my book available for a mere $1,091.50

Posted in Books with tags , , , on May 1, 2016 by xi'an

As I was looking at a link to my Bayesian Choice book on Amazon, I found that one site offered it for the modest sum of $1,091.50, a very slight increase when compared with the reference price of $59.95… I do wonder at the reason (scam?) behind this offer as such a large price is unlikely to attract any potential buyer to the site. (Obviously, if you are interested by this price, feel free to contact me!)

another glorious sunrise in Warwickshire

Posted in pictures, Running, Travel, University life with tags , , , , , , , on April 30, 2016 by xi'an

une vie brève [book review]

Posted in Books, Kids, pictures, University life with tags , , , , , , , on April 30, 2016 by xi'an

This short book is about the equally short life (une vie brève) of the young mathematician Maurice Audin, killed or executed by French special forces (Massu’s paratroopers) in Algiers during the Algerian liberation war. Maurice Audin was 25 when he died and the circumstances of his death remain unknown, since the French army never acknowledged this death and never returned his body to his family, but he presumably died under torture. He was a member of the Algerian communist party which had then been outlawed by the French authorities for supporting Algerian independence. Maurice Audin was arrested on June 11, 1957 for hiding a fugitive and he died in the following days… The book is written by his daughter, Michèle Audin, also a mathematician, and a writer of several novels around mathematics and mathematicians. It does not dwell on the death since so little is known but rather reconstructs the life of Maurice Audin from bits and pieces, family memories, school archives, a few pictures, some grocery bills of the Audin family… The style of Michèle Audin is quite peculiar, almost like written thoughts or half-thoughts at times, with a sort of surgical distanciation that makes the book both strong and touching. Maurice Audin wrote several papers in les Comptes Rendus de l’Académie des Sciences [the French PNAS] but did not live long enough to defend his thesis, which was presented by Laurent Schwartz the following year and defended in absentia… The French State never acknowledged its responsability in Audin’s death. (Another book on this death is L’Affaire Audin by the historian Pierre-Vidal Naquet, which appeared in 1958.)

asking for a rebate and getting it!

Posted in Books, pictures, University life with tags , on April 29, 2016 by xi'an

Following my discussion of Ron Gallant’s paper, I received an email from the Global Journal of Management And Business Research

I came across your research paper entitled, “Comment on: Reflections on the Probability Space Induced by Moment Conditions with Implications for Bayesian Inference” and feel that your research is having a very good impact.

With a view to beginning a fruitful, long-term association with you, I invite you to submit your upcoming research articles/papers for publication in the Global Journal of Management and Business Research (GJMBR), an international, double-blind, peer-reviewed research journal.

Global Journals Inc. (US) is well known – the leading fastest growing research publishing organization in the world. We encourage research activities all around the globe with online, 3D and print versions. We also follow an open journal system.

Dr. R. K. Dixit
Chief Author (Hon.)
(Fellow of Association of Research in Business)
Global Journals Incorporated

While I was not in the least interested in publishing in a journal of management and business, I went and check on the journal website for the small prints and in particular for the cost of publishing there, which was not mentioned in the email. Bingo! The publication charge is listed as $420 for a six page paper. I thus replied politely to this Dr. R.K. Dixit who does not seem to exist anywhere but as a signature for this journal (and neither does the mentioned association!) enquiring about whether publication were waived. The very next day I received a reply offering me a 50% rebate on the cost (which is supposed to cover referees’ fees and hard copy printing). Most revealing, in that getting even a mere $210 seems enough to make a profit for such predatory journals (referring to the list set by Jeffrey Beall).

gap frequencies [& e]

Posted in Kids, R with tags , , , on April 29, 2016 by xi'an

A riddle from The Riddler where brute-force simulation does not pay:

For a given integer N, pick at random without replacement integers between 1 and N by prohibiting consecutive integers until all possible entries are exhausted. What is the frequency of selected integers as N grows to infinity?

