ON RUIN PROBABILITY AND AGGREGATE CLAIM REPRESENTATIONS FOR PARETO CLAIM SIZE DISTRIBUTIONS

Transcription

1 ON RUIN PROBABILITY AND AGGREGATE CLAIM REPRESENTATIONS FOR PARETO CLAIM SIZE DISTRIBUTIONS Hansjörg Albrecher a Dominik Kortschak a,b a Institute of Actuarial Science, University of Lausanne, Quartier UNIL-Dorigny, Batiment Extranef, CH-5 Lausanne, Switzerland b Radon Institute for Computational and Applied Mathematics, Austrian Academy of Sciences, Altenbergerstrasse 69, A-44 Linz, Austria Abstract We generalize an integral representation for the ruin probability in a Crámer-Lundberg risk model with shifted or also called US-Pareto claim sizes, obtained by Ramsay [4], to classical Paretoa claim size distributions with arbitrary real values a > and derive its asymptotic expansion. Furthermore an integral representation for the tail of compound sums of Pareto-distributed claims is obtained and numerical illustrations of its performance in comparison to other aggregate claim approximations are provided. Introduction Consider the classical compound Poisson model of collective risk theory, where the surplus process Rt at time t is given by Nt Rt = u + ct X i, with Nt denoting a homogeneous Poisson process with intensity λ, the claim sizes X i are i.i.d. distributed non-negative random variables with distribution function F and finite mean µ = EX i, c is a constant premium intensity and the net profit condition c > λµ holds. The ruin probability for a given initial surplus level u is denoted by i= ψu = Pr[Rt < for some t > R = u] and its properties are a classical object of study in risk theory see for instance Asmussen [3]. In many situations it turns out that heavy-tailed distributions provide an appropriate fit to actual claim data and the focus of this paper will be on this case. Asymptotic properties of ruin probabilities ψu for large u in the case of subexponential claim size distribution functions F have been studied quite extensively in the literature see for instance Teugels & Veraverbeke [8], Embrechts & Veraverbeke [9] and for higher-order asymptotic expansions Grübel [], Omey & Willekens [3], Willekens & Teugels [], Supported by the Austrian Science Foundation Project P839-MAT

2 Baltrūnas [5] and Barbe et al. [6]. On the other hand, apart from asymptotic results, also explicit expressions for ψu for heavy-tailed claim sizes are of interest, not the least for checking the accuracy of asymptotic estimates when applied to approximate the ruin probability ψu for moderate values of u. Among the heavy-tailed claim size distributions, the Pareto distribution is very popular in actuarial practice. For claims with so-called shifted or US-Pareto distribution function Fx = + m x m, x >, m N, Ramsay [4] derived an interesting integral representation of ψu, which only involves a single non-oscillating integral along the positive real line. In [5] this integral representation was generalized to arbritrary non-integer m >, in [9] to a renewal risk model with Erlang interclaim times and in [6] to the case of compound shifted Pareto sums. In all these cases the used method is akin to the one used in [7] and [8] to determine the density function of the finite sum of certain Pareto variables. It is natural to ask whether one can establish a similar formula for classical Pareto claims with density function fx = addx a+, x >, a >, d, d which is often preferred to the US-Pareto distribution by practitioners for modelling purposes. In this paper, we will show that for Pareto distributions a slightly weaker result holds. In Section a refined analysis of singularities in the complex domain is used to establish an integral representation of ψu for Pareto claim sizes with distribution function and arbitrary positive parameter a >. Two methods of proof are provided. Section. is an extension of the proof technique of Ramsay [4] to our situation. Section. gives a somewhat different proof which later on will allow us to extend the integral representation to other compound sums of Pareto random variables. In Section 3 we show how this new expression can be used to quickly obtain higher-order asymptotics of ψu for large u. While the ruin probability, by the Pollaczek-Khintchine formula cf. [3, Chapter 3., p.6], can be interpreted as a geometric compound of the integrated tail distribution, we indicate in Section 4 how to use the proof method of Section. to derive an integral representation for the tail of a compound sum of Pareto random variables. We especially highlight the case of compound Poisson sums and compound negative binomial sums of Pareto random variables i.e. aggregate claims in the compound Poisson model or the compound negative binomial model, respectively. In Section 5 some numerical illustrations of the performance of the integral representations in comparison to other aggregate claim approximations are provided. An integral representation for the ruin probability with Pareto claims Since the parameter d in is just a scale parameter, we will without loss of generality assume d = in the sequel.

