During the week-end, Philippe Rivière made me discover an interesting problem,
“Everyone in a room keeps giving dollars to random others.
You’ll never guess what happens next.”f
It was coming from a post, a few years ago on decisionsciencenews.com… This problem was mentioned in recent post since it is related to an article published in the American Scientist in november 2019, Is Inequality Inevitable? (that was translated in French last week, for Pour la Science in a section wrongly entitled Economics since it is only a physicist vision of an (old) economic problem) – see also Brewster Kahle’s post.
(for those really interested in mathematics of inequalities, with a (mathematical) economic perspective, there are countless interesting articles…. see at least Thony Atkinson‘s book or several articles published in Econometrica – references are given in the slides of the course I gave a few years ago on that topic).
I wanted to try, on my own, because I did not understood most of the posts. Because my first thought is that the problem is ill-posed. First of all, what is this “giving dollars”? is it a fixed amount or a random one ? Let us start by assuming that it is fixed. Now, if you know a little bit about gambling and ruin, you guess that it’s very likelely that some one will get banckrupt (at least on a very very long range)… what should we do with that person? Actually, those points were clarified in Jordan’s post
“Imagine a room full of 100 people with 100 dollars each. With every tick of the clock, every person with money gives a dollar to one randomly chosen other person. After some time progresses, how will the money be distributed?”
A well-posed problem states that only people with money can give (everyone can receive) and the amount of money given is fixed.
- A first model (with possible bankruptcy)
First of all, assume that everyone has a fixed amont of money, say 100 (as discussed above). And that each one must give 1 to someone, picked randomly, or more precisely
“every person gives a dollar to one randomly chosen other person”
So, the other people of person i means sampling in \{1,2,\cdots,n\}\backslash\{i\}
n = 2000
ns = 20000
init = 100
x = rep(init,n)
VX = x
VR = c(x[1],x[1])
for(s in 1:ns){
r = function(i) sample((1:n)[-i],size=1)
other = Vectorize(r)(1:n)
dx = table(other)
dx = as.numeric(dx[as.character(1:n)])
dx[is.na(dx)]=0
x = x -rep(1,n)+dx
VR=cbind(VR,range(x))
if(s %% 200 ==0) VX=cbind(VX,x)
}
Here, I store the range of the wealth of my 2000 people, and every 200 rounds, I also keep tracks of the wealths. The plot of the evolution of the range is the following,
As expected, some people will be ruined… and so far, I did nothing, they keep playing… An easy solution would have been to given them an initial endowment of 1000, and not 100. But that’s only a temporary solution: over 20,000 rounds, there might have no bankruptcy, but over 200,000 there will be ! Before moving to the reflected problem (where only people with money give a dollar), just look at the evolution of the distribution of wealths,
or the evolution of the cumulative distribution
We clearly have more variability as we play. Here, I cannot compute any inequality indices (Lorenz curve is constructed only for positive wealths for instance).
I did not look at analytical results here. The only thing that I know for sure is that about (if there are enough people sharing money) one third (actually 36.78\% i.e. e^{-1}) will give one dollar, and receive nothing… that’s the law of small numbers (that result was mentioned in Jordan’s post).
- The reflected problem (with no bankruptcy)
Consider now the reflected problem
“Imagine a room full of people with the same amount of money. With every tick of the clock, every person with money gives a dollar to one randomly chosen other person. After some time progresses, how will the money be distributed?”
