What is $\mathbb{E}(X|Z)$ where $Z = X^2 + Y^2$

Question

Pick a point $(X,Y)$ inside the unit square $[0,1]^2$ uniformly at random. Let $Z = X^2 + Y^2$. What is $\mathbb{E}(X|Z)$?

I have got $\mathbb{E}(X^2|Z) = Z/2$ but not sure where to go after that.

Answer 1

(This answer is for the benefit of @PantelisSopasakis)

If $Z\leq1$ is given we may assume that $(X,Y)$ is lying on the quarter circle with radius $\sqrt{Z}$ in the first quadrant, uniformly distributed with respect to arc length, resp., to the polar angle $\phi={\rm Arg}(x,y)$.

(Think of $(X,Y)$ lying in a sector of the annulus $Z\leq X^2+Y^2\leq Z+dZ$. "In the limit" area measure on this annulus becomes arc length.)

It follows that in this case $$E(X\>|\>Z)={2\over\pi}\int_0^{\pi/2}\sqrt{Z}\cos\phi\>d\phi={2\over\pi}\sqrt{Z}\qquad(Z\leq 1)\ .\tag{1}$$ If $Z\geq1$ then the quarter circle becomes a circular arc of radius $\sqrt{Z}$ with polar angle $\phi$ bounded by $$\alpha\leq\phi\leq{\pi\over2}-\alpha\ ,$$ whereby $$\cos\alpha={1\over\sqrt{Z}}\ .$$ Instead of $(1)$ we now have $$E(X\>|\>Z)={1\over{\pi\over2}-2\alpha}\int_\alpha^{\pi/2-\alpha}\sqrt{Z}\>\cos\phi\>d\phi\qquad(1\leq Z\leq 2)\ .$$ I may leave it to you to simplify the result.