3 Extending the approach of Ramsay [4] to the case of general Pareto claim sizes with density, we can derive an integral representation not for ψs itself, but for a function that has the same asymptotic expansion as ψs. In Remark.4 we will indicate why strict equality between the integral representation and ψs can not hold in the case of general Pareto claims. Define the Laplace transform of a function g by ˆL g s = e st gtdt. From the usual integro-differential equation for the ruin probability in the Cramér-Lundberg model, it is not difficult to see that ˆL ψ s = s c λµ cs λ ˆL f s 3 see e.g. Rolski et al. [7, Equation 5.3.4, p.65]. Further note that for the Pareto density fx = ax a ˆLf s = as a Γ a,s, 4 where Γa,s = s e x x a dx is the incomplete Gamma function see e.g. [, Chapter 6.5, Page 6]. For the ruin probability ψu in the compound Poisson model we then get Theorem.. Consider a compound Poisson model with Pareto claim size density fx = ax a. If ρ := λµ/c = λa/a c and F I x = /µ x Ftdt, then we have ψu ρ ρ F c λµ Iu Γa c + λu/xre x a e x + πλx/u a dx. 5 ˆL f x u where the notation ax bx means that there exists a δ > with ax bx = Oe δx as x. Remark.. Recall that lim u ψu/ ρ ρ F Iu = see e.g. [3, Theorem.,p.59]. Remark.. The integrand in 5 is positive non-oscillating and of exponential decay. As in [4] the integral in 5 can be interpreted as the expected value of some function with respect to a Gamma distribution. Remark.3. Note that all common asymptotic expansions of ψu see e.g. [6] have an error term that asymptotically is of the order x k for a k >, which is asymptotically worse than the exponential convergence rate of the error of the approximation 5. In that sense approximation 5 is more accurate than any asymptotic expansion for a numerical illustration, cf. Section 5. Γa 3

4 Remark.4. From Theorem. we get for some integrable function gx ψu e ux gxdx. 6 A heuristic argument why we do not have equality in 6 is as follows: Due to the support of F starting in x =, FI n x is not differentiable at x = n, hence by the Pollaczek- Khintchine formula ψu is not holomorphic in the points u = n. On the other hand the integral in 6 is a Laplace transform and hence holomorphic for Reu > so that we can not have equality, whereas in the US-Pareto case in [4], the support of F was the entire postive half-line and consequently FI n x was holomorphic in the entire right half plane and equality in 6 was feasible.. Proof of Theorem., Method Proof of Theorem.. A standard inversion formula of 3 yields ψu = πι s +ι s ι e us ˆLψ sds = πι s +ι s ι e us c λµ ds 7 sc λ ˆL f s for an s >, where ι =. From the definition of ˆL f s it follows that the integrand e us c λµ sc λ ˆL f s in 7 as a function of s C is meromorphic in the sliced plane D = C\,]. Hence, for the evaluation of 7 we can use a complex contour in D, taking care of its enclosed poles, which are located at the zeroes of Ns := c λ s ˆL f s. Note that there will not be any pole inside the contour with positive real part because ˆL ψ s is bounded on that area except possibly for a branch point at s =. Since Ns = Ns where s denotes the complex conjugate of s, we will concentrate our analysis on Ims. We will use the contour given in Figure and we have to make sure that no zero of Ns lies on this contour. We will now state a series of lemmata, the proofs of which are postponed to the Appendix. The first two lemmata bound Ns from below. Lemma. establishes that for every π/ < φ < π there exists an r such that /Ns is bounded on the set {s = re ιφ : r > r and φ < φ π}: Lemma.. For all ǫ > and π/ < φ < π there exists an r such that for all r > r and φ φ π N rcosφ + ιsinφ > ǫ. The next lemma gives us a family of horizontal lines on which /Ns can be uniformly bounded and hence can be part of our contour. Lemma.3. For all ǫ >, there exists a k > such that for all even integers k > k and all s := α + ιβ α,β R with α > and β = k + π/ Ns > c ǫ 8 4

5 Figure : The contour integral To evaluate the integral in 7 we now use the contour given in Figure. At first we choose a sequence φ n with 3π/ < φ n < π/ and lim n φ n = π/. Then we choose an even integer k n such that for s k n + π/ and for φ n args π, Ns > δ this can be done because of Lemma.. Further we assume that k n > k as defined in Lemma.3 where we choose ǫ < c δ and define R n = k n + π/sinφ n. The contour consists of the line s +ιx for x [ k n +π/,k n +π/], the lines x±ιk n +π/ for x [R n cosφ n,s ], the semi-circle C Rn = {z : z = R n,φ n argz π φ n }, the line from [ R n, r] lying above the branch cut of the negative real axis with r < R n, the circle C r = {z : z = r,z r} and the line from [ r, R n ] lying below the branch cut of the negative real axis see Figure. From the residue theorem one obtains πι = πι s +ιr s ιr C Rn + M n Res si i= e us c λµ cs λ ˆL f s ds R n cosφ n+ι kn+π s +ι kn+π + e us c λµ cs λ ˆL f s s ι kn+π R n cosφ n ι kn+π, + r R + + C r R r c λµ eus cs λ ˆL f s ds where s i, i =,...,M n are the zeros of Ns that lie inside the contour. By using Lemma. and the asymptotic behavior of the Gamma function, it is straightforward to see that Furthermore, lim n c λµ πι C Rn s eus c λ s ˆL ds =. f s lim = e us r πι C r s c λµ c λ s ˆL ds = f s is a consequence of lim s ˆL f s/s = µ, cf. 5 below. From Lemma.3 we get: Rn cosφ n+ι kn+π e us c λµ s Ns ds c λµ δ k n + π s +ι kn+π s = c λµ δ k n + π R n cosφ n e us ds e us e urn cosφn, which tends to as n. For x R define ˆL f x = lim y + ˆL f x + ιy. With ˆL f x = ˆL f x, we get as in [4] that the sum of the two remaining integrals can be expressed as π R r e ux λc λµimˆl f x cx + λ ˆL f x dx. 5

