AGGREGATE CLAIMS, SOLVENCY AND REINSURANCE. David Dickson, Centre for Actuarial Studies, University of Melbourne. Cherry Bud Workshop

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1 AGGREGATE CLAIMS, SOLVENCY AND REINSURANCE David Dickson, Centre for Actuarial Studies, University of Melbourne Cherry Bud Workshop Keio University, 27 March 2006

2 Basic General Insurance Risk Model where S= N X i i=1 Srepresentstheaggregateamountofclaimsinafixedperiod,e.g. oneyear N isacountingvariablerepresentingthenumberofclaims X i =amountofthei-thclaim {X i } i=1 isasequenceofi.i.d. randomvariables

3 Standard Problem- find the distribution function of S Reasons Premium setting Setting appropriate levels of reinsurance Solvency,e.g. foragivenpremiump,findthecapitalusuchthat Pr(u+P >S)=0.99

4 Howtofitourmodelforaggregateclaims? Model for the number of claims, e.g. Poisson, negative binomial, or zeromodified versions Model for claim amounts, e.g. lognormal, Pareto Parameter estimation by maximum likelihood, standard goodness of fit tests Excellent reference: Loss Models- from Data to Decisions by Klugman, Panjer &Willmot

5 Computational Issues Distribution function of S is Pr(S x)= n=0 Pr(N =n)f n (x) wheref(x)=pr(x i x),andf n isthen-foldconvolution Exact computation is difficult Approximations are often used- especially moment based approximationsbut estimation of counting and claim amount distributions is important

6 Insurer s surplus process before reinsurance u = initial surplus, U(t)=u+ct N(t) i=1 X i c=rateofpremiumincomeperunittime(year), N(t)=numberofclaimsin[0,t],N(t) Poisson(λt), X i =amountofthei-thclaim.

7 Time Surplus

8 Let θ denote a general reinsurance arrangement. Then the insurer s surplus process after reinsurance is where U θ (t)=u+c θ t S θ (t) U θ (t)=insurer ssurplusattimet, c θ =rateofpremiumincometotheinsurernetofreinsurance, S θ (t)=netaggregateclaimsfortheinsurerin[0,t], S θ (t)hascdfg θ (x;t)andpdfg θ (x;t)forx>0.

9 Time Surplus

10 Probability of ruin- definitions Discrete time: ψ θ (u,t)=pr[u θ (s)<0forsomes,s=1,2,...,t] Continuous time: ψ θ (u,t)=pr[u θ (s)<0forsomes,s (0,t]]

11 Time Surplus

12 The problem: Determine: where: ˆψ(u,t)= inf θ Θ ψ θ(u,t) Θ is a set of admissible reinsurance arrangements. In our study, admissible means: either proportional or excess of loss reinsurance throughout the period, the insurer s annual net income exceeds net expected claims. θcanbechangedatthestartofeachyear. Ruincanbeindiscreteorcontinuoustime.

13 Proportional reinsurance Let X i denote the amount of the i-th individual claim. Under a proportional reinsurance arrangement with proportion retained a, theinsurerpaysax i thereinsurerpays(1 a)x i.

14 Excess of Loss Reinsurance LetX i denotetheamountofthei-thindividualclaim. Underanexcessofloss reinsurance arrangement with retention level M, theinsurerpaysmin(x i,m) thereinsurerpaysmax(0,x i M).

15 Discrete time formulae ˆψ(u,1) = inf θ Θ (1 G θ(u+c θ ;1)) ψ θ (u,n) = ψ θ (u,1) + u+cθ 0 g θ (u+c θ x;1)ˆψ(x,n 1)dx +e λˆψ(u+cθ,n 1) ˆψ(u,n) = inf θ Θ ψ θ(u,n) (Bellman Principle)

16 Calculating ruin probabilities Translatedgammadistribution: Foreachθcalculateα θ,β θ,κ θ suchthat: E[(Y θ +κ θ ) i ]=E[(S θ (1)) i ] i=1,2,3 wherey θ Γ(α θ,β θ )withcdf Γ θ (x;α θ,β θ )andpdf γ θ (x;α θ,β θ ). Thenforx 0 In particular, G θ (x;1) Pr[Y θ +κ θ x] = Γ θ (x κ θ ;α θ,β θ ). G θ (0;1)=e λ Pr[Y θ +κ θ 0]=Γ θ ( κ θ ;α θ,β θ ). Thepdfapproximationisg θ (x;1) γ θ (x κ θ ;α θ,β θ ).

