Mathematics, 25.04.2020 04:22, mrsburrus
We determined that f(y1, y2) = 6(1 β y2), 0 β€ y1 β€ y2 β€ 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 β y1)2.) E(Y1) =
Answers: 1
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Graph the equationy=x^2 -[tex]y = x^{2} - 2[/tex]
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We determined that f(y1, y2) = 6(1 β y2), 0 β€ y1 β€ y2 β€ 1, 0, elsewhere is a valid joint probability...
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