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 
\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?
 A: 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.
