The MLE of distribution with pdf $f(x;\theta)=\frac{1}{2} e^{-|x-\theta|}$

Question

I am given that the order statistics $X_{(1)}, ..., X_{(n)}$ are iid from a distribution with pdf

$$f_X(x;\theta) = \frac{1}{2}e^{-|x-\theta|}, x\in \Bbb R, \theta\in \Bbb R$$

I approached with a likelihood function

$$L(\theta) = 2^{-n}\exp\left[-\sum_{i=1}^n |X_i-\theta|\right]$$

and the log likelihood

$$l(\theta) = -n \ln2-\sum_{i=1}^n |X_i-\theta|$$

I am not comfortable taking the derivative from here since the range of $X_i$ and $\theta$ does not let me guarantee that it will be $1$ or $-1$ and it also bothers me that the pdf is not the distribution of the order statistics that I am given.

My ultimate goals is to do a likelihood ratio test where

$$H_0: \theta = \theta_0 \quad vs \quad H_1: \theta \ne \theta_0$$

so that is why I am trying to find the MLE.

My notes suggest that the median is the MLE but I have no idea why that would be true.

I appreciate your input.

Answer 1

Differentiating wrt $\theta$ gives: $$\frac{\partial l}{\partial \theta}=-\sum_{i=0}^n\operatorname{sgn}\left(\theta-X_i\right)$$

This is zero iff the number of $X_i$'s that are larger than $\theta$ is equal to the number of $X_i$'s that are less than $\theta$. That is, when $\theta$ is the median of the $X_i$'s.