Finding best critical region

Question

X is a single observation from p(X). Test Ho: Poisson(1) vs H1: geometric with p= 1/2.

I used the Neyman Pearson formula, f(x|H1)/f(x|Ho). Is the critical region X!/2^(X+1)>c? If so, how do I prove that the test is unbiased?

Thank you

Answer 1

Your test should be the likelihood ratio test if you're using a Bayesian or Minimax or Neyman-Pearson formulation: $\frac{p_1(x)}{p_0(x)} > c$ declares $H_1$, $<c$ declares $H_0$ and equality can optionally be randomized between the two hypothesis with probability $\gamma$ (declare $H_1$) if you need to meet some false alarm constraint or are designing the test to be best in a minimax setting or something. In this case, $p_1(x)$ will be the pmf of a Poission(1) distribution, and $p_0(x)$ will be the pmf of a geometric distribution. Your test statistic will be thus $\frac{2^x}{x!}>c$ declare $H_1$, $=$ declare $H_0$ with probability $(1-\gamma)$ and $H_1, otherwise, $

As for the test being unbiased, you want to see that the probability of declaring $H_1$ under $H_1$ is higher than the probability of declaring $H_1$ under $H_0$: $\gamma P_1 [\frac{p_1(x)}{p_0(x)} = c] + P_1 [\frac{p_1(x)}{p_0(x)} > c] \geq (1-\gamma) P_0 [\frac{p_1(x)}{p_0(x)} = c] + P_0 [\frac{p_1(x)}{p_0(x)} > c]$.

That is, the power under $H_1$ (LHS) is bigger than the false alarm probability (RHS).