1

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 ?

David G. Stork
  • 29,774
idm
  • 11,824
  • 2
    Why do you think $X$ is uniform on $[0,r]$? – 5xum Nov 30 '18 at 09:13
  • @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 09:14
  • 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 09:23
  • We just had a question like this half a day ago: Average distance from center of circle –  Nov 30 '18 at 09:25
  • 1
    $\mathbb P{X\leq x} = \frac{\pi x^2}{\pi r^2}$ 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 09:30
  • @idm To visualise this, try a simpler version. Suppose score is $1$ if arrow lands within a distance $\frac{r}{2}$ of the centre and $0$ otherwise. Only one quarter of the area of the target is within a distance $\frac{r}{2}$ of the centre, so expected score of an arrow that lands at a random point in the target is $\frac{1}{4}$, not $\frac{1}{2}$. – gandalf61 Nov 30 '18 at 09:55

3 Answers3

3

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

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

\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}

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
  • I don't get why $F_X(x)=\frac{x^2}{r^2}$ – idm Nov 30 '18 at 09:29
  • 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 09:30
  • 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 09:41
  • the assumption that $R$ is uniform in $[0,r]$ is not valid. – Siong Thye Goh Nov 30 '18 at 09:45
  • ok, stange... thank you :) – idm Nov 30 '18 at 09:46
  • @idm say there are 2 points 'a' and 'b' where a is at a distance x1 from center of the circle and b is at a distance of x2 where x1 > x2. Now, the circumference of the circle with radius x1 > than the one with radius x2. Therefore, there are a greater number of points with a distance x1 from the center of the circle than with a distance of x2. So, R is not uniform in [0,r], rather it is more of a probability distribution which increases as we move away from the center of the circle (say f(x) = m*x + c where m>0 and f(x) is the pdf of a point 'x' distance from center of the circle.) – Faraz Gerrard Jamal Apr 30 '23 at 16:06
1

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

trancelocation
  • 32,243
-1

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

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.

AmbretteOrrisey
  • 981