Théorèmes limites et tests statistiques 1 Théorèmes limites - IMJ-PRG

X
ε > 0
P(|X−E(X)| ≥ ε)≤V(X)
ε2
P(|X−E(X)|< ε)≥1−V(X)
ε2.
X f X
V(X) = E(X−E(X))2
=
+∞
Z
−∞
(x−E(X))2f(x)dx
≥Z|x−E(X)|≥ε (x−E(X))2f(x)dx
≥Z|x−E(X)|≥ε ε2f(x)dx =ε2P|X−E(X)| ≥ ε.
P(|X−E(X)|< ε)=1−P(|X−E(X)| ≥ ε)
≥1−V(X)
ε2.
(Xj)j∈N∗m
σ2Mn=1
n
n
X
j=1
XjMnm ε > 0
lim
n7→+∞
P(|Mn−m|< ε)=1.
Mn
E(Mn) = 1
n
n
X
j=1
E(Xj) = 1
n
n
X
j=1
m=m.
V(Mn) =
⊥⊥
1
n2
n
X
j=1
V(Xj) = 1
n2
n
X
j=1
σ2=σ2
n.
ε > 0
1−V(Mn)
ε2≤P(|Mn−E(Mn)|< ε)≤1.

1−σ2
nε2≤P(|Mn−m|< ε)≤1.
n−→ +∞
1 = lim
n7→+∞1−σ2
nε2≤lim
n7→+∞
P(|Mn−m|< ε)≤1.
(Xj)j∈N∗m
Mnm A
∀ω∈A, lim
n7→+∞Mn(ω) = m.
50
n50
(Xj)j∈N∗
Sn=
n
P
j=1
Xjx y x ≤y
P x≤Sn−E(Sn)
pV(Sn)≤y!−→ P(x≤Z≤y),
Z
P(x≤Z≤y) = 1
√2πZy
x
e−z2
2dz.
Xj∼ B(p)Sn∼b(n, p)
Sn∼b(n, p)n−→ +∞
P x≤Sn−np
pnp(1 −p)) ≤y!−→ 1 √2πZy
x
e−z2
2dz.
x1,··· , xn
X1,··· , Xn
(Xk)1≤k≤n(xk)1≤k≤n
xkxk
k

(xk)k
(Xk)1≤k≤n
{Pθ, θ ∈Θ}
(Xk)1≤k≤nXkE
E=R Pθθ∈Θθ
Xk
θc θnFn:En→Θ
c
θn=F n(X1, X2,··· , Xn)
θ
(Xk)1≤k≤n
Fnθ θ Θ
EθPθ
c
θnθ θ c θn−→ θ
Pθn
θc θnMn
c
θnθ
B(c θn, θ) := Eθ(c θn)−θ.
c
θnB(c θn, θ) = 0 θΘ
B(c θn, θ)c θnFnc θn
Mn:= 1
nPn
k=1 Xk
E(X)
ˆσ2
n:= 1
n
n
X
k=1
X2
k−M2 n=1
n
n
X
k=1
(Xk−Mn)2
Var(X)

Var(X)
Vn:= 1
n−1
n
X
k=1
(Xk−Mn)2.
c
θnθ
In⊂Θθ In
In(Xk)1≤k≤n
Pθθ∈Θ
a In
X1,··· , Xnθ θ Θ
Pθ(θ∈In)≥a.
1−a Inθ
EA p =P(A)p
n n Xj
A jeme Sn=
n
X
j=1
Xj
Mn=Sn
nMn
p p
n A S
S
n
A n = 100 S= 54
p=P(A)α= 0,95
Sn∼b(n, p)E(Sn) = np V (Sn) = np(1 −p)Mn=Sn
nE(Mn) = p
V(Mn) = p(1 −p)
n
ε > 0
P(|Mn−E(Mn)| ≤ ε)≥1−V(Mn)
ε2
P(|Mn−p| ≤ ε)≥1−p(1 −p)
nε2.

P(|Mn−p| ≤ ε)≥1−p(1 −p)
nε2≥1−1 4nε2.
n Mn
S
n=54 100 ε1−1 4nε2=α
ε2=1
4n(1 −α).
ε=1
2pn(1 −α)=1
2√5
P
S
n−p≤ε≥α
PS
n−ε≤p≤S n+ε≥α.
P54
100 −1
2√5≤p≤54
100 +1
2√5≥α.
p
054 −1
2√5; 0,54 + 1 2√5⊂[0,31; 0,77].
0,54 −1
2√5; 0,54 + 1 2√5
Sn∼b(n, p)p
α= 0,95 n= 100 S= 54
n
Zn=Sn−np
pnp(1 −p)'Y∼ N(0,1).
ZnN(0,1) n Zn
Y∼ N(0,1) ε > 0
P 
Sn−np
pnp(1 −p)≤ε!= 2 P(0 ≤Y≤ε)=2
ε
Z
0
1
√2πe−x2/2dx.
ε2P(0 ≤Y≤ε)≥α ε = 1,96
2P(0 ≤Y≤1,96) = 0,95
P(|Sn−np| ≤ 1,96pnp(1 −p)) = 0,95.
p(1 −p)≤1
4
P|Sn−np| ≤ 1,96rn
4≥0,95.
P Sn−1,96pn
4
n≤p≤Sn+ 1,96pn
4
n!≥0,95.
p
54 −1,96 ×5
100 ;54 + 1,96 ×5 100 = [0,442; 0,638].