## another riddle

Posted in Books, Kids, Statistics with tags , , , on August 9, 2016 by xi'an

A riddle from The Riddler, a few weeks ago, to ponder while waiting for Godot another plane:

Consider a game played with a coin, between you and a friend, on the integer line. You are assigned a negative integer -X, and your friend is assigned a positive integer, +Y. A marker is placed at zero on the number line. Then the coin is repeatedly flipped. Every time the coin lands heads, the marker is moved one integer in a positive direction. Every time the coin lands tails, the marker moves one integer in a negative direction. You win if the coin reaches -X first, while your friend wins if the coin reaches +Y first.

What is the expected number of coin flips in a complete game?

Not much to say there since this is the classical Feller’s gambler ruin problem that I used to analyse in my practical classes at Polytechnique. The winning probability is X/(X+Y). Assuming the coin is fair. And the distribution of the hitting times (win or loose) is easily analysed by the reflection principle. But a simpler recursion shows that the expected number of steps till hitting one boundary is XY, which is rather surprising as a multiplicative formula.

Posted in Books, Kids, R, Statistics, University life with tags , , , , , , , , , , , on May 27, 2015 by xi'an 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]


Posted in Books, Kids, R, Statistics, University life with tags , , , , , , , , on May 13, 2015 by xi'an Pierre Druilhet arXived a note a few days ago about the Flatland paradox (due to Stone, 1976) and his arguments against the flat prior. The paradox in this highly artificial setting is as follows:  Consider a sequence θ of N independent draws from {a,b,1/a,1/b} such that

1. N and θ are unknown;
2. a draw followed by its inverse and this inverse are removed from θ;
3. the successor x of θ is observed, meaning an extra draw is made and the above rule applied.

Then the frequentist probability that x is longer than θ given θ is at least 3/4—at least because θ could be zero—while the posterior probability that x is longer than θ given x is 1/4 under the flat prior over θ. Paradox that 3/4 and 1/4 clash. Not so much of a paradox because there is no joint probability distribution over (x,θ).

The paradox was actually discussed at length in Larry Wasserman’s now defunct Normal Variate. From which I borrowed Larry’s graphical representation of the four possible values of θ given the (green) endpoint of x. Larry uses the Flatland paradox hammer to fix another nail on the coffin he contemplates for improper priors. And all things Bayes. Pierre (like others before him) argues against the flat prior on θ and shows that a flat prior on the length of θ leads to recover 3/4 as the posterior probability that x is longer than θ.

As I was reading the paper in the métro yesterday morning, I became less and less satisfied with the whole analysis of the problem in that I could not perceive θ as a parameter of the model. While this may sound a pedantic distinction, θ is a latent variable (or a random effect) associated with x in a model where the only unknown parameter is N, the total number of draws used to produce θ and x. The distributions of both θ and x are entirely determined by N. (In that sense, the flatland paradox can be seen as a marginalisation paradox in that an improper prior on N cannot be interpreted as projecting a prior on θ.) Given N, the distribution of x of length l(x) is then 1/4N times the number of ways of picking (N-l(x)) annihilation steps among N. Using a prior on N like 1/N , which is improper, then leads to favour the shortest path as well. (After discussing the issue with Pierre Druilhet, I realised he had a similar perspective on the issue. Except that he puts a flat prior on the length l(x).) Looking a wee bit further for references, I also found that Bruce Hill had adopted the same perspective of a prior on N.

## Feller’s shoes and Rasmus’ socks [well, Karl’s actually…]

Posted in Books, Kids, R, Statistics, University life with tags , , , , on October 24, 2014 by xi'an Yesterday, Rasmus Bååth [of puppies’ fame!] posted a very nice blog using ABC to derive the posterior distribution of the total number of socks in the laundry when only pulling out orphan socks and no pair at all in the first eleven draws. Maybe not the most pressing issue for Bayesian inference in the era of Big data but still a challenge of sorts!

Rasmus set a prior on the total number m of socks, a negative Binomial Neg(15,1/3) distribution, and another prior of the proportion of socks that come by pairs, a Beta B(15,2) distribution, then simulated pseudo-data by picking eleven socks at random, and at last applied ABC (in Rubin’s 1984 sense) by waiting for the observed event, i.e. only orphans and no pair [of socks]. Brilliant!

The overall simplicity of the problem set me wondering about an alternative solution using the likelihood. Cannot be that hard, can it?! After a few computations rejected by opposing them to experimental frequencies, I put the problem on hold until I was back home and with access to my Feller volume 1, one of the few [math] books I keep at home… As I was convinced one of the exercises in Chapter II would cover this case. After checking, I found a partial solution, namely Exercice 26:

A closet contains n pairs of shoes. If 2r shoes are chosen at random (with 2r<n), what is the probability that there will be (a) no complete pair, (b) exactly one complete pair, (c) exactly two complete pairs among them?

This is not exactly a solution, but rather a problem, however it leads to the value $p_j=\binom{n}{j}2^{2r-2j}\binom{n-j}{2r-2j}\Big/\binom{2n}{2r}$

as the probability of obtaining j pairs among those 2r shoes. Which also works for an odd number t of shoes: $p_j=2^{t-2j}\binom{n}{j}\binom{n-j}{t-2j}\Big/\binom{2n}{t}$

as I checked against my large simulations. So I solved Exercise 26 in Feller volume 1 (!), but not Rasmus’ problem, since there are those orphan socks on top of the pairs. If one draws 11 socks out of m socks made of f orphans and g pairs, with f+2g=m, the number k of socks from the orphan group is an hypergeometric H(11,m,f) rv and the probability to observe 11 orphan socks total (either from the orphan or from the paired groups) is thus the marginal over all possible values of k: $\sum_{k=0}^{11} \dfrac{\binom{f}{k}\binom{2g}{11-k}}{\binom{m}{11}}\times\dfrac{2^{11-k}\binom{g}{11-k}}{\binom{2g}{11-k}}$

so it could be argued that we are facing a closed-form likelihood problem. Even though it presumably took me longer to achieve this formula than for Rasmus to run his exact ABC code!

## the anti-Bayesian moment and its passing online

Posted in Statistics, University life with tags , , on March 8, 2013 by xi'an Our rejoinder “the anti-Bayesian moment and its passing” with Andrew Gelman has now been put online on the webpage of The American Statistician. While this rejoinder is freely available, the paper that generated the discussion and this rejoinder, ““Not Only Defended But Also Applied”: The Perceived Absurdity of Bayesian Inference” is only available to subscribers to The American Statistician. Or through arXiv.