17 Translatedgammadistributionapproximationtoψ θ (u,n) 1. Original(compound Poisson) process: ψ θ (u,1) + u+cθ 0 g θ (u+c θ x;1)ˆψ(x,n 1)dx +e λˆψ(u+c θ,n 1) 2. Translated gamma distribution approximation: 1 Γ θ (u+c θ κ θ ;α θ,β θ ) + u+cθ 0 γ θ (u+c θ κ θ x;α θ,β θ )ˆψ(x,n 1)dx +Γ θ ( κ θ ;α θ,β θ )ˆψ(u+c θ,n 1)

18 Computational Issues Need to use numerical integration. Can reduce the amount of calculation required by introducing a truncation procedure- essentially we set very small ruin probabilities to zero. A grid search is required for the optimal retention level - we limit our set of possible retention levels, e.g. 0.01,0.02,...,0.99,1 for proportional reinsurance. Computations for discrete time ruin are considerably faster than for continuous time- factor may depend on programming language. It does not appear possible to find explicit solutions!

19 Questions to be answered To what extent is a dynamic reinsurance policy better than a static policy? To what extent does the time horizon affect the optimal reinsurance strategy? Is minimising the probability of ruin(in finite time) a sensible optimisation criterion?

20 Numerical example Individual claim amount distribution is Exponential(1) Premium loading factors: 0.1(insurer)& 0.2(reinsurer) Poissonparameter: λ=100 Initialsurplusis u=23ingraphicalillustration

21 Surplus, Remaining term, t u M ˆψ(u, t) M ˆψ(u, t) M ˆψ(u, t) M ˆψ(u, t) Optimal strategy: Exponential claims, loadings 10%/20%, excess loss, discrete time ruin, distributional approximation.

22 Ruin probability No reinsurance Static ppn Dynamic ppn Static XL Dynamic XL Time

23 Questions to be answered To what extent is a dynamic reinsurance policy better than a static policy? Significantly. To what extent does the time horizon affect the optimal reinsurance strategy? Significantly. Is minimising the probability of ruin(in finite time) a sensible optimisation criterion? Answered later!

24 Continuous time formulae Letδ θ (u,t)=1 ψ θ (u,t)bethesurvivalprobabilitytotimet(years). Prahbu s formula is with δ θ (u,1)=g θ (u+c θ,1) c θ 1 δ θ (0,t)= 1 c θ t 0 δ θ(0,1 s)g θ (u+c θ s,s)ds cθ t 0 G θ (y,t)dy.

25 Thenforn>1wehave ˆψ(u,n)=inf ψ θ(u,n), θ where,foragivenvalueofθ, ψ θ (u,n) = ψ θ (u,1)+e λˆψ(u+cθ,n 1)+ and with + u+cθ c θ cθ g θ (u+c θ y,1)ˆψ(y,n 1)dy δ θ (u,1,y)=g θ (u+c θ y,1) c θ 1 y/cθ 0 δ θ (0,t,y)= y c θ t g θ(c θ t y,t). 0 δ θ (u,1,y)ˆψ(y,n 1)dy g θ (u+c θ s,s)δ θ (0,1 s,y)ds

26 Translated gamma distribution approximation Thiscanbeappliedwithease Formulae are relatively straightforward. Numerical example Initialsurplusisu=49 Individual claim amount distribution is Pareto(4,3)(mean is 1) Premium loading factors: 0.1(insurer)& 0.2 or 0.3(reinsurer) Poissonparameter: λ=100

27 Ruin probability No reinsurance Dynamic ppn Static ppn Dynamic XL (10%/20%) Static XL (10%/20%) Dynamic XL (10%/30%) Static XL (10%/30%) Time

28 Questions to be answered To what extent is a dynamic reinsurance policy better than a static policy? Clearly better for proportional reinsurance, but only at the longer time horizons. Marginally better for excess loss reinsurance for both reinsurance loadings. To what extent does the time horizon affect the optimal reinsurance strategy? Significantly: for example, with excess of loss reinsurance and a 30% reinsuranceloadingfactorwegetthefollowingvaluesofm forthefirstyear: Term Dynamic Static

29 Questions to be answered Is minimising the probability of ruin(in finite time) a sensible optimisation criterion? Yes,butitcanleadtostrategiesthatmightnotappearsensible. Forexample,intheprevioustable,withu=49andaremainingtermof1year, whyshouldtheinsurercedesomuchrisk? Alternative criteria, e.g. minimise the ruin probability over a fixed but rolling time horizon? Cannot reduce the ruin probability over the original time horizon.

30 Concluding Remarks A dynamic strategy appears to have a greater effect if we consider discrete time ruin. If we consider discrete time only, the translated gamma approximation still applies for other models, e.g. in the case of a compound negative binomial distribution for claim numbers. It is computationally much faster to compute the optimal static strategy, especially in continuous time. Ruin probabilities under a static strategy bound those under a dynamic strategy.

31 Our study contained a second approach to calculation using translated gamma processes. Results were virtually identical - confidence in robustness of approaches. Acknowledgement: co-author is Prof Howard Waters, Heriot-Watt University, Edinburgh Paper is at

α α λ α = = λ λ α ψ = = α α α λ λ ψ α = + β = > θ θ β > β β θ θ θ β θ β γ θ β = γ θ > β > γ θ β γ = θ β = θ β = θ β = β θ = β β θ = = = β β θ = + α α α α α = = λ λ λ λ λ λ λ = λ λ α α α α λ ψ + α =

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