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Unformatted text preview: 0 , θ1 , . . . produced
by the MetropolisHastings algorithm has a unique stationary distribution,
then the stationary distribution is p(θy ).
Proof. We will use the fact given in (23) that if the posterior p(θy ) satisﬁes
the detailed balance equation, then it is the stationary distribution. Thus we
wish to show:
K (θ, θ' )p(θy ) = K (θ' , θ)p(θ' y ), for all θ, θ' .
where the transition kernel comes from MetropolisHastings:
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(θy ) ≥ J (θ' , θ)p(θ' y ),
so that J (θ' , θ)p(θ' y )
α(θ, θ ) =
,
J (θ, θ' )p(θy )
and α(θ' , θ) = 1. Now let’s plug:
' K (θ, θ' )p(θy ) =J (θ, θ' )α(θ, θ' )p(θy )
J (θ' , θ)p(θ' y )
'
=J (θ, θ )p(θy )
J (θ, θ' )p(θy )
=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.
 Spring '12
 CynthiaRudin

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