Chapter 5, Problem 101SE

a.

To determine

Show that the following function satisfies the properties of a joint probability mass function.

Compute $P\left(X<0.5,Y<1.5\right)$.

Answer to Problem 101SE

$P\left(X<0.5,Y<1.5\right)=\frac{3}{8}$ .

Explanation of Solution

Calculation:

It is given that,

x	y	$f_{\text{XY}}\left(x,y\right)$
0	0	$\frac{1}{4}$
0	1	$\frac{1}{8}$
1	0	$\frac{1}{8}$
1	1	$\frac{1}{4}$
2	2	$\frac{1}{4}$

The joint probability mass function of the discrete random variable X and Y is defined as $f_{\text{XY}}\left(x,y\right)$.

The properties of joint probability mass function is,

• $f_{\text{XY}}\left(x,y\right)\ge 0$
• ${\displaystyle \sum _x{\displaystyle \sum _y{f}_{\text{XY}}\left(x,y\right)}}=1$
• $f_{\text{XY}}\left(x,y\right)=P\left(X=x,Y=y\right)$

According to given information it is clear that, $f_{\text{XY}}\left(x,y\right)>0$.

Let R denote the range of (X, Y).

Thus,

$\begin{array}{c}{\displaystyle \sum _x{\displaystyle \sum _y{f}_{\text{XY}}\left(x,y\right)}}={\displaystyle \sum _R{f}_{\text{XY}}\left(x,y\right)}\\ =\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{4}+\frac{1}{4}\\ =1\end{array}$

Hence, the function satisfies all the properties of joint probability mass function.

The discrete random variable X takes only some discrete values and the range of X is $\left\{0,1,2\right\}$.

The discrete random variable Y takes only some discrete values and the range of Y is $\left\{0,1,2\right\}$.

Thus,

$P\left(X<0.5,Y<1.5\right)=f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(0,0\right)$.

Using the given information,

$\begin{array}{c}P\left(X<0.5,Y<1.5\right)=f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(0,0\right)\\ =\frac{1}{8}+\frac{1}{4}\\ =\frac{3}{8}\end{array}$

Thus, $P\left(X<0.5,Y<1.5\right)=\frac{3}{8}$.

b.

To determine

Compute $P\left(X\le 1\right)$.

Answer to Problem 101SE

$P\left(X\le 1\right)=\frac{3}{4}$.

Explanation of Solution

Calculation:

The discrete random variable X takes only some discrete values and the range of X is $\left\{0,1,2\right\}$.

The discrete random variable Y takes only some discrete values and the range of Y is $\left\{0,1,2\right\}$.

Now, $\left(X\le 1\right)$ implies that X takes value 0, and 1 whereas Y take values 0, 1 and 2.

Thus,

$P\left(X\le 1\right)=f_{\text{XY}}\left(0,0\right)+f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)$

Using the given information,

$\begin{array}{c}P\left(X\le 1\right)=f_{\text{XY}}\left(0,0\right)+f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)\\ =\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{4}\\ =\frac{6}{8}\\ =\frac{3}{4}\end{array}$

Thus, $P\left(X\le 1\right)=\frac{3}{4}$.

c.

To determine

Compute $P\left(X<1.5\right)$.

Answer to Problem 101SE

$P\left(X<1.5\right)=\frac{3}{4}$.

Explanation of Solution

Calculation:

The discrete random variable X takes only some discrete values and the range of X is $\left\{0,1,2\right\}$.

The discrete random variable Y takes only some discrete values and the range of Y is $\left\{0,1,2\right\}$.

Now, $X<1.5$ implies that X takes value 0, and 1 whereas Y take values 0, 1 and 2.

Thus,

$P\left(X<1.5\right)=f_{\text{XY}}\left(0,0\right)+f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)$

Using the given information,

$\begin{array}{c}P\left(X<1.5\right)=f_{\text{XY}}\left(0,0\right)+f_{\text{XY}}\left(0,1\right)+f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)\\ =\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{4}\\ =\frac{6}{8}\\ =\frac{3}{4}\end{array}$

Thus, $P\left(X<1.5\right)=\frac{3}{4}$.

d.

To determine

Compute $P\left(X>0.5,Y<1.5\right)$.

Answer to Problem 101SE

$P\left(X>0.5,Y<1.5\right)=\frac{3}{8}$.

Explanation of Solution

Calculation:

The discrete random variable X takes only some discrete values and the range of X is $\left\{0,1,2\right\}$.

The discrete random variable Y takes only some discrete values and the range of Y is $\left\{0,1,2\right\}$.

Now, $\left(X>0.5,Y<1.5\right)$ implies that X takes value 1 and 2, whereas Y take values 0 and 1.

