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Chapter 7: Problem 27

Suppose a measurement is made on some physical characteristic whose value is known, and let \(X\) denote the resulting measurement error. For an unbiased measuring instrument or technique, the mean value of \(X\) is 0 . Assume that any particular measurement error is normally distributed with variance \(\sigma^{2}\). Let \(X_{1}, \ldots . X_{n}\) be a random sample of measurement errors. a. Obtain the method of moments estimator of \(\sigma^{2}\). b. Obtain the maximum likelihood estimator of \(\sigma^{2}\).

Short Answer

Expert verified
The method of moments estimator for \(\sigma^2\) is \(s^2\); the MLE is \(\frac{1}{n}\sum_{i=1}^{n} X_i^2\).

Step by step solution

01

Define the method of moments

The method of moments involves equating the sample moments to the population moments. Since we are given a random sample \(X_1, X_2, \ldots, X_n\), we denote the sample variance as \(s^2 = \frac{1}{n}\sum_{i=1}^{n} (X_i - \bar{X})^2\), where \(\bar{X}\) is the sample mean.
02

Express the population moment for variance

For the normal distribution, the population variance is \(\sigma^2\). According to the method of moments, we equate the sample moment (sample variance) to the population moment (\(\sigma^2\)) to get:\\[s^2 = \sigma^2\]
03

Solve for the method of moments estimator

From \(s^2 = \sigma^2\), the method of moments estimator for \(\sigma^2\) is simply the sample variance, denoted as \(\hat{\sigma}^2_{MM} = s^2\).
04

Setup likelihood function for maximum likelihood estimation

For maximum likelihood estimation, we start with the likelihood function for the normal distribution. The joint likelihood function for \(X_1, X_2, \ldots, X_n\) is:\\[L(\sigma^2; X_1, X_2, \ldots, X_n) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{X_i^2}{2\sigma^2}\right)\]
05

Take the natural log of the likelihood function

To simplify calculations, take the log of the likelihood function (the log-likelihood):\\[\ln L(\sigma^2) = -\frac{n}{2}\ln(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^{n} X_i^2\]
06

Differentiate the log-likelihood

Differentiate the log-likelihood function with respect to \(\sigma^2\):\\[\frac{\partial}{\partial \sigma^2} \ln L(\sigma^2) = -\frac{n}{2\sigma^2} + \frac{1}{2(\sigma^2)^2} \sum_{i=1}^{n} X_i^2\]
07

Solve for the estimator by setting the derivative to zero

Set the derivative equal to zero to find the estimator:\\[ -\frac{n}{2\sigma^2} + \frac{1}{2(\sigma^2)^2} \sum_{i=1}^{n} X_i^2 = 0\] \( \Rightarrow n\sigma^2 = \sum_{i=1}^{n} X_i^2\) \( \Rightarrow \hat{\sigma}^2_{MLE} = \frac{1}{n} \sum_{i=1}^{n} X_i^2\) Thus, the maximum likelihood estimator of \(\sigma^2\) is \(\hat{\sigma}^2_{MLE} = \frac{1}{n} \sum_{i=1}^{n} X_i^2\).

