Tag Archives: R-english

Optimal transport on large networks

With Alfred Galichon and Lucas Vernet, we recently uploaded a paper entitled optimal transport on large networks on arxiv.

This article presents a set of tools for the modeling of a spatial allocation problem in a large geographic market and gives examples of applications. In our settings, the market is described by a network that maps the cost of travel between each pair of adjacent locations. Two types of agents are located at the nodes of this network. The buyers choose the most competitive sellers depending on their prices and the cost to reach them. Their utility is assumed additive in both these quantities. Each seller, taking as given other sellers prices, sets her own price to have a demand equal to the one we observed. We give a linear programming formulation for the equilibrium conditions. After formally introducing our model we apply it on two examples: prices offered by petrol stations and quality of services provided by maternity wards (only the later is described here for privacy issues). These examples illustrate the applicability of our model to aggregate demand, rank prices and estimate cost structure over the network. We insist on the possibility of applications to large scale data sets using modern linear programming solvers such as Gurobi.

Demand for gas in gas stations in Britanny, and demand for maternity in France (with border correction)

In addition to this paper we released a R toolbox to implement our results and an online tutorial, optimalnetwork.github.io.

On my way to Manizales (Colombia)

Next week, I will be in Manizales, Colombia, for the Third International Congress on Actuarial Science and Quantitative Finance. I will be giving a lecture on Wednesday with Jed Fress and Emilianos Valdez.

I will give my course on Algorithms for Predictive Modeling on Thursday morning (after Jed and Emil’s lectures). Unfortunately, my computer locked itself last week, and I could not unlock it (could not IT team at the university, who have the internal EFI password). So I will not be able to work further on the slides, so it will be based on the version as-at now (clearly in progress).

 

Pareto Models for Top Incomes

With Emmanuel Flachaire, we uploaded on hal a paper on Pareto Models for Top Incomes,

Top incomes are often related to Pareto distribution. To date, economists have mostly used Pareto Type I distribution to model the upper tail of income and wealth distribution. It is a parametric distribution, with an attractive property, that can be easily linked to economic theory. In this paper, we first show that modelling top incomes with Pareto Type I distribution can lead to severe over-estimation of inequality, even with millions of observations. Then, we show that the Generalized Pareto distribution and, even more, the Extended Pareto distribution, are much less sensitive to the choice of the threshold. Thus, they provide more reliable results. We discuss different types of bias that could be encountered in empirical studies and, we provide some guidance for practice. To illustrate, two applications are investigated, on the distribution of income in South Africa in 2012 and on the distribution of wealth in the United States in 2013.

This paper was presented at and UCSB and in several workshops this spring, and this Summer, Emmanuel will present it at ECINEQ.

Note that a R package is also available on github, TopIncomes.

Estimates on training vs. validation samples

Before moving to cross-validation, it was natural to say “I will burn 50% (say) of my data to train a model, and then use the remaining to fit the model”. For instance, we can use training data for variable selection (e.g. using some stepwise procedure in a logistic regression), and then, once variable have been selected, fit the model on the remaining set of observations. A natural question is usually “does it really matter ?”.

In order to visualize this problem, consider my (simple) dataset

MYOCARDE=read.table(
  "http://freakonometrics.free.fr/saporta.csv",
  head=TRUE,sep=";")

Let us generate 100 training samples (where we keep about 50% of the observations). On each of them, we use a stepwise procedure, and we keep the estimates of the remaining variables (and their standard deviation actually)

n=nrow(MYOCARDE)
M=matrix(NA,100,ncol(MYOCARDE))
colnames(M)=c("(Intercept)",names(MYOCARDE)[1:7])
S1=S2=M1=M2=M
for(i in 1:100){
idx = which(sample(0:1,size=n, replace=TRUE)==1)
reg=step(glm(PRONO=="DECES"~.,data=MYOCARDE[idx,]))
nm=names(reg$coefficients)
M1[i,nm]=reg$coefficients
S1[i,nm]=summary(reg)$coefficients[,2]
f=paste("PRONO=='DECES'~",paste(nm[-1],collapse="+"),sep="")
reg=glm(f,data=MYOCARDE[-idx,])
M2[i,nm]=reg$coefficients
S2[i,nm]=summary(reg)$coefficients[,2]
}

Then, for the 7 covariates (and the constant) we can look at the value of the coefficient in the model fitted on the training sample, and the value on the model fitted on the validation sample (of course, only when they were remaining)

for(j in 1:8){
idx=which(!is.na(M1[,j]))
plot(M1[idx,j],M2[idx,j])
abline(a=0,b=1,lty=2,col="gray")
segments(M1[idx,j]-2*S1[idx,j],M2[idx,j],M1[idx,j]+2*S1[idx,j],M2[idx,j])  
segments(M1[idx,j],M2[idx,j]-2*S2[idx,j],M1[idx,j],M2[idx,j]+2*S2[idx,j])  
}

For instance, with the intercept, we have the following

 

where horizontal segments are confidence intervals of the parameter on the model fitted on the training sample, the vertical on the validation sample. The green part means some sort of consistency, while the red one means that actually, the coefficient was negative with one model, positive with the other one. Which is odd (but in that case, observe that coefficients are rarely significant).

We can also visualize the joint distribution of the two estimators,

for(j in 1:8){
library(ks)
idx = which(!is.na(M1[,j]))
Z = cbind(M1[idx,j],M2[idx,j])
H = Hpi(x=Z)
fhat = kde(x=Z, H=H)
image(fhat$eval.points[[1]],
fhat$eval.points[[2]],fhat$estimate)
abline(a=0,b=1,lty=2,col="gray")
abline(v=0,lty=2)
abline(h=0,lty=2)
}

which are here, almost on the diagonal,

meaning that the intercept on the two samples is (more or less) the same. We can then look at other parameters (which is actually more interesting).

On that variable, it seems that it is significant on the training dataset (somehow, it is consistent with the fact that it is remaining in the model after the stepwise procedure) but not on the validation sample (or hardly significant).

Others are much more consistent (with some possible outliers)

 

 

On the next one, we have again significance on the training sample, but not on the validation sample,

 

 

and probably more interesting

where the two are very consistent.

What it the interpretation of the diagonal for a ROC curve

Last Friday, we discussed the use of ROC curves to describe the goodness of a classifier. I did say that I will post a brief paragraph on the interpretation of the diagonal. If you look around some say that it describes the “strategy of randomly guessing a class“, that it is obtained with “a diagnostic test that is no better than chance level“, even obtained by “making a prediction by tossing of an unbiased coin“.

Let us get back to ROC curves to illustrate those points. Consider a very simple dataset with 10 observations (that is not linearly separable)

x1 = c(.4,.55,.65,.9,.1,.35,.5,.15,.2,.85)
x2 = c(.85,.95,.8,.87,.5,.55,.5,.2,.1,.3)
y = c(1,1,1,1,1,0,0,1,0,0)
df = data.frame(x1=x1,x2=x2,y=as.factor(y))

here we can check that, indeed, it is not separable

plot(x1,x2,col=c("red","blue")[1+y],pch=19)

Consider a logistic regression (the course is on linear models)

reg = glm(y~x1+x2,data=df,family=binomial(link = "logit"))

but any model here can be used… We can use our own function

Y=df$y
S=predict(reg)
roc.curve=function(s,print=FALSE){
  Ps=(S>=s)*1
   
  FP=sum((Ps==1)*(Y==0))/sum(Y==0)
     
  TP=sum((Ps==1)*(Y==1))/sum(Y==1)if(print==TRUE){print(table(Observed=Y,Predicted=Ps))}
   
vect=c(FP,TP)names(vect)=c("FPR","TPR")return(vect)}

or any R package actually

library(ROCR)

perf=performance(prediction(S,Y),"tpr","fpr")

We can plot the two simultaneously here

plot(performance(prediction(S,Y),"tpr","fpr"))
V=Vectorize(roc.curve)(seq(-5,5,length=251))points(V[1,],V[2,])segments(0,0,1,1,col="light blue")

So our code works just fine, here. Let us consider various strategies that should lead us to the diagonal.

