/*! 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 59 Suppose \(X_{1}, X_{2}, \ldots, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 59

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample of size \(n\) drawn from a Poisson pdf where \(\lambda\) is an unknown parameter. Show that \(\hat{\lambda}=\bar{X}\) is unbiased for \(\lambda\). For what type of parameter, in general, will the sample mean necessarily be an unbiased estimator? (Hint: The answer is implicit in the derivation showing that \(\bar{X}\) is unbiased for the Poisson \(\lambda\).)

Short Answer

Expert verified
The sample mean \(\hat{\lambda}=\bar{X}\) is an unbiased estimator for \(\lambda\) in the Poisson distribution because \(E[\hat{\lambda}] = \lambda\). The sample mean will necessarily be an unbiased estimator for parameters that are equal to the expected value of the distribution.

Step by step solution

01

Compute the sample mean

Firstly, need to compute the sample mean given by \(\hat{\lambda}=\bar{X}\) where \(\bar{X}\) is the sample mean of \(X_{1}, X_{2}, \ldots, X_{n}\) which equals to \(\frac{1}{n} \sum_{i=1}^n X_{i}\).
02

Calculate the expected value

Then, need to calculate the expected value of \(\hat{\lambda}\). This is given by \(E[\hat{\lambda}]\) which equals to \(E[\bar{X}]\). Using the linearity of expectation we can then write \(E[\bar{X}]\) as \(\frac{1}{n}E[\sum_{i=1}^n X_{i}]\) which simplifies to \(E[X]\). From the characteristics of the Poisson distribution, we know that the expectation of \(X\) is equal to the parameter \(\lambda\).
03

Discuss the result

Since \(E[\hat{\lambda}] = \lambda\), we say that \(\hat{\lambda}\) is an unbiased estimator of \(\lambda\). This is because by definition, an estimator is said to be unbiased if its expected value is equal to the parameter it is estimating.
04

Answer the second part of the question

The sample mean is necessarily an unbiased estimator for parameters that are equal to the expected value of the distribution. This is because of the properties of estimators. For example, the sample mean is an unbiased estimator of the population mean in many distributions (Normal, Uniform, Binomial, Poisson etc.), and in this case, for the Poisson distribution, it is an unbiased estimator of \(\lambda\).

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

. Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be a random sample from \(f_{Y}(y ; \theta)=\frac{1}{3} e^{-y / \theta}, y>0\). Compare the Cram茅r-Rao lower bound for \(f_{Y}(y ; \theta)\) to the variance of the maximum likelihood estimator for \(\theta, \hat{\theta}=\frac{1}{n} \sum_{k=1}^{n} Y_{i}\). Is \(\bar{Y}\) a best estimator for \(\theta\) ?

For the negative binomial pdf \(p x(k ; p, r)=\) \(\left(\begin{array}{c}k+r-1 \\\ k\end{array}\right)(1-p)^{k} p^{\prime}\), find the maximum likelihood estimator for \(p\) if \(r\) is known.

Consider, again, the scenario described in Example \(5.8 .2\) - a binomial random variable \(X\) has parameters \(n\) and \(\theta\), where the latter has a beta prior with integer parameters \(r\) and \(s\). Integrate the joint pdf \(p_{X}(k \mid \theta) f_{\theta}(\theta)\) with respect to \(\theta\) to show that the marginal pdf of \(X\) is given by $$ p_{X}(k)=\frac{\left(\begin{array}{c} k+r-1 \\ k \end{array}\right)\left(\begin{array}{c} n-k+s-1 \\ n-k \end{array}\right)}{\left(\begin{array}{c} n+r+s-1 \\ n \end{array}\right)}, \quad k=0,1, \ldots, n $$

Suppose that \(X\) is a geometric random variable, where \(p_{X}(k \mid \theta)=(1-\theta)^{k-1} \theta, k=1,2, \ldots\). Assume that the prior distribution for \(\theta\) is the beta pdf with parameters \(r\) and \(s\). Find the posterior distribution for \(\theta\).

(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 \(\theta_{2}\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.