Corrigé des exercices - Dunod

400
uniforme.
P(Z = (i,j)) = 2(n−2)!
n!
= 2
n(n−1) .
Pour tout entier i de [1,n − 1], P(X = i) =
Pour tout entier j de [2,n], P(Y = j) = j−1 ∑
i=1
i=1
i=1
n ∑
j=i+1
P(X = i,Y = j) = 2(n−i)
n(n−1) .
P(X = i,Y = j) = 2(j−1)
n(n−1) .
2. Comme X et Y sont des variables aléatoires finies, X, Y et XY admettent une espérance.
E(X) = n−1 ∑
iP(X = i) = n−1 ∑
( )
2(n−i)i
n(n−1) = 2 n 2 (n−1)
n(n−1) 2
− n(n−1)(2n−1)
6
.
∑
E(Y ) = n ∑
jP(X = j) = n
j=2
j=2
Donc E(X) = n + 1
3
( )
2(j−1)j
n(n−1) = 2 n(n+1)(2n+1)
n(n−1) 6
− n(n+1)
2
.
Donc E(Y ) =
2(n + 1)
3
E(XY ) =
=
=
n−1
∑
n∑
i=1 j=i+1
ijP(X = i,Y = j) =
n−1
∑
n−1
2 ∑ (n − i)(n + i + 1)
i
n(n − 1) 2
1
n(n − 1)
i=1
( n 2 (n − 1)(n + 1)
−
2
Donc E(XY ) =
Ainsi cov(X,Y ) = E(XY ) − E(X)E(Y ) = (n+1)(n+2)
E(X 2 ) = n−1 ∑
i 2 P(X = i) = n−1 ∑
i=1
i=1
E(Y 2 ∑
) = n j 2 ∑
P(X = j) = n
j=2
Finalement ρ X,Y = 1 2 .
j=2
2(n−i)i 2
n(n−1)
= 2
n(n−1)
Donc V (X) =
n∑
i=1 j=i+1
2ij
n(n − 1)
n(n − 1)(2n − 1)
6
(n + 1)(3n + 2)
12
36
.
(
n 2 (n−1)(2n−1)
6
− n2 (n−1) 2
4
(n + 1)(n − 2)
18
− n2 (n − 1) 2 )
.
4
( )
2(j−1)j 2
n(n−1)
= 2 n 2 (n+1) 2
n(n−1) 4
− n(n+1)(2n+1)
6
.
Donc E(Y ) =
(n + 1)(n − 2)
18
)
.