The Wikipedia article on order statistics mentions the following result on the order statistics of an exponential distribution with rate parameter, $\lambda$: $X_{(i)} = \frac{1}{\lambda}\sum\limits_{j=1}^i \frac{Z_j}{n-j+1} \tag{1}$ It provides no proof of this. How do I prove it? My attempt: We know that to get X_{(i)} for a distribution with inverse CDF $F_X^{-1}(x)$, we first get the corresponding order statistic of the uniform ($U_{(i)}$) and then apply the inverse CDF to it. We know that $U_{(i)} \sim B(i,n-i+1)$. And the inverse CDF of the exponential distribution is: $F_X^{-1}(x) = -\frac{\log(1-x)}{\lambda}$. This means that the distribution of $X_{(i)}$ should be: $-\frac{\log(1-U_{(i)})}{\lambda}$ Also, $1-U_{(i)} \sim U_{(n-i)}$. So, the distribution of the order statistic becomes: $X_{(i)}\sim -\frac{\log(U_{(n-i)})}{\lambda}$ We have a Beta inside a logarithm. Don't see a path to equation (1) except maybe expressing the Beta as a Gamma and then noting that the Gamma is a sum of exponentials?
