# Actuarial Pricing Game, with Reinsurance

The Third Actuarial Pricing Game is still open, the deadline for submission is still February 25th. As mentioned in the instructions, for those willing to play in a market where reinsurance is available, here are the prices offered by some reinsurance company.

As mentioned in the description, the price is per insurance policy, per year. Players should send me their premiums in a csv file, gross of reinsurance, and mention in the email that they want to purchase treaty (A) – for instance (and mention explicitly in the object of the email that they want to play in this specific market, where reinsurance is available).

# Pricing reinsurance contracts, another case study

A reinsurance case study for tomorrow’s class. The goal will be to price some nonproportional reinsurance contract, for business interruption claims. Consider the following dataset,

> library(gdata)
+ "https://perso.univ-rennes1.fr/arthur.charpentier/SIN_1985_2000-PE.xls",
+  sheet=1)
Content type 'application/vnd.ms-excel' length 183808 bytes (179 Kb)
open URL
==================================================
downloaded 179 Kb

As for any (standard) insurance contract, there are two parts in the pricing

• the expected number of claims
• the average cost of individual claims

Here, we do not have covariates (but it might be possible to use some, like the kind of industry, the location, etc).

Let us start with the expected number of claims, per year. Here is the daily frequency,

The data are rather old… but somehow, it is a good thing since after ten years, we can expect that most of the claims have been settled (we’ll discuss claims dynamic starting next week). To plot the graph above, we use

> date=db$DSUR > D=as.Date(as.character(date),format="%Y%m%d") > vD=seq(min(D),max(D),by=1) > sD=table(D) > d1=as.Date(names(sD)) > d2=vD[-which(vD%in%d1)] > vecteur.date=c(d1,d2) > vecteur.cpte=c(as.numeric(sD),rep(0,length(d2))) > base=data.frame(date=vecteur.date,cpte=vecteur.cpte) > plot(vecteur.date,vecteur.cpte,type="h",xlim=as.Date(as.character( + c(19850101,20111231)),format="%Y%m%d")) Then, we can get a prediction of the daily number of business interruption claims, e.g. for any day in 2010 (assume that we had to price a reinsurance contract a few years ago), using a (standard) Poisson regression > regdate=glm(cpte~date,data=base,family=poisson(link="log")) > nd2010=data.frame(date=seq(as.Date(as.character(20100101),format="%Y%m%d"), + as.Date(as.character(20101231),format="%Y%m%d"),by=1)) > pred2010 =predict(regdate,newdata=nd2010,type="response") > sum(pred2010) [1] 159.4757 Observe that using old data has drawbacks, since we got much more uncertainty if we use a regression on time (to include some possible trend) Say we have something like 160 claims over a given year, on average. > plot(D,db$COUTSIN,type="h")

Let us now focus on the cost of those claims. We have 2,400 claims in our dataset, to fit a model (or at least estimate how much a reinsurance contract might cost us). Assume that we would like to purchase a reinsurance contract for our very large claims. Like the two largest per year. Over 16 years, the decutible should be close to the cost of the 32nd largest claim, which was close to 15 million.

> quantile(db$COUTSIN,1-32/2400)/1e6 98.66667% 15.34579 > abline(h=quantile(db$COUTSIN,1-32/2400),col="blue")

So consider some reinsurance contract with a deductible of 15 million. Unfortunately, we cannot find unlimited covers. So let us assume that a reinsurance company agrees for such a deductible, but with a limited cover of 35 million. The average cost (for the reinsurance company) is $\mathbb{E}(g(X))$ where

$g(x)=\min\{35,\max\{x-15,0\}\}$

A first idea is to look at the first cost, i.e. the empirical average of that indemnity, on our portfolio. The indemnity function is

> indemn=function(x) pmin((x-15)*(x>15),50-15)

we can check on a few losses that it is actually what we wish to compute

> indemn(5)
[1] 0
> indemn(20)
[1] 5
> indemn(50)
[1] 35

Now, if the compute the average repayment by the reinsurance company, over 16 years, we get

> mean(indemn(db$COUTSIN/1e6)) [1] 0.1624292 So, per claim, the reinsurance company will pay, on average 162,430. With 160 claims per year, the pure premium should be close to 26 million > mean(indemn(db$COUTSIN/1e6))*160
[1] 25.98867

(again, for a 35 million cover, for some claims that should occur, on average, twice a year). As we will see, a standard model in reinsurance is the Pareto distribution (or to be more specific, a Generalized Pareto one),

