/*! 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 38 Each of \(n\) specimens is to be... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 38

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 one another, 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 \(\left.\Sigma\left(z_{i}-\bar{z}\right)^{2}=\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\).

Short Answer

Expert verified
The MLE of \( \sigma^2 \) is \( \hat{\sigma}^2 = \frac{1}{4n} \sum (X_i - Y_i)^2 \); it is biased. An unbiased estimator is \( \tilde{\sigma}^2 = \frac{1}{2n} \sum (X_i - Y_i)^2 \).

Step by step solution

01

Understand the Problem

We have two independent measurements for each of the \( n \) specimens: \( X_i \) and \( Y_i \). Both \( X_i \) and \( Y_i \) are normally distributed with mean \( \mu_i \) and variance \( \sigma^2 \). We need to estimate \( \sigma^2 \) using maximum likelihood estimation (MLE).
02

Define the Likelihood Function

The likelihood function for the \( i \)-th specimen, given \( X_i \) and \( Y_i \), under the assumption of independence, is \( L_i(\sigma^2) = f(X_i; \mu_i, \sigma^2) \times f(Y_i; \mu_i, \sigma^2) \). For a normal distribution, the probability density function is:\[f(x; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).\] Thus the likelihood for each \( i \) is a product of these densities.
03

Form the Total Likelihood Function

The total likelihood function for all specimens is the product over all \( n \):\[L(\sigma^2) = \prod_{i=1}^{n} L_i (\sigma^2) = \prod_{i=1}^{n} \left( \frac{1}{2\pi \sigma^4} \right) \exp\left( -\frac{(X_i - \mu_i)^2 + (Y_i - \mu_i)^2}{2\sigma^2} \right).\]
04

Calculate the Log-Likelihood

To simplify, take the natural logarithm of the likelihood function:\[\log L(\sigma^2) = -2n \log (2\pi \sigma^4) - \sum_{i=1}^{n}\frac{(X_i - \mu_i)^2 + (Y_i - \mu_i)^2}{2\sigma^2}.\] This results from properties of logarithms.
05

Differentiate the Log-Likelihood

Differentiate \( \log L(\sigma^2) \) with respect to \( \sigma^2 \) to find the MLE:\[\frac{d}{d\sigma^2} \log L(\sigma^2) = \frac{4n}{\sigma^2} - \sum_{i=1}^{n} \frac{(X_i - \mu_i)^2 + (Y_i - \mu_i)^2}{\sigma^4}.\]
06

Solve for MLE

Set the derivative equal to zero to find the critical points:\[\frac{4n}{\sigma^2} = \sum_{i=1}^{n} \frac{(X_i - \mu_i)^2 + (Y_i - \mu_i)^2}{\sigma^4}.\] Solving for \( \sigma^2 \) gives:\[\hat{\sigma}^2 = \frac{1}{4n} \sum_{i=1}^{n} (X_i - Y_i)^2.\]
07

Evaluate Bias of the Estimator

The expected value of the MLE, \( E[\hat{\sigma}^2] = E\left[\frac{1}{4n} \sum_{i=1}^{n} (X_i - Y_i)^2 \right] \), is evaluated using the fact that both \( X_i \) and \( Y_i \) have variance \( \sigma^2 \). However, because \( E[(X_i-Y_i)^2] = 2\sigma^2 \), the MLE is biased upwards: \( E[\hat{\sigma}^2] = \frac{2\sigma^2 n}{4n} = \frac{\sigma^2}{2} eq \sigma^2 \).
08

Find Unbiased Estimator

To adjust for bias, multiply the MLE by 2. Therefore, an unbiased estimator \( \tilde{\sigma}^2 \) for \( \sigma^2 \) is:\[\tilde{\sigma}^2 = \frac{1}{2n} \sum_{i=1}^{n} (X_i - Y_i)^2.\]

