Let $X$ be a random variable that takes values in $\{-1,0,1\}$. We want to test
$H_{0}:Pr(X = -1|\Theta = 0) = Pr(X = 0|\Theta = 0) = Pr(X = 1| \Theta = 0) = 1/3$
versus
$H_{1}=Pr(X = -1 |\Theta = 1) = Pr(X = 1|\Theta = 1) = 1/4,\,Pr(X = 0|\Theta = 1) = 1/2.$
Exercise: Calculate the power of this test and show that the UMP test of size $\alpha=1/3$ based on $X$ leads to rejecting the null hypothesis when $X=0$.
My attempt: I'm new to this subject and am not sure what a general approach is to these types of questions so any help regarding that would also be much appreciated. Anyway, I figured it would be good to use the Neyman-Pearson Lemma about uniformly most powerful tests but that was about as far as I got.
Thanks!
Calculate likelihood ratio $\frac{L_0(X)}{L_1(X)}$ for $X=-1,0,1$ separately: $$ \dfrac{L_0(-1)}{L_1(-1)}=\dfrac{L_0(1)}{L_1(1)}=\dfrac{1/3}{1/4}=\frac43, $$ $$ \dfrac{L_0(0)}{L_1(0)}=\dfrac{1/3}{1/2}=\frac23. $$ Likelihood ratio for $X=0$ is smaller than for $X=\pm 1$. UMP test rejects null hypothesis for small values of likelihood ratio, so we can simply try to find size of a test $$ \delta(X) = \begin{cases}H_0, & \frac{L_0(X)}{L_1(X)} >\frac23,\cr H_1, & \frac{L_0(X)}{L_1(X)} \leq \frac23 \end{cases} = \begin{cases}H_0, & X = \pm 1,\cr H_1, & X=0\end{cases} $$ If it is equal to $\frac13$, we construct the UMP test of size $\alpha=\frac13$. $$\alpha=\mathbb P_{H_0}(\delta=H_1)=\mathbb P_{H_0}(X=0)=\frac13$$ as required.