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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

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a gamma distribution with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \({\rm{\alpha }}\) and \({\rm{\beta }}\). Do you think they can be solved explicitly? b. Show that the mle of \({\rm{\mu = \alpha \beta }}\) is \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).

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 \({\rm{n = 20}}\)?

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

Prior to obtaining data, denote the beam strengths by X1, 鈥� ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi 鈥檚 constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi 鈥檚 form a random sample (independent of the Xi 鈥檚) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). Calculate the estimate for the given data.

b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error.

c. Calculate a point estimate of the ratio \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)of the two standard deviations.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \({\rm{X - Y}}\) between beam strength and cylinder strength.
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