Sum of two random variables uniformly distributed on circles

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?

• 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 $U_1+U_2$. – 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 $$R and divide by the total area expressed in terms of $$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 $$\Pr[R. This is the same as the probability that $$\cos\phi<-1/2$$, which is $$1/3$$. If $$U$$ had been uniform on the disc, it should have been $$1/4$$.

• I have a quick question. Why is $R^2$ uniform and not $R$? – Lisa Apr 23 at 13:14
• @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 $[(r_1-r_2)^2,(r_1+r_2)^2]$. – Einar Rødland Apr 23 at 19:14
• Thanks. Do you know how this extends to higher dimensions? Or should I ask a separate question? – Lisa Apr 23 at 19:16
• 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 $2r^2$. – Einar Rødland Apr 23 at 21:15
• Thanks. In case you are intersted, I posted this question here: math.stackexchange.com/questions/3198694/… – Lisa Apr 24 at 1:20