Mean and variance of the Potts model in the hottest state

August 20, 2018

I’ve just arXiv’d 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(y) in the hottest state, when the inverse temperature $\beta = 0$ . 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 $y_i$ can take values in the set $\{1, \dots, q\}$ . The Ising model can be viewed as a special case, when q=2. The size of the configuration space $| \mathcal{Y} |$ is therefore $q^n$ , where n is the number of nodes. The dual lattice $\mathcal{E}$ defines undirected edges between neighbouring nodes $i \sim \ell$ . If the nodes in a 2D lattice with c columns are indexed row-wise, the nearest (first-order) neighbours $\partial(i)$ are $\ell \in \{ i-1, i-c, i+c, i+1 \}$ , 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 $\#\mathcal{E} = 2(n - \sqrt{n})$ for a square lattice, or $2n - r - c$ if the lattice is rectangular.

The sufficient statistic of the Potts model is the sum of all like neighbour pairs:
$S(\mathbf{y}) = \sum_{i \sim \ell \in \mathcal{E}} \delta(y_i, y_\ell)$
where $\delta(a,b)$ is the Kronecker delta function, which equals 1 if a = b and equals 0 otherwise. $S(\mathbf{y})$ ranges from 0, when all of the nodes form a chequerboard pattern, to $\#\mathcal{E}$ when all of the nodes have the same value. The likelihood of the Potts model is thus:
$p(\mathbf{y} \mid \beta) = \exp\{\beta S(\mathbf{y}) - \log \mathcal{C}(\beta)\}$
The normalising constant of the Potts model is intractable for any non-trivial lattice, since it requires a sum over the configuration space:
$\mathcal{C}(\beta) = \sum_{\mathbf{y} \in \mathcal{Y}} \exp\{\beta S(\mathbf{y})\}$
When the inverse temperature $\beta = 0$ , $p(\mathbf{y} \mid \beta=0)$ simplifies to $\left(\sum_{\mathbf{y} \in \mathcal{Y}} \exp\{0\}\right)^{-1} = q^{-n}$ , hence the labels $y_i$ 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

$\sum_{\mathbf{y} \in \mathcal{Y}} \sum_{i \sim \ell \in \mathcal{E}} \delta(y_i,y_\ell) = q^{n-1} \#\mathcal{E}$ .

Proof

For a q=2 state Potts model on a lattice with n=4 nodes and $\#\mathcal{E} =4$ edges, $\mathcal{Y}$ contains 16 possible configurations:

$\sum_{\mathbf{y} \in \mathcal{Y}} \mathrm{S}(\mathbf{y}) = 2 \times 4 + 12 \times 2 + 2 \times 0 = 32$ . This can also be written as $2^{3} \times 4 = 32$ .

Now consider a rectangular lattice with r > 1 rows and c > 1 columns, so that $n^\circ = r \times c$ and the dual lattice $\#\mathcal{E}^\circ = 2 r \cdot c - r - c$ . The size of the configuration space is $|\mathcal{Y}^\circ| = q^{n^\circ}$ . Assume that the sum over configuration space is equal to $q^{n^\circ - 1} \#\mathcal{E}^\circ$ . This sum can be decomposed into $(q^c)^r r(c-1)/q$ within each row, plus $(q^c)^r c(r-1)/q$ between rows.

If this lattice is extended by adding another row (or equivalently, another column), then $n' = n^\circ + c$ (or otherwise, $n' = n^\circ + r$ ) and the dual lattice $\#\mathcal{E}' = 2(n^\circ + c) - r - c - 1 = \#\mathcal{E}^\circ + 2c - 1$ . The nodes in this new row can take $q^c$ possible values, so the size of the configuration space is now $|\mathcal{Y}'| = q^c q^{n^\circ} = q^{n'}$ . $\sum_{\mathbf{y} \in \mathcal{Y'}} S(\mathbf{y})$ will increase proportional to $\frac{r+1}{r}q^c$ for the new row, plus $\frac{r}{r-1}q^c$ for the connections with its adjacent row:
$\sum_{\mathbf{y} \in \mathcal{Y'}} S(\mathbf{y}) = \frac{r+1}{r}q^c (q^c)^r \frac{r}{q}(c-1) + \frac{r}{r-1}q^c (q^c)^r \frac{c}{q}(r-1)$
$= \frac{r+1}{q} (q^c)^{r+1}(c-1) + \frac{r c}{q} (q^c)^{r+1} = q^{rc + c - 1} (2rc + c - r - 1)$
$= q^{n^\circ + c - 1} (\#\mathcal{E}^\circ + 2c - 1) = q^{n' - 1} \#\mathcal{E}'$
Q.E.D.

