Chapter 3.5, Problem 85E

Find E[Y(Y − 1)] for a geometric random variable Y by finding $d^2/d{q}^2\left({\displaystyle {\sum }_{y=1}^{\infty }{q}^y}\right)$. Use this result to find the variance of Y.

To determine

Find $E\left[Y\left(Y-1\right)\right]$.

Find the variance of Y.

Answer to Problem 85E

The value of $E\left[Y\left(Y-1\right)\right]$ is $\frac{2q}{p^2}$.

The variance of Y is $\frac{q}{p^2}$.

Explanation of Solution

Calculation:

The geometric random variable Y has the probability distribution,

$p\left(y\right)=q^{y-1}p;y=1,2,\mathrm{3...},0\le p\le 1$, Where $q=1-p$.

The expected value for Y is $E\left(Y\right)=\frac{1}{p}$

Consider, ${\displaystyle \sum _{y=1}^{\infty }{q}^y}$.

$\begin{array}{c}{\displaystyle \sum _{y=1}^{\infty }{q}^y}=q^1+q^2+\mathrm{....}+\infty \\ =\frac{q}{1-q}\left[\text{Since}q=1-p\right]\\ =\frac{1-p}{1-\left(1-p\right)}\\ =\frac{1-p}{p}\end{array}$

         $\begin{array}{c}=\frac{1}{p}-\frac{p}{p}\\ =\frac{1}{1-q}-1\left[\text{Since}p=1-q\right]\end{array}$

Take derivative with respect to q on both sides of the expression.

$\begin{array}{c}\frac{d}{dq}\left({\displaystyle \sum _{y=1}^{\infty }{q}^y}\right)=\frac{d}{dq}\left(\frac{1}{1-q}-1\right)\\ {\displaystyle \sum _{y=1}^{\infty }y{q}^{y-1}}=\frac{1}{{\left(1-q\right)}^2}-0\\ {\displaystyle \sum _{y=1}^{\infty }y{q}^{y-1}}=\frac{1}{{\left(1-q\right)}^2}\end{array}$

Again take derivative with respect to q on both sides of the expression.

$\begin{array}{c}\frac{d}{dq}\left({\displaystyle \sum _{y=1}^{\infty }y{q}^{y-1}}\right)=\frac{d}{dq}\left(\frac{1}{{\left(1-q\right)}^2}\right)\\ {\displaystyle \sum _{y=1}^{\infty }y\left(y-1\right)q^{y-2}}=\frac{2}{{\left(1-q\right)}^3}\\ {\displaystyle \sum _{y=0}^{\infty }y\left(y+1\right)q^{y-1}}=\frac{2}{p^3}\\ {\displaystyle \sum _{y=1}^{\infty }y\left(y+1\right)\frac{q^y}{q}}=\frac{2}{p^3}\left[\text{Since}\text{for}y=0\text{the}\text{product}\text{is}0\right]\\ {\displaystyle \sum _{y=1}^{\infty }y\left(y+1\right)q^y}=\frac{2q}{p^3}\end{array}$

The value of $E\left[Y\left(Y-1\right)\right]$ is,

$\begin{array}{c}E\left[Y\left(Y-1\right)\right]={\displaystyle \sum _{y=1}^{\infty }y\left(y-1\right)q^{y-1}p}\\ ={\displaystyle \sum _{y=0}^{\infty }y\left(y+1\right)q^{y}p}\\ ={\displaystyle \sum _{y=1}^{\infty }y\left(y+1\right)q^{y}p}\left[\text{Since}\text{for}y=0\text{the}\text{product}\text{is}0\right]\\ {\displaystyle \sum _{y=1}^{\infty }y\left(y+1\right)q^y}=\frac{2q}{p^3}\left(p\right)\left[{\displaystyle \sum _{y=1}^{\infty }y\left(y+1\right)q^y}=\frac{2q}{p^3}\right]\\ =\frac{2q}{p^2}\end{array}$

Hence, the value of $E\left[Y\left(Y-1\right)\right]$ is $\frac{2q}{p^2}$.

The value of variance of Y is,

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

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

Hence, the variance of Y is $\frac{q}{p^2}$.