Point Process Techniques in Non-Life Insurance Models

Transcription

1 1 Point Process Techniques in Non-Life Insurance Models Thomas Mikosch University of Copenhagen 1 Conference in Honor of Jan Grandell, Stockholm, June 13,

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3 What is a point process? 3 Consider the arrivals of claims in a portfolio: 0 < T 1 < T 2 <. Count the number of arrivals in a set A N T (A) = #{i 1 : T i A}, finite if A is bounded. This defines a point process N T = (N T (A)) A E on the state space E = (0, ). It is a random counting measure.

4 4 The principle of marking Marking with the claim sizes. If the T i s constitute a point process, so do the pairs (T i, X i ) where X i > 0 are the claim sizes corresponding to the claim arriving at T i : for sets A, B E, N T,X (A B) = #{i 1 : T i A, X i B} The state space, where the points (T i, X i ) live, is E = (0, ) (0, ). The process N T,X is a marked point process: X i is the mark of T i.

5 Marking with the claim sizes and reporting delays. 5 Claims are not reported at the times when they incur. They are reported with some positive delay in reporting, D i > 0. The points (T i, X i, D i ) constitute a marked point process N T,X,D (A B C) = #{i 1 : T i A, X i B, D i C}. The state space is E = (0, ) (0, ) (0, ).

6 6 Marking with the claim sizes, reporting delays and settlement times. Claims are not settled when they are reported. It usually takes some settlement time S i > 0 before the amount of the claim size X i is paid to the insured. The points (T i, X i, D i, S i ) constitute a marked point process N T,X,D,S (A B C D) = #{i 1 : T i A, X i B, D i C, S i D}. The state space is E = (0, ) (0, ) (0, ) (0, ).

7 One could also mark with a (random) function describing the 7 process of settlement. Then the state space becomes more interesting. The notion of point process is too general to be useful. One needs assumptions on the distribution of the point process. One such assumption is that the process be Poisson.

8 8 Who invented the Poisson process?

9 9 Who invented the Poisson process? Filip Lundberg (1903)

10 10 Who invented the Poisson process? Filip Lundberg (1903) according to A.N. Shiryaev Figure 1. Chairman Albert

11 What is a Poisson process (Poisson random measure)? 11 It is a point process on a state space E. (Random measure) There exists a Radon 2 mean measure µ on (E, E) such that 1. N(A) is Poisson(µ(A)) distributed 2. N(A 1 ),..., N(A m ) are independent for A 1,..., A m disjoint. We say that N is PRM(µ) on E. Example. The claim arrivals 0 < T 1 < T 2 < constitute a homogeneous Poisson process with intensity λ > 0 if EN(a, b] = λ (a, b] = λ (b a). 2Finite on bounded sets.

12 12 Marked Poisson processes are Poisson processes If T i constitute PRM(µ) on E and (Y i ) is an iid mark sequence with common distribution F Y, independent of (T i ), the marked process of the points (T i, Y i ) is PRM(µ F Y ). Example. If T i are the Poisson claim arrivals in a portfolio, independent of the iid vectors Y i = (X i, D i, S i ) of claim sizes X i, delays in reporting D i and settlement times S i, then (T i, Y i ) constitute PRM(µ F Y ).

13 Transformations of the points of a PRM yield a PRM 13 If T i are the points of a PRM(µ) on E then f(t i ) are the points of a PRM(µ) on f(e) with mean measure µ f (A) = µ({x E : f(x) A}). One needs the additional restrictions that for bounded sets B f(e), f 1 (B) is bounded, or that µ is finite. Example. Consider the marked PRM(µ F X,D,S ) on (0, ) 4 of the points (T i, X i, D i, S i ). The function f(t, x, d, s) = (t, t + d, t + d + s, x) is measurable and defines a PRM on (0, ) 4 with points f(t i, X i, D i, S i ) = (T i, T i + D i, T i + D i + S i, X i )

14 14 The IBNR process as a model for teletraffic Consider a homogeneous Poisson arrival process on R T 2 < T 1 < 0 < T 1 < T 2 <. with intensity λ and an iid delay in reporting sequence (D i ), independent of (T i ). The number M T = N IBNR at time T = #{i Z 0 : T i T < T i + D i } = N T,D ({(t, d) : t T < t + d}). counts the (non-observable at time T) incurred but not reported (IBNR) claims.

15 For fixed T, M T is Poisson distributed, but the IBNR counting process (M T ) T 0 is not a Poisson process. (M T ) T 0 is a strictly stationary process with covariance function γ(h) = cov(n IBNR at time T, N IBNR at time T + h ) = λ h P(D > x) dx, h 0. Example. Assume D i is Pareto distributed: P(D > x) = x α, x 1, α > 1. Then γ(h) = λ (α 1) 1 h α+1, h 1. 15

16 16 This function is not integrable for α (1, 2): 1 γ(h) dh =. Then the process M is said to have long memory with Hurst coefficient H = (3 α)/2 (0.5, 1). The process M is a model for the activity in large computer networks (Internet), called the infinite source Poisson model. T i is the arrival of an activity (packet) to the network and D i describes the amount of work brought into the system. The main object of interest is the workload W T = T 0 M s ds.

