corrigé - Jean

E= 8 ×1
6+ (−2) ×5 6=−1 3.
n
n
E0= 8 ×11
36 + (−2) ×25 36 =38 36.
38
36 −n≥ −1 3n < 50 36 ≈1,39
k > 1kPar(k)
f]− ∞,1[ [1,+∞]
f(x) = k
xk+1 .
X∼ Par(k)
f
fR+∞
−∞ f(x)dx =R+∞ 1
k
xk+1 dx = [−1 xk]+∞
1= 1
X
E[X] = Z+∞
−∞
xf(x)dx =Z+∞
1
k
xkdx = [−k (k−1)xk−1]+∞
1=k
k−1.
k > 1

a > 1P(X > a)
P(X > a) = Z+∞
a
k
xk+1 dx = [−1 xk]+∞
a=1
ak.
X m P(X < m) = P(X >
m)
mP(X > a) = 1
2
1
mk=1
2m= 21
k
b > 1P(X > a +b|X > a)
a
P(X > a+b|X > a) = P(X > a +b∩X > a)
P(X > a)=P(X > a +b) P(X > a)=ak (a+b)k= ( a a+b)k.
a7→ a
a+b
b
(a+b)2
k > 1 ( a
a+b)ka
X
a b
a
b
Y= ln(X)Y k
Y x > 0
FY(x) = P(Y≤x) = P(ln(X)≤x) = P(X≤ex)=1−1
(ex)k= 1 −e−kx.
fY(x) = F0
Y(x) = ke−kx.
k
n≥30 n Yj
Y Y 1/k 1/k2
Pn
j=1 Yj−n/k
√n/k =1 √n(k
n
X
j=1
Yj−n)∼ N(0,1).
X
k
500000
k Y
x1, . . . , x50 Y1/k
ln(x1),...,ln(x50)Y
1/k
ln(x1) + ··· + ln(x50
50 =ln(x1x2···x50) 50 ≈1 k.

k
k≈50
ln(500000) ≈3,81.
95% k
P(−1,96 <1
√50(k
50
X
j=1
Yj−50) <1,96) ≈0,95.
P50
j=1 Yj= ln(500000) ≈13,12 k
95%
[(50 −√501,96)/13,12; (50 + √501,96)/13,12] = [2,75; 4.87].
f t = 0 t= 1
R1
0f(t)dt
f ε [0,1]
g=f+ε1
nn∈N∗
g(k
n)k= 1 . . . n f
1
nPn
k=1 g(k
n)f
|R1
0f(t)dt −1
nPn
k=1 f(k
n)|<10 n
ε Xk=ε(k
n)
k= 1 . . . n N(0,1)
n= 100
Xk
XkN(0,1)
n= 100 >30
Pn
k=1 Xk−0
1×√nN(0,1).
p|1
nPn
k=1 g(k
n)−1 nPn
k=1 f(k
n)|<0,1
|1
nPn
k=1 g(k
n)−1 nPn
k=1 f(k
n)|=|1 nPn
k=1 ε(k
n)|=|Pn
k=1 Xk
n|
p=P(|Pn
k=1 Xk
n|<0,1) = P(−0,1√n < Pn
k=1 Xk
√n<0,1√n) = P(−1<Pn
k=1 Xk
√n<1)
n= 100
p≈FN(1) −FN(−1) = 2FN(1) −1≈0,68.

n
p
n
p=P(−0,1√n < Pn
k=1 Xk
√n<0,1√n)=2FN(0,1√n)−1.
p= 0,99 FN(0,1√n) = 0,995
0,1√n= 2,58 n= 665,64 n= 666
10% 1%
f
f
|Z1
0
f−1
n
n
X
k=0
g(k
n)|≤|Z1
0
f−1
n
n
X
k=0
f(k
n)|+|1 n
n
X
k=0
f(k
n)−1 n
n
X
k=0
g(k
n)|
n= 666 10/666 ≈0,002 0,1
1% 0,102
n= 100
σ
N(0, σ2)p0,99
n= 100
P100
k=1 Xk
10σN(0,1).
p=P(−0,1<P100
k=1 Xk
100 <0,1) = P(−1 σ<P100
k=1 Xk
10σ<1 σ)p= 0,99
2FN(1
σ)−1 = 0,99 FN(1 σ)=0,995 1 σ≈2,58
σ= 0,39