/*! 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 26 Consider randomly selecting \(n\... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider randomly selecting \(n\) segments of pipe and determining the corrosion loss \((\mathrm{mm})\) in the wall thickness for each one. Denote these corrosion losses by \(Y_{1}, \ldots, Y_{n}\). The article "A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains" (Reliability Engr. and System Safety (2013:270-279) proposes a linear corrosion model: \(Y_{i}=t_{l} R\), where \(t_{i}\) is the age of the pipe and \(R\), the corrosion rate, is exponentially distributed with parameter \(\lambda\). Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). [Hint: If \(c>0\) and \(X\) has an exponential distribution, so does \(c X\).]

Short Answer

Expert verified
The maximum likelihood estimator of \( \lambda \) is \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \).

Step by step solution

01

Understand the Problem

We are asked to find the maximum likelihood estimator (MLE) of the parameter \( \lambda \) for the corrosion model \( Y_i = t_i R \), where \( R \) follows an exponential distribution with parameter \( \lambda \). The key idea is to use the property that a scaled exponential random variable (\( cX \)) is also exponential.
02

Set Up the Likelihood Function

Given \( Y_i = t_i R \) and \( R \sim \text{Exp}(\lambda) \), \( Y_i \) is effectively a scaled version of \( R \) and thus follows an exponential distribution with rate \( \lambda/t_i \). The likelihood function for \( n \) independent observations is: \[ L(\lambda) = \prod_{i=1}^n \left( \frac{\lambda}{t_i} \right) \exp\left( - \frac{\lambda}{t_i} Y_i \right) \] Simplifying, we get: \[ L(\lambda) = \left( \prod_{i=1}^n \frac{\lambda}{t_i} \right) \exp\left( -\lambda \sum_{i=1}^n \frac{Y_i}{t_i} \right) \]
03

Take the Logarithm of the Likelihood Function

The next step is to take the natural logarithm of the likelihood function to facilitate maximization: \[ \ln L(\lambda) = \sum_{i=1}^n \ln \left( \frac{\lambda}{t_i} \right) - \lambda \sum_{i=1}^n \frac{Y_i}{t_i} \] Simplifying further: \[ \ln L(\lambda) = n \ln(\lambda) - \sum_{i=1}^n \ln(t_i) - \lambda \sum_{i=1}^n \frac{Y_i}{t_i} \]
04

Find the Maximum Likelihood Estimator

To find the MLE of \( \lambda \), differentiate \( \ln L(\lambda) \) with respect to \( \lambda \) and set to zero: \[ \frac{d}{d\lambda}\ln L(\lambda) = \frac{n}{\lambda} - \sum_{i=1}^n \frac{Y_i}{t_i} = 0 \] Solving for \( \lambda \): \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \] This is the MLE for the parameter \( \lambda \).

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Key Concepts

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

Exponential Distribution
The exponential distribution is fundamental when modeling the time between events in a Poisson process. It often describes situations where we’re interested in the time until a particular event occurs, such as the time a machine part will last until it fails. In the context of the corrosion model, the corrosion rate \( R \) of pipes is assumed to be exponentially distributed. This implies that each unit of time has a constant probability of contributing to the corrosion rate.

The exponential distribution is characterized by the parameter \( \lambda \), which defines the rate at which events occur. A larger value of \( \lambda \) indicates more frequent events, whereas a smaller \( \lambda \) suggests they are rare. When dealing with random variables that are scaled versions of a true exponential variable, as we have with \( Y_i = t_i R \), the resulting distribution maintains its exponential nature while adjusting its rate to \( \lambda/t_i \). This property allows us to model the corrosion loss in pipe segments effectively.
Corrosion Model
The corrosion model provided in the exercise describes how the material of the pipe degrades over time. Specifically, it sets up a mathematical relationship where the corrosion loss \( Y_i \) is a function of the age of the pipe \( t_i \) and the corrosion rate \( R \). This model is useful for predicting the remaining life of the pipe, assessing risks, and planning maintenance.

In practical terms, the model helps in understanding how long a pipe can be expected to last before it becomes too thin to safely contain the gas transported within. It's a linear model in nature because of the direct multiplication between \( t_i \) and \( R \). The simplicity of the linear relationship makes it especially appealing for engineering applications where straightforward predictions are desirable.
Probabilistic Model
A probabilistic model is used to incorporate uncertainty and variability when analyzing real-world systems, like the corrosion process. By using probability to model the corrosion rate \( R \), we acknowledge that there are many unknown variables affecting how quickly a pipe will corrode, such as environmental conditions, material inconsistencies, and manufacturing variations.

By using the exponential distribution within the probabilistic model, predictions can be made about the average corrosion loss, which provides insights into the overall probabilities of different levels of corrosion occurring over time. This allows decision-makers to strategize around risks and uncertainties, improving the safety and reliability of systems.
Parameter Estimation
Parameter estimation involves determining the parameters of a statistical model that best fit the observed data. In this context, we sought the maximum likelihood estimator (MLE) for \( \lambda \), which tells us the rate of the exponential distribution modeling the corrosion rate.

The MLE approach selects the value of \( \lambda \) that maximizes the likelihood function, thus making the observed data most probable under the model. We do this by setting up the likelihood function based on our exponential model and finding \( \hat{\lambda} \), the equation that results in the highest likelihood of observing the data we have.

This process provides a statistically grounded method of parameter estimation, which is preferable over simple guesswork as it uses all available data to make informed predictions that are consistent and replicable.

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Most popular questions from this chapter

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are \(103,156,118,89,125\), \(147,122,109,138,99\). Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

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

Of \(n_{1}\) randomly selected male smokers, \(X_{1}\) smoked filter cigarettes, whereas of \(n_{2}\) randomly selected female smokers, \(X_{2}\) smoked filter cigarettes. Let \(p_{1}\) and \(p_{2}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. a. Show that \(\left(X_{1} / n_{1}\right)-\left(X_{2} / n_{2}\right)\) is an unbiased estimator for \(p_{1}-p_{2}\) - [Hint: \(E\left(X_{i}\right)=n_{i} p_{i}\) for \(i=1,2\).] b. What is the standard error of the estimator in part (a)? c. How would you use the observed values \(x_{1}\) and \(x_{2}\) to estimate the standard error of your estimator? d. If \(n_{1}=n_{2}=200, x_{1}=127\), and \(x_{2}=176\), use the estimator of part (a) to obtain an estimate of \(p_{1}-p_{2}\). e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solutions yield 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 \(\hat{\mu}=\bar{X}\).

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

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