I am self-thaught in advanced probability and one of the exercises which I found is:
"Let $\xi$ and $\eta$ be independent rvs uniformly distributed on $(0,1)$. Let $X=\xi\eta$ and $Y=\xi/\eta$. Calculate $E(X|Y)$"
So, the general formula is $\int_{x \in range(X)} xf_X(x|Y=y)dx$. We can find $f_X(x|Y=y)$ using $\frac{f_{X,Y}(x,y)}{f_Y(y)}$
I don't know how to compute pdf's $f_{X,Y}(x,y)$ and $f_Y(y)$.
And in our case we have $x=\xi\eta$, so how we can write $dx$ then.
Can anybody help me with it?
Inverting $X=\Xi H$ and $Y=\Xi/H$ yields $\Xi=\sqrt{XY}$ and $H=\sqrt{X/Y}$, so
\begin{eqnarray*} f_{X,Y}(x,y) &=& f_{\Xi,H}(\xi(x,y),\eta(x,y))\frac{\partial(\xi,\eta)}{\partial(x,y)} \\ &=& I_{[0,1]^2}\left(\sqrt{xy},\sqrt{x/y}\right)\left|\begin{array}{cc}\frac12\sqrt{\frac yx}&\frac12\sqrt{\frac1{xy}}\\\frac12\sqrt{\frac xy}&-\frac12\sqrt{\frac x{y^3}}\end{array}\right| \\ &=& I_{[0,1]}(x)I_{[x,1/x]}(y)\cdot\frac1{2y}\;. \end{eqnarray*}
Then
$$ f_Y(y)=\int_0^{\infty}f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac12&y\le1\;,\\\frac1{2y^2}&y\gt1\;,\end{cases} $$
and thus
\begin{eqnarray*} f_X(x\mid Y=y) &=&\frac{f_{X,Y}(x,y)}{f_Y(y)} \\ &=& I_{[0,1]}(x)I_{[x,1/x]}(y)\cdot\begin{cases}\frac1y&y\le1\\y&y\gt1\end{cases} \\ &=& I_{[0,1]}(x)\left(I_{[x,1]}(y)\cdot\frac1y+I_{[1,1/x]}(y)\,y\right) \\ &=&I_{[0,1]}(y)I_{[0,y]}(x)\cdot\frac1y+I_{[1,\infty]}(y)I_{[0,1/y]}(x)\,y \\ &=&I_{[0,1]}(y)\mathsf{Uniform}_{[0,y]}(x)+I_{[1,\infty]}(y)\mathsf{Uniform}_{[0,1/y]}(x)\;. \end{eqnarray*}
That is, $X$ is uniformly distributed conditional on $Y$, on $[0,y]$ if $y\le1$ and on $[0,1/y]$ if $y\gt1$.