As I was studying for my exam I came across the following problem:
Let $X$ be a discrete random variable with probability mass function (pmf) given by $f(x) = P(X = x)$. Further, consider the situation in which $f(x)$ can take on one of two forms which are given in the following table:
Determine the uniformly most powerful (UMP) test for $H_0 : f = f_0$ vs. $H_1 : f = f_1$ at the $\alpha = 0.3$ level.
I know that to find the UMP of a test you must start by setting up the likelihood ratio. However, I am not sure how to do so in this case. Any suggestions are appreciated.
After receiving a hint from @StubbornAtom and doing some more reading, here my attempt to solve this problem. Since our hypothesis test is a simple hypothesis test we can use Neymann-Pearson Lemma, which states the LRT is the UMP of such hypothesis tests. We let $\lambda(x)=\frac{f_1(x)}{f_0(x)}$ and we reject $H_0$ if $\lambda(x)>k$.
In our case the LRT is as follows: $\frac{f_1(1)}{f_0(1)}=3/2$, $\frac{f_1(2)}{f_0(2)}=1/3$, $\frac{f_1(3)}{f_0(3)}=3$, $\frac{f_1(5)}{f_0(5)}=2/3$, $\frac{f_1(7)}{f_0(7)}=1$. From here we see that the greatest value of the LRT is $3$ and it happens when $x=3$. Therefore, our critical region is $\{x=3\}$. Thus, we reject $H_0$ if $X = 3$.
The significance of this test $\alpha=P(\text{reject } H_0\mid H_0)=P(X=3\mid f_0)=0.1$. But we need it to be 0.3
I would appreciate a thumbs up or down for my attempt to solve this problem. Thank you!