Essec1 E 2018 corrige

Telechargé par ALPHA OUMAR BARRY
x
(Ynx) = n
i=1(Xix)
X1, . . . , XnX
Gn(x) =
n
Y
i=1
P(Xix) = F(x)n.
FC1α
GnnC1
Yn
Yngn
x, gn(x) = nf(x)F(x)n1.
Yn]0, α[
X1(ω), . . . , Xn(ω)
x
Yn(ω)x
(Znx) = n
i=1
j∈{1,...,n},j6=i(Xjx)
n
n x
(Znx) =
i∈{1,...,n}(Xix)
| {z }
=(Ynx)
i∈{1,...,n}(Xi> x)
j6=i(Xjx)
(n+ 1)
Hn(x)=P(Ynx)+
n
X
i=1
P(Xi> x)
j6=i(Xjx).
i∈ {1, . . . , n}n(Xi> x) (Xjx)j∈ {1, . . . , n}
j6=i
Hn(x) = F(x)n+
n
X
i=1
P(Xi> x)Y
j6=i
P(Xjx)
X1, . . . , XnX
x, Hn(x) = F(x)n+n1F(x)F(x)n1.
HnC1
α Zn
hn
x, hn(x) = nf(x)F(x)n1+n(n1)f(x)F(x)n2n2f(x)F(x)n1
=nf(x)F(x)n2[F(x)+(n1) nF (x)]
=nf(x)F(x)n2(n1)1F(x)
k3
f
x, f(x) =
1
αx]0, α[
0
X
x, F (x) =
0x0
x
α0xα
1xα
.
x, gn(x) =
nxn1
αnx]0, α[
0
x, hn(x) =
n(n1)(αx)xn2
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(n1)
αnααn
nαn+1
n+ 1=n(n1)α1
n1
n+ 1=n1
n+ 1α.
f\ {0, α}
Z+
−∞
f(x)dx =λ
αλZα
0
xλ1dx =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
αλ
.
x, F (x) =
0x0
x
αλ
0xα
1xα
.
E(X) = Zα
0
λxλ
αλdx =λ
αλ×αλ+1
λ+ 1 =λ
λ+ 1α.
Yn
x, Gn(x) = F(x)n=
0x0
x
α
0xα
1xα
.
α nλ Yn
E(Yn) =
+ 1α.
E(X)λ λn
ZnE(Zn)
E(Zn) = n(n1)λZα
0x
αλ1x
αλx
α(n2)λ
dx.
u=x
α
E(Zn) = n(n1)λα Z1
0
uλ1uλu(n2)λdu
=n(n1)λα Z1
0
u(n1)λdu Z1
0
udu
=n(n1)λα 1
1+(n1)λ1
1 +
=n(n1)λα 1 + 1(n1)λ
(1 + (n1)λ)(1 + )=n(n1)λ2α
(1 + (n1)λ)(1 + )
X]0, α[YnZn
ZnYn
0E(Zn)E(Yn)α
E(Zn) = α
+ 1
| {z }
=E(Yn)
×(n1)λ
1+(n1)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 σ1
y]0, β[,σ10(y) = 1
σ0σ1(y).
(x, y)]0, α[×]0, β[
2(γ)(x, y) = Gn1σ1(y)+ (xy)σ10(y)Gn10σ1(y)
=Gn1σ1(y)+xy
σ0(σ1(y))gn1σ1(y)
x]0, α[y7→ γ(x, y)σ(x)
x]0, α[, ∂2(γ)(x, σ(x)) = 0.
σ1(σ(x)) = x
Gn1(x) + xσ(x)
σ0(x)gn1(x)=0
σ0(x)Gn1(x) + σ(x)gn1(x) = xgn1(x).
u0v+uv0u=σ v =Gn1
x]0, α[,σ(x)Gn1(x)0(x) = xgn1(x)
σGn1
x]0, α[, σ(x)Gn1(x) = Zx
0
tgn1(t)dt
x]0, α[, σ(x) = 1
Gn1(x)Zx
0
tgn1(t)dt.
