Maximum likelihood estimates for multivariate distributions

Consider our loss-ALAE dataset, and – as in Frees & Valdez (1998) – let us fit a parametric model, in order to price a reinsurance treaty. The dataset is the following,

> library(evd)
> data(lossalae)
> Z=lossalae
> X=Z[,1];Y=Z[,2]

The first step can be to estimate marginal distributions, independently. Here, we consider lognormal distributions for both components,

> Fempx=function(x) mean(X<=x)
> Fx=Vectorize(Fempx)
> u=exp(seq(2,15,by=.05))
> plot(u,Fx(u),log="x",type="l",
+ xlab="loss (log scale)")
> Lx=function(px) -sum(log(Vectorize(dlnorm)(
+ X,px[1],px[2])))
> opx=optim(c(1,5),fn=Lx)
> opx$par [1] 9.373679 1.637499 > lines(u,Vectorize(plnorm)(u,opx$par[1],
+ opx$par[2]),col="red") The fit here is quite good, For the second component, we do the same, > Fempy=function(x) mean(Y<=x) > Fy=Vectorize(Fempy) > u=exp(seq(2,15,by=.05)) > plot(u,Fy(u),log="x",type="l", + xlab="ALAE (log scale)") > Ly=function(px) -sum(log(Vectorize(dlnorm)( + Y,px[1],px[2]))) > opy=optim(c(1.5,10),fn=Ly) > opy$par
[1] 8.522452 1.429645
> lines(u,Vectorize(plnorm)(u,opy$par[1], + opy$par[2]),col="blue")

It is not as good as the fit obtained on losses, but it is not that bad,

Now, consider a multivariate model, with Gumbel copula. We’ve seen before that it worked well. But this time, consider the maximum likelihood estimator globally.

> Cop=function(u,v,a) exp(-((-log(u))^a+
+ (-log(v))^a)^(1/a))
> phi=function(t,a) (-log(t))^a
> cop=function(u,v,a) Cop(u,v,a)*(phi(u,a)+
+ phi(v,a))^(1/a-2)*(
+ a-1+(phi(u,a)+phi(v,a))^(1/a))*(phi(u,a-1)*
+ phi(v,a-1))/(u*v)
> L=function(p) {-sum(log(Vectorize(dlnorm)(
+ X,p[1],p[2])))-
+ sum(log(Vectorize(dlnorm)(Y,p[3],p[4])))-
+ sum(log(Vectorize(cop)(plnorm(X,p[1],p[2]),
+ plnorm(Y,p[3],p[4]),p[5])))}
> opz=optim(c(1.5,10,1.5,10,1.5),fn=L)
> opz$par [1] 9.377219 1.671410 8.524221 1.428552 1.468238 Marginal parameters are (slightly) different from the one obtained independently, > c(opx$par,opy$par) [1] 9.373679 1.637499 8.522452 1.429645 > opz$par[1:4]
[1] 9.377219 1.671410 8.524221 1.428552

And the parameter of Gumbel copula is close to the one obtained with heuristic methods in class.

Now that we have a model, let us play with it, to price a reinsurance treaty. But first, let us see how to generate Gumbel copula… One idea can be to use the frailty approach, based on a stable frailty. And we can use Chambers et al (1976)to generate a stable distribution. So here is the algorithm to generate samples from Gumbel copula

