Consider translated Weibull distribution with probability density function:

$$ f(x ; k, \lambda, \theta) = \frac{k}{\lambda} \left( \frac{x-\theta}{\lambda} \right)^{k-1} \exp\left( - \left(\frac{x-\theta}{\lambda} \right)^k \right) \chi_{x \ge \theta}(x) $$

Consider $ \mathbb{E}\left( -\partial_\theta^2 \log f(x; k, \lambda, \theta) \right) = \mathbb{E}\left( \frac{k-1}{(x-\theta)^2 } \left(1 + \left( \frac{x-\theta}{\lambda} \right)^k \right) \right)$. Notice that the expectation is only convergent for $k > 2$.

Question: What is the significance, or interpretation, of infinite matrix element of the Fisher information matrix ? Does it mean that the maximum likelihood estimator for $\theta$ parameter is not asymptotically normal ?

Thank you for any light shed on the subject.


1 Answer 1


A rough answer is that the MLE of a location parameter of distribution with non-translation-invariant support can converge faster than $O(1/\sqrt{n})$. In this case, the divergent entry in the information matrix can be informally interpreted as "asymptotic zero variance" of $\sqrt{n}(\hat{\theta} - \theta)$.

The simplest example is the $\text{Uniform}(\theta-1,\theta)$ distribution. The maximum likelihood estimator is simply $\hat{\theta} = \max\{x_1,...,x_n\}$. Setting $\theta=1$ without loss of generality, we see that the density function of $y=\sqrt{n}\hat{\theta}$ is given by

$$f(y) = I(y \in [0,\sqrt{n}])n^{1-n/2} y^{n-1}$$

Using the following code we plot the density of $y$ for $n = 10,20,30,...,100$:

f <- function(n,y) { sqrt(n)^(2-n)  * (y)^(n-1) * (y/sqrt(n) < 1) * (y > 0) }
y <- (0:1000)/100
plot(NA, xlim=c(0,10), ylim=c(0,10))
for (n in 10*(1:10)) { lines(y,f(n,y),col=hsv(h=n/100,v=0.5)) }

density of MLE of uniform location parameter


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