mcmc - 5 Markov Chain Monte Carlo 5.1 Comparing Two Poisson...

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

View Full Document Right Arrow Icon
Markov Chain Monte Carlo 5.1 Comparing Two Poisson Means Let’s revisit the problem of comparing the means from two independent Pois- son samples. Counts { y Ai } from the weekend days are assumed Poisson with mean λ A and counts { y Bj } from the weekday days are assumed Poisson with mean λ B . We are interested in learning about the ratio of means γ = λ B λ A . We showed that the likelihood function in terms of the Frst Poisson mean θ = λ A and γ is given by L ( θ, γ ) = exp( - n A θ ) θ s A exp( - n B ( θγ ))( θγ ) s B . Assuming that θ and γ are independent with θ Gamma ( a 0 , b 0 ) , γ Gamma ( a g , b g ) , Then the posterior density of ( θ, γ ) is given, up to a proportionality constant, by g ( θ, γ | data) exp( - n A θ ) θ s A exp( - n B ( θγ ))( θγ ) s B × θ a 0 - 1 exp( - b 0 θ ) γ a g - 1 exp( - b g γ ) By combining terms, we obtain the expression g ( θ, γ | data) exp ( - ( b 0 + n A + n B γ ) θ ) θ a 0 + s A + s B - 1 × exp( - b g γ ) γ
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

mcmc - 5 Markov Chain Monte Carlo 5.1 Comparing Two Poisson...

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

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