Dombry Clément
email: clement.dombry@univ-fcomte.fr

Rabehasaina Landy
email: landy.rabehasaina@univ-fcomte.fr

High order expansions for renewal functions and applications to ruin theory

Keywords: MSC 2010 subject classifications: Primary 60K05, ; secondary 60K10 renewal process, renewal function, ruin probability, Cramér-Lundberg approximation, compound Poisson model

A high order expansion of the renewal function is provided under the assumption that the inter-renewal time distribution is light tailed with finite moment generating function g on a neighborhood of 0. This expansion relies on complex analysis and is expressed in terms of the residues of the function 1/(1 -g). Under the assumption that g can be extended into a meromorphic function on the complex plane and some technical conditions, we obtain even an exact expansion of the renewal function. An application to risk theory is given where we consider high order expansion of the ruin probability for the standard compound Poisson risk model. This precises the well known Crámer-Lundberg approximation of the ruin probability when the initial reserve is large.

Introduction

Let (X k ) k∈AE be an i.i.d. sequence of non negative random variables with common cumulative distribution F . The arrival times (S n ) n∈AE are defined by S 0 = 0 and, for n ≥ 1, S n = n k=1 X k . We consider the counting process N defined by N (x) := n≥0 1 {Sn≤x} , x ≥ 0, and the associated renewal function

U (x) := E[N (x)] = ∞ n=0 F * n (x), x ≥ 0.
The renewal Theorem states that if the inter-arrival distribution F has a finite first moment µ := E(X 1 ), then

U (x) ∼ x µ as x → +∞.
Recall that the distribution F is called lattice if F is supported by hN = {0, h, 2h, . . .} for some mesh h > 0. It is well known (see for example Asmussen [START_REF] Asmussen | Applied probability and queues[END_REF]Proposition 6.1]) that if F has a finite second moment µ 2 := E(X 2 1 ), then

U (x) = x µ + µ2+µ 2µ 2 + o(1) if F is lattice x µ + µ2 2µ 2 + o(1) otherwise as x → +∞. (1)
When F has infinite first or second moment, Teugels [START_REF] Teugels | Renewal theorems when the first or the second moment is infinite[END_REF] provides asymptotics for the renewal function U under some regular variation conditions. In this paper, we focus on the case when F is light-tailed and we assume that X 1 has some finite exponential moment so that

R := sup r ≥ 0; ∞ 0 e rx dF (x) < ∞ > 0. ( 2
)
Then the moment generating function

g(z) := ∞ 0 e zx dF (x)
is well defined and holomorphic on S R = {z ∈ C, ℜ(z) < R}.

Using complex analysis, Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF] proved that under the strong non lattice condition lim sup

θ→±∞ 1 1 -g(iθ) < ∞, (3)
there exists some r > 0 such that

U (x) = x µ + µ 2 2µ 2 + o(e -rx ) as x → ∞. (4)
Up to now, relatively few results concern either expansions or closed form expressions for U (x). It appears that the only case where closed form expansions are available are when F is Matrix exponential distributed, in which case an expression of U (x) is given in Asmussen and Bladt [START_REF] Asmussen | Renewal theory and queueing algorithms for matrixexponential distributions[END_REF]. Mitov and Omey [START_REF] Mitov | Intuitive approximations for the renewal function[END_REF] provide heuristics on many terms asymptotics of U (x) that are verified on the already known cases. However, as the authors point out, those interesting expansions are only given formally and are not proved. Other expansions are available in [START_REF] Doering | Non) differentiability and asymptotics for potential densities and subordinators[END_REF]Theorem 4] in the context of potential densities of Lévy processes in terms of the associated Lévy jump distribution. The approach by Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF] for obtaining Expansion (4) is mainly based on complex analysis and proved fruitful for obtaining expansions involving survival functions of random sums, see Blanchet and Glynn [START_REF] Blanchet | Uniform renewal theory with applications to expansions of random geometric sums[END_REF]. The approach in [START_REF] Stone | On moment generating functions and renewal theory[END_REF] was later generalized for spread out distributions in [START_REF] Stone | On absolutely continuous components and renewal theory[END_REF]; note that the generalization of our results to spread out distribution is not available in the present paper because the main technical assumption that enable us to obtain higher expansions (namely, Assumption (6) thereafter), which is stronger than the non lattice condition (3), is not verified for spread out distributions. This is not really surprising, as it is on the other hand known that a spreadout distribution is strongly non lattice (see e.g. Proposition 1.6 p.189 of [START_REF] Asmussen | Applied probability and queues[END_REF]), which is why many results in renewal theory that hold for strongly non lattice distributions also hold for spreadout distributions. We consider in this paper both the lattice and the non lattice cases and, in view of Equation [START_REF] Asmussen | Applied probability and queues[END_REF], we introduce the function

v(x) := U (x) -x µ -µ2+µ 2µ 2 if F is lattice U (x) -x µ -µ2 2µ
2
if F is non-lattice .

Following Stone's approach but with more detailed computations, we obtain higher order expansions for the function v both in the lattice and non-lattice cases (Theorems 1 and 3 respectively). When g has a meromorphic extension to the whole complex plane and under some technical conditions, we are even able to provide exact expansions for v (Corollary 4). The paper is structured as follows. Section 2 presents our results on high order expansions for the renewal function as well as some examples. Section 3 is devoted to applications to ruin theory: we provide asymptotics of the ruin probability in the setting of continuous or discrete time risk processes and also consider a two dimensional risk process. Proofs are gathered in Sections 4 and 5.

High order expansions for the renewal function

Main results

In the sequel, the solutions of the equation g(z) = 1 play a major role. Note that 0 is the unique real solution in S R and that other solutions satisfy ℜ(z) ≥ 0 and come in pairs, i.e. if z is a solution then so is z.

