Feuille TD 4 : espérance, variance

X{1,...,5}
P{X= 1}= 1/4,P{X= 2}= 1/6,P{X= 3}= 1/12,
P{X= 4}= 1/3,P{X= 5}= 1/6.
E(X) = 3
2 3 4 5 6 7 8 9 10 11 12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
7
X1, X2E(X1+X2) = E(X1)+E(X2)=7
. . .
λ > 0
λ λ
{1, . . . , N}
n
X
k=1
k2=n(n+ 1)(2n+ 1)
6.
(N+ 1)/2 (N2−1)/12 N= 1
1/4p= 1/2N= 2
2N−1N≥1

N
N
1+2+22+. . . + 2N−1= 2N−1
N
N N
X1, . . . , XNX1, . . . , XN
N
Xi= 1 Xi= 0
N
{X1= 0}∩{X2= 0} ∩ . . . ∩ {Xn= 0}
Xi(1/2)N
S N
1−2N+ 1 S
S1−2N+ 1
P(S=−2N+ 1) = (1/2)N;P(S= 1) = 1 −(1/2)N.
S
E(S) = 0
Y{0, . . . , N}
N≥1
p= 1/10
X
P(X=k) =
+∞
X
n=0
P(Y=n)P(X=k|Y=n)
E(X) =
+∞
X
k=0
∞
X
n=0
P(Y=n)kP(X=k|Y=n)

k, n ≥0P{X=k|Y=n}
k≤n k > n k > n P{X=k|Y=n}0k≤n
P{X=k|Y=n}(n, p)
k n E(X) = pE(Y)
E(X) =
+∞
X
k=0
∞
X
n=0
P(Y=n)kP(X=k|Y=n)
=
+∞
X
n=0
P(Y=n)
∞
X
k=0
kP(X=k|Y=n)
=
+∞
X
n=0
P(Y=n)
n
X
k=0
kP(X=k|Y=n)
=
+∞
X
n=0
P(Y=n)np
=pE(Y)
Y
n
1n i
Xi0
S=X1+· · · +Xn
XiS
n!i
(n−1)! XiB(1/n)S
n×1/n = 1 Xi
E(XiXj)i6=j S
Yij =XiXj0 1
Yij = 1 (n−2)! Yij B(1/[n(n−1)]) E(XiXj) = 1/[n(n−1)]
E(S2) = E(X2
1) + . . . +E(X2 n) + X
i6=j
E(XiXj) = 2
V(S) = 1
Xi
iE(XiXj)
i j
18/37
19/37 n Xii Sn=X1+. . . +Xn
Yn=Sn/n

Xi
P(X1= 1) = 18/37 ; P(X1=−1) = −19/37
E(Xi) = −1/37
Yn
E(Yn) = −1/37 ; V(Yn) = [1 −(1/37)2]/n
gngn
n
gn=P(Yn>0) = P((Yn−E(Xi)) >1/37) ≤P(|Yn−E(Xi)|>1/37) →0
gn0