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Exposure with Binomial Responses

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Exposure with Binomial Responses

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Last week, we’ve seen how to take into account the exposure to compute nonparametric estimators of several quantities (empirical means, and empirical variances) incorporating exposure. Let us see what can be done if we want to model a binomial response. The model here is the following: ,

  • the number of claims http://latex.codecogs.com/gif.latex?N_i on the period http://latex.codecogs.com/gif.latex?[0,1>
  • the number of claims http://latex.codecogs.com/gif.latex?Y_i on http://latex.codecogs.com/gif.latex?[0,E_i>)

that can be visualize below


Consider the case where the variable of interest is not the number of claims, but simply the indicator of the occurrence of a claim. Then we wish to model the event http://latex.codecogs.com/gif.latex?\{N=0\} versus http://latex.codecogs.com/gif.latex?\{N%3E0\}, interpreted as non-occurrence and occurrence. Given the fact that we can only observe http://latex.codecogs.com/gif.latex?\{Y=0\} versus http://latex.codecogs.com/gif.latex?\{Y%3E0\}. Having an inclusion is not enough to derive a model. Actually, with a Poisson process model, we can get easily that


With words, it means that the probability of not having a claim in the first six months of the year is the square root of not have a claim over a year. Which makes sense. Assume that the probability of not having a claim can be explained by some covariates, denoted http://latex.codecogs.com/gif.latex?\boldsymbol{X}, through some link function (using the GLM terminology),


Now, since we do observe http://latex.codecogs.com/gif.latex?Y – and not http://latex.codecogs.com/gif.latex?N – we have

The dataset we will use is always the same

> sinistre=read.table("http://freakonometrics.free.fr/sinistreACT2040.txt",
+ header=TRUE,sep=";")
> sinistres=sinistre[sinistre$garantie=="1RC",]
> sinistres=sinistres[sinistres$cout>0,]
> contrat=read.table("http://freakonometrics.free.fr/contractACT2040.txt",
+ header=TRUE,sep=";")
> T=table(sinistres$nocontrat)
> T1=as.numeric(names(T))
> T2=as.numeric(T)
> nombre1 = data.frame(nocontrat=T1,nbre=T2)
> I = contrat$nocontrat%in%T1
> T1= contrat$nocontrat[I==FALSE]
> nombre2 = data.frame(nocontrat=T1,nbre=0)
> nombre=rbind(nombre1,nombre2)
> sinistres = merge(contrat,nombre)
> sinistres$nonsin = (sinistres$nbre==0)

The first model we can consider is based on the standard logistic approach, i.e.


That’s nice, but difficult to handle with standard functions. Nevertheless, it is always possible to compute numerically the maximum likelihood estimator of http://latex.codecogs.com/gif.latex?\boldymbol{\beta} given http://latex.codecogs.com/gif.latex?(Y_i,\boldsymbol{X}_i,E_i).

> Y=sinistres$nonsin
> X=cbind(1,sinistres$ageconducteur)
> E=sinistres$exposition
> logL = function(beta){
+ 	pi=(exp(X%*%beta)/(1+exp(X%*%beta)))^E
+ 	-sum(log(dbinom(Y,size=1,prob=pi)))
+ }
> optim(fn=logL,par=c(-0.0001,-.001),
+ method="BFGS")
[1] 2.14420560 0.01040707
[1] 7604.073
function gradient 
      42       10 
[1] 0
> parametres=optim(fn=logL,par=c(-0.0001,-.001),
+ method="BFGS")$par

Now, let us look at alternatives, based on standard regression models. For instance a binomial-log model. Because the exposure appears as a power of the annual probability, everything would be fine if http://latex.codecogs.com/gif.latex?h was the exponential function (or http://latex.codecogs.com/gif.latex?h^{-1} was the log link function), since

Now, if we try to code it, it starts quickly to be problematic,

> reg=glm(nonsin~ageconducteur+offset(exposition),
+ data=sinistresI,family=binomial(link="log")) 
Error: no valid set of coefficients has been found: please supply starting values