We consider first the lattice case and we assume without loss of generality that the mesh of the distribution is equal to h = 1, i.e. F is supported by N. In this case, the moment generating function z → g(z) is 2iπ-periodic on S R = {z ∈ C, ℜ(z) < R} and we introduce the fundamental domain S f R = {z ∈ ; ℜ(z) < R, -π ≤ ℑ(z) ≤ π}.

Theorem 1. Let R 0 ∈ (0, R) be such that Equation g(z) = 1 has no root verifying ℜ(z) = R 0 . Let us denote by z 0 = 0, z 1 , . . . , z N the solutions of this equation in S f R0 . Then, v(k) has the asymptotic expansion

v(k) = N j=1
Res e -kz (e z -1)(1 -g(z))

; z j + o(e -R0k ), k → +∞, k ∈ AE, [START_REF] Biard | Asymptotic multivariate finite-time ruin probabilities with heavy-tailed claim amounts: Impact of dependence and optimal reserve allocation[END_REF] where the notation Res(f (z); z j ) denotes the residue of the meromorphic function f at pole z j .

If z j is a simple zero of g -1, i.e. g ′ (z j ) = 0, then the j-th term in [START_REF] Biard | Asymptotic multivariate finite-time ruin probabilities with heavy-tailed claim amounts: Impact of dependence and optimal reserve allocation[END_REF] has the simple form e -kzj (e zj -1)g ′ (z j ) .

Next we consider the case when F is non-lattice. It is well known that a distribution is non-lattice if and only if θ = 0 is the unique real solution of the equation g(iθ) = 1. We will need here the following stronger technical assumption: for all R 0 < R,

lim sup θ→±∞ sup 0≤r≤R0 1 1 -g(r + iθ) < ∞. (6)
Assumption ( 6) is stronger than the strong lattice condition (3) from Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF]. It is however not too restrictive and satified by a large class of distributions as shown by the following propositon.

Proposition 2. Suppose that the distribution F is absolutely continuous with respect to the Lebesgue measure. Then Assumption [START_REF] Blanchet | Uniform renewal theory with applications to expansions of random geometric sums[END_REF] is satisfied for all R 0 < R.

Our main result in the non-lattice case is the following theorem.

Theorem 3. Consider a strictly non-lattice distribution F satisfying assumption [START_REF] Blanchet | Uniform renewal theory with applications to expansions of random geometric sums[END_REF]. Then for all R 0 ∈ (0, R), the equation g(z) = 1 has a finite number of solutions in S R0 = {z ∈ C; ℜ(z) < R 0 } that we denote by z 0 = 0, z 1 , . . . , z N . Then, supposing that R 0 is such that there is no solution to that equation verifying

ℜ(z) = R 0 , v(x) has the asymptotic expansion v(x) = N j=1
Res e -xz z(1 -g(z))

;

z j + o(e -R0x ) as x → +∞. ( 7
)
If z j is simple zero of g -1, i.e. g ′ (z j ) = 0, then the j-th term in [START_REF] Breiman | Probability[END_REF] has the simple form

- e -xzj z j g ′ (z j )
.

It is worth noting that Theorems 1 and 3 can be extended to obtain asymptotics of higher order, i.e. of order e -rx with r > R, if we assume that the moment generating function g has a meromorphic extension to S R for some R > R. Theorems 1 and 3 and their proofs extend in a straightforward way, but not Proposition 2. In the case when the moment generating function g has a meromorphic extension to the whole complex plane, i.e. R = +∞, it is even possible under some technical assumption to get an exact expansion for the renewal function.

Corollary 4. Assume that g(z) has a meromorphic extension to the whole complex plane.

• In the lattice case, we assume furthermore that

lim inf r→+∞ sup -π≤θ≤π 1 e r (1 -g(r + iθ)) = 0. ( 8
)
We denote by z 0 = 0, z 1 , . . . , z N (possibly N = +∞) the solution of g(z) = 1 in the fundamental domain

S f = {z ∈ ; -π ≤ ℑ(z) ≤ π} with ℜ(z 0 ) ≤ ℜ(z 1 ) ≤ • • • . Then, we have the exact expansion v(k) = N j=1 Res e -kz (e z -1)(1 -g(z)) ; z j , k ≥ 0. ( 9
)
• In the non-lattice case, we suppose that assumption [START_REF] Blanchet | Uniform renewal theory with applications to expansions of random geometric sums[END_REF] holds for all R 0 > 0. Let x > 0. Let us furthermore suppose that -One has an infinite number of roots (z n ) n∈AE of Equation g(z) = 1, all of which are simple, real, and such that the following series converges:

∞ j=1 Res e -xz z(1 -g(z)) ; z j = ∞ j=1 e -xzj z j g ′ (z j ) < +∞. ( 10
)
-The following holds:

lim inf r→+∞ sup θ∈R 1 r(1 -g(r + iθ)) = 0. ( 11
)
Then, we have the exact expansion

v(x) = ∞ j=1 Res e -xz z(1 -g(z)) ; z j . ( 12
)
Remark 5. One of the assumptions of Corollary 4 in the non-lattice case is that there is an infinite number of roots of Equation g(z) = 1. One may wonder what happens in the case when those roots are in finite number N . In fact, one shows in this case, and thanks to Condition [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF], that the moment generating function g(z) is rational, in which case a finite expansion for v(x) can be deduced almost straightforwardly.

In the previous results, a kind of dichotomy arises between the lattice and non-lattice cases. Interestingly, a unified statement can be deduced for the renewal mass function or the renewal density function in the non-lattice and lattice case respectively.

