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Consider the Weibull distribution $$ f(x)=\left\\{\begin{array}{ll} \frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta-1} e^{-\left(\frac{x}{\delta}\right)^{\beta}}, & 0

Short Answer

Expert verified
The likelihood is the product of PDFs, giving a log-likelihood. Maximizing involves nonlinear equations requiring numerical methods.

Step by step solution

01

Understand the Distribution

We have a Weibull distribution with probability density function (PDF):\[ f(x) = \begin{cases} \frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta-1} e^{-\left(\frac{x}{\delta}\right)^{\beta}}, & x > 0 \ 0, & \text{otherwise} \end{cases} \]where \(\beta\) is the shape parameter and \(\delta\) is the scale parameter.
02

Derive the Likelihood Function

For a random sample \(x_1, x_2, \ldots, x_n\) from the Weibull distribution, the likelihood function \(L(\beta, \delta)\) is the product of individual PDFs:\[ L(\beta, \delta) = \prod_{i=1}^{n} \frac{\beta}{\delta}\left(\frac{x_i}{\delta}\right)^{\beta-1} e^{-\left(\frac{x_i}{\delta}\right)^{\beta}} \]Simplifying, this becomes \[ L(\beta, \delta) = \left(\frac{\beta}{\delta}\right)^n \prod_{i=1}^{n} x_i^{\beta-1} e^{-\sum_{i=1}^{n}\left(\frac{x_i}{\delta}\right)^{\beta}} \]
03

Find the Log-Likelihood Function

The log-likelihood function \(\ell(\beta, \delta)\) is obtained by taking the natural logarithm of the likelihood function:\[ \ell(\beta, \delta) = n\ln\beta - n\ln\delta + (\beta - 1)\sum_{i=1}^{n}\ln(x_i) - \sum_{i=1}^{n}\left(\frac{x_i}{\delta}\right)^{\beta} \]
04

Setup the Equations for Maximization

To maximize the log-likelihood, differentiate \(\ell(\beta, \delta)\) with respect to \(\beta\) and \(\delta\), set the derivatives equal to zero, and solve:\[ \frac{\partial \ell}{\partial \beta} = n\frac{1}{\beta} + \sum_{i=1}^{n}\ln(x_i) - \sum_{i=1}^{n}\left(\frac{x_i}{\delta}\right)^{\beta}\ln\left(\frac{x_i}{\delta}\right) = 0 \] \[ \frac{\partial \ell}{\partial \delta} = -\frac{n\beta}{\delta} + \beta\sum_{i=1}^{n}\left(\frac{x_i}{\delta}\right)^{\beta-1}\frac{1}{\delta} = 0 \]These equations simplify to the ones provided in the exercise.
05

Address Complications in Solutions

Solving the two equations involves iterative numerical methods because both parameters \(\beta\) and \(\delta\) appear nonlinearly. The expressions involve summations that depend on the values of \(\beta\), so closed-form solutions are challenging to obtain without approximation methods (e.g., Newton-Raphson).

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Key Concepts

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

Likelihood Function
In statistics, when dealing with a Weibull distribution or any probability distribution, the **Likelihood Function** is a fundamental concept for parameter estimation. The likelihood function, denoted as \( L(\beta, \delta) \), is constructed from a set of observed data points. It provides a measure of how well different values of the parameters (in this case, \( \beta \) and \( \delta \)) explain the observed data.

For the Weibull distribution with a probability density function (PDF) given above, if we have a random sample \( x_1, x_2, \ldots, x_n \), the likelihood function is the product of the PDFs for each observed data point.

  • Each PDF corresponds to the probability of observing a specific data point given the parameters \( \beta \) and \( \delta \).
  • The product operator \( \prod \) across all data points synthesizes how probable the entire dataset is under those parameters.
Understanding the likelihood function helps us set the scene for identifying the best-fitting parameters for our model.
Log-Likelihood Maximization
Once we have the **Likelihood Function**, the next step is to transform it using the natural logarithm. This process gives us the **Log-Likelihood Function**. The reason for this transformation is simple: it turns products into sums, which are easier to work with in calculus.

The log-likelihood function for a Weibull distribution, denoted as \( \ell(\beta, \delta) \), is expressed as:
\[ \ell(\beta, \delta) = n\ln\beta - n\ln\delta + (\beta - 1)\sum_{i=1}^{n}\ln(x_i) - \sum_{i=1}^{n}\left(\frac{x_i}{\delta}\right)^{\beta} \]

This expression retains all the information from the likelihood function but in a manipulated form that facilitates further mathematical handling.

  • The transformation by logarithm simplifies the math by turning products into sums.
  • It helps in deriving partial derivatives, necessary for maximizing the log-likelihood.
Maximizing this function involves finding the best values of \( \beta \) and \( \delta \) that result in the highest likelihood of observing the given data. This step is crucial for precision in parameter estimation.
Maximum Likelihood Estimation
**Maximum Likelihood Estimation (MLE)** is a method used to find estimates of distribution parameters that make the observed data most probable under a specified model. In the case of the Weibull distribution, MLE involves solving for the parameters \( \beta \) and \( \delta \) that maximize the log-likelihood function.