Thus,

$P\left(X>0.5,Y<1.5\right)=f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)$

Using the given information,

$\begin{array}{c}P\left(X>0.5,Y<1.5\right)=f_{\text{XY}}\left(1,0\right)+f_{\text{XY}}\left(1,1\right)\\ =\frac{1}{8}+\frac{1}{4}\\ =\frac{3}{8}\end{array}$

Thus, $P\left(X>0.5,Y<1.5\right)=\frac{3}{8}$.

e.

To determine

Compute $E\left(X\right)$, $E\left(Y\right)$, $V\left(X\right)$ and $V\left(Y\right)$.

Answer to Problem 101SE

$E\left(X\right)=0.875$, $E\left(Y\right)=0.875$, $V\left(X\right)=0.61$ and $V\left(Y\right)=0.61$.

Explanation of Solution

Calculation:

Mean:

Mean of a discrete random variable X is defined as,

$E\left(X\right)={\displaystyle \sum _{x}xf\left(x\right)}$, where $f\left(x\right)$ be the probability mass function of the random variable X.

Variance:

Variance of a discrete random variable X is defined as,

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-E^2\left(X\right)\\ ={\displaystyle \sum _x{x}^{2}f\left(x\right)}-{\left({\displaystyle \sum _{x}xf\left(x\right)}\right)}^2\end{array}$,

where $f\left(x\right)$ be the probability mass function of the random variable X.

Thus, mean of the random variable X is,

$\begin{array}{c}E\left(X\right)={\displaystyle \sum _{x}xf\left(x\right)}\\ =\left(0\right)\left(\frac{1}{4}\right)+\left(0\right)\left(\frac{1}{8}\right)+\left(1\right)\left(\frac{1}{8}\right)+\left(1\right)\left(\frac{1}{4}\right)+\left(2\right)\left(\frac{1}{4}\right)\\ =0+0+0.125+0.25+0.5\\ =0.875\end{array}$

The variance of the random variable X is,

$\begin{array}{c}V\left(X\right)={\displaystyle \sum _x{x}^{2}f\left(x\right)}-{\left({\displaystyle \sum _{x}xf\left(x\right)}\right)}^2\\ ={\left(0\right)}^2\left(\frac{1}{4}\right)+{\left(0\right)}^2\left(\frac{1}{8}\right)+{\left(1\right)}^2\left(\frac{1}{8}\right)+{\left(1\right)}^2\left(\frac{1}{4}\right)+{\left(2\right)}^2\left(\frac{1}{4}\right)-{\left(0.875\right)}^2\\ =\left(0+0+0.125+0.25+1\right)-0.765\\ =1.375-0.765\\ =0.61\end{array}$

Thus, mean of the random variable Y is,

$\begin{array}{c}E\left(Y\right)={\displaystyle \sum _{y}yf\left(y\right)}\\ =\left(0\right)\left(\frac{1}{4}\right)+\left(1\right)\left(\frac{1}{8}\right)+\left(0\right)\left(\frac{1}{8}\right)+\left(1\right)\left(\frac{1}{4}\right)+\left(2\right)\left(\frac{1}{4}\right)\\ =0+0.125+0+0.25+0.5\\ =0.875\end{array}$

The variance of the random variable Y is,

$\begin{array}{c}V\left(Y\right)={\displaystyle \sum _y{y}^{2}f\left(y\right)}-{\left({\displaystyle \sum _{y}yf\left(y\right)}\right)}^2\\ ={\left(0\right)}^2\left(\frac{1}{4}\right)+{\left(1\right)}^2\left(\frac{1}{8}\right)+{\left(0\right)}^2\left(\frac{1}{8}\right)+{\left(1\right)}^2\left(\frac{1}{4}\right)+{\left(2\right)}^2\left(\frac{1}{4}\right)-{\left(0.875\right)}^2\\ =\left(0+0.125+0+0.25+1\right)-0.765\\ =1.375-0.765\\ =0.61\end{array}$

Thus, $E\left(X\right)=0.875$, $E\left(Y\right)=0.875$, $V\left(X\right)=0.61$ and $V\left(Y\right)=0.61$.

f.

To determine

Find marginal probability distribution of X.

Answer to Problem 101SE

The marginal probability distribution of X is,

x	$f\left(x\right)$
0	$\frac{3}{8}$
1	$\frac{3}{8}$
2	$\frac{1}{4}$

Explanation of Solution

Calculation:

Marginal probability mass function:

If the joint probability mass function for random variable X and Y is $f_{XY}\left(x,y\right)$, then the marginal probability mass function of X and Y are,

$f_X\left(x\right)={\displaystyle \sum _y{f}_{XY}\left(x,y\right)}$ and $f_Y\left(y\right)={\displaystyle \sum _x{f}_{XY}\left(x,y\right)}$, respectively.

