WSC 20[1]1

This morning I attended the “Bruce Schmeiser session” at WSC 2011. I had once a meeting with Bruce (and Jim Berger) in Purdue to talk about MCMC methods but I never interacted directly with him. The first two talks were about batch methods, which I did not know previously, and I had trouble understanding what was the problem: for a truly iid normal sample, building an optimal confidence interval on the mean relies on the sufficient statistic rather than on the batch mean variance… It is only through the second talk that I understood that neither normality nor independence was guaranteed, hence the batches. I still wonder whether or a bootstrap strategy could be used instead, given the lack of confidence in the model assumptions. The third talk was about a stochastic approximation algorithm developed by Bruce Schmeiser, called retrospective approximation, where successive and improving approximations of the target to maximise are used in order not to waste time at the beginning. I thus found the algorithm had a simulated annealing flavour, even though the connection is rather tenuous…

The second session of WSC 2011 I attended was about importance sampling, The first talk was about mixtures of importance sampling distributions towards improved efficiency for cross-entropy, à la Rubinstein and Kroese. Its implementation seemed to depend very much on some inner knowledge of the target problem. The second talk was on zero-variance approximations for computing the probability that two notes are connected in a graph, using clever collapsing schemes. The third talk of the session was unrelated with the theme since it was about cross-validated non-parametric density estimation.

My own session was not terribly well attended and, judging from some questions I got at the end I am still unsure I had chosen the right level. Nonetheless, I got interesting discussions afterwards which showed that ABC was also appealing to some members of the audience. And I had a long chat with Enlu Zhou, a nice assistant professor from Urbana-Champaign who was teaching out of Monte Carlo Statistical Method, and had challenging questions about restricted support MCMC. Overall, an interesting day, completed with a light conference dinner in the pleasant company of Jingchen Liu from Columbia and some friends of his.

WSC 2[0]11

I have now registered for the WSC 2011 conference and I am looking forward the first day of talks tomorrow. Especially since, reading from the abstracts to the talks, it sounds as if many participants have a different understanding of the word simulation than I have. (I had the same impression this summer when taking part in a half-day of talks in Lancaster.) I am however slightly worried at having prepared my (advanced) tutorial for the right crowd, being unable to judge the background of the audience. Some of the talks are highly technical, others seem much more elementary… (I spent the whole night and morning, except for a fairly long and great run in the hills at sunrise, collating and adapting my slides from my graduate course and from different talks. The outcome is on slideshare.)

Le Monde puzzle [#752]

After a loooong break, here is one Le Monde mathematical puzzle I had time to look at, prior to going to Dauphine for a Saturday morning class (in replacement of my R class this week)! The question is as follows:

A set of numbers {1,…,N} is such that multiples of 4 are tagged C and multiples of 5 and of 11 are tagged Q. Numbers that are not multiples of 4, 5, or 11, and numbers that are multiples of both 4 and 5 or of both 4 and 11 are not tagged. Find N such that the number of C tags is equal to the number of Q tags.

This is a plain enumeration problem.

N=0
noco=TRUE
nbC=nbQ=0

while (noco){
 N=N+1
 divF=FALSE
 if (trunc(N/4)*4==N){
    nbC=nbC+1
    divF=TRUE
    }
 if ((trunc(N/5)*5==N)||(trunc(N/11)*11==N)){
   if (divF){
     nbC=nbC-1
     }else{ nbQ=nbQ+1}
   }
 noco=(nbC!=nbQ)
 }

When I ran the code, I found many solutions

[1] 1 0 0
[1] 2 0 0
[1] 3 0 0
[1] 5 1 1
[1] 6 1 1
[1] 7 1 1
[1] 10  2  2
[1] 12  3  3
[1] 13  3  3
[1] 14  3  3
[1] 16  4  4
[1] 17  4  4
[1] 18  4  4
[1] 19  4  4
[1] 20  4  4
[1] 21  4  4
[1] 24  5  5
[1] 28  6  6
[1] 29  6  6
[1] 32  7  7
[1] 64 12 12

with no value further than 64 (testing all the way to 3,500,000). This seems in line with the fact that there are more multiples of 5 or 11 than of 4 when N is large enough. This can be seen by drawing the curves of the (approximate) number of multiples:

curve((trunc(x/4)-trunc(x/20)-trunc(x/44)),
  from=10,to=250,n=500)
curve((trunc(x/5)+trunc(x/11)-trunc(x/55)-
  trunc(x/20)-trunc( /44)),from=10,to=250,add=TRUE,n=500)