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?

  • $\begingroup$ One way is to use the fact that the only distributions on the circles and the annulus that are invariant under rotations are the uniform distributions. If $T$ is a rotation of $\mathbb R^{2}$ is is obvious that $T(U_1+U_2)=T(U_1)+T(U_2)$ has same distribution as $U_1+U_2$. $\endgroup$ – Kavi Rama Murthy Apr 22 at 23:50

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 $1/4$.

  • $\begingroup$ I have a quick question. Why is $R^2$ uniform and not $R$? $\endgroup$ – Lisa Apr 23 at 13:14
  • $\begingroup$ @Lisa: Because the area of a disc of radius $R$ is proportional to $R^2$. For a uniform distribution, the probability of a given region is proportional to the area of that region. For a uniform distribution on a disc of radius $r$, the probability density is $1/\pi r^2$ since the area is $\pi r^2$, the likelihood that $R<s$ will be $\pi s^2/\pi r^2=s^2/r^2$, which would make $R^2$ uniform on $[0,r^2]$. For an annulus, it's the same, except the interval is $[(r_1-r_2)^2,(r_1+r_2)^2]$. $\endgroup$ – Einar Rødland Apr 23 at 19:14
  • $\begingroup$ Thanks. Do you know how this extends to higher dimensions? Or should I ask a separate question? $\endgroup$ – Lisa Apr 23 at 19:16
  • $\begingroup$ The approach should generalise to higher dimensions, but the angle between the two vectors $U_1$ and $U_2$ will no longer be uniform: in dimension $n$, it will instead be the angle between two random points on the $n-1$-sphere. And $U_1+U_2$ will still not be uniform as far as I can tell: I'm checking $r_1=r_2=r$ and using that there's a 50-50 chance of the angle between them being above or below $\pi/2$ (with angle given in the interval $[0,\pi]$), which would give a 50-50 chance of $R^2$ being above or below $2r^2$. $\endgroup$ – Einar Rødland Apr 23 at 21:15
  • $\begingroup$ Thanks. In case you are intersted, I posted this question here: math.stackexchange.com/questions/3198694/… $\endgroup$ – Lisa Apr 24 at 1:20

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