Solution: In Exercise 5.9, we determined that f (y1, y2) = 6(1 y2), 0 y1 y2 1, 0
Chapter 5, Problem 5.106(choose chapter or problem)
In Exercise 5.9, we determined that f (y1, y2) = 6(1 y2), 0 y1 y2 1, 0, elsewhere is a valid joint probability density function. In Exercise 5.76, we derived the fact that E(Y13Y2) = 5/4; in Exercise 5.92, we proved that Cov(Y1, Y2) = 1/40. Find V(Y13Y2).
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