Maximum Likelihood 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.  
 A: 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 $|x|'=\text{sign }x,x\ne0$
A: Note that $f(x|\theta)\:$denotes the $\text{pdf}\:$such that$\:x\:$takes on any value belonging to the set  of real numbers with$\:\theta\:$viewed as a fixed probability.
The likelihood function$\:L(\theta|x)\:$is such that$\:x\:$represents an observed sample point, and$\:\theta\:$varying over all parameter values.
Where$\:X=\{X_1,...,X_n\}\:$and$\:X=x\:\:$the observed value
Therefore both functions aren't exactly the same and aren't written in the same fashion.
$\:L(\theta|x)=\large\prod_\limits{i=1}^n$$\frac{1}{2}e^{-|x_i-\theta|}=\frac{1}{2^n}e^{-\sum_{i=1}^n|x_i-\theta|}\:\:$and the MLE minimizes$\:\:\sum\limits_{i=1}^{n}|x_i-\theta|=\sum\limits_{i=1}^{n}|x_{(i)}-\theta|$
with order statistics $\:x_{(1)},...,x_{(n)}$
For$\:\:\theta\in[x_{(j)},x_{(j+1)}],$
$\mathcal{S}=\sum\limits_{i=1}^{n}|x_{(i)}-\theta|=\sum\limits_{i=1}^{j}(\theta-x_{(i)})+\sum\limits_{i=j+1}^{n}(x_{(i)}-\theta)=(2j-n)\theta-\sum\limits_{i=1}^{j}x_{(i)}+\sum\limits_{i=j+1}^{n}x_{(i)}$
$\mathcal{S}\:$is a linear function depending on$\:\theta\:$that increases for $2j-n>0,\:$and decreases for$\:2j-n<0$
If$\:n=2k+1; k\in\mathbb{N},\:$the likelihood is minimized at$\:\theta^{\star}=x_{(\frac{n+1}{2})},\:$wich is the median order value.
If$\:n=2k; k\in\mathbb{N},\:$the likelihood is minimized for $\:\theta^{\star}\in(x_{(\frac{n}{2})},x_{(\frac{n+1}{2})}),\:$because it is constant over this interval.