The first one is : everyone has the same probability (say 50%)

S=rep(.5,10)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(0,1,length=251))points(V[1,],V[2,])

Indeed, we have the diagonal. But to be honest, we have only two points here : (0,0) and (1,1). Claiming that we have a straight line is not very satisfying… Actually, note that we have this situation whatever the probability we choose

S=rep(.2,10)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(0,1,length=251))points(V[1,],V[2,])

We can try another strategy, like “making a prediction by tossing of an unbiased coin“. This is what we obtain

set.seed(1)

S=sample(0:1,size=10,replace=TRUE)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(0,1,length=251))points(V[1,],V[2,])segments(0,0,1,1,col="light blue")

We can also try some sort of “random classifier”, where we choose the score randomly, say uniform on the unit interval

set.seed(1)

S=runif(10)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(0,1,length=251))points(V[1,],V[2,])segments(0,0,1,1,col="light blue")

Let us try to go further on that one. For convenience, let us consider another function to plot the ROC curve

V=Vectorize(roc.curve)(seq(0,1,length=251))

roc_curve=Vectorize(function(x) max(V[2,which(V[1,]<=x)]))

We have the same line as previously

x=seq(0,1,by=.025)

y=roc_curve(x)lines(x,y,type="s",col="red")

But now, consider many scoring strategies, all randomly chosen

MY=matrix(NA,500,length(y))for(i in 1:500){
  
S=runif(10)
  
V=Vectorize(roc.curve)(seq(0,1,length=251))
  
MY[i,]=roc_curve(x)}plot(performance(prediction(S,df$y),"tpr","fpr"),col="white")for(i in 1:500){lines(x,MY[i,],col=rgb(0,0,1,.3),type="s")}lines(c(0,x),c(0,apply(MY,2,mean)),col="red",type="s",lwd=3)segments(0,0,1,1,col="light blue")

The red line is the average of all random classifiers. It is not a straight line, be we observe oscillations around the diagonal.

Consider a dataset with more observations


myocarde = read.table("http://freakonometrics.free.fr/myocarde.csv",head=TRUE, sep=";")

myocarde$PRONO = (myocarde$PRONO=="SURVIE")*1

reg = glm(PRONO~.,data=myocarde,family=binomial(link = "logit"))

Y=myocarde$PRONO

S=predict(reg)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(-5,5,length=251))points(V[1,],V[2,])segments(0,0,1,1,col="light blue")

Here is a “random classifier” where we draw scores randomly on the unit interval

S=runif(nrow(myocarde)plot(performance(prediction(S,Y),"tpr","fpr"))

V=Vectorize(roc.curve)(seq(-5,5,length=251))points(V[1,],V[2,])segments(0,0,1,1,col="light blue")

And if we do that 500 times, we obtain, on average

MY=matrix(NA,500,length(y))for(i in 1:500){
  
S=runif(length(Y))
  
V=Vectorize(roc.curve)(seq(0,1,length=251))
  
MY[i,]=roc_curve(x)}plot(performance(prediction(S,Y),"tpr","fpr"),col="white")for(i in 1:500){lines(x,MY[i,],col=rgb(0,0,1,.3),type="s")}lines(c(0,x),c(0,apply(MY,2,mean)),col="red",type="s",lwd=3)segments(0,0,1,1,col="light blue")

So, it looks like me might say that the diagonal is what we have, on average, when drawing randomly scores on the unit interval…

I did mention that an interesting visual tool could be related to the use of the Kolmogorov Smirnov statistic on classifiers. We can plot the two empirical cumulative distribution functions of the scores, given the response Y

score=data.frame(yobs=Y,
                 ypred=predict(reg,type="response"))

f0=c(0,sort(score$ypred[score$yobs==0]),1)

f1=c(0,sort(score$ypred[score$yobs==1]),1)plot(f0,(0:(length(f0)-1))/(length(f0)-1),col="red",type="s",lwd=2,xlim=0:1)lines(f1,(0:(length(f1)-1))/(length(f1)-1),col="blue",type="s",lwd=2)

we can also look at the distribution of the score, with the histogram (or density estimates)

S=score$ypred

hist(S[Y==0],col=rgb(1,0,0,.2),
     probability=TRUE,breaks=(0:10)/10,border="white")hist(S[Y==1],col=rgb(0,0,1,.2),
     probability=TRUE,breaks=(0:10)/10,border="white",add=TRUE)lines(density(S[Y==0]),col="red",lwd=2,xlim=c(0,1))lines(density(S[Y==1]),col="blue",lwd=2)

The underlying idea is the following : we do have a “perfect classifier” (top left corner)

is the supports of the scores do not overlap

otherwise, we should have errors. That the case below

we in 10% of the cases, we might have misclassification

or even more missclassification, with overlapping supports

Now, we have the diagonal

when the two conditional distributions of the scores are identical

Of course, that only valid when n is very large, otherwise, it is only what we observe on average….

On the poor performance of classifiers in insurance models

Each time we have a case study in my actuarial courses (with real data), students are surprised to have hard time getting a “good” model, and they are always surprised to have a low AUC, when trying to model the probability to claim a loss, to die, to fraud, etc. And each time, I keep saying, “yes, I know, and that’s what we expect because there a lot of ‘randomness’ in insurance”. To be more specific, I decided to run some simulations, and to compute AUCs to see what’s going on. And because I don’t want to waste time fitting models, we will assume that we have each time a perfect model. So I want to show that the upper bound of the AUC is actually quite low ! So it’s not a modeling issue, it is a fondamental issue in insurance !

By ‘perfect model’ I mean the following : \Omega denotes the heterogeneity factor, because people are different. We would love to get \mathbb{P}[Y=1|\Omega]. Unfortunately, \Omega  is unobservable ! So we use covariates (like the age of the driver of the car in motor insurance, or of the policyholder in life insurance, etc). Thus, we have data (y_i,\boldsymbol{x}_i)‘s and we use them to train a model, in order to approximate \mathbb{P}[Y=1|\boldsymbol{X}]. And then, we check if our model is good (or not) using the ROC curve, obtained from confusion matrices, comparing y_i‘s and \widehat{y}_i‘s where \widehat{y}_i=1 when \mathbb{P}[Y_i=1|\boldsymbol{x}_i] exceeds a given threshold. Here, I will not try to construct models. I will predict \widehat{y}_i=1 each time the true underlying probability \mathbb{P}[Y_i=1|\omega_i] exceeds a threshold ! The point is that it’s possible to claim a loss (y=1) even if the probability is 3% (and most of the time \widehat{y}=0), and to not claim one (y=0) even if the probability is 97% (and most of the time \widehat{y}=1). That’s the idea with randomness, right ?

So, here p(\omega_1),\cdots,p(\omega_n) denote the probabilities to claim a loss, to die, to fraud, etc. There is heterogeneity here, and this heterogenity can be small, or large. Consider the graph below, to illustrate,

In both cases, there is, on average, 25% chance to claim a loss. But on the left, there is more heterogeneity, more dispersion. To illustrate, I used the arrow, which is a classical 90% interval : 90% of the individuals have a probability to claim a loss in that interval. (here 10%-40%), 5% are below 10% (low risk), and 5% are above 40% (high risk). Later on, we will say that we have 25% on average, with a dispersion of 30% (40% minus 10%). On the right, it’s more 25% on average, with a dispersion of of 15%. What I call dispersion is the difference between the 95% and the 5% quantiles.

