Chapter 5.8, Problem 106E

In Exercise 5.9, we determined that

$f\text{}\left(y_1,y_2\right)=\{\begin{array}{l}6\left(1-y_2\right),0\le y_1\le y_2\le 1,\\ 0,elsewhere\end{array}$

is a valid joint probability density function. In Exercise 5.76, we derived the fact that E (Y₁ – 3Y₂) = –5/4; in Exercise 5.92, we proved that Cov (Y₁, Y₂) = 1/40. Find V (Y₁ − 3Y₂).

5.9 Let Y₁ and Y₂ have the joint probability density function given by

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

1. a Find the value of k that makes this a probability density function.
2. b Find P (Y₁ ≤ 3/4, Y₂ ≥ 1/2).

5.76 In Exercise 5.8, we derived the fact that

$f\text{}\left(y_1,y_2\right)=\{\begin{array}{l}4y_1{y}_2,0\le y_1\le 1,0\le y_2\le 1,\\ 0,elsewhere.\end{array}$

1. a Find E (Y₁).
2. b Find V (Y₁).
3. c Find E (Y₁ - Y₂).

5.92 In Exercise 5.9, we determined that

$f\text{}\left(y_1,y_2\right)=\{\begin{array}{l}6\left(1-y_2\right),0\le y_1\le y_2\le 1,\\ 0,elsewhere\end{array}$

is a valid joint probability density function. Find Cov (Y_1, Y₂). Are Y₁ and Y₂ independent?