I recently got a comment on a post I published one year ago, here, about the fact that in September 2009, on the 6th and the 10th, the 6 same numbers came out at the lottery, in Bulgaria (but I do not understand the question: the author of the comment ask about the order the numbers came out…)
Xi’an published also a post on that topic, there, since last week, the same thing happened in Israel.
All that reminded me a discussion I had with a colleague about another post (here) where I mentioned that I found a strange distribution of numbers in the French lottery (the old one actually). For those who want to check, all historical events are here, in a zip file. My colleague was wondering if I found the martingale to win the lottery…
First, I do not like that term, since martingale is something different from a mathematical point of view… Second, let us look if it would have been possible to make some money… (free lunch ?)
> loto=read.table("D:\\loto.csv",dec=",",header=TRUE,sep=";")
> ntirage=nrow(loto)
> loto=loto[51:ntirage,]
> ntirage=nrow(loto)
> N=as.matrix(loto[,c("boule_1","boule_2","boule_3","boule_4","boule_5","boule_6")])
> n=as.vector(N)
> length(n)
[1] 28848
> (TN=table(n))
n
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
607 576 571 618 579 598 608 582 588 590 562 577 577 580 591 630 558 567 594 608
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
578 562 579 583 574 589 602 572 550 598 604 582 545 646 597 618 599 636 609 588
41 42 43 44 45 46 47 48 49
576 589 577 585 618 596 560 571 604
So, it might look nice, but we have to compare that distribution with the one we should have with “independent” draws. It is not possible to look at a discrete uniform distribution: the six numbers are not independent. Each day, the 49 balls are back in the urn, but within a day, we do not have independent draws (it is a sample without replacement of balls). Hence, with 4808 lottery draws, each number cannot be obtained more than 4808 times. So, let us use monte carlo techniques to look at the theoretical distribution,
> M=matrix(NA,49,1000)
> for(s in 1:1000){
+ B=NA
+ for(i in 1:ntirage){B=c(B,sample(1:49,size=6,replace=FALSE))}
+ B=B[-1]
+ M[,s]=sort(table(B))
+ }
> q50=function(x){quantile(x,.5)}
> Q50=apply(M,1,q50)
> lines(1:49,Q50,col="red",lwd=2)
> q10=function(x){quantile(x,.1)}
> Q10=apply(M,1,q10)
> q90=function(x){quantile(x,.9)}
> Q90=apply(M,1,q90)
> polygon(c(1:49,49:1),c(Q10,rev(Q90)),col="light blue",border=NA)
> lines(1:49,Q10,col="red",lty=2)
> lines(1:49,Q90,col="red",lty=2)
> lines(1:49,Q50,col="red",lwd=2)
> points(1:49,sort(TN),pch=19,type="b")
Looking at the graph, it looks like some numbers appeared too frequently, especially the ones that did not appear frequently (bottom left). So, since I have removed the last 50 draws, let us see if we could have used that information, somehow…
> nb=names(sort(TN))
> loto=read.table("D:\\loto.csv",dec=",",header=TRUE,sep=";")
> loto=loto[1:50,]
> N=as.matrix(loto[,c("boule_1","boule_2","boule_3","boule_4","boule_5","boule_6")])
> n=as.vector(N)
> TN=table(n)
> TN[nb]
> barplot(TN[nb])
Unfortunately, numbers that came out too frequently over 4800 draws did not appear that frequently of the last 50. Playing top number might not have been a great strategy.
(numbers that came out frequently are on the right, while those we did not see much are on the left)… What about worst numbers: if I had decided to play the 6 that did not come out very frequently (we’ve seen earlier that they should have appeared even less, actually), would it have been interesting ? As we can see, our top 2 numbers were numbers that did not appear frequently earlier (29 and 47 appears respectively 10 and 11 times over 50 draws)….
Over 50 draws of 6 balls, the expected frequency of 6 given number is around 36.7,..
> S=rep(NA,10000)
> for(s in 1:10000){
+ B=NA
+ for(i in 1:50){B=c(B,sample(1:49,size=6,replace=FALSE))}
+ B=B[-1]
+ S[s]=sum(B%in%(1:6))
+ }
> mean(S)
[1] 36.7694
But here for the top 6, we have
> z=TN[nb] > sum(rev(z)[1:6]) [1] 29
i.e. the top 6 appeared 29 times over 50 draw of 6 balls (which looks low) and for the worst 6, it is a bit higher,
> sum(z[1:6]) [1] 38
If we look at the theoretical density of the frequency of 6 given number, we have
i.e. our worst 6 is a nice average (in green) while top 6 did not appear frequently this time (here in blue) ! So we could not have used that information….
Anyway, if some of you are interesting using statistics to get a free lunch, with the nouveau loto, I did not see any strange pattern (data can be downloaded here).
