/*! 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 46 Let \(X_{1}, X_{2}, \ldots, X_{n... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 46

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution with known mean \(\mu\) but with the variance \(\sigma^{2}\) as the unknown parameter. a. Find the information in a single observation and the Cram茅r-Rao lower bound. b. Find the mle of \(\sigma^{2}\). c. Find the distribution of the mle. d. Is the mle an efficient estimator? Explain. e. Is the answer to part (c) in conflict with the asymptotic distribution of the mle given by the second theorem? Explain.

Short Answer

Expert verified
a. Fisher information is \( \frac{1}{2(\sigma^2)^2} \); CRLB is \( \frac{2\sigma^4}{n} \). b. MLE is \( \hat{\sigma}^2 = \frac{1}{n} \sum (X_i - \mu)^2 \). c. Distribution: \( \frac{\sigma^2}{n} \cdot \chi^2(n) \). d. Yes, MLE is efficient. e. No conflict with asymptotic results.

Step by step solution

01

Understanding Fisher Information

In the normal distribution, each observation \(X_i\) has a distribution \(N(\mu, \sigma^2)\). The Fisher information in a single observation \(X_i\) with respect to \(\sigma^2\) is given by:\[ I(\sigma^2) = -E\left[ \frac{\partial^2 \log L}{\partial (\sigma^2)^2} \right] \]where \(\log L\) is the log-likelihood function. Calculating the second derivative, we find\[ I(\sigma^2) = \frac{1}{2(\sigma^2)^2} \].Hence, the Fisher information in a single observation is \( \frac{1}{2(\sigma^2)^2} \).
02

Calculating Cram茅r-Rao Lower Bound

The Cram茅r-Rao lower bound for an unbiased estimator of \(\sigma^2\) is the reciprocal of the Fisher information from a single observation:\[ \text{Cram茅r-Rao lower bound} = \frac{2(\sigma^2)^2}{n} \].
03

Finding the Maximum Likelihood Estimator (MLE) of \(\sigma^2\)

The likelihood function based on a sample \(X_1, X_2, \ldots, X_n\) is\[ L(\sigma^2) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right) \].Taking the log of this function gives\[ \log L = -\frac{n}{2}\log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \mu)^2 \].Maximizing this with respect to \(\sigma^2\) results in the estimator:\[ \hat{\sigma^2}_{MLE} = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \].
04

Determining the Distribution of the MLE

The distribution of \( \hat{\sigma^2}_{MLE} \) can be determined by considering it as a scaled sum of squared normal variables. If \(X_i\) are from \(N(\mu, \sigma^2)\), then \((X_i - \mu)^2 / \sigma^2\) follows a \(\chi^2(1)\) distribution. The sum \(\sum (X_i - \mu)^2 / \sigma^2\) follows a \(\chi^2(n)\) distribution. Thus,\[ \hat{\sigma^2}_{MLE} \sim \frac{\sigma^2}{n} \cdot \chi^2(n) \].
05

Analyzing the Efficiency of the MLE

The MLE is efficient if its variance achieves the Cram茅r-Rao lower bound. From Step 4, we have \( \hat{\sigma^2}_{MLE} \sim \frac{\sigma^2}{n} \cdot \chi^2(n) \) and its variance is \(\frac{2\sigma^4}{n}\). This equals the Cram茅r-Rao lower bound, so the MLE is efficient.
06

Comparing with Asymptotic Distribution

The asymptotic distribution of the MLE is given by the central limit theorem, which predicts that the MLE should be approximately normal for large \(n\). However, we found that \( \hat{\sigma^2}_{MLE} \) follows a scaled chi-squared distribution which matches with the finite-sample result and implies no conflict. For large \(n\), its distribution approximates normality.

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.

