I currently have a problem with a problem set I'm working.
Suppose $U$ and $V$ are independent with exponential distribution with parameter $A$ ($T$ is exponentially distributed with parameter if its density is given by $A\exp(-At)$ for $t>0$).
Define $X=U+V$ and $Y=UV$.
Find $\operatorname E(X)$, $\operatorname E(Y)$, $\operatorname{Cov}(X,Y)$, $\operatorname{Var}X$ and $\operatorname E(Y\mid X)$.
I think I managed to find everything except for $\operatorname E(Y\mid X)$. I know that $\operatorname E(Y\mid X)$ is equal to the integral of $yf(y\mid x)\mathrm dy$ and that $f(y\mid x)= f(y,x)/f(x)$ so I need to find $f(y,x)$.
Problem is I'm not sure how to find $f(y,x)$. I tried using Jacobian transformation and ended up with V/A and I'm not sure if I did it right.
A shortcut is to note that $X=U+V$ and $Y=UV$ hence $Y=U(X-U)$ and that, conditionally on $X$, $U$ is uniformly distributed on $(0,X)$, that is, $U=XW$ where $W$ is independent on $X$ and uniform on $(0,1)$. Thus, $Y=U(X-U)=W(1-W)X^2$ and $E(Y\mid X)=X^2E(W(1-W))$ where $$ E(W(1-W))=\int_0^1w(1-w)\mathrm dw=\frac16, $$ that is, $$ E(Y\mid X)=\frac{X^2}6. $$ On the long road, one computes the conditional density, based on the standard change of variables $(U,V)\to(X,Y)$, since $$X\gt0,\quad0\lt Y\lt\frac{X^2}4,\quad U,V=\frac{X\pm\sqrt{X^2-4Y}}2.$$ The Jacobian is $\mathrm dx\mathrm dy=\sqrt{x^2-4y}\mathrm dx\mathrm dy$ and two points $(U,V)$ are mapped to the same point $(X,Y)$ hence the density of $(X,Y)$ is $$ f_{X,Y}(x,y)=\frac{2A^2\mathrm e^{-Ax}}{\sqrt{x^2-4y}}\,\mathbf 1_{x\gt0}\,\mathbf 1_{0\lt4y\lt x^2}. $$ Thus, for every $x\gt0$, $$ f_{Y\mid X}(y\mid x)=\frac{2}{x\sqrt{x^2-4y}}\,\mathbf 1_{0\lt4y\lt x^2}, $$ and $E(Y\mid X)=u(X)$ where, for every $x\gt0$, $$ u(x)=\int yf_{Y\mid X}(y\mid x)\mathrm dy=\int_0^{x^2/4}\frac{2y}{x\sqrt{x^2-4y}}\mathrm dy=\frac{x^2}8\int_0^{1}\frac{z}{\sqrt{1-z}}\mathrm dz=\frac{x^2}6, $$ hence $$ E(Y\mid X)=\frac{X^2}6. $$