Chapter 5, Problem 77SE

a.

To determine

Draw the region of positive density.

Find the value of k.

Answer to Problem 77SE

The region of positive density is:

The area between the two lines $x+y=30$ and $x+y=20$ is the region of positive density.

The value of k is $\underset{\_}{\frac{3}{81,250}}$.

Explanation of Solution

Given info:

The random variables X and Y denote the amounts (lb) of grains of brand A and B, respectively, on hand, at a health-food store. The joint pdf is given as:

$f\left(x,y\right)=\{\begin{array}{l}kxy,x\ge 0,y\ge 0,20\le x+y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Calculation:

Graphical procedure:

• Label the horizontal axis as x.
• Label the vertical axis as y.
• Draw the line $x+y=30$.
• Draw the line $x+y=20$.

The area between the two lines $x+y=30$ and $x+y=20$ is the region of positive density.

The figure is given below:

It is known that the joint pdf, $f\left(x,y\right)$, of two random variables is such that:

${\displaystyle \int {\displaystyle \int f\left(x,y\right)dxdy}}=1$.

Now, both X and Y are positive and $20\le x+y\le 30$. As a result, $20-x\le y\le 30-x$.

Thus,

$\begin{array}{c}{\displaystyle \int {\displaystyle \int f\left(x,y\right)dxdy}}={\displaystyle \underset{20}{\overset{30}{\int }}{\displaystyle \underset{20-x}{\overset{30-x}{\int }}kxydxdy}}\\ =1.\end{array}$

Now,

$\begin{array}{c}1={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}kxydy}dx}\\ ={\displaystyle \underset{0}{\overset{20}{\int }}{\displaystyle \underset{20-x}{\overset{30-x}{\int }}kxydy}dx}+{\displaystyle \underset{20}{\overset{30}{\int }}{\displaystyle \underset{0}{\overset{30-x}{\int }}kxydy}dx}\left(\text{both }X\text{, }Y\text{ being positive and }20\le x+y\le 30\right)\\ =k\left[{\displaystyle \underset{0}{\overset{20}{\int }}x{\displaystyle \underset{20-x}{\overset{30-x}{\int }}ydy}dx}+{\displaystyle \underset{20}{\overset{30}{\int }}x{\displaystyle \underset{0}{\overset{30-x}{\int }}ydy}dx}\right]\\ =k\left[{\displaystyle \underset{0}{\overset{20}{\int }}x{\left[\frac{y^2}{2}\right]}_{20-x}^{30-x}dx}+{\displaystyle \underset{20}{\overset{30}{\int }}x{\left[\frac{y^2}{2}\right]}_0^{30-x}dx}\right]\end{array}$

$\begin{array}{c}=k\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[\frac{{\left(30-x\right)}^2-{\left(20-x\right)}^2}{2}\right]dx}+{\displaystyle \underset{20}{\overset{30}{\int }}x\left[\frac{{\left(30-x\right)}^2}{2}\right]dx}\right]\\ =\frac{k}{2}\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[\left(30-x+20-x\right)\left(30-x-20+x\right)\right]dx}+{\displaystyle \underset{20}{\overset{30}{\int }}x\left(900-60x+x^2\right)dx}\right]\\ =\frac{k}{2}\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[10\left(50-2x\right)\right]dx}+{\displaystyle \underset{20}{\overset{30}{\int }}\left(900x-60x^2+x^3\right)dx}\right]\\ =\frac{k}{2}\left[10{\displaystyle \underset{0}{\overset{20}{\int }}\left(50x-2x^2\right)dx}+{\displaystyle \underset{20}{\overset{30}{\int }}\left(900x-60x^2+x^3\right)dx}\right]\end{array}$

$\begin{array}{c}=\frac{k}{2}\left({\left[\frac{500x^2}{2}-\frac{20x^3}{3}\right]}_0^{20}+{\left[\frac{900x^2}{2}-\frac{60x^3}{3}+\frac{x^4}{4}\right]}_{20}^{30}\right)\\ =\frac{k}{2}\left(\frac{500{\left(20\right)}^2}{2}-\frac{20{\left(20\right)}^3}{3}+\frac{900\left({30}^2-{20}^2\right)}{2}-\frac{60\left({30}^3-{20}^3\right)}{3}+\frac{\left({30}^4-{20}^4\right)}{4}\right)\\ =\frac{k}{2}\left(\frac{500\times 400}{2}-\frac{20\times 8,000}{3}+\frac{900\left(900-400\right)}{2}-\frac{60\left(27,000-8,000\right)}{3}+\frac{\left(810,000-160,000\right)}{4}\right)\\ =\frac{k}{2}\left(100,000-\frac{160,000}{3}+225,000-380,000+162,500\right)\end{array}$$\begin{array}{c}=\frac{k}{2}\left(\frac{162,500}{3}\right)\\ =\frac{81,250}{3}k\end{array}$

Thus,

$k=\underset{\_}{\frac{3}{81,250}}.$

b.

