Question

Problem 1. Let X and Y be continuous random variables with joint probability density function f(x,y) distributions for X and Y are (i/3) (x +y), for (x, y) in the rectangular region 0ss1,0Sys 2. The two marginal Ix(x)- (z+1), if 0 251 fy(y) = (1+2y), if0 y 2 Calculate E(x IY -v) and Var (X |Y ) for each y l0,2).

Answer 1

Conditional distribution of X|Y is given by :-

$f_{X|Y=y}(x)=\frac{f(x,y)}{f_{Y}(y)}$

$\!\!\therefore f_{X|Y=y}(x)=\frac{(1/3)(x+y)}{(1/6)(1+2y)}=\frac{2(x+y)}{(1+2y)} \;\;\;\;\forall \;x\in[0,1]\;\;for\;each\;y\in[0,2]$

$E(X|Y=y)=\int_{0}^{1}xf_{X|Y=y}(x)dx=\int_{0}^{1}\frac{2x(x+y)}{1+2y}dx=\frac{3y+2}{3(1+2y)}$

for each y$\in$[0,2]

Var(X|Y=y) = E(X²|Y=y) - [E(X|Y=y)]²

$E(X^2|Y=y)=\int_{0}^{1}x^2f_{X|Y=y}(x)dx=\int_{0}^{1}\frac{2x^2(x+y)}{1+2y}dx=\frac{3+4y}{6(1+2y)}$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!Var(X|Y=y)=\frac{3+4y}{6(1+2y)}-[\frac{3y+2}{3(1+2y)}]^2=\frac{3+4y}{6(1+2y)}-\frac{9y^2+12y+4}{9(1+2y)^2}=\frac{6y^2+6y+1}{18(1+2y)^2}$

for each y$\in$[0,2]