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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 Test statistic for testing \(H_{0}: \theta=\theta_{0}\) against \(H_{1}: \theta \neq \theta_{0}\) is \(\Lambda=\left(Y_{n} / \theta_{0}\right)^{n}\) for \(Y_{n} \leq \theta_{0}\) and \(\Lambda=0\) for \(Y_{n}>\theta_{0}\). And under the null hypothesis, the test statistic \(-2 \log \Lambda\) follows an exact \(\chi^{2}(2)\) distribution.

Step by step solution

01

Formulate the likelihood function

We know the pdf for a uniform distribution on \((0, \theta)\) is \(f(x;\theta)=1/\theta\) for \(0<x<\theta\) and 0 otherwise. So, given a sample \(Y_{1}, Y_{2}, ..., Y_{n}\), we can write down the joint density function \(L(\theta)=\prod_{i=1}^{n}f(Y_{i};\theta)\). Since all \(Y_{i}\)'s are from a uniform distribution, \(L(\theta)=1/\theta^{n}\) if \(0<Y_{1}<...<Y_{n}<\theta\) and \(L(\theta)=0\) otherwise.
02

Express the Likelihood Ratio Test

The Likelihood Ratio Test statistic \(\Lambda\) can be formulated as \(\Lambda=\frac{supL(\theta|H_{0}:\theta=\theta_{0})}{supL(\theta|H_{1}:\theta \neq \theta_{0})}\). For given order statistics, the likelihood function under \(H_{1}\) is maximized at \(Y_{n}\) since all \(Y_{i}\)'s are less than or equal to \(Y_{n}\). So, the denominator of the ratio is \(1/Y_{n}^{n}\). Under the null hypothesis, the model is completely specified, so the likelihood function is \(1/\theta_{0}^{n}\) and thus forms the numerator of \(\Lambda\). Hence \(\Lambda=\left(Y_{n} / \theta_{0}\right)^{n}\). This holds only if \(Y_{n} \leq \theta_{0}\), else \(\Lambda\) is 0 as per the laws of likelihood under uniform distribution.
03

Proof that the test statistic has \(\chi^{2}(2)\)

Under the null hypothesis, \(Y_{n}/\theta_{0}\) is a random value from the uniform (0,1) distribution. We can write \(\Lambda\) as \(\left(U_{n}\right)^{n}\), where \(U_{n}=Y_{n}/\theta_{0}\). Taking \(-2log(\Lambda)\) of both sides, we get \(-2log\left(U_{n}^{n}\right)=-2nlog\left(U_{n}\right)\). Now, under \(H_{0}\), we know that \(U_{n}\) has a beta (n,1) distribution which implies that \(n.logU_{n}\) has a exponential (1) distribution. We know that sum of two exponential (1) random variables has a chi-square distribution with 2 degrees of freedom. Hence, \(-2 \log \Lambda\) 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
Understanding order statistics is vital when dealing with samples from any distribution, including the uniform distribution. Order statistics refer to the statistics obtained by arranging a sample in ascending order and picking out the smallest, second smallest, and so on until the largest value. For a given sample set, we denote these sorted values as \( Y_1, Y_2, ..., Y_n \).

In the context of the uniform distribution on \((0, \theta)\), the smallest value \(Y_1\) and the largest value \(Y_n\) play a special role in hypothesis testing and likelihood ratio tests. For instance, when testing a hypothesis about the parameter \(\theta\), \(Y_n\) is essential as it maximizes the likelihood function under the alternative hypothesis, and it is directly used to calculate the likelihood ratio test statistic, \(\Lambda\).

It is interesting to note that the spacing between these ordered values also provides information about the underlying population from which the sample was drawn. For instance, for a uniform distribution, these spacings are expected to be relatively even, whereas in other distributions, such as the exponential distribution, spacings would be more variable. Understanding these nuances helps us to better interpret our test results and to scrutinize the assumptions made in our hypotheses.
Uniform Distribution

Characteristics and Use in Hypothesis Testing

The uniform distribution is one of the simplest probability distributions in statistics. It assumes that every event has an equal chance of occurring within a certain interval, with no event being more likely than another. For a uniform distribution on an interval \((0, \theta)\), any value within this interval is equally likely, and values outside of it are impossible.

