Chapter 5.2, Problem 13E

a.

To determine

Find the value of $F\left(\frac{1}{2},\frac{1}{2}\right)$.

Answer to Problem 13E

The value of $F\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{9}{16}$.

Explanation of Solution

Calculation:

According to the given question using the range of $Y_1\text{and }{Y}_2$, the required probability is obtained as follows:

$\begin{array}{c}F\left(\frac{1}{2},\frac{1}{2}\right)=P\left(Y_1<\frac{1}{2},Y_2<\frac{1}{2}\right)\\ ={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{\displaystyle \underset{y_1-1}{\overset{\frac{1}{2}}{\int }}30y_1{y}_2^{2}d{y}_2}d{y}_1}\\ ={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}30y_1\left[{\displaystyle \underset{y_1-1}{\overset{\frac{1}{2}}{\int }}{y}_2^{2}d{y}_2}\right]d{y}_1}\\ ={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}30y_1{\left[\frac{y_2^3}{3}\right]}_{y_1-1}^{\frac{1}{2}}d{y}_1}\\ =10{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_1\left[\frac{1}{8}-{\left(y_1-1\right)}^3\right]d{y}_1}\\ =10{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_1\left[\frac{1}{8}-y_1^3+3y_1^2-3y_1+1\right]d{y}_1}\\ =10\left[\frac{1}{8}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_{1}d{y}_1}-{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_1^{4}d{y}_1}+3{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_1^{3}d{y}_1}-3{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_1^{2}d{y}_1}+{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_{1}d{y}_1}\right]\\ =10\left[\frac{1}{8}{\left[\frac{y_1^2}{2}\right]}_0^{\frac{1}{2}}-{\left[\frac{y_1^5}{5}\right]}_0^{\frac{1}{2}}+3{\left[\frac{y_1^4}{4}\right]}_0^{\frac{1}{2}}-3{\left[\frac{y_1^3}{3}\right]}_0^{\frac{1}{2}}+{\left[\frac{y_1^2}{2}\right]}_0^{\frac{1}{2}}\right]\\ =10\left[\frac{1}{16}\left[\frac{1}{4}-0\right]-\frac{1}{5}\left[\frac{1}{32}-0\right]+\frac{3}{4}\left[\frac{1}{16}-0\right]-\frac{3}{3}\left[\frac{1}{8}-0\right]+\frac{1}{2}\left[\frac{1}{4}-0\right]\right]\\ =\frac{9}{16}\end{array}$

Thus, the value of $F\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{9}{16}$.

b.

To determine

Find the value of $F\left(\frac{1}{2},2\right)$.

Answer to Problem 13E

The value of $F\left(\frac{1}{2},2\right)$ is $\frac{13}{16}$.

Explanation of Solution

Calculation:

From Part (a), it is obtained that value of $F\left(\frac{1}{2},\frac{1}{2}\right)$ is $\frac{9}{16}$.

According to the given question using the range of $Y_1\text{and }{Y}_2$, the required probability is obtained as follows:

$\begin{array}{c}F\left(\frac{1}{2},\frac{1}{2}\right)=P\left(Y_1\le \frac{1}{2},Y_2<\frac{1}{2}\right)+P\left(Y_1\le \frac{1}{2},Y_2>\frac{1}{2}\right)\\ =F\left(\frac{1}{2},\frac{1}{2}\right)+P\left(Y_1\le \frac{1}{2},Y_2>\frac{1}{2}\right)\end{array}$

Thus,

$\begin{array}{c}P\left(Y_1\le \frac{1}{2},Y_2>\frac{1}{2}\right)={\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1-y_2}{\int }}30y_1{y}_2^{2}d{y}_2}d{y}_1}\\ ={\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}30y_2^2\left[{\displaystyle \underset{0}{\overset{1-y_2}{\int }}{y}_{1}d{y}_1}\right]d{y}_2}\\ ={\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}30y_2^2{\left[\frac{y_1^2}{2}\right]}_0^{1-y_2}d{y}_2}\\ =15{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_2^2\left[{\left(1-y_2\right)}^2-0\right]d{y}_2}\\ =15{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_2^2\left[1-2y_2+y_2^2\right]d{y}_2}\\ =15\left[{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_2^{2}d{y}_2}-2{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_2^{3}d{y}_2}+{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_2^{4}d{y}_2}\right]\\ =15\left[{\left[\frac{y_2^3}{3}\right]}_{\frac{1}{2}}^1-2{\left[\frac{y_2^4}{4}\right]}_{\frac{1}{2}}^1+{\left[\frac{y_2^5}{5}\right]}_{\frac{1}{2}}^1\right]\\ =15\left[\frac{1}{3}\left[1-\frac{1}{27}\right]-2\left[1-\frac{1}{16}\right]+\frac{1}{5}\left[1-\frac{1}{32}\right]\right]\\ =\frac{1}{4}\end{array}$

Thus,

$\begin{array}{c}F\left(\frac{1}{2},\frac{1}{2}\right)=F\left(\frac{1}{2},\frac{1}{2}\right)+P\left(Y_1\le \frac{1}{2},Y_2>\frac{1}{2}\right)\\ =\frac{9}{16}+\frac{1}{4}\\ =\frac{13}{16}\end{array}$

Thus, the value of $F\left(\frac{1}{2},2\right)$ is $\frac{13}{16}$.

c.

To determine

Find the value of $P\left(Y_1>Y_2\right)$.

Answer to Problem 13E

The value of $P\left(Y_1>Y_2\right)$ is $\frac{21}{32}$.

Explanation of Solution

Calculation:

According to the given question using the range of $Y_1\text{and }{Y}_2$, the required probability is obtained as follows:

$\begin{array}{c}P\left(Y_1>Y_2\right)=1-P\left(Y_1\le Y_2\right)\\ =1-{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{\displaystyle \underset{y_1}{\overset{1-y_1}{\int }}30y_1{y}_2^{2}d{y}_2}d{y}_1}\\ =1-{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}30y_1\left[{\displaystyle \underset{y_1}{\overset{1-y_1}{\int }}{y}_2^{2}d{y}_2}\right]d{y}_1}\\ =1-{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}30y_1{\left[\frac{y_2^3}{3}\right]}_{y_1}^{1-y_1}d{y}_1}\\ =1-10{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_1\left[{\left(1-y_1\right)}^3-y_1^3\right]d{y}_1}\\ =1-10{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_1\left[1-3y_1+3y_1^2-y_1^3-y_1^3\right]d{y}_1}\\ =1-10\left[{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_{1}d{y}_1}-3{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_1^{2}d{y}_1}+3{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_1^{3}d{y}_1}-2{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}{y}_1^{4}d{y}_1}\right]\\ =1-10\left[{\left[\frac{y_1^2}{2}\right]}_{\frac{1}{2}}^1-3{\left[\frac{y_1^3}{3}\right]}_{\frac{1}{2}}^1+3{\left[\frac{y_1^4}{4}\right]}_{\frac{1}{2}}^1-2{\left[\frac{y_1^5}{5}\right]}_{\frac{1}{2}}^1\right]\\ =1-10\left[\frac{1}{2}\left[1-\frac{1}{4}\right]-\left[1-\frac{1}{8}\right]+\frac{3}{4}\left[1-\frac{1}{16}\right]-\frac{2}{5}\left[1-\frac{1}{32}\right]\right]\\ =1-\frac{11}{32}\\ =\frac{21}{32}\end{array}$

Thus, the value of $P\left(Y_1>Y_2\right)$ is $\frac{21}{32}$.