MLE estimates are derived by setting the derivative of the log-likelihood function equal to zero and solving the resulting system of equations. For the Weibull distribution, the key maximization equations are:

\[ \beta = \left[\frac{\sum_{i=1}^{n} x_{i}^{\beta} \ln(x_{i})}{\sum_{i=1}^{n} x_{i}^{\beta}}-\frac{\sum_{i=1}^{n} \ln(x_{i})}{n}\right]^{-1} \]

\[ \delta = \left[\frac{\sum_{i=1}^{n} x_{i}^{\beta}}{n}\right]^{1 / \beta} \]

MLE is favored because it is consistent and efficient, often providing robust estimates with large samples, making it particularly useful in statistical modeling.

  • MLE takes advantage of calculus tools to find parameter estimates that best explain the data.
  • These estimates tend to have desirable properties like unbiasedness and minimum variance.
Through MLE, statisticians can deduce the key parameters characterizing the data's underlying distribution.
Iterative Numerical Methods
Solving for parameters in **Maximum Likelihood Estimation** often requires dealing with complex equations that might not be solvable by straightforward algebraic means. This is the point where **Iterative Numerical Methods** come into play, especially in the context of nonlinear parameter estimation like with the Weibull distribution.

Naturally, the complexity arises because both parameters \( \beta \) and \( \delta \) are involved nonlinearly in the maximization equations. Iterative methods like the Newton-Raphson or Expectation-Maximization algorithms are particularly handy for this purpose.

  • Newton-Raphson: This involves using derivatives to iteratively approach the maximum point of the log-likelihood function.
  • Expectation-Maximization: Useful when data is incomplete or has hidden variables, helping in the estimation of parameters by iterating between expectation and maximization steps.
In essence, these methods help bypass the inability to solve complex equations analytically by providing an approximate yet highly accurate solution through repeated iterations.

The utilization of these methods requires initial parameter guesses and then commences with iterations until convergence, where the parameter estimates stabilize to a set level of precision needed for the analysis.

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Most popular questions from this chapter

Wayne Collier designed an experiment to measure the fuel efficiency of his family car under different tire pressures. For each run, he set the tire pressure and then measured the miles he drove on a highway (I-95 between Mills River and Pisgah Forest, NC) until he ran out of fuel using 2 liters of fuel each time. To do this, he made some alterations to the normal flow of gasoline to the engine. In Wayne's words, "I inserted a T-junction into the fuel line just before the fuel filter, and a line into the passenger compartment of my car, where it joined with a graduated 2 -liter Rubbermaid \(^{\circ}\) bottle that I mounted in a box where the passenger seat is normally fastened. Then I sealed off the fuel-return line, which under normal operation sends excess fuel from the fuel pump back to the fuel tank." Suppose that you call the mean miles that he can drive with normal pressure in the tires \(\mu\). An unbiased estimate for \(\mu\) is the mean of the sample runs, \(\bar{x}\). But Wayne has a different idea. He decides to use the following estimator: He flips a fair coin. If the coin comes up heads, he will add 5 miles to each observation. If tails come up, he will subtract 5 miles from each observation. a. Show that Wayne's estimate is, in fact, unbiased. b. Compare the standard deviation of Wayne's estimate with the standard deviation of the sample mean. c. Given your answer to (b), why does Wayne's estimate not make good sense scientifically?

Suppose that \(\hat{\Theta}_{1}, \hat{\Theta}_{2},\) and \(\hat{\Theta}_{3}\) are estimators of \(\theta .\) We know that \(E\left(\hat{\Theta}_{1}\right)=E\left(\hat{\Theta}_{2}\right)=\theta, E\left(\hat{\Theta}_{3}\right) \neq \theta, V\left(\hat{\Theta}_{1}\right)=12, V\left(\hat{\Theta}_{2}\right)=\) 10 , and \(E\left(\hat{\Theta}_{3}-\theta\right)^{2}=6\). Compare these three estimators. Which do you prefer? Why?

\(\bar{X}_{1}\) and \(S_{1}^{2}\) are the sample mean and sample variance from a population with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2} .\) Similarly, \(\bar{X}_{2}\) and \(S_{2}^{2}\) are the sample mean and sample variance from a second independent population with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2} .\) The sample sizes are \(n_{1}\) and \(n_{2},\) respectively. a. Show that \(\bar{X}_{1}-\bar{X}_{2}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). b. Find the standard error of \(\bar{X}_{1}-\bar{X}_{2} .\) How could you estimate the standard error? c. Suppose that both populations have the same variance; that is, \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2} .\) Show that $$ S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Data on pull-off force (pounds) for connectors used in an automobile engine application are as follows: 79.3,75.1 , \(78.2,74.1,73.9,75.0,77.6,77.3,73.8,74.6,75.5,74.0,74.7,\) 75.9,72.9,73.8,74.2,78.1,75.4,76.3,75.3,76.2,74.9,78.0 75.1,76.8 a. Calculate a point estimate of the mean pull-off force of all connectors in the population. State which estimator you used and why. b. Calculate a point estimate of the pull-off force value that separates the weakest \(50 \%\) of the connectors in the population from the strongest \(50 \%\) c. Calculate point estimates of the population variance and the population standard deviation. d. Calculate the standard error of the point estimate found in part (a). Interpret the standard error. e. Calculate a point estimate of the proportion of all connectors in the population whose pull-off force is less than 73 pounds.

Suppose that the random variable \(X\) has the continuous uniform distribution $$f(x)=\left\\{\begin{array}{ll}1, & 0 \leq x \leq 1 \\\0, & \text { otherwise }\end{array}\right.$$ Suppose that a random sample of \(n=12\) observations is selected from this distribution. What is the approximate probability distribution of \(\bar{X}-6\) ? Find the mean and variance of this quantity.

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