Hypothesis test to decide between two discrete distributions (classical inference)

Question

Let $\chi = \{1,2,3\}$ and consider two distributions:

$$P_0(x) = \begin{cases} 0.009 & \text{if $x = 1$} \\ 0.001 & \text{if $x = 2$} \\ 0.990 & \text{if $x = 3$} \end{cases}$$

$$P_1(x) = \begin{cases} 0.001 & \text{if $x = 1$} \\ 0.989 & \text{if $x = 2$} \\ 0.010 & \text{if $x = 3$} \end{cases}$$

Given that we have observed the value $x = 1$, I'm asked to solve following a classical methodology the following hypothesis test:

$$\begin{cases} H_0: & P_0(x) \\ H_1: & P_1(x) \end{cases}$$

for a size $\alpha = 0.01$. I'm also asked to compute the power of the test.

Answer 1

The idea is to follow Neyman-Pearson theorem with the following computations:

\begin{array}{|c|c|c|c|} \hline x & f(x|P_1)/f(x|P_0) \\ \hline 1 & 0.111... \\ \hline 2 & 989 \\ \hline 3 & 0.0101... \\ \hline \end{array}

So if I take $k = 0.1$ then I get a rejection region:

$\{x:f(x|P_1)/f(x|P_0) \ge k\} = \{1,2\}$

which has size $\alpha = P(X \in R|P_0) = P_0(1) + P_0(2) = 0.01$ as requested.

The power function is:

$$\eta(P) = P(X \in R|P) = \begin{cases} 0.01 & \text{if } P = P_0 \\ 0.990 & \text{if } P = P_1 \end{cases}$$

And given that $x = 1$ then the value is in the rejection region and the null hypothesis should be rejected.