Question

Suppose X and Yhave the joint density function f(x,y) =t0, otherwise Find Var(Y| X = 1). A0.056 B0.08o 0.102 D 0.156 E 0.2 22

Answer 1

First we need to find the value of c.

$\int_{0}^{2} \int_{0}^{y} cxy~ dy ~dx = 1$

$\Rightarrow c\int_{0}^{2} y * (y^2/2)~ dy = 1$

$\Rightarrow c\int_{0}^{2}(y^3/2)~ dy = 1$

$\Rightarrow c (2^4/8) = 1$

$\Rightarrow c = 1/2$

Therefore,

$f(x, y ) = xy/2$ for 0 < x < y < 2

$f(x) = \int_{x}^{2} xy/2~dy = (x/2) * ((2^2/2)-(x^2/2)) = (x/2) * (2 - x^2/2)$

$f(x) = x - x^3/4$

$f(y | x)= f(x, y) / f(x) =((xy)/2) / (x - x^3/4) = y / 2(1 - x^2/4)$

$f(y | x= 1)= y / 2(1 - 1^2/4) = 2y/3$

$E[y | x = 1] = \int_{1}^{2}yf(y | x = 1) ~dy= \int_{1}^{2}(2y^2/3)~dy = (2/9) (2^3 - 1^3) = 14/9$

$E[y^2 | x = 1] = \int_{1}^{2}y^2f(y | x = 1) ~dy= \int_{1}^{2}(2y^3/3)~dy$

$= (2/12) (2^4 - 1^4) = 15/6$

$Var[Y = X = 1] = E[y^2 | x = 1] - \left ( E[y | x = 1] \right )^2 = (15/6)- (14/9)^2 = 0.08024691$

So, the correct answer is B 0.080