91Ó°ÊÓ

The principle of unbiasedness (prefer an unbiased estimator to any other) has been criticized on the grounds that in some situations the only unbiased estimator is patently ridiculous. Here is one such example. Suppose that the number of major defects \(X\) on a randomly selected vehicle has a Poisson distribution with parameter \(\lambda\). You are going to purchase two such vehicles and wish to estimate \(\theta=P\left(X_{1}=0, \quad X_{2}=0\right)=e^{-2 \lambda}\), the probability that neither of these vehicles has any major defects. Your estimate is based on observing the value of \(X\) for a single vehicle. Denote this estimator by \(\hat{\theta}=\delta(X)\). Write the equation implied by the condition of unbiasedness, \(E[\delta(X)]=e^{-2 \lambda}\), cancel \(e^{-\lambda}\) from both sides, then expand what remains on the right-hand side in an infinite series,and compare the two sides to determine \(\delta(X)\). If \(X=200\), what is the estimate? Does this seem reasonable? What is the estimate if \(X=199 ?\) Is this reasonable?

Step by step solution

01

Understanding the Requirement

We are to find an unbiased estimator \(\delta(X)\) such that the expected value \(E[\delta(X)] = e^{-2 \lambda}\).

02

Setting up the Equation

The condition for unbiased estimation requires \(E[\delta(X)] = e^{-2\lambda}\). We replace \(\delta(X)\) with its expression based on a single observation: \(\delta(X)\).

03

Expected Value Expression

For a Poisson distribution with parameter \(\lambda\), the probability mass function is \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\). We express the expectation:\[E[\delta(X)] = \sum_{k=0}^{\infty} \delta(k) \frac{\lambda^k e^{-\lambda}}{k!} = e^{-2\lambda}.\]

04

Cancel \(e^{-\lambda}\) and Rearrange

Cancel \(e^{-\lambda}\) from both sides of the expectation equation:\[\sum_{k=0}^{\infty} \delta(k) \frac{\lambda^k}{k!} = e^{-\lambda}.\]

05

Infinite Series Expansion

Expand the right-hand side of \(e^{-\lambda}\) in an infinite series:\[e^{-\lambda} = \sum_{m=0}^{\infty} \frac{(-\lambda)^m}{m!}.\]

06

Equating Terms from Series Expansion

Equate the terms of both series, use trial and error or creative insight to find \(\delta(k)\) such that:- \(\delta(0) = 1\)- \(\delta(k) = 0\) for all \(k eq 0\).Therefore, \(\delta(X) = \mathbb{I}(X=0)\), where \(\mathbb{I}\) is the indicator function that equals 1 if \(X=0\) and 0 otherwise.

07

Evaluating \(\delta(X)\) for Given Values

With the solution \(\delta(X) = \mathbb{I}(X=0)\), evaluate for \(X=200\) and \(X=199\):- If \(X=200\), \(\delta(X) = 0\), indicating it is "impossible" or unreasonable for this value to represent no major defects.- If \(X=199\), \(\delta(X) = 0\), the same reasoning applies.

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Key Concepts

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

Poisson Distribution

The Poisson distribution is a popular statistical concept used to model rare events over a fixed interval of time or space. For instance, it can describe the number of major defects in vehicles manufactured during a particular period. The key characteristic of a Poisson distribution is its parameter \(\lambda\), which represents the average rate of occurrence of the event.
Each outcome in a Poisson process is independent of the others, and the probability of observing exactly \(k\) events (such as defects) is given by the formula:

• \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \)

This formula is known as the probability mass function of a Poisson distribution, explaining how probable it is for any specific number of events to occur. In the exercise, when estimating the number of vehicles with zero defects, we express our findings using this distribution. Understanding the mechanics of Poisson distribution helps in effective modeling of data that follows this pattern.

Expected Value

Expected value is a fundamental concept in probability and statistics. It serves as the weighted average of all possible outcomes of a random variable, where each outcome is weighted by its probability of occurrence. For a discrete random variable, it is calculated by summing the products of each possible value and its probability.

• The formula is \( E[X] = \sum x_i p(x_i) \)

In the context of a Poisson distribution, calculating the expected value involves integrating the probability mass function over all possible values of \(X\). In the exercise, the expected value of the estimator \(\delta(X)\) should equal \(e^{-2\lambda}\), an indicator of no major defects in two vehicles. This criterion of unbiasedness ensures that the estimator truly reflects the underlying parameter it’s trying to estimate.

Indicator Function

The indicator function, denoted by \( \mathbb{I}(condition) \), is a simple yet powerful mathematical tool. It acts similar to a binary switch, returning 1 when a specified condition is true and 0 otherwise. In statistics, it's often used to model outcomes that can only be in one of two states: happening or not happening.

• Example: \( \mathbb{I}(X = 0) \) returns 1 if \(X = 0\), otherwise returns 0.

In our exercise, the indicator function \(\delta(X) = \mathbb{I}(X=0)\) checks whether there are no major defects on a vehicle. When \(X\) (number of defects) is not zero, \(\delta(X)\) becomes 0, marking it unreasonable for vehicles with multiple defects to have none at all. Using such a function simplifies comparison and clarifies how certain calculations hinge on specific conditions.

Probability Mass Function

A probability mass function (PMF) provides the probabilities of outcomes for a discrete random variable. The sum of all probabilities in a PMF equals 1. In a Poisson distribution, the PMF helps determine the likelihood of different numbers of events occurring.

• The PMF formula for a Poisson distribution is: \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \)

In this exercise, the PMF determines the likelihood of observing a certain number of defects in the vehicles, with parameter \(\lambda\) controlling the average number of occurrences. By setting \(E[\delta(X)] = e^{-2\lambda}\), the exercise highlights how the PMF contributes to estimating the probability of zero defects in two vehicles. The PMF is essential since it directly impacts our understanding of the estimator's performance and towards unveiling unbiased estimations.