160a09_hw3sol

160a09_hw3sol - Pstat160A Winter 2009 Solution Homework #3...

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Unformatted text preview: Pstat160A Winter 2009 Solution Homework #3 Problem 1. Form discussion session #2, problem 2 we have: G X ( s ) = G X,Y ( s, 1) so G X ( s ) = exp { λ 1 ( s- 1) } (1) G Y ( s ) = G X,Y (1 ,s ) so G Y ( s ) = exp { λ 2 ( s- 1) } (2) G X + Y ( s ) = G X,Y ( s,s ) so G X + Y ( s ) exp { ( λ 1 + λ 2 )( s- 1) } (3) From (1) we recognize that X ∼ P ( λ 1 ), from (2) that Y ∼ P ( λ 2 ), and from (3) that X + Y ∼ P ( λ 1 + λ 2 ). However, X and Y are NOT independent (can’t split the generating function as product) unless γ = 0. If we compare to property as seen in class: sum of two independent Poisson is Poisson (with same parameters as here). The independence is not a necessary condition, or put differently, the converse is not true: If we know only the marginals are Poisson and the sum is Poisson (even with the “right” parameter), it doesn’t mean that the joint distribution is independent Poisson....
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160a09_hw3sol - Pstat160A Winter 2009 Solution Homework #3...

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