Compute the conditional expectation E( U |UV ) I have $U$ and $V$ independent uniform on $[0,1]$ random variables
I have to compute $\mathbb{E}(U | UV)$.
Is there a specific trick to do this?
The basic way I know leads me find a function, $f$, such that for any measurable function, $g$, I have:
$\mathbb{E}(g(UV)*U) = \mathbb{E}(g(UV)*f(UV))$ , which gives me:
$\displaystyle \int_0^1 \int_0^1 g(xy)*x*dx*dy = \int_0^1 \int_0^1g(xy)*f(xy)*dx*dy$
but then I have no way to proceed and identity $f$... any ideas?
thanks!
edit : I tagged it as homework but it's actual an exercise from a previous exam I am trying to do. 
 A: As you rightly point out, the goal is to identify a function $f$ such that the identity
$$
\iint g(xy)x\mathbf 1_{0\leqslant x,y\leqslant 1}\mathrm dx\mathrm dy=\iint g(xy)f(xy)\mathbf 1_{0\leqslant x,y\leqslant 1}\mathrm dx\mathrm dy,
$$
holds for every (bounded measurable) function $g$.

Hint: Both sides are linear functionals of $g$, hence they are integrals of $g$ with respect to some measures. Equate those measures.

Concretely, this means using some change of variables to reach integrals of $g$. Here the change of variables $(x,y)\to(z,t)=(xy,x)$ seems a good idea. One gets $\mathrm dz\mathrm dt=x\mathrm dx\mathrm dy$, hence
$$
\mathrm{LHS}=\iint g(z)\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dz\mathrm dt,
$$
and
$$
\mathrm{RHS}=\iint g(z)f(z)\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dz\mathrm dt/t.
$$
Thus, $\mathrm{LHS}=\mathrm{RHS}$ for every (bounded measurable) function $g$ if and only if (almost everywhere)
$$
f(z)\int\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dt/t=\int\mathbf 1_{0\leqslant z\leqslant t\leqslant 1}\mathrm dt,
$$
that is,
$$
f(z)\mathbf 1_{0\leqslant z\leqslant 1}(-\log z)=\mathbf 1_{0\leqslant z\leqslant 1}(1-z).
$$
To sum up, $f(z)$ is undefined when $z\lt0$ or $z\geqslant1$ (as was to be expected), and, if $0\leqslant z\lt 1$, $f(z)$ must make true (almost everywhere) the identity above. An example of such a function $f$ is
$$
f(z)=\mathbf 1_{0\lt z\lt 1}(1-z)/(-\log z),
$$
hence

$$
\mathbb E(U\mid UV)=f(UV)=(1-UV)/(-\log UV).
$$

Note that while the function $f$ is not unique, even almost everywhere, since, for example, $f(z)$ for $z\lt0$ can be anything one wants, the random variable $f(UV)$ is unique almost surely.
Sanity checks: $\mathbb E(U\mid UV)\geqslant UV$ (why?) and $f(z)\to1$ when $z\to1$, $z\lt1$ (why?).
A: You can use a change of variables to obtain the joint density of $(U,UV)$, and in particular the density of $UV$. Dividing gives you the conditional density of $U$ given $UV$, from which all you have to do is integrate.
A: I don't have a good background in measure-theory so this may not be a very rigorous solution.

It can be seen that
$$\int\limits_k^{\sqrt{k}} \rho(u) du = \int\limits_{\sqrt{k}}^{1} \rho(u) du$$
or more generally
$$\int\limits_{a}^{b}\rho(u)du = \int\limits_\frac{k}{b}^\frac{k}{a}\rho(u)du$$
This is because, $\Pr[b < U < a | UV = k] = \Pr[1/b < V < 1/a | UV = k]$ and $U$ and $V$ are symmetric and interchangable.
Differentiating both sides w.r.t. $b$ we get,
$$\rho(b) = -\rho\left(\frac{k}{b}\right) * \frac{d \frac{k}{b}}{d b} = \rho\left(\frac{k}{b}\right) \frac{k}{b^2}$$
If we assume the a power-law expression for $\rho$, after normalising, we get
$$\rho(u) = -\frac{1}{\log(k)}\frac{1}{u}$$
which given an expectation value of $\frac{k-1}{\log(k)}$.
