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Suppose we have the following general setup where we wish to test the hypothesis, $$ H_0 : \theta\in\Theta_0 ~~~~~~~~~~~~~~~~~~~~~~~~~H_1 : \theta\not\in\Theta_0. $$ It seems to be pretty well established that under some regularity conditions that a likelihood ratio test, $$ \Lambda\left(\mathbf{X}\right) = \frac{\sup_{\theta\in\Theta_0} \mathcal{L}\left(\theta;\mathbf{X}\right)}{\sup_{\theta\in\Theta} \mathcal{L}\left(\theta;\mathbf{X}\right)} $$ has the property that $-2\log\Lambda\left(\mathbf{X}\right)\sim\chi^2_k$, where $k = \dim\left(\Theta\right) - \dim\left(\Theta_0\right)$ asymptotically. I don't understand why $\Lambda\left(\mathbf{X}\right) \to 1$ in the limit of infinite data when the null hypothesis is true.

Allow me to explain. We know that maximum likelihood is asymptotically consistent such that $$ \lim_{n\to\infty} \Pr\left[\theta = \hat{\theta}\right] = 1. $$ Therefore, if the null is true (i.e. $\theta\in\Theta_0$) it should also be the case that with probability one $\hat{\theta} \in\Theta_0$. By definition, $\hat{\theta}$ maximizes the denominator of the likelihood ratio, but because $\hat{\theta}\in\Theta_0$, it is clear that $\hat{\theta}$ also maximizes the numerator. Additionally, because they are maximized by the same point, the numerator and denominator are equal so that, $$ \Lambda\left(\mathbf{X}\right) = \frac{\mathcal{L}\left(\hat{\theta};\mathbf{X}\right)}{\mathcal{L}\left(\hat{\theta};\mathbf{X}\right)} = 1. $$ Moreover, $-2\log\Lambda\left(\mathbf{X}\right)=0$.

I know there is a problem with this reasoning, but I don't know what it is. It seems that the correction must have to do with the additional $k$ free parameters that were assumed to be present in the unrestricted parameter space $\Theta$.

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Let's take a counter-example of sampling a normal distribution $n$ times, where it has known variance $1$

Suppose the null hypothesis is that $\mu=0$ and this is in fact true

We will not be at all surprised when the mean of the sample is at least one standard error away from the population mean, i.e. $\pm\frac{1}{\sqrt{n}}$, and we expect results more extreme than this over $30\%$ of the time, no matter how large $n$ is

But in this or more extreme cases your likelihood ratio will be no more than $e^{-1/2}\lt 0.607$, no matter how large $n$ is. Thus the likelihood ratio will not converge to $1$ as $n$ increases

A more general answer to your original question is that small random effects consistent with the null hypothesis can still have substantial effects on the likelihood ratio when $n$ is large

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  • $\begingroup$ This was very helpful. For educational purposes I decided to simulate the conditions you describe. Sure enough, the log likelihood ratio has precisely the chi-squared distribution that it should (with one degree of freedom). BUT! I also decided to modify your null hypothesis to mu <= 0. When I do this, the likelihood ratio statistic DOES get crushed to one under the null with large sample sizes. This makes me think there is something going on when dim(Theta_0) = dim(Theta). Do you know why this happens? $\endgroup$ – user1936768 Sep 15 '16 at 15:53
  • $\begingroup$ If you widen your null hypothesis that way, you should be getting your likelihood ratio equal to $1$ half the time (when the sample mean is zero or negative), no matter what the value of $n$ $\endgroup$ – Henry Sep 15 '16 at 16:31
  • $\begingroup$ That's true when mu = 0 and the null hypothesis space is mu <= 0. But when the true mu is less than 0, then for large enough sample sizes I claim that the LR will be one almost surely. $\endgroup$ – user1936768 Sep 15 '16 at 16:43
  • $\begingroup$ OK, you are correct when in fact $\mu =k$ for some $k \lt 0$. So to get there, you are in effect having a null hypothesis that $\mu \lt 0$ with probability $1$ rather than $\mu \le 0$. My guess is that most null hypotheses do not look like that. $\endgroup$ – Henry Sep 15 '16 at 17:03
  • $\begingroup$ I believe you could relate this phenomenon to the asymptotic normality of MLE so that if the null hypothesis is true, and for large enough sample sizes, the MLE will almost surely be contained in an ellipsoid centered around the true value of the parameter. Therefore, if the true parameter is on the strict interior of the null hypothesis set, then the MLE will almost certainly be contained in a small ellipsoid that is also fully contained within the null hypothesis space. Hence, a likelihood ratio statistic of one. $\endgroup$ – user1936768 Sep 15 '16 at 17:14

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