Chapter 6, Problem 104SE

a.

To determine

Find $P\left(Y_1=Y_2\right)=P\left(Y_1-Y_2=0\right)$.

Answer to Problem 104SE

The value of $P\left(Y_1=Y_2\right)=P\left(Y_1-Y_2=0\right)$ is $\frac{p}{2-p}$.

Explanation of Solution

Calculation:

From the given information, Y₁ and Y₂ are independent geometric random variables.

The probability mass function for geometric distribution is $p\left(y\right)=p{\left(1-p\right)}^{y-1}y=1,2,\mathrm{...}$.

Given event $\left\{Y_1=Y_2\right\}$ occurs if $\left\{\left(Y_1=1,Y_2=1\right),\left(Y_1=2,Y_2=2\right),\left(Y_1=3,Y_2=3\right),\mathrm{...}\right\}$.

Therefore, the probability of the event $\left\{Y_1=Y_2\right\}$ is given by,

$\begin{array}{c}P\left(Y_1=Y_2\right)={\left[p\left(1\right)\right]}^2+{\left[p\left(2\right)\right]}^2+{\left[p\left(3\right)\right]}^2+\mathrm{...}\\ =p^2+p^2{\left(1-p\right)}^2+p^2{\left(1-p\right)}^4+\mathrm{...}\\ =p^2\left[1+{\left(1-p\right)}^2+{\left(1-p\right)}^4+\mathrm{...}\right]\\ =p^2{\displaystyle \sum _{i=0}^{\infty }{\left(1-p\right)}^{2i}}\end{array}$

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

b.

To determine

Find $P\left(Y_1-Y_2=1\right)$.

Answer to Problem 104SE

The value of $P\left(Y_1-Y_2=1\right)$ is $\frac{p\left(1-p\right)}{2-p}$.

Explanation of Solution

Calculation:

Given event $\left\{Y_1-Y_2=1\right\}=\left\{Y_1=1+Y_2\right\}$ occurs if

$\left\{\left(Y_1=2,Y_2=1\right),\left(Y_1=3,Y_2=2\right),\left(Y_1=4,Y_2=3\right),\mathrm{...}\right\}$.

Therefore, the probability of the event $\left\{Y_1-Y_2=1\right\}$ is given by,

$\begin{array}{c}P\left(Y_1-Y_2=1\right)=\left[p\left(1\right)p\left(2\right)\right]+\left[p\left(3\right)p\left(2\right)\right]+\left[p\left(4\right)p\left(3\right)\right]+\mathrm{...}\\ =p^2\left(1-p\right)+p^2{\left(1-p\right)}^3+p^2{\left(1-p\right)}^5+\mathrm{...}\\ =p^2\left(1-p\right)\left[1+{\left(1-p\right)}^2+{\left(1-p\right)}^4+\mathrm{...}\right]\\ =p^2\left(1-p\right){\displaystyle \sum _{i=0}^{\infty }{\left(1-p\right)}^{2i}}\end{array}$

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

c.

To determine

Find the discrete probability function for $U=Y_1-Y_2$.

Answer to Problem 104SE

The discrete probability function for $U=Y_1-Y_2$ is,

$P\left(U=u\right)=\frac{p{\left(1-p\right)}^{\left|u\right|}}{2-p},u=0,\pm 1,\pm 2,\mathrm{...}$.

Explanation of Solution

Calculation:

Assume that u>0:

Consider,

$\begin{array}{c}P\left(U=u\right)=P\left(Y_1-Y_2=u\right)\\ ={\displaystyle \sum _{y_2=1}^{\infty }P\left(Y_1=u+y_2\right)P\left(Y_2=y_2\right)}\\ ={\displaystyle \sum _{y_2=1}^{\infty }p{\left(1-p\right)}^{u+y_2-1}p{\left(1-p\right)}^{y_2-1}}\\ =p^2{\left(1-p\right)}^u{\displaystyle \sum _{y_2=1}^{\infty }{\left(1-p\right)}^{2\left(y_2-1\right)}}\end{array}$

                   $\begin{array}{c}=p^2{\left(1-p\right)}^u{\displaystyle \sum _{x\left(y_2-1\right)=0}^{\infty }{\left(1-p\right)}^{2x}}\\ =\frac{p{\left(1-p\right)}^u}{2-p}\end{array}$        (1)

If u<0,

$\begin{array}{c}P\left(U=u\right)=P\left(Y_1-Y_2=u\right)\\ ={\displaystyle \sum _{y_1=1}^{\infty }P\left(Y_2=y_1-u\right)P\left(Y_1=y_1\right)}\\ ={\displaystyle \sum _{y_1=1}^{\infty }p{\left(1-p\right)}^{y_1-u-1}p{\left(1-p\right)}^{y_1-1}}\\ =p^2{\left(1-p\right)}^{-u}{\displaystyle \sum _{y_1=1}^{\infty }{\left(1-p\right)}^{2\left(y_1-1\right)}}\end{array}$

                $\begin{array}{c}=p^2{\left(1-p\right)}^{-u}{\displaystyle \sum _{x\left(y_1-1\right)=0}^{\infty }{\left(1-p\right)}^{2x}}\\ =\frac{p{\left(1-p\right)}^{-u}}{2-p}\end{array}$        (2)

Combine the equations (1) and (2),

$P\left(U=u\right)=\frac{p{\left(1-p\right)}^{\left|u\right|}}{2-p},u=0,\pm 1,\pm 2,\mathrm{...}$