$X_1,X_2,X_3,...X_i,$ are random variables and
$$f(x,a)= 0.5e^{-|x-a|}$$
with $$-∞ < x < +∞$$ $$-∞ < a < +∞$$
Find maximum likelihood estimator of $a$.
I solved this problem as follows
$$L(a) = \prod_{i=1}^{n} 0.5e^{-|xi-a|}$$
$$L(a) = 0.5^n \prod_{i=1}^{n} e^{-|x-a|}$$
$$LL(a) = -n ln2 + \sum-|x-a|$$
But when I take this derivative, I do not achieve anything.
First well copy the text of your exercise...say $(X_1,X_2,X_3)$ is a simple random sample from the following density....
Then, as suggested, the function $\hat{a}$ that minimize the quantity $\Sigma_i |X_i-a|$ is the Sample median: $X_{(2)}$
The proof is very simple:
Let's start with the definition of $E|X-a|$; without loss of generality we can do the brainstoming for a continuous rv
$$E|X-a|=\psi(a)=\int_{a}^{\infty}(x-a)f(x)dx+\int_{-\infty}^{a}(a-x)f(x)dx$$
$$\psi'(a)=-\int_{a}^{\infty}f(x)dx+\int_{-\infty}^{a}f(x)dx$$
(using Leibnitz integral rule)
That is
$$-[1-F_X(a)]+F_X(a)$$
Setting it $=0$ you get
$$F_X(a)=1/2$$
Thus
$$a=Me$$
The results can be found here, as your density is a known one: the Laplace distribution