Consider now some dataset, with Bernoulli variables y, drawn with those probabilities p(\omega). Then, let us assume that we are able to get a perfect model : I do not estimate a model based on some covariates, here, I assume that I know perfectly the probability (which is true, because I did generate those data). More specifically, to generate a vector of probabilities, here I use a Beta distribution with a given mean, and a given variance (to capture the heterogeneity I mentioned above)

a=m*(m*(1-m)/v-1)
b=(1-m)*(m*(1-m)/v-1)
p=rbeta(n,a,b)

from those probabilities, I generate occurences of claims, or deaths,

Y=rbinom(n,size = 1,prob = p)

Then, I compute the AUC of my “perfect” model,

auc.tmp=performance(prediction(p,Y),"auc")

And then, I will generate many samples, to compute the average value of the AUC. And actually, we can do that for many values of the mean and the variance of the Beta distribution. Here is the code

library(ROCR)
n=1000
ns=200
ab_beta = function(m,inter){
  a=uniroot(function(a) qbeta(.95,a,a/m-a)-qbeta(.05,a,a/m-a)-inter,
            interval=c(.0000001,1000000))$root
  b=a/m-a
  return(c(a,b))
}
Sim_AUC_mean_inter=function(m=.5,i=.05){
  V_auc=rep(NA,ns)
  b=-1
  essai = try(ab<-ab_beta(m,i),TRUE) if(inherits(essai,what="try-error")) a=-1 if(!inherits(essai,what="try-error")){ a=ab[1] b=ab[2] } if((a>=0)&(b>=0)){
    for(s in 1:ns){
      p=rbeta(n,a,b)
      Y=rbinom(n,size = 1,prob = p)
      auc.tmp=performance(prediction(p,Y),"auc")
      V_auc[s]=as.numeric(auc.tmp@y.values)}
    L=list(moy_beta=m,
           var_beat=v,
           q05=qbeta(.05,a,b),
           q95=qbeta(.95,a,b),
           moy_AUC=mean(V_auc),
           sd_AUC=sd(V_auc),
           q05_AUC=quantile(V_auc,.05),
           q95_AUC=quantile(V_auc,.95))
    return(L)}
  if((a<0)|(b<0)){return(list(moy_AUC=NA))}}
Vm=seq(.025,.975,by=.025)
Vi=seq(.01,.5,by=.01)
V=outer(X = Vm,Y = Vi, Vectorize(function(x,y) 
Sim_AUC_mean_inter(x,y)$moy_AUC))
library("RColorBrewer")
image(Vm,Vi,V,
      xlab="Probability (Average)",
      ylab="Dispersion (Q95-Q5)",
      col=
        colorRampPalette(brewer.pal(n = 9, name = "YlGn"))(101))
contour(Vm,Vi,V,add=TRUE,lwd=2)

On the x-axis, we have the average probability to claim a loss. Of course, there is a symmetry here. And on the y-axis, we have the dispersion : the lower, the less heterogeneity in the portfolio. For instance, with a 30% chance to claim a loss on average, and 20% dispersion (meaning that in the portfolio, 90% of the insured have between 20% and 40% chance to claim a loss, or 15% and 35% chance), we have on average a 60% AUC. With a perfect model ! So with only a few covariates, having 55% should be great !

My point here is that with a low dispersion, we cannot expect to have a great AUC (again, even with a perfect model). In motor insurance, from my experience, 90% of the insured are between 3% chance and 20% chance to claim a loss ! That’s less than 20% dispersion ! and in that case, even if the (average) probability is rather small, it is very difficult to expect an AUC above 60% or 65% !

Random thoughts on econometric models with (pure) random features

For my lectures on applied linear models, I wanted to illustrate the fact that the R^2 is never a good measure of the goodness of the model, since it’s quite easy to improve it. Consider the following dataset

n=100
df=data.frame(matrix(rnorm(n*n),n,n))
names(df)=c("Y",paste("X",1:99,sep=""))

with one variable of interest y, and 99 features x_j. All of them being (by construction) independent. And we have 100 observations… Consider here the regression on the first k features, and compute R_k^2 of that regression

reg=function(k){
  frm=paste("Y~",paste("X",1:k,collapse="+",sep=""))
  model=lm(frm,data=df)
  summary(model)$adj.r.squared}

Let us see what’s going on…

plot(1:99,Vectorize(reg)(1:99))

(actually, it’s not exactly what we have on the graph…. we have the average obtained over 1,000 samples randomly generated, with 90% confidence bands). Oberve that \mathbb{E}[R^2_k]=k/n, i.e. if we add some pure random noise, we keep increasing the R^2 (up to 1, actually).

Good news, as we’ve seen in the course, the adjusted R^2 – denoted \bar R^2-might help. Observe that \mathbb{E}[\barR^2_k]=0, so, in some sense, adding features does not help here…

reg=function(k){
  frm=paste("Y~",paste("X",1:k,collapse="+",sep=""))
  model=lm(frm,data=df)
  summary(model)$r.squared}
plot(1:99,Vectorize(reg)(1:99))

We can actually do the same with Akaike criteria AIC_k and Schwarz (bayesian) criteria BIC_k.

reg=function(k){
  frm=paste("Y~",paste("X",1:k,collapse="+",sep=""))
  model=lm(frm,data=df)
  AIC(model)}
plot(1:99,Vectorize(reg)(1:99))

For the AIC, the intitial increase makes sense : we should not prefer the model with 10 covariates, compared with nothing. The strange thing is the far right behavior : we prefer here 80 random noise features to none ! Which I find hard to interprete… For the BIC the code is simply

reg=function(k){
  frm=paste("Y~",paste("X",1:k,collapse="+",sep=""))
  model=lm(frm,data=df)
  BIC(model)}
plot(1:99,Vectorize(reg)(1:99))

and here also, we have the same pattern, where we prefer a big model with juste pure noise to nothing…

A last one to conclude (or not) : what about the leave-one-out cross validation mean squared error ? More precisely, CV=\frac{1}{n}\sum_{i=1}\widehat{\varepsilon}^2_{-i}where \widehat{\varepsilon}^2_{-i}=y_i-\widehat{y}_{-i} where \widehat{y}_{-i} is the predicted value obtained with the model is estimated when the ith observation is deleted. One can prove that \widehat{\beta}_{-i}=\widehat{\beta}-(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{x}_i\hat\varepsilon_i(1-H_{i,i})^{-1}where H is the classical hat matrix, thus\widehat{\varepsilon}_{-i}=(1-H_{i,i})^{-1}\hat\varepsilon_ii.e. we do note have to estimate (at each round) n models

reg=function(k){
  frm=paste("Y~",paste("X",1:k,collapse="+",sep=""))
  model=lm(frm,data=df)
  h=lm.influence(model)$hat/2
  mean( (residuals(model)/1-h)^2 ))}
plot(1:99,Vectorize(reg)(1:99))

Here, it make sense : adding noisy features yields overfit ! So the mean squared error is decreasing !

That’s all nice, but it might not be very realistic… Here, for my model with only one variable, I just pick one, at random…. In practice, we try to get the “best one”… So a more natural idea would be to order the variables according to their correlations with y,

df=data.frame(matrix(rnorm(n*n),n,n))
  df=df[,rev(order(abs(cor(df)[1,])))]
  names(df)=c("Y",paste("X",1:99,sep=""))}

and as before, we can plot the evolution of R^2_k as a function of k the number of features considered,

which is increasing, with a higher slope at the beginning… For the \bar R^2_k we might actually prefer a correlated noise to nothing (which makes sense actually). So here since we somehow chose our variables, \bar R^2_k seems to be always positive…

For the AIC_k here also, there is an improvement. Before coming back to the original situation (with about 80 features) and here also, we observe the drop on the far right part of the graph

The BIC_k might like the top three features, but soon, we have a deterioration…. even if here also, we have the drop at the far right (with more than 95 features… for 100 observations).

Finally, observe that here again, our (leave-one-out) cross-validation has not been mesled by our noisy variables : it is always decreasing !

So it seems that cross-validation techniques are more robust than the AIC and BIC (even if we mentioned in a previous post connexions between all those concepts) when we have a lot a noisy (non-relevent) features.

