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Chapter 6: Q24E (page 273)

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?

Short Answer

Expert verified

\({\rm{\hat p = }}\frac{{\rm{r}}}{{\rm{n}}}\)If a random sample of \({\rm{20}}\) mechanics results in \({\rm{3}}\) correct diagnoses: \({\rm{\hat p = }}\frac{{\rm{3}}}{{{\rm{20}}}}{\rm{ = 0}}{\rm{.15}}\).

The numerator and denominator are increased by \({\rm{1}}\)

Step by step solution

01

Introduction

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

02

Explanation

Given:

\({\rm{r = 3 n = 20}}\)

We are interested in the number of successes \({\rm{X}}\)within the \({\rm{20}}\) trials, then \({\rm{X}}\)needs to have a binomial distribution with \({\rm{n = 20}}\)and \({\rm{p}}\)(unknown).

Definition binomial probability:

\(\begin{array}{c}{\rm{f(r) = P(X = r)}}\\{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{ \times }}{{\rm{p}}^{\rm{r}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - r}}}}\\{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{r}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - r}}}}\end{array}\)

The value of \({\rm{p}}\)for which the probability distribution is maximized is the maximum likelihood estimator. The maximum of the probability distribution is also the maximum of the probability distribution's logarithm.

\(\begin{array}{l}{\rm{lnf(r) = ln}}\left( {\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{r}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - r}}}}} \right)\\{\rm{ = ln}}\left( {\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}} \right){\rm{ + ln}}{{\rm{p}}^{\rm{r}}}{\rm{ + ln(1 - p}}{{\rm{)}}^{{\rm{n - r}}}}\\{\rm{ = ln}}\left( {\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}} \right){\rm{ + rlnp + (n - r)ln(1 - p)}}\end{array}\)

Determine the derivative to the parameter\({\rm{p}}\):

\(\begin{array}{c}\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{lnf(r)}}\\{\rm{ = }}\frac{{\rm{d}}}{{{\rm{dp}}}}\left( {{\rm{ln}}\left( {\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}} \right){\rm{ + rlnp + (n - r)ln(1 - p)}}} \right)\\{\rm{ = 0 + }}\frac{{\rm{r}}}{{\rm{p}}}{\rm{ - }}\frac{{{\rm{n - r}}}}{{{\rm{1 - p}}}}{\rm{ = }}\frac{{\rm{r}}}{{\rm{p}}}{\rm{ - }}\frac{{{\rm{n - r}}}}{{{\rm{1 - p}}}}\end{array}\)

The maximum is the value of \({\rm{p}}\)for which the probability distribution is maximum and for which the derivative to \({\rm{p}}\)of the probability distribution is then 0 (\({\rm{r}}\)and \({\rm{n}}\)are independent of \({\rm{p}}\)):

\(\frac{{\rm{d}}}{{{\rm{dp}}}}{\rm{lnf(r) = }}\frac{{\rm{r}}}{{\rm{p}}}{\rm{ - }}\frac{{{\rm{n - r}}}}{{{\rm{1 - p}}}}{\rm{ = 0}}\)

03

Calculation

Add \(\frac{{{\rm{n - r}}}}{{{\rm{1 - p}}}}\)of each side:

\(\frac{{\rm{r}}}{{\rm{p}}}{\rm{ = }}\frac{{{\rm{n - r}}}}{{{\rm{1 - p}}}}\)

Multiply each side by\({\rm{p(1 - p)}}\):

\({\rm{r(1 - p) = (n - r)p}}\)

Use the distributive property:

\({\rm{r - rp = np - rp}}\)

Add \({\rm{rp}}\)to each side:

\({\rm{r = np}}\)Divide each side by\({\rm{n}}\):

\({\rm{p = }}\frac{{\rm{r}}}{{\rm{n}}}\)This expression for \({\rm{p}}\)is then the maximum likelihood estimate:

\({\rm{\hat p = }}\frac{{\rm{r}}}{{\rm{n}}}\)The maximum likelihood estimates of a random sample of \({\rm{20}}\) mechanics that results in \({\rm{3}}\) correct diagnoses is then the expression evaluated at \({\rm{r = 3}}\)and\({\rm{n = 20}}\):

\(\begin{array}{c}{\rm{\hat p = }}\frac{{\rm{r}}}{{\rm{n}}}\\{\rm{ = }}\frac{{\rm{3}}}{{{\rm{20}}}}\\{\rm{ = 0}}{\rm{.15}}\end{array}\)

The estimate of the unbiased estimator in the previous exercise is\({\rm{\hat p = }}\frac{{{\rm{r - 1}}}}{{{\rm{x + r - 1}}}}\), which we note is a different estimator than the one in this exercise \({\rm{\hat p = }}\frac{{\rm{r}}}{{\rm{n}}}\)(the difference is that both the numerator and denominator are increased by \({\rm{1}}\) ).

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Most popular questions from this chapter

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?

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: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

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.

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 \({\rm{cX}}\).)

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)
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