Consider the following problem.
Let $X_1,\dots, X_n$ be i.i.d. $\sim P$ taking values from $\mathbb{X}=\{-1,0,1\}$. Consider the hypothesis test $H_1: P=P_1$ vs $H_2: P=P_2$, where $P_1(-1)=1/2, P_1(0)=1/4, P_1(1)=1/4$ and $P_2(-1)=1/4, P_2(0)=1/4, P_2(1)=1/2$. Find the probability of error in accepting $H_2$ when actually $H_1$ is true while keeping the other probability of error $\le 1/2$ in the best hypothesis test.
First I will apply the Neyman-Pearson lemma to find the required probability of error. Let $$ A(T)=\left\{x\in\mathbb{X}:\frac{P_1(x)}{P_2(x)}\ge T\right\} $$ Then, when $T=1$, we have $A(1)=\{-1,0\}$ and $\text{Pr}(\text{Accepting } H_2|H_1 \text{ is true}) = P_1(A(1)^c) = 1/4$ while other probability of error is $P_2(A(1)) = 1/2$.
On the other hand, when $T=2$, $A(2) = \{-1\}$ and $\text{Pr}(\text{Accepting } H_2|H_1 \text{ is true}) = P_1(A(1)^c) = 1/2$ while other probability of error is $P_2(A(1)) = 1/4\le 1/2$.
So in both cases, the condition of other probability of error being $\le 1/2$ is satisfied. So, in a practical situation, what threshold one should take? Is there any result that talks about choosing the best threshold in a practical situation?
Also, in this situation, if we use Stein's lemma, we get $\text{Pr}(\text{Accepting } H_2|H_1 \text{ is true})\approx \exp (-nD(P_2\|P_1)) = \exp (-n/4)$. So, what is preferred, the probability of error that we calculated using Neyman-Pearson lemma or the one calculated using Stein's lemma?
Any help is greatly appreciated.
PS: I asked the same question in stat.stackexchange where it didn't seem to attract many users. So I'm asking here.