NSERC – Discovery Grants Program, over the past 5 years

In a previous post, I discussed how it was possible to scrap the NSERC website to get stats about discovery grants. Since we just got the new 2018 figures, I thought it would be a good opportunity to update my graphs,

library(XML)
library(stringr)
url="http://www.nserc-crsng.gc.ca/NSERC-CRSNG/FundingDecisions-DecisionsFinancement/ResearchGrants-SubventionsDeRecherche/ResultsGSC-ResultatsCSS_eng.asp"
download.file(url,destfile = "GSC.html")
library(XML)
tables=readHTMLTable("GSC.html")
GSC=tables[[1]]$V1
GSC=as.character(GSC[-(1:2)])
namesGSC=tables[[1]]$V2
namesGSC=as.character(namesGSC[-(1:2)])
Correction = function(x) as.numeric(gsub('[$,]', '', x))
YEAR=2013:2018
for(i in 1:length(YEAR)){
y=YEAR[i]
grants= function(gsc){
  url=paste("http://www.nserc-crsng.gc.ca/NSERC-CRSNG/FundingDecisions-DecisionsFinancement/ResearchGrants-SubventionsDeRecherche/ResultsGSCDetail-ResultatsCSSDetails_eng.asp?Year=",y,"&GSC=",gsc,sep="")
  download.file(url,destfile = "GSC.html")
  library(XML)
  tables=readHTMLTable("GSC.html")
  X=as.character(tables[[1]]$"Awarded Amount")
  A=as.numeric(Vectorize(Correction)(X))
  return(c(median(A),mean(A),as.numeric(quantile(A,(1:99)/100))))
}
M=Vectorize(grants)(GSC[1:12])
plot(M[3:101,8],(1:99)/100,type="s",xlim=c(0,130000),xlab=
paste("Annual Discovery Grant (CAN) - ",y,sep=""),ylab="")
lines(M[3:101,5],(1:99)/100,type="s",col="red")
lines(M[3:101,4],(1:99)/100,type="s",col="blue")
abline(v=M[3,5],lty=2,col=rgb(1,0,0,.4))
idx=which(M[3:101,8]<M[3,5])
lines(M[2+idx,8],(idx)/100,type="s",lwd=4)
legend("bottomright",c("maths","physics","chemestry"),
col=c("black","red","blue"),lty=1,bty="n")}

With those functions, I plot the cumulative distribution functions for three disciplines, manely maths, physics and chemistry. I added a line for the lowest value in physics (the vertical line), and the bold line shows the proportion of researchers in maths who got less than the lowest amount in physics,

Hence, in 2013, 60% of the researchers in maths get less than any researcher in physics (and more than 90% in maths get less than any researcher in chemistry). Then, from 2014 to 2018, we get

It is rather constant : 50% of the researchers in mathematics in Canada get less than any researcher in physics, or in chemistry. I don’t understand why, but it’s interesting to observe that this is very stable…

The “probability to win” is hard to estimate…

Real-time computation (or estimation) of the “probability to win” is difficult. We’ve seem that in soccer games, in elections… but actually, as a professor, I see that frequently when I grade my students.

Consider a classical multiple choice exam. After each question, imagine that you try to compute the probability that the student will pass. Consider here the case where we have 50 questions. Students pass when they have 25 correct answers, or more. Just for simulations, I will assume that students just flip a coin at each question… I have n students, and 50 questions

set.seed(1)
n=10
M=matrix(sample(0:1,size=n*50,replace=TRUE),50,n)

Let X_{i,j} denote the score of student i at question j. Let S_{i,j} denote the cumulated score, i.e. S_{i,j}=X_{i,1}+\cdots+X_{i,j}. At step j, I can get some sort of prediction of the final score, using \hat{T}_{i,j}=50\times S_{i,j}/j. Here is the code

SM=apply(M,2,cumsum)
NB=SM*50/(1:50)

We can actually plot it

plot(NB[,1],type="s",ylim=c(0,50))
abline(h=25,col="blue")
for(i in 2:n) lines(NB[,i],type="s",col="light blue")
lines(NB[,3],type="s",col="red")


But that’s simply the prediction of the final score, at each step. That’s not the computation of the probability to pass !

Let’s try to see how we can do it… If after j questions, the students has 25 correct answer, the probability should be 1 – i.e. if S_{i,j}\geq 25 – since he cannot fail. Another simple case is the following : if after j questions, the number of points he can get with all correct answers until the end is not sufficient, he will fail. That means if S_{i,j}+(50-i+1)< 25 the probability should be 0. Otherwise, to compute the probability to sucess, it is quite straightforward. It is the probability to obtain at least 25-S_{i,j} correct answers, out of 50-j questions, when the probability of success is actually S_{i,j}/j. We recognize the survival probability of a binomial distribution. The code is then simply

PB=NB*NA
for(i in 1:50){
  for(j in 1:n){
    if(SM[i,j]&gt;=25) PB[i,j]=1
    if(SM[i,j]+(50-i+1)&lt;25)   PB[i,j]=0
    if((SM[i,j]&lt;25)&amp;(SM[i,j]+(50-i+1)&gt;=25)) PB[i,j]=1-pbinom(25-SM[i,j],size=(50-i),prob=SM[i,j]/i)
  }}

So if we plot it, we get

plot(PB[,1],type="s",ylim=c(0,1))
abline(h=25,col="red")
for(i in 2:n) lines(PB[,i],type="s",col="light blue")
lines(PB[,3],type="s",col="red")

which is much more volatile than the previous curves we obtained ! So yes, computing the “probability to win” is a complicated exercice ! Don’t blame those who try to find it hard to do !

Of course, things are slightly different if my students don’t flip a coin… this is what we obtain if half of the students are good (2/3 probability to get a question correct) and half is not good (1/3 chance),

If we look at the probability to pass, we usually do not have to wait until the end (the 50 questions) to know who passed and who failed

PS : I guess a less volatile solution can be obtained with a Bayesian approach… if I find some spare time this week, I will try to code it…

Solving the chinese postman problem

Some pre-Halloween post today. It started actually while I was in Barcelona : kids wanted to go back to some store we’ve seen the first day, in the gothic part, and I could not remember where it was. And I said to myself that would be quite long to do all the street of the neighborhood. And I discovered that it was actually an old problem. In 1962, Meigu Guan was interested in a postman delivering mail to a number of streets such that the total distance walked by the postman was as short as possible. How could the postman ensure that the distance walked was a minimum?

A very close notion is the concept of traversable graph, which is one that can be drawn without taking a pen from the paper and without retracing the same edge. In such a case the graph is said to have an Eulerian trail (yes, from Euler’s bridges problem). An Eulerian trail uses all the edges of a graph. For a graph to be Eulerian all the vertices must be of even order.

An algorithm for finding an optimal Chinese postman route is:

  1. List all odd vertices.
  2. List all possible pairings of odd vertices.
  3. For each pairing find the edges that connect the vertices with the minimum weight.
  4. Find the pairings such that the sum of the weights is minimised.
  5. On the original graph add the edges that have been found in Step 4.
  6. The length of an optimal Chinese postman route is the sum of all the edges added to the total found in Step 4.
  7. A route corresponding to this minimum weight can then be easily found.

For the first steps, we can use the codes from Hurley & Oldford’s Eulerian tour algorithms for data visualization and the PairViz package. First, we have to load some R packages

require(igraph)
require(graph)
require(eulerian)
require(GA)

Then use the following function from stackoverflow,

make_eulerian = function(graph){
  info = c("broken" = FALSE, "Added" = 0, "Successfull" = TRUE)
  is.even = function(x){ x %% 2 == 0 }
  search.for.even.neighbor = !is.even(sum(!is.even(degree(graph))))
  for(i in V(graph)){
    set.j = NULL
    uneven.neighbors = !is.even(degree(graph, neighbors(graph,i))) 
if(!is.even(degree(graph,i))){ 
if(sum(uneven.neighbors) == 0){ 
if(sum(!is.even(degree(graph))) &gt; 0){
          info["Broken"] = TRUE
          uneven.candidates &lt;- !is.even(degree(graph, V(graph)))
          if(sum(uneven.candidates) != 0){
            set.j &lt;- V(graph)[uneven.candidates][[1]]
          }else{
            info["Successfull"] &lt;- FALSE
          }
        }       
      }else{
        set.j &lt;- neighbors(graph, i)[uneven.neighbors][[1]]
      }
    }else if(search.for.even.neighbor == TRUE &amp; is.null(set.j)){
      info["Added"] &lt;- info["Added"] + 1     
      set.j &lt;- neighbors(graph, i)[ !uneven.neighbors ][[1]]
      if(!is.null(set.j)){search.for.even.neighbor &lt;- FALSE}
    }
    if(!is.null(set.j)){
      if(i != set.j){
        graph &lt;- add_edges(graph, edges=c(i, set.j))
        info["Added"] &lt;- info["Added"] + 1
      }
    }
  }
  (list("graph" = graph, "info" = info))}

Then, consider some network, with 12 nodes

g1 = graph(c(1,2, 1,3, 2,4, 2,5, 1,5, 3,5, 
4,7, 5,7, 5,8, 3,6, 6,8, 6,9, 9,11, 8,11, 
8,10, 8,12, 7,10, 10,12, 11,12), directed = FALSE)

