## Dorfman’s group testing

Posted in Books, Kids, Statistics, Travel with tags , , , , , , , , on April 9, 2020 by xi'an

A recent note by CREST researchers insists on using group testing to compensate for the shortage of test packages and testing personal, as done in several countries, towards deconfining individuals who are not infected. Or who are exhibiting the right antibodies.  Reminding me of my first entry to the notion, in Feller’s book, of the method implemented by Robert Dorfman to test for syphilis prior to enlisting potential WW II soldiers. I would deem the idea useful for surveys, in identifying the proportion of infected or immunised persons, maybe less for giving the green light to leave one’s house as the logistics of merging the tests while keeping track of every individual could prove impossible.

## complex Cauchys

Posted in Books, pictures, Statistics, Travel, University life with tags , , , , , , , , , , , on February 8, 2018 by xi'an

During a visit of Don Fraser and Nancy Reid to Paris-Dauphine where Nancy gave a nice introduction to confidence distributions, Don pointed out to me a 1992 paper by Peter McCullagh on the Cauchy distribution. Following my recent foray into the estimation of the Cauchy location parameter. Among several most interesting aspects of the Cauchy, Peter re-expressed the density of a Cauchy C(θ¹,θ²) as

f(x;θ¹,θ²) = |θ²| / |x-θ|²

when θ=θ¹+ιθ² [a complex number on the half-plane]. Denoting the Cauchy C(θ¹,θ²) as Cauchy C(θ), the property that the ratio aX+b/cX+d follows a Cauchy for all real numbers a,b,c,d,

C(aθ+b/cθ+d)

[when X is C(θ)] follows rather readily. But then comes the remark that

“those properties follow immediately from the definition of the Cauchy as the ratio of two correlated normals with zero mean.”

which seems to relate to the conjecture solved by Natesh Pillai and Xiao-Li Meng a few years ago. But the fact that  a ratio of two correlated centred Normals is Cauchy is actually known at least from the1930’s, as shown by Feller (1930, Biometrika) and Geary (1930, JRSS B).

## easy riddle

Posted in Books, Kids, R with tags , , , , , on July 12, 2017 by xi'an

From the current Riddler, a problem that only requires a few lines of code and a few seconds of reasoning. Or not.

N households each stole the earnings from one of the (N-1) other households, one at a time. What is the probability that a given household is not burglarised? And what are the expected final earnings of each household in the list, assuming they all start with \$1?

The first question is close to Feller’s enveloppe problem in that

$\left(1-\frac{1}{N-1}\right)^{N-1}$

is close to exp(-1) for N large. The second question can easily be solved by an R code like

N=1e3;M=1e6
fina=rep(1,N)
for (v in 1:M){
ordre=sample(1:N)
vole=sample(1:N,N,rep=TRUE)
while (min(abs(vole-(1:N)))==0)
vole[abs(vole-(1:N))==0]=sample(1:N,
sum(vole-(1:N)==0))
cash=rep(1,N)
for (t in 1:N){
cash[ordre[t]]=cash[ordre[t]]+cash[vole[t]];cash[vole[t]]=0}
fina=fina+cash[ordre]}


which returns a pretty regular exponential-like curve, although I cannot figure the exact curve beyond the third burglary. The published solution gives the curve

${\frac{N-2}{N-1}}^{999}\times 2+{\frac{1}{N-1}}^{t-1}\times{\frac{N-1}{N}}^{N-t}\times\frac{N}{N-1}$

corresponding to the probability of never being robbed (and getting on average an extra unit from the robbery) and of being robbed only before robbing someone else (with average wealth N/(N-1)).

## sampling by exhaustion

Posted in Books, Kids, R, Statistics with tags , , , , on November 25, 2016 by xi'an

The riddle set by The Riddler of last week sums up as follows:

Within a population of size N, each individual in the population independently selects another individual. All individuals selected at least once are removed and the process iterates until one or zero individual is left. What is the probability that there is zero individual left?

