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Chapter 8: Problem 5

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution having pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Short Answer

Expert verified
Following the steps derived from likelihood ratio test, and by properly comparing the critical region criterion, the given set \(C=\{ (x_{1}, x_{2}, \ldots, x_{n}): c \leq \prod_{i=1}^{n} x_{i}\}\) shown on the problem indeed is a best critical region when testing the hypothesis \(H_{0}: \theta=1\) against \(H_{1}: \theta=2\).

Step by step solution

01

Defining Things

First, let's define a few key components. Let's say our null hypothesis \(H_{0}: \theta=1\) and alternative hypothesis \(H_{1}: \theta=2\). The likelihood function for a chosen hypothesis, \(H_{i}\), can be expressed as \(L(\theta; x) = \prod_{i=1}^{n} f(x_{i}; \theta)\) which is simply the product of the pdf, \(f(x;\theta)\), for all observed samples.
02

Compute likelihood ratio test

Computing the likelihood ratio entails that we divide the likelihood under the null hypothesis by that of the alternative hypothesis as: \(LR(x) = \frac{L(\theta=1;x)}{L(\theta=2;x)}\)
03

Apply pdf into likelihood ratio test

We substitute the given pdf into the likelihood ratio to get: \(LR(x) = \frac{1 \cdot \prod_{i=1}^{n} x_{i}^{1-1}}{2 \cdot \prod_{i=1}^{n} x_{i}^{2-1}} = \frac{1}{2^n \cdot \prod_{i=1}^{n} x_{i}}\)
04

Applying critical region criterion

A best critical region for a likelihood ratio test is a criterion that accepts \(H_{0}\) if and only if \(LR(x) \geq k\) for some constant \(k\). Thus, by substituting the expression we got above, \(H_{0}\) will be accepted if and only if \(\frac{1}{2^n \cdot \prod_{i=1}^{n} x_{i}} \geq k\). This simplifies to \(\prod_{i=1}^{n} x_{i} \leq \frac{1}{2^n \cdot k}\).
05

Given critical region comparison

The given critical region is of the shape \(C=\{ (x_{1}, x_{2}, \ldots, x_{n}): c \leq \prod_{i=1}^{n} x_{i}\}\). We can see from the inequality above that such a shape can indeed exist, and thus the given set is a best critical region.

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Most popular questions from this chapter

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 distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the central limit theorcu? to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\).

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 the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2}\). Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

. Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$\mu_{1}^{\prime}=\mu_{2}^{\prime}=0,\quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2}$$ or $$\mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1,\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \rho^{\prime \prime}=\frac{1}{2}$$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).
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