To plot that network, use

V(g1)$name=LETTERS[1:12]
V(g1)$color=rgb(0,0,1,.4)
ly=layout.kamada.kawai(g1)
plot(g1,vertex.color=V(newg)$color,layout=ly)

Then we convert it to some traversable graph by adding 5 vertices

eulerian = make_eulerian(g1)
eulerian$info
     broken       Added Successfull 
          0           5           1 
g = eulerian$graph

as shown below

ly=layout.kamada.kawai(g)
plot(g,vertex.color=V(newg)$color,layout=ly)

We cut those 5 vertices in two part, and therefore, we add 5 artificial nodes

A=as.matrix(as_adj(g))
A1=as.matrix(as_adj(g1))
newA=lower.tri(A, diag = FALSE)*A1+upper.tri(A, diag = FALSE)*A
for(i in 1:sum(newA==2)) newA = cbind(newA,0)
for(i in 1:sum(newA==2)) newA = rbind(newA,0)
s=nrow(A)
for(i in 1:nrow(A)){
  Aj=which(newA[i,]==2)
  if(!is.null(Aj)){
      for(j in Aj){
        newA[i,s+1]=newA[s+1,i]=1
        newA[j,s+1]=newA[s+1,j]=1
        newA[i,j]=1
        s=s+1
      }}}

We get the following graph, where all nodes have an even number of vertices !

newg=graph_from_adjacency_matrix(newA)
newg=as.undirected(newg)
V(newg)$name=LETTERS[1:17]
V(newg)$color=c(rep(rgb(0,0,1,.4),12),rep(rgb(1,0,0,.4),5))
ly2=ly
transl=cbind(c(0,0,0,.2,0),c(.2,-.2,-.2,0,-.2))
for(i in 13:17){
  j=which(newA[i,]&gt;0)
  lc=ly[j,]
  ly2=rbind(ly2,apply(lc,2,mean)+transl[i-12,])
}
plot(newg,layout=ly2)

Our network is now the following (new nodes are small because actually, they don’t really matter, it’s just for computational reasons)

plot(newg,vertex.color=V(newg)$color,layout=ly2,
     vertex.size=c(rep(20,12),rep(0,5)),
     vertex.label.cex=c(rep(1,12),rep(.1,5)))

Now we can get the optimal path

n &lt;- LETTERS[1:nrow(newA)]
g_2 &lt;- new("graphNEL",nodes=n) for(i in 1:nrow(newA)){ for(j in which(newA[i,]&gt;0)){
    g_2 &lt;- addEdge(n[i],n[j],g_2,1) 
  }}
etour(g_2,weighted=FALSE)
 [1] "A" "B" "D" "G" "E" "A" "C" "E" "H" "F" "I" "K" "H" "J" "G" "P" "J" "L" "K" "Q" "L" "H" "O" "F" "C"
[26] "N" "E" "B" "M" "A"

or

edg=attr(E(newg), "vnames")
ET=etour(g_2,weighted=FALSE)
parcours=trajet=rep(NA,length(ET)-1)
for(i in 1:length(parcours)){
  u=c(ET[i],ET[i+1])
  ou=order(u)
  parcours[i]=paste(u[ou[1]],u[ou[2]],sep="|")
  trajet[i]=which(edg==parcours[i])
}
parcours
 [1] "A|B" "B|D" "D|G" "E|G" "A|E" "A|C" "C|E" "E|H" "F|H" "F|I" "I|K" "H|K" "H|J" "G|J" "G|P" "J|P"
[17] "J|L" "K|L" "K|Q" "L|Q" "H|L" "H|O" "F|O" "C|F" "C|N" "E|N" "B|E" "B|M" "A|M"
trajet
 [1]  1  3  8  9  4  2  6 10 11 12 16 15 14 13 26 27 18 19 28 29 17 25 24  7 22 23  5 21 20

Let us try now on a real network of streets. Like Missoula, Montana.

I will not try to get the shapefile of the city, I will just try to replicate the photography above.

If you look carefully, you will see some problem : 10 and 93 have an odd number of vertices (3 here), so one strategy is to connect them (which explains the grey line).

But actually, to be more realistic, we start in 93, and we end in 10. Here is the optimal (shortest) path which goes through all vertices.

Now, we are ready for Halloween, to go through all streets in the neighborhood !

Monte Carlo techniques to create counterfactuals

In the previous STT5100 course, last week, we’ve seen how to use monte carlo simulations. The idea is that we do observe in statistics a sample \{y_1,\cdots,y_n\}, and more generally, in econometrics \{(y_1,\mathbf{x}_1),\cdots,(y_n,\mathbf{x}_n)\}. But let’s get back to statistics (without covariates) to illustrate. We assume that observations y_i are realizations of an underlying random variable Y_i. We assume that Y_i are i.id. random variables, with (unkown) distribution F_{\theta}. Consider here some estimator \widehat{\theta} – which is just a function of our sample \widehat{\theta}=h(y_1,\cdots,y_n). So \widehat{\theta} is a real-valued number like . Then, in mathematical statistics, in order to derive properties of the estimator \widehat{\theta}, like a confidence interval, we must define \widehat{\theta}=h(Y_1,\cdots,Y_n), so that now, \widehat{\theta} is a real-valued random variable. What is puzzling for students, is that we use the same notation, and I have to agree, that’s not very clever. So now, \widehat{\theta} is .

There are two strategies here. In classical statistics, we use probability theorem, to derive properties of \widehat{\theta} (the random variable) : at least the first two moments, but if possible the distribution. An alternative is to go for computational statistics. We have only one sample, \{y_1,\cdots,y_n\}, and that’s a pity. But maybe we can create another one \{y_1^{(1)},\cdots,y_n^{(1)}\}, as realizations of F_{\theta}, and another one \{y_1^{(2)},\cdots,y_n^{(2)}\}, anoter one \{y_1^{(3)},\cdots,y_n^{(3)}\}, etc. From those counterfactuals, we can now get a collection of estimators, \widehat{\theta}^{(1)},\widehat{\theta}^{(2)}, \widehat{\theta}^{(3)}, etc. Instead of using mathematical tricks to calculate \mathbb{E}(\widehat{\theta}), compute \frac{1}{k}\sum_{s=1}^k\widehat{\theta}^{(s)}That’s what we’ve seen last friday.

I did also mention briefly that looking at densities is lovely, but not very useful to assess goodness of fit, to test for normality, for instance. In this post, I just wanted to illustrate this point. And actually, creating counterfactuals can we a good way to see it. Consider here the height of male students,

Davis=read.table(
  "http://socserv.socsci.mcmaster.ca/jfox/Books/Applied-Regression-2E/datasets/Davis.txt")
Davis[12,c(2,3)]=Davis[12,c(3,2)]
X=Davis$height[Davis$sex=="M"]

We can visualize its distribution (density and cumulative distribution)

u=seq(155,205,by=.5)
par(mfrow=c(1,2))
hist(X,col=rgb(0,0,1,.3))
lines(density(X),col="blue",lwd=2)
lines(u,dnorm(u,178,6.5),col="black")
Xs=sort(X)
n=length(X)
p=(1:n)/(n+1)
plot(Xs,p,type="s",col="blue")
lines(u,pnorm(u,178,6.5),col="black")

Since it looks like a normal distribution, we can add the density a Gaussian distribution on the left, and the cdf on the right. Why not test it properly. To be a little bit more specific, I do not want to test if it’s a Gaussian distribution, but if it’s a \mathcal{N}(178,6.5^2). In order to see if this distribution is relevant, one can use monte carlo simulations to create conterfactuals

hist(X,col=rgb(0,0,1,.3))
lines(density(X),col="blue",lwd=2)
  Y=rnorm(n,178,6.5)
  hist(Y,col=rgb(1,0,0,.3))
  lines(density(Y),col="red",lwd=2)
Ys=sort(Y)
plot(Xs,p,type="s",col="white",lwd=2,axes=FALSE,xlab="",ylab="",xlim=c(155,205))
polygon(c(Xs,rev(Ys)),c(p,rev(p)),col="yellow",border=NA)
lines(Xs,p,type="s",col="blue",lwd=2)
lines(Ys,p,type="s",col="red",lwd=2)

