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Chapter 8: Problem 12

Let \(Y_{1}

Short Answer

Expert verified
\(\Lambda\) will always be equal to 1.

Step by step solution

01

Likelihood under \(H_{0}\)

Under \(H_{0}: \theta=\theta_{0}\), the likelihood function for the sample data is \[L(\theta_{0} ; y)=\prod_{i=1}^{5}\frac{1}{2} e^{-|y_{i}-\theta_{0}|}\] Specifying Substituting the given order of statistics, we have: \[L(\theta_{0})=\frac{1}{2^{5}}e^{-|\theta_{0}-y_{1}|}e^{-|\theta_{0}-y_{2}|}e^{-|\theta_{0}-y_{3}|}e^{-|\theta_{0}-y_{4}|}e^{-|\theta_{0}-y_{5}|}\]
02

Likelihood under \(H_{1}\)

Under \(H_{1}: \theta \neq \theta_{0}\), the likelihood function is maximized when \(\theta\) is such that \(y_3=\theta\), because the absolute value will be minimized and as a result, the exponential will be maximized. So, \[L(\theta) =\frac{1}{2^{5}}e^{-|y_3-y_{1}|}e^{-|y_3-y_{2}|}e^{-|y_3-y_{3}|}e^{-|y_3-y_{4}|}e^{-|y_3-y_{5}|}\]
03

Formulating the Likelihood Ratio Test Statistic \(\Lambda\)

\(\Lambda\) is a ratio of likelihood under \(H_{0}\) to likelihood under \(H_{1}\), here given as \[\Lambda=\frac{L(\theta_{0})}{L(\theta)}\] Substituting for \(L(\theta_{0})\) and \(L(\theta)\) gives: \[\Lambda=\frac{\frac{1}{2^{5}}e^{-|\theta_{0}-y_{1}|}e^{-|\theta_{0}-y_{2}|}e^{-|\theta_{0}-y_{3}|}e^{-|\theta_{0}-y_{4}|}e^{-|\theta_{0}-y_{5}|}}{\frac{1}{2^{5}}e^{-|y_3-y_{1}|}e^{-|y_3-y_{2}|}e^{-|y_3-y_{3}|}e^{-|y_3-y_{4}|}e^{-|y_3-y_{5}|}}\] Simplifying gives: \[\Lambda=e^{-2\{|y_3-y_{1}|+|y_3-y_{2}|+|y_3-y_{4}|+|y_3-y_{5}|-|y_3-y_{1}|+|y_3-y_{2}|+|y_3-y_{4}|+|y_3-y_{5}|\}}\] Further simplification provides final form for \(\Lambda\): \[\Lambda = 1\]

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Key Concepts

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

Order Statistics
Order statistics play a critical role in statistics, particularly when dealing with samples. Imagine you collect some data points, say the ages of participants in a study. The order statistics would be these ages arranged from the youngest to the oldest participant. Mathematically, if we have a random sample of size n, the order statistics are those same observations sorted in increasing order.

For example, with a sample size of five, denoted as \(Y_1, Y_2, Y_3, Y_4, Y_5\), \(Y_1\) would be the smallest observation (minimum), and \(Y_5\) the largest (maximum). Order statistics are essential in non-parametric statistics and inference because they underpin many other statistical methods, such as finding the median (which would be \(Y_3\), the third order statistic, in our sample of five).

They are also crucial in determining properties like the range of the data, interquartile range, and for performing various statistical tests. In the context of maximum likelihood estimation and likelihood ratio tests, order statistics can help pinpoint the parameter values that best explain the observed data.
Probability Density Function
The probability density function (PDF) is a foundational concept in statistics, providing a function that describes the relative likelihood for a continuous random variable to take on a given value. Unlike a cumulative distribution function (CDF), which shows the probability that a random variable is less than or equal to a certain value, the PDF describes the probability per unit on the x-axis.

In the exercise, the PDF given is \(f(x ; \theta)=\frac{1}{2} e^{-|x-\theta|}\), which signifies an exponential-type distribution centered at \theta. The absolute value ensures the distribution is symmetric around \theta, making it a double exponential or Laplace distribution.

