My goal is to find $\hat{\theta}_{MLE}$ for $$X_1, ..., X_n \sim_{iid}f_X(x;\theta)=\frac{1}{3\theta}, \quad -\theta < x < 2\theta, \quad \theta \in (0,\infty)$$
I am looking at the likelihood function and I have
$$L(\theta) = \prod_{i=1}^n \frac{1}{3\theta} \Bbb I_{-\theta<x_i<2\theta}=\frac{1}{(3\theta)^n} \Bbb I_{-\theta<X_{(1)}<X_{(n)}<2\theta}$$
This is where I find things a little tricky.
Intuitively speaking since $(3\theta)^{-n}$ is a decreasing function for $0<\theta$ I would want to say $X_{(1)}$ maximizes $L(\theta)$
However, that only happens when $0<X_{(1)}$.
If all $X_i$s are negative and $n$ is even, then I want to say that $X_{(n)}/2$ will maximize $L$.
Is there a particular way to deal with this problem or am I on the right ball park?
I appreciate your help.
Let me rephrase your observations in a way that may make things clearer to you.
You want $\theta$ to be as small as possible while satisfying $\theta > 0$ and $\theta \ge -X_{(1)}$ and $\theta \ge X_{(n)}/2$.