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Chapter 6: Problem 22

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 \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86\), \(x_{6}=.47, x_{7}=.73, x_{9}=.97, x_{9}=.94, x_{10}=.77\). 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
\(\hat{\theta}_{MOM} = 0.25\), \(\hat{\theta}_{MLE} \approx 2.85\).

Step by step solution

01

Understand the PDF

The provided probability density function (PDF) is for a Beta distribution: \(f(x; \theta) = (\theta + 1)x^{\theta}\) for \(0 \leq x \leq 1\). The condition \(-1 < \theta\) is given to ensure \(f(x; \theta)\) integrates to 1 over \([0, 1]\).
02

Calculate the Sample Mean

The method of moments equates sample moments to population moments. Calculate the sample mean, \(\bar{x}\), using the data: \[\bar{x} = \frac{1}{10}(0.92 + 0.79 + 0.90 + 0.65 + 0.86 + 0.47 + 0.73 + 0.97 + 0.94 + 0.77)\].
03

Sample Mean Calculation

Using the given data:\[\bar{x} = \frac{1}{10}(0.92 + 0.79 + 0.90 + 0.65 + 0.86 + 0.47 + 0.73 + 0.97 + 0.94 + 0.77) = 0.8\].
04

Method of Moments Estimator

The expected value of a beta distribution is \(E[X] = \frac{1}{\theta + 2}\). Equating this to the sample mean, \(\bar{x} = 0.8\), we get:\[\frac{1}{\theta + 2} = 0.8\]This resolves to \(\theta = 0.25\).
05

Likelihood Function

Formulate the likelihood function \(L(\theta)\) from the PDF:\[L(\theta) = \prod_{i=1}^{10}((\theta + 1)x_i^{\theta})\]. Take the natural log to simplify: \[\ln L(\theta) = 10\ln(\theta + 1) + \theta \sum_{i=1}^{10} \ln x_i\].
06

Maximum Likelihood Estimator

Differentiate \(\ln L(\theta)\) with respect to \(\theta\) and set the derivative to zero:\[0 = \frac{10}{\theta + 1} + \sum_{i=1}^{10} \ln x_i\].Solve for \(\theta\): \[\hat{\theta}_{MLE} = -\frac{10}{\sum_{i=1}^{10} \ln x_i} - 1\].
07

Compute MLE for the Data

Calculate \(\sum_{i=1}^{10} \ln x_i\), then plug this into the MLE formula:\[\sum_{i=1}^{10} \ln(x_i) = \ln(0.92) + \ln(0.79) + \ln(0.90) + \ln(0.65) + \ln(0.86) + \ln(0.47) + \ln(0.73) + \ln(0.97) + \ln(0.94) + \ln(0.77) = -2.60\] (approx.).Thus, \( \hat{\theta}_{MLE} = -\frac{10}{-2.60} - 1 \approx 2.85\).
08

Final Step: Summary of Estimators

From the method of moments, the estimator for \(\theta\) is \(\hat{\theta}_{MOM} = 0.25\). From the maximum likelihood method, the estimator for \(\theta\) is \(\hat{\theta}_{MLE} \approx 2.85\).

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Key Concepts

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

Maximum Likelihood Estimation
When you want to estimate an unknown parameter from a dataset, maximum likelihood estimation, or MLE for short, is a popular and powerful technique. It's all about finding the value of the parameter that makes the observed data most probable under the assumed statistical model. Think of it like trying to find the 'best fit' for the data.

To use MLE, you start by writing down the likelihood function. This function measures how likely it is to observe your data for different values of the parameter. For example, if you have some experimental data and you're assuming a beta distribution, your likelihood function will include this distribution and your unknown parameter. The goal is to find the parameter value that maximizes this likelihood function.