The PDF helps in determining the likelihood function, which is the product of the PDFs for all observations in the sample, used in methods of statistical inference such as maximum likelihood estimation and likelihood ratio tests.
Hypothesis Testing
Hypothesis testing is a method by which statisticians test an assumption regarding a population parameter. The approach involves two competing hypotheses: the null hypothesis, denoted \(H_0\), and the alternative hypothesis, denoted \(H_1\) or \(H_a\). The null hypothesis typically represents a theory of no effect or no difference, which in our exercise is \(\theta=\theta_0\); it maintains the status quo.

Conversely, the alternative hypothesis represents a theory that there is an effect, or there is a difference, which in our case is \(\theta eq \theta_0\). Through the hypothesis testing process, we collect evidence (data) and measure how compatible the null hypothesis is with the observed data. Based on this compatibility, represented through a p-value, we decide whether to reject the null hypothesis in favor of the alternative. In the likelihood ratio test used in this exercise, we compare the likelihood of the null hypothesis to the likelihood of the alternative to make this decision.
Maximum Likelihood Estimation
Maximum likelihood estimation (MLE) is a method for estimating the parameters of a statistical model. Given a sample and a statistical model, MLE finds the parameter values that make the observed sample most probable. It does so by maximizing a likelihood function, which expresses the probability of the observed sample as a function of the parameters of the model.

The procedure involves taking the PDF, applying it to each observation in the sample, and finding the product of these values 鈥� this product is the likelihood function. For complex models or large samples, the likelihood function may become unwieldy, so statisticians often maximize the natural logarithm of the likelihood function, which is mathematically equivalent but computationally simpler.

In the given exercise, the MLE would be the value of \theta that maximizes the likelihood function. It's shown that when \theta is equal to the median of the sample, here corresponding to \(y_3\), the likelihood is maximized for our specific PDF. MLE is widely used for parameter estimation in many fields such as finance, medicine, and ecological modeling.

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Most popular questions from this chapter

Suppose \(X_{1}, \ldots, X_{n}\) is a random sample on \(X\) which has a \(N\left(\mu, \sigma_{0}^{2}\right)\) distribution, where \(\sigma_{0}^{2}\) is known. Consider the two-sided hypotheses $$ H_{0}: \mu=0 \text { versus } H_{1}: \mu \neq 0 $$ Show that the test based on the critical region \(C=\left\\{|\bar{X}|>\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\right\\}\) is an unbiased level \(\alpha\) test.

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size 10 from a Poisson distribution with parameter \(\theta\). Let \(L(\theta)\) be the joint pdf of \(X_{1}, X_{2}, \ldots, X_{10}\). The problem is to test \(H_{0}: \theta=\frac{1}{2}\) against \(H_{1}: \theta=1\). (a) Show that \(L\left(\frac{1}{2}\right) / L(1) \leq k\) is equivalent to \(y=\sum_{1}^{n} x_{i} \geq c\). (b) In order to make \(\alpha=0.05\), show that \(H_{0}\) is rejected if \(y>9\) and, if \(y=9\), reject \(H_{0}\) with probability \(\frac{1}{2}\) (using some auxiliary random experiment). (c) If the loss function is such that \(\mathcal{L}\left(\frac{1}{2}, \frac{1}{2}\right)=\mathcal{L}(1,1)=0\) and \(\mathcal{L}\left(\frac{1}{2}, 1\right)=1\) and \(\mathcal{L}\left(1, \frac{1}{2}\right)=2\), show that the minimax procedure is to reject \(H_{0}\) if \(y>6\) and, if \(y=6\), reject \(H_{0}\) with probability \(0.08\) (using some auxiliary random experiment).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with pmf \(f(x ; p)=p^{x}(1-p)^{1-x}, x=0,1\), zero elsewhere. Show that \(C=\left\\{\left(x_{1}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i} \leq c\right\\}\) is a best critical region for testing \(H_{0}: p=\frac{1}{2}\) against \(H_{1}: p=\frac{1}{3} .\) Use the Central Limit Theorem to find \(n\) and \(c\) so that approximately \(P_{H_{0}}\left(\sum_{1}^{n^{3}} X_{i} \leq c\right)=0.10\) and \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

Consider a distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the Central Limit Theorem to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\)
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