6 Taking the limit R and r yields ψu = e ux λc λµimˆl f x π cx + λ ˆL f x dx Res si e us c λµ i= cs λ ˆL. 9 f s We have to show that Res si e us c λµ cs λ ˆL = Oe δu. f s i= For a single s i it is quite obvious that for every ǫ > Res si e us c λµ cs λ ˆL = Oe Resi+ǫu. f s Hence we have to ensure that by summing over all s i this property remains to hold. For this we are only interested in s i with s i large. To determine the location of the zeros of Ns and hence the s i needed to evaluate the residuals of the integrand, the following lemma gives an asymptotic relation between Res and Ims of solutions of Ns =. Lemma.4. For every < ǫ < minc,, there exists a β > such that for each zero s := α + ιβ of Ns with β > β, its negative real part α > satisfies c ǫβ + ǫc + ǫβ log α log. aλ + ǫ aλ ǫ The next lemma provides expressions for Ims for Ns =. Together with Lemma.4 we can find expressions for the zeros of Ns. Lemma.5. For every < ǫ < minc, there exists a β > such that for each zero s := α + ιβ of Ns with β > β, sin β < ǫ and cos β < ǫ. After we approximately know where the zeros of Ns are located we have to make sure that there are not too many zeros close to each other in order to guarantee. Further we will show that there exist zeros of Ns. Lemma.6. There exists a constant c such that for every k > the number of zeros of Ns inside the circle K s k around s k = logckπ /aλ + ιkπ with radius is less than c. Moreover, there exists a k such that for every k > k the number of zeros of Ns inside the circle K s k is. Next one can bound the order of the zeros of Nx which is needed to bound the residuals. Lemma.7. s D with Ns = N s = N s = and there exists an M > such that for s > M, Ns = N s = has no solution. The next lemma provides an exponential bound for a single residual at a pole s i of the integrand 8 with s i large. 6

7 Lemma.8. For every ǫ > there exists a β > such that for s i = α+ιβ with β > β and Ns i = N α + ιβ = and every k > Res s i e us c λµ s c λ s ˆL f s e u kα c λµ aλ c ǫ aλ+ǫ aλ+ǫ c ǫ k α + β β α +β / ρ β k e u kα β k+. Hence is proven and the integral representation follows by algebraic manipulations of 9.. Proof of Theorem., Method : Before we identify an alternative proof of Theorem., we will need an auxiliary result. Lemma.9. Let gs be any meromorphic function on D {s : Res > t } for a t >, with a branch cut on the line t,] for < x < t define g x := lim y + g x+ιy. If further gs = gs, gs has no pole for Res, lim s gs =, and if there exists a c and an ǫ > such that, uniformly for α in bounded intervals, c gα + ιβ = O β ǫ, as β, then there exists a δ > with πι s +ι s ι e us s gsds = π t e ux x Img xdx + Oe δu. If further g x can be defined for x,t ] and there exists a δ > with e δ x x Im g x dx <, then there exists a δ > with πι s +ι s ι e us s gsds = π e ux x Img xdx + Oe δu. 3 Proof. At first choose an < r < t and < R < t such that for a c <, for all α [ r,s ] and all β > R c gα + ιβ c β ǫ. For R >, r > we will use the contour given in Figure consisting of the line s + ιx, x [ R,R ], the lines x ± ιr for x [ r,s ], the lines r ± x for x [R,R ], the complex paths γ ± t for t [,] connecting the points { r ± ιr, R} with t < Reγ ± t r, where γ ± t does not cross any poles of gs and the length Lγ ± is 7

8 Figure : The contour integral bounded, the line from [ R, r ] lying above the branch cut of the negative real axis, the circle C r = {s : s = r,s r } and the line from [ r, R] lying below the branch cut of the negative real axis. Note that r±ιr e us s gsds s + re s u + c R s ±ιr which tends to as R. Further, from lim s e us gs = it follows that lim e us gsds =. r C r s From gs = gs we have r e ux R πι R x gx + ιdx + gx ιdx e ux r x = r e ux πι R x gx e ux x gxdx = π By the Cauchy Residual Theorem it follows that πι s +ι s ι = πι e us s gsds π γ + + R r+ιr + γ r+ι + e ux x Img xdx r ι r ιr R R ǫ e uss M R,r gsds + r e ux x Img xdx. i= Res si e us s gs where s i, i =,...,M R,r are the poles of gs inside the contour. Since gs has no poles with Res, there exists a δ > with δ > Res i for all i =,...,M R,r and hence Res si e us s gs = Oe δu. We have e us γ ± s gs Lγ ± max t [,] euγ ±t γ ± t gγ ±t e ru Lγ ± max t [,] γ ± t gγ ±t. Further r+ι e us s gsds = c Note that r+ι r+ιr r+ιr r+ι r+ιr e us r+ι s ds + e us s gs c ds. r+ιr e us s gs c ds e ru r + ιx g r + ιx c ds R e ru R c x +ǫdx = c ǫr ǫe ru. 8

