Chapter 5.6, Problem 78E

a.

To determine

Calculate the values of $E\left(Y_1\right)$ and $E\left(Y_2\right)$.

Answer to Problem 78E

The value of $E\left(Y_1\right)$ is $\underset{\_}{\frac{4}{3}}$.

The value of $E\left(Y_2\right)$ is $\underset{\_}{\frac{1}{3}}$.

Explanation of Solution

Calculation:

Consider that $Y_1$ and $Y_2$ are two continuous real valued random variables with joint probability density function of $f\left(y_1,y_2\right)$.

Then, the marginal probability functions of $Y_1$ and $Y_2$ are defined as, $f_1\left(y_1\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y_1,y_2\right)d{y}_2}$ and $f_2\left(y_2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y_1,y_2\right)d{y}_1}$, respectively.

In the given problem $Y_1$ denotes the amount of pollutant per sample collected above the stack than has no cleaning device and $Y_2$ denotes the amount of pollutant per sample collected above the stack than has cleaning device.

Hence, using the joint probability distribution the marginal probability distribution of $Y_1$ is obtained below:

$\begin{array}{c}{f}_1\left(y_1\right)={\displaystyle \underset{0}{\overset{\frac{y_1}{2}}{\int }}d{y}_2}\\ ={\left[y_2\right]}_0^{\frac{y_1}{2}}\\ =\left[\frac{y_1}{2}-0\right]\\ =\frac{y_1}{2}\end{array}$.

Thus, the marginal density function for $Y_1$ is $f_1\left(y_1\right)=\frac{y_1}{2},\text{ for }0\le y_1\le 2$.

Similarly, using the joint probability distribution the marginal probability distribution of $Y_2$ is obtained below,

$\begin{array}{c}{f}_2\left(y_2\right)={\displaystyle \underset{2y_2}{\overset{2}{\int }}d{y}_1}\\ ={\left[y_1\right]}_{2y_2}^2\\ =\left[2-2y_2\right]\\ =2\left(1-y_2\right)\end{array}$

Thus, the marginal density function for $Y_2$ is $f_2\left(y_2\right)=2\left(1-y_2\right),0\le y_2\le 1$.

The expectation of $Y_1$ is calculated as follows:

$\begin{array}{c}E\left(Y_1\right)={\displaystyle \underset{0}{\overset{2}{\int }}{y}_1{f}_1\left(y_1\right)d{y}_1}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\left(y_1\right)\left(\frac{y_1}{2}\right)d{y}_1}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{2}{\int }}{y}_1^{2}d{y}_1}\\ =\frac{1}{2}{\left[\frac{y_1^3}{3}\right]}_0^2\\ =\frac{1}{2}\left(\frac{8}{3}-0\right)\\ =\frac{4}{3}.\end{array}$

Thus, the value of $E\left(Y_1\right)$ is $\underset{\_}{\frac{4}{3}}$.

The expectation of $Y_2$ is calculated as follows:

$\begin{array}{c}E\left(Y_2\right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}_2{f}_2\left(y_2\right)d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(y_2\right)\left[2\left(1-y_2\right)\right]d{y}_2}\\ =2\left\{{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{2}d{y}_1}-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_2^{2}d{y}_2}\right\}\\ =2\left\{{\left[\frac{y_2^2}{2}\right]}_0^1-{\left[\frac{y_2^3}{3}\right]}_0^1\right\}\\ =2\left\{\left[\frac{1}{2}-0\right]-\left[\frac{1}{3}-0\right]\right\}\\ =2\times \frac{1}{6}\\ =\frac{1}{3}.\end{array}$

Thus, the value of $E\left(Y_2\right)$ is $\underset{\_}{\frac{1}{3}}$.

b.

To determine

Calculate the values of $V\left(Y_1\right)$ and $V\left(Y_2\right)$.

Answer to Problem 78E

The value of $V\left(Y_1\right)$ is $\underset{\_}{\frac{2}{9}}$.

