Compute the conditional expectation $E(X^2|XY)$ when $(X,Y)$ is standard normal

Question

Let $(X,Y)$ be a 2-dimensional standard normal random vector. Is there a specific trick to compute $E(X^2|XY)$?

Answer 1

1. Reduction step: By the usual polar representation of standard normal distribution, $$(X,Y)=\sqrt{2S}\cdot(\cos U,\sin U)$$ where $S$ is standard exponential, $U$ is uniform on $(0,2\pi)$, and $(U,S)$ is independent, hence $$E(X^2\mid XY)=\tfrac12E(X^2+Y^2\mid XY)=E(S\mid SV)$$ where $V=\sin(2U)$ is independent of $S$ and the PDFs of $S$ and $V$ are $$f_S(s)=e^{-s}\mathbf 1_{s>0}\qquad f_V(v)=\frac{\mathbf 1_{|v|<1}}{\pi\sqrt{1-v^2}}$$ 2. Computation step: Now, $E(S\mid SV)=g(SV)$ where $E(Sh(SV))=E(g(SV)h(SV))$ for every measurable bounded function $h$, that is, $$\iint sh(sv)f_S(s)f_V(v)dsdv=\int g(w)h(w)f_{SV}(w)dw$$ The LHS is $$\iint h(w)f_S(s)f_V(s^{-1}w)dsdw$$ hence, by identification, $$g(w)f_{SV}(w)=\int f_S(s)f_V(s^{-1}w)ds$$ that is, $$g(w)\cdot\int_{|w|}^\infty e^{-s}\frac{ds}{\pi\sqrt{s^2-w^2}}=\int_{|w|}^\infty e^{-s}\frac{sds}{\pi\sqrt{s^2-w^2}}$$ The change of variable $s=|w|\cdot\cosh t$ yields $$g(w)\cdot\int_0^\infty e^{-|w|\cosh t}dt=|w|\int_0^\infty e^{-|w|\cosh t}\cosh t\,dt$$ which can be expressed in terms of modified Bessel functions of the second kind $K_n$ as $$g(w)\cdot K_0(|w|)=|w|\cdot K_1(|w|)$$ 3. Conclusion: Finally,

$$E(X^2\mid XY)=g(XY)=|XY|\frac{K_1(|XY|)}{K_0(|XY|)}$$