Hessian of Laplace distribution

The density of the Laplace distribution is given by:

$$f(x;\mu,\sigma)=\frac{1}{2\sigma}\exp\left(-\frac{\vert x- \mu\vert}{\sigma}\right).$$

It is easy to see that this function is not differentiable at $\mu$. However, I am interested on some asymptotic normality results of the MLE $(\hat{\mu},\hat{\sigma})$, which require double-differentiability of the log likelihood function:

$${\mathcal l}(\mu,\sigma) = \sum_{j=1}^n \log f(x_j;\mu,\sigma),$$

with respect to $(\mu,\sigma)$, for a random sample $(x_1,\dots,x_n)$. What I basically need is the existence of the Hessian matrix of the log-likelihood evaluated at the MLE, which means the existence of: $\frac{\partial^2}{\partial \mu^2}{\mathcal l}(\mu,\sigma) \Big\vert_{\mu=\hat{\mu}}$, $\frac{\partial^2}{\partial \sigma^2}{\mathcal l}(\mu,\sigma) \Big\vert_{\sigma=\hat{\sigma}}$, $\frac{\partial^2}{\partial \mu\partial\sigma}{\mathcal l}(\mu,\sigma) \Big\vert_{\mu=\hat{\mu},\sigma=\hat{\sigma}}$.

Is there any reference to justify differentiability of the log-likelihood, or it is not differentiable at the MLE?

• I do not understand the question. I believe you have tried indicate that the log-likelihood cannot be differentiated with respect to $\mu$ at any of the points in the set $\{x_j\}$, so assuming differentiability everywhere cannot legitimately be "justified." Could you clarify what you are asking?
– whuber
Aug 8 '14 at 17:06
• @whuber Done. I hope this is clearer. Aug 8 '14 at 17:14
• Thank you. May I point out that your question seems to assume that the MLE coincides with one of the $x_j$. Otherwise, differentiability at the MLE would not be an issue, would it?
– whuber
Aug 8 '14 at 17:20
• @whuber Yeah, I agree with that. So, I guess I could just say that the Hessian matrix exists for almost any sample (excluding those where at least one observation coincides with the MLE, which is an event of probability zero). Aug 8 '14 at 17:22
• Why do you say that the MLE coinciding with an observation is an event of probability 0? Aug 9 '14 at 1:26

In Huber and Ronchetti "Robust Statistics" (2009 2nd ed), ch. 4.4 it is proven that the finiteness of the Fisher Information matrix is equivalent to the density being absolutely continuous, without needing differentiability. To achieve it, the authors start by providing a generalized definition of Fisher information in terms of the distribution function "when the classical expression does not make sense" (the "classical expression" being $I(f) = \int (f'/f)^2fdx = E(f'/f)^2$, $f$ being the density) and then prove that the generalized expression will coincide with the classical expression iff $f$ is absolutely continuous. The proof is not trivial.