Chapter 5.3, Problem 27E

a.

To determine

Find the marginal density functions for $Y_1$ and $Y_2$.

Answer to Problem 27E

The marginal density function for $Y_1$ is $f_1\left(y_1\right)=3{\left(1-y_1\right)}^2,\text{for }0\le y_1\le 1$ and the marginal density function for $Y_2$ is $f_2\left(y_2\right)=6y_2\left(1-y_2\right),0\le y_2\le 1$.

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 $f_2\left(y_2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y_1,y_2\right)d{y}_1}$.

The range of $Y_1$ and $Y_2$ is given as $0\le y_1\le y_2\le 1$. Hence, the range of $Y_1$ is $0\le y_1\le y_2$ and the range of $Y_2$ is $y_1\le y_2\le 1$.

Hence, the marginal probability density function for $Y_1$ is calculated below:

$\begin{array}{c}{f}_1\left(y_1\right)={\displaystyle \underset{y_1}{\overset{1}{\int }}6\left(1-y_2\right)d{y}_2}\\ =6{\left[\frac{{\left(1-y_2\right)}^2}{2}\right]}_1^{y_1}\\ =3\left[{\left(1-y_1\right)}^2-{\left(1-1\right)}^2\right]\\ =3{\left(1-y_1\right)}^2\end{array}$

Thus, the marginal density function for $Y_1$ is $f_1\left(y_1\right)=3{\left(1-y_1\right)}^2,\text{for }0\le y_1\le 1$.

In similar way, the marginal probability density function for $Y_2$ is calculated below:

$\begin{array}{c}{f}_2\left(y_2\right)={\displaystyle \underset{0}{\overset{y_2}{\int }}6\left(1-y_2\right)d{y}_1}\\ =6\left(1-y_2\right){\left[y_1\right]}_0^{y_2}\\ =6\left(1-y_2\right)\left[y_2-0\right]\\ =6y_2\left(1-y_2\right)\end{array}$.

Thus, the marginal density function for $Y_2$ is $f_2\left(y_2\right)=6y_2\left(1-y_2\right),0\le y_2\le 1$.

b.

To determine

Find the value of $P\left(Y_2\le \frac{1}{2}|Y_1\le \frac{3}{4}\right)$.

Answer to Problem 27E

The value of $P\left(Y_2\le \frac{1}{2}|Y_1\le \frac{3}{4}\right)$ is $\frac{32}{63}$.

Explanation of Solution

Conditional distribution and density function:

Consider that $Y_1$ and $Y_2$ are two discrete real valued random variables with joint probability mass function of $p\left(y_1,y_2\right)$. In addition, the marginal densities of $Y_1$ and $Y_2$ are $f_1\left(y_1\right)$ and $f_2\left(y_2\right)$, respectively.

Now, the conditional distribution function of $Y_1$ given $Y_2=y_2$ is obtained as,

$F\left(y_1|y_2\right)=P\left(Y_1\le y_1|Y_2=y_2\right)$.

Now, for any $y_2$ the conditional density of $Y_1$ given $Y_2=y_2$ is given as,

$f\left(y_1|y_2\right)=\frac{f\left(y_1,y_2\right)}{f_2\left(y_2\right)}$, where $f_2\left(y_2\right)>0$.

Similarly, for any $y_1$ the conditional density of $Y_2$ given $Y_1=y_1$ is given as,

$f\left(y_2|y_1\right)=\frac{f\left(y_1,y_2\right)}{f_1\left(y_1\right)}$, where $f_1\left(y_1\right)>0$.

Hence,

$\begin{array}{c}P\left(Y_2\le \frac{1}{2}|Y_1\le \frac{3}{4}\right)=\frac{{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{\displaystyle \underset{0}{\overset{y_2}{\int }}6\left(1-y_2\right)d{y}_{1}d{y}_2}}}{{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}3{\left(1-y_1\right)}^{2}d{y}_1}}\\ =\frac{6{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\left(1-y_2\right)\left[{\displaystyle \underset{0}{\overset{y_2}{\int }}d{y}_1}\right]d{y}_2}}{3{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(1-2y_1+y_1^2\right)d{y}_1}}\\ =\frac{6{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\left(1-y_2\right)y_{2}d{y}_2}}{3{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(1-2y_1+y_1^2\right)d{y}_1}}\\ =2\frac{{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_{2}d{y}_2-{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_2^{2}d{y}_2}}}{{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}d{y}_1-2{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}{y}_{1}d{y}_1}+{\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}{y}_1^{2}d{y}_1}}}\\ =2\frac{{\left[\frac{y_2^2}{2}\right]}_0^{\frac{1}{2}}-{\left[\frac{y_2^3}{3}\right]}_0^{\frac{1}{2}}}{{\left[y\right]}_0^{\frac{3}{4}}-2{\left[\frac{y_1^2}{2}\right]}_0^{\frac{3}{4}}+{\left[\frac{y_1^3}{3}\right]}_0^{\frac{3}{4}}}\\ =2\frac{\frac{1}{2}\left[\frac{1}{2^2}-0\right]-\frac{1}{3}\left[\frac{1}{2^3}-0\right]}{\left[\frac{3}{4}-0\right]-\left[\frac{9}{16}-0\right]+\frac{1}{3}\left[\frac{27}{64}-0\right]}\\ =2\frac{\frac{1}{8}-\frac{1}{24}}{\frac{3}{4}-\frac{9}{16}+\frac{9}{64}}\\ =2\frac{\frac{1}{12}}{\frac{21}{64}}\\ =\frac{\left(2\right)\left(64\right)}{\left(21\right)\left(12\right)}\\ =\frac{32}{63}\end{array}$