We can see on the left that it is hard to assess normality from the density (histogram and also kernel based density estimator). One can hardly think of a valid distance, between two densities. But if we look at graph on the right, we can compare the empirical distribution cumulative distribution \widehat{F} obtained from \{y_1,\cdots,y_n\} (the blue curve), and some conterfactual, \widehat{F}^{(s)} obtained from \{y_1^{(s)},\cdots,y_n^{(s)}\} generated from F_{\theta_0} – where \theta_0 is the value we want to test. As suggested above, we can compute the yellow area, as suggest in Cramer-von Mises test, or the Kolmogorov-Smirnov distance.

d=rep(NA,1e5)
for(s in 1:1e5){
d[s]=ks.test(rnorm(n,178,6.5),"pnorm",178,6.5)$statistic
}
ds=density(d)
plot(ds,xlab="",ylab="")
dks=ks.test(X,"pnorm",178,6.5)$statistic
id=which(ds$x&gt;dks)
polygon(c(ds$x[id],rev(ds$x[id])),c(ds$y[id],rep(0,length(id))),col=rgb(1,0,0,.4),border=NA)
abline(v=dks,col="red")

If we draw 10,000 counterfactual samples, we can visualize the distribution (here the density) of the distance used a test statistic \widehat{d}^{(1)}, \widehat{d}^{(2)}, etc, and compare it with the one observe on our sample \widehat{d}. The proportion of samples where the test-statistics exceeded the one observed

mean(d&gt;dks)
[1] 0.78248

is the computational version of the p-value

ks.test(X,"pnorm",178,6.5)
 
	One-sample Kolmogorov-Smirnov test
 
data:  X
D = 0.068182, p-value = 0.8079
alternative hypothesis: two-sided

I thought about all that a couple of days ago, since I got invited for a panel discussion on “coding”, and why “coding” helped me as professor. And this is precisely why I like coding : in statistics, either manipulate abstract objects, like random variables, or you actually use some lines of code to create counterfactuals, and generate fake samples, to quantify uncertainty. The later is interesting, because it helps to visualize complex quantifies. I do not claim that maths is useless, but coding is really nice, as a starting point, to understand what we talk about (which can be very usefull when there is a lot of confusion on notations).

October, grant proposal season

In 2012, Danielle Herbert, Adrian Barnett, Philip Clarke and Nicholas Graves published an article entitled “on the time spent preparing grant proposals: an observational study of Australian researchers“, whose conclusions had been included in Nature under a more explicit title, “Australia’s grant system wastes time” ! In this study, they included 3700 grant applications sent to the National Health and Medical Research Council, and showed that each application represented 37 working days: “Extrapolating this to all 3,727 submitted proposals gives an estimated 550 working years of researchers’ time (95% confidence interval, 513-589)“. But in these times when I have to write my funding application, I find that losing 37 days of work is huge. Because it’s become the norm! And somehow, it’s sad.

Forget about the crazy idea that I would rather, in fact, spend more time doing my research. In fact, the thought I had this morning was that it is rather sad that in the Faculty of Science, mathematicians are asked to spend a considerable amount of time, comparable to that required of physicists or chemists, for often smaller amounts of funding… And I thought it could be easily verified. We start by retrieving the discipline codes

url="http://www.nserc-crsng.gc.ca/NSERC-CRSNG/FundingDecisions-DecisionsFinancement/ResearchGrants-SubventionsDeRecherche/ResultsGSC-ResultatsCSS_eng.asp"
download.file(url,destfile = "GSC.html")
library(XML)
tables=readHTMLTable("GSC.html")
GSC=tables[[1]]$V1
GSC=as.character(GSC[-(1:2)])
namesGSC=tables[[1]]$V2
namesGSC=as.character(namesGSC[-(1:2)])

We’re going to need a small function, to remove the $ and other symbols that pollute the data (and prevent them from being treated as numbers)

library(stringr)
Correction = function(x) as.numeric(gsub('[$,]', '', x))

We will now read the 12 pages, and harvest (we will just take the 2017 data, but we could go back a few years before)

grants= function(gsc){
     url=paste("http://www.nserc-crsng.gc.ca/NSERC-CRSNG/FundingDecisions-DecisionsFinancement/ResearchGrants-SubventionsDeRecherche/ResultsGSCDetail-ResultatsCSSDetails_eng.asp?Year=2017&amp;GSC=",gsc,sep="")
    download.file(url,destfile = "GSC.html")
    library(XML)
    tables=readHTMLTable("GSC.html")
    X=as.character(tables[[1]]$"Awarded Amount")
    A=as.numeric(Vectorize(Correction)(X))
return(c(median(A),mean(A),as.numeric(quantile(A,(1:99)/100))))
}
M=Vectorize(grants)(GSC[1:12])

The average amounts of individual grants can be compared,

barplot(M[2,])

In mathematics, the average grant amount is $24400. If we normalize by this quantity, we obtain

barplot(M[2,]/M[2,8])

In other words, the average amount of a (individual) grant in chemistry (to pay for students, conferences, etc.) is twice that in mathematics, 60% higher in physics than in maths…

We can also look at the median values (rather than the averages)

barplot(M[1,])

Here again, it is in mathematics that it is the weakest….

barplot(M[1,]/M[1,8])

in comparable proportions. If we think that the time spent writing should be proportional to the amount allocated, we should spend half as much time in math as in chemistry.

Cumulative functions can also be ploted,

plot(M[3:101,8],(1:99)/100,type="s",xlim=range(M))
lines(M[3:101,5],(1:99)/100,type="s",col="red")
lines(M[3:101,4],(1:99)/100,type="s",col="blue")

with math in black, physics in red, and chemistry in blue. What is surprising is the bottom part: a “bad” researcher in chemistry or physics will earn more than the median researcher in mathematics…

Now that my intuition is confirmed, I have to go back, writing my proposal… and explain to my coauthors that I have to postpone some research projects because, well, you know…

Combining automatically factor levels in R

Each time we face real applications in an applied econometrics course, we have to deal with categorial variables. And the same question arise, from students : how can we combine automatically factor levels ? Is there a simple R function ?

I did upload a few blog posts, over the pas years. But so far, nothing satistfying. Let me write down a few lines about what could be done. And if some wants to write a nice R function, that would be awesome. To illustrate the idea, consider the following (simulated dataset)

n=200
set.seed(1)
x1=runif(n)
x2=runif(n)
y=1+2*x1-x2+rnorm(n,0,.2)
LB=sample(LETTERS[1:10])
b=data.frame(y=y,x1=x1,
             x2=cut(x2,breaks=
             c(-1,.05,.1,.2,.35,.4,.55,.65,.8,.9,2),
             labels=LB))
str(b)
'data.frame':	200 obs. of  3 variables:
 $ y : num  1.345 1.863 1.946 2.481 0.765 ...
 $ x1: num  0.266 0.372 0.573 0.908 0.202 ...
 $ x2: Factor w/ 10 levels "I","A","H","F",..: 4 4 6 4 3 6 7 3 4 8 ...
table(b$x2)[LETTERS[1:10]]
 
 A  B  C  D  E  F  G  H  I  J 
11 12 23 34 23 36 12 32  3 14

There is one (continuous) dependent variable y, one continuous covariable x_1 and one categorical variable x_2, with here ten levels. We can plot the data using

plot(b$x1,y,col="white",xlim=c(0,1.1))
text(b$x1,y,as.character(b$x2),cex=.5)

The output of a linear regression yield the following predictions

for(i in 1:10){
p=function(x) predict(lm(y~x1+x2,data=b),newdata=data.frame(x1=x,x2=LETTERS[i]))
u=seq(-1,1.065,by=.01)
v=Vectorize(p)(u)
lines(u,v)}

the slope for x_1 is the same, we simply add a different constant for each level. As we can see, some levels are very very close, so it seems legitimate to combine them into one single category. Here is the output of the linear regression,

summary(lm(y~x1+x2,data=b))
Coefficients:
             Estimate Std. Error t value Pr(&gt;|t|)    
(Intercept)  0.843802   0.119655   7.052 3.23e-11 ***
x1           1.992878   0.053838  37.016  &lt; 2e-16 ***
x2A          0.055500   0.131173   0.423   0.6727    
x2H          0.009293   0.121626   0.076   0.9392    
x2F         -0.177002   0.121020  -1.463   0.1452    
x2B         -0.218152   0.130192  -1.676   0.0955 .  
x2D         -0.206970   0.121294  -1.706   0.0896 .  
x2G         -0.407417   0.129999  -3.134   0.0020 ** 
x2C         -0.526708   0.123690  -4.258 3.24e-05 ***
x2J         -0.664281   0.128126  -5.185 5.54e-07 ***
x2E         -0.816454   0.123625  -6.604 3.94e-10 ***
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1
 
