Prove that joint density function of Y 1 and Y 2 is same as given in Exercise 5.65.

Question

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Answer 1

Question

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

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Answer 2

Question

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

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Answer 3

Question

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

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Answer 4

Question

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

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Answer 5

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

Answer 6

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

Answer 7

Chapter 5, Problem 163SE

a.

To determine

Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

Answer 8

a.

Expert Solution

Explanation of Solution

Calculation:

In Exercise 5.65 the joint density is given as follows:

$f(y1,y2)={[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2, y1≥0,y2≥00 , otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow exponential distribution with mean 1 and $−1≤α≤1$ .

It is known that the cumulative density function of $Y1$ is $F1(y1)=1−e−y1$ and the cumulative density function of $Y2$ is $F2(y2)=1−e−y2$ .

Now, substitute $F1(y1)=1−e−y1$ and $F2(y2)=1−e−y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(1−e−y1)(1−e−y2)[1−α{1−(1−e−y2)}{1−(1−e−y2)}]=(1−e−y1)(1−e−y2)[1−αe−y1e−y2]$

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[(1−e−y1)(1−e−y2)[1−αe−y1e−y2]]∂y1∂y2=∂2[(1−e−y1)(1−e−y2)]∂y1∂y2−α∂2[(1−e−y1)(1−e−y2)e−y1e−y2]∂y1∂y2=e−y1e−y2−e−y1e−y2{α(1−2e−y1)(1−2e−y2)}=e−y1−y2−e−y1−y2{α(1−2e−y1)(1−2e−y2)}=[1−α{(1−2e−y1)(1−2e−y1)}]e−y1−y2$

Hence, it is proved that the joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Answer 13

b.

To determine

Evaluate $F(y1,y2)$ for any $α$ where $−1≤α≤1$ .

Answer 14

b.

Expert Solution

Answer to Problem 163SE

The joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Calculation:

Consider that $Y1 and Y2$ follow Uniform distribution over the interval $(0,1)$ .

Hence, the $F1(y1)=y1 and F2(y2)=y2$ .

Now, substitute $F1(y1)=y1 and F2(y2)=y2$ in the given joint cumulative density function.

That is,

$F(y1,y2)=(y1)(y2)[1−α{1−(y1)}{1−(y2)}]=y1y2[1−α(1−y1)(1−y2)]$

Hence, the joint cumulative density function of $Y1 and Y2$ is $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

Answer 19

c.

To determine

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

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Answer 20

c.

To determine

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

Answer 21

c.

Expert Solution

Answer to Problem 163SE

The joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Answer 22

c.

Expert Solution

Answer 23

c.

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

From Part (b), the joint cumulative density function of $Y1 and Y2$ is obtained as $F(y1,y2)=y1y2[1−α(1−y1)(1−y2)], −1≤α≤1$ .

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

$f(y1,y2)=∂2F(y1,y2)∂y1∂y2=∂2[y1y2[1−α(1−y1)(1−y2)]]∂y1∂y2=∂2(y1y2)∂y1∂y2−α∂2[y1y2(1−y1)(1−y2)]∂y1∂y2=1−α∂2[(y1−y12)(y2−y22)]∂y1∂y2=1−α(1−2y1)(1−2y2)$

Thus, the joint density function is,

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ .

Answer 26

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

Answer 27

d.

To determine

Provide two specific and different joint densities that yield marginal densities for $Y1 and Y2$ both of which are Uniform over the interval $(0,1)$ .

Answer 28

d.

Expert Solution

Explanation of Solution

Calculation:

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

$f(y1,y2)={1−α(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$ ,

Where marginal densities of $Y1 and Y2$ follow Uniform over the interval $(0,1)$ .

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

Thus, as $α∈(0,1)$ , one can take any two values of $α$ within that range to obtain two different densities.

Consider $α=−0.5$ and substitute $α=−0.5$ in the given probability density function.

$f(y1,y2)={1+0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

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

$f(y1,y2)={1−0.5(1−2y1)(1−2y2), 0≤y1≤1,0≤y2≤10, Otherwise$

Hence, these two specific and different joint densities yield marginal densities for $Y1 and Y2$ that are both Uniform over the interval $(0,1)$ .

Answer 29

d.

Expert Solution

Answer 30

d.

Expert Solution

Answer 31

Expert Solution

Answer 32

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Prove that joint density function of Y 1 and Y 2 is same as given in Exercise 5.65.

Prove that joint density function of Y1 and Y2<m:math><m:mrow><m:msub><m:mi>Y</m:mi><m:mn>1</m:mn></m:msub><m:mtext> </m:mtext><m:mtext>and </m:mtext><m:msub><m:mi>Y</m:mi><m:mn>2</m:mn></m:msub></m:mrow></m:math> is same as given in Exercise 5.65.

Explanation of Solution

Answer to Problem 163SE

Explanation of Solution

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

Answer to Problem 163SE

Explanation of Solution

Explanation of Solution

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Prove that joint density function of $Y1 and Y2$ is same as given in Exercise 5.65.