Find conditional expectation $E(\xi|\xi^2)$ The problem is to find the conditional expectation $E(\xi | \xi^2)$, where $\xi$ is a uniform random value on $[-1, 1]$.
The approach I tried to implement is to prove somehow that $E(\xi|\xi^2)=E(-\xi|\xi^2)$, since $\xi$ is from $[-1,1]$, that is, symmetrical, but I don't know how to continue the solution.
 A: Let's reason from a more intuitive standpoint.  Suppose I draw a realization of $\xi$, and I tell you that $\xi^2 = 0.57314$.  You correctly conclude that I must have either observed $\xi = -0.75706$, or $\xi = 0.75706$.  These are the only two possibilities that could have generated that realization.  Is either one equally likely to have been the value I realized?  Of course--and this is the intuitive part--because $\xi$ was drawn from a uniform distribution.  Since both possibilities are equally likely from your perspective, the expected value of $\xi$ conditioned on your knowledge of $\xi^2$, must be zero.
As the other comments correctly point out, it isn't uniformity of $\xi$ in particular that makes it equally likely for either sign to have been realized.  It is the symmetry of the distribution about $0$.  That is precisely what is meant by $\xi \sim -\xi$.  Indeed, had $\xi \sim \operatorname{Normal}(0, \sigma^2)$, your finding of the conditional expectation $\operatorname{E}[\xi \mid \xi^2] = 0$ would be unchanged.
A: You have that $\mathsf E(\xi\mid \xi^2)=\mathsf E(-\xi\mid \xi^2)$ by an argument from symmetry.   That is okay.
Now the Linearity of Expectation says $\mathsf E((-1)\xi\mid \xi^2)=(-1)\mathsf E(\xi\mid \xi^2)$.
Put them together, what do they mean?

The argument is that if you know any supported value of $\xi^2$, then you know that $\xi$ may be either the positive or negative square root of that value with equal weight, since the distribution of $\xi$ is symmetric over the two intervals which fold symmetrically into the support of $\xi^2$.
$$\begin{align}\mathsf P(\xi={+}\surd x\mid \xi^2=x) ~&=~ \dfrac{f_X({+}\surd x)~\lVert \partial ({+}\surd x)/\partial x\rVert}{f_X({+}\surd x)~\lVert \partial ({+}\surd x)/\partial x\rVert+f_X({-}\surd x)~\lVert \partial ({-}\surd x)/\partial x\rVert} \\[1ex] &=~\mathsf P(\xi={}-\surd x\mid \xi^2=x) \\[1ex] &= \tfrac 12\end{align}$$
From here the conditional expectation should be immediately obvious.
