 Solution

The independent Laplace distribution responses have log likelihoodFor every fixed positive $\sigma$, maximizing this expression means minimizing the sum of absolute residuals. Thus $\hat{\beta }$ is a least absolute deviations estimate, not a least-squares estimate. Such a minimizer exists: minimize the distance in the finite-dimensional closed column space of the design matrix. It need not be unique if the data or design allow multiple minimizers.

Write $s=S_1(\hat{\beta })$. For $s>0$, differentiating the profiled log likelihood gives $-n/\sigma +s/{\sigma }^2=0$. Its unique maximum isIf the fitted residual sum is zero, the likelihood instead increases without bound as $\sigma \downarrow 0$: there is no joint maximum with the stated positive scale parameter. The displayed estimate then denotes the boundary limit, not an admissible positive-scale MLE.

An ordinary exponential dispersion family has log density $[y\theta -b(\theta )]/a(\sigma )+c(y,\sigma )$, so changing the mean changes the log density by an affine function of $y$. For two distinct Laplace locations ${\mu }_1,{\mu }_2$, their log-density difference is $(|y-{\mu }_2|-|y-{\mu }_1|)/\sigma$, with different slopes on the intervals separated by the two locations. It is not affine in $y$. Equivalently the kink at $y=\mu$ moves with the parameter. The Laplace location family is not an ordinary exponential dispersion family.