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A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $\frac{r}{3}$, whereas I found $\frac{r}{2}$ as follow

We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$\mathbb E[Y]=\int_0^r (r-x)f_X(x)dx=\frac{1}{r}\int_0^r (r-x)dx=\frac{r}{2}.$$

Maybe there is a subtlety than I don't see ?

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    $\begingroup$ Why do you think $X$ is uniform on $[0,r]$? $\endgroup$ – 5xum Nov 30 '18 at 9:13
  • $\begingroup$ @5xum : I set $Z=(R\cos \Theta, R\sin \Theta)$ with $R$ uniform on $[0,r]$ and $\Theta$ uniform on $[0,2\pi)$. Then $\mathbb P\{X\leq x\}=\mathbb P\{R\leq x, \Theta \in [0,2\pi]\}=\frac{x}{r}.$ It doesn't work ? $\endgroup$ – idm Nov 30 '18 at 9:14
  • $\begingroup$ Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$\dfrac{\mathsf d ~~}{\mathsf d~x}\mathsf P(X\leqslant x)~\propto~\dfrac{1}{x}\mathbf 1_{0< x\leqslant r}$$ $\endgroup$ – Graham Kemp Nov 30 '18 at 9:23
  • $\begingroup$ We just had a question like this half a day ago: Average distance from center of circle $\endgroup$ – Rahul Nov 30 '18 at 9:25
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    $\begingroup$ $\mathbb P\{X\leq x\} = \frac{\pi x^2}{\pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre. $\endgroup$ – Paul Nov 30 '18 at 9:30
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$$F_X(x) = \frac{\pi x^2}{\pi r^2}$$

$$f_X(x)=\frac{2x}{r^2}$$

\begin{align} E[r-X]&=r-E[X] \\ &=r - \frac1{r^2}\int_0^r 2x^2\, dx\\ &= r - \frac1{r^2}\frac{2r^3}3\\ &= \frac{r}{3} \end{align}

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  • $\begingroup$ I don't get why $F_X(x)=\frac{x^2}{r^2}$ $\endgroup$ – idm Nov 30 '18 at 9:29
  • $\begingroup$ That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them. $\endgroup$ – Siong Thye Goh Nov 30 '18 at 9:30
  • $\begingroup$ Ok I see, thank you. But if $Z=(R\cos\Theta,R\sin\Theta)$ with $R$ uniform in $[0,r]$ and $\Theta$ uniform on $[0,2\pi]$, why $$\mathbb P\{|Z|\leq x\}=\mathbb P\{R\leq x,\Theta\in [0,2\pi]\}=\mathbb P\{R\leq x\}=\frac{x}{r}$$ is not true ? I really don't get this point $\endgroup$ – idm Nov 30 '18 at 9:41
  • $\begingroup$ the assumption that $R$ is uniform in $[0,r]$ is not valid. $\endgroup$ – Siong Thye Goh Nov 30 '18 at 9:45
  • $\begingroup$ ok, stange... thank you :) $\endgroup$ – idm Nov 30 '18 at 9:46
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You may first construct the probability density as follows:

  • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of $$\frac{1}{\pi r^2}\cdot 2\pi \cdot x \cdot dx$$

So, you get $$E(Y) = \frac{1}{\pi r^2} \int_0^r (r-x)2\pi \cdot x\; dx = \cdots = \frac{r}{3}$$

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$$\frac{\int_0^r(r-x)x.dx}{\int_0^r x.dx}$$

This gives $r/3$.

The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2\pi x.dx$ ... but then the $2\pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.

Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.

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