Cross Validated Asked by harisf on December 1, 2021
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?
If possible, suppose there exists a UMP test $phi^*$ (say) of level $alpha$ for testing $H_0:theta=theta_0$ vs $H_1:thetane theta_0$. 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.
Answered by StubbornAtom on December 1, 2021
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