∆ευτέρα, 19 Οκτωβρίου 2009. Άσκηση 1 Έστω X 1 , X 2 τ.µ. από τριωνυµική κατανοµή µε σ.π.: N! N −x −x f ( x1 , x2 , p1 , p2 ) = p1x1 p2 x2 (1 − p1 − p2 ) 1 2 , µε x1 ! x2 !( N − x1 − x2 )! x1 , x2 = 0,1,..., N , x1 + x2 ≤ N , 0 < p1 , p2 < 1 και p1 + p2 < 1 . α) να δειχθεί ότι ανήκει στην Ε.Ο.Κ. β) να µετατραπεί στην κανονική µορφή, γ) να βρεθούν οι E ( X j ) ,V ( X j ) , j = 1,2 και Cov ( X 1 , X 2 ) . Λύση: α) Στήριγµα S f = { x ∈ ℝ : f ( x1 , x2 , p1 , p2 ) > 0} = 0,1, 2,..., N , δηλαδή ανεξάρτητο των παραµέτρων p1 , p2 . N! N −x −x f ( x1 , x2 , p1 , p2 ) = p1x1 p2 x2 (1 − p1 − p2 ) 1 2 = x1 ! x2 !( N − x1 − x2 )! N! (1 − p1 − p2 ) = p1x1 p2 x2 x +x x1 ! x2 !( N − x1 − x2 )! (1 − p1 − p2 ) 1 2 N
N! p1x1 p2 x2 N 1 − p1 − p2 ) = x1 x2 ( x1 ! x2 !( N − x1 − x2 )! (1 − p1 − p2 ) (1 − p1 − p2 ) x1
x2
N! p1 p2 N (1 − p1 − p2 ) = x1 ! x2 !( N − x1 − x2 )! 1 − p1 − p2 1 − p1 − p2 x1 x2 N! p1 p2 N exp log + log + log (1 − p1 − p2 ) = x1 ! x2 !( N − x1 − x2 )! 1 − p − p 1 − p − p 1 2 1 2 exp x1 log 1− exp x1 log 1−
p1 + x2 log p1 − p2 1− p1 + x2 log p1 − p2 1−
p2 N! = + N log (1 − p1 − p2 ) p1 − p2 x1 ! x2 !( N − x1 − x2 )! p2 N! + N log 1 − p − p ( ) 1 2 p1 − p2 x1 ! x2 !( N − x1 − x2 )!
άρα ανήκει στην Ε.Ο.Κ. µε:
p1 , T1 ( x1 , x2 ) = x1 , 1 − p1 − p2
η1 ( p1 , p2 ) = log
p2 , T2 ( x1 , x2 ) = x2 , 1 − p1 − p2
η 2 ( p1 , p2 ) = log
B ( p1 , p2 ) = − N log (1 − p1 − p2 ) και h ( x1 , x2 ) =
1
N! x1 ! x2 !( N − x1 − x2 )!
β)
Κανονική Μορφή:
p1 = η1 ⇒ 1 − p1 − p2
η1 ( p1 , p2 ) = log
p1 = eη1 ⇒ 1 − p1 − p2
p1 = eη1 − eη1 p1 − eη1 p2 ⇒
eη1 − eη1 p2 (1) 1 + eη1
p1 + eη1 p1 = eη1 − eη1 p2 ⇒ p1 (1 + eη1 ) = eη1 − eη1 p2 ⇒ p1 =
p2 p2 = eη2 ⇒ = η2 ⇒ 1 − p1 − p2 1 − p1 − p2
η 2 ( p1 , p2 ) = log
(1)
p2 (1 + eη2 ) = eη2 − eη2 p1 ⇒
p2 = eη2 − eη2 p1 − eη2 p2 ⇒ p2 + eη2 p2 = eη2 − eη2 p1 ⇒
eη1 − eη1 p2 p2 (1 + e ) = e − e ⇒ 1 + eη1 p2 (1 + eη2 )(1 + eη1 ) = (1 + eη1 ) eη2 − eη2 ( eη1 − eη1 p2 ) ⇒ η2
η2
η2
p2 (1 + eη1 + eη2 + eη1 eη2 ) = eη2 + eη1 eη2 − eη1 eη2 + eη1 eη2 p2 ⇒
p2 (1 + eη1 + eη2 ) + p2eη1 eη2 = eη2 + eη1 eη2 p2 ⇒ p2 (1 + eη1 + eη2 ) = eη2 ⇒ p2 =
eη2 (2) 1 + eη1 + eη2
(1),(2) ⇒ p1 =
eη2 eη1 1 + eη1 + eη2 − eη1 eη2 η1 η2 1+ e + e = = 1 + eη1 1 + eη1 1 + eη1 + eη2
eη1 − eη1
(
(
)(
)
)
eη1 (1 + eη1 ) eη1 eη1 + e2η1 + eη1 eη2 − eη1 eη2 = = (1 + eη1 )(1 + eη1 + eη2 ) (1 + eη1 )(1 + eη1 + eη2 ) 1 + eη1 + eη2
Άρα η σ.π. γίνεται: eη1 eη2 f ( x1 , x2 ,η1 ,η 2 ) = exp x1η1 + x2η2 + N log 1 − − η1 η2 1 + e + e 1 + eη1 + eη2
N! = x1 ! x2 !( N − x1 − x2 )!
