Problem source

Each of my $n$ students has to hand in an essay to me. Let $T_{i}$
be the time at which the $i$th essay is handed in and suppose that
$T_{1},T_{2},\ldots,T_{n}$ are independent, each with probability
density function $\lambda\mathrm{e}^{-\lambda t}$ ($t\geqslant0$).
Let $T$ be the time I receive the first essay to be handed in and
let $U$ be the time I receive the last one. 
\begin{questionparts}
\item Find the mean and variance of $T_{i}.$
\item Show that $\mathrm{P}(U\leqslant u)=(1-\mathrm{e}^{-\lambda u})^{n}$
for $u\geqslant0,$ and hence find the probability density function
of $U$. 
\item Obtain $\mathrm{P}(T>t),$ and hence find the probability density
function of $T$. 
\item Write down the mean and variance of $T$. 
\end{questionparts}

Solution source

\begin{questionparts}
\item $T_i \sim \textrm{Exp}(\lambda)$ so $\E[T_i] = \lambda^{-1}, \var[T_i] = \lambda^{-2}$ 
\item $\,$ \begin{align*}
&& \mathbb{P}(U \leq u) &= \mathbb{P}(T_i \leq u\quad \forall i) \\
&&&= \prod \mathbb{P}(T_i \leq u) \\
&&&= \prod \int_0^u \lambda e^{-\lambda t} \d t \\
&&&= (1-e^{-\lambda u})^n \\
\\
\Rightarrow && f_U(u) &= n\lambda e^{-\lambda u}(1-e^{-\lambda u})^{n-1}
\end{align*}
\item $\,$ \begin{align*}
&& \mathbb{P}(T > t) &= \mathbb{P}(T_i > t \quad \forall i) \\
&&&= \prod \mathbb{P}(T_i > t) \\
&&&= e^{-n\lambda t} \\
\Rightarrow && f_T(t) &= n\lambda e^{-n\lambda t} 
\end{align*}

\item Therefore $\E[T] = \frac{1}{n\lambda}, \var[T] = \frac{1}{(n\lambda)^2}$

\end{questionparts}