I am terribly sorry, but I cannot help anyone winning at the French Lottery….
is a spline function (just to make it as general as possible, since in insurance ratemaking, we include continuous variates that do not influence claims frequency linearly in the score). Yes, there might be also useless variables, including one of them which is strongly correlated with one that has an impact in the regression. The code to generate the dataset is simply
) of the dataset.
subsamples (of equal size), and run 
, as function of 
and a confidence interval, defined as
the previous one being now the clear blue line in the back, we see that taking average of estimators on subsamples is not bad at all, on the contrary,
of an i.i.d. sample 
had a finite mean (based on extreme value results). Since I just used it on a small dataset (yes, with real data), I decided to post the R code, since it is rather simple to use. But instead of working on that dataset, let us see what happens on simulated samples. Consider 
=200 observations generated from a Pareto distribution
=2, as a start)
>1.
(against the assumption that the expected value is finite, i.e. 
), it is natural to consider the likelihood ratio
should be chi square distribution with one degree of freedom. As mentioned 
, and thus, to fit a GPD on the 
largest values. And then to plot all that on a graph (like the Hill plot)
the likelihood ratio statistics has a chi-square distribution).
and 
are perfectly known, and the mixture parameter is the only one we care about.
(that cannot be observed), taking value when 
is drawn from 
and 
.
is known, denoted 
. Then I can predict the value of 
. And I can iterate from here.
is the best predictor of 
given my observations (as well as my belief in 
. Recall that we had 
was in 
, then we could have considered mean and standard deviations of observations such that 
before),
proportional to
, and 
being a parameter that will change, from 0 to 4.
or 
, depending whether 
or 
is the smallest parameter.
and 
et 
ont la même loi et que l’on suppose les variables indépendantes, alors
tel que
est la copule de survie associée à 
, au sens où
Samedi après midi, à l’heure de la sieste pour certains, nous discutions avec Magali, Jean Baptiste et Florent des machines qui pouvaient trier les cartes, de là, la discussion a dévié sur les algorithmes de tri, et sur la difficulté d’expliquer l’importance de la vitesse d’un algorithme, ou plus généralement, la vitesse de convergence. Bref, avant de parler de vitesse de convergence en statistique – ou peut être avant sur les méthodes de calcul d’intégrale (ou d’espérance) – je voulais faire un petit billet rapide sur les algorithmes de tri. Je commence par des choses simples. De mémoire c’est des choses que je me souviens avoir vu au lycée dans les années 90, dans les cours d’informatique… Mais l’exemple peut être intéressant pour comprendre la différence, par exemple, entre 
, 
et 
.
. On va balayer autant de fois la base que l’on a d’observations, et à chaque passage, on récupère le maximum (ou le minimum, peu importe). L’algorithme ressemble à ça,
.Bon, j’ai été un peu brutal dans mon algorithme. En fait, on voit que les derniers points sont rapidement bien classés, donc inutile que la seconde boucle aille jusqu’au bout… on peut donc faire du 
, en faisant
. L’idée est de couper, de trier, puis de fusionner. L’idée étant que trier un tableau dont la première moitié des éléments sont ordonnés, et où la dernière moitié est ordonné doit être plus rapide que quand tout est dans le désordre. A chaque étape, l’algorithme est simple, on compare le minimum des deux listes, puis on le vire à l’étape suivante. Sous R, ça donne, quand on commence
(en noir), en 
(en 
(en 
le nombre de carte total (ici 200 car je suppose qu’il existe 200 personnages). Je note 
le coût d’une carte cachée (50 centimes ici) et 
le coût d’une carte non-cachée (1 euro). Supposons que j’ai à en ma possession les cartes 
(et que je n’ai pas les suivantes). Si je veux récupérer une nouvelle carte (que je noterais 
), en achetant au hasard i.e. dans des sachets hermétiques, le nombre de cartes à acheter est 
, qui suit une loi géométrique, de paramètre 
, où
cartes, et qu’il m’en manque simplement une, alors la probabilité de la tirer est
), le coût moyen de ma stratégie est
I wanted to upload here a small problem I started to work on…. unfortunatley, I could not find (yet) a proper answer. Any comments and suggestions are welcomed. The problem is simple: 
and 
. From Fréchet-Hoeffding bounds, we know that
. Further, correlation matrices are necessarily positive-semidefinite matrices.
will be a function of 
.
and 
) are functions of 
,
and 
.
, respectively describing global dependence (that can be related to 
from 
(but actually any positive distribution should work) and 
)
with a Bernoulli 
distribution
from copula 
if 
and 
if 
and 
, and 
In order to get a better understanding, I look at slices, for instance when 
. On the graph on the right, to dark area are values that can not be obtained since correlation have to be positive-semidefinite. Dots are points that have been obtained in a particular scenario. I have also included the convex hull if the points obtained with all the scenarios (here 50,000 scenarios), since I assume that the set of positive correlation is necessarily convex. Graphs below have been obtained when 
and 
.
dépend du niveau des taux d’intérêt. L’ensemble des tables de décallage sont en ligne 
, applied to 
). This upper bound can be far away from the true probability, see e.g. the gaussian case below, i.e. if 
,
, i.e. 
, we have an upper bound for the 
Value-at-Risk,
, then 
, i.e. 
. But since, 
distribution, then