> alpha=opz$par[5] > invphi=function(t,a) exp(-t^(1/a)) > n=500 > x=matrix(rexp(2*n),n,2) > angle=runif(n,0,pi) > E=rexp(n) > beta=1/alpha > stable=sin((1-beta)*angle)^((1-beta)/beta)* + (sin(beta*angle))/(sin(angle))^(1/beta)/ + (E^(alpha-1)) > U=invphi(x/stable,alpha) > plot(U) Here, we consider only 500 simulations, Based on that copula simulation, we can then use marginal transformations to generate a pair, losses and allocated expenses, > Xloss=qlnorm(U[,1],opz$par[1],opz$par[2]) > Xalae=qlnorm(U[,2],opz$par[3],opz$par[4]) In standard reinsurance treaties – see e.g. Clarke (1996) – allocated expenses are splited prorata capita between the insurance company, and the reinsurer. If $X$ denotes losses, and $Y$ the allocated expenses, a standard excess treaty can be has payoff $\begin{cases} 0\cdot \boldsymbol{1}(X\leq R)\\ \displaystyle{\left(X-R+ \frac{X-R}{X}\cdot Y\right) \right )}\cdot\boldsymbol{1}(R where $L$ denotes the (upper) limit, and $R$ the insurer’s retention. Using monte carlo simulation, it is then possible to estimate the pure premium of such a reinsurance treaty. > L=100000 > R=50000 > Z=((Xloss-R)+(Xloss-R)/Xloss*Xalae)* + (R<=Xloss)*(Xloss<L)+ + ((L-R)+(L-R)/R*Xalae)*(L<=Xloss) > mean(Z) [1] 12596.45 Now, play with it… it is possible to find a better fit, I guess… (nonparametric) copula density estimation Today, we will go further on the inference of copula functions. Some codes (and references) can be found on a previous post, on nonparametric estimators of copula densities (among other related things). Consider (as before) the loss-ALAE dataset (since we’ve been working a lot on that dataset) > library(MASS) > library(evd) > X=lossalae > U=cbind(rank(X[,1])/(nrow(X)+1),rank(X[,2])/(nrow(X)+1)) The standard tool to plot nonparametric estimators of densities is to use multivariate kernels. We can look at the density using > mat1=kde2d(U[,1],U[,2],n=35) > persp(mat1$x,mat1$y,mat1$z,col="green",
+ xlab="",ylab="",zlab="",zlim=c(0,7))

or level curves (isodensity curves) with more detailed estimators (on grids with shorter steps)

> mat1=kde2d(U[,1],U[,2],n=101)
> image(mat1$x,mat1$y,mat1$z,col= + rev(heat.colors(100)),xlab="",ylab="") > contour(mat1$x,mat1$y,mat1$z,add=
+ TRUE,levels = pretty(c(0,4), 11))

Kernels are nice, but we clearly observe some border bias, extremely strong in corners (the estimator is 1/4th of what it should be, see another post for more details). Instead of working on sample $(U_i,V_i)$ on the unit square, consider some transformed sample $(Q(U_i),Q(V_i))$, where $Q:(0,1)\rightarrow\mathbb{R}$ is a given function. E.g. a quantile function of an unbounded distribution, for instance the quantile function of the $\mathcal{N}(0,1)$ distribution. Then, we can estimate the density of the transformed sample, and using the inversion technique, derive an estimator of the density of the initial sample. Since the inverse of a (general) function is not that simple to compute, the code might be a bit slow. But it does work,

> gaussian.kernel.copula.surface <- function (u,v,n) {
+   s=seq(1/(n+1), length=n, by=1/(n+1))
+   mat=matrix(NA,nrow = n, ncol = n)
+ sur=kde2d(qnorm(u),qnorm(v),n=1000,
+ lims = c(-4, 4, -4, 4))
+ su<-sur$z + for (i in 1:n) { + for (j in 1:n) { + Xi<-round((qnorm(s[i])+4)*1000/8)+1; + Yj<-round((qnorm(s[j])+4)*1000/8)+1 + mat[i,j]<-su[Xi,Yj]/(dnorm(qnorm(s[i]))* + dnorm(qnorm(s[j]))) + } + } + return(list(x=s,y=s,z=data.matrix(mat))) + } Here, we get Note that it is possible to consider another transformation, e.g. the quantile function of a Student-t distribution. > student.kernel.copula.surface = + function (u,v,n,d=4) { + s <- seq(1/(n+1), length=n, by=1/(n+1)) + mat <- matrix(NA,nrow = n, ncol = n) + sur<-kde2d(qt(u,df=d),qt(v,df=d),n=5000, + lims = c(-8, 8, -8, 8)) + su<-sur$z
+ for (i in 1:n) {
+     for (j in 1:n) {
+ 	Xi<-round((qt(s[i],df=d)+8)*5000/16)+1;
+ 	Yj<-round((qt(s[j],df=d)+8)*5000/16)+1
+ 	mat[i,j]<-su[Xi,Yj]/(dt(qt(s[i],df=d),df=d)*
+ 	dt(qt(s[j],df=d),df=d))
+     }
+ }
+ return(list(x=s,y=s,z=data.matrix(mat)))
+ }

