Recall:
Metropolis Algorithm
S: big discrete set of states
A probability distribution on S:
( )∝ ∈( )
p x
e
x kT
Algorithm constructs a Markov Chain using “proposal followed by accept/reject.”
Start with symmetric Q
ij
; Q
ij
=prob{given I, propose j}
Accept
j as next state with probability
( , )=
<
( ) (

)
>
( )
α i j
1 if ϵj
ϵ i e ϵj ϵi kT if ϵj
ϵ i
Overall transition probabilities for Markov chain are: P(I,j)=Q(I,j)α(I,j)
Recall detailed balance
ideas given a probabilitiy distribution π on S, their πis stationary distribution for
MC with transition probs P(I,j)
if
, = ,
,
πiPi j Pj iπjfor all i j
(Detailed balance conditions)
Claim
: with P(I,j)= Q(I,j)α(I,j), defined as above, stationary distribution is precisely π
*
, where
*= 
∈

πi
e ϵikTj Se ϵjkT
Ie, “π
*
=P”
Proof
: Check detailed balance conditions:
Case 1:
>
( )
ϵi
ϵ j
.
Then P(I,j)=Q(I,j) and P(j,i)=Q(j,i)
(

)
e ϵi ϵj kT
Since Q(j,i)=Q(I,j), this says
P(I,j)=Q(I,j)=Q(j,i)=P(j,i)
(

)
e ϵi ϵj kT
Note:
(

)
= ( ) ( )= * *
e ϵi ϵj kT p j p i
πj πi
Hence:
* , = ,
*
πi Pi j Pj iπj
Bottom Line
: π
*
, the Boltzmann distribution, is the (unique*) stationary distribution for Markov chain
(easy to argue similarly when
>
.
ϵj
ϵi
*unique because, by construction, the Markov chain is generic
every state positively recurrent, only one
recurrence class, hence only one stationary distribution.
Back off for a reality check…