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Talk at ENSAE, January 2013
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Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
On the convergence properties of theWang-Landau Algorithm
Robin J. RyderCEREMADE, Universite Paris Dauphine
and CREST
ENSAE – 31 January 2012
joint work with Pierre Jacob (National University of Singapore)and Pierre Del Moral (INRIA & Universite de Bordeaux)
R. Ryder (Dauphine) & CREST Wang-Landau 1/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Outline
1 Wang–Landau algorithm
2 Flat Histogram in finite time
3 Parallel Wang–Landau: asymptotic behaviour
R. Ryder (Dauphine) & CREST Wang-Landau 2/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Motivation
X
dens
ity
0.0
0.1
0.2
0.3
0.4
0.5
−5 0 5 10 15
Figure: A mixture of well-separated normal distributions.
R. Ryder (Dauphine) & CREST Wang-Landau 3/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Motivation
X
dens
ity
0.0
0.1
0.2
0.3
0.4
0.5
−5 0 5 10 15
Figure: A mixture of well-separated normal distributions.
R. Ryder (Dauphine) & CREST Wang-Landau 4/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Motivation
X
dens
ity
0.0
0.1
0.2
0.3
0.4
0.5
−5 0 5 10 15
Figure: A mixture of well-separated normal distributions.
R. Ryder (Dauphine) & CREST Wang-Landau 5/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Setting
Partition the state space
X =d⋃
i=1
Xi
Desired frequencies
φ = (φ1, . . . , φd)
Penalized distribution
πθ(x) ∝ π(x)
θ(J(x))
where J(x) such that x ∈ XJ(x).
R. Ryder (Dauphine) & CREST Wang-Landau 6/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
First algorithm
Algorithm 1 Wang-Landau with deterministic schedule (γt)
1: Init θ0 > 0, X0 ∈ X .2: for t = 1 to T do3: Sample Xt from Kθt−1(Xt−1, ·), MH kernel targeting πθt−1 .
4: Update the penalties:
log θt(i)← log θt−1(i) + γt (1IXi(Xt)− φi )
5: end for
R. Ryder (Dauphine) & CREST Wang-Landau 7/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Video
R. Ryder (Dauphine) & CREST Wang-Landau 8/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Video
R. Ryder (Dauphine) & CREST Wang-Landau 8/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Flat Histogram
Issues with the first version
Choice of γ has a huge impact on the results.
Flat Histogram
Define the counters:
νt(i) :=t∑
n=1
1IXi(Xn)
Flat Histogram (FH) is reached when:
maxi∈{1,...,d}
∣∣∣∣νt(i)
t− φi
∣∣∣∣ < c
R. Ryder (Dauphine) & CREST Wang-Landau 9/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Flat Histogram
Idea
Instead of decreasing γt at each time step t, decrease only whenthe Flat Histogram criterion is reached.
In practice
Denote by κt the number of FH criteria reached up to time t.
Use γκt instead of γt at time t.
If FH is reached at time t, reset νt(i) to 0 for all i .
R. Ryder (Dauphine) & CREST Wang-Landau 10/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Wang–Landau with Flat Histogram
Algorithm 2 Wang-Landau with stochastic schedule (γκt )
1: Init θ0 > 0, X0 ∈ X .2: Init κ0 ← 0.3: for t = 1 to T do4: Sample Xt from Kθt−1(Xt−1, ·), MH kernel targeting πθt−1 .5: If (FH) then κt ← κt−1 + 1, otherwise κt ← κt−1.
6: Update the penalties:
log θt(i)← log θt−1(i) + γκt (1IXi(Xt)− φi )
7: end for
R. Ryder (Dauphine) & CREST Wang-Landau 11/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
FH is met in finite time
To be sure that eventually, for any c > 0:
maxi∈{1,...,d}
∣∣∣∣νt(i)
t− φi
∣∣∣∣ < c
we want to prove:
∀i ∈ {1, . . . , d} νt(i)
tP−−−→
t→∞φi
for any fixed γ > 0.