Corollary 6. Under the same assumptions as Corollary 4:

• In the lattice case, the renewal measure has mass function

u(k) = - N j=0

Res

e -kz 1 -g(z)

; z j , k ≥ 0.

As a particular case, if all the poles are simple,

u(k) = N j=0 e -kzj g ′ (z j ) , k ≥ 0.
• In the non-lattice case, the renewal measure has density function

u(x) = - N j=0 Res e -xz 1 -g(z) ; z j , x > 0. ( 13
)
As a particular case, if all the poles are simple,

u(x) = N j=0 e -xzj g ′ (z j ) , x > 0. ( 14
)
To conclude this subsection, we present an informal argument leading to (and motivating) Expansion [START_REF] Breiman | Probability[END_REF] and that may lead to some better comprehension of proof of Theorem 3 given in Section 4.3. One verifies, using Fubini, that the Laplace transform of U (.) is

∞ 0 e -xz U (x)dx = ∞ 0 e -xz ∞ k=0 È(S k ≤ x) dx = ∞ k=0 ∞ S k e -xz dx = ∞ k=0 g(-z) k z = 1 z(1 -g(-z))
, so that, the inversion formula for the Laplace transform leads formally to

U (x) = 1 2iπ c+i∞ c-i∞ e xz z(1 -g(-z)) dz (15)
where c > 0 is such that all singularities of e xz z(1-g(-z)) are on the left of vertical line c + iÊ. The right hand side integral of [START_REF] Kuznetsov | On the Wiener-Hopf factorization for Lévy processes with bounded positive jumps[END_REF] does not necessarily converge, however let us suppose that this is the case. The poles of e xz z(1-g(-z)) satisfying ℜ(z) > -R 0 are -z 0 = 0,..., -z N , and one can verify that the residue of

e xz z(1 -g(-z)) at -z 0 = 0 is x µ + µ 2 2µ 2 .
Thus, using a rectangular contour and the theorem of residues leads to the following

U (x) = N j=0 Res e xz z(1 -g(-z)) ; -z j + 1 2iπ -R0-i∞ -R0+i∞ e xz z(1 -g(-z)) dz = x µ + µ 2 2µ 2 + N j=1 Res e -xz z(1 -g(z)) ; z j + 1 2iπ -R0-i∞ -R0+i∞ e xz z(1 -g(-z)) dz. (16)
The last integral is an o(e -R0x ), so that one would obtain Expansion [START_REF] Breiman | Probability[END_REF]. However, the main failing points in this reasoning are first that the integral in [START_REF] Kuznetsov | On the Wiener-Hopf factorization for Lévy processes with bounded positive jumps[END_REF] is not convergent, and that the contour argument leading to [START_REF] Kyprianou | Introductory lectures on fluctuations of Lévy processes with applications[END_REF] is more delicate than it seems. The convergence issue will be solved by introducing a gaussian kernel (an idea already introduced by Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF][START_REF] Stone | On characteristic functions and renewal theory[END_REF]) which, by inversion, will make the corresponding integral converge, see Step 2 in the proof in forthcoming Section 4.3. The contour argument will involve Assumption (6), which will enable two of the pieces of the contour to vanish in the proof, see again Step 2 in Section 4.3 as well as the corresponding Figure 3.

Examples

We provide some examples that illustrate the results above.

Example 7. In the lattice case, we consider the negative binomial distribution with parameters p ∈ (0, 1) and n ≥ 1 defined by

F (dx) = k≥0 k + n -1 k p k (1 -p) n δ k (dx), p ∈ (0, 1), n ≥ 1.
Its moment generating function is given by For n = 1 the only pole in the fundamental domain S f R is z 0 = 0 so that Theorem 1 implies v(x) = o(e -rx ) for all r > 0. In the general case n ≥ 1, there are exactly n poles in the fundamental domain S f given by z j = r k e iαj with

g(z) = 1 -p 1 -pe z n , ℜ(z) < R = -log p, so that 1 1 -g(z) = (1 -pe z ) n (1 -pe z ) n -(1 -p) n
r j = log 1 -(1 -p)e 2iπ j n p and α j = arg 1 -(1 -p)e 2iπ j n .
Furthermore, Assumption (8) is easily satisfied (comparison between an exponential and a power growth) so that Corollaries 4 and 6 apply. We obtain that the renewal measure has mass function

u(k) = n-1 j=0 1 -pe zj np(1 -p)e zj e -
(z) = α(-zI -T ) -1 s (17)
where s := -T e and e = (1, . . . , 1) T (see [START_REF] Asmussen | Ruin probabilities[END_REF]Theorem 1.5]). The equation g(z) = 1 is a polynomial equation with N + 1 solutions z 0 = 0, . . . , z N and Condition ( 6) is satisfied with R 0 = +∞ and Assumption [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF] is easily satisfied (comparison between an exponential and a power growth). From Corollaries 4 and 6, we obtain a closed formula for the renewal distribution U and the renewal density u. On the other hand, Asmussen and Bladt [3, Theorem 3.1] provide the simple expression

u(x) = αe (sα+T )x s, x > 0. ( 18
)
Let us check that this formula agrees with Corollary 6 in the case when the roots z 0 , . . . , z N are simple. It is easy to check that the -z j 's are exactly the eigenvalues of the matrix sα + T . Denoting by P the (N + 1) × (N + 1) matrix made up with the eigenvectors v 1 , . . . , v j ∈ Ê N +1 corresponding to eigenvalues z 0 , . . . , z N , we have sα + T = P ∆P -1 with ∆ := Diag(-z j , j = 0, ..., N ).

