DM bis no 5 - Mathématiques en ECE 2

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f:R→Rt
f(t) = 
0t60
1
(1 + t)2t > 0
f
f
xZx
−∞
f(t)dt
x≤0x > 0
αZα
0
f(t)dt =1
2
x∈[0,+∞[
ϕx[0; +∞[∀u∈[0,+∞[, ϕx(u) = Zx+u
x−u
f(t)dt
ϕx(0) lim
u→+∞ϕx(u)
∀(u, v)∈([0,+∞[)2, u < v =⇒ϕx(v)−ϕx(u)≥Zx+v
x+u
f(t)dt
ϕx[0; +∞[
ϕx[0; +∞[ϕx(u) = 1
2u
[0; +∞[
U: [0; +∞[→Rx∈[0; +∞[U(x)
ϕx(u) = 1
2
x∈[0; +∞[Zx+U(x)
x−U(x)
f(t)dt =1
2
x∈[0; 1
2[U(x)=1−x
x∈[1
2; +∞[ϕx(x)≥1
2x−U(x)≥0U(x) =
p4+(x+ 1)2−2
U[0; +∞[
U[0; +∞[
U
(an)n∈Na0= 1
∀n∈N, an+1 =U(an)
∀n∈N, an>1
2
(an)n∈N
(an)n∈N
1
2
n∈N
an−1
2
610−6

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n
Un n
U
k i ∈J1; nKXi
i k
i∈J1; nKXiXi
X1, X2, . . . , Xn
(i, j)∈J1; nK2i6=j
Xi+XjXi+Xj
(Xi, Xj)
k Zk
k E(Zk)Zk
Z1Z2
E(Z1)E(Z2)
k
P(Zk= 1) P(Zk=k)
l∈J1; nKP(Zk+1 =l) = l
nP(Zk=l) + n−l+1 nP(Zk=l−1)
E(Zk+1 =n−1
nE(Zk)+1
(vk)k>1vk=E(Zk)−n
k E(Zk) = n1−n−1
nk

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n n
l, 2,··· , n. n
n n
(Ω,B,P) Tn
n n
2n
An
anan= P (An)
an
T2n= 2
k>2 : P (T2=k) = (1 −a2) (a2)k−2.
S2=T2−1
T2a2
T3n= 3
P (T3= 2) P (T3= 3) a2a3
A3, A3k>2
P (T3=k+ 1) = (1 −a3) P (T2=k) + a3P (T3=k)
k>2,P (T3=k) = (1 −a2) (1 −a3)
a3−a2h(a3)k−2−(a2)k−2i.
+∞
X
k=2
P (T3=k).
T3−1E(T3−1)
E(T3)a2a3
T3(T3−1)
a2a3
T3

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Z1Z2
Cov (Z1, Z2) = E(Z1Z2)−E(Z1)E(Z2)
Z1Z2
X U (Ω,A,P) ,
XN(0,1) U{−1,1}.
Y=UX Y
(Ω,A,P) .
P (Y6x) = P ([U= 1] ∩[X6x]) + P ([U=−1] ∩[X>−x])
Y X.
U E (XY )=0
Cov (X, Y )=0
EX2R+∞
0x2e−x2
2=1
2√2π
∀A∈R+:ZA
0
x4e−x2
2dx =−A3e−A2 2+ 3 ZA
0
x2e−x2
2dx
R+∞
0x4e−x2
2dx 3
2√2π.
X4EX4= 3
EX2Y2= 3
Cov X2, Y 2
X2Y2X Y