Expected Distance of Two Points on a Circle

Assume that the unit circle is centered at the origin. The coordinates of the first point do not matter. Then, suppose that the first point is at $(0,1)$ and the second makes an angle $x$ with the x-axis, where $0≤x<2\pi$ . Then, the coordinates of the second point are $(\cos{x},\sin{x})$ , and the distance between them can be calculated as $\sqrt{{(\cos{x}-1)}^2+\sin^2{x}}$ .

Using trig identities, simplify it to $2\left|\sin{\frac{x}{2}}\right|$ .

Then, the answer is $\displaystyle\frac{1}{2\pi}\int_0^{2\pi}2\left|\sin{\frac{x}{2}}\right|\ dx$ .

Since, in the bounds $0≤x<2\pi$ , $\sin{\frac{x}{2}}$ is greater than or equal to 0, the above is essentially

$=\displaystyle\frac{1}{2\pi}\int_0^{2\pi}2\sin{\frac{x}{2}}\ dx$

$=\displaystyle\frac{1}{2\pi}\left[-4\cos{\frac{x}{2}}\right]_0^{2\pi}$

$=\displaystyle\frac{4}{\pi}$

Because of symmetry, we only need to consider distances formed by points placed on half of the circle, i.e on any arbitrary semicircle, to obtain every possible distance that can be formed by two arbitrary points placed anywhere on the circle. The distance between points placed on a semicircle can be calculated by the law of cosines: $\displaystyle d=\sqrt{2-2\cos \theta}$ .

Then the expected distance is a Riemann sum $\displaystyle \lim_{n\to\infty} \left(\frac{\sqrt{2-2\cos 0}}{n} + \frac{\sqrt{2-2\cos 0.001}}{n} + \ldots + \frac{\sqrt{2-2\cos \pi}}{n}\right) = \lim_{n\to\infty} \sum_{\theta = 0}^n\frac{\sqrt{2-2\cos \theta}}{n}$ .

This is equivalent to $\displaystyle \frac{1}{\pi}\int_0^{\pi}\sqrt{2-2\cos \theta}\ d\theta$

$=\displaystyle\frac{2}{\pi}\int_0^{\pi}\sin\frac{\theta}{2}\ d\theta$

$=\displaystyle\frac{2}{\pi}\left[-2\cos\frac{\theta}{2}\right]_0^{\pi}$

$=\boxed{\displaystyle \frac{4}{\pi}}$