R. Ryder (Dauphine) & CREST Wang-Landau 12/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Various updates
Right update
log θt(i)← log θt−1(i) + γ (1IXi(Xt)− φi ) (1)
Using this update, FH is met in finite time.
Wrong update
θt(i)← θt−1(i) [1 + γ (1IXi(Xt)− φi )]
⇔log θt(i)← log θt−1(i) + log [1 + γ (1IXi
(Xt)− φi )] (2)
Here FH is not necessarily met in finite time.Special case: ∀i φi = 1
d .
R. Ryder (Dauphine) & CREST Wang-Landau 13/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Assumptions
In this talk, there are only two bins: d = 2. Additionally:
Assumption
The bins are not empty with respect to µ and π:
∀i ∈ {1, 2} µ(Xi ) > 0 and π(Xi ) > 0
Assumption
The state space X is compact.
R. Ryder (Dauphine) & CREST Wang-Landau 14/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Assumptions
Assumption
The proposal distribution Q(x , y) is such that:
∃qmin > 0 ∀x ∈ X ∀y ∈ X Q(x , y) > qmin
Assumption
The MH acceptance ratio is bounded from both sides:
∃m > 0 ∃M > 0 ∀x ∈ X ∀y ∈ X m <π(y)
π(x)
Q(y , x)
Q(x , y)< M
R. Ryder (Dauphine) & CREST Wang-Landau 15/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Theorem
Theorem
Consider the sequence of penalties θt introduced in the WLalgorithm. We define:
Zt = logθt(1)
θt(2)= log θt(1)− log θt(2)
Then:Zt
t
L1−−−→t→∞
0
and consequently, with update (1) (FH) is reached in finite timefor any precision threshold c, whereas this is not guaranteed forupdate (2).
R. Ryder (Dauphine) & CREST Wang-Landau 16/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = log θt(1)(
1IXX
XX
)− log θt(2)
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = log θ0(1) + γ∑t
n=1(1IX1(Xn)− φ1)(
1IXX
XX
)− log θ0(2) + γ
∑tn=1(1IX2(Xn)− φ2)
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = γ(νt(1)− tφ1)(
1IXX
XX
)−γ(νt(2)− tφ2)
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = γ(νt(1)− tφ1)(
1IXX
XX
)−γ(t − νt(1)− t(1− φ1))
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = γ(νt(1)− tφ1)(
1IXX
XX
)+γ(νt(1)− tφ1))
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Zt = 2γ(νt(1)− tφ1)(
1IXX
XX
)(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Ztt = 2γ
(νt(1)t − φ1
)(1IX
X
XX
)(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Ztt = 2γ
(νt(1)t − φ1
)L1−−−→
t→∞0(
1IXX
XX
)⇒ νt(1)
t
L1−−−→t→∞
φ1
(1IX
X
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Consequence
Using update (1), and starting from Z0 = 0:
Ztt = 2γ
(νt(1)t − φ1
)L1−−−→
t→∞0(
1IXX
XX
)⇒ νt(1)
t
L1−−−→t→∞
φ1
(1IX
X
XX
)Using update (2), these simplifications do not occur: νt(i)/tconverges, but not to φi in general.
R. Ryder (Dauphine) & CREST Wang-Landau 17/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Proof
To prove thatZt
t
L1−−−→t→∞
0,
first notice that
Zt+1 − Zt = 2γ (νt+1(1)− νt(1)− φ1)
can only take two different values, which we note +a and −b.
R. Ryder (Dauphine) & CREST Wang-Landau 18/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Proof
Introduce Ut , the increment of Zt :
Zt+1 = Zt + Ut
Lemma
With the introduced processes Zt and Ut , there exists ε > 0 suchthat for all η > 0, there exists Zhi such that, if Zt ≥ Zhi , we havethe following two inequalities:
P[Ut+1 = −b|Ut = +a,Zt ] > ε
P[Ut+1 = −b|Ut = −b,Zt ] > 1− η.
