Info iconThis preview shows page 1. Sign up to view the full content.

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: 0 , θ1 , . . . produced by the Metropolis-Hastings algorithm has a unique stationary distribution, then the stationary distribution is p(θ|y ). Proof. We will use the fact given in (23) that if the posterior p(θ|y ) satisfies the detailed balance equation, then it is the stationary distribution. Thus we wish to show: K (θ, θ' )p(θ|y ) = K (θ' , θ)p(θ' |y ), for all θ, θ' . where the transition kernel comes from Metropolis-Hastings: K (θ, θ' ) = (probability of proposing θ' )× (probability of accepting θ' given it was proposed) = J (θ, θ' )α(θ, θ' ). To show that the detailed balance equation holds, take any θ and θ' , and without loss of generality, suppose that α is less than or equal to 1 for the transition θ to θ' , which means J (θ, θ' )p(θ|y ) ≥ J (θ' , θ)p(θ' |y ), so that J (θ' , θ)p(θ' |y ) α(θ, θ ) = , J (θ, θ' )p(θ|y ) and α(θ' , θ) = 1. Now let’s plug: ' K (θ, θ' )p(θ|y ) =J (θ, θ' )α(θ, θ' )p(θ|y ) J (θ' , θ)p(θ' |y ) ' =J (θ, θ )p(θ|y ) J (θ, θ' )p(θ|y ) =J...
View Full Document

This note was uploaded on 03/24/2014 for the course MIT 15.097 taught by Professor Cynthiarudin during the Spring '12 term at MIT.

Ask a homework question - tutors are online