# Why the expectation of distance to center of disk is $r/3$ and not $r/2$?

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 ?

• Why do you think $X$ is uniform on $[0,r]$? – 5xum Nov 30 '18 at 9:13
• @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 ? – idm Nov 30 '18 at 9:14
• 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.$$\dfrac{\mathsf d ~~}{\mathsf d~x}\mathsf P(X\leqslant x)~\propto~\dfrac{1}{x}\mathbf 1_{0< x\leqslant r}$$ – Graham Kemp Nov 30 '18 at 9:23
• We just had a question like this half a day ago: Average distance from center of circle – Rahul Nov 30 '18 at 9:25
• $\mathbb P\{X\leq x\} = \frac{\pi x^2}{\pi r^2}$ 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. – Paul Nov 30 '18 at 9:30

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

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

\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}

• I don't get why $F_X(x)=\frac{x^2}{r^2}$ – idm Nov 30 '18 at 9:29
• 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. – Siong Thye Goh Nov 30 '18 at 9:30
• 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 – idm Nov 30 '18 at 9:41
• the assumption that $R$ is uniform in $[0,r]$ is not valid. – Siong Thye Goh Nov 30 '18 at 9:45
• ok, stange... thank you :) – idm Nov 30 '18 at 9:46

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

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

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.