/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Let \(X_{1}, \ldots, X_{n}\) be ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X_{1}, \ldots, X_{n}\) be a random sample from an exponential distribution with the density function \(f(x | \theta)=\theta \exp [-\theta x] .\) Derive a likelihood ratio test of \(H_{0}: \theta=\) \(\theta_{0}\) versus \(H_{A}: \theta \neq \theta_{0},\) and show that the rejection region is of the form \(\left\\{\bar{X} \exp \left[-\theta_{0} \bar{X}\right] \leq c\right\\}.\)

Short Answer

Expert verified
The rejection region is \( \{ \bar{X} \exp(-\theta_0 \bar{X}) \leq c \} \).

Step by step solution

01

Write the likelihood function

The likelihood function for the given random sample from an exponential distribution with parameter \( \theta \) is: \[ L(\theta; x_1, x_2, \ldots, x_n) = \prod_{i=1}^{n} \theta \exp(-\theta x_i) = \theta^n \exp(-\theta \sum_{i=1}^{n} x_i). \]
02

Formulate the likelihood ratio

To derive the likelihood ratio test, calculate the ratio of the likelihoods under the null and alternative hypotheses. The likelihood under \( H_0: \theta = \theta_0 \) is \( L(\theta_0; x_1, \ldots, x_n) \), and the maximized likelihood under the alternative \( H_A: \theta eq \theta_0 \) is obtained by substituting \( \hat{\theta} = \frac{n}{\sum x_i} \). Thus, the likelihood ratio \( \Lambda \) is defined as: \[ \Lambda = \frac{L(\theta_0; x_1, \ldots, x_n)}{L(\hat{\theta}; x_1, \ldots, x_n)} = \frac{\theta_0^n \exp(-\theta_0 \sum x_i)}{\left(\frac{n}{\sum x_i}\right)^n \exp\left(-\frac{n}{\sum x_i} \sum x_i \right)}. \]
03

Simplify the likelihood ratio

Substitute \( \theta_0 \) and the maximum likelihood estimator \( \hat{\theta} = \frac{n}{\sum x_i} \), simplify: \[ \Lambda = \left(\frac{\theta_0}{\frac{n}{\sum x_i}}\right)^n \exp\left(-(\theta_0 - \frac{n}{\sum x_i}) \sum x_i\right) = \left(\frac{\theta_0 \bar{X}}{1}\right)^n \exp\left(-n(\theta_0 \bar{X} - 1) \right). \] Note: \( \bar{X} = \frac{\sum x_i}{n} \) is the sample mean.
04

Determine the form of the rejection region

Focus on the expression \( \Lambda \). Simplifying further, we express that the test statistic involves \( \bar{X} \exp(-\theta_0 \bar{X}) \). To create a test, we reject the null hypothesis when this value is small. The rejection region is given by: \[ \left\{ \bar{X} \exp(-\theta_0 \bar{X}) \leq c \right\} \] where \( c \) is a critical value that defines the level of significance.

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Key Concepts

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

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time until an event occurs. It is characterized by a constant rate, making it a common choice for representing lifespans or waiting times. In the density function given, \(f(x | \theta) = \theta \exp [-\theta x]\), \(\theta\) is a positive constant known as the rate parameter.

The function takes non-negative values of \(x\), showcasing a decrease as \(x\) increases, which captures the intuition of a rapid decline in probability as the event is further away in time. This property makes the exponential distribution memoryless, meaning the probability of an event occurring in the future is independent of any past events.

If you are dealing with situations where events occur continuously and independently at a constant average rate, like radioactive decay or the time between bus arrivals, the exponential distribution could be particularly useful. Key elements to remember are:
  • The mean and standard deviation both equal \(1/\theta\).
  • It is a type of gamma distribution with \(k = 1\).
Understanding the exponential distribution helps ground you in contexts such as survival analysis and reliability engineering, enabling deeper awareness of statistical models that demand constant rates over time.
Hypothesis Testing
Hypothesis testing is a critical method in statistics used to decide whether there is enough evidence in a sample to infer that a certain condition holds for the entire population. This process involves formulating two hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_A\).

