# Combining automatically factor levels with trees

Last year, in a post, I discussed how to merge levels of factor variables, using combinatorial techniques (it was for my STT5100 cours, and trees are not in the syllabus), with an extension on trees at the end of the post.

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

Just by looking at the data (see the previous post), we could easily get the feeling that 10 levels was too much.

Following my post, Przemyslaw sent a comment suggesting to use

library(factorMerger)

It is indeed a nice package (unless you have really really big datasets with a lot of categories in your factor variables – as I experienced recently), and you can get great graphs

MF = mergeFactors(response = b$y, factor = b$x2,
family = "gaussian")
plot(MF)

Here is suggests to create three categories. Recall that with student t-tests (changing the reference), we got

Another interesting package, by Piro Polo, is

library(tree.bins)

To use it, we simply call the following function, and we transform automatically our dataset : the continuous variables remain unchanged, and (possibly) categories of categorical variables are merged

b.bins = tree.bins(data=b, y=y)
str(b.bins)
Classes ‘data.table’ and '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 ...
$x2: chr "Group.4" "Group.4" "Group.4" "Group.4" ... - attr(*, ".internal.selfref")= table(b.bins$x2)

Group.1 Group.2 Group.3 Group.4
23      35      26     116

here in four groups. To get the correspondance, use

tree.bins(data=b, y=y, return = "lkup.list")
[[1]]
x2 Categories
1   E    Group.1
2   G    Group.2
3   C    Group.2
4   B    Group.3
5   J    Group.3
6   I    Group.4
7   A    Group.4
8   H    Group.4
9   F    Group.4
10  D    Group.4

(we have a list with one element, one dataframe, since there is only one factor variable). Cool, isn’t it ? I miss Przemyslaw’s plot, but this is rather quick, and efficient..

# On leverage

Last week, in our STT5100 (applied linear models) class, I’ve introduce the hat matrix, and the notion of leverage. In a classical regression model, $\boldsymbol{y}=\boldsymbol{X}\boldsymbol{\beta}$ (in a matrix form), the ordinary least square estimator of parameter $\boldsymbol{\beta}$ is $$\widehat{\boldsymbol{\beta}}=(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top\boldsymbol{y}$$The prediction can then be written$$\widehat{\boldsymbol{y}}=\boldsymbol{X}\widehat{\boldsymbol{\beta}}=\underbrace{\color{blue}{\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top}}_{\color{blue}{\boldsymbol{H}}}\boldsymbol{y}$$where $\color{blue}{\boldsymbol{H}}$ is called the hat matrix.

The matrix is idempotent, i.e. $$\boldsymbol{H}^2={\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\textcolor{grey}{\boldsymbol{X}^\top{\boldsymbol{X}}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}}\boldsymbol{X}^\top}={\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top}=\boldsymbol{H}$$so it can be interpreted as a projection matrix. Furthermore, since$\boldsymbol{H}\boldsymbol{X}=\boldsymbol{X}$ (just do the maths), the projection is on a subspace that contains all the linear combinations of columns of $\boldsymbol{X}$. One can also observe that $\mathbb{I}-\boldsymbol{H}$ is also a projection matrix. And we can write$$\boldsymbol{y}=\underbrace{\boldsymbol{H}\boldsymbol{y}}_{\widehat{\boldsymbol{y}}}+\underbrace{(\mathbb{I}-\boldsymbol{H})\boldsymbol{y}}_{\widehat{\boldsymbol{\varepsilon}}}$$where $\widehat{\boldsymbol{y}}$ is the orthogonal projection of $\boldsymbol{y}$ on the (linear) space of linear combinations of columns of $\boldsymbol{X}$, and $\widehat{\boldsymbol{y}}\perp\widehat{\boldsymbol{\varepsilon}}$, which gives the classical interpretation of residuals, being unpredictible (at least with a linear model using variables $\boldsymbol{X}$).

Let’s move a bit faster now (we’ve seen many other properties last week), and consider elements on the diagonal of matrix $\boldsymbol{H}$. Recall that we have

so entry $\boldsymbol{H}_{i,i}$ is a measure of the influence of entry $\boldsymbol{y}_i$ on its prediction latex]\widehat{\boldsymbol{y}}_i[/latex].

