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