Problem source

The continuous random variable $X$ is uniformly distributed over the interval $[-c,c].$ Write down expressions for the probabilities that: 
\begin{questionparts}
\item $n$ independently selected values of $X$ are all greater than $k$, 
\item $n$ independently selected values of $X$ are all less than $k$,
\end{questionparts}
where $k$ lies in $[-c,c]$. 
A sample of $2n+1$ values of $X$ is selected at random and $Z$ is the median of the sample. Show that $Z$ is distributed over $[-c,c]$ with probability density function 
\[
\frac{(2n+1)!}{(n!)^{2}(2c)^{2n+1}}(c^{2}-z^{2})^{n}.
\]
Deduce the value of ${\displaystyle \int_{-c}^{c}(c^{2}-z^{2})^{n}\,\mathrm{d}z.}$
Evaluate $\mathrm{E}(Z)$ and $\mathrm{var}(Z).$

Solution source

\begin{questionparts}
\item \begin{align*}
\mathbb{P}(n\text{ independent values of }X > k) &= \prod_{i=1}^n \mathbb{P}(X > k) \\
&= \left ( \frac{c-k}{2c}\right)^n
\end{align*}

\item \begin{align*}
\mathbb{P}(n\text{ independent values of }X < k) &= \prod_{i=1}^n \mathbb{P}(X < k) \\
&= \left ( \frac{k+c}{2c}\right)^n
\end{align*}
\end{questionparts}

\begin{align*}
&&\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta) &= \mathbb{P}(n\text{ values } < z - \delta \text{ and } n \text{ values} > z + \delta) \\
&&&= \binom{2n+1}{n,n,1} \left ( \frac{c-(z+\delta)}{2c}\right)^n\left ( \frac{(z-\delta)+c}{2c}\right)^n \frac{2 \delta}{2 c} \\
&&&= \frac{(2n+1)!}{n! n!} \frac{((c-(z+\delta))(c+(z-\delta)))^n 2\delta}{2^n c^n}  \\
&&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-(z+\delta))(c+(z-\delta)))^n 2\delta \\
\Rightarrow && \lim_{\delta \to 0} \frac{\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta)}{2 \delta} &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-z)(c+z))^n \\
&&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2) \\
\end{align*}

\begin{align*}
&& 1 &= \int_{-c}^c  \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\
\Rightarrow && \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!} &= \int_{-c}^c (c^2-z^2)^n \d z
\end{align*}

\begin{align*}
\mathbb{E}(Z) &= \int_{-c}^c z \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\
&=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z (c^2-z^2)^n \d z \\
&= 0
\end{align*}

\begin{align*}
\mathrm{Var}(Z) &= \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 \\
&=   \mathbb{E}(Z^2) \\
&= \int_{-c}^c z^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\
&=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z^2 (c^2-z^2)^n \d z \\
&=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \left ( \left [ -\frac{1}{2(n+1)}z(c^2-z^2)^{n+1} \right]_{-c}^c + \frac{1}{2(n+1)}\int_{-c}^c (c^2-z^2)^{n+1} \d z \right) \\
&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \frac{1}{2(n+1)} \frac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} \\
&= \frac{(n+1)^2(2c)^2}{(n+1)(2n+2)(2n+3)} \\
&= \frac{2c^2}{2n+3}
\end{align*}