An Application of the Neyman-Pearson Lemma.

Question

Problem: Using the Neyman-Pearson Lemma, determine the most powerful test of size $ 5 \% $.

I know the Neyman-Pearson Lemma says that the test with the critical region $$ \left\{ x \in \{ 1,2,3,4 \} ~ \Bigg| ~ \frac{L(x \mid \theta = 0)}{L(x \mid \theta = 1)} \leq A \right\}, $$ where $ A $ satisfies $$ \mathbf{Pr} \left( \frac{L(X \mid \theta = 0)}{L(X \mid \theta = 1)} \leq A ~ \Bigg| ~ H_{0} \right) = 0.05, $$ is the most powerful test of size $ 5 \% $. However, I’m not sure what the likelihood functions are, as I’m more used to seeing them given by a formula.

Thanks.

Answer 1

According to the data, we have $$ L(x \mid \theta = 0) = \begin{cases} 0.02 & \text{if $ x = 1 $}; \\ 0.02 & \text{if $ x = 2 $}; \\ 0.03 & \text{if $ x = 3 $}; \\ 0.93 & \text{if $ x = 4 $}, \end{cases} \\ L(x \mid \theta = 1) = \begin{cases} 0.10 & \text{if $ x = 1 $}; \\ 0.20 & \text{if $ x = 2 $}; \\ 0.30 & \text{if $ x = 3 $}; \\ 0.40 & \text{if $ x = 4 $}. \end{cases} $$ Therefore, $$ \Lambda(x) \stackrel{\text{def}}{=} \frac{L(x \mid \theta = 0)}{L(x \mid \theta = 1)} = \begin{cases} 0.2 & \text{if $ x = 1 $}; \\ 0.1 & \text{if $ x = 2 $}; \\ 0.1 & \text{if $ x = 3 $}; \\ 2.325 & \text{if $ x = 4 $}. \end{cases} $$ The chosen size is $ 0.05 $; the Neyman-Pearson Lemma says that in order to find the most powerful test of this size, we need to find an $ \eta \in \mathbb{R} $ such that $$ \mathbf{Pr}(\Lambda(X) \leq \eta \mid \theta = 0) = 0.05. $$ The choice of $ \eta = 0.1 $ works, so the most powerful test of size $ 0.05 $ is:

Reject $ H_{0} $ in favor of $ H_{1} $ when the observed value of $ \Lambda(X) $ is $ \leq 0.1 $, or equivalently, when the observed value of $ X $ is either $ 2 $ or $ 3 $.