Exercice

X1X2X3
G(1/2) P(X1=X2=X3)
X1X2G(p)Y= min(X1, X2)
P(X1=k X2> k)
Y
X1→ G(p1)X2→ G(p2) min(X1, X2)
k
p∈]0,1[ q= 1 −p X1, ..., XN
p
16i6N n ∈N∗P(Xi6n)
Y= (X1, ..., XN)P(Y6n)P(Y=n)
Y
p∈]0,1[ q= 1 −p X Y N
∀k∈N,P(X=k) = P(Y=k) = pqk
U= (X, Y )V= min(X, Y )
(U, V )U V
U V
S=U+V
X Y
Eii>1B(p)X(ω)
i>1Ei(ω)=1 Y(ω)i>X(ω)Ei(ω)=1
X Y Ω1Ω2
1
X
Y
E(X)E(Y)E(Y)GY(t)
p∈]0,1[ (Xn)n∈N∗
∀n∈N∗,P(Xn= 1) = 1 −pP(Xn=−1) = p
n∈N∗Zn=
n
Y
k=1
Xk

E(Zn)n+∞
Zn
Z1Z2
(X, Y )
∀(j, k)∈N,P(X=j Y =k) = a(j+k)1
3j+k
k∈Nσk=
+∞
P
j=0
(j+k)1
3j+k·
+∞
P
k=0
σka
X Y
X Y
P(X=Y=k)
(Xn)n∈N
1/2Yn=
n
P
k=1
kXkan=n(n+ 1)
2·
XnYn
P(Yn=K) = P(Yn=an−K)
XNGX:t7→ 1
2−t2·
X
XX
2·
N N → P(λ)
p∈]0,1[ X Y
X N =j
(X, N )
X
X Y
X N

NP(λ)
p
X Y
(N, X)
X Y
X Y
X Y X → P(λ)Y → P(µ)
X X +Y
X Y λ µ
Z=X+Y
P(X=k|Z=n)
X Z =n
X Y (Ω,A,P)Y → P(λ)
X(Ω) = N)m∈NX Y =mB(m, p)
X
XG(p)p∈]0,1[
ZNk∈NZ X =k
[[0, k −1]]
(j, k)∈N×N∗P(Z=j|X=k)
Z
E(Z)
+∞
P
j=1
+∞
P
k=j+1
=
+∞
P
k=2
k
P
j=1
XP(λ)λ > 0
P(X6n) = 1
n!Z+∞
λ
xn−xdx
un=Z+∞
λ
xn−xdx n +∞
GXX GX(1) GX(−1)
X
Y[[1,2]] X
XY
XN
X → P(λ)
GX(t)

∀t>1∀a∈R,P(X>a)6GX(t)
ta·
P(X>2λ)64λ
X → P(λ)λ∈]4,5[
un=P(X=n+ 1)
P(X=n)Pun
nP(X=n)
>9
66
b r
1,2, ..., n =b+r N i ∈[[1, n −1]] i i + 1
N
2n n 0 1 n
n k ∈[[1, n]] Uk1k
0
Uk(Ui, Uj)i6=j
N=U1+· · · +Un
S
S Uk
Z0
J=
0100
0010
0001
1000
ABCD
A n n + 1 B1/2D
1/3A1/6
B n n + 1 C1/2A
1/3B1/6

C n n + 1 D1/2B
1/3C1/6
D n n + 1 A1/2C
1/3D1/6
anA n (an)n∈N
A B 1/3C2/3
B A 1/3C2/3
C B 1/3A2/3
AnA n
BnCnXn=
P(An)
P(Bn)
P(Cn)
M∈ M3(R)n∈N∗Xn+1 =MXn
(Xn)n∈N
n1n
1 3
n
n
X
X
P(X= 2)
X
E(X)
1/2
E F G
E F G
E(1/E)
5%
104 100
101
104
A n r m 0< r 6m
a= (a1, ..., ar)∈Ar
X1, ..., XmA
N k ∈[[1, m −r+ 1]]
∀i∈[[1, r]], Xk−1+i=ai.