Another strategy is to consider kernel that have precisely the unit interval as support. The idea is here to consider the product of Beta kernels, where parameters depend on the location

> beta.kernel.copula.surface=
+  function (u,v,bx=.025,by=.025,n) {
+  s <- seq(1/(n+1), length=n, by=1/(n+1))
+  mat <- matrix(0,nrow = n, ncol = n)
+ for (i in 1:n) {
+     a <- s[i]
+     for (j in 1:n) {
+     b <- s[j]
+ 	mat[i,j] <- sum(dbeta(a,u/bx,(1-u)/bx) *
+     dbeta(b,v/by,(1-v)/by)) / length(u)
+     }
+ }
+ return(list(x=s,y=s,z=data.matrix(mat)))
+ }

On those two graphs, we can clearly observe strong tail dependence in the upper (right) corner, that cannot be intuited using a standard kernel estimator…

Copulas and tail dependence, part 1

As mentioned in the course last week Venter (2003) suggested nice functions to illustrate tail dependence (see also some slides used in Berlin a few years ago).

• Joe (1990)’s lambda

Joe (1990) suggested a (strong) tail dependence index. For lower tails, for instance, consider

i.e

• Upper and lower strong tail (empirical) dependence functions

The idea is to plot the function above, in order to visualize limiting behavior. Define

for the lower tail, and

for the upper tail, where is the survival copula associated with , in the sense that

while

Now, one can easily derive empirical conterparts of those function, i.e.

and

Thus, for upper tail, on the right, we have the following graph

and for the lower tail, on the left, we have

For the code, consider some real data, like the loss-ALAE dataset.

> library(evd)
> X=lossalae

The idea is to plot, on the left, the lower tail concentration function, and on the right, the upper tail function.

> U=rank(X[,1])/(nrow(X)+1)
> V=rank(X[,2])/(nrow(X)+1)
> Lemp=function(z) sum((U<=z)&(V<=z))/sum(U<=z)
> Remp=function(z) sum((U>=1-z)&(V>=1-z))/sum(U>=1-z)
> u=seq(.001,.5,by=.001)
> L=Vectorize(Lemp)(u)
> R=Vectorize(Remp)(rev(u))
> plot(c(u,u+.5-u[1]),c(L,R),type="l",ylim=0:1,
+ xlab="LOWER TAIL          UPPER TAIL")
> abline(v=.5,col="grey")

Now, we can compare this graph, with what should be obtained for some parametric copulas that have the same Kendall’s tau (e.g.). For instance, if we consider a Gaussian copula,

> tau=cor(lossalae,method="kendall")[1,2]
> library(copula)
> paramgauss=sin(tau*pi/2)
> copgauss=normalCopula(paramgauss)
> Lgaussian=function(z) pCopula(c(z,z),copgauss)/z
> Rgaussian=function(z) (1-2*z+pCopula(c(z,z),copgauss))/(1-z)
> u=seq(.001,.5,by=.001)
> Lgs=Vectorize(Lgaussian)(u)
> Rgs=Vectorize(Rgaussian)(1-rev(u))
> lines(c(u,u+.5-u[1]),c(Lgs,Rgs),col="red")

or Gumbel’s copula,

> paramgumbel=1/(1-tau)
> copgumbel=gumbelCopula(paramgumbel, dim = 2)
> Lgumbel=function(z) pCopula(c(z,z),copgumbel)/z
> Rgumbel=function(z) (1-2*z+pCopula(c(z,z),copgumbel))/(1-z)
> u=seq(.001,.5,by=.001)
> Lgl=Vectorize(Lgumbel)(u)
> Rgl=Vectorize(Rgumbel)(1-rev(u))
> lines(c(u,u+.5-u[1]),c(Lgl,Rgl),col="blue")

That’s nice (isn’t it?), but since we do not have any confidence interval, it is still hard to conclude (even if it looks like Gumbel copula has a much better fit than the Gaussian one). A strategy can be to generate samples from those copulas, and to visualize what we had. With a Gaussian copula, the graph looks like