For example, with the exponential distribution exercise, the null hypothesis \(H_0: \theta = \theta_0\) suggests that the parameter \(\theta\) equals some specified value \(\theta_0\), whereas the alternative \(H_A: \theta eq \theta_0\) suggests \(\theta\) is different.

The goal of hypothesis testing is to determine whether to reject the null hypothesis based on the data. Here's the basic procedure:
  • Assume \(H_0\) is true.
  • Calculate a test statistic based on sample data.
  • Compare the test statistic to a critical value or p-value to make a decision.
In likelihood ratio tests, like the one in the problem, the test statistic is a ratio of likelihoods—one assuming \(H_0\) is true and another optimized for the best fit (under \(H_A\)). If the likelihood of \(H_A\) being true is significantly better than \(H_0\), you might reject \(H_0\) in favor of \(H_A\). Remember that a hypothesis test will always be subject to a level of significance, typically 5%, indicating a 5% risk of rejecting \(H_0\) when it is actually true.
Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a method used to estimate the parameters of a statistical model. The estimation is done by maximizing a likelihood function, ensuring that under the proposed statistical model, the observed data is most probable.

In the context of the exponential distribution, the likelihood function serves as a pivotal role. It is expressed by \( L(\theta; x_1, x_2, \ldots, x_n) = \theta^n \exp(-\theta \sum_{i=1}^{n} x_i) \). By finding the value of \(\theta\) that maximizes this function, we identify the best estimate for \(\theta\). The MLE for \(\theta\) in this setup turns out to be \(\hat{\theta} = \frac{n}{\sum x_i}\).

Think of MLE as a spotlight that highlights which parameter values make the data you collected most "expected". MLE takes into account how likely it is to observe your sample under various possible parameter values.

Applications of MLE spread across many fields, such as:
  • Econometrics, for estimating behavioral models.
  • Machine learning, for optimizing classifiers and regressors.
Understanding MLE gives you a robust framework for parameter estimation in complex models, reflecting directly on its significance in hypothesis testing and inference.

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Most popular questions from this chapter

Suppose that a single observation \(X\) is taken from a uniform density on \([0, \theta]\) and consider testing \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2.\) a. Find a test that has significance level \(\alpha=0 .\) What is its power? b. For \(0<\alpha<1,\) consider the test that rejects when \(X \in[0, \alpha] .\) What is its significance level and power? c. What is the significance level and power of the test that rejects when \(X \in\) \([1-\alpha, 1] ?\) d. Find another test that has the same significance level and power as the previous one. e. Does the likelihood ratio test determine a unique rejection region? f. What happens if the null and alternative hypotheses are interchanged \(-H_{0}:\) \(\theta=2\) versus \(H_{1}: \theta=1 ?\)

a. Generate samples of size \(25,50,\) and 100 from a normal distribution. Construct probability plots. Do this several times to get an idea of how probability plots behave when the underlying distribution is really normal. b. Repeat part (a) for a chi-square distribution with 10 df. c. Repeat part (a) for \(Y=Z / U,\) where \(Z \sim N(0,1)\) and \(U \sim U[0,1]\) and \(Z\) and \(U\) are independent. d. Repeat part (a) for a uniform distribution. e. Repeat part (a) for an exponential distribution. f. Can you distinguish between the normal distribution of part (a) and the subsequent nonnormal distributions?

