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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution with known standard deviation \(\sigma\). a. Find the mle of \(\mu\). b. Find the distribution of the mle. c. Is the mle an efficient estimator? Explain. d. How does the answer to part (b) compare with the asymptotic distribution given by the second theorem?

Short Answer

Expert verified
a. MLE is \( \bar{X} \); b. \( N(\mu, \frac{\sigma^2}{n}) \); c. Yes, it's efficient; d. Exact and asymptotic distributions match.

Step by step solution

01

Understand the Problem and Identify the Likelihood Function

We are given a random sample \( X_1, X_2, \ldots, X_n \) from a normal distribution with a known standard deviation \( \sigma \). The task is to find the maximum likelihood estimator (MLE) of \( \mu \). The likelihood function for this problem is the product of the normal densities:\[L(\mu) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right)\]We will work with the log-likelihood for simplicity.
02

Calculate the Log-Likelihood and its Derivative

The log-likelihood is obtained by taking the natural logarithm of the likelihood function:\[\ell(\mu) = -\frac{n}{2} \log(2\pi) - n \log(\sigma) - \frac{1}{2\sigma^2} \sum_{i=1}^{n} (X_i - \mu)^2\]To find the MLE, take the derivative of the log-likelihood with respect to \( \mu \):\[\frac{\partial \ell}{\partial \mu} = \frac{1}{\sigma^2} \sum_{i=1}^{n} (X_i - \mu)\]
03

Find the Maximum Likelihood Estimator (MLE)

Setting the derivative equal to zero to find the critical points:\[\frac{1}{\sigma^2} \sum_{i=1}^{n} (X_i - \mu) = 0\]Solving for \( \mu \) gives:\[\hat{\mu}_{\text{MLE}} = \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\]Therefore, the MLE for \( \mu \) is the sample mean \( \bar{X} \).
04

Determine the Distribution of the MLE

The sample mean \( \bar{X} \) is a linear combination of normal variables, so it is normally distributed. Its mean is \( \mu \), and its variance is \( \frac{\sigma^2}{n} \):\[\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)\]
05

Assess the Efficiency of the MLE

An estimator is considered efficient if it meets the Cramér-Rao Lower Bound. Since \( \bar{X} \) is the MLE of \( \mu \), it is the best unbiased estimator with the minimum variance (equal to \( \frac{\sigma^2}{n} \)), satisfying the Cramér-Rao Lower Bound. Hence, \( \bar{X} \) is an efficient estimator.
06

Compare with Asymptotic Distribution

The second theorem references the asymptotic distribution of the MLE, which states the MLE is asymptotically normally distributed with the true mean and the minimum possible variance. For \( \bar{X} \), its exact sampling distribution matches the asymptotic result \( N(\mu, \frac{\sigma^2}{n}) \), showing complete agreement between the finite and asymptotic results for the normal distribution.

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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, often known as the Gaussian distribution, is a fundamental concept in statistics. It is characterized by its bell-shaped curve that is symmetric around the mean. This distribution is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). These parameters determine the center and the spread of the distribution.

When working with a normal distribution, a random sample will have data points that tend to cluster around the mean. This distribution is particularly important because of the Central Limit Theorem, which states that, under certain conditions, the sum of many random variables will have a distribution close to a normal distribution, regardless of the original distribution of the variables.

In the context of our exercise, having a known standard deviation simplifies the process of finding the maximum likelihood estimator (MLE) because it allows us to focus solely on estimating the mean (\( \mu \)). The properties of the normal distribution also ensure that the MLE is a linear combination of the sample data, making it easier to analyze.
Cramér-Rao Lower Bound
The Cramér-Rao Lower Bound provides a benchmark for the variance of unbiased estimators. It indicates the smallest possible variance that an unbiased estimator can achieve for a given parameter. If an estimator reaches this bound, it is considered efficient.

This bound depends on the Fisher Information, which measures the amount of information a sample provides about a parameter. Mathematically, the Cramér-Rao Lower Bound is expressed as follows for an unbiased estimator \( \hat{\theta} \) of a parameter \( \theta \):

\[ \text{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} \]

Where \( I(\theta) \) is the Fisher Information.

In our scenario, the sample mean \( \bar{X} \) as the MLE of \( \mu \) achieves this bound. With a variance of \( \frac{\sigma^2}{n} \), it meets the Cramér-Rao criteria, confirming that \( \bar{X} \) is efficient.

The Cramér-Rao Lower Bound ensures statisticians use the most accurate methods for parameter estimation, minimizing errors whenever analyzing or predicting data.
Sample Mean
The sample mean is a basic concept in statistics. It is calculated as the sum of all data points divided by the number of data points in a sample. If \( X_1, X_2, \ldots, X_n \) are random sample observations, then the sample mean \( \bar{X} \) is given by:

\[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \]

The sample mean is an unbiased estimator of the population mean (\( \mu \)) when data is drawn from a normal distribution. This means, on average, it accurately reflects the actual mean of the population.

Because the sample mean in our exercise is derived from a normal distribution with a known standard deviation, it also holds specific properties. It follows a normal distribution itself, with variance reduced in proportion to the sample size \( n \), described as \( N(\mu, \frac{\sigma^2}{n}) \).

The simplicity and effectiveness of the sample mean make it a widely used statistic in various fields, from social sciences to market analytics.
Estimator Efficiency
Estimator efficiency relates to how well an estimator performs compared to the best possible option based on variance. An estimator is efficient if it achieves the minimum variance bound set by the Cramér-Rao inequality. This means it extracts the maximum possible information about a parameter from the available data.

For an estimator to be efficient, it should be:
  • Unbiased: It provides parameter estimates that are correct on average.
  • Minimum variance: It estimates with the least amount of variability across different samples.

In our exercise, the sample mean \( \bar{X} \) acts as an efficient estimator of the population mean \( \mu \).

Because it meets the Cramér-Rao Lower Bound with its variance \( \frac{\sigma^2}{n} \), \( \bar{X} \) has both low variance and it is unbiased, cementing its efficiency.

Efficient estimators ensure maximum precision in statistical predictions, leading to better data-driven decision-making.

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Most popular questions from this chapter

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?] c. Let \(\hat{p}\) be the sample proportion of successes. Is \(\hat{p}^{2}\) an unbiased estimator for \(p^{2} ?\)

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 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 \(\sigma^{2}\).

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 \(\sigma^{2}\).

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 103 , \(156,118,89,125,147,122,109,138\), 99. Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

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