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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

Expert verified
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

01

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 \(H_{1}: P(\omega_1) = p_1\).
02

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.
03

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.
04

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.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random samples from two normal distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma^{2}\right)\), respectively, where \(\sigma^{2}\) is the common but unknown variance. (a) Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0\) against all alternatives. (b) Rewrite \(\Lambda\) so that it is a function of a statistic \(Z\) which has a well-known distribution. (c) Give the distribution of \(Z\) under both null and alternative hypotheses.

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=\theta(1-x)^{\theta-1}, 00\) (a) Find the form of the uniformly most powerful test of \(H_{0}: \theta=1\) against \(H_{1}: \theta>1\) (b) What is the likelihood ratio \(\Lambda\) for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta \neq 1 ?\)

Let \(X_{1}, X_{2}, \ldots, X_{10}\) denote a random sample of size 10 from a Poisson distribution with mean \(\theta .\) Show that the critical region \(C\) defined by \(\sum_{1}^{10} x_{i} \geq 3\) is a best critical region for testing \(H_{0}: \theta=0.1\) against \(H_{1}: \theta=0.5 .\) Determine, for this test, the significance level \(\alpha\) and the power at \(\theta=0.5\).

Illustrative Example \(8.2 .1\) of this section dealt with a random sample of size \(n=2\) from a gamma distribution with \(\alpha=1, \beta=\theta .\) Thus the mgf of the distribution is \((1-\theta t)^{-1}, t<1 / \theta, \theta \geq 2 .\) Let \(Z=X_{1}+X_{2}\). Show that \(Z\) has a gamma distribution with \(\alpha=2, \beta=\theta .\) Express the power function \(\gamma(\theta)\) of Example \(8.2 .1\) in terms of a single integral. Generalize this for a random sample of size \(n .\)

Consider a normal distribution of the form \(N(\theta, 4)\). The simple hypothesis \(H_{0}: \theta=0\) is rejected, and the alternative composite hypothesis \(H_{1}: \theta>0\) is accepted if and only if the observed mean \(\bar{x}\) of a random sample of size 25 is greater than or equal to \(\frac{3}{5}\). Find the power function \(\gamma(\theta), 0 \leq \theta\), of this test.

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