# Uniform distribution over the unit disk

Suppose that $U_1$ and $U_2$ are independent, and identically and uniformly distributed over the unit disk, i.e., for $i = 1,2$, $U_i = (X_i, Y_i)$ and the joint density is $$f_{(X_i,Y_i)}(x,y) = \frac{1}{\pi} \mathbb{1}_{\{(x,y):x^2+y^2 \leq 1\}}(x,y).$$ I want to know how to compute the probability that the Euclidean distance between $U_1$ and $U_2$ is smaller than, say, $\frac{1}{2}$, that is, $\mathbb{P}(\sqrt{(X_1 - X_2)^2 + (Y_1 - Y_2)^2} < \frac{1}{2})$? I am trying to see if the polar coordinates would help. E.g., let $U_i = (R_i \cos \Theta_i, R_i\sin \Theta_i)$ and then relate the distance to the $R_i$. But would $R_i$ be uniform on $[0,1]$? Anyone has an idea? Thanks very much.

• Yes, you must suppose $U_i \sim U[0,1]$ and $\Theta_i\sim U[0,2\pi)$ – sinbadh Jan 4 '16 at 17:57
• You may find this recent, related question interesting about a similar problem in 3-dimensions. math.stackexchange.com/q/1599950/74357 – CommonerG Jan 4 '16 at 19:32

We are computing the probability that two random points in the unit disk (here random means with respect to a uniform probability distribution) lie within a distance $\leq\frac{1}{2}$. We may assume without loss of generality that the first point lies on the segment joining the origin with the point $(1,0)$, with a probability density function supported on $[0,1]$ and given by $f(z)=2z$. If we assume to know the position of the first point, the second point lies within a distance $\leq\frac{1}{2}$ iff it belongs to the original disk and to a smaller disk. If $z\in[0,1]$ and $g(z)$ is the Lebesgue measure of the set: $$E(z) = \left\{(x,y): x^2+y^2\leq 1, (x-z)^2+y^2\leq\frac{1}{4}\right\},$$ the wanted probability is just: $$P=\frac{1}{\pi}\int_{0}^{1} f(z)g(z)\,dz = \frac{2}{\pi}\int_{0}^{1} z\cdot g(z)\,dz.$$ So the problem boils down to computing $g(z)$, that has a rather ugly closed form in terms of the $\arcsin$ function. Numerically, $$P\approx 19.728\%.$$