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.
 A: 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
$$
\mathrm E(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1i,\qquad
\mbox{Var}(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1{i^2}.
$$
A: 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.
