Example of an inconsistent Maximum likelihood estimator I'm reading a comment to a paper, and the author states that sometimes, even though the estimators (found by ML or maximum quasilikelihood) may not be consistent, the power of a likelihood ratio or quasi-likelihood ratio test can still converge to 1 as the number of data observed tends to infinity (test consistency). How and when does this happen? Do you know of some bibliography?
 A: Let $(X_n)$ be drawn iid from a Normal$(\mu, 1)$ distribution.  Consider the estimator
$$T(x_1, \ldots, x_n) = 1 + \bar{x} = 1 + \frac{1}{n}\sum_{i=1}^n x_n.$$
The distribution of $T(X_1,\ldots,X_n)=1+\bar{X}$ is Normal$(\mu+1, 1/\sqrt{n})$.  It converges to $\mu+1\ne \mu$, showing it is inconsistent.
In comparing a null hypothesis $\mu=\mu_0$ to a simple alternative, say $\mu=\mu_A$, the log likelihood ratio will be exactly the same as the LLR based on $\bar{X}$ instead of $T$.  (In effect, $T$ is useful for comparing the null hypothesis $\mu+1=\mu_0+1$ to the alternative hypothesis $\mu+1=\mu_A+1$.)  Since the test based on the mean has power converging to $1$ for any test size $\alpha\gt 0$ and any effect size, the power of the test using $T$ itself also converges to $1$.
A: [I think this might be an example of the kind of situation under discussion in your question.]
There are numerous examples of inconsistent ML estimators. Inconsistency is commonly seen with a variety of slightly complicated mixture problems and censoring problems.
[Consistency of a test is basically just that the power of the test for a (fixed) false hypothesis increases to one as $n\to\infty$.]
Radford Neal gives an example in his blog entry of 2008-08-09 Inconsistent Maximum Likelihood Estimation: An “Ordinary” Example. It involves estimation of the parameter $\theta$ in:
$$X\ |\ \theta\ \ \sim\ \ (1/2) N(0,1)\ +\ (1/2) N(\theta,\exp(-1/\theta^2)^2) $$
(Neal uses $t$ where I have $\theta$) where the ML estimate of $\theta$ will tend to $0$ as $n\to\infty$ (and indeed the likelihood can be far higher in a peak near 0 than at the true value for quite modest sample sizes). It is nevertheless the case that there's a peak near the true value $\theta$, it's just smaller than the one near 0.
Imagine now two cases relating to this situation:
a) performing a likelihood ratio test of $H_0: \theta=\theta_0$ against the alternative $H_1: \theta<\theta_0$;
b) performing a likelihood ratio test of $H_0: \theta=\theta_0$ against the alternative $H_1: \theta\neq\theta_0$.
In case (a), imagine that the true $\theta<\theta_0$ (so that the alternative is true and $0$ is the other side of the true $\theta$). Then in spite of the fact that the likelihood very close to 0 will exceed that at $\theta$, the likelihood at $\theta$ nevertheless exceeds the likelihood at $\theta_0$ even in small samples, and the ratio will continue to grow larger as $n\to\infty$, in such a way as to make the rejection probability in a likelihood ratio test go to 1.
Indeed, even in case (b), as long as $\theta_0$ is fixed and bounded away from $0$, it should also be the case that the likelihood ratio will grow in such a way as to make the rejection probability in a likelihood ratio test also approach 1.
So this would seem to be an example of inconsistent ML estimation, where the power of a LRT should nevertheless go to 1 (except when $\theta_0=0$). 
[Note that there's really nothing to this that's not already in whuber's answer, which I think is an exemplar of clarity, and is far simpler for understanding the difference between test consistency and consistency of an estimator. The fact that the inconsistent estimator in the specific example wasn't ML doesn't really matter as far as understanding that difference - and bringing in an inconsistent estimator that's specifically ML - as I have tried to do here - doesn't really alter the explanation in any substantive way. The only real point of the example here is that I think it addresses your concern about using an ML estimator.]
