I have been given a problem in which I am given a set of independent random variables, X1, X2, ... Xn with the distribution:
$f_x(x ; b, m) = \frac{1}{2b}e^{-|x-m|/b}$ for $-∞ < x < ∞$
and I am tasked with finding the maxiumum likelihood estimators for b and m given the sample: $-2.1,\: -1.5,\: -.6,\: .1,\: 1.2,\: 1.7,\: 2.3,\: 3.1,\: 4.4$
I followed the standard way of solving these questions as taught by my professor by first taking the product of the distribution over the set of variables:
$\prod_{i=1}^{n} (\frac{1}{2b}e^{-|x_i-m|/b})$ = $\frac{1}{(2b)^n}e^{-1/b\sum_{i=1}^{n}|x_i-m|}$
Taking the natural log of this equation gives:
$-n\ln{2b}\:-\:\frac{1}{b}\sum_{i=1}^{n}|x_i-m|$
So then I must take the derivative in respect to m and in respect to f, set each of these equations to zero and solve for m and f, respectively. I have found the equations for each of these as follows:
for f: $\frac{-n}{b}\:+\:\frac{1}{b^2}\sum_{i=1}^{n}|x_i-m| = 0$
for m: $\frac{-1}{b}\sum_{i=1}^{n}sign(m-x_i) = 0$
With the equation for m having the sum of the signum function. After this I am stuck as I have never worked with an estimator with 2 variables nor have I ever had to actually solve an equation for a set of real values. I am also unsure of how I would work out the sum of the signum function so I could solve for m as I have never worked with this function either. If anyone could help solve the equation or point me in the right direction that would be great.
Your distribution is a Laplace distribution
In your equation for $m$, you will get $\sum_{i=1}^{n}\text{sign}(m-x_i) = 0$ when ($n$ is even) half the signs are $+1$ and half $-1$ or when ($n$ odd) one is $0$ and half the rest $+1$ and half $-1$. This will make MLE $m$ the median
Knowing the MLE for $m$, this gives a fixed value for $\sum_{i=1}^{n}|x_i-m|$ and so you can solve $\frac{-n}{b}\:+\:\frac{1}{b^2}\sum_{i=1}^{n}|x_i-m| = 0$ to get the MLE estimate $b=\frac1n \sum_{i=1}^{n}|x_i-m|$, which is the mean absolute deviation about the median