A simple implementation of the random experiment is as follows

  while (max(frei)==1){

It is however quite slow and does not exploit the recursive feature of the sampling, namely that the first draw breaks the set {1,…,N} into two sets:

  if (N<2){ return((N>0))}else{

But even this faster solution takes a while to run for large values of N:

  for (t in 1:1e3) space=space+generipe(N)

as for instance

>  microbenchmark(frqns(100),time=10)
Unit: nanoseconds
       expr       min        lq         mean    median        uq       max
 frqns(100) 178720117 185020903 212212355.77 188710872 205865816 471395620
       time         4         8        26.13        32        37       102

Hence this limits the realisation of simulation to, say, N=10⁴. Thinking further about the recursive aspect of the process however leads to a recursion on the frequencies qN, as it is straightforward to prove that

q_N=\frac{1}{N}+\frac{2}{N^2}\,\sum_{i=1}^{N-2} iq_i

with q1=1 and q2=1/2. This recursion can be further simplified into


which allows for a much faster computation

s=seq(1,1e7) #s[n]=n*q[n]
for (n in 3:1e7) s[n]=(1+2*s[n-2]+(n-1)*s[n-1])/n

and a limiting value of 0.4323324… Since storing s does not matter, a sliding update works even better:

for (n in 3:1e8){ c=(1+2*a+(n-1)*b)/n;a=b;b=c}

still producing a final value of 0.4323324, which may mean we have reached some limit in the precision.

As I could not figure out a way to find the limit of the sequence (1) above, I put it on the maths forum of Stack Exchange and very quickly got the answer (obtained by a power series representation) that the limit is (rather amazingly!)

\dfrac{1 - e^{-2}}{2}

which is 0.432332358.., hence very close to the numerical value obtained for n=3×10⁸. (Which does not change from n=10⁸, once again for precision reasons.) Now I wonder whether or not an easier derivation of the above is feasible, but we should learn about it in a few hours on The Riddler. [Update: The solution published by The Riddler is exactly that one, using a power series expansion to figure out the limit of the series, unfortunately. I was hoping for a de Montmort trick or sort of…]


Posted in pictures, Statistics, Travel, University life with tags , , , , , , , , , , , , on April 28, 2016 by xi'an

Le Monde puzzle [#960]

Posted in Kids, R with tags , , , , , on April 28, 2016 by xi'an

An arithmetic Le Monde mathematical puzzle:

Given an integer k>1, consider the sequence defined by F(1)=1+1 mod k, F²(1)=F(1)+2 mod k, F³(1)=F²(1)+3 mod k, &tc. [With this notation, F is not necessarily a function.] For which value of k is the sequence the entire {0,1,…,k-1} set?

This leads to an easy brute force resolution, for instance writing the R function

crkl<-function(k) return(unique(cumsum(1:(2*k))%%k))

where 2k is a sufficient substitute for ∞. Then the cases where the successive images of 1 visit the entire set {0,1,…,k-1} are given by

> for (i in 2:550) if (length(crkl(i))==i) print(i)
[1] 2
[1] 4
[1] 8
[1] 16
[1] 32
[1] 64
[1] 128
[1] 256
[1] 512

which suspiciously looks like the result that only the powers of 2 k=2,2²,2³,… lead to a complete exploration of the set {0,1,…,k-1}. Checking a few series in the plane back from Warwick, I quickly found that when k is odd, (1) the sequence is of period k and (2) there is symmetry in the sequence, which means it only takes (k-1)/2 values. For k even, there is a more complicated symmetry, with the sequence being of period 2k, symmetric around its two middle values, and taking the values 1,2,..,1+k(2k+1)/4,..,1+k(k+1)/2. Those values cannot cover the set {0,1,…,k-1} if two are equal, which means an i(i+1)/2 congruent to zero modulo k, hence equal to k. This is clearly impossible when k is a power of 2 because i and i+1 cannot both divide a power of 2. I waited for the published solution as of yesterday’s and the complete argument was to show that when N=2p, the corresponding sequence [for N] is made (modulo p) of the sequence for p plus the same sequence translated by p. The one for N is complete only if the one for p is complete, which by recursion eliminates all cases but the powers of 2…


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