9 Since there exists a c such that for all u > u e ιux r + ιx dx c, we have R r+ι e us s ds c e ur. r+ιr The assertion finally follows by u > δ : e ux x Img xdx e Ru δ R where g x is interpreted as where it is not defined. R e δx Img x dx, x Proof of Theorem.. From 7 we see that we have to apply Lemma.9 with gs = c λµ c λ s ˆL f s. Clearly gs = gs. For Res we get ˆL f s = e sx at a dx = e s + s e sx x a dx = s e sx Fxdx s Fxdx = s µ, so that gs has no poles with Res, due to the net profit condition c > λµ alternatively, one could also argue except for a possible branch point at s =, ˆL ψ s has no poles with Res. An asymptotic expansion of Γ a,s cf. [, Chapter 4] gives n Γ a,s = s a e s + i= i j= a j s i + ǫ n s, 4 where s n ǫ n s can be uniformly bounded for large s. It follows that ˆL f s a s e s, from which we get that, uniformly for r Res s r,s >, ˆL f s as s. Hence c λµ gα + ιβ c = O. β From [, Equation 6.5.9, Page 6] we get that for a N ˆL f s = as a Γ a aγ aγ a,s = as a Γ a a On the other hand, for a N define E a x = e xt t a dt n= s n n an!. 5 9

10 leading to cf. [, Equation 5.., Page 9] ˆL f s = ae a+ s = sa logs γ + a! a n= n a where γ = is Euler s constant. In both cases it follows Note that and cf. 5, 6 lim s s ˆL f s = n= n a a = µ, or equivalently lim gs =. a s x Img x = ImˆL f x c + λ/xre ˆL f x + λ x Im ˆLf x λ s n n an!, 6 ImˆL f x = πxa Γa, 7 so that for every δ > holds. The theorem finally follows with Lemma.9 and some algebraic manipulations. 3 Asymptotic expansion of the ruin probability Expansions for ruin probabilities in the Cramér-Lundberg model can be derived from expansions for compound distributions whenever the ladder height distribution of the claim size distribution F I u = µ u Fxdx admits expansions for convolutions, see e.g [6] or []. For instance, if the claim size distribution Fu has a density fu such that its negative derivative f u is regularly varying with index a,a > 3 such that the moments µ k := a/a k = E [ X k] are finite for k =,, 3, then the expansion for the infinite horizon ruin probability ψu of third order reads cf. [] ψu = ρ ρ F Iu + ρ ρ µ µ Fu + ρ µ 3 ρ 3µ + ρ 3 µ ρ 3 4µ 3 fs + ofs. 8 In this section we are going to show that an asymptotic expansion of 5 indeed retains 8. An advantage of this alternative method to derive expansions is that one can also derive the asymptotic expansion for values a < 3. In particular, we will see that for every noninteger a > the expansion ψu = ρ ρ F ρ µ Iu + ρ µ Fu + ρ ρ 3 3µ ρ 3 4µ 3 + ρ µ 3 ρ 3µ π Γa tanaπ ρ ΓaΓa F Iu + O fu u min{a+,a }, 9 where for a < 3 the first term in the second line of 9 will dominate the O-term.

11 Remark 3.. In 9, one observes that for a = n + / n N, the constant in front of F I u in the expansion vanishes, which illustrates the special role played by these values of the shape parameter a. From now on we will assume that a N. At first we want to look for an expansion of the denominator of the integral 5. Define c = π Γa tanaπ and a n = a n + an +!, to get With we obtain Re ˆLf x x c = π Γa = c x a + and b n = a n x n. n= n a k a n k k= Define λ c + λ/x ReˆLf x + ˆLf x Im x = c ReˆLf x ˆLf + cλ + λ Re x x x = c + λ c + c x a + λc c x a + a n x n + λ c x a a n x n + λ b n x n. n= n= n= c :=λ c + c, { λ c a + λcc n =, d n := λ c a n n, { c + λ b + λca n =, e n := λ b n + λca n n. Note that if a = k + / for k N, then c =. We get + λ ˆLF x Im x λ c + λ/x ReˆLf x + ˆLf x Im x = c x a + x a d n x n + e n x n. n= n=

12 For ease of notation let N u x := c + uλ x Re ˆLf x + u λ ˆLf u x Im x, u and Z u x,n,n,n := I {n =}c x u a x a + u x n + u x n. u d n e n n=n n=n Note that N u x = Z u x,,,. To find an asymptotic expansion for 5 we first give an expansion for integrals of the form where n,m and n,n,n,k. Lemma 3.. We have Z u x,n,n,n x ma +n e x dx, N u x Z u x,n,n,n x ma +n e x dx N u x = I {n =}c u a x m+a +n e x dx N u x + d i u a i x m+a +n+i e x dx i=n N u x + e i u i x ma +n+i e x dx. i=n N u x Proof. There exists an M > such that M /N u x for all x. We get for β > and i x β +i e x dx M x β +i e x dx = MΓβ + i. N u x It then suffices to show that i=n d i u i Γβ + i and are absolutely convergent series. We have for i > i=n e i u i Γβ + i, d i Γβ + i = λ c aγβ + i i + aγi +.

