By Olive D.J.

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Yn are independent normal N(µi , σi2), then n n (ai + bi Yi ) ∼ N( i=1 n b2i σi2). (ai + bi µi ), i=1 i=1 Here ai and bi are fixed constants. , Yn are iid N(µ, σ), then Y ∼ N(µ, σ 2 /n). , Yn are independent Poisson POIS(θi ), then n n Yi ∼ POIS( θi). , Yn are iid P OIS(θ), then n Yi ∼ POIS(nθ). 14: Common problem. 17, find the biYi by finding the mgf or characteristic distribution of W = Yi or W = function of W and recognizing that it comes from a brand name distribution. Tips: a) in the product, anything that does not depend on the product index i is treated as a constant.

MULTIVARIATE DISTRIBUTIONS 41 A hierarchical model models a complicated process by a sequence of models placed in a hierarchy. Interest might be in the marginal expectation E(Y ) and marginal variance VAR(Y ). One could find the joint pmf from f(y, x) = f(y|x)f(x), then find the marginal distribution fY (y) and then find E(Y ) and VAR(Y ). 11. 9. Suppose Y |X ∼ BIN(X, ρ) and X ∼ Poisson (λ). Then E(Y |X) = Xρ, VAR(Y |X) = Xρ(1 − ρ) and E(X) = VAR(X) = λ. Hence E(Y ) = E[E(Y |X)] = E(Xρ) = ρE(X) = ρλ and VAR(Y ) = E[VAR(Y |X)] + VAR[E(Y |X)] = E[Xρ(1 − ρ)] + VAR(Xρ) = λρ(1 − ρ) + ρ2 VAR(X) = λρ(1 − ρ) + ρ2 λ = λρ.

For example, Y2 = X2 may work. Finding the support Y can be difficult, but if the joint pdf of X1 , X2 is g(x1, x2 ) = h(x1 , x2) I[(x1, x2) ∈ X ], then the joint pdf of Y1 , Y2 is −1 −1 −1 f(y1 , y2) = h(t−1 1 (y), t2 (y)) I[(t1 (y), t2 (y)) ∈ X ] |J |, −1 and using I[(t−1 1 (y), t2 (y)) ∈ X ] can be useful for finding Y. Also sketch X with x1 on the horizontal axis and x2 on the vertical axis, and sketch Y with y1 on the horizontal axis and y2 on the vertical axis. CHAPTER 2. 14: the Multivariate Transformation Method.

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