Residual standard error: 0.2014 on 189 degrees of freedom
Multiple R-squared:  0.8995,	Adjusted R-squared:  0.8942 
F-statistic: 169.1 on 10 and 189 DF,  p-value: &lt; 2.2e-16
AIC(lm(y~x1+x2,data=b))
[1] -60.74443
BIC(lm(y~x1+x2,data=b))
[1] -21.16463

Here the reference category is “I”. And it looks like we could actually combine that category with several others. One strategy here would be to select all categories that seem to be not significantly different, and to run a (multiple) test

library(car)
linearHypothesis(lm(y~x1+x2,data=b), c("x2A = 0", "x2H = 0", "x2F = 0"))
 
Hypothesis:
x2A = 0
x2H = 0
x2F = 0
 
Model 1: restricted model
Model 2: y ~ x1 + x2
 
  Res.Df    RSS Df Sum of Sq      F Pr(&gt;F)    
1    192 8.4651                               
2    189 7.6654  3   0.79971 6.5726  3e-04 ***
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1

It seems that we can combine those four categories together.

Here, we can see what’s going on when we change the reference category (actually, loop on all categories)

P=matrix(NA,nlevels(b$x2),nlevels(b$x2))
colnames(P)=rownames(P)=LETTERS[1:10]
plot(1:nlevels(b$x2),1:nlevels(b$x2),col="white",xlab="",ylab="",axes=F,xlim=c(0,10.5),
     ylim=c(0,10.5))
text(1:10,0,LETTERS[1:10])
text(0,1:10,LETTERS[1:10])
for(i in 1:nlevels(b$x2)){
#levels(b$x2)=LETTERS[1:10]
b$x2=relevel(b$x2,LETTERS[i])
p=summary(lm(y~x1+x2,data=b))$coefficients[-(1:2),4]
names(p)=substr(names(p),3,3)
P[LETTERS[i],names(p)]=p
p=P[LETTERS[i],]
idx=which(p&gt;.05)
points(((1:10))[idx],rep(i,length(idx)),pch=1,cex=2)
idx=which(p&gt;.1)
points(((1:10))[idx],rep(i,length(idx)),pch=19,cex=2)}

We are glad to see that it is symmetric : if “H” should be combined with “I”, “I” should also be combined with “H”.

Here black points are related with the 10% p-value, and white points the 5% p-value. This graph is actually hard to read… And actually, this reminds us of  Bertin (1967).

Here, we can predefine manually some ordering (we will see below how it might be automatised)

LETTERSord=c("I","A","H","F","B","D","G","C","J","E")
P=matrix(NA,nlevels(b$x2),nlevels(b$x2))
colnames(P)=rownames(P)=LETTERSord
plot(1:nlevels(b$x2),1:nlevels(b$x2),col="white",xlab="",ylab="",axes=F,xlim=c(0,10.5),
     ylim=c(0,10.5))
ct=c(3,3,2,1,1)
abline(v=.5+c(0,cumsum(ct)),lty=2)
abline(h=.5+c(0,cumsum(ct)),lty=2)
text(1:10,0,LETTERSord)
text(0,1:10,LETTERSord)
for(i in 1:nlevels(b$x2)){
  #levels(b$x2)=LETTERS[1:10]
  b$x2=relevel(b$x2,LETTERSord[i])
  p=summary(lm(y~x1+x2,data=b))$coefficients[-(1:2),4]
  names(p)=substr(names(p),3,3)
  P[LETTERSord[i],names(p)]=p
  p=P[LETTERSord[i],]
  idx=which(p&gt;.05)
  points(((1:10))[idx],rep(i,length(idx)),pch=1,cex=2)
  idx=which(p&gt;.1)
  points(((1:10))[idx],rep(i,length(idx)),pch=19,cex=2)
}

Here we get the following

It looks like we have our combined categories…

Actually, it is possible to use another strategy. We start from some level, say “A”. Then, we merge it with all non-significantly different levels. If “B” is not one of them, we use it as the new reference. Etc.

for(i in 1:nlevels(b$x2)){
  if(LETTERS[i]%in%levels(b$x2)){
  b$x2=relevel(b$x2,LETTERS[i])
  p=summary(lm(y~x1+x2,data=b))$coefficients[-(1:2),4]
  names(p)=substr(names(p),3,nchar(p))
  idx=which(p&gt;.05)
  mix=c(LETTERS[i],names(p)[idx])
  b$x2=recode(b$x2, paste("c('",paste(mix,collapse = "','"),"')='",paste(mix,collapse = "+"),"'",sep=""))
}}

The final categories are

table(b$x2)
 
A+I+H B+D+F   C+G     E     J 
   46    82    35    23    14

with the following regression output

summary(lm(y~x1+x2,data=b))
 
Coefficients:
            Estimate Std. Error t value Pr(&gt;|t|)    
(Intercept)  0.86407    0.03950  21.877  &lt; 2e-16 ***
x1           1.99180    0.05323  37.417  &lt; 2e-16 ***
x2B+D+F     -0.21517    0.03699  -5.817 2.44e-08 ***
x2C+G       -0.50545    0.04528 -11.164  &lt; 2e-16 ***
x2E         -0.83617    0.05128 -16.305  &lt; 2e-16 ***
x2J         -0.68398    0.06131 -11.156  &lt; 2e-16 ***
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1
 
Residual standard error: 0.2008 on 194 degrees of freedom
Multiple R-squared:  0.8975,	Adjusted R-squared:  0.8948 
F-statistic: 339.6 on 5 and 194 DF,  p-value: &lt; 2.2e-16
AIC(lm(y~x1+x2,data=b))
[1] -66.76939
BIC(lm(y~x1+x2,data=b))
[1] -43.68117

Which is consistent with the group we got before. But actually, if we change the order, we can get different combinations. For instance, if we go from “J” to “A”, instead of “A” to “J”, we obtain

for(i in nlevels(b$x2):1){
  #levels(b$x2)=LETTERS[1:10]
  if(LETTERS[i]%in%levels(b$x2)){
  b$x2=relevel(b$x2,LETTERS[i])
  p=summary(lm(y~x1+x2,data=b))$coefficients[-(1:2),4]
  names(p)=substr(names(p),3,nchar(p))
  idx=which(p&gt;.05)
  mix=c(LETTERS[i],names(p)[idx])
  b$x2=recode(b$x2, paste("c('",paste(mix,collapse = "','"),"')='",paste(mix,collapse = "+"),"'",sep=""))
}}
table(b$x2)
 
          E         G+C I+A+B+D+F+H           J 
         23          35         128          14

with different information criteria here

AIC(lm(y~x1+x2,data=b))
[1] -36.61665
BIC(lm(y~x1+x2,data=b))
[1] -16.82675

I guess it would be necessary to run randomly the order we go through the levels. Last, but not least, one can use regression trees (even if it not per se in the syllabus of the course). The problem is that there is another explanatory variable that might interphere. So I would suggest (1) to fit a linear model y=\beta_0+\beta_1x_1+u_i, to calculate the residuals, \widehat{u}_i (2) to run a regression tree, to explain \widehat{u}_i with categorical variable x_2 (I did explain how trees are build when the explanatory variable is a categorical one in a previous post)

library(rpart)
library(rpart.plot)
b$e=residuals(lm(y~x1,data=b))
arbre=rpart(e~x2,data=b)
prp(arbre,type=2,extra=1)

Observe that the leaves have the same groups as the one we got.

arbre
n= 200 
 
node), split, n, deviance, yval
      * denotes terminal node
 
1) root 200 22.563500  7.771561e-18  
  2) x2=G,C,J,E 72  4.441495 -3.232525e-01  
    4) x2=J,E 37  1.553520 -4.578492e-01 *
    5) x2=G,C 35  1.509068 -1.809646e-01 *
  3) x2=I,A,H,F,B,D 128  6.366628  1.818295e-01  
    6) x2=F,B,D 82  2.983381  1.048246e-01 *
    7) x2=I,A,H 46  2.030229  3.190993e-01 *

I guess that it should be possible to put all that in an R function, to suggest combinations of level that might improve the regression.