The value of $V\left(Y_2\right)$ is $\underset{\_}{\frac{1}{18}}$.

Explanation of Solution

Calculation:

The variance of any random variable, $Y_1$, can be expressed as, $V\left(Y_1\right)=E\left(Y_1^2\right)-{\left[E\left(Y_1\right)\right]}^2$.

Now, $E\left(Y_1^2\right)$ is calculated below:

$\begin{array}{c}E\left(Y_1^2\right)={\displaystyle \underset{0}{\overset{2}{\int }}{y}_1{f}_1\left(y_1\right)d{y}_1}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\left(y_1^2\right)\left(\frac{y_1}{2}\right)d{y}_1}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{2}{\int }}{y}_1^{3}d{y}_1}\\ =\frac{1}{2}{\left[\frac{y_1^4}{4}\right]}_0^2\\ =\frac{1}{2}\left(\frac{16}{4}-0\right)\\ =2.\end{array}$

Thus, the variance is calculated below:

$\begin{array}{c}V\left(Y_1\right)=2-{\left[\frac{4}{3}\right]}^2\\ =2-\frac{16}{9}\\ =\frac{2}{9}.\end{array}$

Thus, the value of $V\left(Y_1\right)$ is $\underset{\_}{\frac{2}{9}}$.

Now, $E\left(Y_2^2\right)$ is calculated below:

$\begin{array}{c}E\left(Y_2^2\right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}_2^2{f}_2\left(y_2\right)d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(y_2^2\right)\left[2\left(1-y_2\right)\right]d{y}_2}\\ =2\left\{{\displaystyle \underset{0}{\overset{1}{\int }}{y}_2^{2}d{y}_1}-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_2^{3}d{y}_2}\right\}\\ =2\left\{{\left[\frac{y_2^3}{3}\right]}_0^1-{\left[\frac{y_2^4}{4}\right]}_0^1\right\}\\ =2\left\{\left[\frac{1}{3}-0\right]-\left[\frac{1}{4}-0\right]\right\}\\ =2\times \frac{1}{12}\\ =\frac{1}{6}.\end{array}$

Thus, the variance is calculated below:

$\begin{array}{c}V\left(Y_2\right)=\frac{1}{6}-{\left[\frac{1}{3}\right]}^2\\ =\frac{1}{6}-\frac{1}{9}\\ =\frac{1}{18}.\end{array}$

Thus, the value of $V\left(Y_2\right)$ is $\underset{\_}{\frac{1}{18}}$.

c.

To determine

Find the value of $E\left(Y_1-Y_2\right)$.

Answer to Problem 78E

The value of $E\left(Y_1-Y_2\right)$ is 1.

Explanation of Solution

Calculation:

For any two random variables, $Y_1$ and $Y_2$, irrespective of their distributions, it is known that, $E\left(Y_1-Y_2\right)=E\left(Y_1\right)-E\left(Y_2\right)$.

Thus, using the values obtained in Part a, the value of $E\left(Y_1-Y_2\right)$ is $\underset{\_}{1}\left(=\frac{4}{3}-\frac{1}{3}\right)$.

d.

To determine

Calculate the value of $V\left(Y_1-Y_2\right)$.

Give the range, within which, the values of $\left(Y_1-Y_2\right)$ are expected to fall.

Answer to Problem 78E

The value of $V\left(Y_1-Y_2\right)$ is $\underset{\_}{\frac{1}{6}}$.

The values of $\left(Y_1-Y_2\right)$ are expected to fall within the interval $\underset{\_}{\left(0.1836,1.8164\right)}$.