$F_{(\xi,\mu,\sigma)}(x) = \begin{cases}1 - \left(\ds{1+ \frac{\xi(x-\mu)}{\sigma}\right)^{-1/\xi}} & \text{for }\xi \neq 0, \\1 - \exp \left(-\frac{x-\mu}{\sigma}\right) & \text{for }\xi = 0.\end{cases}$

There are three parameters here

• the threshold $\mu$ (that we will consider as fixed, but we will see its impact on reinsurance pricing)
• the scale parameter $\sigma$ (called $\beta$ in R)
• the tail index $\xi$

The strategy is to consider a threshold below our deductible, e.g. 12 million. Then, given that the loss exceed 12 million, we can fit a Generalized Pareto distribution,

> gpd.PL <- gpd(db$COUTSIN,12e6)$par.ests
> gpd.PL
xi         beta
7.004147e-01 4.400115e+06

and compute

$\int_{d}^{d+c}(x-d)\ dF_{(\xi,\mu,\sigma)}(x)$

>  E <- function(yinf,ysup,xi,beta,threshold){
+    as.numeric(integrate(function(x) (x-yinf)*dgpd(x,xi,mu=threshold,beta),
+    lower=yinf,upper=ysup)$value+ + (1-pgpd(ysup,xi,mu=threshold,beta))*(ysup-yinf)) + } Here, given that a claim exceeds 12 million, the average repayment is close to 6 million > E(15e6,50e6,gpd.PL[1],gpd.PL[2],12e6) [1] 6058125 Now, we have to take into account the probability to reach 12 million, which is here > mean(db$COUTSIN>12e6)
[1] 0.02639296

So, if we summarize, we have on average 160 claims per year,

> p
[1] 159.4757

Only 2.6% will exceed 12 million

> mean(db$COUTSIN>12e6) [1] 0.02639296 So, the yearly frequency of claism larger than 12 million is 4.2 claims > p*mean(db$COUTSIN>12e6)
[1] 4.209036

And for a claim that exceed 12 million, the average repayment is

> E(15e6,50e6,gpd.PL[1],gpd.PL[2],12e6)
[1] 6058125

So, the pure premium should be close to

> p*mean(db$COUTSIN>12e6)*E(15e6,50e6,gpd.PL[1],gpd.PL[2],12e6) [1] 25498867 which (hopefully) is close to the empirical value we got. Actually, it is also possible to look at the impact of the threshold parameter, since it is clearly and intermediate value that could be changed. I mean, why 12 and not 10? Consider > esp=function(threshold=12e6,p=sum(pred2010)){ + (gpd.PL <- gpd(db$COUTSIN,threshold)$par.ests) + return(p*mean(db$COUTSIN>threshold)*E(15e6,50e6,gpd.PL[1],gpd.PL[2],threshold))
+  }

We can plot the pure premium as a function of that threshold,

> seuils=seq(1e6,15e6,by=1e6)
> plot(seuils,Vectorize(esp)(seuils),type="b",col="red")

which is between 24 and 26 for large thresholds. Again, that is only the first step, and we can price a higher reinsurance layer, like a reinsurance contract with a deductible of 50 million (we have our previous reinsurance contract for claims below that threshold), and a cover of 50 million, for instance. For those high layers, it become interesting to have a parametric model, which should be more robust than the empirical average.

# Pricing Reinsurance Contracts

In order to illustrate the next section of the non-life insurance course, consider the following example1, inspired from http://sciencepolicy.colorado.edu/…. This is the so-called “Normalized Hurricane Damages in the United States” dataset, for the period 1900-2005, from Pielke et al. (2008). The dataset is available in xls format, so we have to spend some time to import it,

> library(gdata)
+ sheet=1)

Content type 'application/vnd.ms-excel' length 119296 bytes (116 Kb)
opened URL
==================================================

perl: warning: Setting locale failed.
LANGUAGE = "fr_CA:fr",
LC_ALL = (unset),
LANG = "fr_CA.UTF-8"
are supported and installed on your system.
perl: warning: Falling back to the standard locale ("C").

