Average distance between two points in a circular disk

See the answer to this question. The expected distance is $$ d= {128 r\over 45\pi}. $$

Here is another demonstration of this result.

Following Grimmett & Stirzaker 's "One thousand exercises in probability", we can approach this way:

Let $ f(r) := E \rho $, where $\rho $ is the distance between points P1 and P2 which are independently and uniformly distributed on the disk $D$ of a radius $r$.

Let's consider the problem for the disk of a radius $r + h$ for little $h$. If points distributed over a disk $D_h$ of a radius $r + h$: $$ P\{P1 \in D, P2 \in D \} = (\frac{\pi r^2}{\pi (r+h)^2})^2 = 1 - \frac{4 h}{r} + o(h) $$ $$ P\{P1 \in D, P2 \in D_h \backslash D \} = (\frac{\pi r^2}{\pi (r+h)^2})(1 - \frac{\pi r^2}{\pi (r+h)^2}) = \frac{2 h}{r} + o(h) $$ And finally: $$ P\{P1 \in D_h \backslash D, P2 \in D_h \backslash D \} = o(h) $$

Now let's rewrite $ f(r+h) $ using conditional expectation: $$ f(r+h) = E \rho = E(\rho 1_{P1 \in D, P2 \in D}) + 2 E(\rho 1_{P1 \in D, P2 \in D_h \backslash D}) + o(h) $$

It's easy to see, that $$ \frac{E(\rho 1_{P1 \in D, P2 \in D})}{P\{P1 \in D, P2 \in D \}} = f(r) $$ $$ E(\rho 1_{P1 \in D, P2 \in D}) = f(r)(1 - \frac{4 h}{r} + o(h)) $$

And also we can rewrite $$ E(\rho 1_{P1 \in D, P2 \in D_h \backslash D}) = \frac{E(\rho 1_{P1 \in D, P2 \in D_h \backslash D})}{P\{P1 \in D, P2 \in D_h \backslash D \} } (\frac{2 h}{r} + o(h)) $$ In which $\frac{E(\rho 1_{P1 \in D, P2 \in D_h \backslash D})}{P\{P1 \in D, P2 \in D_h \backslash D \} } $ is exact (with ac. $o(h)$) average distance between points on the disk of radius $r$ and on the edge. Which is easy to calculate in polar coordinates with the center on the edge of the disk. $$ \frac{1}{\pi r^2} \int^{\pi/2}_{-\pi/2} \int^{2r cos(\phi)}_{0} s^2 ds d\phi = \frac{32 r}{9 \pi} + o(h) $$

So we have: $$ f(r + h) = f(r)(1 - \frac{4 h}{r} + o(h)) + 2 (\frac{32 r}{9 \pi} + o(h))(\frac{2 h}{r} + o(h)) $$ $$ f(r+h) - f(r) = - \frac{4 h}{r} f(r) + \frac{128 h}{9 \pi} + o(h)*... $$ $$ f'(r) = - \frac{4}{r} f(r) + \frac{128}{9 \pi} + o(1) $$ With $f(0) = 0$ we get: $$ E \rho = f(r) = \frac{128 r}{45 \pi} $$

Some notes about the integral: disk scheme in geogebra

point F -- is the point in disk,

point C -- is the point on the edge,

distance between them is $s$, the angle between Ox and CF is $\phi$.

And since we transformed coordinates to polar, we should multiply by Jacobian matrix determinant, which is $s$.

And finally $ \frac{1}{\pi r^2}$ is density of point in the disk.