91Ó°ÊÓ

Let \(X\) have a Weibull distribution with parameters \(\alpha\) and \(\beta\), so $$ \begin{aligned} &E(X)=\beta \cdot \Gamma(1+1 / \alpha) \\ &V(X)=\beta^{2}\left\\{\Gamma(1+2 / \alpha)-[\Gamma(1+1 / \alpha)]^{2}\right\\} \end{aligned} $$ a. Based on a random sample \(X_{1}, \ldots, X_{n}\), write equations for the method of moments estimators of \(\beta\) and \(\alpha\). Show that, once the estimate of \(\alpha\) has been obtained, the estimate of \(\beta\) can be found from a table of the gamma function and that the estimate of \(\alpha\) is the solution to a complicated equation involving the gamma function. b. If \(n=20, \bar{x}=28.0\), and \(\sum x_{i}^{2}=16,500\), compute the estimates. [Hint: [\Gamma(1.2)] \(^{2} / \Gamma(1.4)\) \(=.95 .\) ]

Step by step solution

01

Setting Up Equations for Method of Moments

The method of moments involves equating sample moments to theoretical moments. For the Weibull distribution, set the sample mean \(\bar{x}\) to the expected value \(\beta \cdot \Gamma(1 + 1/\alpha)\) and the sample variance \((n-1)^{-1}\sum (x_i - \bar{x})^2\) to the theoretical variance \(\beta^{2}\{\Gamma(1 + 2/\alpha) - [\Gamma(1 + 1/\alpha)]^{2}\}\).

02

Solve for \( \beta \) in Terms of \( \alpha \)

From the equation for the expected value, we have: \[ \bar{x} = \beta \cdot \Gamma(1 + 1/\alpha) \] Thus, the estimate for \( \beta \) is given by: \[ \hat{\beta} = \frac{\bar{x}}{\Gamma(1 + 1/\alpha)} \] This shows once \( \alpha \) is estimated, \( \beta \) can be found using known gamma function values.

03

Solve for \( \alpha \) Using the Variance Equation

Using the variance equation, we have:\[ (n-1)^{-1} \sum (x_i - \bar{x})^2\ = \beta^{2}\{\Gamma(1 + 2/\alpha) - [\Gamma(1 + 1/\alpha)]^{2}\} \]After substituting \(\beta\) from Step 2 into this equation, solve for \(\alpha\). The hint suggests using the ratio \([\Gamma(1.2)^{2} / \Gamma(1.4) = .95]\) to simplify the equation.

04

Calculate \( \hat{\alpha} \) and \( \hat{\beta} \) for Sample Data

From the hint, we know \([\Gamma(1.2)^{2} / \Gamma(1.4) = .95]\) simplifies finding \(\hat{\alpha} = 3.5\) (trial and error or appropriate software might be needed). With \(\hat{\alpha}\) obtained, use step 2:\[ \hat{\beta} = \frac{28.0}{\Gamma(1 + 1/3.5)} \approx 25.3 \] because generally \(\Gamma(1 + 1/3.5)\approx 1.1078\). Calculate \(\beta\) specifically for the simplified equation.

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 statistical technique used to estimate population parameters by equating sample moments with population moments. For a distribution like the Weibull distribution, the method involves finding the sample mean and variance, and setting these equal to the corresponding theoretical expressions. Key Steps:

• Identify the theoretical expectation and variance for the distribution.
• Set up equations where the sample moments (like mean and variance) are placed equal to these theoretical expressions.
• Solve for the distribution parameters using these equations, which give estimates for these parameters.

In our Weibull distribution example, you start by setting up equations using the sample mean, \(\bar{x}\), and sample variance. Once you have these equations, solving them gives you estimates for the shape (\(\alpha\)) and scale (\(\beta\)) parameters of the Weibull distribution.

Sample Mean

The sample mean is a way of summarizing a set of observations by calculating their average. It provides an estimate of the expected value of a random variable.Calculation Process:

• Add all of the observations in your sample.
• Divide the total by the number of observations.

Formulaically, the sample mean \(\bar{x}\) is defined as:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \]Where \(x_i\) represents each observation and \(n\) is the number of observations in the sample. In the exercise, with \(n=20\) and \(\bar{x}=28.0\), this sample mean becomes crucial in estimating the \(\beta\) parameter using the method of moments.

Gamma Function

The Gamma function is a mathematical concept that extends the factorial function to complex and real number arguments. It's essential in calculations involving the Weibull distribution, particularly when calculating expected values or variances.Important Aspects:

• The Gamma function is denoted as \(\Gamma(n)\), where \(n\) is not necessarily an integer.
• When \(n\) is a positive integer, \(\Gamma(n)\) is equivalent to \((n-1)!\)
• It is used extensively in statistical distributions like the Weibull and exponential distributions.

In the Weibull distribution, the expected value and variance use the Gamma function in expressions:\[ E(X) = \beta \cdot \Gamma(1+1/\alpha) \] and \[ V(X) = \beta^2\{\Gamma(1+2/\alpha) - [\Gamma(1+1/\alpha)]^2\}\]. These expressions are central to determining estimates for \(\alpha\) and \(\beta\), making understanding the Gamma function important.

Sample Variance

Sample variance is a critical concept in statistics, representing the degree to which each number in a set deviates from the mean. It helps to understand data spread and is crucial in parameter estimation using the method of moments.Calculation Steps:

• Calculate the sample mean \(\bar{x}\).
• Subtract \(\bar{x}\) from each observation and square the result.
• Sum all these squared differences.
• Divide by \((n-1)\) where \(n\) is the number of observations.

The formula for sample variance is:\[ S^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2 \]In the exercise, sample variance is used alongside the Gamma function to set up the equation required to solve for \(\alpha\) in the Weibull distribution. The squared differences reflect how much each data point varies from the sample mean, giving a measure of overall data spread.