Chapter 3.6, Problem 98E

a

To determine

Show that if $y\ge r+1,\frac{p\left(y\right)}{p\left(y-1\right)}=\left(\frac{y-1}{y-r}\right)q$.

Explanation of Solution

Calculation:

The negative binomial distribution for a random variable Y is,

$p\left(y\right)=\left(\begin{array}{l}y-1\\ r-1\end{array}\right)p^r{q}^{y-r},y=r,r+1,r+2,\mathrm{...},0\le p\le 1$ where $q=1-p$

The negative binomial distribution for a random variable $Y-1$ is,

$\begin{array}{c}p\left(y-1\right)=\left(\begin{array}{l}\left(y-1\right)-1\\ r-1\end{array}\right)p^r{q}^{\left(y-1\right)-r}\\ =\left(\begin{array}{l}y-2\\ r-1\end{array}\right)p^r{q}^{y-r-1}\end{array}$ where $q=1-p$

Take the ratio of $p\left(y\right)$ and $p\left(y-1\right)$.

$\begin{array}{c}\frac{p\left(y\right)}{p\left(y-1\right)}=\frac{\left(\begin{array}{l}y-1\\ r-1\end{array}\right)p^r{q}^{y-r}}{\left(\begin{array}{l}y-2\\ r-1\end{array}\right)p^r{q}^{y-r-1}}\\ =\frac{\frac{\left(y-1\right)!}{\left(r-1\right)!\left(y-1-\left(r-1\right)\right)!}{p}^r{q}^{y-r}}{\frac{\left(y-2\right)!}{\left(r-1\right)!\left(y-2-\left(r-1\right)\right)!}{p}^r{q}^{y-r}{q}^{-1}}\\ =\frac{\left(\frac{\left(y-1\right)\left(y-2\right)!}{\left(y-r\right)!}\right)}{\left(\frac{\left(y-2\right)!}{\left(y-r-1\right)!}\right)q^{-1}}\\ =\frac{\left(\frac{\left(y-1\right)}{\left(y-r\right)\left(y-r-1\right)!}\right)}{\left(\frac{1}{\left(y-r-1\right)!}\right)\left(\frac{1}{q}\right)}\\ =\left(\frac{y-1}{y-r}\right)q\end{array}$

Hence, it sis showed that if $y\ge r+1,\frac{p\left(y\right)}{p\left(y-1\right)}=\left(\frac{y-1}{y-r}\right)q$.

b

To determine

Show that $\frac{p\left(y\right)}{p\left(y-1\right)}>1$ if $y<\frac{r-q}{1-q}$, similarly $\frac{p\left(y\right)}{p\left(y-1\right)}<1$ if $y>\frac{r-q}{1-q}$.

Explanation of Solution

Calculation:

Consider the inequality, $\frac{p\left(y\right)}{p\left(y-1\right)}>1$

$\begin{array}{c}\left(\frac{y-1}{y-r}\right)q>1\\ \left(y-1\right)q>\left(y-r\right)\\ yq-q>y-r\\ yq-y>q-r\end{array}$

$\begin{array}{c}y\left(q-1\right)>q-r\left[\text{multiply}\text{inequality}\text{with}-\text{1}\right]\\ -y\left(q-1\right)<-\left(q-r\right)\\ y\left(1-q\right)<r-q\\ y<\frac{r-q}{1-q}\end{array}$

Hence, it is showed that $\frac{p\left(y\right)}{p\left(y-1\right)}>1$ if $y<\frac{r-q}{1-q}$.

Consider the inequality, $\frac{p\left(y\right)}{p\left(y-1\right)}<1$

$\begin{array}{c}\left(\frac{y-1}{y-r}\right)q<1\\ \left(y-1\right)q<\left(y-r\right)\\ yq-q<y-r\\ yq-y<q-r\end{array}$

$\begin{array}{c}y\left(q-1\right)<q-r\left[\text{multiply}\text{inequality}\text{with}-\text{1}\right]\\ -y\left(q-1\right)>-\left(q-r\right)\\ y\left(1-q\right)>r-q\\ y>\frac{r-q}{1-q}\end{array}$

Hence, it is showed that $\frac{p\left(y\right)}{p\left(y-1\right)}<1$ if $y>\frac{r-q}{1-q}$.

c

To determine

Find the values of y for $r=7$ and $p=0.5$.

Answer to Problem 98E

The values of y is $y<13$.

Explanation of Solution

Calculation:

From part (b), $p\left(y\right)>p\left(y-1\right)$ if $y<\frac{r-q}{1-q}$.

The value of q is,

$\begin{array}{c}q=1-p\\ =1-0.5\\ =0.5\end{array}$

Substitute, $r=7$ and $q=0.5$ in the inequality

$\begin{array}{c}y<\frac{7-0.5}{1-0.5}\\ y<\frac{6.5}{0.5}\\ y<13\end{array}$

Hence, y takes values less than 13.