> u=seq(.0025,.5,by=.0025); nu=length(u)
> nsimul=500
> MGS=matrix(NA,nsimul,2*nu)
> for(s in 1:nsimul){
+ Xs=rCopula(nrow(X),copgauss)
+ Us=rank(Xs[,1])/(nrow(Xs)+1)
+ Vs=rank(Xs[,2])/(nrow(Xs)+1)
+ Lemp=function(z) sum((Us<=z)&(Vs<=z))/sum(Us<=z)
+ Remp=function(z) sum((Us>=1-z)&(Vs>=1-z))/sum(Us>=1-z)
+ MGS[s,1:nu]=Vectorize(Lemp)(u)
+ MGS[s,(nu+1):(2*nu)]=Vectorize(Remp)(rev(u))
+ lines(c(u,u+.5-u[1]),MGS[s,],col="red")
+ }

(including – pointwise – 90% confidence bands)

> Q95=function(x) quantile(x,.95)
> V95=apply(MGS,2,Q95)
> lines(c(u,u+.5-u[1]),V95,col="red",lwd=2)
> Q05=function(x) quantile(x,.05)
> V05=apply(MGS,2,Q05)
> lines(c(u,u+.5-u[1]),V05,col="red",lwd=2)

while it is

with Gumbel copula. Isn’t it a nice (graphical) tool ?

But as mentioned in the course, the statistical convergence can be slow. Extremely slow. So assessing if the underlying copula has tail dependence, or not, it now that simple. Especially if the copula exhibits tail independence. Like the Gaussian copula. Consider a sample of size 1,000. This is what we obtain if we generate random scenarios,

or we look at the left tail (with a log-scale)

Now, consider a 10,000 sample,

or with a log-scale

We can even consider a 100,000 sample,

or with a log-scale

On those graphs, it is rather difficult to conclude if the limit is 0, or some strictly positive value (again, it is a classical statistical problem when the value of interest is at the border of the support of the parameter). So, a natural idea is to consider a weaker tail dependence index. Unless you have something like 100,000 observations…

Tails of copulas, une lecture graphique

Suite à une formation que je faisais en fin de semaine à Brest (les slides sont ici et ), je voulais revenir sur les histoires de tails of copulas, pour reprendre le titre de l’article (ici) de Gary Venter (et qui correspond à des choses que j’avais pu présenter il y a quelques années à Berlin, les slides étant en ligne ici).

• Quantifier la dépendance de queue

L’idée est de noter qu’il est noter qu’il existe deux manières de quantifier la dépendance de queue. La première est liée à l’approche de Joe (1990, ici, ou 1997 pour le livre), qui a introduit un (strong) tail dependence index. Par exemple pour la queue inférieure,

soit

La seconde est liée à une idée que l’on retrouve dans les travaux de Janet Heffernan, Stuart Coles ou Jonathan Tawn. L’intuition est la suivante (on peut la retrouver en ligne ici). Si  et  ont la même loi et que l’on suppose les variables indépendantes, alors

En revanche, si les variables sont comonotones (c’est à dire égales comme on suppose les lois identiques),

Aussi, on peut supposer qu’il existe un indice tel que

Le soucis est que le cas d’indépendance correspond à =2, alors que le cas de dépendance forte correspond au cas =1. Il est alors usuel de faire une transformation affine pour se ramener sur [0,1], et que la force de la dépendance soit croissante avec , e.g.

Posons alors

qui pourra être interprété comme un (weak) tail dependence index.
Bref, ces deux mesures donnent de l’information sur le comportement dans les queues de distribution.