We have seen that$$\sum_{i=1}^n\boldsymbol{H}_{i,i}=\text{trace}(\boldsymbol{H})=\text{trace}(\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top)$$which can be written$$\sum_{i=1}^n\boldsymbol{H}_{i,i}=\text{trace}\boldsymbol{X}^\top(\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1})=\text{trace}(\mathbb{I})=p$$where classically $p=k+1$, where $k$ is the number of explanatory variables. Further, since $\boldsymbol{H}$ is idempotent, we can write (from $\boldsymbol{H}=\boldsymbol{H}^2$) that$$\boldsymbol{H}_{i,i}=\boldsymbol{H}_{i,i}^2 + \sum_{j\neq i}\boldsymbol{H}_{i,j}\boldsymbol{H}_{j,i}=\boldsymbol{H}_{i,i}^2 + \sum_{j\neq i}\boldsymbol{H}_{i,j}^2$$One the one hand, since the second term is positive $\boldsymbol{H}_{i,i}\geq\boldsymbol{H}_{i,i}^2$, i.e. $1\geq\boldsymbol{H}_{i,i}$. And since both terms are positive, then $\boldsymbol{H}_{i,i}\in[0,1]$. And there was a question in the course on the sharpeness of the bounds.

Using Anscombe’s dataset, we’ve seen that it was possible to get a leverage of 1. Using something rather similar

df = data.frame(x = c(rep(1,10),6), y = c(1:10,8)) plot(df)

we obtain

model = lm(y~x,data=df) abline(model,col="red",lwd=2) H = lm.influence(model)$hat plot(1:11,H,type="h") The very last observation, the one one the right, is here extremely influencial : if we remove it, the model is completely different ! And here, we reach the upper bound, $\boldsymbol{H}_{11,11}=1$. Observe that all other points are equally influencial, and because on the constraint on the trace of the matrix, $\boldsymbol{H}_{i,i}=1/10$ when $i\in\{1,2,\cdots,10\}$. Now, what about the lower bound ? In order to have some sort of “non-influencial” observations, consider the two following case. • the case where one observation (below the first one) is such that $\widehat{\boldsymbol{y}}_{i}=\boldsymbol{y}_{i}$ (perfect prediction) • the case where one observation (below the tenth one) is such that $\boldsymbol{x}_{i}=\overline{\boldsymbol{x}}$ and $\boldsymbol{y}_{i}=\overline{\boldsymbol{y}}$ (from the first order condition – or normal equation), the fitted regression line always go through point $(\overline{\boldsymbol{x}},\overline{\boldsymbol{y}})$ Let us move two observations from our dataset, mean(c(4,rep(1,8),6)) [1] 1.8 df = data.frame(x = c(4,rep(1,8),6,1.8), y = c(predict(model,newdata=data.frame(x=4)), 2:9,8, predict(model,newdata=data.frame(x=1.8)))) We now have If we compute the leverages, we obtain model = lm(y~x,data=df) H = lm.influence(model)$hat plot(1:11,H,type="h")

so, for the first observation, its leverage actually increased (the blue part), and for the tenth one, we have the lowest influence, but it is not zero. Is it possible to reach zero ?

Here, observe that for the tenth observation, $\boldsymbol{H}_{i,i}=1/n$. And actually, that’s the best we can do… We can prove that, in the case of a simple regression (as above)$$\boldsymbol{H}_{i,i}=\frac{1}{n}+\frac{(x_i-\overline{x})^2}{n\text{Var}(x)}$$which is minimum when $x_i=\overline{x}$, and then $\boldsymbol{H}_{i,i}=1/n$, otherwise $\boldsymbol{H}_{i,i}>1/n$. And this property is also valid in a multiple regression (as soon as an intercept is included in the regression – which should always be the case). To prove that result, let $\tilde{\boldsymbol{X}}$ denote the matrix of centered variables $\boldsymbol{X}$, then we can prove that $$\boldsymbol{H}_{i,i}=\frac{1}{n}+\big[\tilde{\boldsymbol{X}}(\tilde{\boldsymbol{X}}^\top\tilde{\boldsymbol{X}})^{-1}\tilde{\boldsymbol{X}}^\top\big]_{i,i}$$(which is basically a matrix version of the previous equation).