(and the symmetric corollary)
R. Ryder (Dauphine) & CREST Wang-Landau 19/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Proof
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Zs
Zs+T
Z~
s+T~
5 10 15 20 25 30 35
Figure: We prove that Zt returns below a given horizontal bar wheneverit goes above it, and it does so in finite time. This implies Zt/t → 0.
R. Ryder (Dauphine) & CREST Wang-Landau 20/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Proof
Introduce a crossing time s: Zs−1 ≤ Zhi and Zs ≥ Zhi .Introduce a new process Zt :
Zs = Zs and Zt+1 = Zt + Vt
We can choose Vt such that Zt ≥ Zt and
Lemma
(Vt) is a Markov chain over the space {+a,−b} with transitionmatrix (
1− ε εη 1− η
)where the first state corresponds to +a and the second state to−b.
R. Ryder (Dauphine) & CREST Wang-Landau 21/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Proof
time
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Zs
Zs+T
Z~
s+T~
5 10 15 20 25 30 35
Figure: We prove that Zt returns below a given horizontal bar wheneverit goes above it, and it does so in finite time. This implies Zt/t → 0.
R. Ryder (Dauphine) & CREST Wang-Landau 22/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Case d = 2
This shows that Zt/t → 0, hence (FH) is met in finite time.For the case d > 2, we shall use only update 1.
R. Ryder (Dauphine) & CREST Wang-Landau 23/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Case d > 2
To prove that (FH) is met in finite time in the case with d > 2, weneed an extra assumption:
Assumption
The desired frequencies are all rational numbers:
∀i , φi ∈ Q
R. Ryder (Dauphine) & CREST Wang-Landau 24/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Irreducible Markov chain
Under the assumption of rationality, we have the following lemma:
Lemma
Let Θ be the following subset of Rd :
Θ = {z ∈ Rd : ∃(n1, . . . , nd) ∈ Nd zi = ni − φid∑
j=1
nj}
Then denoting by λ the product of the Lebesgue measure µ on Xand of the counting measure on Θ, (Xt , log θt) is λ-irreducible.
Proof: Bezout’s lemma.
R. Ryder (Dauphine) & CREST Wang-Landau 25/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
A limit exists
Since (Xt , log θt) is λ-irreducible, the proportion of visits to anyλ-measurable set of X ×Θ converges to a limit in [0, 1]. Hencethe vector (ν(i)/t) converges to some limit pi . We now need toprove that this limit corresponds to (FH).
R. Ryder (Dauphine) & CREST Wang-Landau 26/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Reductio ad absurdum
Suppose (reductio ad absurdum) that p 6= φ. Then there exist iand j such that
pi − φi < pj − φjand hence
Z j ,it = −νt(i) + νt(j) + t(φi − φj) ∼ t(−pi + φi + pj − φj)→∞
As before, we can construct Ut and Ut and show that Ut decreaseson average, which contradicts the assumption that Z j ,i
t →∞.
R. Ryder (Dauphine) & CREST Wang-Landau 27/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Conclusion for d > 2
Conclusion: using update 1, (FH) is met in finite time for anynumber of bins d .
R. Ryder (Dauphine) & CREST Wang-Landau 28/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Illustration
X
dens
ity
0.0
0.1
0.2
0.3
0.4
0.5
−4 −2 0 2 4
(a) Histogram of the gen-erated sample
iterations
Pro
port
ions
of v
isits
to e
ach
bin
0.0
0.2
0.4
0.6
0.8
1.0
0 1000 2000 3000 4000 5000
(b) Convergence of theproportions of visits toeach bin, using the rightupdate
iterations
Pro
port
ions
of v
isits
to e
ach
bin
0.0
0.2
0.4
0.6
0.8
1.0
0 1000 2000 3000 4000 5000
(c) Convergence of theproportions of visits toeach bin, using the wrongupdate
R. Ryder (Dauphine) & CREST Wang-Landau 29/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Parallel chains
N chains (X(1)t , . . . ,X
(N)t ) instead of one.
targeting the same biased distribution πθt at iteration t,
sharing the same estimated bias θt at iteration t.
