Given $X_1, X_2, ..., X_n \sim X$ where
$$f_X(x;\theta) = \theta^2 xe^{-\theta x}, \quad x>0,\quad\theta >0$$
I was able to find the expectation $E[X]=\frac{2}{\theta }$ and that the MLE is $\hat{\theta} = \frac{2}{\bar{X}}$.
I am asked to see if this MLE is unbiased、so my instinct is to evaluate
$$E\left[\frac{2}{\bar X}\right]$$ but I am not comfortable with the $\bar X$ being on the denominator like that . . .
Can someone help me out with how to find more information about this?
Using Jensen inequality you can deduce that $$ \mathbb{E} [2/\bar{X} ] \ge2/\mathbb{E}[\bar{X}] = \theta, $$ i.e., $$ \mathbb{E} \hat{\theta}-\theta\ge0. $$ And if you are interested in explicit calculations then note that $2/\bar{X} = 2n/\sum X_i $, where $\sum X_i \sim Erlang(2n, \theta) $, thus $$ \mathbb{E}[\hat{\theta}]=2n \int\frac{\theta^{2n}}{y \Gamma(2n)} e^{-\theta y} y ^ {2n-1}dy = \theta\frac{2n}{2n-1} \ge \theta $$