So I am trying to do this past term exam from my institution:
We pick a random point $M$ in the unit circle. Picking any point has the same probability. We define $X$ as the length from the center to $M$, and $Y$ the angle. What is the cumulative distribution fonction of $X$? Show that $X$ and $Y$ are independent. And calcule their probability density function.
My attempt:
So if $X \leq t$ for a $t \in [0,1]$ then that means that all the points $(x,y)$ in the unit circle such that their length from the center is smaller than $t$, which can expressed as $x^2 + y^2 \leq t$. I first need to calculate the volume of the unit circle, I get: $$\int _{0}^{2\pi}\int_{0}^{1}rdrd\theta = \pi $$
Now then $$F_X(t) = P(X \leq t) =\frac{1}{\pi} \int_{0}^{2\pi}\int_{0}^{t}rdrd\theta$$ $$= \frac{1}{\pi} \int_{0}^{2\pi} (\frac{t^2}{2})d\theta = \frac{t^2}{2\pi} \int_{0}^{2\pi}d\theta = t^2$$
Now to show that $X$ and $Y$ are independent, I can show that $F_{X,Y}(t, \psi) = F_X(t)F_Y(\psi)$ .
So I have $$(1) \text{ }F_{X,Y}(t, \psi) = \frac{1}{\pi } \int_{0}^{\psi}\int_{0}^{t}rdrd\theta = \frac{1}{\pi}\int_{0}^{\psi}d\theta \int_{0}^{t}rdr = \frac{\psi t^2}{2\pi} $$ By Fubini-Tonelli. As $$F_Y(\psi) = \frac{1}{2\pi} \psi $$ We have: $$F_{X,Y}(t, \psi) =F_X(t)F_Y(\psi) $$ Thus the two variables are independent.
Now, their density function are $$\frac{dF_x(t)}{dt}= 2t $$ and $$\frac{dF_Y(\psi)}{dt} = \frac{1}{2\pi} \mathbb{1}_{[0,2\pi]}(\psi)$$
Is my solution correct?