$\phi$ is a random phase angle that is distributed uniformly over the range 0 to $2\pi$, $a$ is a constant, and
$$x = a \cos \phi \quad y = a \sin\phi $$.
Calculate (a) the probability distributions of x and y; (b) the joint probability distribution of x and y; (c) the covariance of x and y. Are the variates x and y statistically independent?
My guess:
I found that the PDF gives the probability distribution, and there are many techniques to compute it. I follow this site to find the PDF's http://ece-research.unm.edu/bsanthan/ece340/transrv.pdf.
for x, I got the PDF:
$$f_X(x)= \frac{1}{\pi a} \frac{1}{\sqrt{a^2-x^2}} \quad \quad \quad -a < x < a$$
and for y:
$$f_Y(y)= \frac{1}{\pi a} \frac{1}{\sqrt{a^2-y^2}} \quad \quad \quad -a < y < a$$
For the other questions, I don't know how to proceed
Covariance is straightforward. $Cov(x,y)=E(xy)-E(x)E(y)$. $E(x)=\frac{a}{2\pi}\int\limits_0^{2\pi}cos(\phi)d\phi=0$, $E(y)=\frac{a}{2\pi}\int\limits_0^{2\pi}sin(\phi)d\phi=0$, and $E(xy)=\frac{a^2}{2\pi}\int\limits_0^{2\pi}cos(\phi)sin(\phi)d\phi=0$. Therefore $cov(xy)=0$.
The important point is that a zero covariance does NOT imply independence.
Derivation of joint density - uses Dirac $\delta$.
The random variable is one dimensional, uniform on a circle of radius $a$. To get the density as a function of $x$ and $y$, first get the density in polar coordinates $dg(r,\phi)=\frac{1}{2\pi a}\delta(r-a)rdrd\phi$. Convert to rectangular coordinates and get $df(x,y)=\frac{1}{2\pi a}\delta(\sqrt{x^2+y^2}-a)dxdy$.
To check $\frac{1}{2\pi a}\int\limits_{-a}^a\delta(\sqrt{x^2+y^2}-a)dx=\frac{1}{\pi a}\int\limits_0^a \delta(\sqrt{x^2+y^2}-a)dx$. This last integal can be evaluated with $s=\sqrt{x^2+y^2}$ or $x=\sqrt{s^2-y^2}$ and the integral becomes $f_y(y)=\frac{1}{\pi a} \int\limits_0^a\frac{s\delta(s-a)}{\sqrt{s^2-y^2}}ds=\frac{1}{\pi\sqrt{a^2-y^2}}$.