Gn1]0, α[gn1
u:t7→ t v :t7→ Gn1(t)C1]0, α[gn1
]0, α[ [A, x]
0<A<x<α
Gn1C1
Zx
A
tgn1(t)dt =htGn1(t)ix
AZx
A
Gn1(t)dt
A0Gn1
tGn1(t)
A00
x]0, α[, σ(x) = xZx
0
Gn1(t)
Gn1(x)dt.
σ
gn1Yn1
t7→ tgn1(t) ]0, x]
Zx
0
tgn1(t)dt
hx
2, xim > 0
Zx
0
tgn1(t)dt Zx
x
2
x
2mdt =mx2
4>0
()σ]0, α[
Zx
0
Gn1(t)
Gn1(x)dt (∗∗)
[0, α]x7→
Zx
0
Gn1(t)dt C1[0, α] (∗∗)
x]0, α[, σ(x) = x1
Gn1(x)Zx
0
Gn1(t)dt
1
Gn1
C1]0, α[C1
gn]0, α[
σC1]0, α[ ()
x]0, α[, σ0(x) = 1
Gn1(x)0Zx
0
tgn1(t)dt 1
Gn1(x)xgn1(x)
=gn1(x)
Gn1(x)2Zx
0
tgn1(t)dt xgn1(x)
Gn1(x)
=gn1(x)
Gn1(x)(σ(x)x) = gn1(x)
Gn1(x)
| {z }
>0
(xσ(x))
σ0
σ
C1]0, α[
]0, α[lim
0+σ, lim
α
σ
lim
0+σ= 0
Yn1]0, α[Gn1(α) = 1 ()
lim
α
σ=Zα
0
tgn1(t)dt = E(Yn1).
γ(x, y) = (xy)Gn1σ1(y)= (xσ(z))Gn1(z)
= (xz)Gn1(z)+(zσ(z))
|{z }
=Zz
0
Gn1(t)
Gn1(z)dt (∗∗)
Gn1(z)=(xz)Gn1(z) + Zz
0
Gn1(t)dt
y=σ(x)z=x
γ(x, σ(x)) = Zx
0
Gn1(t)dt
y
γ(x, σ(x)) γ(x, y) = Zx
0
Gn1(t)dt (xz)Gn1(z)Zz
0
Gn1(t)dt
= (zx)Gn1(z)Zz
x
Gn1(t)dt.
γ(x, σ(x)) γ(x, y) = Zz
xGn1(z)Gn1(t)dt
Gn1
xz
t[x, z], Gn1(z)Gn1(t)0
γ(x, σ(x)) γ(x, y)0.
xz
t[z, x], Gn1(z)Gn1(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(Yn1) = Zα
0
tgn1(t)dt.
E(ϕx(Yn1)) = Zα
0
ϕx(t)gn1(t)dt =Zx
0
tgn1(t)dt.
P(Yn1x) = Gn1(x) ()
σ(x) = E(ϕx(Yn1))
P(Yn1x).
Yn1
E(ϕ(Yn1))
ϕx(Yn1)
P(Yn1x)
X
α λ (∗∗)
Yn1
x]0, α[, σ(x) = xα
x(n1)λZx
0t
α(n1)λ
dt =xx
(n1)λ+ 1 =(n1)λ
(n1)λ+ 1x.
Yn1
σ(x)
xαβ= E(Yn1)
]0, α[α
x]0, α[, σ(x) = n1
nx.
α= 50 λ= 0,2n= 6
x]0,50[, σ(x) = x
2((n1)λ= 1)
An
En
An
An
An
max σ(X1), . . . , σ(Xn1)< ynEnmax σ(X1), . . . , σ(Xn1)yn
f I
a, b I, max f(a), f(b)=fmax(a, b).
ab a b
σ
max σ(X1), . . . , σ(Xn1)=σmax(X1),...,max(Xn1)=σ(Yn1).
Enσ
Yn1< σ1(yn)EnYn1σ1(yn).
Yn1=σ1(yn)
Yn1En
P(En)=PYn1< σ1(yn).
A∈ A,A:
ω7→ 1ωA
0
An
ω, Rn(ω) = xnynEn
0= (xnyn)En(ω).
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Essec1 E 2018 corrige

Telechargé par ALPHA OUMAR BARRY
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