# Conditional Expectation and Probability Density Function

So given the joint pdf of $(X,Y) = 8xy$, as long as $0 < x < y < 1$ (and 0 otherwise). I got the conditional expectation to be $E(X|Y=y)=2y/3$ (if $0 < x < y <1$). Now I'm a little bit confused as how to get the pdf of the random variable $E(X|Y=y)$. Does this need a 1D change of variable (i.e. Let $Z=E(X|Y=y)$, then do a 1D change of variable as $Z=2Y/3$?) Or can I simply integrate this function from 0 to 1, with respect to the variable $y$? I'm practicing for an exam coming up and I'm struggling on this question. Any help is greatly appreciated. Thanks!

Note: $\mathsf E(X\mid Y=y)$ is not a random variable, rather $\mathsf E(X\mid Y)$ is.
But indeed $\mathsf E(X\mid Y)=\tfrac 23 Y$.   The probability density function of this random variable will be obtained by a change of variables transformation.$$f_{2Y/3}(z)=(3/2)~f_Y(3z/2)$$
And you should have found along the way that :$$\begin{split}f_Y(y)&=\int_0^y 8xy\mathop{\mathsf ds} \\[1ex] &= 4y^3\mathbf 1_{0<y<1}\end{split}$$
• Oh ok I understand. So $f(z)=(3/2)(4(3/2z)^3)=81z^3/4$? Or should I express the answer in terms of y? (i.e. replace all the z with 2y/3?) Jun 18, 2018 at 15:39
• If you have $f_Z(z)$ on the left hand side, then the right hand side should be a function of $z$ too. What you need to do is indicate the support interval. Jun 18, 2018 at 20:50