This question is Find the LRT of a simple $H_0$ versus a simple $H_1$: Is this test equivalent to the one obtained from the Neyman-Pearson Lemma?

In fact, this question is not difficult. But I feel confused on one thing. My attempt:

LRT is, according to definition 8.2.1,

$$\lambda(\textbf{x}) =\frac{L(\theta_0|\textbf{x})}{L(\theta_1|\textbf{x})}$$, if the numerator is less than the denominator. otherwise $\lambda(\textbf{x})$ will be 1. This LRT has a rejection region of the form {$\textbf{x}:\lambda(\textbf{x})\leq c$}, where c is any number satisfying [0,1].

The Neyman-Pearson Lemma says the rejection region R is

$$f(\textbf{x}|\theta_1)>kf(\textbf{x}|\theta_0)$$. This is equivalent to saying, in our simple case,

$$\frac{L(\theta_0|\textbf{x})}{L(\theta_1|\textbf{x})}\leq\frac{1}{k}$$. Then I get my conclusion: when kc=1, they are the same.

However, the solution said "But if c ≥ 1 or k ≤ 1, the tests will not be the same." I don't understand this sentence. I know in LRT, the ratio is bounded above by 1. But I think it's okay if we have c bigger than 1. I don't know why the test will not be the same if c ≥ 1 or k ≤ 1.

The solution is here: enter image description here


1 Answer 1


If $\ c\ge1\ $, then the LRT will always reject the hypothesis $\ H_0\ $ in favour of $\ H_1\ $ because it is always true that $\ \lambda(\mathbf{x})\le1\le c\ $. However, there may nevertheless well be a positive probability that $\ \frac{L\left(\theta_1|\mathbf{x}\right)}{L\left(\theta_0|\mathbf{x}\right)}<\frac{1}{c}=k\ $, in which case the Neyman-Pearson test will not always reject $\ H_0\ $. Therefore the tests aren't equivalent in this case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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