:وﻟﺪي ﺑﻴﺎﻧﺎت ﺗﺘﺒﻊ ﻫﺬا اﻟﺘﻮزﻳﻊ
1 1 F x dt 1 t 2 x
1 tan 1 t
MONTE CARLO SAMPLING
٢٦
1 1 tan x 2
x
1 1 tan x 2 z tan 1 x 2 1 (1-2) z tan x 2 x tan z (1-1) 2 1 1 ﺛﻢ ﻧﻌﻮض ﻓﻲ اﻟﻤﻌﺎدﻟﺔ 0,1 ﻣﻦ ﺗﻮزﻳﻊ ﻣﻨﺘﻈﻢ ﻓﻲ اﻟﻔﺘﺮةZ1, Z2 ,....., Zn ﻧﻘﻮم ﺑﺘﻮﻟﻴﺪ z
1 2 ﻣﻦ اﻟﻤﺸﺎﻫﺪات ﺗﺘﺒﻊ اﻟﺘﻮزﻳﻊ ﻓﻲn ﻋﻠﻰ أن :(٦-٢-١) ﻣﺜﺎل :إذا ﻛﺎﻧﺖ داﻟﺔ ﻛﺜﺎﻓﺔ اﻻﺣﺘﻤﺎل ﻟﻤﺘﻐﻴﺮ ﻋﺸﻮاﺋﻲ ﻋﻠﻰ اﻟﺸﻜﻞ x f x ,0 x 2 2 0 e.w
:اﻟﺤﻞ x
x
t t2 x2 F x dt 2 40 4 0 0, x 0 2 x F x ,0 x 2 4 1, x 2 x2 z F x 4 2 4z x x2 z
٢٧
MONTE CARLO SAMPLING