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Chapter 6: Q17E (page 263)

We defined a negative binomial\({\rm{rv}}\)as the number of failures that occur before the\({\rm{rth}}\)success in a sequence of independent and identical success/failure trials. The probability mass function (\({\rm{pmf}}\)) of\({\rm{X}}\)is\({\rm{nb(x,r,p) = }}\)\(\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\quad {\rm{x = 0,1,2, \ldots }}\)

a. Suppose that. Show that\({\rm{\hat p = (r - 1)/(X + r - 1)}}\)is an unbiased estimator for\({\rm{p}}\). (Hint: Write out\({\rm{E(\hat p)}}\)and cancel\({\rm{x + r - 1}}\)inside the sum.)

b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether\({\rm{(S)}}\)or not\({\rm{(F)}}\)they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate\({\rm{p = }}\)the true proportion who support the candidate.

Short Answer

Expert verified

a) Total probability of a valid probability distribution is equal to \({\rm{1}}{\rm{.}}\)\({\rm{E(\hat p) = p}}\)

b) The estimate is \({\rm{\hat p = 0}}{\rm{.4444}}\)the true proportion.

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

(a)

Given: The pmf of \({\rm{X}}\)is:

\({\rm{nb(x;r,p) = }}\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 1}}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{r}}}{{\rm{(1 - p)}}^{\rm{x}}}\)

The expected value of \({\rm{\hat p}}\)is the sum of the product of each possible value of \(\frac{{{\rm{r - 1}}}}{{{\rm{X + r - 1}}}}\) with its probability\({\rm{nb(x;r,p)}}\):

\(\begin{aligned}E(\hat p) &= \sum\limits_{x = 0}^{ + \infty } {\frac{{r - 1}}{{x + r - 1}}} nb(x;r,p)\\ &= \sum\limits_{x = 0}^{ + \infty } {\frac{{r - 1}}{{x + r - 1}}} \left( {\begin{array}{*{20}{c}}{x + r - 1}\\x\end{array}} \right){p^r}{(1 - p)^x}\\ &= \sum\limits_{x = 0}^{ + \infty } {\frac{{r - 1}}{{x + r - 1}}} \frac{{(x + r - 1)!}}{{x!(x + r - 1 - x)!}}{p^r}{(1 - p)^x}\\ &= \sum\limits_{x = 0}^{ + \infty } {(r - 1)} \frac{{(x + r - 2)!}}{{x!(r - 1)!}}{p^r}{(1 - p)^x}\\ &= \sum\limits_{x = 0}^{ + \infty } {\frac{{(x + r - 2)!}}{{x!(r - 2)!}}} {p^r}{(1 - p)^x}\\ &= \sum\limits_{x = 0}^{ + \infty } {\frac{{(x + r - 2)!}}{{x!((x + r - 2) - x)!}}} {p^r}{(1 - p)^x}\\ &= \sum\limits_{x = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{x + r - 2}\\x\end{array}} \right)} {p^r}{(1 - p)^x}\\ &= p\sum\limits_{x = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{x + r - 2}\\x\end{array}} \right)} \\ &= p{(1 - p)^x}\end{aligned}\)

Note:

\(\sum\limits_{x = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{x + r - 2}\\x\end{array}} \right)} {p^{r - 1}}{(1 - p)^x}\)\(\sum\limits_{{\rm{x = 0}}}^{{\rm{ + \yen}}} {\left( {\begin{array}{*{20}{c}}{{\rm{x + r - 2}}}\\{\rm{x}}\end{array}} \right)} {{\rm{p}}^{{\rm{r - 1}}}}{{\rm{(1 - p)}}^{\rm{x}}}\)which is the negative binomial distribution with the \({\rm{(r - 1)}}\)th success (instead of \({\rm{r}}\)th success) and thus total probability of a valid probability distribution is equal to \({\rm{1}}{\rm{.}}\)

03

Explanation

(b)

Considering the given information:

\({\rm{r = 5}}\)

SFFSFFFSSS

SFFSFFFSSS contains \({\rm{5}}\) successes and \({\rm{5}}\) failures. \({\rm{x}}\)is the number of failures.

\({\rm{x = 5}}\)

Determine the estimator of part (a):

\(\begin{aligned}\hat p &= \frac{{{\rm{r - 1}}}}{{{\rm{x + r - 1}}}}\\ &= \frac{{{\rm{5 - 1}}}}{{{\rm{5 + 5 - 1}}}}\\ &= \frac{{\rm{4}}}{{\rm{9}}}\\{\rm{\gg 0}}{\rm{.4444}}\end{aligned}\)

Therefore, the solution is\({\rm{\hat p = 0}}{\rm{.4444}}\).

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Most popular questions from this chapter

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

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 \({{\rm{\sigma }}^{\rm{2}}}\)

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?

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\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)
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