Exercice sur la loi de Cauchy : Probabilité et Intégrabilité

f:R→R+, x 7→ 1
π(1+x2)
h: [0,1] →R
FXFX:R→[0,1], x 7→ Rx
−∞ f(t)dt
∀x∈R,
FX(x) = Zx
−∞
1
π(1 + t2)dt
=1
π[arctan(t)]x
−∞
=arctan(x)
π+1 2
FXFXC1
FX(R) =]0,1[ RC1F−1
XFX
F−1
X:]0,1[→R
y7→ tan(π(y−1/2))
h=F−1
X∈
∀x∈R,
P(h(U)≤x) = P(U≤FX(x)) (∗)
=FU(FX(x))
=FX(x)cf FU=Id sur [0,1]
{h(U)≤x}={U≤FX(x)}
FX

∀x∈R,
P(Y≤x) = P(1/X ≤x)
=P(0 ≥X≥1/x)si x ≤0
P(X≤0) + P(X≥1/x)sinon
=FX(0) −FX(1/x)si x ≤0
FX(0) + (1 −FX(1/x)) sinon
=(−arctan(1/x)
πsi x ≤0 (prolongement par continuit´e en 0)
1−arctan(1/x)
πsinon
=arctan(x)
π+1 2
FX
arctan(1/x) = arctan(x)−π/2si x < 0
arctan(x) + π/2si x > 0
Λ⊂Rα∈Λ|X|α
E[|X|α]<+∞α∈ΛRπ/2
−π/2|tan(t)|αdt < +∞
α
E[|X|α] = ZR |t|αf(t)dt
=ZR
|t|α
π(1 + t2)dt
t7→ |t|α
π(1+t2)α > 2
u(t)→
±∞ +∞
v:t7→ |t|2−α∼
±∞ u(t)
±∞
±∞
±∞ α−2<−1α < 1
w:t7→ |t|α∼
0u(t)
α > 1
Λ =] −1,1[

α∈Λ
E[|X|α] = ZR
|t|α
π(1 + t2)dt < +∞
=Z0
−∞
(−t)α
π(1 + t2)dt +Z+∞
0
tα
π(1 + t2)dt
=Z0
−π/2
(−tan(u))αdu +Zπ/2
0
tan(u)du(∗)
=Zπ/2
−π/2
|tan(u)|αdu
{tan(u) = x, u = arctan(x), du =1
1+x2dx
C1]−
π/2,0] [0, π/2[ R−R+
]−π/2, π/2[ α < 0
Rπ/2
−π/2|tan(u)|αdu < +∞