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Chapter 8: Problem 5

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 \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with pmf \(f(x ; p)=p^{x}(1-p)^{1-x}, x=0,1\), zero elsewhere. Show that \(C=\left\\{\left(x_{1}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i} \leq c\right\\}\) is a best critical region for testing \(H_{0}: p=\frac{1}{2}\) against \(H_{1}: p=\frac{1}{3} .\) Use the Central Limit Theorem to find \(n\) and \(c\) so that approximately \(P_{H_{0}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.10\) and \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

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 \(\theta=0.5\).

Let \(X_{1}, X_{2}, \ldots, X_{25}\) denote a random sample of size 25 from a normal distribution \(N(\theta, 100)\). Find a uniformly most powerful critical region of size \(\alpha=0.10\) for testing \(H_{0}: \theta=75\) against \(H_{1}: \theta>75\)

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) follow the location model $$\begin{aligned} X_{i} &=\theta_{1}+Z_{i}, \quad i=1, \ldots, n \\ Y_{i} &=\theta_{2}+Z_{n+i}, \quad i=1, \ldots, m\end{aligned}$$ where \(Z_{1}, \ldots, Z_{n+m}\) are iid random variables with common pdf \(f(z)\). Assume that \(E\left(Z_{i}\right)=0\) and \(\operatorname{Var}\left(Z_{i}\right)=\theta_{3}<\infty\) (a) Show that \(E\left(X_{i}\right)=\theta_{1}, E\left(Y_{i}\right)=\theta_{2}\), and \(\operatorname{Var}\left(X_{i}\right)=\operatorname{Var}\left(Y_{i}\right)=\theta_{3}\). (b) Consider the hypotheses of Example \(8.3 .1\); i.e, $$H_{0}: \theta_{1}=\theta_{2} \text { versus } H_{1}: \theta_{1} \neq \theta_{2}$$ Show that under \(H_{0}\), the test statistic \(T\) given in expression \((8.3 .5)\) has a limiting \(N(0,1)\) distribution. (c) Using Part (b), determine the corresponding large sample test (decision rule) of \(H_{0}\) versus \(H_{1}\). (This shows that the test in Example \(8.3 .1\) is asymptotically correct.)

Suppose that a manufacturing process makes about 3 percent defective items, which is considered satisfactory for this particular product. The managers would like to decrease this to about 1 percent and clearly want to guard against a substantial increase, say to 5 percent. To monitor the process, periodically \(n=100\) items are taken and the number \(X\) of defectives counted. Assume that \(X\) is \(b(n=100, p=\theta)\). Based on a sequence \(X_{1}, X_{2}, \ldots, X_{m}, \ldots\), determine a sequential probability ratio test that tests \(H_{0}: \theta=0.01\) against \(H_{1}: \theta=0.05 .\) (Note that \(\theta=0.03\), the present level, is in between these two values.) Write this test in the form $$h_{0}>\sum_{i=1}^{m}\left(x_{i}-n d\right)>h_{1}$$ and determine \(d, h_{0}\), and \(h_{1}\) if \(\alpha_{a}=\beta_{a}=0.02\).
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