The problem with excel spreadsheets is that some columns might have pre-specified format (here, losses are with a format 000,000,000 for instance)

> tail(db)
Year Hurricane.Description State Category Base.Economic.Damage
202 2005                 Cindy    LA        1          320,000,000
203 2005                Dennis    FL        3        2,230,000,000
204 2005               Katrina LA,MS        3       81,000,000,000
205 2005               Ophelia    NC        1        1,600,000,000
206 2005                  Rita    TX        3       10,000,000,000
207 2005                 Wilma    FL        3       20,600,000,000
Normalized.PL05 Normalized.CL05  X X.1
202     320,000,000     320,000,000 NA  NA
203   2,230,000,000   2,230,000,000 NA  NA
204  81,000,000,000  81,000,000,000 NA  NA
205   1,600,000,000   1,600,000,000 NA  NA
206  10,000,000,000  10,000,000,000 NA  NA
207  20,600,000,000  20,600,000,000 NA  NA

To get data in a format we can play with, consider the following function,

> stupidcomma = function(x){
+ x=as.character(x)
+ for(i in 1:10){x=sub(",","",as.character(x))}
+ return(as.numeric(x))}

and let’s convert those values into numbers,

> base=db[,1:4]
> base$Base.Economic.Damage=Vectorize(stupidcomma)(db$Base.Economic.Damage)
> base$Normalized.PL05=Vectorize(stupidcomma)(db$Normalized.PL05)
> base$Normalized.CL05=Vectorize(stupidcomma)(db$Normalized.CL05)

Here is the dataset we will use, from now on,

> tail(base)
Year Hurricane.Description State Category Base.Economic.Damage
202 2005                 Cindy    LA        1             3.20e+08
203 2005                Dennis    FL        3             2.23e+09
204 2005               Katrina LA,MS        3             8.10e+10
205 2005               Ophelia    NC        1             1.60e+09
206 2005                  Rita    TX        3             1.00e+10
207 2005                 Wilma    FL        3             2.06e+10
Normalized.PL05 Normalized.CL05
202        3.20e+08        3.20e+08
203        2.23e+09        2.23e+09
204        8.10e+10        8.10e+10
205        1.60e+09        1.60e+09
206        1.00e+10        1.00e+10
207        2.06e+10        2.06e+10

We can visualize the normalized costs of hurricanes, from 1900 till 2005, with the 207 hurricanes (here the x-axis is not time, it is simply the index of the loss)

> plot(base$Normalized.PL05/1e9,type="h",ylim=c(0,155)) As usual, there are two components when computing the pure premium of an insurance contract. The number of claims (or here hurricanes) and the individual losses of each claim. We’ve seen – above – individual losses, let us focus now on the annual frequency. > TB <- table(base$Year)
> years <- as.numeric(names(TB))
> counts <- as.numeric(TB)
> years0=(1900:2005)[which(!(1900:2005)%in%years)]
> db <- data.frame(years=c(years,years0),
+ counts=c(counts,rep(0,length(years0))))
> db[88:93,]
years counts
88  2003      3
89  2004      6
90  2005      6
91  1902      0
92  1905      0
93  1907      0

On average, we experience about 2 (major) hurricanes per year,

> mean(db$counts) [1] 1.95283 In predictive modeling (here, we wish to price a reinsurance contract for, say, 2014), we need probably to take into account some possible trend in the hurricane occurrence frequency. We can consider either a linear trend, > reg0 <- glm(counts~years,data=db,family=poisson(link="identity"), + start=lm(counts~years,data=db)$coefficients)

or an exponential one,

> reg1 <- glm(counts~years,data=db,family=poisson(link="log"))

We can plot those three predictions, and get a prediction for the number of (major) hurricanes in 2014,

> plot(years,counts,type='h',ylim=c(0,6),xlim=c(1900,2020))
> cpred1=predict(reg1,newdata=data.frame(years=1890:2030),type="response")
> lines(1890:2030,cpred1,col="blue")
> cpred0=predict(reg0,newdata=data.frame(years=1890:2030),type="response")
> lines(1890:2030,cpred0,col="red")
> abline(h=mean(db$counts),col="black") > (predictions=cbind(constant=mean(db$counts),linear=
+ cpred0[126],exponential=cpred1[126]))
constant   linear exponential
126  1.95283 3.573999    4.379822
> points(rep((1890:2030)[126],3),prediction,col=c("black","red","blue"),pch=19)

Observe that changing the model will change the pure premium: with a flat prediction, we expect less than 2 (major) hurricanes, but with the exponential trend, we expect more than 4…

This is for the expected frequency. Now, we should find a suitable model to compute the pure premium of a reinsurance treaty, with a (high) deductible, and a limited (but large) cover. As we will seen in class next week, the appropriate model is a Pareto distribution (see Hagstrœm (1925), Huyghues-Beaufond (1991) or a survey – in French – published a few years ago).

