Question: The [Wikipedia article on order statistics](https://en.wikipedia.org/wiki/Order_statistic#) 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?

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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?

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

1. $ \{ X\_{(i+1)} - X\_{(i)} \}\_{i = 0}^{n-1} $ are mutually independent,
2. $X\_{(i+1)} - X\_{(i)}$ is $\mathrm{Exp}(\lambda (n-i))$ distributed, which in turn is the same as the law of $Z/\lambda(n-i)$ for a standard exponential $Z$.

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}
I think I came up with an approach that might work. We've generated $n$ exponentials with rate $\lambda$. Let's start with $X\_{(1)}$. It's the minimum of $n$ exponentials. We know this is just another exponential with rate $n \lambda$.

$$X\_{(1)} \sim \frac{Z\_1}{n \lambda}$$

Now, we're at $X\_{(1)}$, and there are $n-1$ exponentials to go. By the memory-less property, the additional time it'll take for the next exponential to arrive from here is simply the minimum of $n-1$ exponentials. So,

$$X\_{(2)}\sim \frac{Z\_1}{n \lambda} + \frac{Z\_1}{(n-1) \lambda}$$

And we can take this all the way to $i$.