13 Since for α >, m N, m + + b > α > m + b, b N, and i > m + Γα + i Γb + i = Γα i j= α + j Γb i b + j we obtain that j= Γα = Γb m j= b + j i m j= i=n d i u i Γβ + i α + j b + m + + j m α j + i, converges absolutely for all u >. For the second sum, note that for n >, e n = λ b n + λa n, hence we have to show that i=n b i Γβ + iu i is absolutely convergent. But for u > the latter follows from n a b n = k + ak +!n k + an k +! k= a n min z N z a k!n k! k= a n = min z a. z N n! From x ma +n e x N u x dx = = j= x ma +n e x c λµ dx Zu x,,, c λµ x ma +n e x c λµ dx N u x Γ ma + n c λµ Z u x,,,x ma +n e x c λµ dx N u x 3

14 we get Z u x,n,n,n x ma +n e x dx N u x = c Γ m + a + n I {n =} c λµ u a + d n Γ m + a + n + n c λµ u a n + e n Γ ma + n + n c λµ u n I {n =}c u a Z u x,,,x m+a +n e x c λµ dx N u x d n u a n e n u n + Z u x,,,x m+a +n+n e x c λµ dx N u x Z u x,,,x ma +n+n e x c λµ dx N u x Z u x,n +,n +,n + x ma +n e x dx. N u x The terms with integrals depending on u are of higher order than at least one of the other terms and these integrals are of the same type as the starting integral; hence one can iteratively get the complete asymptotic expansion of the first integral. Since ψu ρ ρ F c λµ x α e x Iu Γa N u x dx, one can in this way obtain the complete asymptotic expansion of ψs in terms of Fs α i, α i R. As a concrete example, we derive the asymptotic expansion for ψu up to the order of 4

15 fu = au a for a >. We get ψu ρ ρ F c λµ Iu Γa = ρ ρ F Iu ρ ρ e u For a > 3 we get with so that ψu = From a = = ρ ρ F ρ Iu e ρ Γa c λµ F I u Γa u Z u x,,,x a e x c λµ dx + N u x + ρ e F I u ρ uγa c λµ a F I u c λµ u Z u x,,,x a e x c λµ dx N u x e Γa c λµ Z u x,,,x a e x dx N u x e Γa + u c λµ e Z u x,,,x a e x u c λµ N u x dx + ρ F I u e Γa + ρ Γa u c λµ e Z u x,,,x a e x u c λµ N u x dx + Z u x,,,x a e x dx N u x Z u x,3,,x a e x dx. N u x F I x = a x a+, f I x = a a x a and f I x = ax a, ρ ρ F Iu ρ e ρ c λµ f Iu ρ e + ρ c λµ 4 ρ ρ a a = µ, a = e c λµ a a = µ, a = f Iu + Ou min{a+,a }. a 63 a = µ 3 6 and and b = a a = µµ, b = a a + a = µµ µ 4, e = λ b + λca = λµ c λµ and e = λ b + λca = λ µ 4 it follows that ρ e ρ c λµ f Iu = ρ λµ ρ c λµ µ µ f Iu = 5 ρ ρ µ µ f Iu = λµ 3 c λµ, ρ ρ µ µ Fu,

16 and ρ e ρ c λµ 4 ρ e ρ c λµ f Iu = ρ3 µ ρ 3 µ + ρ λ µ µ ρ c λµ 4µ ρ λµ µ 3 ρ c λµ f 3µ Iu ρ 3 3µ = ρ 3 4µ 3 + ρ µ 3 ρ 3µ fu. which is in accordance with 8. For a < 3, clearly a < a +, so that we have to add further terms. We get Z u x,3,,x a e x N u x dx = d Γ a c λµ u a d u a Z u x,,,x a e x c λµ dx N u x Z u x,3,,x a e x + dx N u x = d Γ a c λµ u a + O u min{a,a,3}. By defining µ 3 = a/a 3, even if the third moment does not exist, we obtain for every noninteger a > the expansion 9. For a <, one has to insert even further terms to get an asymptotic expansion with the desired accuracy. 4 Integral representations for compound sums In this section we are concerned with the collective risk model, i.e. we assume that S N = X X N, where the X i are iid Pareto and N is an integer-valued random variable independent of the X i. We are interested in Gu = PS N > u. If we denote with Q N z = E [ z N], then ˆL G s = s s Q N ˆLf s, and for an s > we get Gu = πι s +ι s ι e us s Q N ˆLf s ds. Theorem 4.. Let X,X,... be iid Pareto random variables with density fx = ax a and N be an integer-valued random variable independent of X i such that there exists an ǫ > with E [ + ǫ N] = p n + ǫ n <, n= 6

