Sujet du DM 1 - Université Paris 13

AP(A| G) = E[1A| G]
GP(A|X) = E[1A|X]
E[X| G]
E[X|Y]X,Y X
E[X|Y]f(Y)f
g
E[Xg(Y)] = E[f(Y)g(Y)].
σ(Y)Y
X Y (E,E)
X Y E[X|Y] = E[X]
f Xf (Y)E[Xf (Y)|Y] = f(Y)E[X|Y]
X Y f :R×E→R E[f(X,Y )|Y] = ϕ(Y)ϕ(y) =
E[f(X,y)] y
X Y Y X Y
f(Y)
X Y
[0,1]
E[XY |X],E[X−XY 2|Y],Eh1
1−XY 
Xi,Eh1
1−XY 
X, Y i,P(X < Y |Y)
E[· | X,Y ] (X,Y )
X,Y λ µ
Z=X+Y
X Z
E[X|X+Y]
N≥1x∈ {1, . . . ,N −1}(Xn)n≥0
{0, . . . ,N}X0=x n ≥0Xn+1 Xn
N1
NXn
Ex[Xn+1 |Xn]Ex[Xn]n
(Xn)n≥0W{0,N}
Ex[W]WPx

(X,Y )
f(x,y) = 1
πe−(1+x2)y1{y>0}.
Y X Y =y
E[X|Y]E[X2|Y]E[X2Y]
X Y X =x Y X
Np∈[0,1
2]
(Xn)n∈NNP P (n,n + 1) = p= 1 −P(n,n −1)
n∈N∗P(0,1) = 1
Tx= inf{n≥1|Xn=x}, x ∈N.
p≤1/2Tx<∞p < 1/2
x∈NTx(Fn)n≥0Fn=σ(X0, . . . ,Xn)
E0[T0]E1[T0]
E1[T0]E2[T0]
T1E2[T0]E2[T1]E1[T0]
E2[T1] = E1[T0]
p= 1/2E1[T0] = ∞E0[T0] = ∞
p < 1/2E0[T0]<∞E1[T0]E0[T0]
X,Y,Z [0,1]
P(Z∈[X,Y ]) = 1
3[X,Y ]X Y X Y
X Y Z
PZ∈[X,Y ]
X,Y 
E|X−Y|
λ > 0X,Y X x 7→ λ2xe−λx1[0,∞[(x)
Y X [0,X]
(X,Y )
Y
X Y
E[Y2|X]