17 It describes the global performance of the network. The 17 behavior of the workload in the long memory case differs very much from the short memory case (lighter tail of D). Heavy tail components cause long memory in a stochastic system.

18 18 PhD Course Long memory, heavy tails and fractal behavior Gennady Samorodnitsky (Cornell) Copenhagen, Oct

19 Poisson integrals 19 Let N = i ε π i be a PRM(µ) on E. Then the integral E f dn = i ( f(x) i ) ε πi (dx) = i f(π i ) is defined as Lebesgue-Stieltjes integral (under conditions on f). The integrals f i dn, i = 1, 2,... are independent if the f i s have disjoint support. (Inheritance from PRM) If f i 0, the integrals f i dn, i = 1, 2,..., are independent if they are uncorrelated. Poisson integrals on a set of finite µ-measure are compound Poisson.

20 20 Examples. The total claim amount x dn T,X (t, x) = A B X i. i:t i A,X i B The annual total claim amounts above and below a threshold u are independent: X i I {Xi >u} and X i I {Xi u}. i:t i [0,1] i:t i [0,1]

21 Examples. The total claim amount x dn T,X (t, x) = A B X i. i:t i A,X i B The annual total claim amounts above and below a threshold u are 21 independent: i:t i [0,1] X i I {Xi >u} and i:t i [0,1] If N is PRM(Leb µ) on [0, ) R d \{0} with R d \{0} min(1, x 2 ) µ(dx) <, then ( Y t = lim x N(dt, dx) δ 0 + [0,t] {x:δ< x 1} [0,t] {x: x >1} x N(dt, dx), t 0. X i I {Xi u}. [0,t] {x:δ< x 1} is the Lévy-Itô representation of a Lévy process (Y t ). ) x dt µ(dx)

22 22 Settled and outstanding RBNS claims. The ith claim is settled in [T i + D i, T i + D i + S i ] according to a deterministic non-decreasing continuous payment function f with f(0) = 0 and f(1) = 1. The settled reported but not settled (RBNS) claim amount at time T is given by {(t,x,d,s):t+d<t t+d+s} = d = x f ( (T t d)/s ) ( ) X i f (T Ti D i )/S i i::t i +D i <T T i +D i +S i M X i f ( ) Π i. i=1 for a Poisson distributed M, iid (X i ), iid (Π i ), all independent.

23 The distributions of the settled and outstanding RBNS claim 23 amounts are compound Poisson and can be simulated or calculated numerically. In this way, one can also approximate the distribution of reserves (predict). The chain ladder. Assume N i,i+j, j = 0, 1,..., m i are the counts of payments for claims that occurred in the ith year and for which payments were executed in the years i, i + 1,..., m, where m is the present year.

24 24 A simplistic idea of predicting N i,m+1 is to assume that the best predictor (in the mean square sense) has form (Mack) E(N i,m+1 N ii,..., N im ) = f m i N im for some constants f j 1 which are estimated by the chain ladder estimators f j. The quantity N i,m+1 = f m i N im is then not the best predictor of N i,m+1 and determining the prediction error is a problem. Such a model is hard to simulate because of the lack of assumptions. Although counts of points underly the model it is difficult (impossible?) to find a point process model for N i,i+j that satisfies the above conditional expectation equality.

25 Can one predict (in the mean square sense) in a point process model? Assume that claims arrive at the times 0 < T 1 < T 2 < of a homogeneous Poisson process on (0, ) with intensity λ. The ith claim triggers a stream of payments X ik at the times T i + Y ik, k = 1, 2,.... For every i, (Y ik ) k 1 constitutes a homogeneous Poisson process on (0, ) with intensity γ, independent of (X ik ) k 1. The payments X ik are iid. The points ((Y ik ) k 1, (X ik ) k 1 ) constitute an iid sequence, independent of (T i ). 25

26 26 This means that the points (T i, (Y ik ) k 1, (X ik ) k 1 ) constitute a marked PRM N on a suitable state space E. The Poisson integrals M(0, b] = I {t [0,1], t+yk [0,b]} N(dt, d(y k ), d(x k )) = S(0, b] = = E k=1 N T [0,1] i=1 E k=1 N T [0,1] i=1 k=1 I {Ti +Y ik [0,b]} x k I {t [0,1], t+yk [0,b]} N(dt, d(y k ), d(x k )) X ik I {Ti +Y ik [0,b]} k=1 describe the number of payments/paid amounts for claims arriving in the first year [0, 1] and payments being executed in [0, b] for some b 1.

27 For b 1, x > 0, we can calculate the mean square predictors 27 of M(b, b + x] and S(b, b + x] given M(0, b]: E(M(b, b + x] M(0, b] = m) = λ γ x φ(m) 1 (γ) φ (m) 2 (γ) E(S(b, b + x] M(0, b] = m) = EX 11 λ γ x φ(m) 1 (γ) φ (m) 2 (γ), where φ (m) 1, φ (m) 2 are the mth derivatives of the Laplace-Stieltjes transforms of the random variables L+1 i=1 (b U i), L i=1 (b U i), respectively, and L is a Poisson(λ) distributed random variable independent of the iid uniform U(0, 1) sequence (U i ). This model does not satisfy the condition (Mack) of the chain ladder model.

28 28 Til lykke, Jan