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Key Concepts

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

normal distribution
The normal distribution is a fundamental concept in statistics and probability theory. It is often referred to as the Gaussian distribution. This distribution is defined by its bell-shaped curve, which is symmetrical around its mean. The normal distribution is characterized by two parameters: the mean (\( \mu \)) and the variance (\( \sigma^2 \)).
  • The mean is the central value, around which the data points are distributed.
  • The variance measures the spread of the data around the mean.
A special property of the normal distribution is that it fully describes the data using only these two parameters. This makes it a popular choice for modeling natural phenomena and measurement errors. In the given exercise, the weights \(X_i\) and \(Y_i\) of each specimen are assumed to be normally distributed, which means that their variations follow this precise bell-shaped curve.
variance estimation
Variance estimation is a critical operation in statistics, where the goal is to quantify the amount of variation or dispersion present in a set of data. The variance is usually represented by the symbol \( \sigma^2 \). Calculating an accurate estimate of variance helps us understand how much the observed values deviate from the mean.
In the context of our problem, we aim to estimate \( \sigma^2 \), the true variance of the weights using the data collected (\(X_i\) and \(Y_i\)). Different methods can be used for this purpose, with the Maximum Likelihood Estimator (MLE) being a common choice.
To derive an estimator, we leverage properties of sample data, where errors or deviations in recorded weights hint at the underlying true variance. Variance estimation thus plays a role in validating experimental data consistency and making statistical predictions.
unbiased estimator
An unbiased estimator refers to a statistical estimator whose expected value is equal to the true parameter of the population. This means, on average, it accurately reflects the actual parameter's value it is intended to estimate.
In our exercise, we identify that while the MLE for variance (\( \hat{\sigma}^2 \)) is effectively derived, it is biased. Its expected value \( E[\hat{\sigma}^2] \) turns out to be \( \frac{\sigma^2}{2} \) rather than \( \sigma^2 \). This half factor shows the MLE underestimates the true variance. To correct this bias, the estimator is adjusted to \( \tilde{\sigma}^2 \) by multiplying the MLE by 2, thus making it:\[\tilde{\sigma}^2 = \frac{1}{2n} \sum_{i=1}^{n} (X_i - Y_i)^2.\]This transformation provides an unbiased estimate, ensuring that in repeated sampling, the average of the estimator corresponds to the true variance.
log-likelihood function
The log-likelihood function is a tool used in statistics to specify the probability of the observed data given some parameters. It is derived from the likelihood function, which is the joint probability of the data, but expressed in its logarithmic form.
Why use the log form? The main reason is mathematical convenience. The product of probabilities in the likelihood function can be difficult to work with, especially when differentiating. By transforming the likelihood into a summation via a logarithm, it simplifies calculations, especially when finding maximum likelihood estimates (MLE).
In our exercise, the log-likelihood function helps determine the best estimate of the variance by simplifying derivations. For normally distributed data, the log-likelihood is computed, differentiated with respect to the parameter (\( \sigma^2 \)), and set to zero to find the MLE. This systematic approach efficiently optimizes parameter estimation, facilitating more accurate statistical results.

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Most popular questions from this chapter

A sample of 20 students who had recently taken elementary statistics yielded the following information on the brand of calculator owned \((\mathrm{T}=\) Texas Instruments, \(\mathrm{H}=\) Hewlett Packard, C = Casio, \(\mathrm{S}=\) Sharp): a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the 10 students who owned a TI calculator, 4 had graphing calculators. Estimate the proportion of students who do not own a TI graphing calculator.

The mean squared error of an estimator \(\hat{\theta}\) is \(\operatorname{MSE}(\hat{\theta})=E(\hat{\theta}-\theta)^{2}\). If \(\hat{\theta}\) is unbiased, then \(\operatorname{MSE}(\hat{\theta})=V(\hat{\theta})\), but in general \(\operatorname{MSE}(\hat{\theta})=V(\hat{\theta})+(\text { bias })^{2}\). Consider the estimator \(\hat{\sigma}^{2}=K S^{2}\), where \(S^{2}=\) sample variance. What value of \(K\) minimizes the mean squared error of this estimator when the population distribution is normal? [Hint: It can be shown that $$ E\left[\left(S^{2}\right)^{2}\right]=(n+1) \sigma^{4} /(n-1) $$ In general, it is difficult to find \(\hat{\theta}\) to minimize \(\operatorname{MSE}(\hat{\theta})\), which is why we look only at unbiased estimators and minimize \(V(\hat{\theta}) .]\)

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a pdf \(f(x)\) that is symmetric about \(\mu\), so that \(\widetilde{X}\) is an unbiased estimator of \(\mu\). If \(n\) is large, it can be shown that \(V(\widetilde{X}) \approx 1 /\left(4 n[f(\mu)]^{2}\right)\). a. Compare \(V(\widetilde{X})\) to \(V(\bar{X})\) when the underlying distribution is normal. b. When the underlying pdf is Cauchy (see Example 6.7), \(V(\bar{X})=\infty\), so \(\bar{X}\) is a terrible estimator. What is \(V(\widetilde{X})\) in this case when \(n\) is large?

Suppose the true average growth \(\mu\) of one type of plant during a 1-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(\left.E\left(X_{i}\right)=\mu, V\left(X_{i}\right)=\sigma^{2}\right]\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu\right.\), \(\left.V\left(Y_{i}\right)=4 \sigma^{2}\right]\) a. Show that for any \(\delta\) between 0 and 1, the estimator \(\hat{\mu}=\delta \bar{X}+(1-\delta) \bar{Y}\) is unbiased for \(\mu\). b. For fixed \(m\) and \(n\), compute \(V(\hat{\mu})\), and then find the value of \(\delta\) that minimizes \(V(\hat{\mu})\). [Hint: Differentiate \(V(\hat{\mu})\) with respect to \(\delta\).]
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