Chapter 5.4, Problem 65E

a.

To determine

Prove that the marginal distribution of $Y_1$ is exponential with mean 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 $f_2\left(y_2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y_1,y_2\right)d{y}_1}$.

Exponential distribution:

A random variable Y is said to follow the exponential distribution with mean 1, if and only if the probability density function of Y is $f\left(y\right)=e^{-y},y>0$.

Consider that $C_1=1-2e^{-y_1}$.

Thus, using the joint probability density functions of $Y_1$ and $Y_2$, the marginal probability of $Y_1$ is calculated below:

$\begin{array}{c}{f}_1\left(y_1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}\left[1-\alpha \left\{C_1\left(1-2e^{-y_2}\right)\right\}\right]}{e}^{-y_1-y_2}d{y}_2\\ =e^{-y_1}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y_2}d{y}_2}-\alpha C_1{e}^{-y_1}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(1-2e^{-y_2}\right)}{e}^{-y_2}d{y}_2\\ =e^{-y_1}{\left[-e^{-y_2}\right]}_0^{\infty }-\alpha C_1{e}^{-y_1}\left[{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y_2}d{y}_2}-2{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2y_2}d{y}_2}\right]\\ =e^{-y_1}\left[0+1\right]-\alpha C_1{e}^{-y_1}\left[{\left[-e^{-y_2}\right]}_0^{\infty }-{\left[-e^{-2y_2}\right]}_0^{\infty }\right]\\ =e^{-y_1}-\alpha C_1{e}^{-y_1}\left[\left[0+1\right]-\left[0+1\right]\right]\\ =e^{-y_1}-\alpha C_1{e}^{-y_1}\left[0\right]\\ =e^{-y_1}\end{array}$

Thus, it is proved that the marginal distribution of $Y_1$ is exponential with mean 1.

b.

To determine

Find the marginal distribution of $Y_2$.

Answer to Problem 65E

The marginal distribution of $Y_2$ is Exponential with mean 1.

Explanation of Solution

Calculation:

Consider that $C_2=1-2e^{-y_2}$.

Thus, using the joint probability density functions of $Y_1$ and $Y_2$, the marginal probability of $Y_2$ is calculated below:

$\begin{array}{c}{f}_2\left(y_2\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}\left[1-\alpha \left\{\left(1-2e^{-y_1}\right)C_2\right\}\right]}{e}^{-y_1-y_2}d{y}_1\\ =e^{-y_2}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y_1}d{y}_1}-\alpha C_2{e}^{-y_2}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(1-2e^{-y_1}\right)}{e}^{-y_1}d{y}_1\\ =e^{-y_2}{\left[e^{-y_1}\right]}_{\infty }^0-\alpha C_2{e}^{-y_2}\left[{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-y_1}d{y}_1}-2{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2y_1}d{y}_1}\right]\\ =e^{-y_2}\left[1-0\right]-\alpha C_2{e}^{-y_2}\left[{\left[-e^{-y_1}\right]}_0^{\infty }-{\left[-e^{-2y_1}\right]}_0^{\infty }\right]\\ =e^{-y_2}-\alpha C_2{e}^{-y_2}\left[\left[0+1\right]-\left[0+1\right]\right]\\ =e^{-y_2}-\alpha C_2{e}^{-y_2}\left[0\right]\\ =e^{-y_2}\end{array}$

Thus, the marginal distribution of $Y_2$ is Exponential with mean 1.

c.

To determine

Prove that $Y_1$ and $Y_2$ are independent if and only if $\alpha =0$.

Explanation of Solution

The two continuous random variables $Y_1$ and $Y_2$ are said to be independent if and only if $f\left(y_1,y_2\right)=f_1\left(y_1\right)f_2\left(y_2\right)$, where $f\left(y_1,y_2\right)$ is the joint probability density function of $Y_1$ and $Y_2$ and $f_1\left(y_1\right),f_2\left(y_2\right)$ are the marginal probability density functions of $Y_1$ and $Y_2$, respectively.

Substitute $\alpha =0$ in the joint probability distribution.

That is,

$\begin{array}{c}f\left(y_1,y_2\right)=\left[1-\left(0\right)\left\{\left(1-2e^{-y_1}\right)\left(1-2e^{-y_2}\right)\right\}\right]e^{-y_1-y_2}\\ =\left[1-0\right]e^{-y_1-y_2}\\ =e^{-y_1-y_2}\end{array}$

From Part (a) and Part (b), the marginal distribution of $Y_1$ is Exponential with mean 1 and the marginal distribution of $Y_2$ is Exponential with mean 1.

Here, the ranges of $Y_1$ and $Y_2$ are not dependent on each other and according to the rule of independence the probability of joint distribution is equal to the product of two marginal distributions if and only if $\alpha =0$.

That is,

$\begin{array}{c}{f}_1\left(y_1\right)f_2\left(y_2\right)=\left(e^{-y_1}\right)\left(e^{-y_2}\right)\\ =e^{-y_1-y_2}\\ =f\left(y_1,y_2\right)\end{array}$

Thus, it is proved that $Y_1$ and $Y_2$ are independent if and only if $\alpha =0$.