To determine

Decide whether X and Y are independent.

Answer to Problem 77SE

The random variables X and Y are not independent.

Explanation of Solution

Calculation:

The marginal pdf of X, $f_X\left(x\right)$ can be obtained by integrating Y as follows:

• Integrate Y over the range $\left[20-x,30-x\right]$ for $0\le x\le 20$,
• Integrate Y over the range $\left[0,30-x\right]$ for $20\le x\le 30$.

The inner integrals from part a provide the marginal pdf of X. Thus,

$\begin{array}{c}{f}_X\left(x\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}kxydy}\\ ={\displaystyle \underset{20-x}{\overset{30-x}{\int }}kxydy}+{\displaystyle \underset{0}{\overset{30-x}{\int }}kxydy}\\ =k\left[x{\displaystyle \underset{20-x}{\overset{30-x}{\int }}ydy}+x{\displaystyle \underset{0}{\overset{30-x}{\int }}ydy}\right]\\ =\frac{k}{2}\left[x\left[10\left(50-2x\right)\right]+\left(900x-60x^2+x^3\right)\right]\end{array}$

$=k\left(250x-10x^2\right)+k\left(450x-30x^2+\frac{x^3}{2}\right)$.

Thus, the marginal pdf of X is:

$f_X\left(x\right)=\{\begin{array}{l}k\left(250x-10x^2\right),0\le x\le 20\\ k\left(450x-30x^2+\frac{x^3}{2}\right),20\le x\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

The distributions of X and Y being identical, the marginal pdf of Y is the same as that of X. Thus, the marginal pdf of Y is:

$f_Y\left(y\right)=\{\begin{array}{l}k\left(250y-10y^2\right),0\le y\le 20\\ k\left(450y-30y^2+\frac{y^3}{2}\right),20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Any two random variables are independent only if $f_X\left(x\right)\cdot f_X\left(y\right)=f\left(x,y\right)$.

Here,

$f_X\left(x\right)\cdot f_Y\left(y\right)=\{\begin{array}{l}\left[k\left(250x-10x^2\right)\right]\left[k\left(250y-10y^2\right)\right],0\le x\le 20,0\le y\le 20\\ \left[k\left(450x-30x^2+\frac{x^3}{2}\right)\right]\left[k\left(450y-30y^2+\frac{y^3}{2}\right)\right]20\le x\le 30,20\le y\le 30\\ 0,\text{ otherwise}\text{.}\end{array}$

Now, $f\left(x,y\right)=kxy$. Thus, $f_X\left(x\right)\cdot f_X\left(y\right)\ne f\left(x,y\right)$. Hence, X and Y are not independent.

c.

To determine

Find the value of $P\left(X+Y\le 25\right)$.

Answer to Problem 77SE

The value of $P\left(X+Y\le 25\right)$ is 0.355.

Explanation of Solution

Calculation:

The value of the probability is:

$P\left(X+Y\le 25\right)={\displaystyle \underset{0}{\overset{20}{\int }}{\displaystyle \underset{20-x}{\overset{25-x}{\int }}f\left(x,y\right)dy}dx}+{\displaystyle \underset{20}{\overset{25}{\int }}{\displaystyle \underset{0}{\overset{25-x}{\int }}f\left(x,y\right)dy}dx}$.

Using the steps of integration in part a,

$\begin{array}{c}P\left(X+Y\le 25\right)=k\left[{\displaystyle \underset{0}{\overset{20}{\int }}x{\left[\frac{y^2}{2}\right]}_{20-x}^{25-x}dx}+{\displaystyle \underset{20}{\overset{25}{\int }}x{\left[\frac{y^2}{2}\right]}_0^{25-x}dx}\right]\\ =k\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[\frac{{\left(25-x\right)}^2-{\left(20-x\right)}^2}{2}\right]dx}+{\displaystyle \underset{20}{\overset{25}{\int }}x\left[\frac{{\left(25-x\right)}^2}{2}\right]dx}\right]\\ =\frac{k}{2}\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[\left(25-x+20-x\right)\left(25-x-20+x\right)\right]dx}+{\displaystyle \underset{20}{\overset{25}{\int }}x\left(625-50x+x^2\right)dx}\right]\\ =\frac{k}{2}\left[{\displaystyle \underset{0}{\overset{20}{\int }}x\left[5\left(45-2x\right)\right]dx}+{\displaystyle \underset{20}{\overset{25}{\int }}\left(625x-50x^2+x^3\right)dx}\right]\end{array}$

