Ask George Casella question 8.24

Question

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:

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.