Publications by xi'an
a Simpson paradox of sorts
The riddle from The Riddler this week is about finding an undirected graph with N nodes and no isolated node such that the number of nodes with more connections than the average of their neighbours is maximal. A representation of a connected graph is through a matrix X of zeros and ones, on which one can spot the nodes satisfying the above condit...
1657 sym R (1190 sym/2 pcs) 6 img
AISTATS 2016 [#1]
Travelling through Seville, I arrived in Càdiz on Sunday night, along with a massive depression [weather-speaking!]. Walking through the city from the station was nonetheless pleasant as this is an town full of small streets and nice houses. If with less churches than Seville! Richard Samworth gave the first plenary talk of AISTATS 2016 with a...
2764 sym 6 img
reversible chain[saw] massacre
A paper in Nature this week that uses reversible-jump MCMC, phylogenetic trees, and Bayes factors. And that looks at institutionalised or ritual murders in Austronesian cultures. How better can it get?! “by applying Bayesian phylogenetic methods (…) we find strong support for models in which human sacrifice stabilizes social stratification on...
3915 sym 6 img
Using MCMC output to efficiently estimate Bayes factors
As I was checking for software to answer a query on X validated about generic Bayes factor derivation, I came across an R software called BayesFactor, which only applies in regression settings and relies on the Savage-Dickey representation of the Bayes factor when the null hypothesis writes as θ=θ⁰ (and possibly additional nuisance parameter...
4186 sym 8 img
ABC random forests for Bayesian parameter inference
Before leaving Helsinki, we arXived [from the Air France lounge!] the paper Jean-Michel presented on Monday at ABCruise in Helsinki. This paper summarises the experiments Louis conducted over the past months to assess the great performances of a random forest regression approach to ABC parameter inference. Thus validating in this experimental sen...
4309 sym 6 img
occupancy rules
While the last riddle on The Riddler was rather anticlimactic, namely to find the mean of the number Y of empty bins in a uniform multinomial with n bins and m draws, with solution [which still has a link with e in that the fraction of empty bins converges to 1-e⁻¹ when n=m], this led me to some more involved investigation on the distribution...
1596 sym R (534 sym/3 pcs) 10 img
another riddle with a stopping rule
A puzzle on The Riddler last week that is rather similar to an earlier one. Given the probability (1/2,1/3,1/6) on {1,2,3}, what is the mean of the number N of draws to see all possible outcomes and what is the average number of 1’s in those draws? The second question is straightforward, as the proportions of 1’s, 2’s and 3’s in the seque...
1297 sym R (458 sym/2 pcs) 6 img
the random variable that was always less than its mean…
Although this is far from a paradox when realising why the phenomenon occurred, it took me a few lines to understand why the empirical average of a log-normal sample is apparently a biased estimator of its mean. And why the biased plug-in estimator does not appear to present a bias. The picture below compares two estimators of the mean of a log-n...
1496 sym 6 img
Le Monde puzzle [#964]
A not so enticing Le Monde mathematical puzzle: Find the minimal value of a five digit number divided by the sum of its digits. This can formalised as finding the minimum of N/(a+b+c+d+e) when N writes abcde. And solved by brute force. Using a rough approach to finding the digits of a five-digit number, the question can be easily solved as pri...
980 sym R (108 sym/1 pcs) 6 img
the new version of abcrf
A new version of the R package abcrf has been posted on Friday by Jean-Michel Marin, in conjunction with the recent arXival of our paper on point estimation via ABC and random forests. The new R functions come to supplement the existing ones towards implementing ABC point estimation: covRegAbcrf, which predicts the posterior covariance between t...
1686 sym 6 img