MLE for a Laplace center is the median

Question

A pricing model's residuals (in ticks) are modeled as Laplace centered at $\mu$ with known scale $b$ , density

$f(x \mid \mu) = \frac{1}{2b}\, e^{-|x - \mu|/b}, \qquad x \in \mathbb{R}.$

Five residuals are 1, 2, 4, 6, 9.

Derive the maximum likelihood estimator of the center $\mu$ and compute it.

Accepted Answer

The log-likelihood in \mu is \ell(\mu) = -n\ln(2b) - \frac{1}{b}\sumi xi - \mu. Since b 0, maximizing \ell over \mu is the same as minimizing the sum of absolute deviations \sumi xi - \mu. Unlike the squared-error case (which the mean minimizes), the sum of absolute deviations is minimized by the sample median: as you slide \mu, the derivative counts points above minus points below, which is zero exactly when \mu sits at the middle order statistic. For the sorted data 1, 2, 4, 6, 9 the median is the third value: \hat\mu = 	ext{median} = 4. Takeaway. The Laplace center MLE is the sample media…

MLE for a Laplace center is the median

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