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Chapter 6: Q26E (page 273)

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

Short Answer

Expert verified

The maximum likelihood estimator is \({\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {{{\rm{Y}}_{\rm{j}}}} {\rm{/}}{{\rm{t}}_{\rm{j}}}}}\).

Step by step solution

01

Define exponential function

A function that increases or decays at a rate proportional to its present value is called an exponential function.

02

Explanation

Given a set of random variables,

\({{\rm{Y}}_{\rm{i}}}{\rm{ = }}{{\rm{t}}_{\rm{i}}}{\rm{R}}\)

where\({{\rm{t}}_{\rm{i}}}\)is an integer and\({\rm{R}}\)is an exponentially distributed random variable with parameters\({\rm{\lambda }}\). The\({{\rm{Y}}_{\rm{i}}}\)cdf is,

\(\begin{array}{c}{{\rm{F}}_{{{\rm{Y}}_{\rm{i}}}}}{\rm{(y) = P}}\left( {{{\rm{Y}}_{\rm{i}}} \le {\rm{y}}} \right)\\{\rm{ = P}}\left( {{{\rm{t}}_{\rm{i}}}{\rm{R}} \le {\rm{y}}} \right)\\{\rm{ = P}}\left( {{\rm{R}} \le \frac{{\rm{y}}}{{{{\rm{t}}_{\rm{i}}}}}} \right)\\{\rm{ = }}{{\rm{F}}_{\rm{R}}}\left( {\frac{{\rm{y}}}{{{{\rm{t}}_{\rm{i}}}}}} \right)\\{\rm{ = 1 - exp}}\left\{ {{\rm{ - \lambda }}\frac{{\rm{y}}}{{{{\rm{t}}_{\rm{i}}}}}} \right\}\\{\rm{ = 1 - exp}}\left\{ {{\rm{ - }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{i}}}}}{\rm{y}}} \right\}{\rm{,y}} \ge {\rm{0}}\end{array}\)

For\({\rm{y < 0}}\), and is zero. As a result,\({{\rm{Y}}_{\rm{i}}}\); has a parametric exponential distribution.

\({{\rm{\lambda }}_{\rm{i}}}{\rm{ = }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{i}}}}}\)

Allow joint pdf or pmb for random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\).

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right){\rm{, n,m}} \in {\rm{N}}\)

where\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)are unknown parameters. The likelihood function is defined as a function of parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)where function f is a function of parameter. The maximum likelihood estimates (mle's), or values\(\widehat {{{\rm{\theta }}_{\rm{i}}}}\)for which the likelihood function is maximised, are the maximum likelihood estimates,

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{{\rm{\hat \theta }}}_{\rm{1}}}{\rm{,}}{{{\rm{\hat \theta }}}_{\rm{2}}}{\rm{, \ldots ,}}{{{\rm{\hat \theta }}}_{\rm{m}}}} \right) \ge {\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right)\)

As, \({\rm{i = 1,2, \ldots ,m}}\) for every \({{\rm{\theta }}_{\rm{i}}}\). Maximum likelihood estimators are derived by replacing \({{\rm{X}}_{\rm{i}}}\) with \({{\rm{x}}_{\rm{i}}}\).

03

Evaluating the maximum likelihood estimators

Because of the independence, the likelihood function becomes,

\(\begin{array}{c}{\rm{f}}\left( {{{\rm{y}}_{\rm{1}}}{\rm{,}}{{\rm{y}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{y}}_{\rm{n}}}{\rm{;\lambda }}} \right){\rm{ = }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{1}}}}}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{1}}}}}{{\rm{y}}_{\rm{1}}}} \right\}{\rm{ \times }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{2}}}}}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{2}}}}}{{\rm{y}}_{\rm{2}}}} \right\}{\rm{ \times \ldots \times }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{n}}}}}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{n}}}}}{{\rm{y}}_{\rm{n}}}} \right\}\\{\rm{ = }}{{\rm{\lambda }}^{\rm{n}}}\prod\limits_{{\rm{i = 1}}}^{\rm{n}} {\frac{{\rm{1}}}{{{{\rm{t}}_{\rm{i}}}}}} {\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{i}}}}}{{\rm{y}}_{\rm{i}}}} \right\}\\{\rm{ = }}{{\rm{\lambda }}^{\rm{n}}}\left( {\prod\limits_{{\rm{i = 1}}}^{\rm{n}} {\frac{{\rm{1}}}{{{{\rm{t}}_{\rm{i}}}}}} } \right){\rm{ \times exp}}\left\{ {{\rm{ - }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{j}}}}}} {{\rm{y}}_{\rm{j}}}} \right\}\\{\rm{ = }}{{\rm{\lambda }}^{\rm{n}}}{\rm{ \times p \times exp}}\left\{ {{\rm{ - \lambda }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} } \right\}\end{array}\)

Where,\({\rm{p = }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{\rm{\lambda }}}{{{{\rm{t}}_{\rm{j}}}}}} \).

Look at the log likelihood function to determine the maximum.

