Chapter 5.1, Problem 19E

The joint pdf of pressures for right and left front tires is given in Exercise 9.

1. a. Determine the conditional pdf of Y given that X = x and the conditional pdf of X given that Y = y.
2. b. If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire has a pressure of at least 25 psi? Compare this to P(Y ≥ 25).
3. c. If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?

To determine

Find the conditional pdf of Y given that $X=x$ and the conditional pdf of X given that $Y=y$.

Answer to Problem 19E

The conditional pdf of Y given that $X=x$ is:

$f_{Y|X}\left(y|x\right)=\{\begin{array}{l}\frac{K\left(x^2+y^2\right)}{10K{x}^2+0.05},20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

The conditional pdf of X given that $Y=y$ is:

$f_{X|Y}\left(x|x\right)=\{\begin{array}{l}\frac{K\left(x^2+y^2\right)}{10K{y}^2+0.05},20\le x\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Explanation of Solution

Given info:

Denote X is the actual air pressure in the right tire of a vehicle and Y is that in the left tire. The joint pdf of X and Y is:

$f\left(x,y\right)=\{\begin{array}{l}\frac{3}{380,000}\left(x^2+y^2\right)\text{, }20\le x\le 30,20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Calculation:

From Exercise 9, the marginal pdf’s of X and Y are found to be:

$f_X\left(x\right)=\{\begin{array}{l}\frac{3x^2}{38,000}+0.05\text{, }20\le x\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

$f_Y\left(y\right)=\{\begin{array}{l}\frac{3y^2}{38,000}+0.05\text{, }20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

The conditional pdf of Y given that $X=x$ can be obtained by finding the conditional distribution of Y, considering the particular case of $X=x$.

The expression for the conditional pdf for Y given X is:

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

Thus, the conditional pdf of Y given that $X=x$ is:

$\begin{array}{c}{f}_{Y|X}\left(y|x\right)=\frac{f\left(x,y\right)}{f_X\left(x\right)}\\ =\frac{\frac{3}{380,000}\left(x^2+y^2\right)}{\frac{3x^2}{38,000}+0.05}\\ =\frac{\frac{3}{380,000}\left(x^2+y^2\right)}{\frac{3x^2\times 10}{380,000}+0.05}\\ =\frac{\left(\frac{3}{380,000}\right)\left(x^2+y^2\right)}{\left(\frac{3}{380,000}\right)10x^2+0.05}.\end{array}$

Replace $K=\frac{3}{380,000}$.

Thus, the conditional pdf of Y given that $X=x$ is:

$f_{Y|X}\left(y|x\right)=\{\begin{array}{l}\frac{K\left(x^2+y^2\right)}{10K{x}^2+0.05},20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Again the conditional pdf of X given that $Y=y$ is:

$\begin{array}{c}{f}_{X|Y}\left(x|x\right)=\frac{f\left(x,y\right)}{f_Y\left(y\right)}\\ =\frac{\frac{3}{380,000}\left(x^2+y^2\right)}{\frac{3y^2}{38,000}+0.05}\\ =\frac{\left(\frac{3}{380,000}\right)\left(x^2+y^2\right)}{\left(\frac{3}{380,000}\right)10y^2+0.05}\\ =\frac{K\left(x^2+y^2\right)}{10K{y}^2+0.05}\left(\text{where }K=\frac{3}{380,000}\right).\end{array}$

Thus, the conditional pdf of X given that $Y=y$ is:

$f_{X|Y}\left(x|x\right)=\{\begin{array}{l}\frac{K\left(x^2+y^2\right)}{10K{y}^2+0.05},20\le x\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

To determine

Find the probability that the left tire has a pressure of at least 25 psi, if the pressure in the right tire is found to be 22 psi.

Compare the obtained probability to $P\left(Y\ge 25\right)$.

Answer to Problem 19E

The probability that the left tire has a pressure of at least 25 psi, if the pressure in the right tire is found to be 22 psi is 0.5559.

The probability that the left tire has a pressure of at least 25 psi is slightly higher if the right tire already has a pressure of 22 psi.

Explanation of Solution

Calculation:

The probability that the left tire has a pressure of at least 25 psi, if the pressure in the right tire is found to be 22 psi is:

$\begin{array}{c}P\left(Y\ge 25|X=22\right)={\displaystyle \underset{25}{\overset{30}{\int }}{f}_{Y|X=22}\left(y|22\right)dy}\\ ={\displaystyle \underset{25}{\overset{30}{\int }}\frac{K\left({22}^2+y^2\right)}{10K{\left(22\right)}^2+0.05}dy}\\ =\frac{K}{10K{\left(22\right)}^2+0.05}\left[{22}^2{\displaystyle \underset{25}{\overset{30}{\int }}dy}+{\displaystyle \underset{25}{\overset{30}{\int }}{y}^{2}dy}\right]\\ =\frac{1}{\left(10\times 484\right)+\frac{0.05}{K}}\left[484{\left[y\right]}_{25}^{30}+{\left[\frac{y^3}{3}\right]}_{25}^{30}\right]\end{array}$

      $\begin{array}{c}=\frac{1}{4,840+\frac{0.05}{\frac{3}{380,000}}}\left[\left(484\cdot \left(30-25\right)\right)+\frac{{30}^3-{25}^3}{3}\right]\\ =\frac{1}{4,840+\frac{0.05\times 380,000}{3}}\left[\left(484\times 5\right)+\frac{27,000-15,625}{3}\right]\\ =\frac{1}{4,840+\frac{19,000}{3}}\left[2,420+\frac{11,375}{3}\right]\\ =\frac{3}{\left(3\times 4,840\right)+19,000}\left[\frac{\left(2,420\times 3\right)+11,375}{3}\right]\end{array}$

