MIT15_097S12_lec15

MIT15_097S12_lec15

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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...
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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.

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