/*! 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 23 Let \(X\) denote the proportion ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(x_{1}=.92, \quad x_{2}=.79, \quad x_{3}=.90\), \(x_{4}=.65, x_{5}=.86, x_{6}=.47, x_{7}=.73, x_{8}=.97\), \(x_{9}=.94, x_{10}=.77\). 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.

Short Answer

Expert verified
Method of moments gives \(\theta = 8\); Maximum likelihood gives \(\theta \approx 2.967\).

Step by step solution

01

Understand the Problem and Given Data

We have a continuous random variable \(X\) representing the time proportion a student spends on an aptitude test, with a probability density function (pdf) given by \( f(x; \theta) = (\theta + 1)x^\theta \) for \(0 \leq x \leq 1\) and \(\theta > -1\). We have a sample of 10 observed values and need to estimate \(\theta\) using two methods: method of moments and maximum likelihood estimation.
02

Method of Moments for Estimator of \(\theta\)

In the method of moments, we equate the sample mean with the expected value of \(X\). The expected value of \(X\) is \(E[X] = \frac{\theta}{\theta + 2}\) for the given distribution. The sample mean \(\bar{x}\) is computed from the data:\[\bar{x} = \frac{1}{10} \sum_{i=1}^{10} x_i = \frac{1}{10} (0.92 + 0.79 + 0.90 + 0.65 + 0.86 + 0.47 + 0.73 + 0.97 + 0.94 + 0.77)\]Solving \(\bar{x} = \frac{\theta}{\theta + 2}\) for \(\theta\) gives us the method of moments estimator.
03

Calculate Sample Mean

Calculate the sample mean using the data:\[\bar{x} = \frac{0.92 + 0.79 + 0.90 + 0.65 + 0.86 + 0.47 + 0.73 + 0.97 + 0.94 + 0.77}{10} = 0.8\]
04

Solve Method of Moments Equation

Set up the equation for the method of moments: \[0.8 = \frac{\theta}{\theta + 2}\]Solving for \(\theta\),\[0.8(\theta + 2) = \theta\]\[0.8\theta + 1.6 = \theta\]\[1.6 = \theta - 0.8\theta = 0.2\theta\]\[\theta = 8\]
05

Maximum Likelihood Estimator (MLE) for \(\theta\)

The likelihood function for a sample is \[L(\theta) = \prod_{i=1}^{n} f(x_i; \theta) = \prod_{i=1}^{10} (\theta+1)x_i^\theta\]The log-likelihood function is \[\ln(L(\theta)) = n\ln(\theta + 1) + \theta\sum_{i=1}^{n}\ln(x_i)\]Differentiate the log-likelihood with respect to \(\theta\) and equate to zero to find the maximum likelihood estimator.
06

Derive and Solve MLE Equation

Differentiate the log-likelihood:\[\frac{d}{d\theta} \ln(L(\theta)) = \frac{10}{\theta + 1} + \sum_{i=1}^{10} \ln(x_i) = 0\]Solving \(-\frac{10}{\theta + 1} = \sum_{i=1}^{10} \ln(x_i)\) gives us: \[\theta = -1 + \frac{10}{-\sum_{i=1}^{10} \ln(x_i)}\]Compute \(\sum_{i=1}^{10}\ln(x_i)\) and substitute.
07

Compute the Sum for MLE

Calculate \(\sum_{i=1}^{10} \ln(x_i)\):\[\sum_{i=1}^{10} \ln(x_i) = \ln(0.92) + \ln(0.79) + \ln(0.90) + \ln(0.65) + \ln(0.86) + \ln(0.47) + \ln(0.73) + \ln(0.97) + \ln(0.94) + \ln(0.77)\approx -2.522\]
08

Solve for \(\theta\) using MLE

Substitute the value into the equation:\[\theta = -1 + \frac{10}{-(-2.522)} = -1 + \frac{10}{2.522}\]\[\theta \approx 2.967\]

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 91Ó°ÊÓ!

Key Concepts

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

Method of Moments
The Method of Moments is a simple and intuitive technique used for parameter estimation in statistical models. Its essence lies in equating sample moments, like the mean or variance, to theoretical moments of a probability distribution.

Moments are essentially expected values of powers of the random variable. The simplest moment is the mean, which can be seen as the first moment.