Burr (1974) gives the following data on the percentage of manganese in iron made in a blast furnace. For 24 days, a single analysis was made on each of five casts. Examine the normality of this distribution by making a normal probability plot and a hanging rootogram. (As a prelude to topics that will be taken up in later chapters, you might also informally examine whether the percentage of manganese is roughly constant from one day to the next or whether there are significant trends over time.) $$\begin{array}{cccccccccccc} \hline \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } & \text { Day } \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline 1.40 & 1.40 & 1.80 & 1.54 & 1.52 & 1.62 & 1.58 & 1.62 & 1.60 & 1.38 & 1.34 & 1.50 \\ 1.28 & 1.34 & 1.44 & 1.50 & 1.46 & 1.58 & 1.64 & 1.46 & 1.44 & 1.34 & 1.28 & 1.46 \\ 1.36 & 1.54 & 1.46 & 1.48 & 1.42 & 1.62 & 1.62 & 1.38 & 1.46 & 1.36 & 1.08 & 1.28 \\ 1.38 & 1.44 & 1.50 & 1.52 & 1.58 & 1.76 & 1.72 & 1.42 & 1.38 & 1.58 & 1.08 & 1.18 \\ 1.44 & 1.46 & 1.38 & 1.58 & 1.70 & 1.68 & 1.60 & 1.38 & 1.34 & 1.38 & 1.36 & 1.28 \end{array}$$ $$\begin{array}{cccccccccccc} \hline \begin{array}{c} \text { Day } \\ 13 \end{array} & \begin{array}{c} \text { Day } \\ 14 \end{array} & \begin{array}{c} \text { Day } \\ 15 \end{array} & \begin{array}{c} \text { Day } \\ 16 \end{array} & \begin{array}{c} \text { Day } \\ 17 \end{array} & \begin{array}{c} \text { Day } \\ 18 \end{array} & \begin{array}{c} \text { Day } \\ 19 \end{array} & \begin{array}{c} \text { Day } \\ 20 \end{array} & \begin{array}{c} \text { Day } \\ 21 \end{array} & \begin{array}{c} \text { Day } \\ 22 \end{array} & \begin{array}{c} \text { Day } \\ 23 \end{array} & \begin{array}{c} \text { Day } \\ 24 \end{array} \\ \hline 1.26 & 1.52 & 1.50 & 1.42 & 1.32 & 1.16 & 1.24 & 1.30 & 1.30 & 1.48 & 1.32 & 1.44 \\ 1.50 & 1.50 & 1.42 & 1.32 & 1.40 & 1.34 & 1.22 & 1.48 & 1.52 & 1.46 & 1.22 & 1.28 \\ 1.52 & 1.46 & 1.38 & 1.48 & 1.40 & 1.40 & 1.20 & 1.28 & 1.76 & 1.48 & 1.72 & 1.10 \\ 1.38 & 1.34 & 1.36 & 1.36 & 1.26 & 1.16 & 1.30 & 1.18 & 1.16 & 1.42 & 1.18 & 1.06 \\ 1.50 & 1.40 & 1.38 & 1.38 & 1.26 & 1.54 & 1.36 & 1.28 & 1.28 & 1.36 & 1.36 & 1.10 \\ \hline \end{array}$$

It has been suggested that dying people may be able to postpone their death until after an important occasion, such as a wedding or birthday. Phillips and King (1988) studied the patterns of death surrounding Passover, an important Jewish holiday, in California during the years \(1966-1984\). They compared the number of deaths during the week before Passover to the number of deaths during the week after Passover for 1919 people who had Jewish surnames. Of these, 922 occurred in the week before Passover and \(997,\) in the week after Passover. The significance of this discrepancy can be assessed by statistical calculations. We can think of the counts before and after as constituting a table with two cells. If there is no holiday effect, then a death has probability \(\frac{1}{2}\) of falling in each cell. Thus, in order to show that there is a holiday effect, it is necessary to show that this simple model does not fit the data. Test the goodness of fit of the model by Pearson's \(X^{2}\) test or by a likelihood ratio test. Repeat this analysis for a group of males of Chinese and Japanese ancestry, of whom 418 died in the week before Passover and 434 died in the week after. What is the relevance of this latter analysis to the former?

Suppose that the null hypothesis is true, that the distribution of the test statistic, \(T\) say, is continuous with cdf \(F\) and that the test rejects for large values of \(T\). Let \(V\) denote the \(p\) -value of the test. a. Show that \(V=1-F(T).\) b. Conclude that the null distribution of \(V\) is uniform. (Hint: See Proposition \(\mathrm{C}\) of Section \(2.3 .)\) c. If the null hypothesis is true, what is the probability that the \(p\) -value is greater than .1? d. Show that the test that rejects if \(V<\alpha\) has significance level \(\alpha .\)

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