(I call that reflected because if someone hits the zero-barrier, it can only go up : that person gives nothing, and can possibly receive)
for(s in 1:ns){
r = function(i) sample((1:n)[-i],size=1)
other = Vectorize(r)(which(x>0))
dx = table(other)
dx = as.numeric(dx[as.character(1:n)])
dx[is.na(dx)] = 0
x = x -(x>0)*1+dx
VR = cbind(VR,range(x))
if(s %% 200 ==0) VX = cbind(VX,x)
}
Here the range is the following
We are bounded from below (it is not possible to have less than 0) and it seems that extremely reach people are less rich than before. We can look now at the cumulative distribution function (since there is no density here, because of the mass at 0)
(for to get some smooth function, I used a symmetric kernel estimate here, so numerically there are values below 0). Since wealths are positive, we can look at Lorenz curves
It seems that there are more and more inequality, as we play that reallocation game. But here again, I will have to run more simulations (and actually a lot more*) to see if there is a non-degenerated limit with such a game. Here, the distribution of wealth after n rounds is an homogenous Markov chain, taking values in \mathbb{N}_+, and using combinatorials, it should be possible to get the transition matrix…
* in did try (during the night) following the advise of Alex (@AlexSablay) advise, and indeed, there is a limiting distribution, see here
- When the contribution is a fixed part (e.g. 1%) of the wealth
An important issue previously was about additivity : “every person with money gives a dollar“. Inequality measures do not like additive operations, they like multiplicate operations (see Serge Christophe Kohlm’s discussion, for instance), or using other words, changes should be relative, not absolute. What about the following question
“Imagine a room full of 100 people with the same amount of money. With every tick of the clock, every person gives a fixed percentage of his money to one randomly chosen other person. After some time progresses, how will the money be distributed?”
The code will be the following: as previously, we match givers and receivers, but here, we have to compute how people give (here it is 1/100 of the money, at each round). At the very first round, we are strictly equivalent to the previous versions : everyone gives 1. The only thing is that, at the second round, those who got nothing at the first one are required to give “only” 99¢.
frac = 1/100
for(s in 1:ns){
r = function(i) sample((1:n)[-i],size=1)
other = Vectorize(r)(1:n)
df = data.frame(dep = 1:n, arr = other, mont = x*frac)
A = aggregate(df$mont,by=list(df$arr),FUN=sum)
dx = A$x
names(dx) = as.character(A$Group.1)
dx = as.numeric(dx[as.character(1:n)])
dx[is.na(dx)] = 0
x = x*(1-frac)+dx
VR = cbind(VR,range(x))
if(s %% 200 ==0) VX = cbind(VX,x)
}
Here is looks like we have some sort of convergence… at least, no one gets less than 75, and more than 125… The distribution can be visualized below
or via the cumulative distribution function
But to be honest, I don’t know what that distribution is…
To conclude, we can also try something (slightly) different : what if we start with non identical wealths ? Instead of having everyone with wealth 100$, what if it was uniformely distributed between 0$ and 200$ ?
x = seq(0,2*init,length=n)
It looks like we have a convergence towards the same distribution, with clearly less inequality than when we started… Here is the cumulative distribution (that started with the uniform distribution)
Again, if someone know what that limiting distribution is, I’d be glad to know !
Didn’t like the ending, February 10, 2009 By Damien Katz
I found a typo, September 14, 2007 By fanfan
A serious reference work?, October 16, 2006 By BJ
Not Nearly A Million, September 3, 2006 By Liron
chansons.
lectures (on suppose 
) est
, soit
et donc la probabilité s’approche par
, alors
on obtient
, et on part de 0.
Et si on regarde l’histogramme et la fonction d’autocorrélation sur les 10 000 premières valeurs, on a gardé l’uniformité (ou tout du moins on ne doit pas en être trop loin), et on n’a plus d’autocorrélation.
Mais cet algorithme présente de très gros défauts numériques. Je n’aborderais pas ici les problèmes de ce générateur (je l’avais fait dans le chapitre sur les simulations dans le tome 2 du livre avec Michel Denuit).
. Par exemple si on prend comme valeur initiale π-3, on obtient, avec une simple fonction R
Plus généralement sur le caractère aléatoire des décimales de π, je renvoie aux publications de David H. Bailey (
Bref, en l’état cette méthode n’est pas vraiment géniale pour générer du “hasard“… à suivre donc.
sur [0,1].
Bref, on a à notre disposition des gens qui ont tiré des boules dans une urne. Alors rigoureusement, il faut faire attention car les tirages sont fait sans remise pour les 7 premières boules, puis on remet les boules pour le tirage suivant… Bref, les tirages ne sont pas rigoureusement indépendant les uns des autres (on ne peut pas avoir 3 fois consécutives le nombre “13”, alors qu’en théorie cela devrait être possible).