Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Prove that \(\bar{X}\), the mean of a random sample of size \(n\) from a distribution that is \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\), is, for every known \(\sigma^{2}>0\), an efficient estimator of \(\theta\).

Short Answer

Expert verified
The sample mean \(\bar{X}\) serves as an efficient estimator for the population mean \(\theta\) given it is shown to be an unbiased estimator, and its variance achieves the Cramér-Rao lower bound (equal to the reciprocal of the Fisher information), which is the minimal achievable variance for an unbiased estimator of a population parameter. Therefore, \(\bar{X}\) is an efficient estimator of \(\theta\).

Step by step solution

01

Demonstrate that the sample mean is an unbiased estimator

An unbiased estimator is one where its expectation equals the population parameter it seeks to estimate. In this case, we wish to show that the expectation of the sample mean equals \(\theta\). This can be done as follows: since \(X_i\) are i.i.d \(N(\theta, \sigma^2)\), we have \(E(X_i) = \theta\). The expectation of the sample mean, \(E(\bar{X})\), is the same as the expectation of the sum of the \(X_i\)s divided by \(n\). Hence, \(E(\bar{X}) = E(\frac{1}{n}\sum_{i=1}^{n} X_i) = \frac{1}{n} \sum_{i=1}^{n}E(X_i) = \frac{1}{n} n\theta = \theta\).
02

Prove that the sample mean achieves the Cramér-Rao lower bound

For an estimator to be considered efficient, it has to achieve the lower limit on the variance of unbiased estimators, known as the Cramér-Rao lower bound. To prove this, it is necessary to firstly calculate the variance of the sample mean which is \(Var(\bar{X}) = Var(\frac{1}{n}\sum_{i=1}^{n} X_i) = \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i) = \frac{1}{n^2} n\sigma^2 = \frac{\sigma^2}{n}\). The Fisher information \(I(\theta)\) (being the measure of information that an observable random variable \(X\) carries about an unknown parameter upon which the probability of \(X\) depends) for a normal distribution is \(\frac{1}{\sigma^2}\). Then, applying Cramér-Rao theorem, the lower bound \(\frac{1}{I(\theta)} = \sigma^2\), proving that the variance of \(\bar{X}\) achieves the Cramér-Rao lower bound, thus making it an efficient estimator.

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 Vaia!

Key Concepts

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

Unbiased Estimator
When we talk about estimators in the context of statistical inference, an unbiased estimator is a statistical technique that produces estimates that, on average, are equal to the parameter being estimated. In other words, if you were to repeatedly draw samples and calculate the estimator for each sample, the average of all these estimators would be equal to the true parameter value.

Consider the sample mean, \( \bar{X} \), which is the average of all observations in a sample. If the sample is drawn from a normally distributed population with mean \( \theta \) and variance \( \sigma^2 \), the sample mean itself is an unbiased estimator of \( \theta \). This means that the expected value of the sample mean is equal to \( \theta \), as mathematically shown in the textbook solution provided.
Cramér-Rao Lower Bound
In statistics, the Cramér-Rao lower bound gives the minimum variance that an unbiased estimator can have. It's a benchmark to assess the efficiency of estimators. The lower the variance, the closer the estimates are likely to be to the actual parameter value on average. Therefore, an estimator that achieves this bound is considered to be the most precise (or efficient) among all unbiased estimators.

The sample mean's variance being equal to \( \sigma^2/n \) is significant as it matches the Cramér-Rao lower bound, which indicates that within the class of unbiased estimators of \( \theta \) for our normal distribution, the sample mean is as efficient as possible. This is crucial in statistical inference, as having an efficient estimator means we're making the most of our data to draw conclusions about the underlying population parameter.
Statistical Inference
The process of statistical inference involves drawing conclusions about a population's parameters based on a sample of data. This process relies heavily on the concepts of estimators and their properties, such as bias and variance. Efficient, unbiased estimators are critical for making the most accurate inferences.

Using the sample mean as our estimator for the population mean in a normal distribution is not only mathematically sound but also practically advantageous. By proving its efficiency and lack of bias, we bolster the reliability of the statistical inferences that can be drawn from sample data, leading to more robust scientific conclusions and decisions in various fields where data analysis is vital.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is the arithmetic average of all the sample observations. It plays a central role in statistical estimation as it is frequently used to estimate the population mean. In most cases, particularly when dealing with large samples from normally distributed populations, the sample mean is a reliable and straightforward estimator.

