Question

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

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

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.

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 \({\rm{1}}{\rm{.}}\)

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\({\rm{\hat p = 0}}{\rm{.4444}}\).