# MLE of $\beta$ and $\sigma^2$ in linear regression with Laplace errors

Say we have a linear regression model $$y_i = x_i^T\beta + \sigma\varepsilon_i$$ for $$i = 1, \dots, n$$, and $$\varepsilon_i$$ is distributed according to a Laplace distribution $$\mathcal{L}(0, 1)$$ with density $$f(\varepsilon) = 2^{-1}e^{-|\varepsilon|}$$, then what are the MLE estimators for $$\beta$$ and $$\sigma^2$$?

So far for the $$\beta_{MLE}$$, I have gotten down to $$\beta_{MLE} = \underset{\beta}{\operatorname{argmin}} \sum_i |y_i - x_i^T\beta|$$ but now I do not know how to solve it. I attempted to differentiate wrt $$\beta$$ and then set that to 0, where I got $$\sum_i -x_i\operatorname{sign}{(y_i-x_i^T\beta)} = 0$$. I do not know how to proceed from here to find $$\beta_{MLE}$$.

For the $$\sigma^2_{MLE}$$, I have reduced it down to $$\sigma^2_{MLE} = \underset{\sigma^2}{\operatorname{argmin}} \sum_i \sigma^{-1}|y_i-x_i^T\beta|$$. I attempted to differentiate wrt $$\sigma$$ and then set that to 0, where I got $$-\sigma^{-2}\sum_i |y_i-x_i^T\beta| = 0$$. I do not know how to proceed from here to find $$\sigma^2_{MLE}$$.

Guidance would be much appreciated!

• I think there is no closed-form solution, see e.g. en.wikipedia.org/wiki/Least_absolute_deviations "The LAD estimate also arises as the maximum likelihood estimate if the errors have a Laplace distribution...Unlike least squares regression, least absolute deviations regression does not have an analytical solving method." Commented Mar 30 at 21:33
• I think there is no general closed-form solution for $\hat\beta_{MLE}$ like there is when we do maximum likelihood estimation for $iid$ Gaussian errors. I am not sure about a maximum likelihood estimator of $\sigma^2$ but suspect the same story to be the case, a lack of a closed-form solution.
– Dave
Commented Mar 31 at 2:25

The scale of the errors, $$\sigma$$, can be found seperately by using the method of moments, or by maximizing the entire log likelihood function (after filling in the solution for the optimum of $$\beta$$, which can be found seperately from the optimum for $$\sigma$$)
$$\mathcal{L}(\sigma,\boldsymbol{\beta}) = - n \log(2\sigma) -\frac{1}{\sigma}\sum_{i=1}^n |X_i\boldsymbol{\beta} - y_i|$$
Somewhere along the way of your derivations you have lost this $$- n \log(2\sigma)$$ term.