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 > 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.

From → Imaging

ELLA KAYE

Computational Bayesian statistics

Bayes' Food Cake

A bit of statistics, a bit of cakes.

RWeekly.org - Blogs to Learn R from the Community

Computational Bayesian statistics

Richard Everitt's blog

Computational Bayesian statistics

Let's Look at the Figures

David Firth's blog

Nicholas Tierney

Computational Bayesian statistics

Sweet Tea, Science

Two southern scientistas will be bringing you all that is awesome in STEM as we complete our PhDs. Ecology, statistics, sass.

Musings, useful code etc. on R and data science

Darren Wilkinson's blog

Statistics, computing, functional programming, data science, Bayes, stochastic modelling, systems biology and bioinformatics

Computational Bayesian statistics

Igor Kromin

Computational Bayesian statistics

Statisfaction

I can't get no

Xi'an's Og

an attempt at bloggin, nothing more...

Sam Clifford

Postdoctoral Fellow, Bayesian Statistics, Aerosol Science