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Chapter 6: Problem 4

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from the two normal distributions \(N\left(0, \theta_{1}\right)\) and \(N\left(0, \theta_{2}\right)\). (a) Find the likelihood ratio \(\Lambda\) for testing the composite hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) against the composite alternative \(H_{1}: \theta_{1} \neq \theta_{2}\). (b) This \(\Lambda\) is a function of what \(F\) -statistic that would actually be used in this test?

Short Answer

Expert verified
The likelihood ratio \(\Lambda\) is defined as the ratio of the likelihood of the data under the null hypothesis to the likelihood under the alternative. The computed \(\Lambda\) will be a function of the F-statistic, which is a measure comparing the observed sample variances.

Step by step solution

01

Define Likelihood Ratio

The likelihood ratio \(\Lambda\) is used in hypothesis testing to compare the likelihood of the observed data under one statistical model (the null hypothesis \(H_{0}: \theta_{1}=\theta_{2}\)) against another (the alternative hypothesis \(H_{1}: \theta_{1} \neq \theta_{2}\)). It is defined as the ratio of the likelihood of the data under \(H_{0}\) to the likelihood of the data under \(H_{1}\). Presuming that the data are distributed normally, the likelihood of the data given the model parameters can be computed from the normal distribution's probability density function.
02

Calculate the Likelihood Ratio

For the given samples, the likelihood ratio \(\Lambda\) is given by: \[\Lambda = \frac{L(\theta_{1}=\theta_{2}|X, Y)}{L(\theta_{1} \neq \theta_{2}|X, Y)} = \frac{\prod_{i=1}^{n} f(X_{i}|0, \theta_{2}) \prod_{j=1}^{m} f(Y_{j}|0, \theta_{2})} {\prod_{i=1}^{n} f(X_{i}|0, \theta_{1}) \prod_{j=1}^{m} f(Y_{j}|0, \theta_{2})}\] Here, \(L(\$...|X, Y)\) represents the likelihood of the data \(X\) and \(Y\) given model parameters \(\theta_{1}\) and \(\theta_{2}\), and \(f(\$...)\) denotes the probability density function of the normal distribution.
03

Determine the relevant F-statistic

The F-statistic arises in the context of ANOVA where it is used to compare variances. We should note that when \(\Lambda\) is expressed in terms of sample variances, it actually serves as a function of the F-statistic. Therefore, in this context, the F-statistic would be the ratio of the observed sample variances. F-statistic is given by: \[F = \frac{s^2_{X}}{s^2_{Y}}\] where \(s^2_{X}\) and \(s^2_{Y}\) are the sample variances of \(X\) and \(Y\) respectively.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(\alpha, \beta)\) distribution where \(\alpha\) is known and \(\beta>0\). Determine the likelihood ratio test for \(H_{0}: \beta=\beta_{0}\) against \(H_{1}: \beta \neq \beta_{0}\)

Let \(X_{1}, X_{2}\), and \(X_{3}\) have a multinomial distribution in which \(n=25, k=4\), and the unknown probabilities are \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\), respectively. Here we can, for convenience, let \(X_{4}=25-X_{1}-X_{2}-X_{3}\) and \(\theta_{4}=1-\theta_{1}-\theta_{2}-\theta_{3} .\) If the observed values of the random variables are \(x_{1}=4, x_{2}=11\), and \(x_{3}=7\), find the maximum likelihood estimates of \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Bernoulli distribution with parameter \(p .\) If \(p\) is restricted so that we know that \(\frac{1}{2} \leq p \leq 1\), find the mle of this parameter.

Suppose the pdf of \(X\) is of a location and scale family as defined in Example 6.4.4. Show that if \(f(z)=f(-z)\), then the entry \(I_{12}\) of the information matrix is 0 . Then argue that in this case the mles of \(a\) and \(b\) are asymptotically independent.

Consider a location model (Example \(6.2 .2\) ) when the error pdf is the contaminated normal (3.4.14) with \(\epsilon\) as the proportion of contamination and with \(\sigma_{c}^{2}\) as the variance of the contaminated part. Show that the ARE of the sample median to the sample mean is given by $$e\left(Q_{2}, \bar{X}\right)=\frac{2\left[1+\epsilon\left(\sigma_{c}^{2}-1\right)\right]\left[1-\epsilon+\left(\epsilon / \sigma_{c}\right)\right]^{2}}{\pi}$$ Use the hint in Exercise \(6.2 .5\) for the median. (a) If \(\sigma_{c}^{2}=9\), use \((6.2 .34)\) to fill in the following table: $$\begin{array}{|l|l|l|l|l|}\hline \epsilon & 0 & 0.05 & 0.10 & 0.15 \\ \hline e\left(Q_{2}, X\right) & & & & \\\\\hline\end{array}$$ (b) Notice from the table that the sample median becomes the "better" estimator when \(\epsilon\) increases from \(0.10\) to \(0.15 .\) Determine the value for \(\epsilon\) where this occurs [this involves a third-degree polynomial in \(\epsilon\), so one way of obtaining the root is to use the Newton algorithm discussed around expression \((6.2 .32)]\).
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