$\begin{array}{c}=\frac{k}{2}\left[5{\displaystyle \underset{0}{\overset{20}{\int }}\left(45x-2x^2\right)dx}+{\displaystyle \underset{20}{\overset{25}{\int }}\left(625x-50x^2+x^3\right)dx}\right]\\ =\frac{k}{2}\left({\left[\frac{225x^2}{2}-\frac{10x^3}{3}\right]}_0^{20}+{\left[\frac{625x^2}{2}-\frac{50x^3}{3}+\frac{x^4}{4}\right]}_{20}^{25}\right)\\ =\frac{k}{2}\left(\frac{225{\left(20\right)}^2}{2}-\frac{10{\left(20\right)}^3}{3}+\frac{625\left({25}^2-{20}^2\right)}{2}-\frac{50\left({25}^3-{20}^3\right)}{3}+\frac{\left({25}^4-{20}^4\right)}{4}\right)\\ =\frac{k}{2}\left(\begin{array}{l}\frac{225\times 400}{2}-\frac{10\times 8,000}{3}+\frac{625\left(625-400\right)}{2}\\ -\frac{50\left(15,625-8,000\right)}{3}+\frac{\left(390,625-160,000\right)}{4}\end{array}\right)\end{array}$$\begin{array}{l}=\frac{1}{2}\left(\frac{3}{81,250}\right)\left(45,000-\frac{80,000}{3}+\frac{140,625}{2}-\frac{381,250}{3}+\frac{230,625}{4}\right)\\ =\frac{1}{2}\cdot \frac{3}{81,250}\cdot \frac{76,875}{4}\\ \approx \underset{\_}{0.355}.\end{array}$

d.

To determine

Find the expected total amount of grain on hand.

Answer to Problem 77SE

The expected total amount of grain on hand is 25.97.

Explanation of Solution

Calculation:

The expected total amount of grain on hand is $E\left(X+Y\right)$.

Now,

$\begin{array}{c}E\left(X+Y\right)=E\left(X\right)+E\left(Y\right)\left(\text{using sum law of expectation}\right)\\ =2E\left(X\right)\left(\text{due to identical distribution of }X\text{ and }Y\right).\end{array}$

Again,

$\begin{array}{c}2E\left(X\right)=2{\displaystyle \underset{0}{\overset{\infty }{\int }}x{f}_X\left(x\right)dx}\\ =2{\displaystyle \underset{0}{\overset{20}{\int }}xk\left(250x+10x^2\right)dx}+{\displaystyle \underset{20}{\overset{30}{\int }}xk\left(450x-30x^2+\frac{x^3}{2}\right)dx}\\ =2k\left[{\displaystyle \underset{0}{\overset{20}{\int }}\left(250x^2+10x^3\right)dx}+{\displaystyle \underset{20}{\overset{30}{\int }}\left(450x^2-30x^3+\frac{x^4}{2}\right)dx}\right]\\ =2k\left({\left[\frac{250x^3}{3}+\frac{10x^4}{4}\right]}_0^{20}+{\left[\frac{450x^3}{3}-\frac{30x^4}{4}+\frac{x^5}{2\times 5}\right]}_{20}^{30}\right)\end{array}$

$\begin{array}{l}=2k\left(\left[\frac{250\left({20}^3\right)}{3}+\frac{10\left({20}^4\right)}{4}\right]+\left[\frac{450\left({30}^3-{20}^3\right)}{3}-\frac{30\left({30}^4-{20}^4\right)}{4}+\frac{\left({30}^5-{20}^5\right)}{2\times 5}\right]\right)\\ =\frac{3\times 2}{81,250}\left[\frac{2,000,000}{3}+40,000+2,850,000-3,250,000+2,110,000\right]\\ \approx \underset{\_}{25.97}.\end{array}$

e.

To determine

Find the values of $\text{Cov}\left(X,Y\right)$ and $\text{Corr}\left(X,Y\right)$.

Answer to Problem 77SE

The value of $\text{Cov}\left(X,Y\right)$ is –32.2.

The value of $\text{Corr}\left(X,Y\right)$ is 0.89.

Explanation of Solution

Calculation:

It is known that $\text{Cov}\left(X,Y\right)=E\left(XY\right)-E\left(X\right)\cdot E\left(X\right)$.