Explanation of Solution

Calculation:

The value of $V\left(Y_1-Y_2\right)$ can be obtained using the formula as obtained below:

$\begin{array}{c}V\left(Y_1-Y_2\right)=E{\left(Y_1-Y_2\right)}^2-{\left[E\left(Y_1-Y_2\right)\right]}^2\\ =E\left(Y_1^2-2Y_1{Y}_2+Y_2^2\right)-{\left[E\left(Y_1-Y_2\right)\right]}^2\\ =E\left(Y_1^2\right)-2E\left(Y_1{Y}_2\right)+E\left(Y_2^2\right)-{\left[E\left(Y_1-Y_2\right)\right]}^2.\end{array}$

Now, using the values obtained in Parts b and c, it can be said that, $E\left(Y_1^2\right)=2$, $E\left(Y_2^2\right)=\frac{1}{6}$, and $E\left(Y_1-Y_2\right)=1$.

The calculation of $E\left(Y_1{Y}_2\right)$ is shown below:

$\begin{array}{c}E\left(Y_1{Y}_2\right)={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{2y_2}{\overset{2}{\int }}{y}_1{y}_{2}f\left(y_1,y_2\right)d{y}_1}d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{2y_2}{\overset{1}{\int }}{y}_1{y}_2\times \left(1\right)d{y}_1}d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}_2{\displaystyle \underset{2y_2}{\overset{1}{\int }}{y}_{1}d{y}_1}d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}_2{\left[\frac{y_1^2}{2}\right]}_{2y_2}^{2}d{y}_2}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}_2\left[\frac{4}{2}-\frac{4y_2^2}{2}\right]d{y}_2}\\ =2{\displaystyle \underset{0}{\overset{1}{\int }}{y}_2\left(1-y_2^2\right)d{y}_2}\\ =2\left\{{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{2}d{y}_2}-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_2^{3}d{y}_2}\right\}\\ =2\left\{{\left[\frac{y_2^2}{2}\right]}_0^1-{\left[\frac{y_2^4}{4}\right]}_0^1\right\}\\ =2\left\{\left[\frac{1}{2}-0\right]-\left[\frac{1}{4}-0\right]\right\}\\ =2\times \frac{1}{4}\\ =\frac{1}{2}.\end{array}$

Thus, $V\left(Y_1-Y_2\right)$ is calculated below:

$\begin{array}{c}V\left(Y_1-Y_2\right)=E\left(Y_1^2\right)-2E\left(Y_1{Y}_2\right)+E\left(Y_2^2\right)-{\left[E\left(Y_1-Y_2\right)\right]}^2\\ =2-\left(2\times \frac{1}{2}\right)+\frac{1}{6}-{\left(1\right)}^2\\ =\frac{1}{6}.\end{array}$

Thus, the value of $V\left(Y_1-Y_2\right)$ is $\underset{\_}{\frac{1}{6}}$.

Tchebysheff’s theorem:

According to Tchebysheff’s theorem discussed in Exercise 1.32, the fraction of observations lying within the interval $\overline{y}-ks$ and $\overline{y}+ks$ is at least $1-\frac{1}{k^2}$.

For the 2-standard deviations interval about the mean, $k=2$. Then, $1-\frac{1}{k^2}$ is calculated below:

$\begin{array}{c}1-\frac{1}{k^2}=1-\frac{1}{2^2}\\ =1-\frac{1}{4}\\ =\frac{3}{4}\\ =\mathrm{0.75.}\end{array}$

In this case, $\overline{y}=E\left(Y_1-Y_2\right)$. Now, the standard deviation of $\left(Y_1-Y_2\right)$ is calculated below:

$\begin{array}{c}s=\sqrt{V\left(Y_1-Y_2\right)}\\ =\sqrt{\frac{1}{6}}\\ \approx \mathrm{0.4082.}\end{array}$

The interval $\left(\overline{y}\pm 2s\right)$ is calculated below:

$\begin{array}{c}\left(\overline{y}\pm 2s\right)=\left(1-\left(2\times 0.4082\right),1+\left(2\times 0.4082\right)\right)\\ =\underset{\_}{\left(0.1836,1.8164\right)}.\end{array}$

About 75% of the scores must fall in the above interval.

Hence, most of the values of $\left(Y_1-Y_2\right)$ are expected to fall within the interval $\underset{\_}{\left(0.1836,1.8164\right)}$.