# Obtaining the correct Log-likelihood function

$$X_1, ..., X_n$$ 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.

The loglikelihood is not differentiable with respect to $$\mu$$ where $$\mu=x_i$$ (and in particular, not at the MLE). The second derivative with respect to $$\mu$$ is zero everywhere that it is defined. So you can't just do calculus.
For any fixed $$\lambda$$, the loglikelihood is piecewise linear (and concave) in $$\mu$$ with corners at the observations, so there is a maximum (and no other stationary points) and the maximum must be at an observation. It's fairly easy to satisfy yourself that the maximum is at the median if $$n$$ is odd and on the whole median interval if $$n$$ is even. That's true for every fixed $$\lambda$$ so it must be true for varying $$\lambda$$ as well.
Now, fixing $$\mu$$ at the MLE you have a differentiable one-parameter problem in $$\lambda$$. Solving $$\frac{\partial}{\partial \lambda} \ell(\hat\mu, \lambda ; \mathbf{x})=\frac{n}{\lambda}-\sum_{i=1}^{n}\left|x_{i}-\hat\mu\right|$$ gives you $$\hat\lambda$$ as the mean absolute deviation from the median as the only stationary point. You can then check the second derivative wrt $$\theta$$ to make sure it's a maximum. Or you can argue that the maximum must be there or at the endpoints $$\lambda=0$$ or $$\lambda=\infty$$, and it's not either of those.
• I have a couple of questions. 1) You are fixing a parameter and then solving the individual partial derivatives for when they are equal to zero. Since we want the MLE for $(\mu, \lambda)$, wouldn't one have to solve both partial derivatives for 0 simultaneously? By forming a system of 2 equations and solving both. Commented Oct 24, 2020 at 15:04
• 2) After finding the critical point/ MLE estimators $(\tilde{x}, n/\sum_{i=1}^{n} |x_i - \mu|)$ how can I go about proving that this critical point is the global maximum? Yes the function is concave, but how could I prove that? Commented Oct 24, 2020 at 15:06