Exercícios resolvidos de Probabilidade C

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(152159) Probabilidade C
Lista 1 - Entrega ate´ 10/11/2016, 19h
Exerc´ıcio 1. Considere que (X,Y ) tem a seguinte densidade conjunta:
f(X,Y )(x, y) = cI(0 < x < 1)I(|y| < x)
(a) Determine o valor de c.
(b) Determine F(X,Y )(x, y).
(c) Determine fX(x) e fY (y).
(d) Determine f(X|Y )(x|y).
(e) Determine P (Y > 2−1X).
Soluc¸a˜o:
(a)
1 = P ((X,Y ) ∈ R2) =
∫ ∞
−∞
f(X,Y )(x, y)dydx
=
∫ 1
0
∫ x
−x
cdydx
= c
∫ 1
0
2xdx
= cx2
∣∣∣∣1
0
= c
Portanto, c = 1.
(b)
F(X,Y )(u, v) =
∫ u
−∞
∫ v
−∞
f(X,Y )(x, y)dydx
=
∫ u
−∞
∫ v
−∞
I(0 < x < 1)I(|y| < x)dydx
Para determinar a u´ltima integral, e´ necessa´rio considerar 4 casos. Se u < 0 ou v < −|u|, enta˜o a integral e´
0. Caso contra´rio, se v > min(u, 1), enta˜o∫ u
−∞
∫ v
−∞
I(0 < x < 1)I(|y| < x)dydx =
∫ min(u,1)
0
∫ x
−x
dydx
=
∫ min(u,1)
0
2xdx
= min(u, 1)2
1
Caso contra´rio, se v < 0, enta˜o∫ u
−∞
∫ v
−∞
I(0 < x < 1)I(|y| < x)dydx =
∫ min(u,1)
−v
∫ v
−x
dydx
=
∫ min(u,1)
−v
(x+ v)dx
=
(
x2
2
+ vx
) ∣∣∣∣min(u,1)
−v
=
min(u, 1)2
2
− v
2
2
+ min(u, 1)v + v2 =
(min(u, 1) + v)2
2
Caso contra´rio, se 0 < v < min(u, 1), enta˜o∫ u
−∞
∫ v
−∞
I(0 < x < 1)I(|y| < x)dydx =
∫ v
0
∫ x
−x
dydx+
∫ min(u,1)
v
∫ v
−x
dydx
=
∫ v
0
2xdx+
∫ min(u,1)
v
(v + x)dx
= v2 +
(
x2
2
+ vx
) ∣∣∣∣min(u,1)
v
=
min(u, 1)2
2
+ min(u, 1)v − v
2
2
Portanto, obtemos dos casos anteriores que
F(X,Y )(u, v) =

0 , se u < 0 ou v < −|u|
min(u, 1)2 , se u ≥ 0 e v > min(u, 1)
(min(u,1)+v)2
2 , se u ≥ 0 e −|u| < v < 0
min(u,1)2
2 + min(u, 1)v − v
2
2 , se u ≥ 0 e 0 < v < min(u, 1)
(c) Do item anterior, obtemos
FX(u) = lim
v→∞FX,Y (u, v) = min(u, 1)
2I(u > 0)
FY (v) = lim
u→∞FX,Y (u, v) =

