# Proving a hypothesis test is not a UMP test

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 $$\lvert \bar{X} - \theta_0 \rvert \geq \frac{\sigma_0}{\sqrt{n}}z_{\alpha/2},$$ 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?

• 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! Commented Nov 11, 2015 at 14:37
• 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. Commented Nov 11, 2015 at 14:45
• 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? Commented Nov 11, 2015 at 14:47
• 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 $\theta \neq \theta_0$? Commented Nov 11, 2015 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.

• 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 $\theta<\theta_1$
– Remy
Commented Mar 30, 2021 at 22:13
• @Remy I am not sure. Commented Mar 31, 2021 at 6:12