Given the Laplace distribution parametrized by $\mu$ and $b$,
$f(x\mid \mu ,b)={\frac {1}{2b}}\exp \left(-{\frac {|x-\mu |}{b}}\right)\,\!$ ,
I know that $\hat \mu$, the maximum likelihood estimator of $\mu$, is the sample median. But how do we find the standard error of the MLE? I'm not sure how to proceed since the Fisher Information is undefined when $x = \hat \mu$, which, for my purposes, I must assume will happen.
I'm aware of the answers provided to the question here. What I'm looking for in general is a method to compute the standard errors when the log-likelihood is not twice differentiable.
Best Answer
Appeal to the Fisher information gives you an asymptotic approximation to the standard error. As whuber correctly points out in the comments, so long as this function is almost surely differentiable, that should be sufficient to obtain your result.
However, in the present case it is also possible to obtain the exact distribution of the MLE via first principles methods, without appeal to the asymptotic theory of MLEs. Your MLE is the median, so its distribution can be obtained using standard distributional results for order statistics, where the underlying distribution is continuous. To derive the result we will assume an odd number of observations for simplicity in dealing with the median.
In this case we have $n = 2k+1$ for some non-negative integer $k$ and the MLE is $\hat{\mu} = X_{(k+1)}$. We let $f$ and $F$ be the respective density and distribution functions for the (zero-mean) sample distribution. For IID values from a continuous distribution we then have:
$$\begin{equation} \begin{aligned} \mathbb{V}( \hat{\mu}) = \mathbb{E}( (\hat{\mu} - \mu)^2 ) &= \int \limits_{-\infty}^\infty t^2 f(t) \text{Beta}(F(t) | k+1, k+1) dt \\[6pt] &= \frac{(2k+1)!}{k! k!} \int \limits_{-\infty}^\infty t^2 f(t) F(t)^k (1-F(t))^k dt \\[6pt] &= \frac{(2k+1)!}{k! k!} \int \limits_{-\infty}^\infty t^2 f(t) (F(t)-F(t)^2)^k dt \\[6pt] &= \frac{(2k+1)!}{k! k!} \sum_{i=0}^k {k \choose i} (-1)^{k-i} \int \limits_{-\infty}^\infty t^2 f(t) F(t)^i F(t)^{2k-2i} dt \\[6pt] &= \frac{(2k+1)!}{k! k!} \sum_{i=0}^k {k \choose i} (-1)^{k-i} \int \limits_{-\infty}^\infty t^2 f(t) F(t)^{2k-i} dt. \\[6pt] \end{aligned} \end{equation}$$
For the Laplace distribution we have $F(t)(1-F(t)) = \tfrac{1}{2} \exp(- |t|/b) - \tfrac{1}{4} \exp(- 2|t|/b)$ and $f(t) = \tfrac{1}{2b} \exp(- |t|/b)$ so that:
$$\begin{equation} \begin{aligned} \mathbb{V}( \hat{\mu}) &= \frac{(2k+1)!}{k! k!} \sum_{i=0}^k {k \choose i} (-1)^{k-i} \int \limits_{-\infty}^\infty t^2 f(t) F(t)^{2k-i} dt \\[6pt] &= 2 \frac{(2k+1)!}{k! k!} \sum_{i=0}^k {k \choose i} (-1)^{k-i} \int \limits_0^\infty t^2 f(t) F(t)^{2k-i} dt \\[6pt] &= b^2 \frac{(2k+1)!}{k! k!} \sum_{i=0}^k {k \choose i} (-1)^{k-i} \Big( \frac{1}{2} \Big)^{2k-i} \int \limits_0^\infty t^2 \exp (-(2k-i+1)t) dt \\[6pt] &= b^2 \frac{(2k+1)!}{2^{k-1} k! k!} \sum_{i=0}^k {k \choose i} \frac{(-1 /2)^{k-i} }{(2k-i+1)^3}. \\[6pt] \end{aligned} \end{equation}$$
This gives us a closed form (finite sum) expression for the variance of the MLE estimator (i.e., the sample median). (It is important to be careful of rounding error when evaluating the variance expression, since it involves a product of very large and very small terms.) A related expression has been investigated in some excellent analysis by Claude Liebovici in this related question, which shows that $\mathbb{V}( \hat{\mu}) \approx b^2 / n$ as $n \rightarrow \infty$. This limiting result accords with the asymptotic theory of the sample median, which has $\mathbb{V}(\hat{\mu}) \rightarrow 1 / (4n f(\mu)^2) = b^2 / n$.