Let $X_1, \ldots ,X_n$ be an independent and identically distributed sequence of Exponential(1) random variables, where $n \geq 3$. Find the conditional probability density function for the maximum $M = X_n$ given that the second smallest $X_2 = 1$. Use this to verify that, in the case that $n = 4$,
$$E(M\mid X_2 = 1) = \frac{5}{2}$$
Hint: $\int_{a}^{\infty} ue^{-u}du = (1+a)e^{-a}$.
I started by using the “PDF method” to try and find the density for $X_2$ and the joint density for $(M, X_2)$, but I'm not sure what step to take next.
Let $f$ and $F$ be the probability density and cumulative distribution functions for the iid random variables. Let $f_{(k)}$ and $F_{(k)}$ be those of the $k$ eth order statistic $X_{(k)}$, and similarly for joint functions.
When the second smallest order statistic is $s$, then one variable will be at most that, one exactly that, and the remaining $n-2$ variables will at least that. So the pdf for the second (smallest) order statistic is:
$$\begin{align}f_{(2)}(s)&= \dfrac{n!}{1!1!(n-2)!} F(s)f(s)(1-F(s))^{n-2}\\[1ex]&= n(n-1)(1-\mathrm e^{-s})\mathrm e^{s(1-n)}~\mathrm 1_{0\leqslant s} \end{align}$$
Similarly when the the second smallest statistic is $s$ and the greatest statistic is $m$, then one variable is less than $s$, one exactly $s$, one is exactly $m$, and the remaining $n-3$ are betwixt $s$ and $m$. Thus:
$$\begin{align}f_{(2),(n)}(s,m)&= \dfrac{n!}{1!1!1!(n-3)!} F(s)f(s)(F(m)-F(s))^{n-3}f(m) \end{align}$$
etc.
Divide to find the conditional pdf $f_{(n)\mid (2)}(m\mid s)$, solve at $s=1$ when $n=4$, then integrate as required to find the expected value.
$$\mathsf E(M\mid X_{(2)}{=}1)\big\rvert_{n=4}=\int_1^\infty m~f_{(n)\mid(2)}(m\mid 1)\big\rvert_{n{=}4}~\mathrm d m$$