      $\begin{array}{c}=\frac{7,260+11,375}{14,520+19,000}\\ =\frac{18,635}{33,520}\\ =\underset{\_}{0.5559}.\end{array}$

Now,

$\begin{array}{c}P\left(Y\ge 25\right)={\displaystyle \underset{25}{\overset{30}{\int }}{f}_Y\left(y\right)dy}\\ ={\displaystyle \underset{25}{\overset{30}{\int }}\left(\frac{3y^2}{38,000}+0.05\right)dy}\\ =\frac{1}{38,000}{\displaystyle \underset{25}{\overset{30}{\int }}3y^{2}dy}+0.05{\displaystyle \underset{25}{\overset{30}{\int }}dy}\\ =\frac{1}{38,000}{\left[y^3\right]}_{25}^{30}+0.05{\left[y\right]}_{25}^{30}\\ =\frac{1}{38,000}\left({30}^3-{25}^3\right)+0.05\left(30-25\right)\\ =\frac{11,375}{38,000}+\left(0.05\times 5\right)\\ =0.2993+0.25\\ =\underset{\_}{0.5493}.\end{array}$

Thus, the probability that the left tire has a pressure of at least 25 psi is slightly higher if the right tire already has a pressure of 22 psi.

To determine

Find the expected pressure in the left tire, if the pressure in the right tire is found to be 22 psi.

Find the standard deviation of pressure in the left tire under the condition.

Answer to Problem 19E

The expected pressure in the left tire, if the pressure in the right tire is found to be 22 psi is 25.373 psi.

The standard deviation of pressure in the left tire under the condition is 2.87 psi.

Explanation of Solution

Calculation:

The expected pressure in the left tire, if the pressure in the right tire is found to be 22 psi is:

$\begin{array}{c}E\left(Y|X=22\right)={\displaystyle \underset{20}{\overset{30}{\int }}y{f}_{Y|X=22}\left(y|22\right)dy}\\ ={\displaystyle \underset{20}{\overset{30}{\int }}y\frac{K\left({22}^2+y^2\right)}{10K{\left(22\right)}^2+0.05}dy}\\ =\frac{K}{4,840K+0.05}\left[484{\displaystyle \underset{20}{\overset{30}{\int }}ydy}+{\displaystyle \underset{20}{\overset{30}{\int }}{y}^{3}dy}\right]\\ =\frac{1}{4,840+\frac{0.05}{K}}\left[484{\left[\frac{y^2}{2}\right]}_{20}^{30}+{\left[\frac{y^4}{4}\right]}_{20}^{30}\right]\end{array}$

$\begin{array}{c}=\frac{3}{33,520}\left[484\left[\frac{{30}^2-{20}^2}{2}\right]+\left[\frac{{30}^4-{20}^4}{4}\right]\right]\\ =\frac{3}{33,520}\left[484\cdot \frac{900-400}{2}+\frac{810,000-160,000}{4}\right]\\ =\frac{3}{33,520}\left[\frac{484\times 500}{2}+\frac{650,000}{4}\right]\\ =\frac{3}{33,520}\left[121,000+162,500\right]\end{array}$

$\begin{array}{c}=\frac{3\times 283,500}{33,520}\\ =\underset{\_}{25.373\text{ psi}}\text{.}\end{array}$

The variance of pressure in the left tire under the condition is $V\left(Y|X=22\right)=E\left(Y^2|X=22\right)-{\left[E\left(Y|X=22\right)\right]}^2$.

Now,

$\begin{array}{c}E\left(Y^2|X=22\right)={\displaystyle \underset{20}{\overset{30}{\int }}{y}^2{f}_{Y|X=22}\left(y|22\right)dy}\\ ={\displaystyle \underset{20}{\overset{30}{\int }}{y}^2\frac{K\left({22}^2+y^2\right)}{10K{\left(22\right)}^2+0.05}dy}\\ =\frac{K}{4,840K+0.05}\left[484{\displaystyle \underset{20}{\overset{30}{\int }}{y}^{2}dy}+{\displaystyle \underset{20}{\overset{30}{\int }}{y}^{4}dy}\right]\\ =\frac{1}{4,840+\frac{0.05}{K}}\left[484{\left[\frac{y^3}{3}\right]}_{20}^{30}+{\left[\frac{y^5}{5}\right]}_{20}^{30}\right]\end{array}$

$\begin{array}{c}=\frac{3}{33,520}\left[484\left[\frac{{30}^3-{20}^3}{3}\right]+\left[\frac{{30}^5-{20}^5}{5}\right]\right]\\ =\frac{3}{33,520}\left[484\cdot \frac{27,000-8,000}{3}+\frac{24,300,000-3,200,000}{5}\right]\\ =\frac{3}{33,520}\left[\frac{484\times 19,000}{3}+\frac{21,100,000}{5}\right]\\ =\frac{3}{33,520}\left[3,065,333.33+4,220,000\right]\end{array}$

$\begin{array}{c}=\frac{3\times 7,285,333.33}{33,520}\\ =\mathrm{652.0286.}\end{array}$

Thus,

$\begin{array}{c}V\left(Y|X=22\right)=E\left(Y^2|X=22\right)-{\left[E\left(Y|X=22\right)\right]}^2\\ =652.0286-{\left(25.373\right)}^2\\ =652.0286-643.7891\\ =\mathrm{8.2395.}\end{array}$

Hence, the standard deviation is:

$\begin{array}{c}SD\left(Y|X=22\right)=\sqrt{V\left(Y|X=22\right)}\\ =\sqrt{8.2395}\\ =\underset{\_}{2.87\text{ psi}}.\end{array}$