Chapter 2.6, Problem 19E

a.

To determine

Find $Cov\left(X,Y\right)$.

Answer to Problem 19E

The value of $Cov\left(X,Y\right)$ is $-\frac{1}{64}$.

Explanation of Solution

Given info:

The joint probability density function is,

$f\left(x\right)=\{\begin{array}{l}\frac{3}{2}\left(x^2+y^2\right)0<x<1\text{ and }0<y<1\\ 0\text{ Otherwise}\end{array}$

Calculation:

The formula for finding $Cov\left(X,Y\right)$ is,

$Cov\left(X,Y\right)={\mu }_{XY}-{\mu }_X{\mu }_Y$

The value of ${\mu }_{XY}$ can be obtained as follows:

$\begin{array}{c}{\mu }_{XY}={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}xy\frac{3\left(x^2+y^2\right)}{2}}}dxdy\\ ={{\displaystyle {\int }_0^1\left[\frac{3x^{4}y}{8}+\frac{3x^2{y}^3}{4}\right]}}_0^{1}dy\\ ={\displaystyle {\int }_0^1\left[\frac{3\left(1^4-0^4\right)y}{8}+\frac{3\left(1^2-0^2\right)y^3}{4}\right]}dy\\ ={\displaystyle {\int }_0^1\left[\frac{3y}{8}+\frac{3y^3}{4}\right]}dy\end{array}$

$\begin{array}{c}={\left[\frac{3y^2}{16}+\frac{3y^4}{16}\right]}_0^1\\ =\frac{3\left(1^2-0^2\right)}{16}+\frac{3\left(1^4-0^4\right)}{16}\\ =\frac{6}{16}\\ =\frac{3}{8}\end{array}$

Thus, the value of ${\mu }_{XY}$ is $\frac{3}{8}$.

For ${\mu }_X$:

From Exercise 18,

$f_X\left(x\right)=\frac{1+3x^2}{2}$ for $0<x<1$.

The formula for finding ${\mu }_X$ is,

${\mu }_X={\displaystyle {\int }_{-\infty }^{\infty }x{f}_X\left(x\right)dx}$

Therefore,

$\begin{array}{c}{\mu }_X={\displaystyle {\int }_0^{1}x\frac{1+3x^2}{2}dx}\\ ={\left(\frac{x^2}{4}+\frac{3x^4}{8}\right)}_0^1\\ =\frac{1}{4}+\frac{3\left(1\right)}{8}+0+0\\ =\frac{5}{8}\end{array}$

Thus, the Thus, the value of ${\mu }_X$ is $\frac{5}{8}$.

For ${\mu }_Y$:

From exercise 18,

$f_Y\left(y\right)=\frac{1+3y^2}{2}$ for $0<x<1$.

The formula for finding ${\mu }_Y$ is,

${\mu }_Y={\displaystyle {\int }_{-\infty }^{\infty }y{f}_Y\left(y\right)dy}$

Therefore,

$\begin{array}{c}{\mu }_Y={\displaystyle {\int }_0^{1}y\frac{1+3y^2}{2}dx}\\ ={\left(\frac{y^2}{4}+\frac{3y^4}{8}\right)}_0^1\\ =\frac{1}{4}+\frac{3\left(1\right)}{8}+0+0\\ =\frac{5}{8}\end{array}$

Thus, the Thus, the value of ${\mu }_Y$ is $\frac{5}{8}$.

Now $Cov\left(X,Y\right)$ becomes,

$\begin{array}{c}Cov\left(X,Y\right)=\frac{3}{8}-\left(\frac{5}{8}\right)\left(\frac{5}{8}\right)\\ =\frac{24-25}{64}\\ =-\frac{1}{64}\end{array}$

Thus, the value of $Cov\left(X,Y\right)$ is $-\frac{1}{64}$.

b.

To determine

Find the value of ${\rho }_{X,Y}$.

Answer to Problem 19E

The value of ${\rho }_{X,Y}$ is $-\frac{15}{73}$.

Explanation of Solution

Calculation:

The formula for finding the ${\rho }_{X,Y}$ is,

$\begin{array}{c}{\rho }_{X,Y}=\frac{\text{Cov}\left(\text{X,Y}\right)}{{\sigma }_X{\sigma }_Y}\text{ (or)}\\ =\frac{\text{Cov}\left(\text{X,Y}\right)}{\sqrt{{\sigma }_X^2{\sigma }_Y^2}}\end{array}$

From part (a), the value of $Cov\left(X,Y\right)$ is $-\frac{1}{64}$.

The formula for finding ${\sigma }_{{}_X}^2$ is,

${\sigma }_{{}_X}^2={\displaystyle {\int }_{-\infty }^{\infty }{x}^2{f}_X\left(x\right)dx-{\mu }_{{}_X}^2}$

Therefore,

$\begin{array}{c}{\sigma }_{{}_X}^2={\displaystyle {\int }_0^1{x}^2\frac{1+3x^2}{2}dx}-{\left(\frac{5}{8}\right)}^2\\ ={\left(\frac{x^3}{6}+\frac{3x^5}{10}\right)}_0^1-\frac{25}{64}\\ =\frac{1}{6}+\frac{3\left(1\right)}{10}+0+0-\frac{25}{64}\\ =\frac{160+288-375}{960}\end{array}$

$=\frac{73}{960}$

Thus, the Thus, the value of ${\sigma }_{{}_X}^2$ is $\frac{73}{960}$.

