## diffusions, sampling, and transport

Posted in Books, pictures, Statistics, Travel, University life with tags , , , , , , , , , , , , , on November 21, 2022 by xi'an

The third and final day of the workshop was shortened for me as I had to catch an early flight back to Paris (and as I got overly conservative in my estimation for returning to JFK, catching a train with no delay at Penn Station and thus finding myself with two hours free before boarding, hence reviewing remaining Biometrika submission at the airport while waiting). As a result I missed the afternoon talks.

The morning was mostly about using scores for simulation (a topic of which I was mostly unaware), with Yang Song giving the introductory lecture on creating better [cf pix left] generative models via the score function, with a massive production of his on the topic (but too many image simulations of dogs, cats, and celebrities!). Estimating directly the score is feasible via Fisher divergence or score matching à la Hyvärinen (with a return of Stein’s unbiased estimator of the risk!). And relying on estimated scores to simulate / generate by Langevin dynamics or other MCMC methods that do not require density evaluations. Due to poor performances in low density / learning regions a fix is randomization / tempering but the resolution (as exposed) sounded clumsy. (And made me wonder at using some more advanced form of deconvolution since the randomization pattern is controlled.) The talk showed some impressive text to image simulations used by an animation studio!

And then my friend Arnaud Doucet continued on the same theme, motivating by estimating normalising constant through annealed importance sampling [Yuling’s meta-perspective comes back to mind in that the geometric mixture is not the only choice, but with which objective]. In AIS, as in a series of Arnaud’s works, like the 2006 SMC Read Paper with Pierre Del Moral and Ajay Jasra, the importance (!) of some auxiliary backward kernels goes beyond theoretical arguments, with the ideally sequence being provided by a Langevin diffusion. Hence involving a score, learned as in the previous talk. Arnaud reformulated this issue as creating a transportation map and its reverse, which is leading to their recent Schrödinger bridge generative model. Which [imho] both brings a unification perspective to his work and an efficient way to bridge prior to posterior in AIS. A most profitable morn for me!

Overall, this was an exhilarating workshop, full of discoveries for me and providing me with the opportunity to meet and exchange with mostly people I had not met before. Thanks to Bob Carpenter and Michael Albergo for organising and running the workshop!

## optimal Gaussian zorbing

Posted in Books, Kids, R, Statistics with tags , , , , , , on August 30, 2022 by xi'an

A zorbing puzzle from the Riddler: cover the plane with four non-intersecting disks of radius one towards getting the highest probability (under the standard bivariate Normal distribution).

As I could not see a simple connection between the disks and the standard Normal, beyond the probability of a disk being given by a non-central chi-square cdf (with two degrees of freedom), I (once again) tried a random search by simulated annealing, which ended up with a configuration like the above, never above 0.777 using a pedestrian R code like

for(t in 1:1e6){# move the disk centres
Ap=A+vemp*rnorm(2)
Bp=B+vemp*rnorm(2)
while(dist(rbind(Ap,Bp))<2)Bp=B+vemp*rnorm(2)
Cp=C+vemp*rnorm(2)
while(min(dist(rbind(Ap,Bp,Cp)))<2)Cp=C+vemp*rnorm(2)
Dp=D+vemp*rnorm(2)
while(min(dist(rbind(Ap,Bp,Cp,Dp)))<2)Dp=D+vemp*rnorm(2)
#coverage probability
pp=pchisq(1,df=2,ncp=Ap%*%Ap)+pchisq(1,df=2,ncp=Bp%*%Bp)+
pchisq(1,df=2,ncp=Cp%*%Cp)+pchisq(1,df=2,ncp=Dp%*%Dp)
#simulated annealing step
if(log(runif(1))<(pp-p)/sqrt(temp)){
A=Bp;B=Cp;C=Dp;D=Ap;p=pp
if (sol\$val<p) sol=list(val=pp,pos=rbind(A,B,C,D))}
temp=temp*.9999}


I also tried a simpler configuration where all disk centres were equidistant from a reference centre, but this led to a lower “optimal” probability. I was looking forward the discussion of the puzzle, to discover if anything less brute-force was possible! But there was no deeper argument there beyond the elimination of other “natural” configurations (and missing the non-central χ² connection!). Among these options, having two disks tangent at (0,0) were optimal. But the illustration was much nicer:

## not a Bernoulli factory

Posted in Books, Kids, pictures, R with tags , , , , , , , on May 20, 2020 by xi'an

A Riddler riddle I possibly misunderstood:

Four isolated persons are given four fair coins, which can be either flipped once or returned without being flipped. If all flipped coins come up heads, the team wins! Else, if any comes up tails, or if no flip at all is done, it looses. Each person is further given an independent U(0,1) realisation. What is the best strategy?

