# limiting distribution of mle (boundary case)

If $X_1,...,X_n$ follow a Pois($\mu$). Further suppose that parameter space is restricted to [$\mu_0$,$\infty$] where $\mu_0$ is the (unknown)true value of $\mu$. Find the limiting distribution(Under H0: $\mu=\mu_0$) of $\sqrt(n)(\hat \mu-\mu_0)$? and the limiting distribution of the likelihood ratio statistic i.e $R_n$?

Now I have deduced that $\hat \mu$ is max($\bar x_n,\mu_0)$. I do not know how to procedure in figuring out the limiting distribution. If someone can give me hint that would be great!

Edit: Hopefully this makes more sense. I used this post :Maximum likelihood of function of the mean on a restricted parameter space. as a guideline for finding the mle. Maybe since we are working under the null hypothesis that is why the problem is phrased this way? I somewhat see your point.

• These assumptions appear to make no sense: if you know the parameter space, then you know $\mu_0$, and therefore you don't need any data to estimate it and the optimal estimate always is $\hat\mu=\mu_0$, not $\max(\bar x_n, 0)$. (How could $\bar x_n$, assuming it's the mean of the data, possibly be negative anyway?) The language is confusing, too, because you seem to be using "$\mu_0$" as another term for "$\mu$". Could you clarify your situation for us? – whuber Jan 31 '17 at 19:19
• I think I have finally answered your concerns. As for the limiting distribution of say the LRT..my intuition tells me it is some sort of a mixture..with prob 1/2 we have chi-squared distribution with degree of freedom 1 and with prob half it is 0. ..but this is just my intuition..not able to prove it. – NoIdea Feb 1 '17 at 2:56

I'm going to try to see if I understand you correctly. In the future, it is good practice to define not only the null hypothesis but also the alternative hypothesis. Given your comments on the distribution of the likelihood ratio statistic being $$\frac{1}{2}\chi_0^2+\frac{1}{2}\chi_1^2$$ (or $$\frac{1}{2}\delta_0+\frac{1}{2}\chi_1^2$$ where $$\delta_0$$ is the Dirac delta function centered at $$0$$), I am going to assume that the hypotheses are not exhaustive and that the alternative hypothesis is $$H_1: \mu\in\left(\mu_0,\infty\right)$$. When you say that $$\mu_0$$ is unknown (do you mean $$\mu$$?) I'm going to assume that you mean $$\mu$$ is unknown under $$H_1$$ only. As you have described it, $$\mu$$ is known under $$H_0$$. So, $$H_0$$ is the point $$\mu_0$$ and $$H_1$$ is a half-line extending from $$\mu_0$$ to $$\infty$$. If this is incorrect then please let me know what you mean.

From the central limit theorem, the sample mean of the random variables converges in distribution as $$\bar{X}_n\xrightarrow{d}N\left(\mu_0,\frac{\mu_0}{n}\right)$$ under $$H_0$$.

Under $$H_0$$, $$\hat{\mu}=\mu_0$$ and thus $$\sqrt{n}\left(\hat{\mu}-\mu_0\right)=0$$. Asymptotically, $$P\left(\bar{X}_n\leq{}\mu_0\right)=\frac{1}{2}$$ and $$P\left(\bar{X}_n>\mu_0\right)=\frac{1}{2}$$. When $$\bar{X}_n\leq{}\mu_0$$, $$\bar{X}_n$$ is equally extreme under the two hypotheses and the likelihood ratio statitic ($$2\ln\left(L_{H_0\cup{}H_1}\right)-2\ln\left(L_{H_0}\right)$$) is $$0$$. This is because the MLE for $$\mu$$ under $$H_0$$ is $$\mu_0$$ and the supremum of $$L_{H_0\cup{}H_1}$$ also occurs at $$\mu_0$$.

When $$\bar{X}_n>\mu_0$$, $$\bar{X}_n$$ follows a truncated normal distribution, with the support being restricted to positive values. This is the normal distribution with variance $$\frac{\mu_0}{n}$$ truncated at the mean $$\mu_0$$. Thus, under $$H_1$$ the distribution of $$\hat{\mu}$$ is half of a Dirac delta function centered at $$\mu_0$$ and half of the truncated normal distribution, truncated at $$\mu_0$$. Then $$\hat{\mu}-\mu_0$$ has distribution of half of a Dirac delta function centered at $$0$$ and half of the truncated normal distribution, truncated at $$0$$. Finally, $$\sqrt{n}\left(\hat{\mu}-\mu_0\right)$$ has distribution of half of a Dirac delta function centered at $$0$$ and half of a truncated normal distribution, truncated at $$0$$ with variance $$\mu_0$$.

When $$\bar{X}_n$$ is normally distributed, the likelihood ratio statistic is the squared Mahalanobis distance from $$\bar{X}_n$$ to the parameter space for $$H_0$$ minus the squared Mahalanobis distance from $$\bar{X}_n$$ to the parameter space for $$H_0\cup{}H_1$$. Essentially, this means that $$\bar{X}_n$$ can now be treated as the standard normally distributed $$\tilde{X}_n\xrightarrow{d}N\left(0,1\right)$$. There is an affine transformation of the parameter space, which makes the mean the origin and the variance $$1$$. In general, the affine transformation changes the geometry of the parameter space, however points and half-lines remain as points and half-lines, so we need not worry in this case.

The distribution of the square of random variables from the (standard univariate) normal distribution truncated at the mean is the $$\chi_1^2$$ distribution, the same distribution as the square of random variables from the (standard univariate) normal distribution. Thus, the distribution of the likelihood ratio statistic is $$\frac{1}{2}\chi_0^2+\frac{1}{2}\chi_1^2$$ (or $$\frac{1}{2}\delta_0+\frac{1}{2}\chi_1^2$$, where $$\delta$$ is the Dirac delta function centered at $$0$$).

If you want more information, see my pre-print on ArXiv. If I understand you correctly, your example is similar to our Example 2.1, but with $$\theta=\frac{1}{2}$$ for $$H_0$$ and the equality reversed for $$H_1$$ (which makes no difference). If you want a formal proof of the distribution for your example then you can use Theorem 3.1, although if I understand you correctly your intuition is correct. Note that there are many other papers that have similar examples with the same distribution. These distributions come up in phylogenetics, as well as other fields. I think your example is also similar to case 5 on page 608 of this paper.

Hopefully this helps and isn't too late for you.