Let $X_1,...,X_n$ be a sample from a distribution having as a p.d.f: $f(x) = \frac1{\theta} e^{-x/\theta}, x,\theta > 0$ and $0$ elsewhere.
The maximum likelihood estimator of $\theta$ is $\bar{X}=\sum_{i=1}^n X_i /n$. Why isn't the maximum likelihood estimator of $\theta^2$ just $\bar{X} ^2$? Why does it have to be the variance? The distribution is exponential with parameter $\theta$ so the variance is $\theta^2$ I understand that. But why doesn't it make sense to take $\bar{X} ^2$ as the m.l.e of $\theta^2$?