Denoting by J k the Ê (N +1)×(N +1) matrix with 1 at the (k, k)th position and 0 elsewhere, we compute u(x) = αe (sα+T )x s = αP e ∆x P -1 s = N +1 j=1 e -zj-1x αP J j P -1 s.

On the other hand, we deduce from (17)

g ′ (z) = α(-zI -T ) -2 s = α(-zI + sα -P ∆P -1 ) -2 s = αP (-zI + P -1 sαP -∆) -2 P -1 s.
Hence, in order to prove that ( 19) and ( 13) agree, we need to prove

αP J j P -1 s = 1 g ′ (z j-1 )
for all j = 1, . . . , N + 1 or equivalently

αP J j P -1 s × αP (-z j-1 I + P -1 sαP -∆) -2 P -1 s = 1.
This can be easily verified with elementary algebra (the relation J j (∆+z j-1 I) = 0 is useful).

Example 9. Let us consider the simple case of the uniform distribution on [0, 1], i.e. X ∼ U[0, 1]. In that case g(z) = e z -1 z and the equation g(z) = 1 is equivalent to e z = z + 1. The solutions are z j = -W j (-e -1 ) -1, j ∈ , where W j (.) is the jth generalized Lambert function. Proposition 2 together with Theorem 3 provide an asymptotic expansion of v. Using the relations z -j = zj and g ′ (z j ) = 1, we obtain, for all N ≥ 1,

v(x) = -2 N j=1 ℜ 1 z j e -xzj + o(e -rN x ) as x → +∞,
with z j = -W j (-e -1 ) -1 and r N = ℜ(z N ). As N → +∞, r N → +∞ so that the expansion has arbitrary high order.

At this point, there still lacks an example of distribution X such that a meromorphic extension g(z) exists, Equation g(z) = 1 admits an infinite number of solution and infinite expansion [START_REF] Hu | On joint ruin probabilities of a two-dimensional risk model with constant interest rate[END_REF] holds. An example of such an infinite expansion of v(x) will be given in upcoming Example 14, in the context of meromorphic Lévy processes.

Application to ruin theory

As an application of Theorem 3, we provide asymptotic expansions for the ruin probability in risk theory. We consider both a continuous setting (compound Poisson risk process) and a discrete setting (binomial risk process). Estimation of the ruin probability in a two-dimensional model is also investigated.

Ruin theory in continuous time

We consider the following classical continuous time risk process

R x t = x + ct - Nt k=1 Z k = x + Y t , t ≥ 0,
with {N t , t ≥ 0} a Poisson process with intensity α > 0 and (Z k ) k∈AE an i.i.d. sequence of non-negative random variables with common distribution G and finite expectation m, independent from {N t , t ≥ 0}. Such a process models the capital of an insurance company with premium rate c > 0, initial reserve x, and incoming claims (Z k ) k∈AE , see e.g. Asmussen and Albrecher [START_REF] Asmussen | Ruin probabilities[END_REF]. We define Ḡ = 1 -G the tail function and assume the following on the moment generating function

s → ∞ 0 e sx dG(x) is finite for all s > 0. ( 20
)
We are interested in the ruin probability

ψ(x) := P inf t≥0 R x t < 0 , x ≥ 0. ( 21
)
and its asymptotic when the initial reserve x is large. It is well known that ψ(x) < 1 if and only if the safety loading is positive, i.e.

E(Y 1 ) = c -αm > 0. ( 22
)
In the asymptotic analysis, a key role is played by the Lundberg equation

∞ 0 e zy α c Ḡ(y)dy = 1, z ∈ C. (23)
Under Assumption [START_REF] Stone | On moment generating functions and renewal theory[END_REF], this equation restricted to real numbers admits a unique real solution denoted by κ > 0. The Lundberg inequality states that ψ(x) ≤ e -κx for all x > 0, while the Cramér-Lundberg approximation provides the asymptotic behavior as

x → +∞ ψ(x) ∼ Ce -κx with C = c -αm E (Ze κZ ) -c . ( 24
)
We provide high order asymptotic expansions for the ruin probability ψ(x). Similar considerations as well as exact expansions have been proved with different methods by Kuznetsov and Morales [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF] for a so called meromorphic risk process and by Roynette et al. [START_REF] Roynette | Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes[END_REF].

Using the fact that κ > 0 solves the Lundberg Equation ( 23), one can define the probability measure F on [0, +∞) by

F (dx) := e κx α c Ḡ(x)dx.
The moment generating function

g(z) = ∞ 0 e zx F (dx), z ∈ C
is well defined and holomorphic on the complex plane. For future reference, note that the Lundberg equation ( 23) is equivalent to

g(z -κ) = 1, z ∈ C. (25)
Using a renewal equation solved by the ruin probability function ψ and the asymptotic behavior of the renewal function provided by Theorem 3, we can deduce an asymptotic expansion for ψ(x) as x → +∞.

Theorem 10. Assume conditions [START_REF] Roynette | Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes[END_REF] and [START_REF] Stone | On moment generating functions and renewal theory[END_REF] are satisfied. Let r > 0 be fixed and

z 0 = 0, z 1 , . . . , z N the solutions of g(z) = 1 in S r = {z ∈ C; ℜ(z) ≤ r}.
Then, the ruin probability ψ(x) has the asymptotic expansion

ψ(x) = N j=0 Res (αm -c)e -x(z+κ) c(1 -g(z))(z + κ) ; z j + o(e -(r+κ)x ) as x → +∞. ( 26
)
If z j is a simple zero of g -1, i.e. g ′ (z j ) = 0, then the j-th term in (26) has the simple form

- (αm -c) cg ′ (z j )(z j + κ) e -x(zj+κ) = c -αm αE Ze (zj+κ)Z -c e -(zj+κ)x . ( 27
)
The term j = 0 of the asymptotic expansion (39) is exactly the Cramer-Lundberg approximation [START_REF] Stone | On absolutely continuous components and renewal theory[END_REF].

