Chapter 5.3, Problem 25E

a.

To determine

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

Identify the densities.

Answer to Problem 25E

The marginal density function for $Y_1$ is $f_1\left(y_1\right)=e^{-y_1},y_1>0$ and the marginal density function for $Y_2$ is $f_2\left(y_2\right)=e^{-y_2},y_2>0$.

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

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

$\begin{array}{c}{f}_1\left(y_1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\left(y_1+y_2\right)}d{y}_2}\\ =e^{-y_1}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y_2}d{y}_2}\\ =e^{-y_1}{\left[e^{-y_2}\right]}_{\infty }^0\\ =e^{-y_1}\left[e^0-e^{-\infty }\right]\\ =e^{-y_1}\end{array}$

Thus, the marginal density function for $Y_1$ is $f_1\left(y_1\right)=e^{-y_1},y_1>0$.

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

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

Thus, the marginal density function for $Y_2$ is $f_2\left(y_2\right)=e^{-y_2},y_2>0$.

Gamma function:

A random variable Y is said to follow a Gamma distribution with parameter $\alpha \left(>0\right)$ and $\beta \left(>0\right)$, if and only if the density function of Y is,

$f\left(y\right)=\{\begin{array}{l}\frac{e^{-\frac{y}{\beta }}{y}^{\alpha -1}}{{\beta }^a\Gamma \left(\alpha \right)},0\le y<\infty \\ 0\text{elsewhere}\end{array}$ ,

Where $\Gamma \alpha ={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y}{y}^{\alpha -1}dy}$.

Hence, comparing with the probability density function of Gamma function the marginal density function for $Y_1$ follows Gamma distribution with parameters $\alpha =1$ and $\beta =1$. In similar way, the probability density function of $Y_2$ is Gamma distribution with parameters $\alpha =1$ and $\beta =1$.

b.

To determine

Find the values of $P\left(1<Y_1<2.5\right)$ and $P\left(1<Y_2<2.5\right)$.

Answer to Problem 25E

The value of both $P\left(1<Y_1<2.5\right)$ and $P\left(1<Y_2<2.5\right)$ is 0.2858.

Explanation of Solution

From Part (a), marginal density function for $Y_1$ is obtained as $f_1\left(y_1\right)=e^{-y_1},y_1>0$.

Hence,

$\begin{array}{c}P\left(1<Y_1<2.5\right)={\displaystyle \underset{1}{\overset{2.5}{\int }}{e}^{-y_1}d{y}_1}\\ ={\left[e^{-y_1}\right]}_{2.5}^1\\ =\left[e^{-1}-e^{-2.5}\right]\\ \approx 0.2858\end{array}$

Thus, $P\left(1<Y_1<2.5\right)=0.2858$.

From Part (b), marginal density function for $Y_2$ is obtained as $f_2\left(y_2\right)=e^{-y_2},y_2>0$.

Hence,

$\begin{array}{c}P\left(1<Y_2<2.5\right)={\displaystyle \underset{1}{\overset{2.5}{\int }}{e}^{-y_2}d{y}_2}\\ ={\left[e^{-y_2}\right]}_{2.5}^1\\ =\left[e^{-1}-e^{-2.5}\right]\\ \approx 0.2858\end{array}$

Thus, $P\left(1<Y_2<2.5\right)=0.2858$.

c.

To determine

Find the values of $y_2$, for which the conditional density $f\left(y_1|y_2\right)$ is valid.

Answer to Problem 25E

The values of $y_2$ for which the conditional density $f\left(y_1|y_2\right)$ is defined must be $y_2>0$.

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

In the given problem the range of both $Y_1\text{and }{Y}_2$ is $0<y_1$ and $0<y_2$.

Hence, the values of $y_2$ for which the conditional density $f\left(y_1|y_2\right)$ is defined must be $y_2>0$.

d.

To determine

Find the conditional density function of $Y_1$ given $Y_2=y_2$ for any $y_2$, such that, $y_2>0$.

Answer to Problem 25E

The conditional density function of $Y_1$ given $Y_2=y_2$ for any $y_2$ is $f\left(y_1|y_2\right)=e^{-y_1},y_1>0$.

Explanation of Solution

Calculation:

Using the joint probability density function of $Y_1$ and $Y_2$ and the marginal probability function of $Y_1$, the condition probability density function $f\left(y_1|y_2\right)$ is defined as,

$\begin{array}{c}f\left(y_1|y_2\right)=\frac{f\left(y_1,y_2\right)}{f_2\left(y_2\right)}\\ =\frac{e^{-\left(y_1+y_2\right)}}{e^{-\left(y_2\right)}}\\ =e^{-y_1}\\ =f_1\left(y_1\right)\end{array}$.

Hence, the conditional density function of $Y_1$ given $Y_2=y_2$ for any $y_2$ is $f\left(y_1|y_2\right)=e^{-y_1},y_1>0$.

e.

To determine

Find the conditional density function of $Y_2$ given $Y_1=y_1$ for any $y_1$, such that, $y_1>0$.

Answer to Problem 25E

The conditional density function of $Y_2$ given $Y_1=y_1$ for any $y_1$ is $f\left(y_2|y_1\right)=e^{-y_2},y_2>0$.

Explanation of Solution

Calculation:

Using the joint probability density function of $Y_1$ and $Y_2$ and the marginal probability function of $Y_1$, the condition probability density function $f\left(y_2|y_1\right)$ is defined as,

$\begin{array}{c}f\left(y_2|y_1\right)=\frac{f\left(y_1,y_2\right)}{f_1\left(y_1\right)}\\ =\frac{e^{-\left(y_1+y_2\right)}}{e^{-\left(y_1\right)}}\\ =e^{-y_2}\\ =f_2\left(y_2\right)\end{array}$.

Hence, the conditional density function of $Y_2$ given $Y_1=y_1$ for any $y_1$ is $f\left(y_2|y_1\right)=e^{-y_2},y_2>0$.

f.

To determine

Compare the marginal density function $f_1\left(y_1\right)$ obtained in Part (a) with the conditional density function $f\left(y_1|y_2\right)$ obtained in Part (d).

Explanation of Solution

From Part (d), conditional density function of $Y_1$ given $Y_2=y_2$ for any $y_2$ is $f\left(y_1|y_2\right)=e^{-y_1},y_1>0$.

From Part (a), the marginal density function for $Y_1$ is $f_1\left(y_1\right)=e^{-y_1},y_1>0$.

Hence, $f\left(y_1|y_2\right)=f_1\left(y_1\right)$.

Thus, it can be said that $Y_1$ and $Y_2$ are stochastically independent.

g.

To determine

Explain the meaning of the answer in Part (f) regarding the implication of the marginal and conditional probabilities that $Y_1$ falls in any interval.

Explanation of Solution

From Part (f), it is obtained that $Y_1$ and $Y_2$ are stochastically independent and $f\left(y_1|y_2\right)=f_1\left(y_1\right)$.

Hence, it can be said the marginal probability of $Y_1$ and the conditional probability of $Y_1$ given any values of $Y_2$ are same. In addition, the marginal and conditional probabilities that $Y_1$ falls in any interval are same.