Problem source

A target consists of a disc of unit radius and centre $O$. A certain marksman never misses the target, and the probability of any given shot hitting the target within a distance $t$ from $O$ it $t^{2}$, where $0\leqslant t\leqslant1$. 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. Show that the probability density function of $Y$ is $2ny^{2n-1}$ and find the expected area of the circle. 
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 
\[
\left(\frac{n-1}{n+1}\right)\pi.
\]

Solution source

Another way to describe $Y$ is the maximum distance of any shot from $O$. Let $X_i$, $1 \leq i \leq n$ be the $n$ shots then,

\begin{align*}
F_Y(y) &= \mathbb{P}(Y \leq y) \\
&=  \mathbb{P}(X_i \leq y \text{ for all } i) \\
&=  \prod_{i=1}^n \mathbb{P}(X_i \leq y) \tag{each shot independent}\\
&=  \prod_{i=1}^n y^2\\
&= y^{2n} 
\end{align*}

Therefore $f_Y(y) = \frac{\d}{\d y} (y^{2n}) = 2n y^{2n-1}$.

\begin{align*}
\mathbb{E}(\pi Y^2) &= \int_0^1\pi y^2 \f_Y(y) \d y \\
&=\pi \int_0^1 2n y^{2n+1} \d y \\
&=\left ( \frac{n}{n+1} \right )\pi
\end{align*}.

Let $Z$ be the distance of the second furthest shot, then:

\begin{align*}
&& F_Z(z) &= \mathbb{P}(Z \leq z) \\
&&&= \mathbb{P}(X_i \leq z \text{ for at least } n - 1\text{ different } i) \\
&&&= n\mathbb{P}(X_i \leq z \text{ for all but 1}) + \mathbb{P}(X_i \leq z \text{ for all } i) \\
&&&= n \left (  \prod_{i=1}^{n-1} \mathbb{P}(X_i \leq z) \right) \mathbb{P}(X_n > z) + z^{2n} \\
&&&= nz^{2n-2}(1-z^2) + z^{2n} \\
&&&= nz^{2n-2} -(n-1)z^{2n} \\
\Rightarrow && f_Z(z) &= n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \\
\Rightarrow && \mathbb{E}(\pi Z^2) &= \int_0^1 \pi z^2 \left (n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \right) \d z \\
&&&= \pi \left ( \frac{n(2n-2)}{2n} - \frac{2n(n-1)}{2n+2}\right) \\
&&&= \left ( \frac{n-1}{n+1} \right) \pi
\end{align*}