From An Intermediate Course in Probability by Allan Gut
Let $X_{1},X_{2},\ldots,X_{n}$ be independent, $\text{Exp}(a)$-distributed random variables. Determine the distribution of $\sum_{k=1}^{n}X_{(k)}$.
I tried this ():
$L_{Y}(s)=\int_{-\infty}^{\infty}e^{-sy}f_{Y}(y)\, dy$ i.e. the Laplace transform
$X_{(k)}$ is the k:th ordered statistic of $X$. \begin{equation*} \begin{split} f_{X_{(k)}}x)=\frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}(F(x))^{k-1}(1-F(x))^{n-k}f(x)\\ =\frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}(1-e^{-\frac{x}{a}})^{k-1}(e^{-\frac{x}{a}})^{n-k}e^{-\frac{x}{a}}\\ =\frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}(1-e^{-\frac{x}{a}})^{k-1}(e^{-\frac{x}{a}})^{n-k+1}\\ L_{X_{k}}(t)=\frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}\int_{0}^{\infty}e^{-xt}(1-e^{-\frac{x}{a}})^{k-1}(e^{-\frac{x}{a}})^{n-k+1}\, dx\\ u=e^{-\frac{x}{a}}\qquad du=-ae^{-\frac{x}{a}}\\ \frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}\int_{1}^{0}-u^{at}(1-u)^{k-1}u^{n-k}\frac{1}{a}\, du\\ =\frac{\Gamma(n+1)}{\Gamma(k)\Gamma(n+1-k)}\int_{1}^{0}-(1-u)^{k-1}u^{n-k+at}\frac{1}{a}\, du\\ =\frac{\Gamma(n+1)}{a\Gamma(k)\Gamma(n+1-k)}\frac{(k-1)!(n-k+at)!}{(k+n-k+at)!}\\ =\frac{\Gamma(n+1)}{a\Gamma(k)\Gamma(n+1-k)}\frac{(k-1)!(n-k+at)!}{(n+at)!}\\ =\frac{\Gamma(n+1)}{a\Gamma(n+1-k)}\frac{(n-k+at)!}{(n+at)!}\\ =\frac{\Gamma(n+1)}{a(n+at)!}\cdot \frac{(n-k+at)!}{(n-k)!}\\ =\frac{(n-k+1)(n-k+2)\cdots n}{(n-k+1+at)(n-k+2+at)\cdots (n+at)} \end{split} \end{equation*} \begin{equation*} \begin{split} \sum_{k=1}^{n}X_{(k)}=\Pi_{k=1}^{n}\frac{(n-k+1)(n-k+2)\cdots n}{(n-k+1+at)(n-k+2+at)\cdots (n+at)}\\ =\Big((\frac{n}{n+at})\Big)\Big((\frac{n-1}{n-1+at})(\frac{n}{n+at})\Big)\cdots \Big((\frac{1}{1+at})(\frac{2}{2+at})\cdots (\frac{n}{n+at})\Big)\\ \sum_{k=1}^{n}X_{(k)}=\Pi_{k=1}^{n}\frac{(n-k+1)(n-k+2)\cdots n}{(n-k+1+at)(n-k+2+at)\cdots (n+at)}\\ =(\frac{n}{n+at})^{n}(\frac{n-1}{n-1+at})^{n-1}\cdots (\frac{2}{2+at})^{2}(\frac{1}{1+at}) \end{split} \end{equation*} How do I evaluate this last expression to show that $\sum_{k=1}^{n}X_{(k)}$ has a Gamma distribution? A Gamma distribution has a Laplace transform of the form: $L_{Y}(s)=\frac{1}{(1+\alpha s)^{p}},\quad Y\in\Gamma(p,\alpha)$