Show $(X^{1/2}\cos(\Theta), X^{1/2} \sin(\Theta))$ is a multidimensional normal distribution where $X$ is exponential and $\Theta$ uniform

Question

Let $X$ be a random variable with an exponential distribution of parameter $1/2$ and $\Theta$ the uniform law on $[0;2\pi]$. We suppose $X$ and $\Theta$ independent. Show that $Y=(X^{1/2}\cos(\Theta), X^{1/2} \sin(\Theta))$ has a multivariate normal distribution.

What I tried:

I tried to show that for any $u \in \Bbb R^2$, $\langle u \mid Y \rangle$ is a gaussian random variable. So I started to consider $\Bbb E \left[e^{i\langle u \mid Y \rangle}\right]$ but I cannot find how to deal with the integrals.

Answer 1

Pdf of $(X,\Theta)$ is

$$f_{X,\Theta}(x,\theta)=\frac1{4\pi}e^{-x/2}1_{x>0,\,0<\theta<2\pi}$$

Changing variables $(x,\theta)\mapsto (u,v)$ where $u=\sqrt x\cos\theta$ and $v=\sqrt x\sin\theta$.

Then $u^2+v^2=x$ and $\tan\theta=\frac{v}{u}$; moreover, $x>0,0<\theta<2\pi\implies (u,v)\in\mathbb R^2$.

Jacobian (determinant) of the inverse map is

$$J\left(\frac{u,v}{x,\theta}\right)=\frac12$$

This gives the pdf of $(U,V)$ as

\begin{align} f_{U,V}(u,v)&=\frac1{4\pi}e^{-(u^2+v^2)/2}\left|J\left(\frac{x,\theta}{u,v}\right)\right| \\&=\frac1{4\pi}e^{-(u^2+v^2)/2}\times 2 \\&=\frac1{2\pi}e^{-(u^2+v^2)/2}\qquad,\,(u,v)\in\mathbb R^2 \end{align}

Clearly $U$ and $V$ are independent standard normal, so $(U,V)$ is trivially bivariate normal.

Answer 2

This follows from the fact that, since $|Y|^2=X\sim\chi_2^2$, $|Y|\sim\chi_2$.