/*! 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 20 Suppose that \(Y_{1}, Y_{2}, Y_{... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) denote a random sample of size 4 from a population with an exponential distribution whose density is given by $$f(y)\left\\{\begin{array}{ll} (1 / \theta) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Let \(X=\sqrt{Y_{1} Y_{2}}\). Find a multiple of \(X\) that is an unbiased estimator for \(\theta\). [Hint: Use your knowledge of the gamma distribution and the fact that \(\Gamma(1 / 2)=\sqrt{\pi}\) to find \(E(\sqrt{Y_{1}})\). Recall that the variables \(Y_{i}\) are independent.] b. Let \(W=\sqrt{Y_{1} Y_{2} Y_{3} Y_{4}}\). Find a multiple of \(W\) that is an unbiased estimator for \(\theta^{2}\). [Recall the hint for part (a).]

Short Answer

Expert verified
A multiple of \(X\) to estimate \(\theta\) is \(\frac{X}{\sqrt{\pi}}\), and for \(W\) to estimate \(\theta^2\), it's \(\frac{W}{\pi}\).

Step by step solution

01

Understand the Problem

The exercise involves dealing with samples from an exponential distribution. Our goal is to find unbiased estimators for \( \theta \) using transformations of random variables drawn from this distribution. To achieve this, we need to understand how to work with the product of these random variables.
02

Exponential Distribution and Expectation

The exponential distribution with parameter \( \theta \) has mean \( \theta \). For a variable \( Y \sim \text{Exponential}(\theta) \), the expectation is \( E(Y) = \theta \). This fact will be useful when understanding the expected value of transformed variables.
03

Connection to Gamma Distribution

A useful property of the exponential distribution is its connection to the gamma distribution. A sum of exponential variables with parameter \( \theta \) is a gamma \((n, \theta)\) variable. Using transformation properties and knowledge of the gamma distribution, we simplify calculations.
04

Solve Part (a): Find the Expectation of \(X\)

Let \( X = \sqrt{Y_1 Y_2} \). Notice that \( Y_1 \sim \text{Exponential}(\theta) \) and \( Y_2 \sim \text{Exponential}(\theta) \) are independent. Using the hint, find \( E(\sqrt{Y_1}) \). For \( Y \sim \text{Exponential}(\theta) \), we know \( \sqrt{Y} \sim \text{folded half-normal}\), leading us to find \( E(\sqrt{Y_1}) = \sqrt{\frac{\pi \theta}{2}} \). Thus, \( E(X) = E(\sqrt{Y_1 Y_2}) = E(Y_1) = \theta \). The unbiased estimator is \( \frac{X}{\Gamma(1/2)} = \frac{X}{\sqrt{\pi}} \).
05

Solve Part (b): Find the Expectation of \(W\)

Here, \( W = \sqrt{Y_1 Y_2 Y_3 Y_4} \). Each \( Y_i \sim \text{Exponential}(\theta) \). Recognize that \( Y_1 Y_2 Y_3 Y_4 \sim \text{Gamma}(n=4, \theta) \). Consider \( E(Y) = \theta \), then \( E(\sqrt{Y}) \) with multiple variables yields \( E(W) = \theta^2 \). An unbiased estimator becomes multiplying \( W \) with a constant derived similarly as part (a), yielding \( \frac{W}{(\Gamma(1/2))^2} = \frac{W}{\pi} \).

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Key Concepts

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

Exponential Distribution
An exponential distribution is a continuous probability distribution that is widely used to model the time until an event occurs, such as the lifespan of a mechanical system before it fails. It is characterized by a parameter \( \theta \), which is the mean or expected value of the distribution. More formally, if \( Y \sim \text{Exponential}(\theta) \), the probability density function (PDF) is given by:
  • \( f(y) = \frac{1}{\theta} e^{-y/\theta} \) for \( y > 0 \)
  • 0 elsewhere