Importantly, the sample mean is particularly powerful due to its properties as an unbiased estimator and, as demonstrated for the normal distribution, its efficiency. The ease and practicality of calculating the sample mean, combined with its desirable statistical properties, make it a cornerstone of descriptive statistics and inferential methodologies.
Fisher Information
The concept of Fisher information is somewhat abstract but very important in statistical theory. It measures the amount of information that an observable random variable carries about an unknown parameter upon which the probability of the random variable depends. Higher Fisher information means that the sample contains more information about the parameter, which usually results in a lower potential variance for an unbiased estimator.

For a normal distribution, as in our example, the Fisher information for estimating the parameter \( \theta \) is \( \frac{1}{\sigma^2} \). We subsequently use this information to determine the Cramér-Rao lower bound, therefore linking Fisher information directly to the efficiency of an estimator. Understanding Fisher information is key to grasping why some estimators are superior to others and how they provide us with the best possible estimates from observed data.

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

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from the two normal distributions \(N\left(0, \theta_{1}\right)\) and \(N\left(0, \theta_{2}\right)\). (a) Find the likelihood ratio \(\Lambda\) for testing the composite hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) against the composite alternative \(H_{1}: \theta_{1} \neq \theta_{2}\). (b) This \(\Lambda\) is a function of what \(F\) -statistic that would actually be used in this test?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the beta distribution with \(\alpha=\beta=\theta\) and \(\Omega=\\{\theta: \theta=1,2\\}\). Show that the likelihood ratio test statistic \(\Lambda\) for testing \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2\) is a function of the statistic \(W=\) \(\sum_{i=1}^{n} \log X_{i}+\sum_{i=1}^{n} \log \left(1-X_{i}\right)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N(0, \theta)\) distribution. We want to estimate the standard deviation \(\sqrt{\theta}\). Find the constant \(c\) so that \(Y=\) \(c \sum_{i=1}^{n}\left|X_{i}\right|\) is an unbiased estimator of \(\sqrt{\theta}\) and determine its efficiency.

Let \(n\) independent trials of an experiment be such that \(x_{1}, x_{2}, \ldots, x_{k}\) are the respective numbers of times that the experiment ends in the mutually exclusive and exhaustive events \(C_{1}, C_{2}, \ldots, C_{k} .\) If \(p_{i}=P\left(C_{i}\right)\) is constant throughout the \(n\) trials, then the probability of that particular sequence of trials is \(L=p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}}\). (a) Recalling that \(p_{1}+p_{2}+\cdots+p_{k}=1\), show that the likelihood ratio for testing \(H_{0}: p_{i}=p_{i 0}>0, i=1,2, \ldots, k\), against all alternatives is given by $$ \Lambda=\prod_{i=1}^{k}\left(\frac{\left(p_{i 0}\right)^{x_{i}}}{\left(x_{i} / n\right)^{x_{i}}}\right) $$ (b) Show that $$ -2 \log \Lambda=\sum_{i=1}^{k} \frac{x_{i}\left(x_{i}-n p_{i 0}\right)^{2}}{\left(n p_{i}^{\prime}\right)^{2}} $$ where \(p_{i}^{\prime}\) is between \(p_{i 0}\) and \(x_{i} / n\). Hint: Expand \(\log p_{i 0}\) in a Taylor's series with the remainder in the term involving \(\left(p_{i 0}-x_{i} / n\right)^{2}\). (c) For large \(n\), argue that \(x_{i} /\left(n p_{i}^{\prime}\right)^{2}\) is approximated by \(1 /\left(n p_{i 0}\right)\) and hence \(-2 \log \Lambda \approx \sum_{i=1}^{k} \frac{\left(x_{i}-n p_{i 0}\right)^{2}}{n p_{i 0}}\) when \(H_{0}\) is true. Theorem \(6.5 .1\) says that the right-hand member of this last equation defines a statistic that has an approximate chi-square distribution with \(k-1\) degrees of freedom. Note that dimension of \(\Omega-\) dimension of \(\omega=(k-1)-0=k-1\)

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are iid with pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free