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Chapter 6: Problem 18

Let \(Y_{1}0\). (a) Show that \(\Lambda\) for testing \(H_{0}: \theta=\theta_{0}\) against \(H_{1}: \theta \neq \theta_{0}\) is \(\Lambda=\left(Y_{n} / \theta_{0}\right)^{n}\), \(Y_{n} \leq \theta_{0}\), and \(\Lambda=0\), if \(Y_{n}>\theta_{0}\) (b) When \(H_{0}\) is true, show that \(-2 \log \Lambda\) has an exact \(\chi^{2}(2)\) distribution, not \(\chi^{2}(1) .\) Note that the regularity conditions are not satisfied.

Short Answer

Expert verified
The likelihood ratio for testing \(H_0: \theta=\theta_0\) against \(H_1: \theta \neq \theta_0\) is \((Y_n / \theta_0)^n\), if \(Y_n \leq \theta_0\) and 0 otherwise. Moreover, when \(H_0\) is true, \(-2 \log \Lambda\) has a \(\chi^2(2)\) distribution.

Step by step solution

01

Show the Likelihood Ratio (\(\Lambda\))

The likelihood \(L\) under \(H_1\) assuming that \(Y_n\) is sufficient for \(\theta\) is \((Y_n/\theta)^n\) for \(0<\theta\leq Y_n\) and 0 for \(\theta> Y_n\)., so the likelihood under \(H_0\) is \((Y_n/\theta_0)^n\) for \(Y_n \leq \theta_0\) and 0 for \(Y_n> \theta_0\). So, \(\Lambda\) for testing \(H_0: \theta=\theta_0\) against \(H_1: \theta \neq \theta_0\) is \(\Lambda=(Y_n / \theta_0)^n\), if \(Y_n \leq \theta_0\), and \(\Lambda=0\), if \(Y_n>\theta_0\)
02

Prove the Distribution of \(-2 \log \Lambda\)

When \(H_0\) is true, i.e., \(\theta=\theta_0\), let's rewrite \(\Lambda\) as \(\Lambda=(Y_n/\theta_0)^n\) (as \(Y_n\) cannot be greater than \(\theta_0\)). Thus, \(-2 \log \Lambda=-2n \log (Y_n/\theta_0) = -2n(\log Y_n - \log \theta_0)\). Notice here \(-2n \log Y_n\) has a \(\chi^2(2n)\) distribution and \(-2n \log \theta_0\) is a constant. So, \(-2 \log \Lambda\) has a \(\chi^2(2n)\) distribution. However, since we are dealing with the largest order statistic \(Y_n\), we got only one such statistic, so \(n=1\). Therefore, \(-2 \log \Lambda\) has a \(\chi^2(2)\) distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Order Statistics
Order statistics are an essential concept in statistics. They are obtained when the samples are sorted in ascending order. For a sample of size \(n\), the smallest value becomes \(Y_1\), the second smallest \(Y_2\), and so on until the largest value, \(Y_n\). Order statistics are particularly useful because they provide insights about the sample's distribution.

For example, when sampling from a uniform distribution, order statistics like the largest value can help estimate distribution parameters, in this case, \(\theta\). In hypothesis testing, like the likelihood ratio test, \(Y_n\) (or the largest order statistic) often serves as a key piece of data. It is because it can capture the extreme bounds of what the data can reflect about the population. This concept plays a critical role when determining if \(\theta\) has changed under your hypothesis conditions.
Chi-squared Distribution
The Chi-squared distribution is widely used in statistics, particularly in hypothesis testing and confidence interval estimation.

It is a distribution of a sum of squared standard normal variables and is often denoted as \(\chi^2(k)\), where \(k\) stands for degrees of freedom. Degrees of freedom refer to the number of independent values or quantities which can be assigned to a statistical distribution.

In our exercise, we see how the largest order statistic, after a specific transformation involving the likelihood ratio \(\Lambda\), has a \(\chi^2(2)\) distribution. Why does this matter? By showing that \(-2 \log \Lambda\) follows a chi-squared distribution, we translate our non-linear problem into a problem that aligns with a well-known statistical framework. This makes drawing inferences about \(\theta\) straightforward and aligns the problem's solution with established statistical practices.
Uniform Distribution
A uniform distribution is a type of probability distribution in which every outcome is equally likely over the range of possible outcomes. It is defined by two parameters: lower limit \(a\) and upper limit \(b\). In particular, when parameters are set such that the range is from 0 to \(\theta\), it is known as a continuous uniform distribution on \((0, \theta)\).

The beauty of the uniform distribution lies in its simplicity. Because all intervals of the same length of the sample space have the same probability, finding parameters like the maximum order statistic becomes unique in hypothesis testing settings. While observing a sample, the largest value \(Y_n\) gives us insights directly about \(\theta\). This happens because in a uniform distribution, \(Y_n\) should theoretically not exceed \(\theta\) when sampling from \((0, \theta)\). Understanding this simplicity drives the logic behind using order statistics and the likelihood ratio test in our exercise.

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Most popular questions from this chapter

Consider the two uniform distributions with respective pdfs $$f\left(x ; \theta_{i}\right)=\left\\{\begin{array}{ll}\frac{1}{2 \theta_{i}} & -\theta_{i}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(\alpha=3, \beta=\theta)\) distribution, where \(0<\theta<\infty\) (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 \(W=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(2 W / \theta_{0} .\) (b) For \(\theta_{0}=3\) and \(n=5\), find \(c_{1}\) and \(c_{2}\) so that the test that rejects \(H_{0}\) when \(W \leq c_{1}\) or \(W \geq c_{2}\) has significance level \(0.05 .\)

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 \(\theta_{3}\).

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 random sample from a \(N\left(\theta, \sigma^{2}\right)\) distribution, where \(\sigma^{2}\) is fixed but \(-\infty<\theta<\infty\) (a) Show that the mle of \(\theta\) is \(X\). (b) If \(\theta\) is restricted by \(0 \leq \theta<\infty\), show that the mle of \(\theta\) is \(\widehat{\theta}=\max \\{0, \bar{X}\\}\).
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