This post looks at the convergence of the chequerboard Gibbs sampler for the hidden Potts model, in the presence of an external field. This algorithm is implemented as the function mcmcPotts
in my R package, bayesImageS. Previous posts have looked at the convergence of Gibbs and Swendsen-Wang algorithms without an external field, as implemented in mcmcPottsNoData
and swNoData
functions.
The most accurate way to measure convergence is using the coupling time of a perfect sampling algorithm, such as coupling from the past (CFTP). However, we can obtain a rough estimate by monitoring the distribution of the sufficient statistic: \[ \text{S}(\mathbf{z}) = \sum_{\{i,j\} \in \mathcal{E}} \delta(z_i, z_j) \] Where \(\delta(x,y)\) is the Kronecker delta function. Note that this sum is defined over the unique undirected edges of the lattice, to avoid double-counting. Under this definition, the critical temperature of the \(q\)-state Potts model is \(\log\{1 + \sqrt{q}\}\), or \(\approx 0.88\) for the Ising model with \(q=2\) unique labels. Some papers state that the critical temperature of the Ising model is \(0.44\), but this is because they have used a different definition of \(\text{S}(\mathbf{z})\).
We will generate synthetic data for a sequence of values of the inverse temperature, \(\beta = (0.22, 0.44, 0.88, 1.32)\):
library(bayesImageS)
library(doParallel)
set.seed(123)
<- 2
q <- c(0.22, 0.44, 0.88, 1.32)
beta <- matrix(1,nrow=500,ncol=500)
mask <- prod(dim(mask))
n <- getNeighbors(mask, c(2,2,0,0))
neigh <- getBlocks(mask, 2)
block <- getEdges(mask, c(2,2,0,0))
edges <- nrow(edges)
maxS
<- makeCluster(min(4, detectCores()))
cl registerDoParallel(cl)
<- system.time(synth <- foreach (i=1:length(beta),
tm .packages="bayesImageS") %dopar% {
<- list()
gen $beta <- beta[i]
gen# generate labels
<- swNoData(beta[i], q, neigh, block, 200)
sw $z <- sw$z
gen$sum <- sw$sum[200]
gen# now add noise
$mu <- rnorm(2, c(-1,1), 0.5)
gen$sd <- 1/sqrt(rgamma(2, 1.5, 2))
gen$y <- rnorm(n, gen$mu[(gen$z[1:n,1])+1],
gen$sd[(gen$z[1:n,1])+1])
gen
gen
})stopCluster(cl)
## user system elapsed
## 0.326 0.127 16.998
Now let’s look at the distribution of Gibbs samples for the first dataset, using a fixed value of \(\beta\):
<- list()
priors $k <- q
priors$mu <- c(-1,1)
priors$mu.sd <- rep(0.5,q)
priors$sigma <- rep(2,q)
priors$sigma.nu <- rep(1.5,q)
priors$beta <- rep(synth[[1]]$beta, 2)
priors
<- list(algorithm="ex", bandwidth=1, adaptive=NA,
mh auxiliary=1)
<- system.time(res <- mcmcPotts(synth[[1]]$y, neigh,
tm 100, 50)) block, priors, mh,
data("res", package = "bayesImageS")
print(res$tm)
## user system elapsed
## 23.299 3.729 13.109
mean(res$sum[51:100])
## [1] 277581.5
print(synth[[1]]$sum)
## [1] 277664
ts.plot(res$sum, xlab="MCMC iterations", ylab=expression(S(z)))
abline(h=synth[[1]]$sum, col=4, lty=2)
As expected for \(\beta=0.22\) with \(n=400\) pixels, convergence takes only a dozen iterations or so. The same is true for \(\beta=0.66\):
$beta <- rep(synth[[2]]$beta, 2)
priors<- system.time(res2 <- mcmcPotts(synth[[2]]$y,
tm2 100, 50)) neigh, block, priors, mh,
## user system elapsed
## 24.004 4.035 12.776
Now with \(\beta=0.88\):
$beta <- rep(synth[[3]]$beta, 2)
priors<- system.time(res3 <- mcmcPotts(synth[[3]]$y,
tm3 100, 50)) neigh, block, priors, mh,
## user system elapsed
## 23.497 3.992 12.331
So far, so good. Now let’s try with \(\beta=1.32\):
$beta <- rep(synth[[4]]$beta, 2)
priors<- system.time(res4 <- mcmcPotts(synth[[4]]$y,
tm4 100, 50)) neigh, block, priors, mh,
## user system elapsed
## 20.781 3.318 13.196
This doesn’t really count as slow mixing, since the Gibbs sampler has converged within 50 iterations for a lattice with \(500 \times 500\) pixels. Compare how long it takes without the external field:
<- system.time(res5 <- mcmcPottsNoData(synth[[4]]$beta, q,
tm5 5000)) neigh, block,
## user system elapsed
## 124.986 3.804 130.207
This explains why single-site Gibbs sampling should never be used for the auxiliary iterations in ABC or the exchange algorithm, but it is usually fine to use when updating the hidden labels. Note that all of these results have been for a fixed \(\beta\). It is more difficult to assess convergence when \(\beta\) is unknown. A topic for a future post!