/*! 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 75 Let \(Y_{1}, Y_{2}, \ldots, Y_{n... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 75

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be a random sample of size \(n\) from the pdf \(f_{Y}(y ; \theta)=\frac{1}{\theta} e^{-y / \theta}, y>0\). (a) Show that \(\hat{\theta}_{1}=Y_{1}, \hat{\theta}_{2}=\bar{Y}\), and \(\hat{\theta}_{3}=n \cdot Y_{\min }\) are all unbiased estimators for \(\theta\). (b) Find the variances of \(\theta_{1}, \theta_{2}\), and \(\hat{\theta}_{3}\).

Short Answer

Expert verified
The unbiased estimators for \(\theta\) are calculated to be \(Y_{1}, \bar{Y}\) and \(n \cdot Y_{min}\) with variances \(\theta^2\), \(\frac{\theta^2}{n}\), and \(\frac{\theta^2}{n^2}\) respectively.

Step by step solution

01

Calculation of Expectation for \(\hat{\theta}_1\)

Calculate the expectation of \(Y_1\). As \(Y_1\) follows an exponential distribution, by property, the expected value of \(Y_1\) is \(\theta\). Hence, \(E(\hat{\theta}_1) = E(Y_1) = \theta\). So, \(\hat{\theta}_1\) is an unbiased estimator of \(\theta\).
02

Calculation of Expectation for \(\hat{\theta}_2\)

\(\hat{\theta}_2 = \bar{Y} = \frac{1}{n} \sum \limits _{i=1} ^{n} Y_{i}\). Each \(Y_i\) is an IID (Independent and Identically Distributed) random variable from an exponential distribution. Therefore, \(E(\hat{\theta}_2) = E(\bar{Y}) = E(\frac{1}{n} \sum \limits _{i=1} ^{n} Y_{i}) = \frac{1}{n} \sum \limits _{i=1} ^{n} E(Y_{i}) = \frac{1}{n} \sum \limits _{i=1} ^{n} \theta = \theta\). Hence, \(\hat{\theta}_2\) is also an unbiased estimator of \(\theta\).
03

Calculation of Expectation for \(\hat{\theta}_3\)

For the minimum of a sample \(Y_{min}\) from exponential distribution with parameter \(\frac{1}{\theta}\), we have \(E(Y_{min}) = \frac{1}{n\theta}\). Therefore, \(E(\hat{\theta}_3) = E(n \cdot Y_{min}) = n \cdot E(Y_{min}) = n \cdot \frac{1}{n\theta} = \theta\). Thus, \(\hat{\theta}_3\) is an unbiased estimator of \(\theta\).
04

Variance of \(\hat{\theta}_1\)

The variance of an exponential distribution is \(\theta^2\). So, \(\text{Var}(\hat{\theta}_1) = \text{Var}(Y_{1}) = \theta^2\).
05

Variance of \(\hat{\theta}_2\)

The variance of the average of IID (Independent and Identically Distributed) random variables is the variance of the individual variables divided by the number of variables. Therefore, \(\text{Var}(\hat{\theta}_2) = \text{Var}(\bar{Y}) = \frac{1}{n} \cdot \theta^2\).
06

Variance of \(\hat{\theta}_3\)

The variance of the minimum of a sample from an exponential distribution is \(\frac{1}{(n^2)\theta^2}\). Hence, \(\text{Var}(\hat{\theta}_3) = \text{Var}(n \cdot Y_{min}) = \text{Var}(Y_{min}) = \frac{1}{n^2} \cdot \theta^2\). Note that the expectation operator is linear but the variance operator is not, so the variance of a constant times a random variable is not the constant times the variance of the random variable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

suppose that \(Y_{1}=8.3, Y_{2}=4.9, Y_{3}=2.6\), and \(Y_{4}=6.5\) is a random sample of size 4 from the twoparameter uniform pdf, $$ f_{Y}\left(y ; \theta_{1}, \theta_{2}\right)=\frac{1}{2 \theta_{2}}, \quad \theta_{1}-\theta_{2} \leq y \leq \theta_{1}+\theta_{2} $$ Use the method of moments to calculate \(\theta_{1 e}\) and \(\theta_{2 e}\).

. If the random variable \(Y\) denotes an individual's income, Pareto's law claims that \(P(Y \geq y)=\left(\frac{k}{y}\right)^{9}\), where \(k\) is the entire population's minimum income. It follows that \(F_{Y}(y)=1-\left(\frac{k}{y}\right)^{9}\), and, by differentiation, $$ f y(y ; \theta)=\theta k^{3}\left(\frac{1}{y}\right)^{\theta+1}, \quad y \geq k ; \quad \theta \geq 1 $$ Assume \(k\) is known. Find the maximum likelihood estimator for \(\theta\) if income information has been collected on a random sample of 25 individuals.

(a) Based on the random sample \(Y_{1}=6.3, Y_{2}=\) \(1.8, Y_{3}=14.2\), and \(Y_{4}=7.6\), use the method of maximum likelihood to estimate the parameter \(\theta\) in the uniform pdf $$ f_{Y}(y ; \theta)=\frac{1}{\theta}, \quad 0 \leq y \leq \theta $$ (b) Suppose the random sample in part (a) represents the two-parameter uniform pdf $$ f_{Y}\left(y ; \theta_{1}, \theta_{2}\right)=\frac{1}{\theta_{2}-\theta_{1}}, \quad \theta_{1} \leq y \leq \theta_{2} $$ Find the maximum likelihood estimates for \(\theta_{1}\) and \(\theta_{2}\).

A criminologist is searching through FBI files to document the prevalence of a rare double-whorl fingerprint. Among six consecutive sets of 100,000 prints scanned by a computer, the numbers of persons having the abnormality are \(3,0,3,4,2\), and 1, respectively. Assume that double whorls are Poisson events. Use the method of moments to estimate their occurrence rate, \(\lambda\). How would your answer change if \(\lambda\) were estimated using the method of maximum likelihood?

A random sample of size \(2, Y_{1}\) and \(Y_{2}\), is drawn from the pdf $$ f_{Y}(y ; \theta)=2 y \theta^{2}, \quad 0
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.