/*! 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 6 Suppose that \(E\left(\hat{\thet... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(E\left(\hat{\theta}_{1}\right)=E\left(\hat{\theta}_{2}\right)=\theta, V\left(\hat{\theta}_{1}\right)=\sigma_{1}^{2},\) and \(V\left(\hat{\theta}_{2}\right)=\sigma_{2}^{2} .\) Consider the estimator \(\hat{\theta}_{3}=a \hat{\theta}_{1}+(1-a) \hat{\theta}_{2}\) a. Show that \(\hat{\theta}_{3}\) is an unbiased estimator for \(\theta\). b. If \(\hat{\theta}_{1}\) and \(\hat{\theta}_{2}\) are independent, how should the constant \(a\) be chosen in order to minimize the variance of \(\hat{\theta}_{3} ?\)

Short Answer

Expert verified
a) \( \hat{\theta}_{3} \) is unbiased. b) Choose \( a = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}} \).

Step by step solution

01

Verifying Unbiasedness of Estimator

To show that \( \hat{\theta}_{3} \) is an unbiased estimator of \( \theta \), we need to demonstrate that \[ E(\hat{\theta}_{3}) = \theta. \]Given that \( E(\hat{\theta}_{1}) = E(\hat{\theta}_{2}) = \theta \), calculate:\[ E(\hat{\theta}_{3}) = E(a \hat{\theta}_{1} + (1-a) \hat{\theta}_{2}) = a E(\hat{\theta}_{1}) + (1-a) E(\hat{\theta}_{2}) = a\theta + (1-a)\theta. \]Simplifying gives:\[ E(\hat{\theta}_{3}) = (a + 1-a)\theta = \theta. \]Thus, \( \hat{\theta}_{3} \) is an unbiased estimator.
02

Computing the Variance of the New Estimator

The variance of \( \hat{\theta}_{3} \) is given by:\[ V(\hat{\theta}_{3}) = V(a \hat{\theta}_{1} + (1-a)\hat{\theta}_{2}) = a^2 V(\hat{\theta}_{1}) + (1-a)^2 V(\hat{\theta}_{2}) \]Using \( V(\hat{\theta}_{1}) = \sigma_{1}^{2} \) and \( V(\hat{\theta}_{2}) = \sigma_{2}^{2} \), we substitute:\[ V(\hat{\theta}_{3}) = a^2 \sigma_{1}^{2} + (1-a)^2 \sigma_{2}^{2}. \]
03

Minimizing the Variance

To minimize the variance \( V(\hat{\theta}_{3}) \), we need to find the optimal \( a \). We do this by taking the derivative of \( V(\hat{\theta}_{3}) \) with respect to \( a \) and setting it to zero:\[ \frac{d}{da}(a^2 \sigma_{1}^{2} + (1-a)^2 \sigma_{2}^{2}) = 2a \sigma_{1}^{2} - 2(1-a)\sigma_{2}^{2}. \]Setting \( \frac{d}{da} = 0 \) gives:\[ 2a \sigma_{1}^{2} - 2\sigma_{2}^{2} + 2a \sigma_{2}^{2} = 0. \]\[ 2a ( \sigma_{1}^{2} + \sigma_{2}^{2}) = 2 \sigma_{2}^{2}. \]Thus, \[ a = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}}. \]

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 91Ó°ÊÓ!

Key Concepts

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

Variance Minimization
While dealing with multiple estimators, we often aim for variance minimization to enhance accuracy. This involves combining basic estimators to form a new estimator, something that balances their contributions. Here, our goal is to minimize the variance of the combined estimator, \( \hat{\theta}_{3} \), thus making it more precise.
When \( \hat{\theta}_{1} \) and \( \hat{\theta}_{2} \) are independent estimators, their combination \( a \hat{\theta}_{1} + (1-a) \hat{\theta}_{2} \) offers a new way to determine the target parameter \( \theta \). The variance of this combination, given by \( a^2 \sigma_{1}^{2} + (1-a)^2 \sigma_{2}^{2} \), needs optimization.
By calculating the derivative and setting it to zero, we find the value of \( a \) that minimizes this variance. Thus, variance minimization ensures we derive value from both estimators effectively and accurately shift towards a lower error potential.
Estimator Variance
The estimator variance is a crucial metric in assessing estimator effectiveness. It quantifies how much the estimator's measurements might differ from their expected values. Lower variance indicates an estimator that is usually closer to the parameter's actual value, hence more reliable.
In our exercise, \( V(\hat{\theta}_{3}) = a^2 \sigma_{1}^{2} + (1-a)^2 \sigma_{2}^{2} \) represents the variance for our combined estimator. The task is to find the optimal \( a \) to achieve minimal variance, enhancing estimation precision.
It's important to note that when the variances \( \sigma_{1}^{2} \) and \( \sigma_{2}^{2} \) differ, the choice of \( a \) becomes significant in maintaining balance. By optimally choosing \( a = \frac{\sigma_{2}^{2}}{\sigma_{1}^{2} + \sigma_{2}^{2}} \), the balance is achieved, ensuring the new estimator \( \hat{\theta}_{3} \) is robust and reliable.
Statistical Estimators
Statistical estimators play a key role in inferential statistics, where we aim to determine unknown population parameters through sample data. An estimator, like \( \hat{\theta}_{3} \), serves as a tool to estimate the parameter of interest, in this case, \( \theta \). The accuracy of an estimator is defined by two key attributes: unbiasedness and variability.
An estimator is unbiased if the expected value equals the true parameter value, as established in our example with \( E(\hat{\theta}_{3}) = \theta \). This unbiasedness ensures that the estimator neither systematically overstates nor understates the parameter.
Understanding and selecting statistical estimators, like in the case of combining \( \hat{\theta}_{1} \) and \( \hat{\theta}_{2} \), is essential for reliable statistical inference, as it dictates the degree of precision and error in our estimations.

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 \(Y\) have probability density function $$f_{Y}(y)=\left\\{\begin{array}{ll} \frac{2(\theta-y)}{\theta^{2}}, & 0

Suppose that you want to estimate the mean pH of rainfalls in an area that suffers from heavy pollution due to the discharge of smoke from a power plant. Assume that \(\sigma\) is in the neighborhood of \(.5 \mathrm{pH}\) and that you want your estimate to lie within. 1 of \(\mu\) with probability near. 95 Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain.

A survey of 415 corporate, government, and accounting executives of the Financial Accounting Foundation found that 278 rated cash flow (as opposed to earnings per share, etc.) as the most important indicator of a company's financial health. Assume that these 415 executives constitute a random sample from the population of all executives. Use the data to find a \(95 \%\) confidence interval for the fraction of all corporate executives who consider cash flow the most important measure of a company's financial health.

To estimate the proportion of unemployed workers in Panama, an economist selected at random 400 persons from the working class. Of these, 25 were unemployed. a. Estimate the true proportion of unemployed workers and place bounds on the error of estimation. b. How many persons must be sampled to reduce the bound on the error of estimation to.02?

Suppose that two independent random samples of \(n_{1}\) and \(n_{2}\) observations are selected from normal populations. Further, assume that the populations possess a common variance \(\sigma^{2}\). Let $$S_{i}^{2}=\frac{\sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i}\right)^{2}}{n_{i}-1}, \quad i=1,2$$ a. Show that \(S_{p}^{2}\), the pooled estimator of \(\sigma^{2}\) (which follows), is unbiased: $$S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2}$$ b. Find \(V\left(S_{p}^{2}\right)\)

See all solutions

Recommended explanations on Math Textbooks

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.