$F_{(\xi,\mu,\sigma)}(x) = \begin{cases}1 - \left(\ds{1+ \frac{\xi(x-\mu)}{\sigma}\right)^{-1/\xi}} & \text{for }\xi \neq 0, \\1 - \exp \left(-\frac{x-\mu}{\sigma}\right) & \text{for }\xi = 0.\end{cases}$

We can use Hill’s plot to estimate the tail index,

> library(evir)
> hill(base$Normalized.PL05) Clearly, costs of major hurricanes are heavy tailed. Now, consider an insurance company, in the U.S., with 5% market share (just to illustrate). We will consider there \tilde Y_i= Y_i/20. The losses are given below. Consider a reinsurance treaty, with a deductible of 2 (billion) and a limited cover of 4 (billion), For our Pareto model, consider only losses above 500 millions, > threshold=.5 > (gpd.PL <- gpd(base$Normalized.PL05/1e9/20,threshold)$par.ests) xi beta 0.4424669 0.6705315 Keep in mind the 1 hurricane out of 8 reaches that level > mean(base$Normalized.CL05/1e9/20>.5)
[1] 0.1256039

Given that the loss exceeds 500 millions, we can now compute the expected value of the reinsurance contact,

$\int_{d}^{d+c}(x-d)\ dF_{(\xi,\mu,\sigma)}(x)$

To compute it we can use

> E <- function(yinf,ysup,xi,beta){
+   as.numeric(integrate(function(x) (x-yinf)*dgpd(x,xi,mu=threshold,beta),
+   lower=yinf,upper=ysup)$value+ + (1-pgpd(ysup,xi,mu=threshold,beta))*(ysup-yinf)) + } [Nov 5th] there is a typo in the previous function, since the threshold should be used, here, as a parameter in the function, if you want to play with that function an see the impact of the threshold (see a more recent post on the same topic, but a different dataset)… but here, we do not change the threshold, so it is not a big deal. Now, it is probably time to bring all the pieces together. We might expect a bit less than 2 (major) hurricanes per year, > predictions[1] [1] 1.95283 and each hurricane has 12.5% chances to cost more than 500 million for our insurance company, > mean(base$Normalized.PL05/1e9/20>.5)
[1] 0.1256039

and given that an hurricane exceeds 500 million loss, then the expected repayment by the reinsurance company is (in millions)

> E(2,6,gpd.PL[1],gpd.PL[2])*1e3
[1] 330.9865

So the pure premium of the reinsurance contract is simply

> predictions[1]*mean(base$Normalized.PL05/1e9/20>.5)* + E(2,6,gpd.PL[1],gpd.PL[2])*1e3 [1] 81.18538 for a cover of 4 billion, in excess of 2. 1.This example will be found in the Reinsurance and Extremal Events chapter in the forthcoming Computational Actuarial Science with R, by Eric Gilleland and Mathieu Ribatet. # Maximum likelihood estimates for multivariate distributions Consider our loss-ALAE dataset, and – as in Frees & Valdez (1998) – let us fit a parametric model, in order to price a reinsurance treaty. The dataset is the following, > library(evd) > data(lossalae) > Z=lossalae > X=Z[,1];Y=Z[,2] The first step can be to estimate marginal distributions, independently. Here, we consider lognormal distributions for both components, > Fempx=function(x) mean(X<=x) > Fx=Vectorize(Fempx) > u=exp(seq(2,15,by=.05)) > plot(u,Fx(u),log="x",type="l", + xlab="loss (log scale)") > Lx=function(px) -sum(log(Vectorize(dlnorm)( + X,px[1],px[2]))) > opx=optim(c(1,5),fn=Lx) > opx$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…

# Some stylized facts about large risk covers

A couple of weeks ago, David Cummins (here) was giving a talk in Laval University. And we’ve seen a series of extremely interesting graphs and figures about catastrophe reinsurance market, as well as Cat Bonds prices. The first one was the rate one line index for catastrophe reinsurance (the rate on line is the excess of loss premium expressed as a percentage of the reinsurance cover), from Guy Carpenter (2010, page 10 here).

Following hurricane Andrew in 1992, prices went up quite high. But following hurricane Katrina (which is, so far, the most costly insured disaster following the second World War, with a cost exceeding 70 billions US$– 2008$ – while Andrew was only 24 billions – again 2008 \$), the bump is much smaller. I though cycles where much larger in the reinsurance industry.

Then there was a discussion about cat bond pricing, with a graph from Lane Financial (2010, page 13, here) with the ratio premium over expected loss