Chapter 5, Problem 149SE

a.

To determine

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

Answer to Problem 149SE

The marginal probability density function of $Y_1$ is $f_1\left(y_1\right)=3y_1^2,0\le y_1\le 1$ and the marginal density function for $Y_2$ is $f_2\left(y_2\right)=\frac{3}{2}-\frac{3}{2}{y}_2^2,0\le y_2\le 1$.

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

Thus, marginal probability density function of $Y_1$ is obtained as,

$\begin{array}{c}{f}_1\left(y_1\right)=3y_1{\displaystyle \underset{0}{\overset{y_1}{\int }}d{y}_2}\\ =3y_1{\left[y_2\right]}_0^{y_1}\\ =3y_1^2\end{array}$

Thus, the marginal probability density function of $Y_1$ is $f_1\left(y_1\right)=3y_1^2,0\le y_1\le 1$.

Similarly, marginal probability density function of $Y_2$ is obtained as,

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

Thus, the marginal density function for $Y_2$ is $f_2\left(y_2\right)=\frac{3}{2}-\frac{3}{2}{y}_2^2,0\le y_2\le 1$.

b.

To determine

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

Answer to Problem 149SE

The value of $P\left(Y_1\le \frac{3}{4}|Y_2\le \frac{1}{2}\right)$ is $\frac{23}{44}$.

Explanation of Solution

Calculation:

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 $f_1\left(y_1\right)>0$.

From Part (a), it is found that the marginal probability density function of $Y_1$ is $f_1\left(y_1\right)=3y_1^2,0\le y_1\le 1$ and the marginal density function for $Y_2$ is $f_2\left(y_2\right)=\frac{3}{2}-\frac{3}{2}{y}_2^2,0\le y_2\le 1$.

Hence, using the marginal probability functions the required probability is obtained as,

$\begin{array}{c}P\left(Y_1\le \frac{3}{4}|Y_2\le \frac{1}{2}\right)=\frac{{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{\displaystyle \underset{y_2}{\overset{\frac{3}{4}}{\int }}3y_{1}d{y}_{1}d{y}_2}}}{{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\left(\frac{3}{2}-\frac{3}{2}{y}_2^2\right)d{y}_2}}\\ =\frac{\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{\left[y_1^2\right]}_{y_2}^{\frac{3}{4}}d{y}_2}}{\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}d{y}_2}-\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_2^{2}d{y}_2}}\\ =\frac{\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\left(\frac{9}{16}-y_2^2\right)d{y}_2}}{\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}d{y}_2}-\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_2^{2}d{y}_2}}\\ =\frac{\frac{27}{32}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}d{y}_2-\frac{3}{2}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}{y}_2^{2}d{y}_2}}}{\frac{3}{2}{\left[y_2\right]}_0^{\frac{1}{2}}-\frac{1}{2}{\left[y_2^3\right]}_0^{\frac{1}{2}}}\\ =\frac{\frac{27}{32}{\left[y_2\right]}_0^{\frac{1}{2}}-\frac{1}{2}{\left[y_2^3\right]}_0^{\frac{1}{2}}}{\frac{3}{2}{\left[y_2\right]}_0^{\frac{1}{2}}-\frac{1}{2}{\left[y_2^3\right]}_0^{\frac{1}{2}}}\\ =\frac{\frac{27}{64}-\frac{1}{16}}{\frac{3}{4}-\frac{1}{16}}\\ =\frac{\frac{27-4}{64}}{\frac{12-1}{16}}\\ =\frac{23}{44}\end{array}$

Thus, the value of $P\left(Y_1\le \frac{3}{4}|Y_2\le \frac{1}{2}\right)$ is $\frac{23}{44}$.

c.

To determine

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

Answer to Problem 149SE

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

Explanation of Solution

Using the joint probability function of $Y_1$ and $Y_2$ and the marginal probability function of $Y_2$, the conditional density $Y_1$ given $Y_2=y_2$ is obtained as,

$\begin{array}{c}f\left(y_1|y_2\right)=\frac{3y_1}{\frac{3}{2}-\frac{3}{2}{y}_2^2}\\ =\frac{2y_1}{1-y_1^2}\end{array}$

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

d.

To determine

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

Answer to Problem 149SE

The value of $P\left(Y_1\le \frac{3}{4}|Y_2=\frac{1}{2}\right)$ is $\frac{5}{12}$.

Explanation of Solution

Calculation:

From Part (C), it is obtained that the conditional density $Y_1$ given $Y_2=y_2$ is $f\left(y_1|y_2\right)=\frac{2y_1}{1-y_1^2}$ for all $y_2\le y_1\le 1$ and $y_2\ge 0$.

Hence, the required probability is obtained as,

$\begin{array}{c}P\left(Y_1\le \frac{3}{4}|Y_2=\frac{1}{2}\right)={\displaystyle \underset{y_2}{\overset{\frac{3}{4}}{\int }}\frac{2y_1}{1-y_2^2}}d{y}_1\\ ={\frac{1}{1-y_2^2}{\left[y_1^2\right]}_{y_2}^{\frac{3}{4}}|}_{y_2=\frac{1}{2}}\\ ={\frac{1}{1-y_2^2}\left(\frac{9}{16}-y_2^2\right)|}_{y_2=\frac{1}{2}}\\ =\frac{4}{3}\left(\frac{9}{16}-\frac{1}{4}\right)\\ =\frac{5}{12}\end{array}$

Thus, the value of $P\left(Y_1\le \frac{3}{4}|Y_2=\frac{1}{2}\right)$ is $\frac{5}{12}$.