/*! 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 106 Suppose that \(Y_{1}, Y_{2}, \ld... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 106

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a Poisson distribution with mean \(\lambda\). Find the MVUE of \(P\left(Y_{i}=0\right)=e^{-\lambda} .\) [Hint: Make use of the Rao-Blackwell theorem.]

Short Answer

Expert verified
The MVUE of \( P(Y_i=0) = e^{-\lambda} \) is \( e^{-T/n} \), where \( T = \sum Y_i \).

Step by step solution

01

Identify the Estimator

First, note that the parameter of interest is the probability of obtaining zero from a Poisson distribution: \( P(Y_i = 0) = e^{- heta} \). Here, \( \theta \) represents the mean and variance of the Poisson distribution, denoted \( \lambda \).
02

State the Unbiased Estimator

The unbiased estimator for \( \lambda \) for a random sample from a Poisson distribution \( Y_1, Y_2, \ldots, Y_n \) is the sample mean \( \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \). This is an unbiased estimator since \( E(\bar{Y}) = \lambda \).
03

Apply Rao-Blackwell Theorem

To use the Rao-Blackwell theorem, we define a function \( g(\bar{Y}) = e^{-\bar{Y}} \) because \( e^{-\bar{Y}} \) is an unbiased estimator for \( P(Y_i = 0) = e^{-\lambda} \). The Rao-Blackwell theorem states that the MVUE can be derived by conditioning an unbiased estimator on a sufficient statistic.
04

Determine the Sufficient Statistic

For a Poisson distribution, the sufficient statistic for \( \lambda \) is \( T = \sum_{i=1}^{n} Y_i \). Thus, \( \bar{Y} = \frac{T}{n} \) is a function of the sufficient statistic.
05

Find Conditional Expectation

The expectation of \( e^{-\bar{Y}} \) conditioned on \( T \) is the same as using the fact that \( E(e^{-\bar{Y}}) = e^{-\lambda} \). Given that \( T \) holds the total sum from a Poisson distribution, the expected value \( E(e^{-\bar{Y}}|T) = e^{-T/n} \), where \( E(T) = n\lambda \), matches the original unbiased form.
06

Conclude with the Calculated MVUE

Given the derivations, the function of the sufficient statistic \( e^{-T/n} \) is concluded to be the MVUE for \( P(Y_i = 0) = e^{-\lambda} \). This is the minimum variance unbiased estimator sourced by using Rao-Blackwell.

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!

Key Concepts

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

Poisson Distribution
A Poisson distribution is a statistical concept that describes the probability of a given number of events happening in a fixed interval of time or space, provided these events occur with a known constant rate and independently of the time since the last event. Some key points about the Poisson distribution are:
  • The parameter \( \lambda \) represents both the mean and the variance of the distribution. It indicates how often an event is likely to occur.
  • The measure is particularly useful for modeling count data and rare events.
  • The probability of observing exactly \( k \) events in an interval is given by the formula: \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \), where \( k \) is a non-negative integer, \( \lambda \) is the average number of events, and \( e \) is the base of the natural logarithm.
This distribution is commonly used in fields like natural sciences and business, where it helps model phenomena such as the number of emails received in an hour or earthquakes in a region.
Rao-Blackwell Theorem
The Rao-Blackwell Theorem is a fundamental theorem in the field of statistics that helps in obtaining a better estimator, known as the minimum variance unbiased estimator (MVUE). It allows us to improve the quality of an unbiased estimator by conditioning it on a sufficient statistic. Key aspects of the Rao-Blackwell Theorem include:
  • The theorem can be applied when an unbiased estimator and a sufficient statistic are available for the parameter of interest.
  • By deriving the conditional expectation of an unbiased estimator given a sufficient statistic, one can obtain an estimator with a lower or equal variance.
  • This results in an MVUE, which, as the name suggests, has the minimum variance among all unbiased estimators.
The use of Rao-Blackwell Theorem is particularly powerful when working with complex datasets as it simplifies estimations while maintaining precision.
Sufficient Statistic
A sufficient statistic is a crucial concept in statistics, especially in the context of parameter estimation. It contains all the information needed to compute any estimate of a distribution parameter. Here are some important aspects of a sufficient statistic:
  • A statistic \( T \) is said to be sufficient for a parameter \( \theta \) if the conditional distribution of the data given \( T \) does not depend on \( \theta \).
  • For the Poisson distribution with mean \( \lambda \), the statistic \( T = \sum_{i=1}^{n} Y_i \) is sufficient. This means that all the information about \( \lambda \) in the sample is captured by this total sum.
  • Using a sufficient statistic simplifies the estimation process, as estimators based on sufficient statistics are generally more efficient.
The concept of sufficient statistics streamlines computation, making the estimation process rational and focused on reducing variability in estimators.

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 \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of size \(n\) from a uniform distribution on the interval \((0, \theta) .\) If \(Y_{(1)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right),\) the result of Exercise 8.18 is that \(\hat{\theta}_{1}=(n+1) Y_{(1)}\) is an unbiased estimator for \(\theta .\) If \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right),\) the results of Example 9.1 imply that \(\hat{\theta}_{2}=[(n+1) / n] Y_{(n)}\) is another unbiased estimator for \(\theta\). Show that the efficiency of \(\hat{\theta}_{1}\) to \(\hat{\theta}_{2}\) is \(1 / n^{2}\). Notice that this implies that \(\hat{\theta}_{2}\) is a markedly superior estimator.

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample of size \(n\) from a normal distribution with mean \(\mu\) and variance \(1 .\) Consider the first observation \(Y_{1}\) as an estimator for \(\mu\) a. Show that \(Y_{1}\) is an unbiased estimator for \(\mu\). b. Find \(P\left(\left|Y_{1}-\mu\right| \leq 1\right)\) c. Look at the basic definition of consistency given in Definition 9.2. Based on the result of part (b), is \(Y_{1}\) a consistent estimator for \(\mu\) ?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a Weibull distribution with known \(m\) and unknown \(\alpha\). (Refer to Exercise \(6.26 .\) ) Show that \(\sum_{i=1}^{n} Y_{i}^{m}\) is sufficient for \(\alpha\).

20\. Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from the uniform distribution over the interval \(\left(\theta_{1}, \theta_{2}\right)\). Show that \(Y_{(1)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) and \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) are jointly sufficient for \(\theta_{1}\) and \(\theta_{2}\).

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a Poisson distribution with mean \(\lambda\) and define $$ W_{n}=\frac{Y-\lambda}{\sqrt{\bar{Y} / n}} $$ a. Show that the distribution of \(W_{n}\) converges to a standard normal distribution. b. Use \(W_{n}\) and the result in part (a) to derive the formula for an approximate \(95 \%\) confidence interval for \(\lambda\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Mechanics 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.