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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

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:

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

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

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) from the pdf

\({\rm{f(x;\theta ) = }}{\rm{.5(1 + \theta x)}}\quad {\rm{ - 1拢 x拢 1}}\)

where \({\rm{ - 1拢 \theta 拢 1}}\) (this distribution arises in particle physics). Show that \({\rm{\hat \theta = 3\bar X}}\) is an unbiased estimator of\({\rm{\theta }}\). (Hint: First determine\({\rm{\mu = E(X) = E(\bar X)}}\).)

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.

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