# GLM, non-linearity and heteroscedasticity

Last week in the non-life insurance course, we’ve seen the theory of the Generalized Linear Models, emphasizing the two important components

• the link function (which is actually the key component in predictive modeling)
• the distribution, or the variance function

Just to illustrate, consider my favorite dataset

­lin.mod = lm(dist~speed,data=cars)

A linear model means here $Y_i=\beta_0+\beta_1X_i+\varepsilon_i$

where the residuals are assumed to be centered, independent, and with identical variance. If we visualize that linear regression, we usually see something like that

The idea here (in GLMs) is to assume $Y\vertX=x\sim\mathcal{N}(\beta_0+\beta_1x,\sigma^2)$

which will produce the same model as the one describe previously, based on some error term. That model can be visualized below,

attach(cars)
n=2
X= cars$speed Y=cars$dist
df=data.frame(X,Y)
vX=seq(min(X)-2,max(X)+2,length=n)
vY=seq(min(Y)-15,max(Y)+15,length=n)
mat=persp(vX,vY,matrix(0,n,n),zlim=c(0,.1),theta=-30,ticktype ="detailed", box = FALSE)
reggig=glm(Y~X,data=df,family=gaussian(link="identity"))
x=seq(min(X),max(X),length=501)
C=trans3d(x,predict(reggig,newdata=data.frame(X=x),type="response"),rep(0,length(x)),mat)
lines(C,lwd=2)
sdgig=sqrt(summary(reggig)$dispersion) x=seq(min(X),max(X),length=501) y1=qnorm(.95,predict(reggig,newdata=data.frame(X=x),type="response"), sdgig) C=trans3d(x,y1,rep(0,length(x)),mat) lines(C,lty=2) y2=qnorm(.05,predict(reggig,newdata=data.frame(X=x),type="response"), sdgig) C=trans3d(x,y2,rep(0,length(x)),mat) lines(C,lty=2) C=trans3d(c(x,rev(x)),c(y1,rev(y2)),rep(0,2*length(x)),mat) polygon(C,border=NA,col="yellow") C=trans3d(X,Y,rep(0,length(X)),mat) points(C,pch=19,col="red") n=8 vX=seq(min(X),max(X),length=n) mgig=predict(reggig,newdata=data.frame(X=vX)) sdgig=sqrt(summary(reggig)$dispersion)
for(j in n:1){
stp=251
x=rep(vX[j],stp)
y=seq(min(min(Y)-15,qnorm(.05,predict(reggig,newdata=data.frame(X=vX[j]),type="response"), sdgig)),max(Y)+15,length=stp)
z0=rep(0,stp)
z=dnorm(y, mgig[j], sdgig)
C=trans3d(c(x,x),c(y,rev(y)),c(z,z0),mat)
polygon(C,border=NA,col="light blue",density=40)
C=trans3d(x,y,z0,mat)
lines(C,lty=2)
C=trans3d(x,y,z,mat)
lines(C,col="blue")}

We do have two parts here: the linear increase of the average, $\mathbb{E}(Y\vert X=x)=\beta_0+\beta_1x$ and the constant variance of the normal distribution $\text{Var}(Y\vert X=x)=\sigma^2$.

On the other hand, if we assume a Poisson regression,

poisson.reg = glm(dist~speed,data=cars,family=poisson(link="log"))

we have something like

This time, two things have changed simultaneously: our model is no longer linear, it is an exponential one $\mathbb{E}(Y\vert X=x)=e^{\beta_0+\beta_1x}$, and the variance is also increasing with the explanatory variable $\text{Var}(Y\vert X=x)=e^{\beta_0+\beta_1x}$, since with a Poisson regression,
$Y\vert X=x\sim\mathcal{P}(e^{\beta_0+\beta_1x})$

If we adapt the previous code, we get

The problem is that we changed two things when we introduced the Poisson regression from the linear model. So let us look at what happens when we change the two components independently. First, we can change the link function, with a Gaussian model but this time a multiplicative model (with a logarithm link function)

gaussian.reg = glm(dist~speed,data=cars,family=gaussian(link="log"))

which is still, here, an homoscedasctic model, but this time non-linear. Or we can change the link function in the Poisson regression, to get a linear model, but heteroscedastic

poisson.lin = glm(dist~speed,data=cars,family=poisson(link="identity"))

So this is basically what GLMs are about….

# Modélisation des coûts individuels

Cette semaine, même si le réseau de l’UQAM est down, on va continuer le cours et finir la section sur la modélisation de la surdispersion pour la fréquence de sinistres. On devrait ensuite commencer la modélisation des coûts individuels. En particulier, on passera du temps autour de deux points,

• la distinction lognormale et gamma
• l’écrêtement des gros sinistres

Les transparents sont en ligne. Et la base des coûts est celle évoquée au second cours.