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Key Concepts

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

Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a statistical method used for estimating the parameters of a statistical model.
The goal is to find the parameter values that maximize the likelihood that the process described by the model produced the data that was actually observed.
When you deal with normally distributed measurement errors, the variance \( \sigma^2 \) is a key parameter to estimate.
  • Start by considering the likelihood function for your data, which is the probability of observing your data given the parameters of the model.
  • For independent normal observations, the joint likelihood function is the product of the individual likelihoods.
  • To simplify the process, focus on the log of the likelihood function, known as the log-likelihood. This transforms the product of probabilities into a sum, which is easier to handle mathematically.
Differentiation and calculus play a significant role here. By taking the derivative of the log-likelihood with respect to \( \sigma^2 \) and finding where it equals zero, you can derive the MLE for variance.
In this context, \( \hat{\sigma}^2_{MLE} = \frac{1}{n} \sum_{i=1}^{n} X_i^2 \) is the MLE for variance based on your sample errors.
Method of Moments
The Method of Moments is a straightforward technique used to estimate population parameters by matching sample moments to theoretical moments.
This method leverages the principle that sample moments are reflections of the corresponding population moments.
  • A moment in statistics is a quantitative measure used to describe the shape of a set of points.
  • The most common moments are the sample mean and the sample variance.
For the normal distribution, the population variance is denoted by \( \sigma^2 \).
According to the Method of Moments, you equate the sample variance, which is \( s^2 = \frac{1}{n}\sum_{i=1}^{n} (X_i - \bar{X})^2 \), with this population variance.
This gives us the estimator for variance, known as \( \hat{\sigma}^2_{MM} = s^2 \).
This method is favored for its simplicity and reliability when assumptions about the normality of the population hold true.
Normal Distribution
The normal distribution is a key concept in statistics, often used to model random variables that cluster around a mean.
Recognizable by its bell-shaped curve, it is defined by its mean \( \mu \) and variance \( \sigma^2 \).
  • The mean \( \mu \) represents the center of this distribution.
  • The variance \( \sigma^2 \) indicates how spread out the values are.
Measurement errors are typically assumed to follow a normal distribution due to the Central Limit Theorem.
This theorem suggests that the sum of a large number of errors will be normally distributed, regardless of the distribution of the individual errors.
Understanding this allows statisticians to make predictions about the behavior of errors in measurements, estimating them with more confidence and accuracy.
Measurement Error
Measurement error is the difference between a measured value and the true value of what you are trying to measure.
It's crucial in data analysis because it impacts the accuracy of data collection.
When dealing with measurement errors, especially in scientific experiments, knowing their properties helps in making more reliable decisions.
  • Measurement errors can be systematic or random.
  • In theory, the mean of random errors is zero, which implies unbiased measurements.
  • The variance of measurement errors reveals how spread out these errors are in relation to the true value.
When measurement errors follow a normal distribution, the error variance \( \sigma^2 \) becomes a critical component in modeling and decision-making.
Estimating this variance using techniques like MLE or Method of Moments provides insights into the reliability and precision of the measurements.

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Most popular questions from this chapter

The long run proportion of vehicles that pass a certain emissions test is \(p\). Suppose that three vehicles are independently selected for testing. Let \(X_{i}=1\) if the \(i\) th vehicle passes the test and \(X_{i}=0\) otherwise \((i=1,2,3)\), and let \(X=X_{1}+\) \(X_{2}+X_{3}\). Use the definition of sufficiency to show that \(X\) is sufficient for \(p\) by obtaining the conditional distribution of the \(X_{i}\) 's given that \(X=x\) for each possible value \(x\). Then generalize by giving an analogous argument for the case of \(n\) vehicles.

For \(\theta>0\) consider a random sample from a uniform distribution on the interval from \(\theta\) to \(2 \theta\) (pdf \(1 / \theta\) for \(\theta

Here is a result that allows for easy identification of a minimal sufficient statistic: Suppose there is a function \(t\left(x_{1}, \ldots, x_{n}\right)\) such that for any two sets of observations \(x_{1}, \ldots, x_{n}\) and \(y_{1}, \ldots, y_{n}\), the likelihood ratio \(f\left(x_{1}, \ldots, x_{n} ; \theta\right) / f\left(y_{1}, \ldots, y_{n} ; \theta\right)\) doesn't depend on \(\theta\) if and only if \(t\left(x_{1}, \ldots, x_{n}\right)\) \(=t\left(y_{1}, \ldots, y_{n}\right)\). Then \(T=t\left(X_{1}, \ldots, X_{n}\right)\) is a minimal sufficient statistic. The result is also valid if \(\theta\) is replaced by \(\theta_{1}, \ldots, \theta_{k}\), in which case there will typically be several jointly minimal sufficient statistics. For example, if the underlying pdf is exponential with parameter \(\lambda\), then the likelihood ratio is \(\lambda^{\sum x_{i}-\Sigma y_{i}}\), which will not depend on \(\lambda\) if and only if \(\sum x_{i}=\sum y_{i}\), so \(T=\sum x_{i}\) is a minimal sufficient statistic for \(\lambda\) (and so is the sample mean). a. Identify a minimal sufficient statistic when the \(X_{i}\) 's are a random sample from a Poisson distribution. b. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and variance \(\theta\). c. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and standard deviation \(\theta\).

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 \(p\) ?

Let \(p\) denote the proportion of all individuals who are allergic to a particular medication. An investigator tests individual after individual to obtain a group of \(r\) individuals who have the allergy. Let \(X_{i}=1\) if the \(i\) th individual tested has the allergy and \(X_{i}=0\) otherwise ( \(i=1,2,3, \ldots\) ). Recall that in this situation, \(X=\) the number of nonallergic individuals tested prior to obtaining the desired group has a negative binomial distribution. Use the definition of sufficiency to show that \(X\) is a sufficient statistic for \(p\).
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