Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y_1\text{and }{Y}_2$ is same as given in Exercise 5.65.

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

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

Where marginal densities of $Y_1\text{and }{Y}_2$ follow exponential distribution with mean 1 and $-1\le \alpha \le 1$.

It is known that the cumulative density function of $Y_1$ is $F_1\left(y_1\right)=1-e^{-y_1}$ and the cumulative density function of $Y_2$ is $F_2\left(y_2\right)=1-e^{-y_2}$.

Now, substitute $F_1\left(y_1\right)=1-e^{-y_1}$ and $F_2\left(y_2\right)=1-e^{-y_2}$ in the given joint cumulative density function.

That is,

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

Hence, the joint probability density function is obtained as follows:

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

Hence, it is proved that the joint density function of $Y_1\text{and }{Y}_2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F\left(y_1,y_2\right)$ for any $\alpha$ where $-1\le \alpha \le 1$.

Answer to Problem 163SE

The joint cumulative density function of $Y_1\text{and }{Y}_2$ is $F\left(y_1,y_2\right)=y_1{y}_2\left[1-\alpha \left(1-y_1\right)\left(1-y_2\right)\right],-1\le \alpha \le 1$.

Explanation of Solution

Calculation:

Consider that $Y_1\text{and }{Y}_2$ follow Uniform distribution over the interval $\left(0,1\right)$.

Hence, the $F_1\left(y_1\right)=y_1\text{and }{F}_2\left(y_2\right)=y_2$.

Now, substitute $F_1\left(y_1\right)=y_1\text{and }{F}_2\left(y_2\right)=y_2$ in the given joint cumulative density function.

That is,

$\begin{array}{c}F\left(y_1,y_2\right)=\left(y_1\right)\left(y_2\right)\left[1-\alpha \left\{1-\left(y_1\right)\right\}\left\{1-\left(y_2\right)\right\}\right]\\ =y_1{y}_2\left[1-\alpha \left(1-y_1\right)\left(1-y_2\right)\right]\end{array}$

Hence, the joint cumulative density function of $Y_1\text{and }{Y}_2$ is $F\left(y_1,y_2\right)=y_1{y}_2\left[1-\alpha \left(1-y_1\right)\left(1-y_2\right)\right],-1\le \alpha \le 1$.

c.

To determine

Obtain the joint density function associated with the distribution function that is obtained in Part (b).

Answer to Problem 163SE

The joint density function is,

$f\left(y_1,y_2\right)=\{\begin{array}{l}1-\alpha \left(1-2y_1\right)\left(1-2y_2\right),0\le y_1\le 1,0\le y_2\le 1\\ 0\text{,}\text{Otherwise}\end{array}$.

Explanation of Solution

From Part (b), the joint cumulative density function of $Y_1\text{and }{Y}_2$ is obtained as $F\left(y_1,y_2\right)=y_1{y}_2\left[1-\alpha \left(1-y_1\right)\left(1-y_2\right)\right],-1\le \alpha \le 1$.

Hence, the joint probability density function is obtained as follows:

$\begin{array}{c}f\left(y_1,y_2\right)=\frac{{\partial }^{2}F\left(y_1,y_2\right)}{\partial y_1\partial y_2}\\ =\frac{{\partial }^2\left[y_1{y}_2\left[1-\alpha \left(1-y_1\right)\left(1-y_2\right)\right]\right]}{\partial y_1\partial y_2}\\ =\frac{{\partial }^2\left(y_1{y}_2\right)}{\partial y_1\partial y_2}-\alpha \frac{{\partial }^2\left[y_1{y}_2\left(1-y_1\right)\left(1-y_2\right)\right]}{\partial y_1\partial y_2}\\ =1-\alpha \frac{{\partial }^2\left[\left(y_1-y_1^2\right)\left(y_2-y_2^2\right)\right]}{\partial y_1\partial y_2}\\ =1-\alpha \left(1-2y_1\right)\left(1-2y_2\right)\end{array}$

Thus, the joint density function is,

$f\left(y_1,y_2\right)=\{\begin{array}{l}1-\alpha \left(1-2y_1\right)\left(1-2y_2\right),0\le y_1\le 1,0\le y_2\le 1\\ 0\text{,}\text{Otherwise}\end{array}$.

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y_1\text{and }{Y}_2$ both of which are Uniform over the interval $\left(0,1\right)$.

Explanation of Solution

Calculation:

From Part (c), the density function is obtained as follows:

$f\left(y_1,y_2\right)=\{\begin{array}{l}1-\alpha \left(1-2y_1\right)\left(1-2y_2\right),0\le y_1\le 1,0\le y_2\le 1\\ 0\text{,}\text{Otherwise}\end{array}$,

Where marginal densities of $Y_1\text{and }{Y}_2$ follow Uniform over the interval $\left(0,1\right)$.

The marginal density function does not depend on the values of $\alpha$.

Thus, as $\alpha \in \left(0,1\right)$, one can take any two values of $\alpha$ within that range to obtain two different densities.

Consider $\alpha =-0.5$ and substitute $\alpha =-0.5$ in the given probability density function.

$f\left(y_1,y_2\right)=\{\begin{array}{l}1+0.5\left(1-2y_1\right)\left(1-2y_2\right),0\le y_1\le 1,0\le y_2\le 1\\ 0\text{,}\text{Otherwise}\end{array}$

Consider $\alpha =0.5$ and substitute $\alpha =0.5$ in the given probability density function.

$f\left(y_1,y_2\right)=\{\begin{array}{l}1-0.5\left(1-2y_1\right)\left(1-2y_2\right),0\le y_1\le 1,0\le y_2\le 1\\ 0\text{,}\text{Otherwise}\end{array}$

Hence, these two specific and different joint densities yield marginal densities for $Y_1\text{and }{Y}_2$ that are both Uniform over the interval $\left(0,1\right)$.