Old update of the penalties
log θt(i)← log θt−1(i) + γ (1IXi(Xt)− φi )
(1IX
X
XX
)
(L. Bornn, P. Jacob, P. Del Moral, A. Doucet)
R. Ryder (Dauphine) & CREST Wang-Landau 30/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Parallel chains
N chains (X(1)t , . . . ,X
(N)t ) instead of one.
targeting the same biased distribution πθt at iteration t,
sharing the same estimated bias θt at iteration t.
New update of the penalties
log θt(i)← log θt−1(i) + γ(
1N
∑Nk=1 1IXi
(X(k)t )− φi
) (1IX
X
XX
)
(L. Bornn, P. Jacob, P. Del Moral, A. Doucet)
R. Ryder (Dauphine) & CREST Wang-Landau 30/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Infinite number of chains
Suppose φi = 1/d . We can use either update and choose the“wrong” update
θt(i)← θt−1(i)
[1 + γ
(1
N
N∑k=1
1IXi(X
(k)t )− 1
d
)]
Note that if (X(k)t )Nk=1
iid∼ πθt then
1
N
N∑k=1
1IXi(X
(k)t )
a.s.−−−−→N→∞
∫Xi
πθt (x)dx =ψi
θt−1(i)
(∑k
ψk
θ(k)
)−1
R. Ryder (Dauphine) & CREST Wang-Landau 31/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
This corresponds to update
θt(i)← θt−1(i)
[1 + γ
(∫Xi
πθt (x)dx − 1
d
)]
⇔ θt(i)← θt−1(i)
1 + γ
ψi
θt−1(i)
(∑k
ψk
θ(k)
)−1
− 1
d
Normalize:
θt(i)← θt(i)∑j θt(j)
Then the penalties (θt)t≥0 should converge towards ψ.
R. Ryder (Dauphine) & CREST Wang-Landau 32/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)←θt−1(i)
[1+γ
(ψi
θt−1(i)(∑
kψkθ(k))
−1− 1
d
)](1IXX
XX
)∑d
j=1 θt−1(j)
[1+γ
(ψj
θt−1(j)(∑
kψkθ(k))
−1− 1
d
)](1IXX
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)←θt−1(i)
[1+γ
(ψi
θt−1(i) M−1ψ (θt−1)− 1
d
)](1IXX
XX
)∑d
j=1 θt−1(j)[1+γ
(ψj
θt−1(j) M−1ψ (θt−1)− 1
d
)](1IXX
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)←θt−1(i)(1−γd ) + ψiγM−1
ψ (θt−1)
(1IXX
XX
)∑d
j=1 θt−1(j)(1−γd ) + ψjγM−1ψ (θt−1)
(1IXX
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)←θt−1(i)(1−γd ) + ψiγM−1
ψ (θt−1)
(1IXX
XX
)(1−γd )+γM−1
ψ (θt−1)
(1IXX
XX
)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)← θt−1(i)1− γ
d
1− γd
+γM−1ψ (θt−1)
+ ψiγM−1
ψ (θt−1)
1− γd
+γM−1ψ (θt−1)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
θt(i)← θt−1(i)× (1− α(θt−1)) + ψi × α(θt−1)
R. Ryder (Dauphine) & CREST Wang-Landau 33/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Hence the update can be written as a convex combination
θt(i)← θt−1(i)× (1− α(θt−1)) + ψi × α(θt−1)
where
α(θt−1) =γM−1
ψ (θt−1)
1− γd + γM−1
ψ (θt−1)
It can easily be shown that
∀t ≥ 0 0 < αmin ≤ α(θt) ≤ αmax < 1
provided that ∀i , θ0(i) > 0, ψ(i) > 0.
