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?
Best Answer
This can be shown quite directly. Fix an $i$. Then note that $$ P(X_{(i+1)} -X_{(i)}> x| X_{(i)} = y) = P(X_{(i+1)} > x+y|X_{(i) } = y).$$
Now, the latter probability is that of finding that the smallest of $n-k$ exponential random variables is bigger than $x+y$, given that each is at least $y$. But, due to the independence, and the memorylessness of the exponential distribution, this just works out to $n-i$ independent exponentials being bigger than $x$, i.e. $$ P(X_{(i+1)} - X_{(i)} > x| X_{(i)} = y) = e^{-\lambda (n-i) x}. $$ But, since the right hand side is independent of $y$, we can integrate over the conditioning and conclude that $X_{(i+1)} - X_{(i)}$ has the distribution of a $\mathrm{Exp}( \lambda (n-i))$ random variable. Further, this random variable is independent of $X_{(i)}$
In fact, yet more is true - instead of conditioning over only the value of $X_{(i)},$ we could have conditioned on the values of $X_{(0)} := 0, X_{(1)}-X_{(0)}, X_{(2)} - X_{(1)}$ and so on all the way to $X_{(i)} - X_{(i-1)}$ without changing the conclusion or the reasoning (again, this critically relies on the memorylessness of the exponential law). This means that
But now the result is obvious - \begin{align} X_{(i)} &= \sum_{j = 0}^{i-1} X_{(j+1)} - X_{(j)} \\ &\equiv \sum_{j = 0}^{i-1} \frac{Z_j}{\lambda (n - j)}.\end{align}