Consider a population with three kinds of individuals labeled $1, 2$, and $3$ occuring in the Hardy–Weinberg proportions $f(1,\theta)=\theta^2,f(2,\theta)=2\theta(1−\theta),f(3,\theta)=(1−\theta)^2$. For a sample $X_1, . . . , X_n$ from this population, let $N_1$, $N_2$, and $N_3$ denote the number of $X_j$ equal to $1$, $2$, and $3$, respectively. Let $0<\theta_0<\theta_1 <1$. We know that the likelihood ratio is an increasing function of $2N_1+N_2$. (Likelihood ratio of Hardy–Weinberg proportions)

What I want to conclude now is that if $c>0$ and $\alpha\in(0,1)$ satisfy $P_{\theta_0}[2N_1+N_2\geq c]=\alpha$, then the test that rejects $H$ if, and only if, $2N_1 + N_2\geq c$ is most powerful (MP) for testing $H : \theta = \theta_0$ versus $K : \theta = \theta_1$.

Thanks for your help in advance!


1 Answer 1


The Neyman–Pearson Lemma says the most powerful test for two point hypotheses is always the likelihood ratio test. You have a test comparing two point hypotheses, and you are looking at the likelihood ratio test, so it's the most powerful test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.