The mean of an exponential distribution is \( \theta \), and it is often used to represent random time intervals between events in a process that happens continuously and independently. This makes the exponential distribution important in fields like reliability testing and queuing theory. Other notable properties include its memoryless property, which means that the probability of an event occurring in the future is independent of any past events.
Gamma Distribution
Expanding from the exponential distribution, we encounter the gamma distribution, which is a two-parameter family of continuous probability distributions. It is often denoted as \( \text{Gamma}(n, \theta) \), where \( n \) is the shape parameter, sometimes denoted as \( k \), and \( \theta \) is the scale parameter. An important property of this distribution is that it can be interpreted as the distribution of the sum of \( n \) independent exponential variables, each with parameter \( \theta \).
The PDF of the gamma distribution is more complex but builds on the exponential's:
  • \( f(y; n, \theta) = \frac{y^{n-1} e^{-y/\theta}}{\theta^n \Gamma(n)} \)
where \( \Gamma(n) \) is the gamma function. This relationship provides a strategic advantage in dealing with problems of summing exponential samples, such as in queuing theory or risk assessment, since it allows us to use properties of the gamma distribution for simplification. In the context of our exercise, using the gamma distribution helps derive expectations and unbiased estimators of parameters like \( X \) and \( W \).
Random Sample
A random sample refers to a subset of individuals chosen from a larger set or population, ensuring that every individual has an equal probability of being chosen. When dealing with random samples, particularly from well-defined distributions like exponential or gamma, it allows for accurate and unbiased estimates of population parameters.
Random samples from the exponential distribution maintain certain properties through transformations. For example, in our exercise, we work with random variables \( Y_1, Y_2, Y_3, \) and \( Y_4 \), which are assumed to be independent samples from an exponential distribution. Their independence is crucial because it validates the use of mathematical properties and formulas related to sums of product terms and expectations when computing unbiased estimators.
  • Independence ensures that properties like expectations and variances multiply straightforwardly.
  • The randomness and independence provide a foundation for estimates that can extend to the entire population.
Understanding how to properly manage and interpret data from a random sample is fundamental in statistics and directly impacts the validity of the conclusions drawn from any analysis.

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Most popular questions from this chapter

The Mars twin rovers, Spirit and Opportunity, which roamed the surface of Mars in the winter of 2004, found evidence that there was once water on Mars, raising the possibility that there was once life on the plant. Do you think that the United States should pursue a program to send humans to Mars? An opinion poll \(^{\star}\) indicated that \(49 \%\) of the 1093 adults surveyed think that we should pursue such a program. a. Estimate the proportion of all Americans who think that the United States should pursue a program to send humans to Mars. Find a bound on the error of estimation. b. The poll actually asked several questions. If we wanted to report an error of estimation that would be valid for all of the questions on the poll, what value should we use? [Hint: What is the maximum possible value for \(p \times q ?]\)

Suppose that we take a sample of size \(n_{1}\) from a normally distributed population with mean and variance \(\mu_{1}\) and \(\sigma_{1}^{2}\) and an independent of sample size \(n_{2}\) from a normally distributed population with mean and variance \(\mu_{2}\) and \(\sigma_{2}^{2} .\) If it is reasonable to assume that \(\sigma_{1}^{2}=\sigma_{2}^{2},\) then the results given in Section 8.8 apply. What can be done if we cannot assume that the unknown variances are equal but are fortunate enough to know that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) for some known constant \(k \neq 1 ?\) Suppose, as previously, that the sample means are given by \(\bar{Y}_{1}\) and \(\bar{Y}_{2}\) and the sample variances by \(S_{1}^{2}\) and \(S_{2}^{2}\), respectively. a. Show that \(Z^{\star}\) given below has a standard normal distribution. $$Z^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sigma_{1} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}$$ b. Show that \(W^{\star}\) given below has a \(\chi^{2}\) distribution with \(n_{1}+n_{2}-2\) df. $$W^{*}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{\sigma_{1}^{2}}$$ c. Notice that \(Z^{\star}\) and \(W^{\star}\) from parts (a) and (b) are independent. Finally, show that $$T^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{S_{p}^{*} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}, \quad \text { where } S_{p}^{2 *}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{n_{1}+n_{2}-2}$$ has a \(t\) distribution with \(n_{1}+n_{2}-2\) df. d. Use the result in part (c) to give a \(100(1-\alpha) \%\) confidence interval for \(\mu_{1}-\mu_{2},\) assuming that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) e. What happens if \(k=1\) in parts \((\mathrm{a}),(\mathrm{b}),(\mathrm{c}),\) and \((\mathrm{d}) ?\)

In a study to compare the perceived effects of two pain relievers, 200 randomly selected adults were given the first pain reliever, and \(93 \%\) indicated appreciable pain relief. Of the 450 individuals given the other pain reliever, \(96 \%\) indicated experiencing appreciable relief. a. Give an estimate for the difference in the proportions of all adults who would indicate perceived pain relief after taking the two pain relievers. Provide a bound on the error of estimation. b. Based on your answer to part (a), is there evidence that proportions experiencing relief differ for those who take the two pain relievers? Why?

The ages of a random sample of five university professors are \(39,54,61,72,\) and \(59 .\) Using this information, find a \(99 \%\) confidence interval for the population standard deviation of the ages of all professors at the university, assuming that the ages of university professors are normally distributed.

In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children; in a sample of 100 non-graduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference in the proportions of firstborn or only children for the two populations from which these samples were drawn. Give a bound for the error of estimation.

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