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Chapter 8: Problem 9

Let \(Y_{1}

Short Answer

Expert verified
The likelihood ratio test (LRT) for testing \(H_{0}:\theta=\theta_{0}\) against \(H_{1}:\theta \neq \theta_{0}\) is \(\Lambda= e^{-(|Y_{3}-\theta_{0}|-\Sigma_{i=1}^{5}|Y_{i}-Y_{3}|)}\).

Step by step solution

01

Understanding the Distribution and Hypotheses

Identify the given distribution and understand the hypotheses to be tested. The distribution specified is a Laplace or double-exponential distribution with pdf \(f(x; \theta)=\frac{1}{2}e^{-|x-\theta|}\) for all real \(\theta\). The hypotheses to be tested are \(H_{0}: \theta=\theta_{0}\) and \(H_{1}: \theta \neq \theta_{0}\).
02

Calculating Likelihood under Null Hypothesis

Under \(H_{0}\), calculate the likelihood function which is given by the product of the pdfs given the observation and the theory. The likelihood function will be \(L(\theta_{0})=\prod_{i=1}^{5}\frac{1}{2}e^{-|Y_{i}-\theta_{0}|}\).
03

Calculating Likelihood under Alternative Hypothesis

Under \(H_{1}\), the likelihood function is maximized when \(\theta\) equals to the median of the observations. Since the sample size is 5, the median will be the value of the third order statistic \(Y_{3}\). This gives the maximum likelihood \(L(\hat{\theta}_{ML})=\prod_{i=1}^{5}\frac{1}{2}e^{-|Y_{i}-Y_{3}|}\)
04

Calculating the Likelihood Ratio Test

The likelihood ratio test (LRT) statistic, denoted by \(\Lambda\), is defined as the ratio of maximized likelihood under null hypothesis to maximized likelihood under alternative hypothesis. Therefore, \(\Lambda =\frac{L(\theta_{0})}{L(\hat{\theta}_{ML})}\)
05

Simplify the LRT

Simplify the ratio by plugging in the likelihood functions from Steps 2 and 3. After a lot of cancellations and simplifications, we get \(\Lambda= e^{-(|Y_{3}-\theta_{0}|-\Sigma_{i=1}^{5}|Y_{i}-Y_{3}|)}\)

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Most popular questions from this chapter

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 \(X\) have a Poisson distribution with mean \(\theta\). Find the sequential probability ratio test for testing \(H_{0}: \theta=0.02\) against \(H_{1}: \theta=0.07\). Show that this test can be based upon the statistic \(\sum_{1}^{n} X_{i} .\) If \(\alpha_{a}=0.20\) and \(\beta_{a}=0.10\), find \(c_{0}(n)\) and \(c_{1}(n)\)

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample from a distribution that is \(N\left(\theta_{1}, \theta_{2}\right)\). Find a best test of the simple hypothesis \(H_{0}: \theta_{1}=\theta_{1}^{\prime}=0, \theta_{2}=\theta_{2}^{\prime}=1\) against the alternative simple hypothesis \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

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 \(X\) 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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