Let $X_1,...,X_n$~ $n(\theta,\sigma_0^2)$, where $\sigma_0^2$ is known. Given the hypothesis $H_0: \theta = \theta_0$ vs $H_1: \theta \neq \theta_0$, I know that a LRT has rejection region \begin{equation} \lvert \bar{X} - \theta_0 \rvert \geq \frac{\sigma_0}{\sqrt{n}}z_{\alpha/2}, \end{equation} since $\bar{X}$ is a sufficient statistic for $\theta$. How can I show that this is not a UMP test?

So far I'm thinking that you can divide the original hypothesis test into two different tests, \begin{align} H_0: \theta \geq \theta_0 \quad &\text{vs} \quad H_1: \theta < \theta_0 \quad \text{(test 1)}\\ H_0: \theta \leq \theta_0 \quad &\text{vs} \quad H_1: \theta > \theta_0 \quad \text{(test 2)} \end{align} with their respective rejection regions given by \begin{align} \bar{X} &\leq \theta_0 - \frac{\sigma_0}{\sqrt{n}}z_{\alpha/2} \quad \text{(rejection region for test 1)}\\ \bar{X} &\geq \theta_0 + \frac{\sigma_0}{\sqrt{n}}z_{\alpha/2} \quad \text{(rejection region for test 2)}. \end{align} If you fix $\theta_1 < \theta_0$ and $\theta_2 > \theta_0$, it is possible to show that $\beta_2(\theta_2) > \beta_1(\theta_1)$, where $\beta_i$ is the power function for test $i$. Is this result contradictory to the existence of a UMP test for the original hypothesis test? If so, how?

  • 1
    $\begingroup$ It cannot be a UMP test since it is concerned with two-sided alternatives. Try to see what happens to the rejection region for $\theta>\theta_0$ and $\theta<\theta_0$. The best critical regions are different! $\endgroup$ – JohnK Nov 11 '15 at 14:37
  • $\begingroup$ Is it a general result that any two-sided hypothesis test cannot be a UMP test? I would appreciate any counter examples if that is not the case. $\endgroup$ – harisf Nov 11 '15 at 14:45
  • 1
    $\begingroup$ That claim seems reasonable but since it's been some years that I studied these things, I don't want to mislead you. Do you understand why in the present case, this is not a UMP test though? $\endgroup$ – JohnK Nov 11 '15 at 14:47
  • $\begingroup$ I think so. Is it because according to the Neyman-Pearson lemma a UMP test has a fixed critical region for all $\theta$ specified in $H_1$, in this case for every $\theta \neq \theta_0$? $\endgroup$ – harisf Nov 11 '15 at 14:55

If possible, suppose there exists a UMP test $\phi^*$ (say) of level $\alpha$ for testing $H_0:\theta=\theta_0$ vs $H_1:\theta\ne \theta_0$. Then $\phi^*$ will also be UMP level $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1':\theta>\theta_0$ as well as $H_1'':\theta<\theta_0$.

But a UMP level $\alpha$ test for $(H_0,H_1')$ is

$$ \phi_1(\mathbf X)=\begin{cases} 1 &,\text{ if }\frac{\sqrt n(\overline X-\theta_0)}{\sigma_0}>z_{\alpha} \\ 0 &,\text{ otherwise } \end{cases} $$

And that for $(H_0,H_1'')$ is

$$ \phi_2(\mathbf X)=\begin{cases} 1 &,\text{ if }\frac{\sqrt n(\overline X-\theta_0)}{\sigma_0}<-z_{\alpha} \\ 0 &,\text{ otherwise } \end{cases} $$

So the test functions $\phi^*$ and $\phi_1$ should coincide on the sets where $\phi_1$ is zero or one. Same goes for $\phi^*$ and $\phi_2$. Now suppose we observed a data $\mathbf X$ such that the observed value of $\frac{\sqrt n(\overline X-\theta_0)} {\sigma_0}$ exceeds $z_{\alpha}$. Then for such $\mathbf X$, we must have $\phi_1(\mathbf X)=1$ and $\phi_2(\mathbf X)=0$. This means that on the part of the sample space where $\frac{\sqrt n(\overline X-\theta_0)} {\sigma_0}>z_{\alpha}$, the test $\phi^*$ fails to coincide with both $\phi_1$ and $\phi_2$. Hence the contradiction.

This is pretty much the idea behind the nonexistence of a UMP test for $(H_0,H_1)$. Hence the LRT is not a UMP test; however it is a UMP unbiased (UMPU) test.

  • $\begingroup$ Would this same argument work in the case where we test $H:\theta_1 \leq \theta\leq \theta_2$ against the alternative $H_1:\theta>\theta_2$ or $\theta<\theta_1$? We could let $\theta_0\in[\theta_1,\theta_2]$ and consider $\phi^*$ to be UMP ($\alpha$) for $H_0:\theta=\theta_0$ against $\theta>\theta_2$ or $\theta<\theta_1$ $\endgroup$ – Remy Mar 30 at 22:13
  • $\begingroup$ @Remy I am not sure. $\endgroup$ – StubbornAtom Mar 31 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.