Given that $X_1,X_2,...,X_i$ are random variables, find maximum likelihood estimator of $a$ $X_1,X_2,X_3,...X_i,$ 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.
 A: If I asked you for the value of $x$ for which the function $$f(x) = 1 - |x - 2|$$ attains its global maximum, would you try to differentiate $f$ to find it?
So, why would you expect that taking the derivative of the log-likelihood will yield the desired solution?  You need to understand that finding the extrema of a function is not always about differentiation.  Such is the case here:  the sum of absolute deviations is not something you can differentiate and get a critical point.  Instead, think of what it means to look for $a$ that minimizes $$\sum_{k=1}^n |x_k - a|$$ for some set of numbers $(x_1, x_2, \ldots, x_n)$.  For instance, what if $n = 4$ and $(x_1, x_2, x_3, x_4) = (1, 3, 7, 11)$?  What value of $a$ would you pick?  Does a unique value of $a$ exist that minimizes such a sum?
A: 
X1,X2,X3 are random variables and...

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
