Exercise: Let $X$ be a random variable that takes values in $\{-1,0,1\}$. We want to test $$H_0: Pr(X = -1) = Pr(X = 1) = Pr(X = 0) = 1/3$$ versus $$H_1: Pr(X = -1) = Pr(X = 1) = 1/4, \, Pr(X = 0) = 1/2.$$
Show that the uniformly most powerful (UMP) test of size $\alpha = 1/3$ based on $X$ leads to rejecting the null hypothesis when $X = 0$, and compute the power of the test.
What I've tried: I first need to find the critical region $C$ for the UMP test based on the test statistic $T(X) = X$. This can be done in two steps:
1) Choose $C$ such that under $H_0$, $P_\theta(T(X)\in C)\leq \alpha$ (this ensures that the probability of a type I error is no greater than $\alpha$). I choose $C = \{0\}$.
2) Conditional on 1), maximise $P(T(X)\in C)$ under $H_1$ (this ensures minimising the probability a type II error). With my choice of $C = \{0\}$, $P(T(X)\in C)$ is already maximised.
So I have a test $(T,C)$ with $T(X) = X$ and $C = \{0\}$. Since $X = 0$, $X\in C$ and $H_0$ is rejected. The power of the test is equal to the probability of rejecting $H_0$ when $H_1$ is correct. Hence the power of the test is equal to $1/2$.
Question: Is my answer correct? If not; what am I doing wrong? The thing that I'm most uncertain about in my approach is whether or not the region $C$ that I derived is the critical region of a UMP test.
Thanks in advance!
Looks right. I'm not $100\%$ percent about the technical details. But you can check the answer. In order to keep the $\alpha$, i.e., the probability of erroneously rejecting $H_0$ to be equal $1/3$, you must choose one of the values $\{-1,0,1\}$ as a rejection region. Now, you can simply check what would be the power for every value, i.e., the probability of getting the value under $H_1$, that is $\{1/4, 1/2, 1/4\}$ respectively, so the maximal power $(1/2)$ is attained for $C=\{0\}$.