I have been solving some problems regarding finding the likelihood function; i.e determining the MLE of a parameter from a statistical distribution, i.e the Normal distribution.
Now, I have an expression for the likelihood function of the parameter $a, -\infty < a < \infty$; $x_i$ are iid $i = 0,1,...,n$ $$\frac{1}{a^{9n}}I{\{max (x_i) < a, min (x_i) > -a}\}$$
I know that the MLE is where the graph of this likelihood function is maximum. Obviously this is a decreasing function in $a,$ my thoughts are that the mle is $max(x_i)$ if $|max(x_i)| > |min(x_i)|$, and $min(x_i)$ if $|min(x_i)| > |max(x_i)|$, but I'm unsure whether this is true or not as I haven't seen an example where the MLE depends on the values of the extremum.
You are right, as the support of $X$ is an interval with unknown boundaries, the MLE is the closest value to these boundaries, namely $$ -a \le X_{(1)} \le X_2 \le ... \le X_{(n-1)} \le X_{(n)} \le a, $$
where $a > 0$, hence, $$ \hat{a}_{MLE} = \max\{ | X_{(1)}|, | X_{(n)}| \}. $$