Of course the assumptions N →∞ andiid∼ sampling are strong.
What happens in practice?
R. Ryder (Dauphine) & CREST Wang-Landau 34/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
Toy example
X = {1, 2, 3, 4, 5}, π = (π1, . . . , π5)
We take the partition: Xi = {i} and launch the Wang–Landaualgorithm for N = 1, 10, 100 and γ = 1.In this case we know the value of ψi : ψi = πi , so we can computethe deterministic sequence of penalties corresponding to N =∞.
Animation
R. Ryder (Dauphine) & CREST Wang-Landau 35/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
N =∞R. Ryder (Dauphine) & CREST Wang-Landau 36/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
N = 100R. Ryder (Dauphine) & CREST Wang-Landau 36/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
N = 10R. Ryder (Dauphine) & CREST Wang-Landau 36/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
N = 1R. Ryder (Dauphine) & CREST Wang-Landau 36/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Animation
“
R. Ryder (Dauphine) & CREST Wang-Landau 37/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
New take on the Parallel Wang–Landau
Call fψ the function such that:
θt ← fψ(θt−1)
As we saw:fψ(θ) = θ(1− α(θ)) + ψα(θ))
where
α(θ) =γM−1
ψ (θ)
1− γd + γM−1
ψ (θ)
R. Ryder (Dauphine) & CREST Wang-Landau 38/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
New take on the Parallel Wang–Landau (in progress)
In the ideal case we have
θn = fψ ◦ . . . ◦ fψ︸ ︷︷ ︸×n
(θ0) = f nψ (θ0)
If we add perturbations to ψ we also add perturbations to θ, asfollows:
θ1 = f εψ(θ0) = fψ(θ0) + V0
R. Ryder (Dauphine) & CREST Wang-Landau 39/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
New take on the Parallel Wang–Landau (in progress)
In the ideal case we have
θn = fψ ◦ . . . ◦ fψ︸ ︷︷ ︸×n
(θ0) = f nψ (θ0)
If we add perturbations to ψ we also add perturbations to θ, asfollows:
θ1 = f εψ(θ0) = fψ(θ0) + V0
θ2 = f εψ(θ1) = f εψ(fψ(θ0) + V0) = f 2ψ (θ0) + V1
. . .
R. Ryder (Dauphine) & CREST Wang-Landau 39/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
New take on the Parallel Wang–Landau (in progress)
In the ideal case we have
θn = fψ ◦ . . . ◦ fψ︸ ︷︷ ︸×n
(θ0) = f nψ (θ0)
If we add perturbations to ψ we also add perturbations to θ, asfollows:
θ1 = f εψ(θ0) = fψ(θ0) + V0
θ2 = f εψ(θ1) = f εψ(fψ(θ0) + V0) = f 2ψ (θ0) + V1
. . .
The properties of fψ should help control the noise terms(Vn)n≥0. . .
R. Ryder (Dauphine) & CREST Wang-Landau 39/ 40
Wang–Landau algorithmFlat Histogram in finite time
Parallel Wang–Landau: asymptotic behaviour
Bibliography
Atchade, Y. and Liu, J. (2010). The Wang-Landau algorithmin general state spaces: applications and convergence analysis.Statistica Sinica, 20:209–233.
Wang, F. and Landau, D. (2001). Determining the density ofstates for classical statistical models: A random walkalgorithm to produce a flat histogram. Physical Review E,64(5):56101.
An Adaptive Interacting Wang-Landau Algorithm forAutomatic Density Exploration, L. Bornn, P. Jacob, P. Del
Moral, A. Doucet, on arXiv.
The Wang-Landau algorithm reaches the Flat Histogramcriterion in finite time, P. Jacob & RR, on arXiv.
R. Ryder (Dauphine) & CREST Wang-Landau 40/ 40