Find the conditional expectation. Let  (X,Y) be a random vector with a pdf $cx(y-x)e^{-y},$ $0\leq x\leq y \leq \infty$. Find c, $\mathbb{E}(X|Y)$ and $\mathbb{E}(Y|X)$. c = 1, it was easy to calculate ($\int_\mathbb{R}\int_\mathbb{R}cx(y-x)e^{-y}dxdy = 1$). But what can i do with conditional expectations? I kinda have some problems with understanding its definition. Could you please give me any hints? 
 A: The PDF of the random variable $X|Y$ is$$f_{X|Y}(x,y)=\frac{P(X=x,Y=y)}{P(Y=y)}=\frac{f_{XY}(x,y)}{f_Y(y)}$$Find the marginal distribution of $Y$ by integrating the joint distribution with respect to $x$. Correspondingly for $Y|X$. The expectation $\mathbb E[X|Y]$ is defined by$$\int_{-\infty}^\infty xf_{X|Y}(x,y)dx$$Correspondingly for $\mathbb E[Y|X]$.
A: We have $f_{X, Y}(x, y) = x (y - x) e^{-y} [0 < x < y]$, so the marginal pdf $f_X(x)$ is
$$x \hspace {1.5px} [0 < x] \int_{\mathbb R} (y - x) e^{-y} [x < y] dy =
x \hspace {1.5px} [0 < x] \int_{\mathbb R^+} \tau e^{-\tau - x} d\tau.$$
$f_Y(y)$ can be found in the same way. Then we can find $\mathbb E(X \mid Y = y)$, and from that we can construct the random variable $\mathbb E(X \mid Y)$.
Instead, let $(X, Y) = (U, U + V)$. Since the Jacobian of the transformation is $1$, we have $f_{U, V}(u, v) = u v e^{-u - v} [0 < u \land 0 < v]$, which means that $U$ and $V$ are independent and identically distributed. Therefore
$$\mathbb E(Y \mid X) =
\mathbb E(U + V \mid U) =
U + \mathbb E V =
X + \int_{\mathbb R^+} v^2 e^{-v} dv, \\
\mathbb E(X \mid Y) =
\mathbb E(U \mid U + V) =
\frac 1 2 \mathbb E(U + V \mid U + V) =
\frac Y 2.$$