Convex Regression Model

This morning during the lecture on nonlinear regression, I mentioned (very) briefly the case of convex regression. Since I forgot to mention the codes in R, I will publish them here. Assume that y_i=m(\mathbf{x}_i)+\varepsilon_i where m:\mathbb{R}^d\rightarrow \mathbb{R} is some convex function.

Then m is convex if and only if \forall\mathbf{x}_1,\mathbf{x}_2\in\mathbb{R}^d, \forall t\in[0,1], m(t\mathbf{x}_1+[1-t]\mathbf{x}_2) \leq tm(\mathbf{x}_1)+[1-t]m(\mathbf{x}_2)Hidreth (1954) proved that if m^\star=\underset{m \text{ convex}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \big(y_i-m(\mathbf{x_i})\big)^2\right\rbracethen \mathbf{\theta}^\star=(m^\star(\mathbf{x_1}),\cdots,m^\star(\mathbf{x_n})) is unique.

Let \mathbf{y}=\mathbf{\theta}+\mathbf{\varepsilon}, then \mathbf{\theta}^\star=\underset{\mathbf{\theta}\in \mathcal{K}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \big(y_i-\theta_i)\big)^2\right\rbracewhere\mathcal{K}=\{\mathbf{\theta}\in\mathbb{R}^n:\exists m\text{ convex },m(\mathbf{x}_i)=\theta_i\}. I.e. \mathbf{\theta}^\star is the projection of \mathbf{y} onto the (closed) convex cone \mathcal{K}. The projection theorem gives existence and unicity.

For convenience, in the application, we will consider the real-valued case, m:\mathbb{R}\rightarrow \mathbb{R}, i.e. y_i=m(x_i)+\varepsilon_i. Assume that observations are ordered x_1\leq x_2\leq\cdots \leq x_n. Here \mathcal{K}=\left\lbrace\mathbf{\theta}\in\mathbb{R}^n:\frac{\theta_2-\theta_1}{x_2-x_1}\leq \frac{\theta_3-\theta_2}{x_3-x_2}\leq \cdots \leq \frac{\theta_n-\theta_{n-1}}{x_n-x_{n-1}}\right\rbrace

Hence, quadratic program with n-2 linear constraints.

m^\star is a piecewise linear function (interpolation of consecutive pairs (x_i,\theta_i^\star)).

If m is differentiable, m is convex if m(\mathbf{x})+ \nabla m(\mathbf{x})^{\text{T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y})

More generally, if m is convex, then there exists \xi_{\mathbf{x}}\in\mathbb{R}^n such that m(\mathbf{x})+ \xi_{\mathbf{x}}^{\text{ T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y})
\xi_{\mathbf{x}} is a subgradient of m at {\mathbf{x}}. And then \partial m(\mathbf{x})=\big\lbrace m(\mathbf{x})+ \xi^{\text{ T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y}),\forall \mathbf{y}\in\mathbb{R}^n\big\rbrace

Hence, \mathbf{\theta}^\star is solution of \text{argmin}\big\lbrace\|\mathbf{y}-\mathbf{\theta}\|^2\big\rbrace\text{subject to }\theta_i+\xi_i^{\text{ T}}[\mathbf{x}_j-\mathbf{x}_i]\leq\mathbf{\theta}_j,~\forall i,j and \xi_1,\cdots,\xi_n\in\mathbb{R}^n. Now, to do it for real, use cobs package for constrained (b)splines regression,

library(cobs)

To get a convex regression, use

plot(cars)
x = cars$speed
y = cars$dist
rc = conreg(x,y,convex=TRUE)
lines(rc, col = 2)


Here we can get the values of the knots

rc
 
Call:  conreg(x = x, y = y, convex = TRUE) 
Convex regression: From 19 separated x-values, using 5 inner knots,
     7,    8,    9,   20,   23.
RSS =  1356; R^2 = 0.8766;
 needed (5,0) iterations

and actually, if we use them in a linear-spline regression, we get the same output here

reg = lm(dist~bs(speed,degree=1,knots=c(4,7,8,9,,20,23,25)),data=cars)
u = seq(4,25,by=.1)
v = predict(reg,newdata=data.frame(speed=u))
lines(u,v,col="green")

Let us add vertical lines for the knots

abline(v=c(4,7,8,9,20,23,25),col="grey",lty=2)

Game of Friendship Paradox

In the introduction of my course next week, I will (briefly) mention networks, and I wanted to provide some illustration of the Friendship Paradox. On network of thrones (discussed in Beveridge and Shan (2016)), there is a dataset with the network of characters in Game of Thrones. The word “friend” might be abusive here, but let’s continue to call connected nodes “friends”. The friendship paradox states that

People on average have fewer friends than their friends

This was discussed in Feld (1991) for instance, or Zuckerman & Jost (2001). Let’s try to see what it means here. First, let us get a copy of the dataset

download.file("https://www.macalester.edu/~abeverid/data/stormofswords.csv","got.csv")
GoT=read.csv("got.csv")
library(networkD3)
simpleNetwork(GoT[,1:2])

Because it is difficult for me to incorporate some d3js script in the blog, I will illustrate with a more basic graph,

Consider a vertex v\in V in the undirected graph G=(V,E) (with classical graph notations), and let d(v) denote the number of edges touching it (i.e. v has d(v) friends). The average number of friends of a random person in the graph is \mu = \frac{1}{n_V}\sum_{v\in V} d(v)=\frac{2 n_E}{n_V} The average number of friends that a typical friend has is
\frac{1}{n_V}\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)But
\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)=\sum_{v,v' \in G} \left(<br /> \frac{d(v')}{d(v)}+\frac{d(v)}{d(v')}\right)=\sum_{v,v' \in G}\left(\frac{d(v')^2+d(v)^2}{d(v)d(v')}\right)=\sum_{v,v' \in G} \left(\frac{(d(v')-d(v))^2}{d(v)d(v')}+2\right){\color{red}{\succ}}\sum_{v,v' \in G} \left(2\right)=\sum_{v\in V} d(v)
Thus,\frac{1}{n_V}\sum_{v\in V} \left(\frac{1}{d(v)}\sum_{v'\in E_v} d(v')\right)\succ \frac{1}{n_V}\sum_{v\in V} d(v)
Note that this can be related to the variance decomposition \text{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2i.e.\frac{\mathbb{E}[X^2]}{\mathbb{E}[X]} =\mathbb{E}[X]+\frac{\text{Var}[X]}{\mathbb{E}[X]}\succ\mathbb{E}[X](Jensen inequality). But let us get back to our network. The list of nodes is

M=(rbind(as.matrix(GoT[,1:2]),as.matrix(GoT[,2:1])))
nodes=unique(M[,1])

and we each of them, we can get the list of friends, and the number of friends

friends = function(x) as.character(M[which(M[,1]==x),2])
nb_friends = Vectorize(function(x) length(friends(x)))

as well as the number of friends friends have, and the average number of friends

friends_of_friends = function(y) (Vectorize(function(x) length(friends(x)))(friends(y)))
nb_friends_of_friends = Vectorize(function(x) mean(friends_of_friends(x)))

We can look at the density of the number of friends, for a random node,

Nb  = nb_friends(nodes)
Nb2 = nb_friends_of_friends(nodes)
hist(Nb,breaks=0:40,col=rgb(1,0,0,.2),border="white",probability = TRUE)
hist(Nb2,breaks=0:40,col=rgb(0,0,1,.2),border="white",probability = TRUE,add=TRUE)
lines(density(Nb),col="red",lwd=2)
lines(density(Nb2),col="blue",lwd=2)


and we can also compute the averages, just to check

mean(Nb)
[1] 6.579439
mean(Nb2)
[1] 13.94243

So, indeed, people on average have fewer friends than their friends.