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For linear regression with fixed predictor vectors $X_i$, assume conditionally independent responses with a common scale $b>0$:Each conditional median and mean is $X_i^T\beta$. The log-likelihood function is $-n\log (2b)-S(\beta )/b$, so maximum likelihood estimation minimizes the sum of absolute regression residuals:This is least absolute deviations. With only an intercept, it reduces to the Laplace location and scale maximum likelihood problem, so a sample median is a fitted intercept. Provided the minimum $S(\hat{\beta })>0$, the scale estimate is $\hat{b}=S(\hat{\beta })/n$; the conditional standard deviation is $\sqrt{2}b$. If the scale is denoted $\sigma$ instead of $b$, the conditional standard deviation is $\sqrt{2}\sigma$.

Minimizing a sum of absolute losses does not automatically give a unique coefficient vector. Even an intercept-only fit to an even sample size can have an interval of solutions, and redundant predictor columns can introduce further nonuniqueness. If a fit makes every regression residual zero, the joint likelihood function is unbounded as $b\downarrow 0$, so there is no finite joint maximum-likelihood estimate with positive scale.

The linear growth of absolute loss limits sensitivity to an extreme response at fixed predictors compared with squared-error loss. It does not protect against every outlier: an extreme predictor can have large regression leverage. The numerical example above changes a response in an intercept-only problem, and makes no general claim of immunity to such predictor changes.