In Exercise 5.8, we established that f (y1, y2) = 4y1 y2, 0 y1 1, 0 y2 1, 0, elsewhere is a valid joint probability density function. In Exercise 5.52, we established that Y1 and Y2 are independent; in Exercise 5.76, we determined that E(Y1 Y2) = 0 and found the value for V(Y1). Find V(Y1 Y2).

Chapter 2 Notes • Deviation—Given a value y and a data point x then x-‐y represents how far x deviates from y. o Ex. Given x=10 and y=5. 10 deviates by 5 from 5. • Bar Charts—Display where the length of a bar corresponds...