How to compute the probability distributions and the joint probability distribution of continuos variables when you just know the form of funtion?

$$\phi$$ 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

• Answer for covariance is below. Joint distribution is harder to get Oct 2 '21 at 4:06
• The denominator for $f_Y(y)$ seems to have an extra $a$ since it is$\pi a^2\sqrt{1-(\frac{y}{a})^2}$, which integrates to $\frac{arcsin(\frac{y}{a})}{\pi a}$. Set limits and the total integral $=\frac{1}{a}$. Oct 2 '21 at 18:03

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 $$cov(xy)=0$$.
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 $$f_y(y)=\frac{1}{\pi a} \int\limits_0^a\frac{s\delta(s-a)}{\sqrt{s^2-y^2}}ds=\frac{1}{\pi\sqrt{a^2-y^2}}$$.