Cram茅r-Rao Lower Bound
The Cram茅r-Rao Lower Bound (CRLB) provides a theoretical limit on the variance of any unbiased estimator of a parameter. It sets a benchmark for the smallest possible variance that any unbiased estimator can achieve. In this context, it's crucial because it allows us to evaluate whether a given estimator is efficient or optimal.
For an unbiased estimator of \(\sigma^2\) from a normal distribution, the CRLB is determined using the Fisher information. Here, the formula is \[\frac{2(\sigma^2)^2}{n}\], where \(n\) represents the sample size. This bound implies that, no matter how you construct an unbiased estimator for \(\sigma^2\), its variance cannot be lower than this value.
  • This makes the CRLB a vital tool in statistical estimation, acting as a yardstick to measure the estimator's potential accuracy.
  • Knowing the CRLB helps in comparing different estimators and understanding their efficiency.
Maximum Likelihood Estimator (MLE)
The Maximum Likelihood Estimator (MLE) is one of the most commonly used methods for estimating parameters of a statistical model. The idea is to find the parameter values that maximize the likelihood function, which measures how likely it is to observe the given data.
For estimating \(\sigma^2\) in a normal distribution, the MLE is obtained by setting the derivative of the log-likelihood function with respect to \(\sigma^2\) to zero, and solving for \(\sigma^2\). This process yields the MLE:\[ \hat{\sigma^2}_{MLE} = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \]
  • The MLE is a widely favored estimation method because of its desirable properties like bias consistency, asymptotic normality, and efficiency.
  • In this exercise, the MLE gives us an estimator for the unknown variance by utilizing the sample variance formula.
Fisher Information
Fisher Information measures the amount of information that an observable random variable carries about an unknown parameter. It is a key concept in the context of designing efficient estimators and understanding their limitations.
For a single observation from a normal distribution with respect to \(\sigma^2\), the Fisher information is calculated as the expected value of the negative second derivative of the log-likelihood function.
The formula obtained is:\[ I(\sigma^2) = \frac{1}{2(\sigma^2)^2} \]
  • Fisher Information is essential in determining the Cram茅r-Rao Lower Bound, as it appears in the denominator of the CRLB formula.
  • A higher Fisher information implies that the sample provides more information about the parameter, allowing for potentially more accurate estimation.
Normal Distribution
The Normal Distribution, often noted as \(N(\mu, \sigma^2)\), is fundamental in statistics due to its properties and the central limit theorem. It's characterized by its bell-shaped curve, symmetric about the mean \(\mu\), and parameterized by mean \(\mu\) and variance \(\sigma^2\).
In this exercise, the focus is on a normal distribution with a known mean and an unknown variance. The properties of the normal distribution simplify many calculations, making estimations straightforward.
  • The MLE for \(\sigma^2\) and its distribution were identified using the characteristics of the normal distribution.
  • As the sample size \(n\) increases, the distribution of the estimator approaches normality due to the central limit theorem.

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

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). [Hint: For any rv \(Y, E\left(Y^{2}\right)=\) \(V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X}\). b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ?

Consider the following sample of observations on coating thickness for low- viscosity paint ("Achieving a Target Value for a Manufacturing Process: A Case Study," J. Qual. Technol., 1992: 22-26): \(\begin{array}{rrrrrrrr}.83 & .88 & .88 & 1.04 & 1.09 & 1.12 & 1.29 & 1.31 \\\ 1.48 & 1.49 & 1.59 & 1.62 & 1.65 & 1.71 & 1.76 & 1.83\end{array}\) Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption). a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used. b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used. c. Calculate a point estimate of the value that separates the largest \(10 \%\) of all values in the thickness distribution from the remaining \(90 \%\), and state which estimator you used. [Hint: Express what you are trying to estimate in terms of \(\mu\) and \(\sigma\) ] d. Estimate \(P(X<1.5)\), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of \(\mu\) and \(\sigma\), you could calculate this probability. These values are not available, but they can be estimated.] e. What is the estimated standard error of the estimator that you used in part (b)?

Let \(X_{1}, \ldots, X_{n}\) be a random sample of component lifetimes from an exponential distribution with parameter \(\lambda\). Use the factorization theorem to show that \(\sum X_{i}\) is a sufficient statistic for \(\lambda\).

A random sample of \(n\) bike helmets manufactured by a company is selected. Let \(X=\) the number among the \(n\) that are flawed, and let \(p=P\) (flawed). Assume that only \(X\) is observed, rather than the sequence of \(S\) 's and \(F\) 's. a. Derive the maximum likelihood estimator of \(p\). If \(n=20\) and \(x=3\), what is the estimate? b. Is the estimator of part (a) unbiased? c. If \(n=20\) and \(x=3\), what is the mle of the probability \((1-p)^{5}\) that none of the next five helmets examined is flawed?

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 \(p\) ?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

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.