How can one calculate the following conditional expectancy : $\mathbb E[X\mid X^2+Y^2]$ where $X,Y\overset{iid}{\sim}\mathcal N(0,1)$ $?$
What I tried was to write that $\mathbb E[X\mid X^2+Y^2]=h(X^2+Y^2)$ where $\forall s\ge0$, $f(s)=\mathbb E[X\mid X^2+Y^2=s]$. This quantity is calculable as : $$h(s)=\int_\mathbb R xf_{X\mid X^2+Y^2=s}(x)dx$$ where the conditional density $f_{X\mid X^2+Y^2=s}$ verifies $\forall x\in\mathbb R$, $f_{X\mid X^2+Y^2=s}(x)=\dfrac{f_{(X,X^2+Y^2)}(x,s)}{f_{X^2+Y^2}(s)}$.
It is known that in this situation we have $f_{X^2+Y^2}(s)=\frac{1}{2}e^{-\frac s2}\mathbb 1_{s>0}$ ($X^2+Y^2\sim\chi^2(k))$ and by taking $\varphi:\mathbb R^2\rightarrow\mathbb R$ to be continuous and bounded, we have : \begin{eqnarray*} \mathbb E[\varphi(X,X^2+Y^2)]&=&\int_{\mathbb R^2}\varphi(x,x^2+y^2)\frac{1}{2\pi}e^\frac{-(x^2+y^2)}{2}dxdy\\ &=&\int_{\mathbb R}\int_{s^2}^\infty\varphi(s,t)\frac{1}{2\pi}e^{-\frac t2}\frac{dsdt}{2\sqrt{t-s^2}}\quad\text{by letting $(s,t)=(x,x^2+y^2)$} \end{eqnarray*} hence $f_{(X,X^2+Y^2)}(x,s)=\dfrac{1}{4\pi\sqrt{s-x^2}}e^{-\frac s2}\mathbb 1_{s>x^2}$ which yields $f_{X\mid X^2+Y^2=s}(x)=\dfrac{1}{2\pi\sqrt{s-x^2}}\mathbb 1_{s>x^2}$ and finally : $$h(s)=\int_{-\sqrt s}^{\sqrt s}\frac{x}{2\pi\sqrt{s-x^2}}dx=0\quad\text{by imparity}$$ and therefore $\mathbb E[X\mid X^2+Y^2]=0$.
This result does intuitively make sense (having information on $X^2+Y^2$ yields no information about the sign of $X$, and therefore since $X$ is symmetric about $0$, we could expect the conditional expectancy to be $0$). Therefore, one could be tempted to ask the following question : is the result still true if we only assume $X$ and $Y$ to be independent ant and symmetric about $0$? The problem is that with the method presented here, it isn't possible to answer this question. So is there a more general method of calculating this conditional expectancy that would allow one to give an answer to the above question?
By symmetry of their distributions, and independance of $X$ and $Y$: random vectors $\langle X,Y\rangle$ and $\langle-X,Y\rangle$ shall have identical distributions, so:
$$\begin{align}\mathsf E(X\mid X^2+Y^2) &= \mathsf E(-X\mid (-X)^2+Y^2)&&\text{via symmetry and independance}\\&=-\mathsf E(X\mid X^2+Y^2) && \text{by identical sigma algebrae}\end{align}$$
Which entails that $\mathsf E(X\mid X^2+Y^2) =0$
Therefore, the distributions don't have to be iid normal, it is sufficient that they be symmetric about $0$ and independant.