Another trip in the métro today (to work with Pierre Jacob and Lawrence Murray in a Paris Anticafé!, as the University was closed) led me to infer—warning!, this is not the exact distribution!—the distribution of x, namely

$f(x|N) = \frac{4^p}{4^{\ell+2p}} {\ell+p \choose p}\,\mathbb{I}_{N=\ell+2p}$

since a path x of length l(x) will corresponds to N draws if N-l(x) is an even integer 2p and p undistinguishable annihilations in 4 possible directions have to be distributed over l(x)+1 possible locations, with Feller’s number of distinguishable distributions as a result. With a prior π(N)=1/N on N, hence on p, the posterior on p is given by

$\pi(p|x) \propto 4^{-p} {\ell+p \choose p} \frac{1}{\ell+2p}$

Now, given N and  x, the probability of no annihilation on the last round is 1 when l(x)=N and in general

$\frac{4^p}{4^{\ell+2p}}{\ell-1+p \choose p}\big/\frac{4^p}{4^{\ell+2p}}{\ell+p \choose p}=\frac{\ell}{\ell+p}=\frac{2\ell}{N+\ell}$

which can be integrated against the posterior. The numerical expectation is represented for a range of values of l(x) in the above graph. Interestingly, the posterior probability is constant for l(x) large  and equal to 0.8125 under a flat prior over N.

Getting back to Pierre Druilhet’s approach, he sets a flat prior on the length of the path θ and from there derives that the probability of annihilation is about 3/4. However, “the uniform prior on the paths of lengths lower or equal to M” used for this derivation which gives a probability of length l proportional to 3l is quite different from the distribution of l(θ) given a number of draws N. Which as shown above looks much more like a Binomial B(N,1/2).

However, being not quite certain about the reasoning involving Fieller’s trick, I ran an ABC experiment under a flat prior restricted to (l(x),4l(x)) and got the above, where the histogram is for a posterior sample associated with l(x)=195 and the gold curve is the potential posterior. Since ABC is exact in this case (i.e., I only picked N’s for which l(x)=195), ABC is not to blame for the discrepancy! I asked about the distribution on Stack Exchange maths forum (and a few colleagues here as well) but got no reply so far… Here is the R code that goes with the ABC implementation:

#observation:
elo=195
#ABC version
T=1e6
el=rep(NA,T)
N=sample(elo:(4*elo),T,rep=TRUE)
for (t in 1:T){
#generate a path
paz=sample(c(-(1:2),1:2),N[t],rep=TRUE)
#eliminate U-turns
uturn=paz[-N[t]]==-paz[-1]
while (sum(uturn>0)){
uturn[-1]=uturn[-1]*(1-
uturn[-(length(paz)-1)])
uturn=c((1:(length(paz)-1))[uturn==1],
(2:length(paz))[uturn==1])
paz=paz[-uturn]
uturn=paz[-length(paz)]==-paz[-1]
}
el[t]=length(paz)}
#subsample to get exact posterior
poster=N[abs(el-elo)==0]


### 2 Responses to “the Flatland paradox [#2]”

1. Dan Simpson Says:

I am excited about the concept of an “anticafe”!

• More exciting than it actually is! A bit noisy for working there in my opinion…