17 where p n = PN = n. Then there exists a t > with PS N > u = Gu t π x e ux Im Q N ˆLf x dx. If further Q N ˆLf x exists for all x > and for a δ > e δ x Q x Im N ˆLf x <, then PS N > u = Gu π Proof. We would like to apply Lemma.9 with gs = Q N ˆLf s. x e ux Im Q N ˆLf x dx. At first we have to show that gs is holomorphic on D {s : Res > t }. Since n= p n + ǫ n <, it is enough to show that there exists a t such that ˆL f s < + ǫ/ on {s : Res > t }. Since ˆL f s for Res, lim s ˆL f s = on {s : Res > t } for every t > and ˆL f s is continuous on {s : Res > t } for every t >, it follows by continuity arguments that there exists a t > such that {s : Res > t } for every t >. Clearly we have gs = gs. From lim s ˆLf s = and E [ ] + ǫ N] < it follows that lim s gs = lim s E [ˆL f s N =. Denote by m the smallest positive integer with p m, then uniformly for bounded sets of α > t : p gα + ιβ ˆL f α + ιβ m p m+n ˆL f α + ιβ n = O n= β m. 4. The compound Poisson model Assume that N Poissonλ, i.e. PN = n = e λ λn n!. Note that ˆL G s = s s e λ ˆL f s = s s e λ asa Γ a,s, is holomorphic on D. From Theorem 4. we get Gu λx Γ a, x a π π x e ux λre a xa sin dx. Γa Remark 4.. The integral in is oscillating, but it is at least absolutely integrable, which is not the case for the integral in. Further note that Re a x a Γ a, x a x ex, so that the integrand decays very quickly for large x, which makes a numerical evaluation of the integral quite feasible. 7

18 4. The compound negative binomial model For the negative binomial distribution given by r + n PN = n = ρ r ρ n, < ρ <, r > n and Theorem 4. gives PS N > u = Gu π ρ r Q N z =, ρz 5 Numerical examples x e ux Im ρ ρa x a Γ a, x r dx. We now provide numerical illustrations for the performance of the derived approximations to ψu and Gu, respectively. We will compare the approximations with the first order asymptotic approximations given by ρ/ ρf I u and E [N] Fu, respectively. We also include a second order asymptotic approximation containing the next significant asymptotic term. For a compound sum with E [X] < and regularly varying density [ N fx, this extra term is E ] E [X] fu. When fx is regularly varying with index [ N ] α = a + =, then the next term is E fu u Fxdx. In principle, there is also a next asymptotic term when fx is regularly varying with index α = a + <, but in the case a = / which is the case in our examples it turns out that this term is and hence we omit the second order approximation for the compound sums in this case see e.g. [] and [] for further details on these higher order expansions. In the case of the ruin probability, we use the Pollaczek-Khintchine formula to rewrite the ruin probability as a compound sum, for the case a =.5 we use the next term in the expansion given in Section 3. Furthermore we use a Monte Carlo estimate with iterations. We use the estimator NFmaxM N,u S N, where M N = maxx,...,x N of [4]. This estimator is proven to be asymptotically efficient for compound sums. To provide upper and lower bounds for the actual values of ψu and Gu, respectively, we use Panjer recursions with stepsize. for the ruin probabilities and. for the compound sums cf. [, Chapter 6.6]. To compare the efficiency of the approximations, the needed computation times of the Monte Carlo estimate, Panjer bounds and the integral approximations are provided Panjer recursions were implemented in C++, whereas all other calculations where done using Mathematica 6.. We will see that the integral approximations provide excellent results whereas the common asymptotic approximations can be far off the correct values. 5. Approximation of the ruin probability For the risk process we choose the parameters a =.5, λ =, c = 3.5 or a =.5, λ =, c =.5 or a =.5, λ =, c =, which corresponds to ρ =.857, ρ =.8 or ρ =.833, respectively. For each of these examples, the Panjer recursion for the evaluation of upper 8

19 and lower bounds needs 39 seconds. A Monte Carlo simulation of sufficient accuracy takes between 8 and 4 seconds and the numerical evaluation of the integral 5 takes approximately.4 seconds. The estimates are given in Table, or 3 respectively. u upper bound lower bound integral first order second order MC Table : Numerical estimates of ruin probabilities with different methods with a =.5, λ =, c = 3.5 and ρ =.857 u upper bound lower bound integral first order second order MC Table : Numerical estimates of ruin probabilities with different methods with a =, λ =, c =.5 and ρ =.8 u upper bound lower bound integral first order second order MC Table 3: Numerical estimates of ruin probabilities with different methods with a =.5, λ =, c =, and ρ = Compound Poisson Let us now consider an example for the calculation of the tail of a compound Poisson sum of Pareto.5, Pareto and Pareto.5 claims with λ =. In this case the Panjer recursion takes 4 seconds to evaluate upper and lower bounds. The Monte Carlo simulation 9

20 takes 4 seconds and the evaluation of each integral takes between. and.6 seconds. The estimates are given in Tables 4, 5 and 6: u upper bound lower bound integral first order MC Table 4: Numerical estimates of tail probabilities for a compound Poisson sum with a =.5 and λ = with different methods u upper bound lower bound integral first order second order MC Table 5: Numerical estimates of tail probabilities for a compound Poisson sum with a = and λ = with different methods u upper bound lower bound integral first order second order MC Table 6: Numerical estimates of tail probabilities for a compound Poisson sum with a =.5, λ = and E [S N ] = 6 with different methods 5.3 Compound negative binomial Finally, we consider the determination of the tail of a compound negative binomial sum of Pareto.5, Pareto and Pareto.5 claims with ρ =.5 and r = 3. The Panjer recursion takes 4 seconds to evaluate upper and lower bounds. A Monte Carlo simulation takes 7 seconds and the evaluation of the integral takes between. and.4 seconds. The estimates are given in Tables 7, 8 and 9.

