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Chapter 6: Problem 8

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0\) (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(Y=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(Y\). (b) For \(\theta_{0}=2\) and \(n=5\), find the significance level of the test that rejects \(H_{0}\) if \(Y \leq 4\) or \(Y \geq 17\)

Short Answer

Expert verified
The likelihood ratio test statistic for the given Poisson distribution is \(位 = 2(\sum_{i=1}^n X_i \log(\theta_{MLE}) - n\theta_{MLE} - \sum_{i=1}^n X_i \log(\theta_0) + n\theta_0)\). The null distribution of Y is a Poisson with mean \(n\theta_0\). The significance level of the test that rejects \(H_0\) if \(Y \leq 4\) or \(Y\geq 17\) for \(\theta_{0} = 2\) and \(n = 5\) is given by \(伪 = F(4;10)+ 1- F(16;10)\), where \(F\) is the cumulative distribution function of the Poisson distribution.

Step by step solution

01

Derivation of the Likelihood Ratio Test Statistic

First, the likelihood function for the Poisson distribution can be written as: \[L(\theta)= \frac{e^{-n\theta}\theta^{\sum_{i=1}^n X_i}}{\prod_{i=1}^n X_i!}\] Next, under the null hypothesis, \(\theta = \theta_0\), the likelihood function becomes: \[L(\theta_0)= \frac{e^{-n\theta_0}\theta_0^{\sum_{i=1}^n X_i}}{\prod_{i=1}^n X_i!}\] The likelihood ratio test statistic is given by: \[位 = 2(\log(L(\theta_MLE)) - \log(L(\theta_0))\] where, \(\theta_{MLE}\) (maximum likelihood estimate) is derived as \(\frac{\sum_{i=1}^n X_i}{n}\) Therefore, the test statistic, \(位\), becomes: \(位 =2(\sum_{i=1}^n X_i \log(\theta_{MLE}) - n\theta_{MLE} - \sum_{i=1}^n X_i \log(\theta_0) + n\theta_0)\)
02

Obtain the Null Distribution of Y

The test statistic here is \(Y = \sum_{i=1}^{n}X_{i}\). Under the null hypothesis, \(\theta = \theta_0\), the random variable \(X_i\) follows a Poisson distribution with mean \(\theta_0\). Therefore, \(Y\) which is the sum of \(n\) such random variables, also follows a Poisson distribution with mean \(n\theta_0\). This is the null distribution of \(Y\).
03

Calculate the Significance Level

The significance level, \(伪\), is the probability that the null hypothesis is rejected when it is true. This can be calculated as the sum of the probabilities of the two rejection regions defined by \(Y \leq 4\) or \(Y\geq 17\). Under the null hypothesis, \(Y\) follows a Poisson distribution with mean \(n\theta_0 = 2*5 =10\). Using the cumulative distribution function of the Poisson distribution, \(P(Y \leq 4 | \theta_0 = 2) = F(4;10)\) and \(P(Y\geq 17 | \theta_0 = 2) = 1- F(16;10)\). Hence, the significance level \(伪 = P(Y \leq 4 | \theta_0 = 2) + P(Y\geq 17 | \theta_0 = 2) = F(4;10)+ 1- F(16;10)\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\theta \exp \left\\{-|x|^{\theta}\right\\} / 2 \Gamma(1 / \theta),-\infty0 .\) Suppose \(\Omega=\) \(\\{\theta: \theta=1,2\\}\). Consider the hypotheses \(H_{0}: \theta=2\) (a normal distribution) versus \(H_{1}: \theta=1\) (a double exponential distribution). Show that the likelihood ratio test can be based on the statistic \(W=\sum_{i=1}^{n}\left(X_{i}^{2}-\left|X_{i}\right|\right)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\) distributions, respectively. (a) If \(\Omega \subset R^{3}\) is defined by $$\Omega=\left\\{\left(\theta_{1}, \theta_{2}, \theta_{3}\right):-\infty<\theta_{i}<\infty, i=1,2 ; 0<\theta_{3}=\theta_{4}<\infty\right\\}$$ find the mles of \(\theta_{1}, \theta_{2}, \theta_{3}\). (b) If \(\Omega \subset R^{2}\) is defined by $$\Omega=\left\\{\left(\theta_{1}, \theta_{3}\right):-\infty<\theta_{1}=\theta_{2}<\infty ; 0<\theta_{3}=\theta_{4}<\infty\right\\}$$ find the mles of \(\theta_{1}\) and \(\theta_{3}\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the distribution \(N\left(\theta_{1}, \theta_{2}\right)\). Show that the likelihood ratio principle for testing \(H_{0}: \theta_{2}=\theta_{2}^{\prime}\) specified, and \(\theta_{1}\) unspecified, against \(H_{1}: \theta_{2} \neq \theta_{2}^{\prime}, \theta_{1}\) unspecified, leads to a test that rejects when \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq c_{1}\) or \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq c_{2}\), where \(c_{1}

Consider two Bernoulli distributions with unknown parameters \(p_{1}\) and \(p_{2}\). If \(Y\) and \(Z\) equal the numbers of successes in two independent random samples, each of size \(n\), from the respective distributions, determine the mles of \(p_{1}\) and \(p_{2}\) if we know that \(0 \leq p_{1} \leq p_{2} \leq 1\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0 .\) Test \(H_{0}: \theta=2\) against \(H_{1}: \theta \neq 2\) using (a) \(-2 \log \Lambda\). (b) a Wald-type statistic. (c) Rao's score statistic.
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