Theorem 2

The expectation of the q-state Potts model on a rectangular 2D lattice is $\mathbb{E}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})] = \#\mathcal{E}/q$ when the inverse temperature $\beta = 0$ .

Proof

The proof follows from Theorem 1 by noting that $p(\mathbf{y} | \beta = 0) = q^{-n}$ and hence:
$\mathbb{E}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})] = \sum_{\mathbf{y} \in \mathcal{Y}} S(\mathbf{y}) \;p(\mathbf{y} \mid \beta = 0)$
$= q^{n-1} \#\mathcal{E} \;q^{-n} = \frac{\#\mathcal{E}}{q}$
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
$\sum_{\mathbf{y} \in \mathcal{Y}} \mathrm{S}(\mathbf{y})^2 = q^n \#\mathcal{E} \left(q^{-2}\#\mathcal{E} + q^{-1}(1 - q^{-1})\right).$

Proof

For a q=2 state Potts model on a lattice with n=4 nodes and $\#\mathcal{E} =4$ edges, $\sum_{\mathbf{y} \in \mathcal{Y}} \mathrm{S}(\mathbf{y})^2 = 2 \times 4^2 + 12 \times 2^2 + 2 \times 0^2 = 80$ . This can also be written as $2^4 \times 4 (4/2^2 + 0.5(1 - 0.5)) = 2^6(1 + 0.25) = 80$ .

Now assume for a rectangular lattice with r > 1 rows and c > 1 columns that
$\sum_{\mathbf{y} \in \mathcal{Y}^\circ} \mathrm{S}(\mathbf{y})^2 = q^{rc - 2} (2rc - r - c)^2 + q^{rc - 1}\left(1 - \frac{1}{q}\right)(2rc - r - c).$
This can be decomposed into $\left( (q^c)^r r(c-1)/q + (q^c)^r c(r-1)/q \right)^2 + (q^c)^r (1 - q^{-1}) (2rc - r - c)/q$ .

If we extend the lattice by adding another row, then
$\sum_{\mathbf{y} \in \mathcal{Y}'} \mathrm{S}(\mathbf{y})^2 = \left( \frac{r+1}{q} (q^c)^{r+1} (c-1) + \frac{rc}{q}(q^c)^{r+1} \right)^2 + (q^c)^{r+1} \frac{(1 - q^{-1})(2rc - r + c - 1)}{q}$
$= \left( q^{n^\circ + c - 1} (\#\mathcal{E}^\circ + 2c - 1) \right)^2 + q^{n^\circ + c - 1} \left(1 - q^{-1}\right) (\#\mathcal{E}^\circ + 2c - 1)$
$= \left( q^{n' - 1} \#\mathcal{E}' \right)^2 + q^{n' - 1} \left(1 - q^{-1}\right) \#\mathcal{E}'$
$= q^{n'} \#\mathcal{E}' \left(q^{-2}\#\mathcal{E}' + q^{-1}(1 - q^{-1})\right)$
Q.E.D.

Theorem 4

The variance of the q-state Potts model on a rectangular 2D lattice is $\mathbb{V}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})] = \mathcal{I}(0) = \#\mathcal{E} q^{-1} (1 - q^{-1})$ when the inverse temperature $\beta = 0$ .

Proof

The proof follows from Theorems 1 and 3:
$\mathbb{V}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})] = \mathbb{E}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})^2] - \mathbb{E}_{\mathbf{y} | \beta = 0}[\mathrm{S}(\mathbf{y})]^2$
$= \sum_{\mathbf{y} \in \mathcal{Y}} S(\mathbf{y})^2 \;p(\mathbf{y} \mid \beta = 0) - \left(\frac{\#\mathcal{E}}{q}\right)^2$
$= q^n \#\mathcal{E} \left(q^{-2}\#\mathcal{E} + q^{-1}(1 - q^{-1})\right) q^{-n} - \frac{\#\mathcal{E}^2}{q^2}$
$= \frac{\#\mathcal{E}^2}{q^2} + \#\mathcal{E} q^{-1} (1 - q^{-1}) - \frac{\#\mathcal{E}^2}{q^2} = \#\mathcal{E} q^{-1} (1 - q^{-1})$
Q.E.D.