For maximum likelihood estimation we need to set the first derivative of the log-likelihood function equal to $\mathbf{0}$.

The negative expected value of the Hessian matrix (second derivative) is then called the Fisher information matrix.

Is there anything inherent to the definition of a log-likelihood (probability density) function, that guarantees twice differentiabiliy of the log-likelihood? If not, what conditions must I impose to guarantee it?


1 Answer 1


In short: no. Note that, to maximise the log likelihood we frequently use differentiation, but in fact to truly maximise a function we need to consider several types of points

  • Stationary/turning points (when $\frac{\partial \ell}{\partial \theta} = 0$)
  • Singular points (e.g. where the function cannot be differentiated)
  • End points - this only applies on a finite interval $[a,b]$, possibly with one of $a$ or $b$ being infinite in modulus

Of course, that is provided that the parameter of interest is actually continuous.

Let's consider the Laplace distribution with density

$$p(x \mid \mu, b) = \frac{1}{2b} \exp \left\{ -\frac{|x - \mu|}{b} \right\}$$

Then the log-likelihood is, given a sample $\mathbf{x}$ of size $n$

$$ \ell(\mu, b \mid \mathbf{x} ) = -n \log (2b) - \sum_{i=1}^n \frac{|x_i - \mu|}{b}$$

It can be shown that $\hat{b} = \frac{1}{n} \sum_{i=1}^n |x_i - \hat{\mu}|$. The difficult bit is finding $\hat{\mu}$.

Now if we differentiate w.r.t. $\mu$ then we need to differentiate $|x_i - \mu|$. If $\mu \neq x_i$ for any $x_i$ then $\frac{\partial \ell}{\partial \mu} = - \sum_{i=1}^n\text{sign}(x_i - \mu)$ which can be zero only if $n$ is even (but still might be non zero!). At any $\mu \in \mathbf{x}$ the gradient does not exist!.

Now for any $\mu$ that is equal to one of the $x_i$, the log likelihood will not be differentiable at these points. Now assume $n$ is odd, it can be shown that $\hat{\mu}$ is actually the sample median. The sample median will be one of the $x_i$ (the middle $x_i$ when the $x_i$ are in order). Therefore, the m.l.e. is at one of the non-differentiable points - a singular point!

How can we guarantee that the log-likelihood is differentiable? I don't think we can actually force this to be true unless we choose a log-likelihood that is twice differentiable. I'd view this as a modelling choice or an assumption. Rather than something we can guarantee. Other assumptions might imply a twice differentiable log-likelihood, but in general I can't see how we would end up with such a log-likelihood.

  • 2
    $\begingroup$ Thank you very much. I'll just impose it in my Lemma then. No big deal, since all pdfs I consider are actually twice differentiable. $\endgroup$
    – stollenm
    Jun 11, 2021 at 9:18
  • $\begingroup$ Good answer, and in all books I’ve read, the twice differentiable log likelihood is among the regularity conditions (I.e. assumptions) of MLE. In which case it is necessary but not sufficient for the search for an MLE, and not something that can be guaranteed $\endgroup$
    – Rick Hass
    Jun 17, 2021 at 18:14

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