MLE for Laplacian $f (y; α) = 1/2 \exp(−|y − \alpha|)$

Question

Let $Y_1, ..., Y_n$ be iid from a Laplace distribution (also known as a double exponential distribution) with density function $$f (y; α) = 1/2 \exp(−|y − α|), y,\alpha \in \mathbb{R}.$$ Assuming that $n$ is even determine a maximum likelihood estimator of α. Is the MLE unique?

To solve this, do I need to consider different densities depending on whether the data are smaller or greater than $\alpha$? Say I have $n_0$ data where $y<\alpha$ then the density is $f (y; α) = 1/2 \exp((y − \alpha))$ and for remaining $n-n_0$ observations $y>\alpha$ withhaving density $f (y; α) = 1/2 \exp((\alpha − y))$. Then the likelihood is $$ (1/2 \exp((y − \alpha)))^{n_0}( 1/2 \exp((\alpha − y)))^{n-n_0}.$$ Maximizing the loglikelihood $n_0(y-\alpha)+(n-n_0)(\alpha-y)+\text{const}$ for $\alpha$, I get $n=2n_0$ which is independent of $\alpha$. Does it mean I cannot use the derivative and have to try something else

Answer 1

Because of the absolute value, you cannot take a derivative at $x=a$, hence the solution is more analytical. Note that maximizing the likelihood is the same as minimizing $\sum_{i=1}^n | x_i - a|$ w.r.t $\alpha$, namely, what value of $\alpha$ will give you $$ \sum_{i=1}^n \operatorname{sign}( x_i - \alpha) =0 . $$ A value that exactly half of the $x_i$s are larger then it (and the other half lower), i.e., $$ \hat{\alpha} = \operatorname{median} \{x_1,\ldots, x_n\}. $$ Full derivation you can find here.

Answer 2

You don't get $ (\frac 1 2 \exp((y − \alpha)))^{n_0}( \frac 1 2 \exp((\alpha − y)))^{n-n_0}.$ Rather, you get $$ \left(\frac 1 2 \right)^n \exp\left( (-1)\cdot\left(\sum_{i \, \in \, I} (\alpha-y_i) + \sum_{i\,\in\,J} (y_i - \alpha) \right) \right) $$ where $I$ is the set of indices $i$ for which $y_i>\alpha$ and $J$ is the set of indices $i$ for which $y_i<\alpha.$ The $y$-values are different for different indices $i$, and that was neglected in the form that you wrote.

This is a decreasing function of the sum inside the $\displaystyle \left( \vphantom{\sum_I} \text{large parentheses} \right),$ so you need the value of $\alpha$ that minimizes that.

Suppose for a particular value of $\alpha$ you have $|I|>|J|.$ I.e. there are more indices $i$ for which $y_i< \alpha$ than for which $y_i>\alpha.$ Then a small decrease in $\alpha$ decreases the sum over $I$ by more than the amount by which it increases the sum over $J,$ so that decreases the whole sum. Likewise if $|I|< |J|,$ increasing $alpha$ reduces that sum. Therefore the sum is minimized by making $|I|=|J|,$ i.e. $\alpha$ is less than the same number of observations that $\alpha$ is greater than.

In other words, the MLE is the sample median.