In practice, instead of dealing with the very large or very small numbers that can come from the likelihood function, we often use the natural logarithm to simplify computations鈥攖his is called the log-likelihood. Deriving and solving the log-likelihood gives you the maximum likelihood estimator (MLE).
  • Formulate the likelihood function using your probability model.
  • Take the natural logarithm of the likelihood to form the log-likelihood.
  • Differentiate the log-likelihood with respect to the parameter and find where this derivative equals zero.
  • Solve for the maximum likelihood estimate, which is the parameter value maximizing the likelihood.
Finding the spot where the derivative of the log-likelihood equals zero tells you when you're at a peak, which is where the maximum likelihood estimator lies.
Probability Density Function
Probability density functions (PDFs) are crucial components in understanding continuous datasets. PDFs describe the likelihood of a random variable to take on a particular value, though it's worth noting that, strictly speaking, it鈥檚 the probability of landing within a specific range that matters when dealing with continuous data.

In the context of a PDF, the function gives you an idea about how the data are distributed over possible values. The area under the whole PDF curve (over its entire interval) is always equal to 1, representing the total probability.

The shape of a probability density function tells you a lot about the characteristics of the distribution. For instance, when dealing with a beta distribution鈥攚hich our exercise involves鈥攊ts PDF is flexible and can take on different shapes, depending on the parameters. This flexibility makes the beta distribution a versatile choice for modeling the probability of success in a sequence of trials.
  • The PDF is used to calculate probabilities over intervals by integrating over those intervals.
  • The PDF value at any specific point itself does not represent a probability.
  • The primary parameter in our beta distribution PDF is \(\theta\), which influences its shape and characteristics.
Understanding PDFs is essential for correctly applying statistical methods like MLE and the method of moments because they form the basis for approximating real-world processes mathematically.
Beta Distribution
The beta distribution is a versatile and widely-used probability distribution, particularly when dealing with variables that are constrained to the interval [0, 1], like proportions or percentages. In many statistical applications, it helps model outcomes that need to fit within this unit interval.

One of the interesting features of the beta distribution is its flexibility鈥攊ts shape can drastically change depending on its parameters. Although the exercise uses a simplified form of the beta distribution, the standard version has two shape parameters, usually denoted as \(\alpha\) and \(\beta\). These shape parameters allow you to model a variety of continuous distributions, making the beta distribution very adaptable.

In our problem, the PDF is presented as a specific form with parameter \(\theta\), which modifies the power of \(x\). This highlights it as a special type of beta distribution. By understanding how \(\theta\) affects the shape, you can better fit the distribution to your data and enhance your interpretations.
  • The beta distribution is tailored for data bounded between 0 and 1.
  • Its flexibility comes from how the parameters allow different shapes, enhancing modeling versatility.
  • Such a distribution is frequently used in Bayesian statistics, among other applications.
By gaining insight into these characteristics, you're not only calculating statistical estimates but gaining the ability to model real-world processes realistically and effectively with the beta distribution.

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Most popular questions from this chapter

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cc} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example 4.5, in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44\), \(3.42,10.39,8.93,17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).

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 \(p\) ?]

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 \(\mu^{2}\). [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 \(E\left(\bar{X}^{2}-k S^{2}\right)\).]

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 \(Y \sim\) 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? [Hinr: 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 \(p\) ?

The National Health and Nutrition Examination Survey (NHANES) collects demographic, socioeconomic, dietary, and health-related information on an annual basis. Here is a sample of 20 observations on HDL cholesterol level (mg/dl) obtained from the 20092010 survey (HDL is "good" cholesterol; the higher its value, the lower the risk for heart disease): \(\begin{array}{lllllll}35 & 49 & 52 & 54 & 65 & 51 & 51 \\ 47 & 86 & 36 & 46 & 33 & 39 & 45 \\ 39 & 63 & 95 & 35 & 30 & 48 & \end{array}\) a. Calculate a point estimate of the population mean HDL cholesterol level. b. Making no assumptions about the shape of the population distribution, calculate a point estimate of the value that separates the largest \(50 \%\) of HDL levels from the smallest \(50 \%\). c. Calculate a point estimate of the population standard deviation. d. An HDL level of at least 60 is considered desirable as it corresponds to a significantly lower risk of heart disease. Making no assumptions about the shape of the population distribution, estimate the proportion \(p\) of the population having an HDL level of at least 60 .
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