If the MLE is constrained to only two values, ${1,2}$, then what does the MLE become?

Question

Suppose my pdf is some function that requires that $x\geq0$ or else it becomes $0$. Also, the parameter is $\theta>0$. Now, I would like to calculate the MLE of that pdf for $X_1,...,X_n$ i.i.d. random variables.

Now, suppose I differentiated with regards to the log-likelihood and found that the derivative is:

$$\frac{\partial l_x(\theta)}{\partial x}=\frac{n}{\theta}-\sum_{i=1}^{n}x_i^2$$

and so we get that

$$\hat{\theta}^{MLE}=\frac{n}{\sum_{i=1}^{n}x_i^2}$$

Now if we were to constrain the parameter space from $\theta>0$ to {1,2}, meaning the parameter space for $\theta$ is now only 1 or 2, what would my new MLE of $\theta$ be?

Thanks!

Answer 1

Rather than trying to force the notion of a derivative to translate to a discrete parameter space, consider the definition of an MLE. Using your notation for the log-likelihood, we have

$$ \begin{align} \hat{\theta}^{MLE} &= \text{argmax}_{\theta \in \{1,2\}}\ l_x(\theta)\\ &= \text{argmax }\{l_x(1), l_x(2)\}. \end{align} $$

In other words, calculate the log-likelihood at each value of $\theta$ in the (discrete) parameter space and determine the value that maximizes the likelihood.