I tried (almost) everything I could, but I could not get rid of that error message,

> startglm=c(0,-.001)
> names(startglm)=c("(Intercept)","ageconducteur")
> etaglm=rep(-.01,nrow(sinistresI))
> etaglm[sinistresI$nonsin==0]=-10
> muglm=exp(etaglm)
> reg=glm(nonsin~ageconducteur+offset(exposition),
+ data=sinistresI,family=binomial(link="log"),
+ control = glm.control(epsilon=1e-5,trace=TRUE,maxit=50),
+ start=startglm,
+ etastart=etaglm,mustart=muglm)
Deviance = NaN Iterations - 1 
Error: no valid set of coefficients has been found: please supply starting values

So I decided to give up. Almost. Actually, the problem comes from the fact that http://latex.codecogs.com/gif.latex?\mathbb{P}(Y=0) is closed to 1. I guess everything would be nicer if we could work with probability close to 0. Which is possible, since


where http://latex.codecogs.com/gif.latex?\mathbb{P}(N%3E0) is close to 0. So we can use Taylor’s expansion,


Here, the exposure does no longer appears as a power of the probability, but appears multiplicatively. Of course, there are higher order terms. But let us forget them (so far). If – one more time – we consider a log link function, then we can incorporate the exposure, or to be more specific, the logarithm of the exposure.

> regopp=glm((1-nonsin)~ageconducteur+offset(log(exposition)),
+ data=sinistresI,family=binomial(link="log"))

which now works perfectly.

Now, to see a final model, perhaps we should get back to our Poisson regression model since we do have a model for the probability that http://latex.codecogs.com/gif.latex?\mathbb{P}(Y=\cdot).

> regpois=glm(nbre~ageconducteur+offset(log(exposition)),
+ data=sinistres,family=poisson(link="log"))

We can now compare those three models. Perhaps, we should also include the prediction without any explanatory variable. For the second model (actually, it does run without any explanatory variable), we run

>  regreff=glm((1-nonsin)~1+offset(log(exposition)),
+ data=sinistres,family=binomial(link="log"))

so that the prediction is here

> exp(coefficients(regreff))

This value is comparable with the logistic regression,

> logL2 = function(beta){
+ 	pi=(exp(beta)/(1+exp(beta)))^E
+ 	-sum(log(dbinom(Y,size=1,prob=pi)))}
> param=optim(fn=logL2,par=.01,method="BFGS")$par
> 1-exp(param)/(1+exp(param))
[1] 0.06747777

But is quite different from the Poisson model,

> exp(coefficients(glm(nbre~1+offset(log(exposition)),
+ data=sinistres,family=poisson(link="log"))))

Let us produce a graph, to compare those models,

> age=18:100
> yml1=exp(parametres[1]+parametres[2]*age)/(1+exp(parametres[1]+parametres[2]*age))
> plot(age,1-yml1,type="l",col="purple")
> yp=predict(regpois,newdata=data.frame(ageconducteur=age,
+ exposition=1),type="response")
> yp1=1-exp(-yp)
> ydl=predict(regopp,newdata=data.frame(ageconducteur=age,
+ exposition=1),type="response")
> plot(age,ydl,type="l",col="red")
> lines(age,yp1,type="l",col="blue")
> lines(age,1-yml1,type="l",col="purple")
> abline(h=exp(coefficients(regreff)),lty=2)


Observe here that the three models are quite different. Actually, with two models, it is possible to run more complex regression, e.g. with splines, to visualize the impact of the age on the probability of having – or not – a car accident. If we compare the Poisson regression (still in red) and the log-binomial model, with Taylor’s expansion, we get


The next step is to see how to incorporate the exposure in a tree. But that’s another story…

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Published at DZone with permission of Arthur Charpentier, DZone MVB. See the original article here.

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