While I cannot see a clean analytical solution to this problem, it reminds me of an enveloppe-versus-letter (matches) problem I saw in graduate school. Indeed, the expected number of removed (or selected) individuals is given by

$N\left\{1-\frac{N-2}{N-1}\right\}^{N-1}$

which is equivalent to (1-e⁻¹)N for N large, meaning that the population decreases by an average factor of e⁻¹ at each round. And that it takes on average approximately log(N) iterations to reach a single individual. A simulation of the first probabilities of ending with one individual led me to the above curve, which wiggles in an almost periodic way around the probability ½, equal to the average of those probabilities. Using the R code

rad=function(N){#next population size
ut=sample(rep(2:N,2),1)
for (i in 2:N)#sampling
ut=c(ut,sample(rep((1:N)[-i],2),1))
return(N-length(unique(ut))}
sal=rep(0,m);sal[1]=1
for (N in 3:M){
prop=0;
for (t in 1:T){#one single step
if (i>0) prop=prop+sal[i]}
sal[N]=prop/T}


which exploits the previously computed probabilities. The variability is most interesting if unexpected, but looking back at Feller‘s sections and exercises on the classical occupancy problem, I could not find a connection with this problem. If it exists. Still, if N is large enough, the exclusion of one index from the selection becomes negligible and the probability of moving from n to m individuals should be approximately [Feller, eqn (2.4), p.102]

$p_n(m)={n\choose m}\sum_{v=}^{n-m} (-1)^v {n-m\choose v} \left(1-\frac{m+v}{n}\right)^n$

This formula approximates quite well the exact probability, but as in a previous post about the birthday problem, it proves quite delicate to compute. As already noticed by Feller.

## random walk on a torus [riddle]

Posted in Books, Kids, pictures with tags , , , , , , , , , on September 16, 2016 by xi'an

The Riddler of this week(-end) has a simple riddle to propose, namely given a random walk on the {1,2,…,N} torus with a ⅓ probability of death, what is the probability of death occurring at the starting point?

The question is close to William Feller’s famous Chapter III on random walks. With his equally famous reflection principle. Conditioning on the time n of death, which as we all know is definitely absorbing (!), the event of interest is a passage at zero, or any multiple of N (omitting the torus cancellation), at time n-1 (since death occurs the next time). For a passage in zero, this does not happen if n is even (since n-1 is odd) and else it is a Binomial event with probability

${n \choose \frac{n-1}{2}} 2^{-n}$

For a passage in kN, with k different from zero, kN+n must be odd and the probability is then

${n \choose \frac{n-1+kN}{2}} 2^{-n}$

which leads to a global probability of

$\sum_{n=0}^\infty \dfrac{2^n}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}} 2^{-n}$

i.e.

$\sum_{n=0}^\infty \dfrac{1}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}}$

Since this formula is rather unwieldy I looked for another approach in a métro ride [to downtown Paris to enjoy a drink with Stephen Stiegler]. An easier one is to allocate to each point on the torus a probability p[i] to die at position 1 and to solve the system of equations that is associated with it. For instance, when N=3, the system of equations is reduced to

$p_0=1/3+2/3 p_1, \quad p_1=1/3 p_0 + 1/3 p_1$

which leads to a probability of ½ to die at position 0 when leaving from 0. When letting N grows to infinity, the torus structure no longer matters and the probability of dying at position 0 implies returning in position 0, which is a special case of the above combinatoric formula, namely

$\sum_{m=0}^\infty \dfrac{1}{3^{2m+1}} {2m \choose m}$

which happens to be equal to

$\dfrac{1}{3}\,\dfrac{1}{\sqrt{1-4/9}}=\dfrac{1}{\sqrt{5}}\approx 0.4472$

as can be [unnecessarily] checked by a direct R simulation. This √5 is actually the most surprising part of the exercise!