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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 \({\rm{V(\hat \theta )}}\).)

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data:

Assume that X has a Poisson distribution with parameter \({\bf{\mu }}.\)and that X represents the number of scratches on a randomly picked item.

a. Calculate the estimate for the data using an unbiased \({\bf{\mu }}.\)estimator. (Hint: for X Poisson, \({\rm{E(X) = \mu }}\) ,therefore \({\rm{E(\bar X) = ?)}}\)

c. What is your estimator's standard deviation (standard error)? Calculate the standard error estimate. (Hint: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)

A diagnostic test for a certain disease is applied to\({\rm{n}}\)individuals known to not have the disease. Let\({\rm{X = }}\)the number among the\({\rm{n}}\)test results that are positive (indicating presence of the disease, so\({\rm{X}}\)is the number of false positives) and\({\rm{p = }}\)the probability that a disease-free individual's test result is positive (i.e.,\({\rm{p}}\)is the true proportion of test results from disease-free individuals that are positive). Assume that only\({\rm{X}}\)is available rather than the actual sequence of test results.

a. Derive the maximum likelihood estimator of\({\rm{p}}\). If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the estimate?

b. Is the estimator of part (a) unbiased?

c. If\({\rm{n = 20}}\)and\({\rm{x = 3}}\), what is the mle of the probability\({{\rm{(1 - p)}}^{\rm{5}}}\)that none of the next five tests done on disease-free individuals are positive?

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 m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t 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?

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