Corrigé Épreuve Maths ESSEC - Probabilités et Algorithmes

x∈
(Yn≤x) = n ∩
i=1(Xi≤x)
X1, . . . , XnX
Gn(x) =
n
Y
i=1
P(Xi≤x) = F(x)n.
FC1α
GnnC1
Yn
Yngn
∀x∈, gn(x) = nf(x)F(x)n−1.
Yn]0, α[
X1(ω), . . . , Xn(ω)
x
Yn(ω)x
(Zn≤x) = n ∪
i=1 ∩
j∈{1,...,n},j6=i(Xj≤x)
n
n x
(Zn≤x) = ∩
i∈{1,...,n}(Xi≤x)
| {z }
=(Yn≤x)
∪∪
i∈{1,...,n}(Xi> x)∩∩
j6=i(Xj≤x)
(n+ 1)
Hn(x)=P(Yn≤x)+
n
X
i=1
P(Xi> x)∩∩
j6=i(Xj≤x).
i∈ {1, . . . , n}n(Xi> x) (Xj≤x)j∈ {1, . . . , n}
j6=i
Hn(x) = F(x)n+
n
X
i=1
P(Xi> x)Y
j6=i
P(Xj≤x)
X1, . . . , XnX
∀x∈, Hn(x) = F(x)n+n1−F(x)F(x)n−1.
HnC1
α Zn
hn
∀x∈, hn(x) = nf(x)F(x)n−1+n(n−1)f(x)F(x)n−2−n2f(x)F(x)n−1
=nf(x)F(x)n−2[F(x)+(n−1) −nF (x)]
=nf(x)F(x)n−2(n−1)1−F(x)
k≥3

f
∀x∈, f(x) = 
1
αx∈]0, α[
0
X
∀x∈, F (x) = 
0x≤0
x
α0≤x≤α
1x≥α
.
∀x∈, gn(x) = 
nxn−1
αnx∈]0, α[
0
∀x∈, hn(x) = 
n(n−1)(α−x)xn−2
αnx∈]0, α[
0
E(Yn) = Zα
0
xgn(x)dx =n
αnZα
0
xndx =n
αn×αn+1 n+ 1 =n n+ 1α.
E(Zn) = Zα
0
xhn(x)dx =n(n−1)
αnααn n−αn+1 n+ 1=n(n−1)α1 n−1 n+ 1=n−1 n+ 1α.
f\ {0, α}
Z+∞
−∞
f(x)dx =λ
αλZα
0
xλ−1dx =λ
αλ×αλ λ= 1.
f
f]0, α[X
]− ∞,0] [α, +∞[
x∈[0, α]
F(x) = Zx
0
f(t)dt =λ
αλ×xλ λ=x αλ
.
∀x∈, F (x) = 
0x≤0
x
αλ
0≤x≤α
1x≥α
.
E(X) = Zα
0
λxλ
αλdx =λ αλ×αλ+1 λ+ 1 =λ λ+ 1α.
Yn
∀x∈, Gn(x) = F(x)n=
0x≤0
x
αnλ
0≤x≤α
1x≥α
.
α nλ Yn
E(Yn) = nλ
nλ + 1α.
E(X)λ λn
ZnE(Zn)
E(Zn) = n(n−1)λZα
0x
αλ1−x αλx α(n−2)λ
dx.
u=x
α
E(Zn) = n(n−1)λα Z1
0
uλ1−uλu(n−2)λdu
=n(n−1)λα Z1
0
u(n−1)λdu −Z1
0
unλdu

=n(n−1)λα 1
1+(n−1)λ−1 1 + nλ 
=n(n−1)λα 1 + nλ −1−(n−1)λ
(1 + (n−1)λ)(1 + nλ)=n(n−1)λ2α (1 + (n−1)λ)(1 + nλ)
X]0, α[YnZn
Zn≤Yn
0≤E(Zn)≤E(Yn)≤α
E(Zn) = αnλ
nλ + 1
| {z }
=E(Yn)
×(n−1)λ
1+(n−1)λ
| {z }
≤1
E(Yn)−→
n→+∞αE(Zn)−→ n→+∞α
n
n α
n= 2 Y2+Z2=X1+X2E(Y2) + E(Z2) = 2E(X)
E(Y2) + E(Z2) = 2
3α+1 3α=α= 2E(X) ;
E(Y2) + E(Z2) = 2λ
2λ+ 1 +2λ2 (1 + 2λ)(1 + λ)α=2λ 1 + λα= 2E(X).
λ= 1
α
]0, α[
σ σ−1
∀y∈]0, β[,σ−10(y) = 1
σ0σ−1(y).
(x, y)∈]0, α[×]0, β[
∂2(γ)(x, y) = −Gn−1σ−1(y)+ (x−y)σ−10(y)Gn−10σ−1(y)
=−Gn−1σ−1(y)+x−y
σ0(σ−1(y))gn−1σ−1(y)
x∈]0, α[y7→ γ(x, y)σ(x)
∀x∈]0, α[, ∂2(γ)(x, σ(x)) = 0.
σ−1(σ(x)) = x
−Gn−1(x) + x−σ(x)
σ0(x)gn−1(x)=0
σ0(x)Gn−1(x) + σ(x)gn−1(x) = xgn−1(x).
u0v+uv0u=σ v =Gn−1
∀x∈]0, α[,σ(x)Gn−1(x)0(x) = xgn−1(x)
σGn−1
∀x∈]0, α[, σ(x)Gn−1(x) = Zx
0
tgn−1(t)dt
∀x∈]0, α[, σ(x) = 1
Gn−1(x)Zx
0
tgn−1(t)dt.
Gn−1]0, α[gn−1
u:t7→ t v :t7→ Gn−1(t)C1]0, α[gn−1
]0, α[ [A, x]
0<A<x<α
Gn−1C1
Zx
A
tgn−1(t)dt =htGn−1(t)ix
A−Zx
A
Gn−1(t)dt
A→0Gn−1
tGn−1(t)−→
A→00
∀x∈]0, α[, σ(x) = x−Zx
0
Gn−1(t)
Gn−1(x)dt.
σ
gn−1Yn−1
t7→ tgn−1(t) ]0, x]
Zx
0
tgn−1(t)dt

