MLE Interpretation

Question

To show this, I solved for the MLE, but as a verbal explanation, I was wondering if the following would suffice. Since the MLE of $\theta$ is $\bar x$, given the range of the distribution $(0 < x < 2)$, the MLE is $\min(\bar X,2)$.

Answer 1

Well, kinda. MLE is a maximization problem. I.e.

$$ \hat{\theta} = {\arg\max}_{\theta\in(0,\infty)} L(\theta) = \overline{X}.$$

You want to use this to conclude that $$ \hat{\theta}' = {\arg\max}_{\theta\in(0,2]} L(\theta) = \min(\overline{X},2).$$

This conclusion is not valid for arbitrary functions $L$. For example

• The conclusion is not true if $L$ has a local maximum somewhere in $(0,2)$, and its global maximum in $(2,\infty)$.
• The conclusion is true however if $L$ only has a single local maximum (which is therefore the global maximum). I'm pretty sure this is true for the Poisson-Likelihood (as well as most other commonly used distrubtion functions). But it does not need to be true for general MLE.