0 , se v < −1
(1+v)2
2 , se −1 ≤ v < 0
1
2 + v − v
2
2 , se 0 ≤ v < 1
1 , se v ≥ 1
Portanto,
fX(u) =
dFX(u)
du
= 2uI(0 ≤ u ≤ 1)
fY (v) =
dFY (v)
dv
=
(v + 1)I(v ≥ −1) , se v < 0(−v + 1)I(v ≤ 1) , se v ≥ 0
2
(d)
f(X|Y )(x|y) =
f(X,Y )(x, y)
fY (y)
=
I(0 < x < 1)I(|y| < x)
(1− |y|)I(−1 ≤ y ≤ 1)
= (1− |y|)−1I(|y| < x < 1)
(e)
P (Y > 2−1X) =
∫
(x,y):y>2−1x
fX,Y (x, y)dydx
=
∫ 1
0
∫ x
2−1x
dydx
=
∫ 1
0
x
2
dx
=
x2
4
∣∣∣∣1
0
= 4−1
Exerc´ıcio 2. Considere que fX|Y (x|y) = fX(x). Prove que:
(a) fX,Y (x, y) = fX(x)fY (y).
(b) fY |X(y|x) = fY (y).
Soluc¸a˜o:
(a) fX,Y (x, y) = fY (y)fX|Y (x|y) = fY (y)fX(x).
(b) fY |X(y|x) = fX,Y (x,y)fX(x) =
fY (y)fX(x)
fX(x)
= fY (y).
Exerc´ıcio 3. Dado τ2, X ∼ N (0, 1
τ2
)
. Tambe´m, τ2 ∼ Gamma(a, b). Ache a densidade de τ2|X.
Soluc¸a˜o:
fτ2|X(τ2|x) =
fτ2(τ
2)fX|τ2(x|τ2)∫∞
−∞ fτ2(τ
2)fX|τ2(x|τ2)dτ2
=
ba
Γ(a)(τ
2)a−1 exp(−bτ2)I(τ2 > 0) · τ√2pi−1 exp
(
−τ2 x22
)
∫∞
−∞
ba
Γ(a)(τ
2)a−1 exp(−bτ2)I(τ2 > 0) · τ√2pi−1 exp
(
−τ2 x22
)
dτ2
=
(τ2)a+0.5−1 exp
(
−
(
b+ x
2
2
)
τ2
)
I(τ2 > 0)∫∞
0 (τ
2)a+0.5−1 exp
(
−
(
b+ x
2
2
)
τ2
)
dτ2
=
(
b+x
2
2
)(a+0.5)
Γ(a+0.5) (τ
2)a+0.5−1 exp
(
−
(
b+ x
2
2
)
τ2
)
I(τ2 > 0)
∫∞
0
(
b+x
2
2
)(a+0.5)
Γ(a+0.5) (τ
2)a+0.5−1 exp
(
−
(
b+ x
2
2
)
τ2
)
dτ2
=
(
b+ x
2
2
)(a+0.5)
Γ(a+ 0.5)
(τ2)a+0.5−1 exp
(
−
(
b+
x2
2
)
τ2
)
I(τ2 > 0)
3
Portanto, τ2|X ∼ Gamma
(
a+ 0.5, b+ x
2
2
)
.
Exerc´ıcio 4. Considere que (X,Y ) tem densidade dada por
f(X,Y )(x, y) = exp(−(x+ y))I(x > 0)I(y > 0)
(a) Seja z > 0, determine P (X + Y ≤ z).
(b) Determine fX+Y (z). Qual a distribuic¸a˜o de X + Y ?
(c) Mostre que X e Y sa˜o independentes.
Soluc¸a˜o:
(a) Se z ≤ 0, enta˜o como X > 0 e Y > 0, temos que P (X + Y ≤ z) = 0. Tambe´m, se z > 0, enta˜o definimos
Az = {(x, y) ∈ R2 : x+ y ≤ z}. Note que
P (X + Y ≤ z) = P ((X,Y ) ∈ Az)
=
∫
Az
f(X,Y )(x, y)dxdy
=
∫ z
0
∫ z−x
0
exp(−(x+ y))dydx
=
∫ z
0
exp(−x)
∫ z−x
0
exp(−y)dydx
=
∫ z
0
− exp(−x) exp(−y)
∣∣∣∣z−x
0
dx
=
∫ z
0
exp(−x)(1− exp(x− z))dx
=
∫ z
0
exp(−x)− exp(−z)dx
= 1− exp(−z)− z exp(−z) = 1− (z + 1) exp(−z)
Portanto, P (X + Y ≤ z) = (1− (z + 1) exp(−z))I(z > 0)
(b)
fX+Y (z) =
dFX+Y (z)
dz
=
dP (X + Y ≤ z)
dz
=
d((1− (z + 1) exp(−z))I(z > 0))
dz
= z exp(−z)
Como fX+Y (z) = z exp(−z), conclua que X + Y ∼ Gamma(2, 1).
4
(c)
f(X,Y )(x, y) = I(x > 0) exp(−x) · I(y > 0) exp(−y)
= h1(x) · h2(y) h1(x) = exp(−x)I(x > 0)
h2(y) = exp(−y)I(y > 0)
Portanto X e Y sa˜o independentes.
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