hx
2, xim > 0
Zx
0
tgn−1(t)dt ≥Zx
x
2
x
2mdt =mx2 4>0
(∗)σ]0, α[
Zx
0
Gn−1(t)
Gn−1(x)dt (∗∗)
[0, α]x7→
Zx
0
Gn−1(t)dt C1[0, α] (∗∗)
∀x∈]0, α[, σ(x) = x−1
Gn−1(x)Zx
0
Gn−1(t)dt
1
Gn−1
C1]0, α[C1
gn]0, α[
σC1]0, α[ (∗)
∀x∈]0, α[, σ0(x) = 1
Gn−1(x)0Zx
0
tgn−1(t)dt −1
Gn−1(x)xgn−1(x)
=−gn−1(x)
Gn−1(x)2Zx
0
tgn−1(t)dt −xgn−1(x)
Gn−1(x)
=−gn−1(x)
Gn−1(x)(σ(x)−x) = gn−1(x) Gn−1(x)
| {z }
>0
(x−σ(x))
σ0
σ
C1]0, α[
]0, α[lim
0+σ, lim α−
σ
lim
0+σ= 0
Yn−1]0, α[Gn−1(α) = 1 (∗)
lim
α−
σ=Zα
0
tgn−1(t)dt = E(Yn−1).
γ(x, y) = (x−y)Gn−1σ−1(y)= (x−σ(z))Gn−1(z)
= (x−z)Gn−1(z)+(z−σ(z))
|{z }
=Zz
0
Gn−1(t)
Gn−1(z)dt (∗∗)
Gn−1(z)=(x−z)Gn−1(z) + Zz
0
Gn−1(t)dt
y=σ(x)z=x
γ(x, σ(x)) = Zx
0
Gn−1(t)dt
y
γ(x, σ(x)) −γ(x, y) = Zx
0
Gn−1(t)dt −(x−z)Gn−1(z)−Zz
0
Gn−1(t)dt
= (z−x)Gn−1(z)−Zz
x
Gn−1(t)dt.
γ(x, σ(x)) −γ(x, y) = Zz
xGn−1(z)−Gn−1(t)dt
Gn−1
x≤z
∀t∈[x, z], Gn−1(z)−Gn−1(t)≥0
γ(x, σ(x)) −γ(x, y)≥0.
x≥z
∀t∈[z, x], Gn−1(z)−Gn−1(t)≤0
γ(x, σ(x)) −γ(x, y)≥0.
∀x∈]0, α[,∀y∈]0, β[, γ(x, y)≤γ(x, σ(x))
x∈]0, α[y∈]0, β[7→ γ(x, y)
σ(x)

t
x
x
α
0
•E(Yn−1) = Zα
0
tgn−1(t)dt.
E(ϕx(Yn−1)) = Zα
0
ϕx(t)gn−1(t)dt =Zx
0
tgn−1(t)dt.
P(Yn−1≤x) = Gn−1(x) (∗)
σ(x) = E(ϕx(Yn−1))
P(Yn−1≤x).
Yn−1
E(ϕ(Yn−1))
ϕx(Yn−1)
P(Yn−1≤x)
X
α λ (∗∗)
Yn−1
∀x∈]0, α[, σ(x) = x−α
x(n−1)λZx
0t
α(n−1)λ
dt =x−x
(n−1)λ+ 1 =(n−1)λ (n−1)λ+ 1x.
Yn−1
σ(x)−→
x→αβ= E(Yn−1)
]0, α[α
∀x∈]0, α[, σ(x) = n−1
nx.
α= 50 λ= 0,2n= 6
∀x∈]0,50[, σ(x) = x
2((n−1)λ= 1)
An
En
An
An
An
max σ(X1), . . . , σ(Xn−1)< yn⊂En⊂max σ(X1), . . . , σ(Xn−1)≤yn
f I
∀a, b ∈I, max f(a), f(b)=fmax(a, b).
≥
≤a≤b a ≥b
σ
max σ(X1), . . . , σ(Xn−1)=σmax(X1),...,max(Xn−1)=σ(Yn−1).
Enσ
Yn−1< σ−1(yn)⊂En⊂Yn−1≤σ−1(yn).
Yn−1=σ−1(yn)
Yn−1En
P(En)=PYn−1< σ−1(yn).
A∈ A,A:
Ω→
ω7→ 1ω∈A
0
An
∀ω∈Ω, Rn(ω) = xn−ynEn
0= (xn−yn)En(ω).