Sejam X e Y variáveis aleatórias contínuas com densidade conjunta:
f_{X, Y}(x, y)=3(x+y) I_{(0,1)}(x+y) I_{(0,1)}(x) I_{(0,1)}(y)
a. Obtenha E(X \mid Y=y).
b. Determine E(X Y \mid Y=y).
c. Calcule \operatorname{Var}\left(X \left\lvert\, Y=\frac{1}{2}\right.\right).

Passo 1

O enunciado dá a densidade conjunta
f_{X,Y}(x,y)=3(x+y)\,I_{(0,1)}(x)\,I_{(0,1)}(y)\,I_{(0,1)}(x+y)
isto é, f_{X,Y}(x,y)=3(x+y) para x>0, y>0 e x+y<1, e zero caso contrário.
Precisamos de condicionais \,E(X\mid Y=y), E(XY\mid Y=y) e \Var\bigl(X\mid Y=\tfrac12\bigr). Para isso vamos:
  1. calcular a marginal f_Y(y) e a condicional f_{X\mid Y}(x\mid y);
  2. obter E(X\mid Y=y);
  3. usar esse resultado para E(XY\mid Y=y);
  4. calcular E(X^2\mid Y=y) e a variância condicional;
  5. avaliar em y=\tfrac12.

Passo 2

Marginal de Y:
f_Y(y)=\int_0^{1-y}3(x+y)\,dx =3\Bigl[\tfrac{(1-y)^2}{2}+y(1-y)\Bigr] =\tfrac{3}{2}(1-y^2), \quad 0<y<1.
Logo
f_{X\mid Y}(x\mid y) =\frac{3(x+y)}{\tfrac{3}{2}(1-y^2)} =\frac{2(x+y)}{1-y^2}, \quad 0<x<1-y.
Agora
E(X\mid Y=y) =\int_0^{1-y}x\,f_{X\mid Y}(x\mid y)\,dx =\frac{1}{\tfrac{3}{2}(1-y^2)} \int_0^{1-y}3x(x+y)\,dx.
Calcule o numerador:
\int_0^{1-y}3(x^2+xy)\,dx =3\Bigl[\tfrac{(1-y)^3}{3}+y\tfrac{(1-y)^2}{2}\Bigr] =(1-y)^2\frac{2+y}{2}.
Dividindo por f_Y(y)=\tfrac{3}{2}(1-y^2) resulta
E(X\mid Y=y) =\frac{(1-y)^2\frac{2+y}{2}}{\frac{3}{2}(1-y^2)} =\frac{(1-y)(2+y)}{3(1+y)}.

Passo 3

Para E(XY\mid Y=y) basta multiplicar por y, pois Y=y torna Y constante:
E(XY\mid Y=y) =y\,E(X\mid Y=y) =\frac{y\,(1-y)(2+y)}{3(1+y)}.

Passo 4

Calcule E(X^2\mid Y=y):
\int_0^{1-y}x^2\cdot3(x+y)\,dx =3\Bigl[\tfrac{(1-y)^4}{4}+y\tfrac{(1-y)^3}{3}\Bigr] =(1-y)^3\frac{3+y}{4}.
Então
E(X^2\mid Y=y) =\frac{\frac{(1-y)^3(3+y)}{4}}{\frac{3}{2}(1-y^2)} =\frac{(1-y)^2(3+y)}{6(1+y)}.
A variância condicional é
\Var(X\mid Y=y) =E(X^2\mid Y=y)-\bigl[E(X\mid Y=y)\bigr]^2.

Passo 5

Em y=\tfrac12 temos
E(X\mid Y=\tfrac12) =\frac{(1-\tfrac12)(2+\tfrac12)}{3(1+\tfrac12)} =\frac{\tfrac12\cdot\tfrac52}{3\cdot\tfrac32} =\frac{5}{18},
E(X^2\mid Y=\tfrac12) =\frac{(1-\tfrac12)^2(3+\tfrac12)}{6(1+\tfrac12)} =\frac{\tfrac14\cdot\tfrac72}{6\cdot\tfrac32} =\frac{7}{72}.
Logo
\Var\bigl(X\mid Y=\tfrac12\bigr) =\frac{7}{72}-\Bigl(\frac{5}{18}\Bigr)^2 =\frac{7}{72}-\frac{25}{324} =\frac{13}{648}.

Resposta

a. \displaystyle E(X\mid Y=y)=\frac{(1-y)(2+y)}{3(1+y)}.
b. \displaystyle E(XY\mid Y=y)=\frac{y\,(1-y)(2+y)}{3(1+y)}.
c. \displaystyle \Var\bigl(X\mid Y=\tfrac12\bigr)=\frac{13}{648}.

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