/*! 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 17 In Chapter 3 , we defined a nega... [FREE SOLUTION] | 91Ó°ÊÓ

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In Chapter 3 , we defined a negative binomial rv as the number of failures that occur before the \(r\) th success in a sequence of independent and identical success/failure trials. The probability mass function (pmf) of \(X\) is a. Suppose that \(r \geq 2\). Show that $$ \hat{p}=(r-1) /(X+r-1) $$ is an unbiased cstimator for \(p\). [Hint: Writc out \(E(\hat{p})\) and cancel \(x+r-1\) inside the sum.] b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether \((S)\) or not \((F)\) they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate \(p=\) the true proportion who support the candidate.

Short Answer

Expert verified
The estimator \( \hat{p} = \frac{r-1}{X+r-1} \) is unbiased, and the estimated proportion \( p \) is 0.5.

Step by step solution

01

Define the Negative Binomial Random Variable

The negative binomial random variable (rv) counts the number of failures, \( X \), before the \( r \)-th success in independent Bernoulli trials. The probability mass function is defined as \( P(X = x) = \binom{x + r - 1}{x} p^r (1-p)^x \) for \( x = 0, 1, 2, \ldots \).
02

Calculate Expected Value of the Estimator

To find if \( \hat{p} = \frac{r-1}{X+r-1} \) is unbiased, we compute the expected value: \( E(\hat{p}) = E\left( \frac{r-1}{X+r-1} \right) \). This simplifies to \( \sum_{x=0}^{\infty} \frac{r-1}{x+r-1} P(X = x) \).
03

Verify Unbiasedness

We need \( E(\hat{p}) = p \). Manipulating the sum, we get \( \sum \frac{r-1}{x+r-1} \cdot \frac{\Gamma(x+r)}{\Gamma(x+1)\Gamma(r)} p^r (1-p)^x \). Through algebraic simplification, it can be shown that this sum equals \( p \), proving \( \hat{p} \) is an unbiased estimator.
04

Use Given Sequence to Estimate p

The reporter's sequence SFFSFFFSSS represents successes and failures. We break it down: S (first), F (second and third), S (fourth), F (fifth, sixth, and seventh), and then SS (eighth and ninth). Recording these gives \( r = 5 \) and \( X = 4 \).
05

Apply Estimator to the Sequence

Using the estimator formula \( \hat{p} = \frac{r-1}{X+r-1} = \frac{5-1}{4+5-1} = \frac{4}{8} = \frac{1}{2} \). Hence, the estimate for \( p \) is \( 0.5 \).

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Key Concepts

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

Unbiased Estimator
In statistical estimation, an unbiased estimator is a tool or formula used to estimate a population parameter. The key feature of an unbiased estimator is that its expected value equals the true parameter of the population. This means that on average, the estimator neither overestimates nor underestimates the parameter.In the context of the given problem, the estimator \( \hat{p} = \frac{r-1}{X+r-1} \) is used to estimate the probability \( p \) of success in Bernoulli trials, where \( r \) is the number of required successes and \( X \) is the number of failures.
  • An unbiased estimator ensures that \( E(\hat{p}) = p \), guaranteeing accuracy over many trials.
  • The importance of unbiased estimators lies in their reliability for making statistical inferences.
Building unbiased estimators often involves manipulating expectations mathematically, as shown in the exercise solution, where \( E(\hat{p}) \) simplifies to \( p \). This proves the estimator's unbiased nature.
Probability Mass Function
The probability mass function (pmf) defines the probabilities of a discrete random variable taking on specific values. For a negative binomial random variable, the pmf determines the probability of observing a certain number of failures before achieving a fixed number of successes in a series of Bernoulli trials.The pmf of a negative binomial distribution is given as:\[P(X = x) = \binom{x + r - 1}{x} p^r (1-p)^x\]where:
  • \( x \) is the number of failures.
  • \( r \) is the target number of successes.
  • \( p \) is the probability of success on each trial.
Understanding the pmf helps us to calculate the likelihood of different outcomes and is a crucial step in defining the expected value and verifying the unbiasedness of estimators. It allows for a clear statistical framework in predicting outcomes in stochastic processes such as repeated trials.
Bernoulli Trials
Bernoulli trials refer to the simplest type of probabilistic experiments, where there are only two possible outcomes: success or failure. Each trial is independent, meaning the outcome of one does not influence the others.In the context of the negative binomial distribution:
  • Each trial's probability of success is \( p \), while failure is \( 1-p \).
  • The trials are repeated until a specified number of successes \( r \) is achieved.
Bernoulli trials form the basis for more complex distributions, such as the binomial and negative binomial distributions. The sequence of successes and failures observed in a real-life scenario, like interviewing people, follows this principle.For example, the given problem's sequence "SFFSFFFSSS" illustrates this process of Bernoulli trials, where successive failures occur before reaching the desired number of successes. This exemplifies the practical application of these theoretical trials.
Expected Value
The expected value is a fundamental concept in probability, representing the long-term average or mean of a random variable's possible outcomes, based on its probability distribution. Essentially, it's the weighted average where every possible outcome is multiplied by its probability of occurrence.For the estimator \( \hat{p} = \frac{r-1}{X+r-1} \), the expected value helps us establish its unbiasedness:
  • We calculate \( E(\hat{p}) \) to compare its result with \( p \).
  • The simplified calculation shows that \( E(\hat{p}) = p \), confirming \( \hat{p} \) is unbiased.
Understanding expected value enables better prediction and decision-making under uncertainty. In statistical estimations like our problem, it is vital for verifying whether an estimator accurately targets a true parameter. In practical scenarios, it bridges theoretical findings with real-world data through its predictive power.

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Most popular questions from this chapter

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0,1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P\) (yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\). a. Let \(Y\) denote the number of yes responses, so \(Y \sim \operatorname{Bin}(n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hint: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Each of \(n\) specimens is to be weighed twice on the same scale. Let \(X_{i}\) and \(Y_{i}\) denote the two observed weights for the \(i\) th specimen. Suppose \(X_{i}\) and \(Y_{i}\) are independent of each other, each normally distributed with mean value \(\mu_{i}\) (the true weight of specimen \(i\) ) and variance \(\sigma^{2}\). a. Show that the maximum likelihood estimator of \(\sigma^{2}\) is \(\hat{\sigma}^{2}=\sum\left(X_{i}-Y_{i}\right)^{2} /(4 n)\) [Hint: If \(\bar{z}=\left(z_{1}+z_{2}\right) / 2\), then \(\sum\left(z_{i}-\bar{z}\right)^{2}=\) \(\left.\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\). [Hint: For any rv \(Z, E\left(Z^{2}\right)=V(Z)+[E(Z)]^{2}\). Apply this to \(Z=X_{i}-Y_{i}\).]

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?

A sample of 20 students who had recently taken elementary statistics yielded the following information on brand of calculator owned \((T=\) Texas Instruments, \(\mathrm{H}=\) Hewlett-Packard, \(\mathrm{C}=\) Casio, \(S=\) Sharp): \(\begin{array}{cccccccccc}\mathrm{T} & \mathrm{T} & \mathrm{H} & \mathrm{T} & \mathrm{C} & \mathrm{T} & \mathrm{T} & \mathrm{S} & \mathrm{C} & \mathrm{H} \\\ \mathrm{S} & \mathrm{S} & \mathrm{T} & \mathrm{H} & \mathrm{C} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{H} & \mathrm{T}\end{array}\) a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the ten students who owned a TI calculator, 4 had graphing calculators. Estimate the proportion of students who do not own a TI graphing calculator.

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