1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Browse other questions tagged or ask your own question.