• Les fonctions de concentration dans les queues

L’idée est de noter qu’il est possible d’étudier ces fonctions afin de mieux comprendre le comportement dans les queues. En s’inspirant de Gary Venter, on peut définir

pour étudier le comportement dans la queue inférieure, et

pour la queue supérieure,où  est la copule de survie associée à , au sens où

et

Cet outil permettra de modéliser la dépendance forte. On peut également poser, afin d’étudier la dépendance faible,

ou

• Application statistique

L’idée est de noter qu’il est facile d’estimer ces fonctions. Ces outils peuvent être utiles pour mieux comprendre le comportement dans les queues.
Par exemple pour une copule Gaussienne de corrélation 0,5, on a la forme théorique suivante pour les fonctions de concentration (au sens fort)

Statistiquement, il est possible d’estimer ces quantités en comptant simplement le nombre d’observations dans le coin inférieur gauche, ou le coin supérieur droit.  Si on dispose d’un échantillon, on peut alors regarder ce que donnent les versions

et

Pour un échantillon de taille n=500, on obtient les intervalles de confiance à 90% de la forme suivante,

Le code R ressemble à ça

> library(evd); data(lossalae)
> cor(lossalae,method="spearman")
Loss     ALAE
Loss 1.000000 0.451872
ALAE 0.451872 1.000000

avec le code suivant pour la version empirique,

> z=seq(0,.5,by=.001)
> U=rank(v[,1])/(nrow(v)+1)
> V=rank(v[,2])/(nrow(v)+1)
> Lemp=rep(NA,length(z))
> Remp=rep(NA,length(z))
> for(i in 1:length(z)){
+  Lemp[i]=sum((U<=z[i])&(V<=z[i]))/sum(U<=z[i])
+  Remp[i]=sum((U>=1-z[i])&(V>=1-z[i]))/sum(U<=z[i])
+ }

et pour la version théorique,

> Lg=(pcopula(copclayton,cbind(z,z)))/(z)
> Rg=((1-2*(1-z)+pcopula(copclayton,cbind(1-z,1-z))))/(z)
> plot(c(1-z,z),c(Lg,Rg))

De plus, on a des fonctions similaires pour la dépendance au sens faible, avec le code suivant pour la version théorique,

> Lg=log(pcopula(cop,cbind(z,z)))/log(z)
> Rg=log((1-2*(1-z)+pcopula(cop,cbind(1-z,1-z))))/log(z)
> Lg=1/Lg*2-1
> Rg=1/Rg*2-1

et celui là pour la version empirique

> z=seq(0,.5,by=.001)
> v <- lossalae
> U=rank(v[,1])/(nrow(v)+1)
> V=rank(v[,2])/(nrow(v)+1)
> Lemp=rep(NA,length(z))
> Remp=rep(NA,length(z))
> for(i in 1:length(z)){
+  Lemp[i]=log(mean((U<=z[i])&(V<=z[i])))/log(mean(U<=z[i]))
+  Remp[i]=log(mean((U>=1-z[i])&(V>=1-z[i])))/log(mean(U<=z[i]))
+ }
> Lemp=1/Lemp*2-1
> Remp=1/Remp*2-1

Bref, on peut utiliser ces fonctions sur des vrais échantillons. Considérons l’exemple classique loss-alae (où l’on couple les frais dans des sinistres assurés, et les frais payés par l’assureur). On souhaite ajuster une copule, sans trop savoir laquelle. On peut commencer par étudier la dépendance forte, et comparer avec une copule Gaussienne. La copule Gaussienne de référence possède ici le même rho de Spearman que l’échantillon dont on dispose,

> cor(lossalae,method="spearman")
Loss     ALAE
Loss 1.000000 0.451872
ALAE 0.451872 1.000000
> library(copula)
> paramgauss=.47
> paramclayton=.9
> paramgumbel=1.45
> copgauss=normalCopula(paramgauss)
> copclayton=claytonCopula(paramclayton, dim = 2)
> copgumbel=gumbelCopula(paramgumbel, dim = 2)

On obtient ici

La courbe verte est l’intervalle de confiance (ponctuel) à 95% pour une copule Gaussienne et un échantillon de même taille. On voit qu’on modélise mal la structure de dépendance. Avec une copule duale de Clayton, on obtient

et enfin pour une copule de Gumbel,

Bref, la copule de Gumbel semble réellement bien adaptée… Si on creuse en étudiant la dépendance au sens faible, on peut valider là aussi ce modèle. En effet, si la référence est la copule Gaussienne,

ou pour une copule de Clayton,

alors qu’une copule de Gumbel donnerait