Archive for Tofino

beach walk in the fog [jatp]

Posted in pictures, Running, Travel with tags , , , , , , , , , , , on August 22, 2018 by xi'an

riddles on a line

Posted in Books, Kids, R with tags , , , , , , , , , on August 22, 2018 by xi'an

In the Riddler of August 18, two riddles connected with the integer set Ð={2,3,…,10}:

  1. Given a permutation of Ð, what is the probability that the most likely variation (up or down) occurs at each term?
  2. Given three players choosing three different integers in Ð sequentially, and rewards in Ð allocated to the closest of the three players (with splits in case of equal distance), what is the reward for each given an optimal strategy?

For the first question, a simple code returns 0.17…

winofail<-function(permz){ 
  if (length(permz)>1){
    lenoperm=length(permz[-1])/2
    win=(permz[1]<permz[2])*(sum(permz[-1]>permz[1])>lenoperm)+
     (permz[1]>permz[2])*(sum(permz[-1]<permz[1])>lenoperm)+
      (runif(1)<.5)*(sum(permz[-1]>permz[1])==lenoperm)
    win=win&&winofail(permz[-1])
  }else win=TRUE
  return(win)}

(but the analytic solution exhibits a cool Pascal triangular relation!) and for the second question, a quick recursion or dynamic programming produces 20, 19, 15 as the rewards (and 5, 9, 8 as the locations)

gainz<-function(seqz){
  difz=t(abs(outer(2:10,seqz,"-")))
  cloz=apply(difz,2,rank)
  return(apply(rbind(2:10,2:10,2:10)*
   ((cloz==1)+.5*(cloz==1.5)),1,sum))}

timeline<-function(prev){ 
  if (length(prev)==0){ 
   sol=timeline(10);bez=gainz(sol)[1] 
   for (i in 2:9){ 
    bol=timeline(i);comp=gainz(bol)[1] 
    if (comp>bez){
        bez=comp;sol=bol}}}
  if (length(prev)==1){
    bez=-10
    for (i in (2:10)[-(prev-1)]){
      bol=timeline(c(prev,i))
      comp=gainz(bol)[2]
      if (comp>bez){
        bez=comp;sol=bol}}}
  if (length(prev)==2){
    bez=-10
    for (i in (2:10)[-(prev-1)]){
      bol=c(prev,i)
      comp=gainz(bol)[3]
      if (comp>bez){
        bez=comp;sol=bol}}}
  return(sol)}

After reading the solution on the Riddler, I realised I had misunderstood the line as starting at 2 when it was actually starting from 1. Correcting for this leads to the same 5, 9, 8 locations of picks, with rewards of 21, 19, 15.

Tofino from the air [jatp]

Posted in Statistics with tags , , , , , , , , , , on August 21, 2018 by xi'an

a funny mistake

Posted in Statistics with tags , , , , , , , , , , , on August 20, 2018 by xi'an

While watching the early morning activity in Tofino inlet from my rental desk, I was looking at a recent fivethirthyeight Riddle, which consisted in finding the probability of stopping a coin game which rule was to wait for the n consecutive heads if (n-1) consecutive heads had failed to happen when requested, which is

p+(1-p)p²+(1-p)(1-p²)p³+…

or

q=\sum_{k=1}^\infty p^k \prod_{j=1}^{k-1}(1-p^j)

While the above can write as

q=\sum_{k=1}^\infty \{1-(1-p^k)\} \prod_{j=1}^{k-1}(1-p^j)

or

\sum_{k=1}^\infty \prod_{j=1}^{k-1}(1-p^j)-\prod_{j=1}^{k}(1-p^j)

hence suggesting

q=\sum_{k=1}^\infty \prod_{j=1}^{k-1}(1-p^j) - \sum_{k=2}^\infty \prod_{j=1}^{k-1}(1-p^j) =1

the answer is (obviously) false and the mistake in separating the series into a difference of series is that both terms are infinite. The correct answer is actually

q=1-\prod_{j=1}^{\infty}(1-p^j)

which is Euler’s function. Maybe nonstandard analysis can apply to go directly from the difference of the infinite series to the answer!

asymptotics of M³C²L

Posted in Statistics with tags , , , , , , , on August 19, 2018 by xi'an
In a recent arXival, Blazej Miasojedow, Wojciech Niemiro and Wojciech Rejchel establish the convergence of a maximum likelihood estimator based on an MCMC approximation of the likelihood function. As in intractable normalising constants. The main result in the paper is a Central Limit theorem for the M³C²L estimator that incorporates an additional asymptotic variance term for the Monte Carlo error. Where both the sample size n and the number m of simulations go to infinity. Independently so. However, I do not fully perceive the relevance of using an MCMC chain to target an importance function [which is used in the approximation of the normalising constant or otherwise for the intractable likelihood], relative to picking an importance function h(.) that can be directly simulated.