A large target consists of a disc of unit radius and centre O. A certain maksman never misses the target and the probability of any given shot hitting the target within a distance $t$ from $O$ is $t^2$, where $ 0\leq t \leq 1$. The marksman fires n shots independently. The random variable $Y$ is the radius of the smallest circle, with centre $O$, which encloses all the shots.
(1) Show that the probability density function of Y is $2ny^{2n-1}$
(2) The shot which is furthest from $O$ is rejected. Show that the expected area of the smallest circle, with centre $O$, which encloses the remaining $(n-1)$ shots is $(\dfrac {n-1}{n+1})\pi$.

(1) Probability density function - probability that (n-1) of the shots are within a distance $y$ of the centre, and the $n$th is between $y$ and $y+dy$ of the centre. $n \times y^{2n-2}\times ((y+dy)^2 - y^2)= 2ny^{2n-1}dy$
Since $dy^2 = 0$.
What can I do with that $dy$ at the end?

(2) I have problem modelling this part. Naturally I assumed it to be equivalent method to the first part, so this time random variable of say $Z$ and the probability of (n-2) shots within a distance $z$ and (n-1)th between $z$ and $z+dz$ of the centre:
$(n-1) \times z^{2(n-2)} \times ((z+dy)^2-z^2)$

But looking at the solution this modelling is wrong. It says:

If we recject the outermost shot, the porbability that radius of the circle is between $z$ and $z+dz$ is:
$n(n-1) \times z^{2(n-2)} \times (1-z^2) \times 2z dz$

Why is it different? More specifically why do we need $1-z^2$, the probability a shot is further than z from the centre?

Many thanks in advance


(1) The easiest approach is to determine the CDF first, and then derive the PDF by differentiation. In order for $Y < y$, all of the shots must be within $y$, each of which happens with probability $y^2$, so the probability of all doing so is

$$ F_Y(y) = \left(y^2\right)^n = y^{2n} $$

Then the PDF of this is

$$ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}y^{2n} = 2ny^{2n-1} $$

There's nothing wrong with your approach; you simply ended up with

$$ dF = 2ny^{2n-1} \, dy $$

and so you can simply "divide by $dy$" (to articulate an abuse of notation) to derive the PDF.

(2) The distribution of shot locations indicates that the marksman shoots uniformly over the unit disc. (A bit odd, given that the marksman also never misses the target, but those are the givens.) So let us first determine the average radius of the minimum circle containing all $n$ shots:

$$ A_n = \int_{y=0}^1 2\pi y [1-F(y)] \, dy = \frac{n\pi}{n+1} $$

In other words, given an overall area of $\pi$, the minimum circle had an average of $n/(n+1)$ of the area of the maximum circle. Because the shots are uniformly distributed over the target, we can apply the same logic to the innermost $n-1$ shots: The $(n-1)$-circle has an area $(n-1)/n$ of the $n$-circle, no matter how large the $n$-circle is. Thus, the area of the $(n-1)$-circle is

$$ A_{n-1} = \int_{y=0}^1 \frac{2\pi y(n-1)}{n} [1-F_Y(y)] \, dy = \frac{(n-1)\pi}{n+1} $$

  • $\begingroup$ Ok, I seem to lack basic principles here, can you help please? for (1) why is it $dF$ instead of $F$, is it because I have $dy$ that automatically makes $dF$ too? also (2) How did you derive that $A_n$ equation? When I solve it for $A_n$ I used $E(\pi y^2) = \pi \int_0^1 y^2 2n y^{2n-1}dy$ This seems to be vastly different from your one. Thank you $\endgroup$ – zcahfg2 May 6 '15 at 10:03
  • $\begingroup$ (1) By measuring the area in a ring of width $dy$, you are in effect giving the differential area of the circle. (2) $2 \pi y dy$ is the area of the ring again. Integrating with $1-F(y)$ is an old trick for averaging functions by their differentials. One very common application is to integrate just $\int [1-F_Y(y)] \, dy$ to get the average of $Y$. $\endgroup$ – Brian Tung May 6 '15 at 23:13

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