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)

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….

Defense de doctorat d’Amadou Diogo Barry

Ce jeudi matin, Amadou Diogo Barry défendra sa thèse de doctorat sur la régression expectile pour l’analyse de données longitudinales.

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% !

Variance decomposition and price segmentation in Insurance

Today, I was giving a talk at the Economics department, and I got a very interesting question about some tables I keep showing to explain why insurance companies like segmentation. The tables illustrate three different case. Here, $S$ stands for the individual (random) loss.

• the first one is the case where the premium asked is the same for all the insured – i.e. the pure premium $\mathbb{E}[S]$

As explain, the loss is here on an individual basis, so, per policy, the insurer faces the (random) loss $S-\mathbb{E}[S]$, which is, on average, null. That’s the second line. For the last line, I keep saying that we look at the overall loss of the insurer, but that’s not correct. Here, with a factor $n$, we would have the variance of the total loss for the insurance company. We just removed the $n$ factor in the table

• then, we have perfectly observable heterogeneity : insured have a risk factor $\Omega$, obervable, and in that case, the ‘pure’ premium is $\mathbb{E}[S|\Omega]$

That’s what we have below. Here again, on average, the insured should have a null profit. And the total variance (which was $\text{Var}[S]$ in our previous example) is now splitted in two parts (that’s basically Pythagoras theorem).

The interpreration is the following

And then, I usually mention the third and last case, more realistic

• the risk factor $\Omega$ is not obervable, but segmentation is still possible using some proxy of the risk factor, obtained using some covariates, and the ‘pure’ premium is $\mathbb{E}[S|\boldsymbol{X}]$

And here also, there is a nice interpretation, because of the variance decomposition : there is one part that we observed previously, with some ‘perfect pricing’ and an additional part (that is positive) that is related to the fact that the covariates are just a proxy of the risk factor….

The term on the left is then a lower bound, obtained if actually, using our covariate, available for the pricing, we can get the risk factor.

That was my story, but the fact that $n$ (the portfolio size) was not mentioned in the tables was a bit confusing… So I decided to create some graphs to illustrate those three cases

Consider some simple simulations. On the graph on the left, we have on the x-axis the risk factor, and on the y-axis, the loss (going roughly from 0 to 20). The pure premium is the average of those losses. Here, it’s 10. That’s the plain red line (on the left). In the middle, the y-axis is the insured profit/loss per policy. Someone with a loss close to 0 means a gain of 10, someone with a loss close to 20 means a loss of 10. On average, there is no profit (that’s the plain line). And then, on the right, we have the distribution of the profit/loss (per contract). Again, on average it’s 0, with some variance,

Consider here is simple covariate $x$ : assume here that’s we’ve been able to create a binary variable, that can distinguish the low risks and the high risks. Here, there are two levels for the premium. The low premium is close to 6, and the high one is close to 14. That’s again the graph on the left

Then we have the profit/loss per policy for the insured, in the middle. Here, when the loss was close to 0, the gain is smaller : it is 6 (while it was 10 before). When it was close to 10, previously, it meant a 0 profit, but now it’s either a loss of 4, or a gain of 4. The profit/loss distribution is now on the right. There is less dispersion, and less variance. That the decrease of variance we’ve discussed before. To summarize, segmentation does reduce the variability of the result for the insurance company. That’s what we observe on the right.

• premium based on the risk factor

Assume now that $\Omega$ is observable. And that we use it for our pricing. The premium is now continuous, and it is the red line, on the left. The profit/loss (in the middle) is the difference between the loss, and its expected value (conditional on the risk factor). And on the right, we have the distribution.

As expected, there is much less variability on the profit/loss distribution of the insurance company in that case. And actually, that’s a lower bound for the variance of result of the insurance company… I hope that the graph clarify what’s going on here…

Variance of the slope in a regression model

In my “applied linear models” exam, there was a tricky question (it was a multiple choice, so no details were asked). I was simply asking if the following statement was valid, or not

Consider a linear regression with one single covariate, $y=\beta_0+\beta_1x_1+\varepsilon$ and the least-square estimates. The variance of the slope is $\text{Var}[\widehat{\beta}_1]$ Do we decrease this variance if we add one variable, and consider $y=\beta_0+\beta_1x_1+\beta_2x_2+\varepsilon$ ?

For the exam, the expected answer was simply “no”. In a nutshell, there are two cases where we should expect different changes,

• if $x_1$ and $x_2$ are highly correlated, then we should expect the variance to increase
• if $x_1$ and $x_2$ are not correlated, then we should expect the variance to decrease

We did briefly observed (and discussed) those points on examples during the lecture… but I wanted to go a bit further, since I couldn’t find any analytical results. Let us generate a model $y=\beta_0+\beta_1x_1+\beta_2x_2+\varepsilon$, and then compare the variance $\text{Var}[\widehat{\beta}_1]$ on the two fitted modes, depending on the correlation between $x_1$ and $x_2$

 library(mnormt) n=200 s=function(r=0){ S=matrix(c(1,r,r,1),2,2) X=rmnorm(n,c(0,0),S) B=data.frame(y=-2+X[,1]+X[,2]+rnorm(n)/2, x1=X[,1], x2=X[,2]) reg12=lm(y~x1+x2,data=B) reg1=lm(y~x1,data=B) k=summary(reg12)$coefficients[2,2]/summary(reg1)$coefficients[2,2] k}

Let us generate 500 samples for each value of the correlation, from -0.9 to _0.9

 M=NULL for(r in ((-9):9)/10) M=cbind(M,Vectorize(s)(rep(r,500)))

and let us plot the ratio of the two variances

 plot(0:1,0:1,xlim=c(-1,1),ylim=c(0,2),col="white") for(i in 1:19) points(rep((((-9):9)/10)[i],500),M[,i],col="light blue") VM=apply(M,2,mean) lines((((-9):9)/10),VM,col="red",lwd=2) abline(h=1,lty=2)

If the ratio exceeds 1, the variance increases when adding a covariate.

Indeed, here, when the two variables are independent, the variance is divided by two. But when covariates are highly correlated, the variance is multiplied by two…

Now, what if, actually, $x_2$ is not a real explanatory variable : the true model we generate is $y=\beta_0+\beta_1x_1+\varepsilon$. In that case,

 s=function(r=0){ S=matrix(c(1,r,r,1),2,2) X=rmnorm(n,c(0,0),S) B=data.frame(y=-2+X[,1]+rnorm(n)/2, x1=X[,1], x2=X[,2]) reg12=lm(y~x1+x2,data=B) reg1=lm(y~x1,data=B) k=summary(reg12)$coefficients[2,2]/summary(reg1)$coefficients[2,2] k}

we get our samples as previously

 M=NULL for(r in ((-9):9)/10) M=cbind(M,Vectorize(s)(rep(r,500)))

and we plot those ratios

 plot(0:1,0:1,xlim=c(-1,1),ylim=c(0,2),col="white") for(i in 1:19) points(rep((((-9):9)/10)[i],500),M[,i],col="light blue") VM=apply(M,2,mean) lines((((-9):9)/10),VM,col="red",lwd=2) abline(h=1,lty=2)

In the case we add a useless variable $x_2$, whatever the correlation with $x_1$, it will always, on average, increase the variance of $\widehat{\beta}_1$.

Talk at the Economics Seminar at UQAM

Wednesday, at lunch time, I will give a talk on insurance pricing at the (internal) economics seminar, at UQAM. Slides are now online.

In May, I will also give a seminar on the same topic at the Computational Science seminar…