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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. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a uniform distribution on \({\rm{(0,\theta )}}\). Then the mle of \({\rm{\theta }}\) is \({\rm{\hat \theta = Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\). Use the fact that \({\rm{Y}} \le {\rm{y}}\) if each \({{\rm{X}}_{\rm{i}}} \le {\rm{y}}\) to derive the cdf of Y. Then show that the pdf of \({\rm{Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\) is \({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}}&{{\rm{0}} \le {\rm{y}} \le {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

b. Use the result of part (a) to show that the mle is biased but that \({\rm{(n + 1)}}\)\({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{/n}}\) is unbiased.

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 “A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains” (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\({\rm{X}}\)represent the error in making a measurement of a physical characteristic or property (e.g., the boiling point of a particular liquid). It is often reasonable to assume that\({\rm{E(X) = 0}}\)and that\({\rm{X}}\)has a normal distribution. Thus, the pdf of any particular measurement error is

\({\rm{f(x;\theta ) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \theta }}} }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\theta }}}}\quad {\rm{ - \yen < x < \yen}}\)

(Where we have used\({\rm{\theta }}\)in place of\({{\rm{\sigma }}^{\rm{2}}}\)). Now suppose that\({\rm{n}}\)independent measurements are made, resulting in measurement errors\({{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{.}}\)Obtain the mle of\({\rm{\theta }}\).

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 ?)

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)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 \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).
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