Calculate conditional expectation where $U$ is uniformly distributed on $[0,1]$ and $V = U(1-U)$ Suppose that $U$ is uniformly distributed on $[0,1]$ and $V = U(1-U)$. Calculate $E(U | V)$. 
I'm not really sure where to even begin on this one. I guess that I started by rewriting as $E (U | \sigma(U(1-U)))$, but I don't know where to go from here.
 A: Hint: when $V=v$, either $v=1/4$ (in which case $U=1/2$ necessarily), or else there are two values of $U$ which are possible for each value of $v$. What are those values?
Details: doing this calculation rigorously requires some sort of framework to explain how the variable $U|V=v$ is discrete even though $U$ is continuous. The Dirac delta is one way of dealing with this, and is convenient for calculation.
For instance, we can write the calculation like this:
\begin{align*}
E[U|V=v] &= \frac{\int_0^1 x \delta(x-x^2-v) dx}{\int_0^1 \delta(x-x^2-v) dx} \\
& = \frac{\frac{x_1}{|-2x_1+1|} + \frac{x_2}{|-2x_2+1|}}{\frac{1}{|-2x_1+1|}+\frac{1}{|-2x_2+1|}} 
\end{align*}
where $v \in [0,1/4]$ and $x_1,x_2$ are the solutions to $x-x^2-v=0$. Now it is a matter of simplifying. Of course the intuitive calculation, which asserts that $U|V=v$ is uniform on $x_1,x_2$, is simpler. One can check that the intuitive calculation works using this calculation by noting that the roots of a quadratic are symmetric about the vertex, which means that $|-2x_1+1|$ and $|-2x_2+1|$ are the same number. So if we call this number $d$ then we just have
$$\frac{\frac{x_1+x_2}{d}}{\frac{2}{d}}=\frac{x_1+x_2}{2}.$$
