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A stone is thrown into a circular pond of radius 1 meter. Suppose the stone falls uniformly at random on the area of the pond. The expected distance of the stone from the center of the pond is ?. Can anyone help me out with this one, can't seem to reach a proper answer without having $\pi$ in it.

Meera Unni
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  • It would be easier to answer which is the expected square distance $r^2 = x^2 + y^2$. – Andreas May 17 '18 at 06:53
  • If you have a proper answer, why do you doubt it? This is about a circle, so having $\pi$ in it should not be a surprise. – Ingix May 17 '18 at 07:14

1 Answers1

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One could work here with the PDF $$f_{\bf Z}(x,y)={1\over\pi}\qquad\bigl(0\leq\sqrt{x^2+y^2}\leq a^2\bigr)\ ,$$ whereby the random variable ${\bf Z}=(X,Y)$ denotes the exact spot where the stone falls. This then would lead to the expected value $$E(R)={1\over\pi}\int_{B(a,{\bf 0})}\sqrt{x^2+y^2}\>{\rm d}(x,y)$$ of the distance $R:=|{\bf Z}|$. The integral $(1)$ has to be computed via a transformation to polar coordinates. The following setup circumvents this and brings only the variable $R$ into the game.

The distance $R$ of the stone from the center is a random variable supported on the interval $[0,a]$, where $a$ denotes the radius of the pond. For a given $r\in[0,a]$ the cumulated probability that $R\leq r$ amounts to $$F(r)={\pi r^2\over \pi a^2}\ .$$ Therefore the probability density of the random variable $R$ is given by $$f_R(r)=F'(r)={2r\over a^2}\ ,$$ and the expected distance computes to $$E(R)=\int_0^a r\>f_R(r)\>dr={1\over a^2}\int_0^a 2r^2\>dr={2\over3}a\ .\tag{1}$$

Christian Blatter
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