Chapter 3.5, Problem 89E

a

To determine

Derive the mean of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$.

Derive the variance of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$.

Answer to Problem 89E

The mean of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$ is $\frac{1}{p}-1$.

The variance of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$ is $\frac{1-p}{p^2}$.

Explanation of Solution

Calculation:

The mean for geometric random variable Y is, $E\left(Y\right)=\frac{1}{p}$ and variance is $V\left(Y\right)=\frac{1-p}{p^2}$. Also, The random variable $Y^{*}$ is defined as $Y^{*}=Y-1$.

The results from exercise 3.33 are,

• $E\left(aY+b\right)=a\mu +b$
• $V\left(aY+b\right)=a^2{\sigma }^2$

The mean of random variable $Y^{*}$ is,

$\begin{array}{c}E\left(Y^{*}\right)=E\left(Y-1\right)\\ =E\left(Y\right)-1\\ =\frac{1}{p}-1\end{array}$

Hence, the mean of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$ is $\frac{1}{p}-1$.

The variance of random variable $Y^{*}$ is,

$\begin{array}{c}V\left(Y^{*}\right)=V\left(Y-1\right)\\ =V\left(Y\right)\\ =\frac{1-p}{p^2}\end{array}$

Hence, the variance of random variable $Y^{*}$ using relationship $Y^{*}=Y-1$ is $\frac{1-p}{p^2}$.

b

To determine

Derive the mean of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$

Derive the variance of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$

Answer to Problem 89E

The mean of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$ is $\frac{1}{p}-1$.

The variance of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$ is $\frac{1-p}{p^2}$.

Explanation of Solution

Calculation:

The probability distribution from Exercise 3.88 is $P\left(Y^{*}=y\right)=q^{y}p$, where $q=1-p$.

The mean of random variable $Y^{*}$ is,

$\begin{array}{c}E\left(Y^{*}\right)={\displaystyle \sum _{y=1}^{\infty }y{q}^{y}p}\\ =p{\displaystyle \sum _{y=1}^{\infty }y{q}^y}{q}^{-1}q\\ =pq{\displaystyle \sum _{y=1}^{\infty }y{q}^{y-1}}\\ =pq\left(\frac{1}{{\left(1-q\right)}^2}\right)\left[\text{Since}{\displaystyle \sum _{x=1}^{\infty }x{r}^{x-1}}=\frac{1}{{\left(1-r\right)}^2}\right]\end{array}$

           $\begin{array}{c}=\frac{p\left(1-p\right)}{{\left(1-\left(1-p\right)\right)}^2}\\ =\frac{p\left(1-p\right)}{p^2}\\ =\frac{1-p}{p}\left(\text{or}\frac{q}{p}\right)\\ =\frac{1}{p}-1\end{array}$

Hence, the mean of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$ is $\frac{1}{p}-1$.

The variance of random variable $Y^{*}$ is,

$\begin{array}{c}V\left(Y^{*}\right)=E\left[{\left(Y^{*}-\frac{q}{p}\right)}^2\right]\\ ={\displaystyle \sum _{y=1}^{\infty }{\left(y-\frac{q}{p}\right)}^2{q}^{y}p}\\ ={\displaystyle \sum _{y=1}^{\infty }\left(y^2+\frac{q^2}{p^2}-2\frac{yq}{p}\right)q^{y}p}\\ ={\displaystyle \sum _{y=1}^{\infty }{y}^2{q}^{y}p}+{\displaystyle \sum _{y=1}^{\infty }\frac{q^2}{p^2}{q}^{y}p}-{\displaystyle \sum _{y=1}^{\infty }2\left(\frac{yq}{p}\right)q^{y}p}\end{array}$

            $\begin{array}{c}=p{\displaystyle \sum _{y=1}^{\infty }{y}^2{q}^y}+\frac{q^2}{p}{\displaystyle \sum _{y=1}^{\infty }{q}^y}-2q{\displaystyle \sum _{y=1}^{\infty }y{q}^y}\\ =p{\displaystyle \sum _{y=1}^{\infty }{y}^2{q}^y}+\frac{q^2}{p}{\displaystyle \sum _{y=1}^{\infty }{q}^y}-2q{\displaystyle \sum _{y=1}^{\infty }y{q}^y}\left(q^{-1}q\right)\\ =p{\displaystyle \sum _{y=1}^{\infty }{y}^2{q}^y}+\frac{q^2}{p}{\displaystyle \sum _{y=1}^{\infty }{q}^y}-2q{\displaystyle \sum _{y=1}^{\infty }y{q}^{y-1}}\left(q\right)\\ =p\left(-\frac{q\left(q+1\right)}{{\left(q-1\right)}^3}\right)+\frac{q^2}{p}\left(\frac{1}{1-q}\right)-2q\left(\frac{q}{{\left(1-q\right)}^2}\right)\\ =p\left(\frac{-q\left(1-p+1\right)}{{\left(1-p-1\right)}^3}\right)+\frac{q^2}{p\left(p\right)}-\frac{2q^2}{p^2}\\ =p\left(\frac{-q\left(2-p\right)}{-p^3}\right)+\frac{q^2}{p^2}-\frac{2q^2}{p^2}\end{array}$

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

            $\begin{array}{c}=\frac{q+q^2-q^2}{p^2}\\ =\frac{q}{p^2}\\ =\frac{1-p}{p^2}\end{array}$

Hence, the variance of random variable $Y^{*}$ by using the probability distribution $P\left(Y^{*}=y\right)=q^{y}p$ is $\frac{1-p}{p^2}$.