# Expressing distribution of sum of exponential random variables as distribution of order statistic

Let $(U_{mn})_{1\leqslant m,n\leqslant \infty}$ be a family of i.i.d. random variables, each distributed as $\text{Exp}(1/2)$. Define $X_j=\min_{1\leqslant i\leqslant j} U_{ji}$, so that $X_j\sim \text{Exp}(j/2)$.

Tavaré claims (see bottom of pg. 22 here) that the sum \begin{equation*} L_n = \sum_{j=1}^{n-1} X_j \overset{d}{=} \max_{1\leqslant k\leqslant n-1} W_k, \end{equation*}
where $(W_k)$ is another family of i.i.d. r.v.s each distributed as $\text{Exp}(1/2)$, and "$\overset{d}{=}$" denotes equality in distribution. The argument supposedly uses a coupling trick, but I have not been successful at thinking of one. Any hints or solutions are much appreciated!

• I have also tried moment generating functions and induction on the (known) density functions for both random variables $L_n$ and $\max W_k$, but they both look like combinatorial nightmares, so I'm really hoping to find an elegant probabilistic argument! – youngtableaux Jan 24 '17 at 0:24
The characteristic function of an exponential distribution with rate $j/2$ is $\phi_{X_j}(t) = (1 - 2it/j)^{-1}$. The characteristic function of $L_n$ is thus $$\phi_{L_n}(t) = \prod_{j=1}^{n-1} \dfrac{1}{1 - 2 i t/j} = \frac{\Gamma(n) \Gamma(1-2it)}{\Gamma(n-2it)}$$
The maximum of $n-1$ independent exponentials with rate $1/2$ has CDF $(1-\exp(-x/2))^{n-1}$ for $x > 0$, and characteristic function $$\frac{n-1}{2} \int_0^\infty \left(1 - e^{-x/2}\right)^{n-2} e^{itx-x/2}\; dx$$ The change of variables $s = 1 - e^{-x/2}$ makes this into $$(n-1) \int_0^1 s^{n-2} (1-s)^{-2it}\; ds$$ where we can recognize a Beta function $$(n-1) B(n-1,1-2it) = (n-1) \frac{\Gamma(n-1)\Gamma(1-2it)}{\Gamma(n-2it)} = \frac{\Gamma(n) \Gamma(1-2it)}{\Gamma(n-2it)}$$