Maximum Likelihood

Question

Find the maximum likelihood estimator of $f(x|\theta) = \frac{1}{2}e^{(-|x-\theta|\:)}$,

$-\infty < x < \infty$ ; $-\infty < \theta < \infty$.

I am confused of how to deal with the absolute value here.

Answer 1

The likelihood function $$L(x;\theta)=\prod\limits_{i=1}^n\frac{1}{2}e^{-|x_i-\theta|}=\frac{1}{2^n}e^{-\sum\limits_{i=1}^n|x_i-\theta|}$$

$$\ln L(x;\theta)=-n\ln2-\sum_{i=1}^{n}|x_i-\theta|$$

$$\frac{\partial\ln L(x;\theta)}{\partial\theta}=\sum_{i=1}^n \text{sign } (x_i-\theta)$$ because $|x|'=\text{sign }x,x\ne0$