Let $X, Y$ both be independent normal random variables with mean zero and variance $\sigma^2$. Let $U:= X^2 + Y^2$ and $V:= X/U^{\frac12}$.
Find the joint distribution and find out whether $U$ and $V$ are independent.
For the joint distribution I got:
$f_{u,v}(u,v) = \frac{1}{2\pi\sigma} e^{-\frac{v\sqrt{u}}{2\sigma^2}} e^{-\frac{\sqrt{u-v^2u}}{2\sigma^2}}\frac{\sqrt{u}}{2\sqrt{u-v^2u}}$
That does not look like it's independent, is that correct? Also I am not really keen about calculating marginals and therefore integrating that function…
Best Answer
I will assume $\sigma=1$ this will not affect the method of proof. Consider an arbitrary positive bounded Borel function $h$. Then $$\eqalign{\mathbb{E} h(U,V)&=\mathbb{E} h\left(X^2+Y^2,\frac{X}{\sqrt{X^2+Y^2}}\right)\cr &=\frac{1}{2\pi}\int_{\mathbb{R}^2} h\left(x^2+y^2,\frac{x}{\sqrt{x^2+y^2}}\right)e^{-(x^2+y^2)/2}dxdy\cr &=\frac{1}{2\pi}\int_0^\infty\int_0^{2\pi} h\left(r^2,\cos\theta\right)e^{-r^2/2}r d\theta dr\cr &=\frac{1}{2\pi}\int_0^\infty\int_0^{\pi} h\left(u,\cos\theta\right)e^{-u/2} d\theta du\cr &=\frac{1}{2\pi}\int_0^\infty\int_{-1}^1 h\left(u,v\right)\frac{e^{-u/2}}{\sqrt{1-v^2}} dv du\cr &=\int_{\mathbb{R}^2}h(u,v)f_{U,V}(u,v)dudv }$$ and because $h$ is arbitrary we conclude that $$\eqalign{f_{U,V}(u,v)&=\mathbb{I}_{(0,+\infty)\times(-1,1)}(u,v) \frac{1}{2\pi}\frac{e^{-u/2}}{\sqrt{1-v^2}}\cr &=\mathbb{I}_{(0,+\infty)}(u) \frac{1}{2}e^{-u/2}\cdot \mathbb{I}_{(-1,1)}(v) \frac{1}{\pi}\frac{1}{\sqrt{1-v^2}}\cr &=f_U(u)\cdot f_V(v) } $$ This proves that $U$ and $V$ are independent and gives their probability density distributions.