Order statistics of i.i.d. exponentially distributed sample

I have been trying to find the general formula for the $k$th order statistics of $n$ i.i.d exponential distribution random variables with mean $1$. And how to calculate the expectation and the variance of the $k$th order statistics. Can someone give me some general formula? It would be nice if there is any proof.

The minimum $X_{(1)}$ of $n$ independent exponential random variables with parameter $1$ is exponential with parameter $n$. Conditionally on $X_{(1)}$, the second smallest value $X_{(2)}$ is distributed like the sum of $X_{(1)}$ and an independent exponential random variable with parameter $n-1$. And so on, until the $k$th smallest value $X_{(k)}$ which is distributed like the sum of $X_{(k-1)}$ and an independent exponential random variable with parameter $n-k+1$.

One sees that $X_{(k)}=Y_{n}+Y_{n-1}+\cdots+Y_{n-k+1}$ where the random variables $(Y_i)_i$ are independent and exponential with parameter $i$. Each $Y_i$ is distributed like $\frac1iY_1$, and $Y_1$ has expectation $1$ and variance $1$, hence $$\mathrm E(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1i,\qquad \mbox{Var}(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1{i^2}.$$

• Very nice, Didier. I'll remember this trick! – Chris Taylor Nov 10 '11 at 7:45
• @Chris, thanks. – Did Nov 10 '11 at 7:47
• Is there an easy way to generalize this to independent exponential with different rates? – A.S. Oct 6 '15 at 2:31
• @A.S. I see no obvious way to. – Did Oct 6 '15 at 7:33
• Could you please explain why $X_{(2)}$ is distributed like the sum of $X_{(1)}$ and an independent exponential random variable with parameter $n-1$? – PSPACEhard Apr 13 '16 at 10:26

The probability density function for the $k$th order statistic of a sample of size $n$ from a distribution with pdf $f(x)$ and distribution function $F(x)$ is

$$f_k(x) = n {n-1\choose k-1} F(x)^{k-1} (1-F(x))^{n-k} f(x)$$

and, since the exponential distribution with mean 1 has pdf $f(x)=e^{-x}$ and distribution function $F(x)=1-e^{-x}$, we can compute

$$f_k(x) = n {n-1\choose k-1} (1-e^{-x})^{k-1} e^{-(n-k+1)x}$$

Computing the expectation and variance of the $k$th order statistic can now be done via the usual method.

• but how would you calculate the expectation of this? It involves a lot of calculus – geraldgreen Nov 9 '11 at 10:22
• It's quite likely that there isn't a closed form. Since it involves terms like $y^a(1-y)^b$ the integrals can probably be expressed using hypergeometric functions, although maybe you could give it a crack by constructing a recurrence for $n$ and $k$. I'll have a look at that later. – Chris Taylor Nov 9 '11 at 10:25
• "It's quite likely that there isn't a closed form" See other answer. – Did Sep 3 '16 at 7:47