Question

Show that the likelihood ratio principle leads to the same test when testing a simple hypothesis \(H_{0}\) against an alternative simple hypothesis \(H_{1}\), as that given by the Neyman-Pearson theorem. Note that there are only two points in \(\Omega\).

Short answer

The exercise has been solved by establishing the hypotheses based on the given conditions, calculating the likelihood ratios, and comparing the principles of the likelihood ratio and Neyman-Pearson theorem. They result in the same conclusion, proving the statement true.

Step-by-step solution

Set Up the Hypotheses

Let's start by denoting the two points in the sample space \(\Omega\) as \(\omega_1\) and \(\omega_2\). Then, the hypotheses can be presented as \(H_{0}: P(\omega_1) = p_0\) and \(H_{1}: P(\omega_1) = p_1\).

Define the Likelihood Ratio

The likelihood ratio is defined as \(LR = \frac{L(p_1)}{L(p_0)}\), where \(L(p)\) is the likelihood of point \(\omega_1\). For an observed sample \(x\), the likelihoods can be represented as \(L(p_0) = P(X = x|H_{0})\) and \(L(p_1) = P(X = x|H_{1})\). Since the sample space has only two points, these probabilities can be easily evaluated.

Calculation of Likelihood Ratio

For the sample \(x\), the likelihood ratio \(LR = \frac{P(X=x | H_{1})}{P(X=x | H_{0})}\). As per the Neyman-Pearson theorem, we reject \(H_{0}\) if \(LR > k\) for some critically chosen \(k\) to maintain the desired significance level.

Neyman-Pearson Principle

According to the Neyman-Pearson principle, for testing \(H_{0}\) against \(H_{1}\), the most powerful test at the level \(\alpha\) chooses to reject \(H_{0}\) if \(\frac{f(x ; p_1)}{f(x ; p_0)} > k\), which is equivalent to the likelihood ratio. This proves that both Neyman-Pearson and likelihood ratio principles lead to the same conclusion.