I have this form $x=r \cos (\phi)$, where $r$ is Rayleigh distributed RV, $r$ ~ $Ray (\sigma_r)$ and $\phi$ is uniform in $[0,2\pi]$.
Does $x$ follow a Gaussian distribution? Can you please provide the derivation for it? and what is the mean and variance in this case?
Thanks
We have $x=r\cos\phi$, where $r$ is a random variable with support $(0,\infty)$ whose pdf is $p_r(r)=\frac{1}{\sigma_r^2}r \exp(-r^2/(2\sigma_r^2))$ and $\phi$ is uniform between $0$ and $2\pi$. So, the pdf of $x$ is given by $$ f_X(x)=\frac{1}{2\pi}\int_0^\infty dr \frac{r}{\sigma_r^2}\exp(-r^2/(2\sigma_r^2))\int_0^{2\pi}d\phi \delta(x-r\cos\phi)\ .$$
Introducing the integral representation of the delta function $\delta(y)=\int_{-\infty}^\infty \frac{dk}{2\pi}\ e^{\mathrm{i}ky}$, we obtain $$ f_X(x)=\frac{1}{2\pi}\int_0^\infty dr \frac{r}{\sigma_r^2}\exp(-r^2/(2\sigma_r^2))\int_0^{2\pi}d\phi\int_{-\infty}^\infty \frac{dk}{2\pi}\ e^{\mathrm{i}kx}e^{-\mathrm{i}kr\cos\phi}\ . $$ Performing the $\phi$ integral yields $2\pi$ times a Bessel function $$ f_X(x)=\int_0^\infty \frac{dk}{2\pi\sigma_r^2}e^{\mathrm{i}kx}\underbrace{\int_0^\infty dr\ r \exp(-r^2/(2\sigma_r^2)) J_0(kr)}_{\sigma_r ^2 e^{-\frac{1}{2} k^2 \sigma_r ^2}} $$ and the final Gaussian result is immediate (the integral above is just the Fourier transform of a Gaussian). Hence $$ f_X(x)=\frac{1}{\sqrt{2\pi\sigma_r^2}}e^{-x^2/(2\sigma_r^2)}\ . $$