\(\begin{array}{c}{\rm{lnf}}\left( {{{\rm{y}}_{\rm{1}}}{\rm{,}}{{\rm{y}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{y}}_{\rm{n}}}{\rm{;\lambda }}} \right){\rm{ = ln}}\left( {{{\rm{\lambda }}^{\rm{n}}}{\rm{ \times p \times exp}}\left\{ {{\rm{ - \lambda }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} } \right\}} \right)\\{\rm{ = n \times ln\lambda + lnp - \lambda }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} \end{array}\)

The maximum likelihood estimator is generated by taking the derivative of the log likelihood function in regard to\({\rm{\lambda }}\)and equating it to\({\rm{0}}\).

As a result, the derivative,

\(\begin{array}{c}\frac{{\rm{d}}}{{{\rm{d\lambda }}}}{\rm{f}}\left( {{{\rm{y}}_{\rm{1}}}{\rm{,}}{{\rm{y}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{y}}_{\rm{n}}}{\rm{;\lambda }}} \right){\rm{ = }}\frac{{\rm{d}}}{{{\rm{d\lambda }}}}\left( {{\rm{n \times ln\lambda + lnp - \lambda }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} } \right)\\{\rm{ = }}\frac{{\rm{n}}}{{\rm{\lambda }}}{\rm{ + 0 - }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} \\{\rm{ = }}\frac{{\rm{n}}}{{\rm{\lambda }}}{\rm{ - }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} \end{array}\)

As a result, solving equation provides the maximum likelihood estimator.

\(\begin{array}{c}\frac{{\rm{n}}}{{{\rm{\hat \lambda }}}}{\rm{ - }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} {\rm{ = 0}}\\\frac{{\rm{n}}}{{{\rm{\hat \lambda }}}}{\rm{ = }}\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {\frac{{{{\rm{y}}_{\rm{j}}}}}{{{{\rm{t}}_{\rm{j}}}}}} \end{array}\)

for\({\rm{\hat \lambda }}\). Hence, the maximum likelihood estimator is,

\({\rm{\hat \lambda = }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{j = 1}}}^{\rm{n}} {{{\rm{Y}}_{\rm{j}}}} {\rm{/}}{{\rm{t}}_{\rm{j}}}}}\).

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Most popular questions from this chapter

The mean squared error of an estimator \({\rm{\hat \theta }}\) is \({\rm{MSE(\hat \theta ) = E(\hat \theta - \hat \theta }}{{\rm{)}}^{\rm{2}}}\). If \({\rm{\hat \theta }}\) is unbiased, then \({\rm{MSE(\hat \theta ) = V(\hat \theta )}}\), but in general \({\rm{MSE(\hat \theta ) = V(\hat \theta ) + (bias}}{{\rm{)}}^{\rm{2}}}\) . Consider the estimator \({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = K}}{{\rm{S}}^{\rm{2}}}\), where \({{\rm{S}}^{\rm{2}}}{\rm{ = }}\) sample variance. What value of K minimizes the mean squared error of this estimator when the population distribution is normal? (Hint: It can be shown that \({\rm{E}}\left( {{{\left( {{{\rm{S}}^{\rm{2}}}} \right)}^{\rm{2}}}} \right){\rm{ = (n + 1)}}{{\rm{\sigma }}^{\rm{4}}}{\rm{/(n - 1)}}\) In general, it is difficult to find \({\rm{\hat \theta }}\) to minimize \({\rm{MSE(\hat \theta )}}\), which is why we look only at unbiased estimators and minimize \({\rm{V(\hat \theta )}}\).)

Suppose a certain type of fertilizer has an expected yield per acre of \({{\rm{\mu }}_{\rm{2}}}\)with variance \({{\rm{\sigma }}^{\rm{2}}}\)whereas the expected yield for a second type of fertilizer is with the same variance \({{\rm{\sigma }}^{\rm{2}}}\).Let \({\rm{S}}_{\rm{1}}^{\rm{2}}\) and \({\rm{S}}_{\rm{2}}^{\rm{2}}\)denote the sample variances of yields based on sample sizes \({{\rm{n}}_{\rm{1}}}\)and \({{\rm{n}}_{\rm{2}}}\),respectively, of the two fertilizers. Show that the pooled (combined) estimator

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\left( {{{\rm{n}}_{\rm{1}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{1}}^{\rm{2}}{\rm{ + }}\left( {{{\rm{n}}_{\rm{2}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{2}}^{\rm{2}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ - 2}}}}\)

is an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\)

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 ?)

A vehicle with a particular defect in its emission control system is taken to a succession of randomly selected mechanics until\({\rm{r = 3}}\)of them have correctly diagnosed the problem. Suppose that this requires diagnoses by\({\rm{20}}\)different mechanics (so there were\({\rm{17}}\)incorrect diagnoses). Let\({\rm{p = P}}\)(correct diagnosis), so\({\rm{p}}\)is the proportion of all mechanics who would correctly diagnose the problem. What is the mle of\({\rm{p}}\)? Is it the same as the mle if a random sample of\({\rm{20}}\)mechanics results in\({\rm{3}}\)correct diagnoses? Explain. How does the mle compare to the estimate resulting from the use of the unbiased estimator?

When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?
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