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My question is similar to this one but for rectangles instead of lines.

Suppose I have a rectangle with sides of length $L_w$ and $L_h$. What is the average distance between two uniformly-distributed random points inside the rectangle, and why?

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Related: stats.stackexchange.com/q/22488/2970 –  cardinal Oct 7 '12 at 14:32
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2 Answers

up vote 6 down vote accepted

The answer, given here and here, is

$$ \frac1{15} \left( \frac{L_w^3}{L_h^2}+\frac{L_h^3}{L_w^2}+d \left( 3-\frac{L_w^2}{L_h^2}-\frac{L_h^2}{L_w^2} \right) +\frac52 \left( \frac{L_h^2}{L_w}\log\frac{L_w+d}{L_h}+\frac{L_w^2}{L_h}\log\frac{L_h+d}{L_w} \right) \right)\;, $$

where $d=\sqrt{L_w^2+L_h^2}$.

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+1. Good find. $ $ –  Did Oct 7 '12 at 12:02
It seems to be a simple problem. But it's not. Good find. (+1) –  Patrick Li Oct 7 '12 at 14:18
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You can split your two points into coordinates. From the problem you've linked we know that: the expectated distance is $E = \frac{L}{3}$.

Now, assuming that coordinates are independent you can apply the result to the rectangle considering the orthogonal sides as lines, so: $$E_w = \frac{L_w}{3} $$ $$E_h = \frac{L_h}{3} $$ $$d = \sqrt{L_w^2 + L_h^2}$$ $$E_d = \frac{\sqrt{L_w^2 + L_h^2}}{3} $$

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The fact that the coordinates are independent doesn't imply that the square and square root and the expectation commute. –  joriki Oct 7 '12 at 11:41
Sorry, the answer seemed to be so obvious... but it wasn't. –  Blex Oct 7 '12 at 12:25
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