Example 11. Let us consider the case where the claims are of the form X = min(V, d) where d > 0 and V has an exponential distribution with parameter λ > 0. This models a reinsurance scenario where a reinsurance company covers the excess of claim above d only, i.e. according to a stop loss contract with priority d > 0. In that case, Lundberg Equation ( 23) reads

e (z-λ)d = 1 + c α (z -λ), z ∈ , (28)
and the Lundberg exponent is κ = λ. Solutions z j satisfy

z j = - α c - 1 d W j - αd c e -αd c , j ∈ ,
where W j (.) is the jth generalized Lambert function. It is easy to check that

g ′ (z j ) = d + dα/c -1 z j + κ -λ = 0,
so that Theorem 10 entails the asymptotic expansion (26) with jth term given by (27). Since claims here are bounded, the expression of the ruin probability can in fact be made more precise. Indeed

ψ(x) = 1 -E[Y 1 ]W (x)
where W (x) is the so-called scale function associated to Lévy process {Y t , t ≥ 0} (see Expression (8.7) p.215 of [START_REF] Kyprianou | Introductory lectures on fluctuations of Lévy processes with applications[END_REF]). The expression of W (x) is available in Theorem 3 of [START_REF] Kuznetsov | On the Wiener-Hopf factorization for Lévy processes with bounded positive jumps[END_REF] as an infinite series, it yields that (26) can in fact be written as an infinite series, i.e.

ψ(x) = - ∞ j=0 (αm -c) cg ′ (z j )(z j + κ) e -x(zj+κ) .
Remark 12. It is worth comparing Theorem 10 with the results of Kuznetsov and Morales [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF]. They consider a so called meromorphic risk process {R t , t ≥ 0}, which amounts to assume that the claims Z k 's have density

P[Z k ∈ dx] dx = ∞ m=1 b m e -ρmx , x ≥ 0,
for some positive coefficients (b m ) m≥1 and increasing sequence (ρ m ) m≥1 satisfying ρ m → +∞. Corollary 1 of [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF] states that the Laplace exponent Λ(z) := log E e zY1 of the Lévy process {R t , t ≥ 0} admits a meromorphic extension on z ∈ and that all the solutions of the (extended) Lundberg equation ( 25) are real, negative and simple. Furthermore, denoting these solutions by (-ζ n ) n≥1 , the ruin probability ( 21) has expansion

ψ(x) = ∞ n=1 -E[Y 1 ] Λ ′ (-ζ n ) e -ζnx (29)
In this framework, condition [START_REF] Roynette | Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes[END_REF] is not satisfied but we check below that these results are still consistent with Theorem 10. The Laplace exponent satisfies

Λ(z) = cz + α E e -zZ1 -1 and Λ ′ (z) = c -αE Z 1 e -zZ1 .
Elementary computations reveal that

g(z) = α c 1 z + κ E e (z+κ)Z -1 = Λ(-z -κ) c(z + κ) + 1 and g ′ (z) = α c E Ze (z+κ)Z z + κ - α c E e (z+κ)Z -1 (z + κ) 2 . ( 30
)
One can check that the solutions of g(z) = 1 satisfy z n + κ = -ζ n and that

g ′ (z j ) = 1 z j + κ α c E Ze (zj+κ)Z -1 ,
so that Expansion (26), with corresponding terms given by ( 27), corresponds to the N + 1 first terms of (29).

Remark 13. With some more effort, Theorem 10 can be extended to more general Gerber-Shiu functions, e.g. of the form ψ(x, θ, b, a) := E x e -θτ ½ {R τ -≥b; R τ ≥a; τ <+∞} where θ, b, a are non negative, τ := inf{t ≥ 0| R t < 0} is the ruin time of the risk process and R t = inf 0≤s≤t R s is the running minimum at time t, see Theorem 2.8 of [START_REF] Roynette | Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes[END_REF] as well as Theorem 1 of [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF] for example of such expansions. For ease of presentation, we stick in this paper to ψ(x) as defined by [START_REF] Schiff | Normal families[END_REF].

Example 14. We give an example of an infinite expansion in the non lattice case of v(x) as in Corollary 4. Conditions [START_REF] Feller | A renewal theorem[END_REF] and [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF] may look hard to verify in practice. To exhibit such an X, we again use the theory of meromorphic Lévy processes. As in Remark 12, we pick spectrally negative process {Y t , t ≥},

Y t = ct - Nt k=1 Z k where {N t , t ≥} is a Poisson process with intensity α > 0, such that Laplace exponent is of the form Λ(z) = μz + z 2 ∞ m=1 b m ρ 2 m (ρ m + z) ,
for some μ > 0, where sequences of positive real numbers (b m ) m∈AE * and (strictly) increasing (ρ m ) m∈AE * are such that series ∞ m=1 bm ρm converges so that Lévy process {Y t , t ≥} is indeed a compound Poisson process, see (3.2) in [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF]. We will additionally suppose that sequence (ρ m ) m∈AE * grows like a polynomial (in addition to being increasing), i.e. there exists a ≥ 1 such that

