Maximum likelihood estimate of the distribution involving mod

Question

$$f(x)=\frac{1}{2} exp[|x-\theta|],\ \ x\in \mathbb{R},\ \theta \in \mathbb{R}$$ After doing the calculation I got that $$ L=\frac{1}{2^n}+\sum_{i=1}^n|x_i-\theta| $$ Now this function is not differentiate at $ x_i$ so how will I maximize the above function to get the male of $\theta$.

Answer 1

Your pdf is wrong as what you have written is not a pdf. It should be $f(x;\theta)=\dfrac{1}{2}e^{-|x-\theta|}$.

So the joint density is $\dfrac{1}{2^n}e^{-\sum_{i=1}^n |x_i-\theta|}$. You want to maximize this, so essentially you need to minimise $\sum_{i=1}^n |x_i-\theta|$. But the $\theta$ minimising this is $\text{median}(x_i)$. This is your MLE.

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Suppose you are given $n$ points $x_1,x_2,\ldots,x_n$ and you want to find $y$ such that $y$ minimises $\sum_{i=1}^n |x_i-y|$. We will show that $y=\text{median}(x_1,x_2,\ldots,x_n)$.

First note you can always order the $x_i$, so without loss of generality assume $x_1<x_2<\ldots<x_n$. Since $y$ minimises $\sum_{i=1}^n |x_i-y|$, $y$ must lie between $x_1$ and $x_n$. Similarly $y$ must lie between $x_2$ and $x_{n-1}$, and so on, so $y$ must lie between the middle two points if $n$ is even and $y$ is indeed the middle point if $n$ is odd. Hence $y$ is the median of $x_i$.