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Consider a linear regression model: $$ y_i = \mathbf x_i \cdot \boldsymbol \beta + \varepsilon _i, \, i=1,\ldots ,n, $$ where $\varepsilon _i \sim \mathcal L(0, b)$, that is, Laplace distribution with $0$ mean and $b$ scale parameter, are all are mutually independent. Consider a maximum likelihood estimation of unknown parameter $\boldsymbol \beta$: $$ -\log p(\mathbf y \mid \mathbf X, \boldsymbol \beta, b) = n\log (2b) + \frac 1b\sum _{i=1}^n |\mathbf x_i \cdot \boldsymbol \beta - y_i| $$ from which $$ \hat{\boldsymbol \beta}_{\mathrm {ML}} = {\arg\min }_{\boldsymbol \beta \in \mathbb R^m} \sum _{i=1}^n |\mathbf x_i \cdot \boldsymbol \beta - y_i| $$

How can one find a distribution of residuals $\mathbf y - \mathbf X\hat{\boldsymbol \beta}_{\mathrm {ML}}$ in this model?

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  • $\begingroup$ What do you mean by find a distribution of residuals? $\endgroup$ – jlimahaverford Sep 30 '15 at 17:56
  • $\begingroup$ Since residuals can be grouped in a random vector, I'd like to know its distribution. At least first two moments. $\endgroup$ – nmerci Sep 30 '15 at 18:35
  • $\begingroup$ Got it, thanks! Have you considered simulating and plotting? $\endgroup$ – jlimahaverford Sep 30 '15 at 18:39
  • $\begingroup$ Yes, I want to construct a confidence region for residuals. For instance, for Gaussian errors the region is an ellipsoid. $\endgroup$ – nmerci Sep 30 '15 at 19:07
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The residuals (actually called errors) are assumed to be randomly distributed with a double-exponential distribution (Laplace distribution). If you are fitting this x and y data points, do it numerically. You first calculate beta-hat_ML for these points as a whole using the formula you posted above. This will determine a line through the points. Then subtract each point's y value from the y value of the line at that x value. This is is the residual for that point. The residuals of all points can be used to construct a histogram that will give you the distribution of the residuals.

There is a good mathematical article on it by Yang (2014).

--Lee

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    $\begingroup$ The link does not work. $\endgroup$ – Michael R. Chernick Jun 10 '17 at 16:12

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