/*! 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 57 This problem is concerned with t... [FREE SOLUTION] | 91Ó°ÊÓ

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This problem is concerned with the estimation of the variance of a normal distribution with unknown mean from a sample \(X_{1}, \ldots, X_{n}\) of i.i.d. normal random variables. In answering the following questions, use the fact that (from Theorem \(\mathrm{B}\) of Section 6.3) $$\frac{(n-1) s^{2}}{\sigma^{2}} \sim \chi_{n-1}^{2}$$ and that the mean and variance of a chi-square random variable with \(r\) df are \(r\) and \(2 r,\) respectively. a. Which of the following estimates is unbiased? $$s^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2} \quad \hat{\sigma}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}$$ b. Which of the estimates given in part (a) has the smaller MSE? c. For what value of \(\rho\) does \(\rho \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}\) have the minimal MSE?

Short Answer

Expert verified
a. \(s^2\) is unbiased. b. \(\hat{\sigma}^2\) has smaller MSE. c. Optimal \(\rho\) is \(1/(n+1)\).

Step by step solution

01

Understanding Unbiased Estimator

The task is to determine which of the two given estimates, \(s^2\) and \(\hat{\sigma}^2\), is unbiased. An unbiased estimator means that its expected value equals the true parameter being estimated. We know from the properties of chi-square distributions and given \(\frac{(n-1)s^2}{\sigma^2} \sim \chi_{n-1}^2\), that \(E\left(\frac{(n-1)s^2}{\sigma^2}\right) = n-1\). Therefore, \(E(s^2) = \sigma^2\), making \(s^2\) an unbiased estimator. On the other hand, \(\hat{\sigma}^2 = \frac{n-1}{n}s^2\), which is biased because \(E(\hat{\sigma}^2) = \frac{n-1}{n} \sigma^2 eq \sigma^2\).
02

Compare Mean Squared Error (MSE)

The MSE is given by \(MSE(\theta) = \text{Var}(\theta) + (\text{Bias}(\theta))^2\). \(s^2\) is unbiased, so \(MSE(s^2) = \text{Var}(s^2)\). \(\hat{\sigma}^2 = \frac{n-1}{n}s^2\) so it has a bias of \(\frac{-1}{n}\sigma^2\). Calculating MSE for \(\hat{\sigma}^2\), we have \(MSE(\hat{\sigma}^2) = \text{Var}(\hat{\sigma}^2) + \left(\frac{\sigma^2}{n}\right)^2\). Given \(\text{Var}(\hat{\sigma}^2) = \frac{2\sigma^4}{n}\), we need to calculate and compare MSE values to determine which one is smaller. After simplification, it turns out that \(\hat{\sigma}^2\) always has a smaller MSE than \(s^2\) for finite \(n\).
03

Determine Minimal MSE for Scaling Constant \(\rho\)

We are tasked with finding the \(\rho\) that minimizes the MSE of the estimator \(\rho\sum_{i=1}^{n}(X_i-\bar{X})^2\). The estimator can be written as \(\rho(n-1)s^2\). Its bias is \((n-1)(\rho-1)\sigma^2\), and its variance is \(2(n-1)^2\rho^2\sigma^4\). The MSE is \((n-1)^2(\rho-1)^2\sigma^4 + 2(n-1)^2\rho^2\sigma^4\). To minimize MSE, we differentiate with respect to \(\rho\), set it equal to zero, and solve: \(2(n-1)^2(\rho-1) + 4(n-1)^2\rho = 0\). Solving gives the optimal \(\rho = \frac{1}{n+1}\).

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Key Concepts

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

Unbiased Estimator
An unbiased estimator is a statistical measure used to estimate a population parameter, with the distinguishing feature that its expected value is equal to the true parameter value. This property ensures that across many samples, the estimator accurately targets the parameter. In other words, it avoids systematic errors. For example, in the exercise, we have estimates for variance: \(s^2\) and \(\hat{\sigma}^2\). To determine which is unbiased, we need to check if the expected value matches the true variance \(\sigma^2\). Since we know from the properties of the chi-square distribution that \(E(s^2) = \sigma^2\), \(s^2\) is unbiased. Meanwhile, \(\hat{\sigma}^2\) is less than \(\sigma^2\) on expectation, thus it is biased as \(E(\hat{\sigma}^2) eq \sigma^2\). This distinction is crucial in statistical analysis as unbiased estimators, like \(s^2\), provide more reliable estimates than biased ones.
Mean Squared Error (MSE)
The Mean Squared Error (MSE) is a key metric that combines both the variance of an estimator and its bias, squared. This serves as a comprehensive measure of an estimator's quality. Mathematically, it is expressed as \(MSE(\theta) = \text{Var}(\theta) + (\text{Bias}(\theta))^2\). An estimator with a lower MSE is generally preferred, as it indicates that the estimator has less error overall.
  • For \(s^2\), the MSE is simplified to its variance because \(s^2\) is unbiased, meaning its bias is zero.
  • However, \(\hat{\sigma}^2\) is not unbiased, as it is biased by \(\frac{-1}{n}\sigma^2\). Thus, the MSE of \(\hat{\sigma}^2\) includes both its variance and the square of its bias.
  • When comparing \(s^2\) and \(\hat{\sigma}^2\), \(\hat{\sigma}^2\) has a smaller MSE, even though it is biased, because its variance is notably reduced compared to \(s^2\), making it preferable for finite sample sizes.
Chi-Square Distribution
The chi-square distribution is a fundamental probability distribution in statistics, often arising in scenarios involving variance estimation. It is defined by the sum of the squares of independent standard normal variables and is characterized by its degrees of freedom, often denoted as \(r\). The chi-square distribution has important applications, particularly in hypothesis testing and estimating confidence intervals.
  • In our given problem, the variance estimator \(\frac{(n-1)s^2}{\sigma^2}\) is distributed as \(\chi_{n-1}^2\), which tells us that the form of \(s^2\) directly aligns with the properties of the chi-square distribution.
  • The expected value and variance of a chi-square-distributed random variable are \(r\) and \(2r\), respectively.
  • This property helps in constructing unbiased estimators for variance, as seen with \(s^2\), allowing it to precisely estimate the variance under the chi-square framework.
Overall, understanding the chi-square distribution enables statisticians to derive meaningful estimators for variance and execute effective statistical tests.

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Most popular questions from this chapter

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