Let $X_1,..., X_n$ be iid $Exp(\lambda)$.
The MLE for $\lambda$ is $\hat{\lambda}=\frac{1}{\bar{X}}$, where $\bar{X}=1/n \sum^n_{i=1}X_i$
How can I conclude that $E(\hat{\lambda}) = n\lambda/(n-1)$?
This seems to be very simple, but somehow I'm not getting it.
Any help would be appreciated.
Consider the random variable $S = n \bar X = \sum_{i=1}^n X_i$. Then $S \sim \mathrm{Gamma}(n,\lambda)$ where the parametrization is by rate, not scale. What is the distribution of $T = 1/S$? This is inverse gamma: it is easy to see that $F_T(t) = \Pr[T \le t] = \Pr[1/T \ge 1/t] = 1 - F_S(1/t)$, hence $f_T(t) = f_S(1/t)/t^2$. I leave it to you to determine how to use this information to obtain ${\rm E}[\hat \lambda]$.