Sufficiency and MLE of $\theta$ where $f(x;\theta)=\dfrac{1}{2}e^{|x-\theta|},-\infty<\theta<\infty$

Question

Let $X_1,X_2,...,X_n$ be a random sample from the density. $f(x;\theta)=\dfrac{1}{2}e^{|x-\theta|},-\infty<\theta<\infty$.

Discuss the sufficiency and MlE of $\theta$ for this density

I can write this density function as

$ f(x;\theta) = \begin{cases} \dfrac{1}{2}e^{-(x-\theta)} , \theta < x < \infty \\ \dfrac{1}{2}e^{(x-\theta)} , -\infty < x < \theta \end{cases}$

I am skipping the whole proof as I am preparing for time based exam. I just want to know for the first part,we have $\theta<x_{(1)}$ and for second part $\theta>x_{(n)}$

So $x_{(1)}$ and $x_{(n)}$ are jointly sufficient. Am I right ?
and for MLE $\theta$ satisfies $ x_{(n)}<\theta<x_{(1)}$ which seems absurd. Mle doesnt exist in this case?

Answer 1

To find MLE for $\theta$ an easy approach is the usual one

$$L(\theta)\propto e^{-\Sigma_i |X_i-\theta|}$$

$$l(\theta)=-\Sigma_i |X_i-\theta|$$

To be maximized...

Maximizing $-\Sigma_i |X_i-\theta|$ is the same as minimizing

$$ \bbox[5px,border:2px solid black] { \Sigma_i |X_i-\theta| \qquad (1) } $$

It is well known that $\hat{\theta}$ which minimize (1) is the sample median

(the proof is easy and I can show it to you if necessary)