1 + eη1 + eη2 − eη1 − eη2 N! exp x1η1 + x2η2 + N log = η1 η2 1+ e + e x1 ! x2 !( N − x1 − x2 )! 1 N! exp x1η1 + x2η2 + N log = η1 η2 1 + e + e x1 ! x2 !( N − x1 − x2 )!
{
} x !x !( NN−!x − x )! ,
f ( x1 , x2 ,η1 ,η 2 ) = exp x1η1 + x2η 2 − N log (1 + eη1 + eη2 )
1
άρα A (η1 ,η2 ) = N log (1 + eη + eη 1
2
)
2
2
1
2
∂ N log (1 + eη1 + eη2 ) ∂A (η1 ,η 2 ) E (xj ) = , j = 1, 2 ⇒ E ( x j ) = = ∂η j ∂η j
γ)
1
pj
1
η
= p1 p2 1 − p − p 1 2 1+ + 1 − p1 − p2 1 − p1 − p2 pj pj 1 1 N = N = Np j ⇒ 1 − p1 − p2 + p1 + p2 1 − p1 − p2 1 1 − p1 − p2 1 − p1 − p2 1 − p1 − p2
N
1 + eη1 + eη2
e j= N
E ( x j ) = Np j , j = 1,2
∂ 2 A (η1 ,η2 ) ∂ ∂A (η1 ,η 2 ) , j = 1, 2 ⇒ Cov ( X 1 , X 2 ) = Cov ( X 1 , X 2 ) = = ∂η1 ⋅ ∂η 2 ∂η 2 ∂η1 1 ⋅ eη2 Neη1 eη2 ∂ 1 η1 η1 = − = N e = Ne − η1 η2 2 (1 + eη1 + eη2 )2 ∂η2 1 + eη1 + eη2 1 + e + e ( ) p1 p2 p1 p2 2 1 − p1 − p2 ) ( 1 − p1 − p2 1 − p1 − p2 −N = −N = 2 2 p1 p2 1 − p1 − p2 − p1 + p2 + 1 + 1 − p1 − p2 1 − p1 − p2 1 − p1 − p2 p1 p2 2 1 − p1 − p2 ) ( −N = − Np1 p2 ⇒ 2 1 1 − p1 − p2
Cov ( X 1 , X 2 ) = − Np1 p2
∂ 2 A (η1 ,η2 ) V ( x j ) = Cov ( x j , x j ) , j = 1, 2 ⇒ V ( x j ) = Cov ( x j , x j ) = = ∂η j 2 ∂ ∂η j N
∂A (η1 ,η2 ) ∂ = ∂η j ∂η j
1 ∂ ηi N e = N η η 1 + e 1 + e 2 ∂η j
eηi (1 + eη1 + eη2 ) − eηi ⋅ eηi
(1 + e
η1
)
η2 2
+e
=
3
eηi 1 + eη1 + eη2 =
2
pj pj p1 p2 + 1 + − 1 − p1 − p2 1 − p1 − p2 1 − p1 − p2 1 − p1 − p2 N = 2 p1 p2 + 1 + 1 − − 1 − p1 − p2 p p 1 2 2
pj pj 1 − p1 − p2 + p1 + p2 − 1 − p1 − p2 1 − p1 − p2 1 − p1 − p2 N = 2 1 − p1 − p2 + p1 + p2 1 − p1 − p2
N
pj p j2 1 − 1 − p1 − p2 1 − p1 − p2 (1 − p1 − p2 )2 2
1 1 − p1 − p2 N ( p j − p j 2 ) = Np j (1 − p j ) ⇒
pj = N
(1 − p1 − p2 )
V ( x j ) = Np j (1 − p j )
4
2
−
p j2
(1 − p1 − p2 )
1 1 − p1 − p2
2
2
=