ρ m ∼ Cm a , m → ∞ (31)
for some C > 0. Remembering that G is the cdf of the Z k 's, we then consider r.v. X with descending ladder height distribution of Lévy process {Y t , t ≥ 0}, with corresponding moment generating function

g(z) := ∞ 0 e (κ+z)y α c Ḡ(y)dy,
where κ > 0 is solution to Lundberg equation [START_REF] Stone | On characteristic functions and renewal theory[END_REF], see Relation (5.7) p. 87 of [START_REF] Asmussen | Ruin probabilities[END_REF]. We proceed to show that an infinite expansion for the corresponding function v(x) is available. The sole condition for this expansion is (31), which is not too stringent and covers a wide range of processes. The relation between g(z) and Λ(z) is given by ( 30). This has two important consequences. The first one is that z → g(z) is meromorphic, as Λ(z) is. The second one is that z is a solution to g(z) = 1 iff Λ(-z -κ) = 0. By Properties (v) and (vi) of [START_REF] Kuznetsov | Computing the finite-time expected discounted penalty function for a family of Lévy risk processes[END_REF] (see also Theorem 1 (7) of [START_REF] Kuznetsov | Meromorphic Lévy processes and their fluctuation identities[END_REF]), one deduces that roots (z n ) n∈AE are real non negative and verify

z 0 = 0 < ρ 1 -κ < z 1 < ρ 2 -κ < z 2 < ... (32)
We now turn back to Conditions [START_REF] Feller | A renewal theorem[END_REF] and [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF]. We start by [START_REF] Feller | A renewal theorem[END_REF]. We compute from ( 30)

g(z) = - μ c + 1 + (z + κ) ∞ m=1 b m cρ 2 m (ρ m -z -κ) , (33)
hence, for all j ∈ AE, g ′ (z j ) = ∞ m=1 bm cρm(ρm-zj-κ) 2 , which happens to be positive. We now write

z 2 j g ′ (z j ) ≥ j m=1 b m cρ m 1 + ρm-κ zj 2 = ∞ m=1 b m cρ m 1 + ρm-κ zj 2 ½ [m≤j] .
Using (32) and the dominated convergence theorem, one easily shows that

∞ m=1 bm cρm 1+ ρm-κ z j 2 ½ [m≤j] -→ ∞ m=1 bm cρm as j → ∞. One then deduces from the above inequality that lim inf j→∞ z 2 j g ′ (z j ) ≥ ∞ m=1 b m cρ m > 0. ( 34
)
Now, (31) and (32) implies that ∞ m=1 e -xzm is a convergent series for all x > 0 which, combined with (34), implies the convergence [START_REF] Feller | A renewal theorem[END_REF]. We now prove [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF], by establishing that lim n→∞ 1 rn(1-g(rn+iθ)) = 0 with r n := ρ n -κ. Using (33) and g(0) = 1 implies for all θ ∈ Ê

1 -g(r n + iθ) = g(0) -g(r n + iθ) = ∞ m=1 b m cρ 2 m κ ρ m -κ - ρ n + iθ ρ m -ρ n -iθ = ∞ m=1 b m cρ 2 m κ ρ m -κ + 1 -ρ m ρ m -ρ n + iθ (ρ m -ρ n ) 2 + θ 2 := R n (θ) + iI n (θ). Let us set R n,1 (θ) := ∞ m=1 bm cρ 2 m κ ρm-κ + 1 + n m=1 bm cρ 2 m ρ m ρn-ρm (ρm-ρn) 2 +θ 2 and R n,2 (θ) := ∞ m=n+1 bm cρ 2 m ρ m ρn-ρm (ρm-ρn) 2 +θ 2 . As (ρ n ) n∈AE is increasing one gets the following inequalities R n,1 (θ) ≥ ∞ m=1 b m cρ 2 m κ ρ m -κ + 1 := ξ > 0, 0 ≥ R n,2 (θ) ≥ ∞ m=n+1 b m cρ 2 m ρ m ρ n -ρ m (ρ m -ρ n ) 2 = ∞ m=n+1 b m cρ m 1 ρ n -ρ m := χ n ,
so that the real part of 1 -g(r n + iθ) verifies the inequality

|R n (θ)| ≥ |R n,1 (θ)| -|R n,2 (θ)| = R n,1 (θ) + R n,2 (θ) ≥ ξ + χ n . ( 35
) (31) entails that |χ n | ≤ 1 ρn+1-ρn ∞ m=n+1 bm cρm ∼ 1 Can a-1 ∞ m=n+1 bm cρm -→ 0 as n → ∞.
One then deduces that ξ + χ n > 0 for n large enough, and

sup θ∈Ê 1 r n |1 -g(r n + iθ)| ≤ sup θ∈Ê 1 r n |R n (θ)| ≤ 1 r n (ξ + χ n ) -→ 0, n → ∞,
proving [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF]. Hence infinite expansion (12) holds.

Skip free random walks on

Quite unlike its continuous time counterpart, risk theory in discrete time seems to have been less studied. We refer to [START_REF] Li | A review of discrete-time risk models[END_REF] for an overview of such processes, as well as [START_REF] Asmussen | Ruin probabilities[END_REF]Chapter XVI]. This type of process is but a skip free random walk, i.e. a random walk with at most unit upward movement, and is in fact studied in many fields of applied probability. We consider here the so-called binomial discrete time risk model defined by

R n = x + n - n j=1 Z j = x + Y n , n ∈ AE
where x ∈ AE is the initial reserve, the premium rate is assumed w.l.o.g. to be equal to 1, the claims (Z j ) j∈AE form an i.i.d. sequence taking values in AE. We let m 1 = E[Z 1 ] and assume that m 1 ∈ (0, 1).

The discrete ruin probability is defined by

ψ(x) := P inf n∈AE\{0} R n ≤ 0 , x ∈ AE. ( 36
)
This corresponds to the probability that a -valued random walk starting from x ∈ AE eventually becomes nonpositive. The condition m 1 ∈ (0, 1) ensures that the random walk has a positive drift so that ψ(x) < 1.