To apply the Method of Moments:
  • Compute the sample mean from the observed data.
  • Equate this sample mean to the theoretical expected value of the random variable, given in the problem.
  • Solve the resulting equation for the parameter of interest.
In the context of this problem, the sample mean was calculated from ten observed data points. It was then set equal to the expected value formula for the given probability density function (pdf), which is \[E[X] = \frac{\theta}{\theta + 2}.\] By solving for \(\theta\), we found that \(\theta = 8\). This is the method of moments estimator for the parameter \(\theta\).
Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a highly popular method for identifying the parameter values that make the observed data most probable under a given statistical model. It reflects the principle of likelihood, signifying the "best fit" of the model to the data.

Here's how you can understand MLE:
  • Start by establishing a likelihood function, which is the probability of the observed data as a function of the parameter.
  • Calculate the log of this likelihood, because it simplifies the math by turning products into sums.
  • Find the value of the parameter that maximizes this log-likelihood by setting its derivative to zero and solving.
In this exercise, the pdf is used to form the likelihood expression. We then derived the log-likelihood function and took its derivative concerning \(\theta\).

After simplifying, we solve for \(\theta\) and find \(\theta \approx 2.967\). This value is the maximum likelihood estimator for \(\theta\). MLE is powerful because it provides estimates with desirable properties like consistency and asymptotic normality.
Probability Density Function (PDF)
A Probability Density Function (PDF) is fundamental for continuous random variables and gives insight into the probability distribution of the variable. Unlike a probability mass function (PMF) used for discrete variables, a PDF describes probabilities over an interval rather than at a specific point.

Here's what you need to understand about PDFs:
  • The area under the PDF curve over an interval gives the probability that the random variable falls within that interval.
  • The PDF must satisfy two conditions: it must be non-negative at all points, and the total area under the curve must be equal to 1.
  • Specific values cannot have probabilities, but intervals do.
In our problem, the PDF is defined as \(f(x; \theta) = (\theta + 1) x^{\theta}\) for \(0 \leq x \leq 1\), where \(\theta > -1\).
This function shapes the distribution of how students allocate their time on an aptitude test. It helps us understand the spread and shape of the data which is crucial for both the Method of Moments and Maximum Likelihood Estimation.

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

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data: \begin{tabular}{llllllllll} Number of scratches per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \\ \hline \end{tabular} Let \(X=\) the number of scratches on a randomly chosen item, and assume that \(X\) has a Poisson distribution with parameter \(\lambda\). a. Find an unbiased estimator of \(\lambda\) and compute the estimate for the data. [Hint: \(E(X)=\lambda\) for \(X\) Poisson, so \(E(\bar{X}=\) ?)] b. What is the standard deviation (standard error) of your estimator? Compute the estimated standard error. [Hint: \(\sigma_{X}^{2}=\lambda\) for \(X\) Poisson.]

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

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solution yields the maximum likelihood estimators of \(\alpha\) and \(\beta\). Do you think they can be solved explicitly? b. Show that the mle of \(\mu=\alpha \beta\) is \(\hat{\mu}=\bar{X}\).

Components of a certain type are shipped in batches of size \(k\). Suppose that whether or not any particular component is satisfactory is independent of the condition of any other component, and that the long run proportion of satisfactory components is \(p\). Consider \(n\) batches, and let \(X_{i}\) denote the number of satisfactory components in the ith batch ( \(i=1,2, \ldots, n\) ). Statistician A is provided with the values of all the \(X_{i}\) 's, whereas statistician B is given only the value of \(X=\sum X_{i}\). Use a conditional probability argument to decide whether statistician A has more information about \(p\) than does statistician B.

Assume that the number of defects in a car has a Poisson distribution with parameter \(\lambda\). To estimate \(\lambda\) we obtain the random sample \(X_{1}\), \(X_{2}, \ldots, X_{n}\). a. Find the Fisher information in a single observation using two methods. b. Find the Cramér-Rao lower bound for the variance of an unbiased estimator of \(\lambda\). c. Use the score function to find the mle of \(\lambda\) and show that the mle is an efficient estimator. d. Is the asymptotic distribution of the mle in accord with the second theorem? Explain.

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.