Question

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

The likelihood ratio test statistic \(\Lambda\) for testing \(H_0: \theta = \theta_0\) against \(H_1: \theta \neq \theta_0\) when \(Y_n \leq \theta_0\) is \(\left( Y_n / \theta_0 \right)^n\) and \(\Lambda = 0\) if \(Y_n > \theta_0\). Under the null hypothesis, \(-2 \log \Lambda\) has an exact \(\chi^2(2)\) distribution.

Step-by-step solution

Formulate the Likelihood Ratio

The first step involves formulating the likelihood ratio (\(\Lambda\)) for testing the null hypothesis \(H_{0}: \theta=\theta_{0}\) against the alternative \(H_{1}: \theta \neq \theta_{0}\). If the likelihood function for the given data is \(L(\theta)\), then the likelihood ratio \(\Lambda\) is defined as the likelihood of the null hypothesis over the likelihood of the most probable alternative hypothesis. This can be written as \( \Lambda = \frac{L(\theta_0)}{L(\hat{\theta})}\) where \(\hat{\theta}\) is the maximum likelihood estimator.

Simplify the likelihood ratio

Given a uniform distribution, it's known that \(L(\theta) = 1/\theta^n\) for all \(\theta > Y_n\) and 0 otherwise. As such, we can express \(L(\theta_0)\) and \(L(\hat{\theta})\) given the conditions. When \(Y_n \leq \theta_0\), \(L(\theta_0) = 1/\theta_0^n\) and \(L(\hat{\theta}) = 1/Y_n^n\). Under this condition \(\Lambda = \left(Y_{n} / \theta_{0}\right)^{n}\). If \(Y_n > \theta_0\), \(L(\theta_0)\) is not defined and so \(\Lambda = 0\).

Determine the distribution of the test statistic

The next step involves showing the exact distribution of \(-2 \log \Lambda\), under the null hypothesis. That can be done by noting that \(-2 \log \Lambda\) simplifies to be twice the log of \(Y_n / \theta_0\), which is \(2n \log(\theta_0 / Y_n)\). It can be shown that this quantity has a chi-square distribution with two degrees of freedom when multiplied by 2, not \(\chi(1)\).

Key concepts

Order Statistics

Order statistics are key tools in statistical analysis that provide critical insight into the distribution of a sample set. Essentially, when we have a series of data points, order statistics sort these values from the smallest to the largest. In the context of the given exercise, the order statistics are denoted as \(Y_{1}
Understanding order statistics is crucial when performing statistical tests, as it helps us to identify characteristics such as the sample minimum \(Y_{1}\), maximum \(Y_{n}\), or any other position-specific value, which can be considered as a quantile of the underlying population. For instance, the maximum value \(Y_{n}\) plays a pivotal role in the likelihood ratio test, as it is pivotal to determining the value of \(\Lambda\) depending on its relation to \(\theta_0\).

Order statistics can also be involved in more complex analyses like finding confidence intervals or performing hypothesis tests where the distribution of these statistics underlies the testing method's validity.

Uniform Distribution

The uniform distribution is a type of probability distribution where all outcomes are equally likely. For a continuous uniform distribution defined on the interval \((0, \theta)\), any interval of equal length within the bounds has the same probability of containing a drawn value. The importance of this distribution is emphasized in the exercise as it sets the groundwork for the parameters of the likelihood ratio test.

Understanding the uniform distribution is crucial when analyzing the provided exercise. The likelihood function \(L(\theta)\) leverages the uniform distribution's properties since the probability of observing our sample is \(1/\theta^n\) if \(\theta > Y_n\) and is 0 if otherwise. This comes into play when we calculate the likelihood ratio \(\Lambda\), especially in the case where \(Y_n\) is less than or equal to \(\theta_0\), where it simplifies to \((Y_{n} / \theta_{0})^{n}\).

In hypothesis testing, like the scenario portrayed, the uniform distribution simplifies many calculations and enables us to derive properties of test statistics explicitly, due to its straightforward probability density function and cumulative distribution function.

Chi-Square Distribution

The chi-square distribution is a fundamental concept in statistics commonly used in hypothesis testing and in constructing confidence intervals, especially for variance. It is an asymmetric distribution that only takes on positive values and is defined by a single parameter: degrees of freedom, often denoted as \(\text{df}\).

In the exercise, we encounter an application of the chi-square distribution that may seem counterintuitive. Typically, when performing a likelihood ratio test, we expect a chi-square distribution with \(\text{df}\) equal to the difference in the number of parameters estimated by the null and alternative hypothesis. However, the exercise illustrates an exception where \(-2 \log \Lambda\), the test statistic, follows a chi-square distribution with 2 degrees of freedom, not 1 as might be expected given the test only involves one parameter, \(\theta\).

This exceptional case arises due to the non-regularity of the likelihood function, where the usual asymptotic results do not apply, and hence the exact distribution must be used. Getting acquainted with the chi-square distribution allows students to better comprehend the principles behind hypothesis testing and reinforces the need to carefully confirm the conditions under which statistical theorems hold, as they directly impact the resulting inferences.