Problem 8 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The following data represent the number of revolutions to failure (in millions) of 22 deep-groove ball-bearings. $$ \begin{array}{rrrrr} \hline \hline 17.88 & 28.92 & 33.00 & 41.52 & 42.12 \\ 45.60 & 48.48 & 51.84 & 51.96 & 54.12 \\ 55.56 & 67.80 & 68.64 & 68.88 & 84.12 \\ 93.12 & 98.64 & 105.12 & 105.84 & 127.92 \\ 128.04 & 173.40 & & & \\ \hline \hline \end{array} $$ Lieblein and Zelen propose modeling the dataset as a realization of a random sample from a Weibull distribution, which has distribution function $$ F(x)=1-\mathrm{e}^{-(\lambda x)^{\alpha}} \quad \text { for } x \geq 0 $$ and \(F(x)=0\), for \(x<0\), where \(\alpha, \lambda>0\). a. Suppose that \(X\) is a random variable with a Weibull distribution. Check that the random variable \(Y=X^{\alpha}\) has an exponential distribution with parameter \(\lambda^{\alpha}\) and conclude that \(\mathrm{E}\left[X^{\alpha}\right]=1 / \lambda^{\alpha}\). b. Use part a to explain how one can use the data in the table to find an estimate for the parameter \(\lambda\), if it is given that the parameter \(\alpha\) is estimated by \(2.102\).

Step-by-Step Solution

Answer

Estimate \( \lambda \) by calculating the mean of \( X^{2.102} \) and using \( \lambda = (1/\text{mean})^{1/2.102} \).

Step 1: Understand Weibull and Exponential Distributions

The given Weibull distribution has the cumulative distribution function (CDF): \( F(x)=1-\mathrm{e}^{-(\lambda x)^\alpha} \). If \( X \) follows a Weibull distribution, transforming it to \( Y = X^\alpha \) implies \( Y \) follows an exponential distribution with parameter \( \lambda^\alpha \). The expectation or mean of \( Y \) is \( \frac{1}{\lambda^\alpha} \), i.e., \( \mathrm{E}[Y] = \mathrm{E}[X^\alpha] = \frac{1}{\lambda^\alpha} \).

Step 2: Use the Dataset for Estimation

We need to find an estimate for the parameter \( \lambda \). Given \( \alpha = 2.102 \), calculate \( X_i^\alpha \) for each observation \( X_i \) in the dataset. Sum these transformed values and then divide by the number of observations to get the sample mean of \( X^\alpha \).

Step 3: Calculate Expected Value Estimate

Use the expectation formula for \( X^\alpha \). From Step 1, we know \( \mathrm{E}[X^\alpha] = \frac{1}{\lambda^\alpha} \). Therefore, set the sample mean calculated in Step 2 equal to \( \frac{1}{\lambda^\alpha} \) and solve for \( \lambda \).

Step 4: Apply Given Alpha to Find Lambda

Given \( \alpha = 2.102 \), use the equation \( \lambda = \left(\frac{1}{\text{Sample Mean of } X^\alpha}\right)^{\frac{1}{\alpha}} \) to compute \( \lambda \). Insert the calculated sample mean from Steps 2 and 3, ensuring to adjust \( \lambda \) to fit the power of the exponential distribution.

Key Concepts

Exponential Distribution

The Exponential Distribution is a continuous probability distribution often used to model the time until an event occurs, such as failure or arrival times. This distribution is defined by its parameter, known as the rate parameter \( \lambda \), which influences the distribution's shape. The probability density function (PDF) for an exponential distribution is given by \[ f(y; \lambda) = \lambda e^{-\lambda y} \quad \text{for} \quad y \geq 0, \] and is zero for \( y < 0 \). The exponential distribution is particularly known for its memoryless property, meaning that the probability of an event occurring in the next unit of time is the same regardless of how much time has already passed.

• The mean or expected value of an exponential distribution is \( \frac{1}{\lambda} \).
• It is a special case of the Gamma distribution.

In the exercise about the Weibull distribution, we observe how the variable \( Y = X^\alpha \) from a Weibull distribution can transform into an exponential distribution with the parameter \( \lambda^\alpha \). This transformation shows a deep interconnection between these two statistical models.

Parameter Estimation

Parameter Estimation refers to the process of using sample data to estimate the parameters of a probability distribution. It is a key aspect of statistical analysis in many applications, including reliability engineering, econometrics, and more. When given a dataset, the aim is to find the best estimate for our parameters in the chosen model, such as \( \lambda \) and \( \alpha \) for the Weibull distribution.In the given exercise, we focus on estimating the \( \lambda \) parameter. The process follows these steps:

• Calculate the transformed data set \( \{ X_i^\alpha \} \) using the provided \( \alpha = 2.102 \).
• Find the sample mean of these transformed values, which represents the empirical mean of \( X^\alpha \).
• Utilize the formula \( \mathrm{E}[X^\alpha] = \frac{1}{\lambda^\alpha} \) to set up an equation with the sample mean and solve for \( \lambda \).

This step-by-step approach ensures a precise estimation of the \( \lambda \) parameter, reflecting the underlying characteristics of the sample data.

Random Variables

Random Variables are fundamental to probability and statistics. They are functions that assign a numerical value to each outcome in a sample space. Random variables can either be discrete, taking on specific values, or continuous,where they can assume any value within a given range. In this context, the conceptualization of random variables is essential in understanding how data from real-world phenomena are modeled.In relation to the exercise on Weibull and exponential distributions, we deal with random variable transformations. The variable \( X \) follows a Weibull distribution, and by raising it to the power of \( \alpha \), we create a new random variable \( Y = X^\alpha \). This transformation shows how different distributions relate to one another and how statistical properties change under specific operations. Recognizing different types of random variables and their transformations allows statisticians and engineers to model data more effectively:

• Discrete Random Variables have a countable number of possible values.
• Continuous Random Variables can take any value within some range.

Understanding these concepts provides a solid foundation for analyzing complex statistical models like the Weibull distribution used in reliability analysis.