Q:

How to calculate the variance and the expected value of the MLE of stochasts $(X_1,\ldots,X_n)$ with density $f_\mu(x)=e^{x-\mu}1_{[0,\infty)}(x)$

So far i've got that the MLE is $\mu'= \min\{X_1,\ldots,X_n\}$ now i'm supposed to construct $$F_\mu'(x)= P(\mu'\le x)$$
The problem is that i don't understand how to construct this function for a stochast which is in the form of $\mu'$. 
I know that $F_{X_1}(x) = \int_\mu^x e^{-(s-\mu)} \rm ds$. But that's about it, i think i'm supposed to rewrite $F_\mu'(x)$ in a form with only constant terms and $F_{X_i}(x)$.
Anyone a hint/clue?
Kees

A:

Let $Y=\min(X_k)$, then $[Y\geqslant x]=\bigcap\limits_k[X_k\geqslant x]$. The functions $f_\mu$ in the question are not densities hence let us assume instead that $f_\mu$ is given by
$$
f_\mu(x)=\mathrm e^{-x+\mu}\mathbf 1_{[\mu,\infty)}(x),
$$
then $P_\mu[X_k\geqslant x]=\mathrm e^{-x+\mu}$ if $x\geqslant\mu$ hence
$P_\mu[Y\geqslant x]=\mathrm e^{-nx+n\mu}$ if $x\geqslant\mu$, that is,
$$
F_\mu(x)=(1-\mathrm e^{-nx+n\mu})\,\mathbf 1_{[\mu,\infty)}(x).
$$
This is a well-known distribution hence the mean and variance follow.

