How to solve E($2X $| Y)? Where $f(x,y) = 4e^{-2y}$

Question

Help, please!

How to solve:

$E(2X| Y) = ?$

$$f(x,y) = 4e^{-2y}\;,0 < x<y \mbox{ and }\; y>0.$$

After integrating $f(x,y)$ over domain of $y$ we get marginal density of $X$:
$$f_X(x) = 2e^{-2x} \;,\;x>0$$
Similarly, marginal density of $Y$
$$f_Y(y) = 4ye^{-2y}\;,y>0$$

Then I found out the distribution of $2X$ using Jacobian Theorem as
$$f_U(u) = e^{-u}\;,u>0$$

How to calculate $E(2X| Y)$?

Answer 1

Conditional density of $X$ given $Y=y >0$ is $$ f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{4 e^{-2y}\mathbb 1_{\{0<x<y\}}}{4y e^{-2y}} = \frac{\mathbb 1_{\{0<x<y\}}}{y}. $$ This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $\mathbb E(X|Y=y)=\frac{y}2$, $\mathbb E(2X|Y=y)=y$. You can also calulate it directly as $$ \mathbb E(2X|Y=y) = 2\mathbb E(X|Y=y) = 2\int x f_{X|Y}(x|y) \, dx = 2\int_0^y x \frac1y \, dx = y. $$ Then $\mathbb E(2X|Y) =Y$ a.s.