Now,

$\begin{array}{c}E\left(XY\right)={\displaystyle \underset{0}{\overset{20}{\int }}{\displaystyle \underset{20-x}{\overset{30-x}{\int }}xyf\left(x,y\right)dx}dy}\\ ={\displaystyle \underset{0}{\overset{20}{\int }}{\displaystyle \underset{20-x}{\overset{30-x}{\int }}xykxydx}dy}\\ =k{\displaystyle \underset{0}{\overset{20}{\int }}{\displaystyle \underset{20-x}{\overset{30-x}{\int }}{x}^2{y}^{2}dx}dy}.\end{array}$

Using the steps of integration in part a and proceeding in a similar manner,

$\begin{array}{c}E\left(XY\right)=k\left[{\displaystyle \underset{0}{\overset{20}{\int }}{x}^2{\displaystyle \underset{20-x}{\overset{30-x}{\int }}{y}^{2}dy}dx}+{\displaystyle \underset{20}{\overset{30}{\int }}{x}^2{\displaystyle \underset{0}{\overset{30-x}{\int }}{y}^{2}dy}dx}\right]\\ =k\left[{\displaystyle \underset{0}{\overset{20}{\int }}{x}^2{\left[\frac{y^3}{3}\right]}_{20-x}^{30-x}dx}+{\displaystyle \underset{20}{\overset{30}{\int }}{x}^2{\left[\frac{y^3}{3}\right]}_0^{30-x}dx}\right]\\ =\frac{k}{3}\left[{\displaystyle \underset{0}{\overset{20}{\int }}{x}^2\left({\left(30-x\right)}^3-{\left(20-x\right)}^3\right)dx}+{\displaystyle \underset{20}{\overset{30}{\int }}{x}^2{\left(30-x\right)}^{3}dx}\right].\end{array}$

Proceeding in a similar manner as before,

$\begin{array}{c}E\left(XY\right)=\frac{k}{3}\cdot \frac{33,250,000}{3}\\ =\frac{3}{81,250\times 3}\cdot \frac{33,250,000}{3}\\ =\mathrm{136.41.}\end{array}$

Thus,

$\begin{array}{c}\text{Cov}\left(X,Y\right)=E\left(XY\right)-E\left(X\right)\cdot E\left(X\right)\\ =136.41-{\left(\frac{25.97}{2}\right)}^2\\ =136.41-168.61\\ =\underset{\_}{-32.2}.\end{array}$

Now, $\text{Corr}\left(X,Y\right)=\frac{\text{Cov}\left(X,Y\right)}{{\sigma }_X\cdot {\sigma }_Y}$, where ${\sigma }_X$ and ${\sigma }_Y$ are the standard deviations of X and Y.

Now,

$\begin{array}{c}V\left(X\right)={\sigma }_X^2\\ =E\left(X^2\right)-{\left(E\left(X\right)\right)}^2.\end{array}$

Using integration as before,

$\begin{array}{c}E\left(X^2\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2{f}_X\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{20}{\int }}{x}^{2}k\left(250x-10x^2\right)dx}+{\displaystyle \underset{20}{\overset{30}{\int }}{x}^{2}k\left(450x-30x^2+\frac{x^3}{2}\right)dx}\\ =\mathrm{204.6154.}\end{array}$

Thus,

$\begin{array}{c}V\left(X\right)=204.6154-{\left(\frac{25.97}{2}\right)}^2\\ =204.6157-168.61\\ \approx 36.\end{array}$

Due to identical distribution of X and Y, $V\left(Y\right)=V\left(X\right)=36$.

Thus, ${\sigma }_X={\sigma }_Y=\sqrt{36}$, that is, ${\sigma }_X={\sigma }_Y=6$.

Hence,

$\begin{array}{c}\text{Corr}\left(X,Y\right)=\frac{\text{Cov}\left(X,Y\right)}{{\sigma }_X\cdot {\sigma }_Y}\\ =\frac{-32.2}{6\times 6}\\ =\underset{\_}{-0.89}.\end{array}$

f.

To determine

Find the variance of the total amount of grain on hand.

Answer to Problem 77SE

The variance of the total amount of grain on hand is 7.6.

Explanation of Solution

Calculation:

It is known that variance of the sum of two variables that are not independent, is:

$V\left(X+Y\right)=V\left(X\right)+V\left(Y\right)+2\text{Cov}\left(X,Y\right)$.

Using the results in part e, the variance of the total amount of grain on hand is:

$\begin{array}{c}V\left(X+Y\right)=V\left(X\right)+V\left(Y\right)+2\text{Cov}\left(X,Y\right)\\ =36+36+\left[2\times \left(-32.2\right)\right]\\ =36+36-64.4\\ =\underset{\_}{7.6}.\end{array}$