Suppose we have two independent random variables $U_1$ and $U_2$ unfiorm on
\begin{align}
S_i = \left\{ (s_1,s_2) \in \mathbb{R}: \sqrt{s_1^2+s_2^2} =r_i \right\}
\end{align}
respectily. Assume $r_1 \ge r_2$.
Question: How to find the pdf of $U_1+U_2$?
We know that it would be distributed on
\begin{align}
S_3 = \left\{ (s_1,s_2) \in \mathbb{R}: r_1-r_2 \le \sqrt{s_1^2+s_2^2} \le r_1+r_2 \right\}
\end{align}
In other words, show that the sum of two random variables on the circles results in a random variable distributed on an annulus.
The question now is how to find the pdf of $U_1+U_2$?
Can this, for example, be done by using characteristic functions?
Best Answer
It is not the case.
One way to go about deriving the distribution of the sum is to note that the points may be written $U_i=(r_i\sin\theta_i,r_i\cos\theta_i)$ where $r_i$ are fixed and $\theta_i$ are uniform on $(0,2\pi)$.
Now, write $U=U_1+U_2=(R\sin\theta,R\cos\theta)$. As Kavi has already pointed out, the angle $\theta$ will be uniformly distributed due to rotational symmetry, so we can ignore it and focus on the distribution of $R$.
For $U$ to be uniform on the annulus, $R^2$ would have had to be uniformly distributed between $(r_1-r_2)^2$ and $(r_1+r_2)^2$: just write down the area in the annulus with $R<r$ and divide by the total area expressed in terms of $r^2$.
We now derive the distribution of $R^2$ by use of the relation $R^2=r_1^2+r_2^2+2r_1r_2\cos\phi$ where $\phi=\theta_1-\theta_2$. Note that $\phi$ will also be uniform on $(0,2\pi)$ provided we consider angles modulo $2\pi$.
When $\phi$ is uniform on $(0,2\pi)$, the random variable $X=\cos\phi$ will have probability density $\frac{dx}{\pi\sqrt{1-x^2}}$. So, as you can see, the density is higher around the outer and inner edges of the annulus.
The simplest counter-example comes when $r_1=r_2=r$ and you compute $\Pr[R<r]$. This is the same as the probability that $\cos\phi<-1/2$, which is $1/3$. If $U$ had been uniform on the disc, it should have been $1/4$.