21 u upper bound lower bound integral first order MC Table 7: Numerical estimates of tail probabilities for a compound negative binomial sum with a =.5, ρ =.5 and r = 3 with different methods u upper bound lower bound integral first order second order MC Table 8: Numerical estimates of tail probabilities for a compound negative binomial sum with a =, ρ =.5 and r = 3 with different methods References [] M. Abramowitz and I. A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55 of National Bureau of Standards Applied Mathematics Series. Washington, D.C., 964. [] H. Albrecher, C. Hipp, and D. Kortschak. Higher order expansions for compound distributions and ruin probabilities with subexponential claims. Scand. Actuar. J., 9, To appaer. [3] S. Asmussen. Ruin probabilities, volume of Advanced Series on Statistical Science & Applied Probability. World Scientific Publishing Co. Inc., River Edge, NJ,. [4] S. Asmussen and D. P. Kroese. Improved algorithms for rare event simulation with heavy tails. Adv. in Appl. Probab., 38: , 6. [5] A. Baltrūnas. Second order behaviour of ruin probabilities. Scand. Actuar. J., : 33, 999. [6] P. Barbe and W. P. McCormick. Asymptotic expansions for infinite weighted convolutions of heavy tail distributions and applications. American Mathematical Society, 9. [7] M. Blum. On the sums of independently distributed Pareto variates. SIAM J. Appl. Math., 9:9 98, 97.

22 u upper bound lower bound integral first order second order MC Table 9: Numerical estimates of tail probabilities for a compound negative binomial sum with a =.5, ρ =.5, r = 3 and E [S N ] = 9 with different methods [8] L. Brennan, I. Reed, and W. Sollfrey. A comparison of average-likelihood and maximum-likelihood ratio tests for detecting radar targets of unknown Doppler frequency. IEEE Trans. Information Theory, 4:4, 968. [9] P. Embrechts and N. Veraverbeke. Estimates for the probability of ruin with special emphasis on the possibility of large claims. Insurance Math. Econom., :55 7, 98. [] R. Grübel. On subordinated distributions and generalized renewal measures. Ann. Probab., 5:394 45, 987. [] S. A. Klugman, H. H. Panjer, and G. E. Willmot. Loss models. Wiley Series in Probability and Statistics: Applied Probability and Statistics. John Wiley & Sons Inc., New York, 998. [] F. W. J. Olver. Asymptotics and special functions. AKP Classics. A K Peters Ltd., Wellesley, MA, 997. [3] E. Omey and E. Willekens. Second-order behaviour of distributions subordinate to a distribution with finite mean. Comm. Statist. Stochastic Models, 33:3 34, 987. [4] C. M. Ramsay. A solution to the ruin problem for Pareto distributions. Insurance Math. Econom., 33:9 6, 3. [5] C. M. Ramsay. Exact waiting time and queue size distributions for equilibrium M/G/ queues with Pareto service. Queueing Syst., 574:47 55, 7. [6] C. M. Ramsay. The distribution of compound sums of pareto distributed losses. Scandinavian Actuarial Journal, page To Appear, 9. [7] T. Rolski, H. Schmidli, V. Schmidt, and J. Teugels. Stochastic processes for insurance and finance. Wiley Series in Probability and Statistics. John Wiley & Sons Ltd., Chichester, 999. [8] J. Teugels and N. Veraverbeke. Cramér-type estimates for the probability of ruin. Technical report, C.O.R.E. Discussion Paper No 736, 973.

23 [9] L. Wei and H.-l. Yang. Explicit expressions for the ruin probabilities in Erlang risk processes with Pareto individual claim distributions. Acta Math. Appl. Sin. Engl. Ser., 3:495 56, 4. [] E. Willekens. Asymptotic approximations of compound distributions and some applications. Bull. Soc. Math. Belg. Sér. B, 4:55 6, 989. [] E. Willekens and J. L. Teugels. Asymptotic expansions for waiting time probabilities in an M/G/ queue with long-tailed service time. Queueing Systems Theory Appl., 4:95 3, 99. A Proofs of the auxiliary lemmata Proof of Lemma.. The lemma is a direct consequence of the asymptotic behaviour of the incomplete Gamma function 4. Proof of Lemma.3. Clearly, cos β = and sin β =, hence we have s e s = αβ ια β α + β e α, and Res e s >. Using the asymptotic behaviour of the incomplete Gamma function 4 we get Ns = c + aλs e s aλs e s Γ a,s s a e s λ s. Now choose k > such that for k > k s = α + ιk + π/ Γ a,s s a e s < ǫ λ and c s < ǫ. This gives Ns c + aλs e s aλs e s Γ a,s s a e s λ s c + aλs e s ǫ c Proof of Lemma.4. Choose β > such that for β > β both λ/ s ǫ, Γ a,s s a e s ǫ ǫ c ǫ. hold and further for a δ > and for all β > β and all α > ǫβ it is valid that N α + ιβ > δ compare Lemma.. Since Ns = we get that α ǫβ and c = aλs e s aλs e s Γ a,s s a e s λ s, 3