A closed form expression for ψ(x) may be found in [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF], however this expression requires computing an infinite number of convolutions of distribution of Z 1 . We are here interested in finding a simple expansion of ψ(x) as x → ∞. Similarly to Condition [START_REF] Roynette | Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes[END_REF] in the continuous case, we assume here that the moment generating function E e sZ is finite for all s > 0.

(37)

In this discrete setting, the Lundberg equation writes

∞ k=0 e zk P[Z > k] = 1, z ∈ . ( 38
)
Restricted to the real numbers, this equation has, thanks to convexity of mean generating function of Z, a unique solution κ > 0. We define the probability mass function f defined by f (k) = e κk P[Z > k], k ∈ N, with moment generating function

g(z) = ∞ k=0 e zk f (k) = 1 -E(e (z+κ)Z ) 1 -e z+κ , z ∈ .
The following theorem provides an asymptotic expansion of ψ(x) as x → ∞, x ∈ AE and is the discrete analog of Theorem 10.

Theorem 15. Assume conditions (37) holds. Let r > 0 be fixed and z 0 = 0, z 1 , . . . , z N the solutions to Equation g(z) = 1 in S f r = {z ∈ C; ℜ(z) < r, -π ≤ ℑ(z) ≤ π}. Then, the ruin probability ψ(x) has the asymptotic expansion

ψ(x) = - N j=0 Res 1 1 -g(z)
m -g(z)e κ+z 1 -e κ+z e -(z+κ)x ; z j +o(e -(r+κ)x ) as x → ∞.

(39) If z j is a simple zero of g -1, i.e. g ′ (z j ) = 0, then the j-th term in (39) has the simple form m -e κ+zj e κ+zj -E Ze (κ+zj)Z e -(zj +κ)x .

(40)

A two dimensional ruin problem

We consider a two dimensional ruin problem motivated by applications in reinsurance. The capitals of two insurance companies are modeled by the risk pro-cesses

R j t = x j + c j t - N j t k=1 Z j k = x j + Y j t , t ≥ 0, j = 1, 2, (41)
where, x j ≥ 0, c j > 0 are the respective initial reserves and premium rates, {N j t , t ≥ 0} are Poisson processes with intensities α j > 0, and (Z j k ) k∈AE are the corresponding claims with mean m j . For each j = 1, 2, independence between {N j t , t ≥ 0} and (Z j k ) k∈AE is assumed. However no independence is required between processes {R 1 t , t ≥ 0} and {R 2 t , t ≥ 0}. We suppose that the mean drifts E(Y j ) = c j -α j m j , j = 1, 2, are positive, and then define the eventual ruin probabilities for each company

ψ j (x j ) := P inf t≥0 R j t < 0 , j = 1, 2.
We also consider the probability that (at least) one of the companies is eventually ruined

ψ or (x 1 , x 2 ) := P inf t≥0 R 1 t < 0 or inf t≥0 R 2 t < 0 ,
i.e. the probability that the two dimensional process

{(R 1 t , R 2 t ), t ≥ 0} exits the first quadrants [0, +∞) 2 .
We are interested here in the asymptotics of ψ or (x 1 , x 2 ) as (x 1 , x 2 ) tend to infinity along a fixed direction x 2 /x 1 = q ∈ (0, +∞). We refer to [START_REF] Palmowski | Exit problem of a twodimensional risk process from the quadrant: Exact and asymptotic results[END_REF][START_REF] Rabehasaina | A Markov additive risk process in dimension 2 perturbed by a fractional Brownian motion[END_REF] for related results that concern light tailed claims, or [START_REF] Biard | Asymptotic multivariate finite-time ruin probabilities with heavy-tailed claim amounts: Impact of dependence and optimal reserve allocation[END_REF][START_REF] Hu | On joint ruin probabilities of a two-dimensional risk model with constant interest rate[END_REF] for models featuring heavy tailed claims.

For j = 1, 2, letting g j (z) := E(e zZ j 1 ), we suppose that Equation g j (z) = 1 has solutions z j 0 = 0, z j 1 , z j 1 in S r = {z ∈ C, 0 ≤ ℜ(z) < r} for some r > 0, and that those z j 1 , z j 1 are simple zeros of g -1. Thus ψ j (x j ) has the following 2 terms expansion from Theorem 10 ψ j (x j ) = C j 0 e -κjxj +ℜ C j 1 e -(κj+z 1 j )xj +ε j (x j )e -(r+κj)xj , x j → +∞, j = 1, 2, (42) where ε j (x j ) -→ 0 as x j → +∞, and

C j k := c j -α j m j α j E Z j e (zj+κ)Z j -c j , k = 0, 1, j = 1, 2.
The main result of this subsection is the following theorem.

Proposition 16. A two term asymptotic for ψ or (x 1 , x 2 ) as (x 1 , x 2 ) tend to infinity along the fixed direction x 2 /x 1 = q ∈ (0, +∞), is given by

ψ or (x, qx) = ℜ D 0 e -d0(q)x + ℜ D 1 e -d1(q)x + η q (x)e -ℜ(d1(q))x , x → ∞, (43)
where d 0 (q) > 0, ℜ(d 1 (q)) > d 0 (q), 0 ≤ lim sup x→∞ |η q (x)| ≤ 1 and where x → ℜ D 0 e -d0(q)x and x → ℜ D 1 e -d1(q)x are the two first dominant func-

tions among x → C 1 0 e -κ1x , x → C 2 0 e -qκ2x , x → ℜ C 1 1 e -(κ1+z 1 1 )x and x → ℜ C 2 1 e -q(κ2+z 2 1 )x .
Four different cases occur in the asymptotic described in Proposition 16, depending on the asymptotic direction u = (1, q) :

ℜ(κ 1 + z 1 1 ) > qκ 2 , κ 1 > qκ 2 > ℜ(κ 1 + z 1 1 ), qκ 2 > κ 1 > qℜ(κ 2 + z 2 1 ), or qℜ(κ 2 + z 2 1 ) > κ 1 .
To each case corresponds a different two terms expansion for ψ or (x, qx) as summarized in Figure 1. Proposition 16 generalizes the one term expansion given in Theorem 3 of [START_REF] Palmowski | Exit problem of a twodimensional risk process from the quadrant: Exact and asymptotic results[END_REF]. The last term in (43) is only O(e -ℜ(d1(q))x ) but the condition lim sup x→∞ |η q (x)| ≤ 1 provides information on how fast this term tends to 0.

