{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

160a09_hw3sol

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

  • Spring '09
  • Probability theory, Inverse transform sampling, inverse transform method, Pstat160A Solution Homework, Form discussion session, right distribution Problem

{[ snackBarMessage ]}

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online