I can maybe add another comment on Anscombe’s data. We’ve seen that on the right that we did reach 1. But I did not prove it. One way to prove it is actually to focus on the remaining $n-1$ points, on the left. Those have all the same $x$ values. We can prove that if $\boldsymbol{X}_{i_1}=\boldsymbol{X}_{i_2}$, then $$\boldsymbol{H}_{i_1,i_2}=\boldsymbol{X}_{i_1}^\top(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}_{i_2}=\boldsymbol{H}_{i_1,i_1}$$hence, using the relationship obtained since the hat matrix is idempotent$$\boldsymbol{H}_{i_1,i_1}=2\boldsymbol{H}_{i_1,i_1}^2+\sum_{j\notin\{i_1,i_2\}}\boldsymbol{H}_{i_1,j}^2$$thus, we now have$$\boldsymbol{H}_{i_1,i_1}\big(1-2\boldsymbol{H}_{i_1,i_1}\big)>0$$i.e. $\boldsymbol{H}_{i_1,i_1}\in[0,1/2]$, where the upper bound becomes $1/(n-1)$ “duplicates”. So for $n-1$ $\boldsymbol{H}_{i,i}$‘s, we have values below $1/(n-1)$, the last one should be below $1$ and the sum has to be $k=2$ . So we have the value of the $n$ $\boldsymbol{H}_{i,i}$‘s.

# Insurance data science : Networks

At the Summer School of the Swiss Association of Actuaries, in Lausanne, I will start talking about networks and insurance this Friday. Slides are available online

# Insurance data science : Text

At the Summer School of the Swiss Association of Actuaries, in Lausanne, I will start talking about text based data and NLP this Thursday. Slides are available online

Ewen Gallic (AMSE) will present a tutorial on tweets. I can upload a few additional slides on LSTM (recurrent neural nets)

# Insurance data science : Pictures

At the Summer School of the Swiss Association of Actuaries, in Lausanne, following the part of Jean-Philippe Boucher (UQAM) on telematic data, I will start talking about pictures this Wednesday. Slides are available online

Ewen Gallic (AMSE) will present a tutorial on satellite pictures, and a simple classification problem, related to Alzeimher detection.

We will try to identify what is on the following pictures, starting with the car

(we will see that the car is indeed identified)

a skier,

and a fire,

We will also discuss previous pictures from the summer school

# Insurance data science : use and value of unusual data #1

Next week, with , I will be at the Summer School of the Swiss Association of Actuaries, in Lausanne, with Jean-Philippe Boucher (UQAM) and Ewen Gallic (AMSE).

I will give an introductionary talk on Monday morning, and the slides are now available

There will be some hands-on applications, on R. I will share some codes in the slides.

# 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&gt;=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,]&lt;=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))$root b=a/m-a return(c(a,b)) } Sim_AUC_mean_inter=function(m=.5,i=.05){ V_auc=rep(NA,ns) b=-1 essai = try(ab&lt;-ab_beta(m,i),TRUE) if(inherits(essai,what="try-error")) a=-1 if(!inherits(essai,what="try-error")){ a=ab[1] b=ab[2] } if((a&gt;=0)&amp;(b&gt;=0)){ for(s in 1:ns){ p=rbeta(n,a,b) Y=rbinom(n,size = 1,prob = p) auc.tmp=performance(prediction(p,Y),"auc") V_auc[s]=as.numeric(auc.tmp@y.values)} L=list(moy_beta=m, var_beat=v, q05=qbeta(.05,a,b), q95=qbeta(.95,a,b), moy_AUC=mean(V_auc), sd_AUC=sd(V_auc), q05_AUC=quantile(V_auc,.05), q95_AUC=quantile(V_auc,.95)) return(L)} if((a&lt;0)|(b&lt;0)){return(list(moy_AUC=NA))}} Vm=seq(.025,.975,by=.025) Vi=seq(.01,.5,by=.01) V=outer(X = Vm,Y = Vi, Vectorize(function(x,y) Sim_AUC_mean_inter(x,y)$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 $i$th 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$$\widehat{\varepsilon}_{-i}=(1-H_{i,i})^{-1}\hat\varepsilon_i$$i.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,"&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]) 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]&lt;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…