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?

  • $\begingroup$ What are LR & QLR? $\endgroup$ Commented Sep 26, 2014 at 14:42
  • $\begingroup$ Likelihood ratio and quasilikelihood ratio test ;) $\endgroup$ Commented Sep 26, 2014 at 19:17
  • $\begingroup$ Power should go to 1 everywhere except at one point. What you won't have is the nominal type 1 error rate. $\endgroup$
    – Glen_b
    Commented Sep 26, 2014 at 22:59
  • $\begingroup$ @Glen_b, could you please elaborate more on your comment? Thanks ;) $\endgroup$ Commented Sep 27, 2014 at 21:02
  • $\begingroup$ @Glen_b, unfortunately no, and wiki doesn't seem to have an entry on it... $\endgroup$ Commented Sep 27, 2014 at 22:42

2 Answers 2


[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.]

  • $\begingroup$ Thanks Glen for your answer.I still have one question though. The thing is that usually in the proof for the limiting distribution of the LRT to be chi-squared, it is assumed that the ML estimators are consistent. In your case, how would you justify that a growing likelihood ratio will make the rejection probability go to 1, when the limiting distribution is unknown? Or is it known? $\endgroup$ Commented Oct 1, 2014 at 8:06
  • $\begingroup$ All you need have for the likelihood ratio test statistic to grow without bound is that the likelihood at the $\theta$ value in the numerator to grow more quickly than the one in the denominator. My understanding from the linked discussion was that Neal was implying it did, but I've made no actual check of the details. I don't think there's any good reason to assert the test would have the chi-square distribution though; my assumption from what little information you gave in the question was that the test described was being done as if it was asymptotically chi-square, but ... (ctd) $\endgroup$
    – Glen_b
    Commented Oct 1, 2014 at 23:30
  • $\begingroup$ (ctd)... you'd have to ask the author of the comment you described whether that was what they meant. $\endgroup$
    – Glen_b
    Commented Oct 1, 2014 at 23:32
  • $\begingroup$ Actually, what I said is not quite right, since it's possible for the numerator to grow faster than the denominator but the ratio not to grow without bound (in the sense that the ratio of the two might grow but be bounded). I should have said something like "sufficiently faster". $\endgroup$
    – Glen_b
    Commented Oct 2, 2014 at 9:53

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 $1$.

  • $\begingroup$ thank your for your interest in this question. How can we in a more general setting, be sure of the consistency of the test? I was looking for a more general answer, and not a specific case. And also some bibliography if available. Thanks ;) $\endgroup$ Commented Sep 27, 2014 at 20:58
  • 1
    $\begingroup$ Also, I maybe wrong, but the estimator T doesn't seem to be the ML estimator. The question is «when do we have test consistency, when the ML estimators, or the maximum quasilikelihood estimators are not consistent?» $\endgroup$ Commented Sep 27, 2014 at 21:05
  • $\begingroup$ I edited the question, since it might not had clearly what I wanted. Sorry ;) $\endgroup$ Commented Sep 27, 2014 at 23:19

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