Since the players are separated, I would presume the same procedure is used by all. Meaning that a coin is tossed with probability p, ie if the uniform is less than p, and untouched otherwise. The probability of winning is then

4(1-p)³p½+6(1-p)³p½²+4(1-p)p³½³+p⁴½⁴

which is maximum for p=0.3420391, with a winning probability of 0.2848424.

solve $$x⌊x⌊x⌊x⌋⌋⌋=2020$$

Where the integral part is the integer immediately below x. Puzzle that I first fail solving by brute force, because I did not look at negative x’s… Since the fourth root of 2020 is between 6 and 7, the solution is either x=6+ε or x=-7+ε, with ε in (0,1). The puzzle then becomes either

$$(6+ε)⌊(6+ε)⌊(6+ε)⌊6+ε⌋⌋⌋ = (6+ε)⌊(6+ε)⌊36+6ε⌋⌋ = (6+ε)⌊(6+ε)(36+⌊6ε⌋)⌋ = 2020$$

where there are 6 possible integer values for $$⌊6ε⌋, with only ⌊6ε⌋=5 being possible, turning the equation into$$

$$(6+ε)⌊41(6+ε)⌋ = (6+ε)(246+⌊41ε⌋) = 2020$$

where again only $$⌊42ε⌋$$=40 being possible, ending up with

$$1716+286ε = 2020$$

which has no solution in (0,1). In the second case

(-7+$$ε$$)$$⌊(-7+ε)⌊(-7+ε)⌊-7+ε⌋⌋⌋$$ = (-7+$$ε$$)$$⌊(-7+ε)(49+⌊-7ε⌋)⌋ = 2020$$

shows that only $$⌊-7ε⌋$$=-3 is possible, leading to

(-7+$$ε$$)$$⌊46(-7+ε))⌋$$ = (-7+$$ε$$) ($$-322+⌊46ε⌋)=2020$$

with only $$⌊46ε⌋$$=17 possible, hence

2135-305$$ε$$=2020

and

$$ε$$=115/305.

A brute force simulated annealing resolution returns x=-6.622706 after 10⁸ iterations. A more interesting question is to figure out the discontinuity points of the function

$$ℵ(x) = x$$$$⌊x⌊x⌊x⌋⌋⌋$$

as they seem to be numerous:

For instance, only 854 of the first 2020 integers enjoy a solution to $$ℵ(x)$$=n.

## Le Monde puzzle [#1127]

Posted in Books, Kids, R, Statistics with tags , , , , on January 17, 2020 by xi'an

A permutation challenge as Le weekly Monde current mathematical puzzle:

When considering all games between 20 teams, of which 3 games have not yet been played, wins bring 3 points, losses 0 points, and draws 1 point (each). If the sum of all points over all teams and all games is 516, was is the largest possible number of teams with no draw in every game they played?

The run of a brute force R simulation of 187 purely random games did not produce enough acceptable tables in a reasonable time. So I instead considered that a sum of 516 over 187 games means solving 3a+2b=516 and a+b=187, leading to 142 3’s to allocate and 45 1’s. Meaning for instance this realisation of an acceptable table of game results

games=matrix(1,20,20);diag(games)=0
while(sum(games*t(games))!=374){
games=matrix(1,20,20);diag(games)=0
games[sample((1:20^2)[games==1],3)]=0}
games=games*t(games)
games[lower.tri(games)&games]=games[lower.tri(games)&games]*
sample(c(rep(1,45),rep(3,142)))* #1's and 3'
(1-2*(runif(20*19/2-3)<.5)) #sign
games[upper.tri(games)]=-games[lower.tri(games)]
games[games==-3]=0;games=abs(games)


Running 10⁶ random realisations of such matrices with no constraint whatsoever provided a solution with] 915,524 tables with no no-draws, 81,851 tables with 19 teams with some draws, 2592 tables with 18 with some draws and 3 tables with 17 with some draws. However, given that 10*9=90 it seems to me that the maximum number should be 10 by allocating all 90 draw points to the same 10 teams, and 143 3’s at random in the remaining games, and I reran a simulated annealing version (what else?!), reaching a maximum 6 teams with no draws. Nowhere near 10, though!

## riddle on a circle

Posted in Books, Kids, R, Travel with tags , , , , , , , on December 22, 2019 by xi'an

The Riddler’s riddle this week provides another opportunity to resort to brute-force simulated annealing!

Given a Markov chain defined on the torus {1,2,…,100} with only moves a drift to the right (modulo 100) and a uniformely random jump, find the optimal transition matrix to reach 42 in a minimum (average) number of moves.

Which I coded in my plane to Seattle, under the assumption that there is nothing to do when the chain is already in 42. And the reasoning that there is not gain (on average) in keeping the choice between right shift and random jump random.

dure=min(c(41:0,99:42),50)
temp=.01
for (t in 1:1e6){
i=sample((1:100)[-42],1)
dura=1+mean(dure)
if (temp*log(runif(1))<dure[i]-dura) dure[i]=dura
if(temp*log(runif(1))<dure[i]-(dura<-1+dure[i*(i<100)+1]))
dure[i]=dura
temp=temp/(1+.1e-4*(runif(1)>.99))}


In all instances, the solution is to move at random for any position but those between 29 and 41, for an average 13.64286 number of steps to reach 42. (For values outside the range 29-42.)