/*! 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 13 Consider a random sample \(X_{1}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 13

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 \(\theta\). [Hint: First determine \(\mu=E(X)=E(\bar{X})\).]

Short Answer

Expert verified
\(\hat{\theta} = 3 \bar{X}\) is an unbiased estimator of \(\theta\).

Step by step solution

01

Calculate Expected Value of X

To prove that \(\hat{\theta} = 3 \bar{X}\) is an unbiased estimator for \(\theta\), we need to find \(E(X)\). The expected value of \(X\) is given by the integral: \[ E(X) = \int_{-1}^{1} x \, f(x; \theta) \, dx. \] Substitute the given probability density function, \[ f(x; \theta) = 0.5(1 + \theta x). \] We have: \[ E(X) = \int_{-1}^{1} x \, 0.5(1+\theta x) \, dx. \] Split the integral: \[ E(X) = 0.5 \left( \int_{-1}^{1} x \, dx + \theta \int_{-1}^{1} x^2 \, dx \right). \]
02

Evaluate the Integrals

Evaluate the first integral: \[ \int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0. \]Evaluate the second integral: \[ \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]Substitute back into the expression for \(E(X)\): \[ E(X) = 0.5 \left( 0 + \theta \cdot \frac{2}{3} \right) = \frac{\theta}{3}. \]
03

Calculate Expected Value of Sample Mean

Since \(\bar{X}\) is the sample mean, \(E(\bar{X}) = \frac{1}{n} \sum_{i=1}^{n} E(X_i) = E(X)\). Therefore, \[ E(\bar{X}) = \frac{\theta}{3}. \]
04

Check Unbiasedness of \(\hat{\theta}\)

\(\hat{\theta}\) is defined as \(3 \bar{X}\). Check if \(E(\hat{\theta}) = \theta\):\[ E(\hat{\theta}) = E(3 \bar{X}) = 3 E(\bar{X}) = 3 \times \frac{\theta}{3} = \theta. \]Since \(E(\hat{\theta}) = \theta\), \(\hat{\theta}\) is an unbiased estimator of \(\theta\).

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 91影视!

Key Concepts

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

Expected Value
The expected value, often represented as \(E(X)\), is a fundamental concept in probability and statistics. It is used to determine the average or mean outcome of a random variable over a large number of experiments. For a continuous random variable with a probability density function (PDF) \(f(x)\), the expected value is calculated by integrating the product of \(x\) and \(f(x)\) over all possible values. This can be expressed as follows:\[ E(X) = \int_{-\infty}^{\infty} x \, f(x) \, dx \]In the context of the exercise, we used the range \(-1\) to \(1\) because it represents the domain of our specific PDF. Calculating the expected value helps us derive properties such as the unbiased estimator, which ensures the estimator delivers accurate parameter estimates on average.
Random Sample
A random sample consists of a collection of independent and identically distributed (i.i.d.) random variables. These samples come from the same probability distribution, which allows them to collectively yield meaningful statistical properties.
  • Independence implies that each sample does not influence the others.
  • Identically distributed means each sample follows the same probability distribution.
In this exercise, the random sample is \(X_1, X_2, \ldots, X_n\). By examining the properties of the sample鈥攁s in determining the sample mean \(\bar{X}\)鈥攚e can construct estimators such as \(\hat{\theta} = 3\bar{X}\) to make inferences about unknown parameters. Ensuring our estimator is unbiased means these inferences remain accurate on average across many samples.
Probability Density Function
The probability density function (PDF) is a function that describes the relative likelihood of a continuous random variable taking on a particular value. It is critical in computing probabilities for continuous random variables, such as those concerning physical phenomena.
  • The PDF must fulfill the condition: \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \), ensuring the total probability is 1.
  • The specific form given, \(f(x; \theta) = 0.5(1 + \theta x)\), shows how probabilities depend on the parameter \(\theta\).
Understanding PDFs enables us to compute expectations and variances, as seen when we determined \(E(X)\) in the solution to show the unbiased nature of \(\hat{\theta}\). By manipulating the integral involving the PDF, we gauge important characteristics of the random variable.
Integral Calculus
Integral calculus is a branch of mathematics that deals with integrating functions to calculate areas, volumes, and other quantities. It's especially useful in probability for finding expected values and determining other statistical measures.In this problem, integral calculus was used to compute the expected value \(E(X)\) of the random variable. We split the integral:
  • First part: \(\int_{-1}^{1} x \, dx = 0\)
  • Second part: \(\int_{-1}^{1} x^2 \, dx = \frac{2}{3}\)
By resolving these integrals, we could derive the expected value \(E(X) = \frac{\theta}{3}\). This result then allowed further investigation into the unbiased nature of the estimator. Integral calculus provided a method for handling the continuous nature of \(X\) within the specified range.

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

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2. \(\begin{array}{rrrrrrr}5.9 & 7.2 & 7.3 & 6.3 & 8.1 & 6.8 & 7.0 \\ 7.6 & 6.8 & 6.5 & 7.0 & 6.3 & 7.9 & 9.0 \\ 8.2 & 8.7 & 7.8 & 9.7 & 7.4 & 7.7 & 9.7 \\\ 7.8 & 7.7 & 11.6 & 11.3 & 11.8 & 10.7 & \end{array}\) a. Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. b. Calculate a point estimate of the strength value that separates the weakest \(50 \%\) of all such beams from the strongest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? [Hint: \(\left.\sum x_{i}^{2}=1860.94 .\right]\) d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds \(10 \mathrm{MPa}\). [Hint: Think of an observation as a "success" if it exceeds 10.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86, x_{6}=.47\), \(x_{7}=.73, x_{8}=.97, x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

The article from which the data of Exercise 1 was extracted also gave the accompanying strength observations for cylinders: \(\begin{array}{rrrrrrrrrr}6.1 & 5.8 & 7.8 & 7.1 & 7.2 & 9.2 & 6.6 & 8.3 & 7.0 & 8.3 \\ 7.8 & 8.1 & 7.4 & 8.5 & 8.9 & 9.8 & 9.7 & 14.1 & 12.6 & 11.2\end{array}\) Prior to obtaining data, denote the beam strengths by \(X_{1}, \ldots, X_{m}\) and the cylinder strengths by \(Y_{1}, \ldots, Y_{n^{*}}\). Suppose that the \(X_{i}\) s constitute a random sample from a distribution with mean \(\mu_{1}\) and standard deviation \(\sigma_{1}\) and that the \(Y_{i} \mathrm{~s}\) form a random sample (independent of the \(X_{i} \mathrm{~s}\) ) from another distribution with mean \(\mu_{2}\) and standard deviation \(\sigma_{2^{\circ}}\) a. Use rules of expected value to show that \(\bar{X}^{2}-\bar{Y}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). Calculate the estimate for the given data. b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error. c. Calculate a point estimate of the ratio \(\sigma_{1} / \sigma_{2}\) of the two standard deviations. d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \(X-Y\) between beam strength and cylinder strength.

Suppose the true average growth \(\mu\) of one type of plant during a l-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 \(E\left(X_{i}\right)=\mu\), \(V\left(X_{i}\right)=\sigma^{2}\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu, V\left(Y_{i}\right)=4 \sigma^{2}\right]\). 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 \(\delta\).]

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). \(\left[\right.\) Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X}\).] b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ? [Hint: Compute \(E\left(\bar{X}^{2}-k S^{2}\right)\).]
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.