Thus, for $x=0$ the marginal probability distribution of X is obtained as,

$\begin{array}{c}{f}_X\left(0\right)={\displaystyle \sum _{y=0}^1{f}_{XY}\left(0,y\right)}\\ =f_{XY}\left(0,0\right)+f_{XY}\left(0,1\right)\\ =\frac{1}{4}+\frac{1}{8}\\ =\frac{3}{8}\end{array}$

Thus, for $x=1$ the marginal probability distribution of X is obtained as,

$\begin{array}{c}{f}_X\left(1\right)={\displaystyle \sum _{y=0}^1{f}_{XY}\left(1,y\right)}\\ =f_{XY}\left(1,0\right)+f_{XY}\left(1,1\right)\\ =\frac{1}{8}+\frac{1}{4}\\ =\frac{3}{8}\end{array}$

Thus, for $x=2$ the marginal probability distribution of X is obtained as,

$\begin{array}{c}{f}_X\left(2\right)=f_{XY}\left(2,y\right)\\ =f_{XY}\left(2,2\right)\\ =\frac{1}{4}\end{array}$

Therefore, the marginal probability distribution of X is,

x	$f\left(x\right)$
0	$\frac{3}{8}$
1	$\frac{3}{8}$
2	$\frac{1}{4}$

g.

To determine

Find the conditional probability distribution of Y given that $X=1$.

Answer to Problem 101SE

The conditional probability distribution of Y given that $X=1.5$ is,

y	$f_{Y|1}\left(y\right)$
0	$\frac{1}{3}$
1	$\frac{2}{3}$

Explanation of Solution

Calculation:

Conditional probability mass function:

If the joint probability mass function for random variable X and Y is $f_{XY}\left(x,y\right)$, then the conditional probability mass function of Y given $X=x$ is,

$f_{Y|x}\left(y\right)=\frac{f_{XY}\left(x,y\right)}{f_X\left(x\right)}$ .

Similarly, the conditional probability mass function of X given $Y=y$ is,

$f_{X|y}\left(x\right)=\frac{f_{XY}\left(x,y\right)}{f_Y\left(y\right)}$

It is given that, for $\left(X=1,Y=0\right)$ the joint probability mass function is $f_{\text{XY}}\left(1,0\right)=\frac{1}{8}$.

It is also given that, for $\left(X=1,Y=1\right)$ the joint probability mass function is $f_{\text{XY}}\left(1,1\right)=\frac{1}{4}$.

From part (f) it is found that marginal probability distribution of X is,

x	$f\left(x\right)$
0	$\frac{3}{8}$
1	$\frac{3}{8}$
2	$\frac{1}{4}$

Thus, for $y=0$ the conditional probability distribution of Y given that $X=1$ is,

$\begin{array}{c}{f}_{Y|1}\left(0\right)=\frac{f_{XY}\left(1,0\right)}{f_X\left(1\right)}\\ =\frac{\frac{1}{8}}{\frac{3}{8}}\\ =\frac{1}{3}\end{array}$

Thus, for $y=1$ the conditional probability distribution of Y given that $X=1$ is,

$\begin{array}{c}{f}_{Y|1}\left(1\right)=\frac{f_{XY}\left(1,1\right)}{f_X\left(1\right)}\\ =\frac{\frac{1}{4}}{\frac{3}{8}}\\ =\frac{2}{3}\end{array}$

Therefore, the conditional probability distribution of Y given that $X=1$ is,

y	$f_{Y|1}\left(y\right)$
0	$\frac{1}{3}$
1	$\frac{2}{3}$

h.

To determine

Find $E\left(Y|X=1\right)$.

Answer to Problem 101SE

$E\left(Y|X=1\right)=\frac{2}{3}$.

Explanation of Solution

Calculation:

If the joint probability mass function for random variable X and Y is $f_{XY}\left(x,y\right)$, then the conditional mean of Y given $X=x$ is,

$E\left(Y|x\right)={\displaystyle \sum _{y}y{f}_{Y|x}\left(y\right)}$ , where $f_{Y|x}\left(y\right)$ be the conditional distribution of Y given x.

From part (g) it is found that conditional probability distribution of Y given that $X=1.5$ is,

y	$f_{Y|1}\left(y\right)$
0	$\frac{1}{3}$
1	$\frac{2}{3}$

Thus,

$\begin{array}{c}E\left(Y|X=1\right)=\left(0\right)\left(\frac{1}{3}\right)+\left(1\right)\left(\frac{2}{3}\right)\\ =0+\frac{2}{3}\\ =\frac{2}{3}\end{array}$

Thus, $E\left(Y|X=1\right)=\frac{2}{3}$.

i.

To determine

Explain whether X and Y are independent or not.

Answer to Problem 101SE

The random variable X and Y are not independent.

