## master project?

Posted in Books, Kids, Statistics, University life with tags , , , , , , , on July 25, 2022 by xi'an

A potential master project for my students next year inspired by an X validated question: given a Gaussian mixture density

$f(x)\propto\sum_{i=1}^m \omega_i \sigma^{-1}\,\exp\{-(x-\mu_i)^2/2\sigma^2\}$

with m known, the weights summing up to one, and the (prior) information that all means are within (-C,C), derive the parameters of this mixture from a sufficiently large number of evaluations of f. Pay attention to the numerical issues associated with the resolution.  In a second stage, envision this problem from an exponential spline fitting perspective and optimise the approach if feasible.

## dice and sticks

Posted in Books, Kids, R with tags , , , , , , on November 19, 2021 by xi'an

A quick weekend riddle from the Riddler about the probability of getting a sequence of increasing numbers from dice with an increasing number of faces, eg 4-, 6-, and 8-face dice. Which happens to be 1/4. By sheer calculation (à la Gauss) or simple enumération (à la R):

> for(i in 1:4)for(j in (i+1):6)F=F+(8-j)
> F/4/6/8
[1] 0.25

The less-express riddle is an optimisation problem related with stick breaking: given a stick of length one, propose a fraction a and win (1-a) if a Uniform x is less than one. Since the gain is a(1-a) the maximal average gain is associated with a=½. Now, if the remaining stick (1-a) can be divided when x>a, what is the sequence of fractions one should use when the gain is the length of the remaining stick? With two attempts only, the optimal gain is still ¼. And a simulation experiment with three attempts again returns ¼.

## data assimilation and reduced modelling for high-D problems [CIRM]

Posted in Books, Kids, Mountains, pictures, Running, Statistics, University life with tags , , , , , , , , , , , , , , , , , on February 8, 2021 by xi'an

Next summer, from 19 July till 27 August, there will be a six week program at CIRM on the above theme, bringing together scientists from both the academic and industrial communities. The program includes a one-week summer school followed by 5 weeks of research sessions on projects proposed by academic and industrial partners.

Confirmed speakers of the summer school (Jul 19-23) are:

• Albert Cohen (Sorbonne University)
• Masoumeh Dashti (University of Sussex)
• Eric Moulines (Ecole Polytechnique)
• Anthony Nouy (Ecole Centrale de Nantes)
• Claudia Schillings (Mannheim University)

Junior participants may apply for fellowships to cover part or the whole stay. Registration and application to fellowships will be open soon.

## Fermat’s Riddle

Posted in Books, Kids, R with tags , , , , , , , , , , on October 16, 2020 by xi'an

·A Fermat-like riddle from the Riddler (with enough room to code on the margin)

An  arbitrary positive integer N is to be written as a difference of two distinct positive integers. What are the impossible cases and else can you provide a list of all distinct representations?

Since the problem amounts to finding a>b>0 such that

$N=a^2-b^2=(a-b)(a+b)$

both (a+b) and (a-b) are products of some of the prime factors in the decomposition of N and both terms must have the same parity for the average a to be an integer. This eliminates decompositions with a single prime factor 2 (and N=1). For other cases, the following R code (which I could not deposit on tio.run because of the packages R.utils!) returns a list

library(R.utils)
library(numbers)
bitz<-function(i,m) #int2bits
c(rev(as.binary(i)),rep(0,m))[1:m]
ridl=function(n){
a=primeFactors(n)
if((n==1)|(sum(a==2)==1)){
print("impossible")}else{
m=length(a);g=NULL
for(i in 1:2^m){
b=bitz(i,m)
if(((d<-prod(a[!!b]))%%2==(e<-prod(a[!b]))%%2)&(d<e))
g=rbind(g,c(k<-(e+d)/2,l<-(e-d)/2))}
return(g[!duplicated(g[,1]-g[,2]),])}}

For instance,

> ridl(1456)
[,1] [,2]
[1,]  365  363
[2,]  184  180
[3,]   95   87
[4,]   59   45
[5,]   40   12
[6,]   41   15

Checking for the most prolific N, up to 10⁶, I found that N=6720=2⁶·3·5·7 produces 20 different decompositions. And that N=887,040=2⁸·3²·5·7·11 leads to 84 distinct differences of squares.

## another easy Riddler

Posted in Books, Kids, R with tags , , , , on January 31, 2020 by xi'an

A quick riddle from the Riddler

In a two-person game, Abigail and Zian both choose between a and z. Abigail win one point with probability .9 if they choose (a,a) and with probability 1 if they choose (a,z), and two points with probability .4 if they choose (z,z) and with probability .6 if they choose (z,a). Find the optimal probabilities δ and ς of choosing a for both Abigail and Zian when δ is known to Zian.

Since the average gain for Abigail is δ(1-.1ς)+2(1-δ)(.4+.2ς) the riddle sums up as solving the minmax problem

$\max_\delta \min_\varsigma\delta(1-.1\varsigma)+2(1-\delta)(.4+.2\varsigma)$

the solution in ς is either 0 or 1 depending on δ being smaller or larger than 12/22, which leads to this value as the expected gain. The saddlepoint is hardly visible in the above image. While ς is either 0 or 1 in the optimal setting,  a constant choice of 1 or 0 would modify the optimal for δ except that Abigail must declare her value of δ!