$X_1, ..., X_n$ is a random sample from a population with pdf given by
$$ f(x; \mu, \lambda) = \frac{\lambda}{2}\operatorname{exp}(- \lambda |x - \mu|) $$
where $\mu \in \mathbb{R}$ is the location parameter, and $\lambda > 0$ is the scale parameter.
I'm trying to obtain the log-likehood function to this problem. I have,
$$ L(\mu, \lambda; \mathbf{x}) = \prod_{i=1}^{n} f(x_i; \mu, \lambda) = \prod_{i=1}^{n} \frac{\lambda}{2}\operatorname{exp}(- \lambda |x - \mu|) = \frac{\lambda^n}{2^n}\operatorname{exp}\Bigl(- \lambda \sum_{i=1}^{n} |x - \mu|\Bigr). $$
Let $ \ell(\mu, \lambda; \mathbf{x}) = \text{log}[L(\mu, \lambda; \mathbf{x})] $. Then we have
$$ \ell(\mu, \lambda; \mathbf{x}) = n\text{log}(\lambda/2) - \lambda \sum_{i=1}^{n} |x - \mu|. $$
Is this the correct form of the sample log-likelihood? Because if it is... then trying to maximize it results in a Hessian determinant which is negative, even before evaluation at the critical point.
Assuming the log-likehood function is correct, you'd get:
$$ \frac{\partial}{\partial \mu} \ell(\mu, \lambda; \mathbf{x}) = \lambda \sum_{i=1}^{n} \frac{x_i - \mu}{|x_i - \mu|} $$
$$ \frac{\partial}{\partial \lambda} \ell(\mu, \lambda; \mathbf{x}) = \frac{n}{\lambda} - \sum_{i=1}^{n} |x_i - \mu| $$
$$ \frac{\partial^2}{\partial \lambda^2} \ell(\mu, \lambda; \mathbf{x}) = \frac{-n}{\lambda^2}. $$
$$ \frac{\partial^2}{\partial \mu^2} \ell(\mu, \lambda; \mathbf{x}) = 0. $$
$$ \frac{\partial^2}{\partial\lambda \partial \mu} \ell(\mu, \lambda; \mathbf{x}) = \frac{\partial^2}{\partial\mu \partial \lambda} \ell(\mu, \lambda; \mathbf{x}) = \sum_{i=1}^{n} \frac{x_i - \mu}{|x_i - \mu|}. $$
Also I got the (single) critical point $ (\mu, \lambda) = (\bar{x}, n/\sum_{i=1}^{n} |x_i - \mu|) $. With this information you'd get $\text{det}(Hessian) < 0 \Rightarrow $ saddle point. So there's something wrong.