In hypothesis testing, particularly with the likelihood ratio test, the characteristics of the uniform distribution play a critical role. For example, the calculation of the likelihood for a set of order statistics from a uniform distribution is straightforward 鈥� it is simply \(1/\theta^n\) if the observed values fall within the interval and zero otherwise.

The calculation of the test statistic for the likelihood ratio \(\Lambda\) relies entirely on the behavior of the uniform distribution, especially considering the largest order statistic \(Y_n\). Should \(Y_n\) exceed the hypothesized parameter \(\theta_0\), the likelihood under the alternative hypothesis becomes zero, reflecting the impossibility of such an event under the specified model. This characteristic enforces the rigor of the testing procedure and assures the specificity of statistical inference.
Chi-Square Distribution

Understanding the Chi-Square Distribution in the Context of a Likelihood Ratio Test (LRT)

The chi-square distribution is a fundamental distribution in statistics, especially for hypothesis testing. It is traditionally used in the goodness-of-fit test, tests of independence in contingency tables, and for the variance of a normally distributed sample. This distribution is characterized by its degrees of freedom, which essentially define the shape of its probability density function.

In the context of a likelihood ratio test, the chi-square distribution serves a key purpose in deducing p-values from the LRT test statistic. Specifically, from the exercise provided, we saw that the \(-2 \log \Lambda\) statistic has an exact chi-square distribution with 2 degrees of freedom, not 1, when the null hypothesis holds true. This is somewhat counterintuitive, as we often expect a chi-square distribution with degrees of freedom equal to the number of parameters being estimated.

However, due to the specific conditions under which the exercise's null hypothesis is tested, regularity conditions do not apply here, and the chi-square distribution we arrive at has 2 degrees of freedom. This stresses the importance of understanding the underlying assumptions and conditions of statistical tests, as deviations from the norm can have significant implications for the interpretation of our results and the conclusions we draw from statistical analyses.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) be independent random samples from the distributions \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\), respectively. (a) Show that the likelihood ratio for testing \(H_{0}: \theta_{1}=\theta_{2}, \theta_{3}=\theta_{4}\) against all alternatives is given by $$\frac{\left[\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} / n\right]^{n / 2}\left[\sum_{1}^{m}\left(y_{i}-\bar{y}\right)^{2} / m\right]^{m / 2}}{\left\\{\left[\sum_{1}^{n}\left(x_{i}-u\right)^{2}+\sum_{1}^{m}\left(y_{i}-u\right)^{2}\right] /(m+n)\right\\}^{(n+m) / 2}},$$ where \(u=(n \bar{x}+m \bar{y}) /(n+m)\). (b) Show that the likelihood ratio test for testing \(H_{0}: \theta_{3}=\theta_{4}, \theta_{1}\) and \(\theta_{2}\) unspecified, against \(H_{1}: \theta_{3} \neq \theta_{4}, \theta_{1}\) and \(\theta_{2}\) unspecified, can be based on the random variable $$F=\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} /(n-1)}{\sum_{1}^{m}\left(Y_{i}-\bar{Y}\right)^{2} /(m-1)}$$

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

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 \(H_{1}: \beta \neq \beta_{0}\)

\left\\{X_{i}>\theta_{0}\right\\}\(. (b) Show that the scores test for this model is equival鈥� # Consider Example \)6.3 .4\(. (a) Show that we can write \)S^{*}=2 T-n\(, where \)T=\\#\left\\{X_{i}>\theta_{0}\right\\}\(. (b) Show that the scores test for this model is equivalent to rejecting \)H_{0}\( if \)Tc_{2}\( (c) Show that under \)H_{0}, T\( has the binomial distribution \)b(n, 1 / 2) ;\( hence, determine \)c_{1}\( and \)c_{2}\( so the test has size \)\alpha\(. (d) Determine the power function for the test based on \)T\( as a function of \)\theta$.

Let the table $$\begin{array}{c|cccccc}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline \text { Frequency } & 6 & 10 & 14 & 13 & 6 & 1 \end{array}$$ represent a summary of a sample of size 50 from a binomial distribution having \(n=5 .\) Find the mle of \(P(X \geq 3)\).
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