The formula for finding ${\sigma }_{{}_Y}^2$ is,

${\sigma }_{{}_Y}^2={\displaystyle {\int }_{-\infty }^{\infty }{y}^2{f}_Y\left(y\right)dx-{\mu }_Y^2}$

Therefore,

$\begin{array}{c}{\sigma }_Y^2={\displaystyle {\int }_0^1{y}^2\frac{1+3y^2}{2}dx}-{\left(\frac{5}{8}\right)}^2\\ ={\left(\frac{y^3}{6}+\frac{3y^5}{10}\right)}_0^1-\frac{25}{64}\\ =\frac{1}{6}+\frac{3\left(1\right)}{10}+0+0-\frac{25}{64}\\ =\frac{160+288-375}{960}\end{array}$

$=\frac{73}{960}$

Thus, the Thus, the value of ${\sigma }_{{}_Y}^2$ is $\frac{73}{960}$.

Substitute $Cov\left(X,Y\right)=-\frac{1}{64}$, ${\sigma }_X=\frac{73}{960}$ and ${\sigma }_Y=\frac{73}{960}$ in ${\rho }_{X,Y}$

$\begin{array}{c}{\rho }_{X,Y}=\frac{-\frac{1}{64}}{\sqrt{\left(\frac{73}{960}\right)\left(\frac{73}{960}\right)}}\\ =-\frac{\frac{1}{64}}{\frac{73}{960}}\\ =-\frac{15}{73}\end{array}$

Thus, the value of ${\rho }_{X,Y}$ is $-\frac{15}{73}$.

c.

To determine

Find the conditional probability density function $f_{Y|X}\left(y|0.5\right)$.

Answer to Problem 19E

The conditional probability density function $f_{Y|X}\left(y|0.5\right)$ is,

$f_{Y|X}\left(y|0.5\right)=\frac{3+12y^2}{7}\text{ for }0<y<1$

Explanation of Solution

Calculation:

The formula for finding $f_{Y|X}\left(y|0.5\right)$ is,

$f_{Y|X}\left(y|0.5\right)=\frac{f_{X,Y}\left(0.5,y\right)}{f_X\left(0.5\right)}$

The value of $f_{X,Y}\left(0.5,y\right)$ is,

$\begin{array}{c}{f}_{X,Y}\left(0.5,y\right)=\frac{3\left({\left(\frac{1}{2}\right)}^2+y^2\right)}{2}\\ =\frac{3+12y^2}{8}\langle \text{for }0<y<1\rangle \end{array}$

The value of $f_X\left(0.5\right)$ is,

$\begin{array}{c}{f}_X\left(0.5\right)=\frac{1+3{\left(\frac{1}{2}\right)}^2}{2}\\ =\frac{7}{8}\langle \text{for }0<y<1\rangle \end{array}$

Now the conditional probability density function $f_{Y|X}\left(y|0.5\right)$ becomes,

$\begin{array}{c}{f}_{Y|X}\left(y|0.5\right)=\frac{\frac{3+12y^2}{8}}{\frac{7}{8}}\\ =\frac{3+12y^2}{7}\langle \text{for }0<y<1\rangle \end{array}$

Thus, the conditional probability density function $f_{Y|X}\left(y|0.5\right)$ is,

$f_{Y|X}\left(y|0.5\right)=\frac{3+12y^2}{7}\text{ for }0<y<1$

d.

To determine

Find the conditional expectation $E\left(Y|X=0.5\right)$.

Answer to Problem 19E

The conditional expectation $E\left(Y|X=0.5\right)$ is $\frac{9}{14}$.

Explanation of Solution

Calculation:

The formula for finding $E\left(Y|X=0.5\right)$ is,

$E\left(Y|X=0.5\right)={\displaystyle {\int }_0^{1}y{f}_{Y|X}\left(y|0.5\right)}dy$

From part (c), the conditional probability density function $f_{Y|X}\left(y|0.5\right)$ is,

$f_{Y|X}\left(y|0.5\right)=\frac{3+12y^2}{7}\text{ for }0<y<1$.

Therefore,

$\begin{array}{c}E\left(Y|X=0.5\right)={\displaystyle {\int }_0^{1}y\frac{3+12y^2}{7}}dy\\ ={\left[\frac{3y^2+6y^4}{14}\right]}_0^1\\ =\frac{3{\left(1\right)}^2+6{\left(1\right)}^4+0+0}{14}\\ =\frac{9}{14}\end{array}$

Thus, the conditional expectation $E\left(Y|X=0.5\right)$ is $\frac{9}{14}$.