Explanation of Solution

Calculation:

It is known that the random variables X and Y are said to be independent if $f_{\text{XY}}\left(x,y\right)=f_{\text{X}}\left(x\right)f_Y\left(y\right)$.

From part (f) it is already found that for $x=1$ the marginal probability distribution of X is obtained as,$f_X\left(1\right)=\frac{3}{8}$.

Now, for $y=0$ the marginal probability distribution of Y is obtained as,

$\begin{array}{c}{f}_Y\left(0\right)={\displaystyle \sum _{x=0}^1{f}_{XY}\left(x,0\right)}\\ =f_{XY}\left(0,0\right)+f_{XY}\left(1,0\right)\\ =\frac{1}{4}+\frac{1}{8}\\ =\frac{3}{8}\end{array}$

Thus, for $y=1$ the marginal probability distribution of Y is obtained as,

$\begin{array}{c}{f}_Y\left(1\right)={\displaystyle \sum _{x=0}^1{f}_{XY}\left(x,1\right)}\\ =f_{XY}\left(0,1\right)+f_{XY}\left(1,1\right)\\ =\frac{1}{8}+\frac{1}{4}\\ =\frac{3}{8}\end{array}$

Thus, for $y=2$ the marginal probability distribution of Y is obtained as,

$\begin{array}{c}{f}_Y\left(2\right)=f_{XY}\left(x,2\right)\\ =f_{XY}\left(2,2\right)\\ =\frac{1}{4}\end{array}$

Therefore, the marginal probability distribution of Y is,

y	$f\left(y\right)$
0	$\frac{3}{8}$
1	$\frac{3}{8}$
2	$\frac{1}{4}$

It is given that, for $\left(X=1,Y=1\right)$ the joint probability mass function is $f_{\text{XY}}\left(1,1\right)=\frac{1}{4}$.

Thus,

$\begin{array}{c}{f}_X\left(1\right)f_Y\left(1\right)=\left(\frac{3}{8}\right)\left(\frac{3}{8}\right)\\ =\frac{9}{64}\end{array}$

Hence, it is clear that, $f_{\text{XY}}\left(1,1\right)\ne f_X\left(1\right)f_Y\left(1\right)$.

Thus, X and Y are not independent.

j.

To determine

Find the correlation between X and Y.

Answer to Problem 101SE

The correlation for the joint probability distribution is 0.794, respectively.

Explanation of Solution

Calculation:

Thus, $E\left(X\right)$, $E\left(Y\right)$ and $E\left(XY\right)$ is obtained as,

x	y	$f_{XY}\left(x,y\right)$	$xf\left(x,y\right)$	$yf\left(x,y\right)$	$x^{2}f\left(x,y\right)$	$y^{2}f\left(x,y\right)$	$xyf\left(x,y\right)$
0	0	$\frac{1}{4}$	0	0	0	0	0
0	1	$\frac{1}{8}$	0	0.125	0	0.125	0
1	0	$\frac{1}{8}$	0.125	0	0.125	0	0
1	1	$\frac{1}{4}$	0.25	0.25	0.25	0.25	0.25
2	2	$\frac{1}{4}$	0.5	0.5	1	1	1
Total			0.875	0.875	1.375	1.375	1.25

Thus $E\left(XY\right)=1.25$

From part (e), it is found that $E\left(X\right)=0.875$, $E\left(Y\right)=0.875$, $V\left(X\right)=0.61$ and $V\left(Y\right)=0.61$.

The covariance between two random variables X and Y id defined as,

$\begin{array}{c}\mathrm{cov}\left(X,Y\right)={\sigma }_{XY}\\ =E\left[\left(X-{\mu }_X\right)\left(Y-{\mu }_Y\right)\right]\\ =E\left(XY\right)-{\mu }_X{\mu }_Y\end{array}$

Thus, the covariance between X and Y is obtained as,

$\begin{array}{c}\mathrm{cov}\left(X,Y\right)=E\left(XY\right)-{\mu }_X{\mu }_Y\\ =1.25-\left(0.875\right)\left(0.875\right)\\ =1.25-0.765\\ =0.484375\end{array}$

The correlation between two random variables X and Y is denoted as,

$\begin{array}{c}{\rho }_{\text{XY}}=\frac{\mathrm{cov}\left(X,Y\right)}{\sqrt{V\left(X\right)V\left(Y\right)}}\\ =\frac{{\sigma }_{XY}}{{\sigma }_X{\sigma }_Y}\end{array}$

Thus, the correlation between X and Y is obtained as,

$\begin{array}{c}{\rho }_{\text{XY}}=\frac{{\sigma }_{XY}}{{\sigma }_X{\sigma }_Y}\\ =\frac{0.484375}{\sqrt{\left(0.61\right)\left(0.61\right)}}\\ =\frac{0.484375}{0.61}\\ =\underset{\_}{0.794}.\end{array}$