Thus, value of $P\left(Y_2\le \frac{1}{2}|Y_1\le \frac{3}{4}\right)$ is $\frac{32}{63}$.

c.

To determine

Find the conditional density function of $Y_1$ given $Y_2=y_2$.

Answer to Problem 27E

The conditional density function of $Y_1$ given $Y_2=y_2$ is $f\left(y_1|y_2\right)=\frac{1}{y_2},0\le y_1\le y_2\le 1$.

Explanation of Solution

Calculation:

From Part (a), the marginal density function for $Y_2$ is $f_2\left(y_2\right)=6y_2\left(1-y_2\right),0\le y_2\le 1$.

Hence, using the joint probability density function of $Y_1$ and $Y_2$ and the marginal density function of $Y_2$, the conditional density function of $Y_1$ given $Y_2=y_2$ is obtained below:

$\begin{array}{c}f\left(y_1|y_2\right)=\frac{6\left(1-y_2\right)}{6y_2\left(1-y_2\right)}\\ =\frac{1}{y_2}\end{array}$

Thus, the conditional density function of $Y_1$ given $Y_2=y_2$ is $f\left(y_1|y_2\right)=\frac{1}{y_2},0\le y_1\le y_2\le 1$.

d.

To determine

Find the conditional density function of $Y_2$ given $Y_1=y_1$.

Answer to Problem 27E

The conditional density function of $Y_2$ given $Y_1=y_1$ is $f\left(y_2|y_1\right)=\frac{2\left(1-y_2\right)}{{\left(1-y_1\right)}^2},0\le y_1\le y_2\le 1$.

Explanation of Solution

Calculation:

From Part (a), the marginal density function for $Y_1$ is $f_1\left(y_1\right)=3{\left(1-y_1\right)}^2,\text{for }0\le y_1\le 1$.

Hence, using the joint probability density function of $Y_1$ and $Y_2$ and the marginal density function of $Y_1$, the conditional density function of $Y_2$ given $Y_1=y_1$ is obtained below:

$\begin{array}{c}f\left(y_2|y_1\right)=\frac{6\left(1-y_2\right)}{3{\left(1-y_1\right)}^2}\\ =\frac{2\left(1-y_2\right)}{{\left(1-y_1\right)}^2}\end{array}$

Thus, the conditional density function of $Y_2$ given $Y_1=y_1$ is $f\left(y_2|y_1\right)=\frac{2\left(1-y_2\right)}{{\left(1-y_1\right)}^2},0\le y_1\le y_2\le 1$.

e.

To determine

Find the value of $P\left(Y_2\ge \frac{3}{4}|Y_1=\frac{1}{2}\right)$.

Answer to Problem 27E

The value of $P\left(Y_2\ge \frac{3}{4}|Y_1=\frac{1}{2}\right)$ is $\frac{1}{4}$.

Explanation of Solution

Using the joint probability density function of $Y_1$ and $Y_2$ and the marginal density function of $Y_1$, the required probability is obtained below:

Hence,

$\begin{array}{c}P\left(Y_2\ge \frac{3}{4}|Y_1=\frac{1}{2}\right)=\frac{{\displaystyle \underset{\frac{3}{4}}{\overset{1}{\int }}6\left(1-y_2\right)d{y}_2}}{3{\left(1-\frac{1}{2}\right)}^2}\\ =8{\displaystyle \underset{\frac{3}{4}}{\overset{1}{\int }}\left(1-y_2\right)d{y}_2}\\ =8{\displaystyle \underset{\frac{3}{4}}{\overset{1}{\int }}d{y}_2}-8{\displaystyle \underset{\frac{3}{4}}{\overset{1}{\int }}{y}_{2}d{y}_2}\\ =8{\left[y_2\right]}_{\frac{3}{4}}^1-8{\left[\frac{y_2^2}{2}\right]}_{\frac{3}{4}}^1\\ =8\left[1-\frac{3}{4}\right]-4\left[1-\frac{9}{16}\right]\\ =\left(8\right)\frac{1}{4}-\left(4\right)\frac{7}{16}\\ =2-\frac{7}{4}\\ =\frac{8-7}{4}\\ =\frac{1}{4}\end{array}$

Thus, value of $P\left(Y_2\ge \frac{3}{4}|Y_1=\frac{1}{2}\right)$ is $\frac{1}{4}$.