We determined that f(y1, y2) = 6(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere is a valid joint probability density function. (a) Find
E(Y1|Y2 = y2). E(Y1|Y2 = y2) = (b) Use the answer derived in part (a) to find E(Y1). (Compare this with E(Y1) found using the marginal density function f1(y1) = 3(1 − y1)2.) E(Y1) =
1 answer:
You might be interested in