24 from which we can deduce c + ǫ aλ ǫ s e s c ǫ aλ + ǫ. Since we get and s e s = e α α + β, c ǫ α log aλ + ǫ α + β log c + ǫ α log aλ ǫ α + β log c ǫβ aλ + ǫ + ǫc + ǫβ aλ ǫ. Proof of Lemma.5. At first choose an ǫ > such that for all x with cos x < ǫ it follows that sin x < ǫ, note that this is possible since sin x and the inverse of cos x are continuous. In view of, choose β > such that for all β > β aλs e s Γ a,s s a e s + λ s < cǫ. We have Ns =c + aλs e s aλs e s Γ a,s s a e s λ s, ReNs =c + aλeα α β αβ α + β Re aλs e s Γ a,s s a e s ImNs = aλeα αβ α + β α + β cosβ α β α + β sinβ Im aλs e s Γ a,s s a e s + λ, s α + β cosβ + α + β sinβ + λ, s and ReNs c ccosβ c aλβ e α α + β cosβ aλe α α α + β α + β cosβ + αβ α + β sinβ cǫ. Since from Lemma.4, it follows that lim β c aλβ e α α + β cosβ + lim aλe α α β α + β α + β cosβ + αβ α + β sinβ =, 4

25 there exists a β such that for β > β = ReNx c cosβ cǫ, i.e. cosβ ǫ. Proof of Lemma.6. Let k be large. Note that all zeros of Ns are poles of order one of the function hs = N s/ns and the corresponding residual is. Hence the number of zeros is We have πι K s k hs = aa λsa Γ a,s + ae s λs c λ s asa Γ a,s htdt. as a+ Γ a,s s a e s Γ a,s c aλs a Γ a,s as a Γ a,s + = a s + Note that s a Γ a,s as s and s K s k for a k >. For φ [,π and s φ,k = s k + e ιφ we have: c aλs a φ,k Γ a,s φ,k = cs φ,k aλe s cs φ,k φ,k aλe s s a φ,k e s φ,k, φ,k Γ a,s φ,k and for α k = logckπ /aλ, β k = kπ cs φ,k aλe s φ,k = ecosφ e ιsinφ+β k hence we get uniformly in φ and it follows that lim k cosφ α k sinφ + β k + cosφ α k sinφ + β k ι βk, lim hs φ,k = k e cosφ e ιsinφ, htdt = π πι K s k π Note that πι K s k htdt is an integer. Proof of Lemma.7. Recall Ns = c λ s asa Γ a,s. e ιφ e cosφ dx =. eιsinφ N s = λ s asa Γ a,s aλs a Γ a,s = λ s asa Γ a,s + a λs a Γ a,s aλs e s =aa λs a Γ a,s + ae s λs, 5

26 N s =a aa λs a 3 Γ a,s aa λs 3 e s ae s λs 3 + aλs e s =aa a λs a 3 Γ a,s λs 3 aa 3e s + + aλs e s. For the existence of an s with Ns = N s = N s =, the following three equations would have to be fulfilled at the same time: From 3 and 4 we get From 4 and 5 we get Γ a,s = s a a cs a+ aλ, 3 Γ a,s = ae s aa s a, 4 Γ a,s = aa 3e s + ase s aa a s a. 5 s a a cs a+ aλ = ae s aa s a, cs λ = a a e s a, s ρ e s =. 6 ae s aa s a = aa 3e s + ase s aa a s a, ae s aa 3 = a e s + a ase s a, a a e s a a = a a se s, Substituting this into 6 one obtains e s = s. s ρ s =, s s s =, ρ ss /ρ =, { s, + ρ }, ρ but N + /ρ and N. For the second part of the assertion choose β as in Lemma.4 and s = α + ιβ with Ns = to obtain s ρ e s e s s ρ c ǫ α aλ + ǫ β + β, ρ 6

27 which tends to infinity as β and the result follows from 6. Proof of Lemma.8. Choose β > maxβ,m where β is according to Lemma.4 and M is according to Lemma.7 such that for every β > β. c λµ aλ c ǫ aλ+ǫ aλ+ǫ k α +β c ǫ β β α +β / ρ. Since N s i one has Res /Ns i = N s i = aa λs i a Γ a,s i + ae s i λsi. From 3 it follows that N s i = aa λs i a si a a cs a+ i + ae s i λs i aλ = aλ s i s i ρ e s i aλeα s i /ρ s i aλ c ǫ aλ+ǫ β α +β / ρ α + β, so that the assertion is implied by e us i aλ + ǫ k = e u kα e kα e u kα β k. c ǫ 7