Mean and variance of the Potts model in the hottest state

I’ve just arXived another revision of my paper on the PFAB algorithm: arXiv:1503.08066v3 [stat.CO]. It includes a rather elegant proof of the exact mean and variance of the sufficient statistic S(z) in the hottest state, when the inverse temperature β=0. The proof by my co-author Geoff Nicholls holds for any Potts model with first-order neighbours. That is, the nearest 4 neighbours in a 2D lattice (or 6 neighbours in 3D). For posterity, I present my rather clunkier proof below, which involves induction on dimension for a rectangular lattice.

The Potts (1952) model is an example of a Gibbs random field on a regular lattice, where each node yi can take values in the set {1,,q}. The Ising model can be viewed as a special case, when q=2. The size of the configuration space |Y| is therefore qn, where n is the number of nodes. The dual lattice E defines undirected edges between neighbouring nodes i. If the nodes in a 2D lattice with c columns are indexed row-wise, the nearest (first-order) neighbours (i) are {i1,ic,i+c,i+1}, except at the boundary. Nodes situated on the boundary of the domain have less than four neighbours. The total number of unique edges is thus #E=2(nn) for a square lattice, or 2nrc if the lattice is rectangular.

The sufficient statistic of the Potts model is the sum of all like neighbour pairs: S(y)=iEδ(yi,y) where δ(a,b) is the Kronecker delta function, which equals 1 if a=b and equals 0 otherwise. S(y) ranges from 0, when all of the nodes form a chequerboard pattern, to #E when all of the nodes have the same value. The likelihood of the Potts model is thus: p(yβ)=exp{βS(y)logC(β)} The normalising constant of the Potts model is intractable for any non-trivial lattice, since it requires a sum over the configuration space: C(β)=yYexp{βS(y)}

When the inverse temperature β=0, p(yβ=0) simplifies to (yYexp{0})1=qn, hence the labels yi are independent and uniformly-distributed.

Theorem 1

The sum over configuration space of the sufficient statistic of the q-state Potts model on a rectangular 2D lattice is yYiEδ(yi,y)=qn1#E.

Proof

For a q=2 state Potts model on a lattice with n=4 nodes and #E=4 edges, Y contains 16 possible configurations: yYS(y)=2×4+12×2+2×0=32. This can also be written as 23×4=32.

Now consider a rectangular lattice with r>1 rows and c>1 columns, so that n=r×c and the dual lattice #E=2rcrc. The size of the configuration space is |Y|=qn. Assume that the sum over configuration space is equal to qn1#E. This sum can be decomposed into (qc)rr(c1)/q within each row, plus (qc)rc(r1)/q between rows.

If this lattice is extended by adding another row (or equivalently, another column), then n=n+c (or otherwise, n=n+r) and the dual lattice #E=2(n+c)rc1=#E+2c1. The nodes in this new row can take qc possible values, so the size of the configuration space is now |Y|=qcqn=qn. yYS(y) will increase proportional to r+1rqc for the new row, plus rr1qc for the connections with its adjacent row: yYS(y)=r+1rqc(qc)rrq(c1)+rr1qc(qc)rcq(r1)=r+1q(qc)r+1(c1)+rcq(qc)r+1=qrc+c1(2rc+cr1)=qn+c1(#E+2c1)=qn1#E

Q.E.D.

Theorem 2

The expectation of the q-state Potts model on a rectangular 2D lattice is Ey|β=0[S(y)]=#E/q when the inverse temperature β=0.

Proof

The proof follows from Theorem 1 by noting that p(y|β=0)=qn and hence: Ey|β=0[S(y)]=yYS(y)p(yβ=0)=qn1#Eqn=#Eq

Q.E.D.

Theorem 3

The sum over configuration space of the square of the sufficient statistic of the q-state Potts model on a rectangular 2D lattice is yYS(y)2=qn#E(q2#E+q1(1q1)).

Proof

For a q=2 state Potts model on a lattice with n=4 nodes and #E=4 edges, yYS(y)2=2×42+12×22+2×02=80. This can also be written as 24×4(4/22+0.5(10.5))=26(1+0.25)=80.

Now assume for a rectangular lattice with r>1 rows and c>1 columns that yYS(y)2=qrc2(2rcrc)2+qrc1(11q)(2rcrc). This can be decomposed into ((qc)rr(c1)/q+(qc)rc(r1)/q)2+(qc)r(1q1)(2rcrc)/q. If we extend the lattice by adding another row, then yYS(y)2=(r+1q(qc)r+1(c1)+rcq(qc)r+1)2+(qc)r+1(1q1)(2rcr+c1)q=(qn+c1(#E+2c1))2+qn+c1(1q1)(#E+2c1)=(qn1#E)2+qn1(1q1)#E=qn#E(q2#E+q1(1q1))

Q.E.D.

Theorem 4

The variance of the q-state Potts model on a rectangular 2D lattice is Vy|β=0[S(y)]=I(0)=#Eq1(1q1) when the inverse temperature β=0.

Proof

The proof follows from Theorems 1 and 3: Vy|β=0[S(y)]=Ey|β=0[S(y)2]Ey|β=0[S(y)]2=yYS(y)2p(yβ=0)(#Eq)2=qn#E(q2#E+q1(1q1))qn#E2q2=#E2q2+#Eq1(1q1)#E2q2=#Eq1(1q1)

Q.E.D.

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