ℜ(κ1 + z 1 1 ) > qκ2 R 1 t R 2 t asymptotics direction u = (1, q) C 1 0 e -κ1x + ℜ C 1 1 e -(κ1+z 1 1 )x C 1 0 e -κ1x + C 2 0 e -qκ2x C 2 0 e -qκ2x + C 1 0 e -κ1x qℜ(κ2 + z 2 1 ) > κ1 C 2 0 e -qκ2x + ℜ C 2 1 e -q(κ2+z 2 1 )x qκ2 > κ1 > qℜ(κ2 + z 2 1 )
κ1 > qκ2 > ℜ(κ1 + z

2 = α 1 P(V > d) = α 1 e -λd .
As the Z 2 n 's are exponentially distributed,

ψ 2 (x 2 ) = C 2 0 e -κ2x2 , κ 2 = λ -α 2 /c 2 , C 2 0 = α 2 c 2 λ , (44)
see Corollary 3.2 p.78 of [START_REF] Asmussen | Ruin probabilities[END_REF]. The two terms expansion for ψ 1 (x 1 ) is given by

ψ 1 (x 1 ) = C 1 0 e -κ1x1 + ℜ C 1 1 e -(κ1+z 1 1 )x + o(e -(r+κ1)x1 )
where z 1 1 is the solution to Equation (28) with smallest real part ℜ(z 1 1 ) > κ 1 and r > ℜ(z 1 1 ). Note that even though {R 1 t , t ≥ 0} and {R 2 t , t ≥ 0} are correlated, Proposition 16 may be applied and, since C 2 1 = 0, only three cases occur :

ℜ(κ 1 + z 1 1 ) > qκ 2 , κ 1 > qκ 2 > ℜ(κ 1 + z 1 1 ) and qκ 2 > κ 1 .
This is summarized in Figure 2.

q < ℜ(κ1+z 1 1 ) λ-α2/c2 R 1 t R 2 t asymptotics direction u = (1, q) C 1 0 e -κ1x + ℜ C 1 1 e -(κ1+z 1 1 )x C 1 0 e -κ1x + C 2 0 e -qκ2x q ∈ κ1 λ-α2/c2 , ℜ(κ1+z 1 1 ) λ-α2/c2 C 2 0 e -qκ2x + C 1 0 e -κ1x q > κ1 λ-α2/c2

Proofs for section 2

Proof of Theorem 1

For the proof of Theorem 1, we need the following lemma.

Lemma 18. Let f : [-π, π] -→ be a continuous function satisfying f (0) ∈ Ê and ℑ(f (θ)) = O(θ) as θ → 0. Then we have the following convergence as r → 1-:

π -π ℜ f (θ) 1 1 -rg(iθ) dθ -→ π -π ℜ f (θ) 1 1 -g(iθ) dθ + π f (0) µ . ( 45
)
The above lemma is akin to the preliminary result of Stone [23, p.330], see also Breiman [START_REF] Breiman | Probability[END_REF] and Feller and Orey [START_REF] Feller | A renewal theorem[END_REF]. In these references, only the case of a real valued function f is considered. In the complex case, we give a short proof inspired by Lemma 10.11 p.221 of [START_REF] Breiman | Probability[END_REF].

Proof of Lemma 18. Inspecting the proof of [START_REF] Breiman | Probability[END_REF], one can see that it is sufficient to check that θ → ℜ f (θ)

1 1-g(iθ)
is integrable at θ = 0. The rest of the proof may be applied similarly (with minor modification) in order to prove (45). Since

ℜ f (θ) 1 1 -g(iθ) = θ 2 |1 -g(iθ)| 2 ℜ f (θ)(1 -g(iθ)) θ 2 and θ 2 |1 -g(iθ)| 2 -→ 1 µ 2 as θ → 0, it is sufficient to prove local integrability of ℜ(f (θ)(1-g(iθ))) θ 2 at θ = 0. We compute further ℜ f (θ)(1 -g(iθ)) = ℜ (f (θ)) ℜ 1 -g(iθ) -ℑ (f (θ)) ℑ 1 -g(iθ) .
The first term is integrable at 0 since

ε -ε ℜ (f (θ)) ℜ 1 -g(iθ) θ 2 dθ ≤ sup θ∈[-π,π] |f (θ)|× ε -ε 1 -E(cos(θX 1 )) θ 2 dθ < +∞.
For the integrability of the second term, we need the assumption ℑ(f (θ)) = O(θ) which implies the existence of a constant C > 0 such that |ℑ(f (θ))| ≤ C|θ| for |θ| ≤ ε. We use also the inequality | sin(x)| ≤ x, x ∈ R. Using this, we have

ε -ε ℑ (f (θ)) ℑ 1 -g(iθ) θ 2 dθ ≤ E ε -ε ℑ (f (θ)) θ sin(θX 1 ) θ dθ ≤ E ε -ε C|X 1 |dθ = 2εCE(|X 1 |) < +∞.
Proof of Theorem 1. Let us define

u k := U (k) -U (k -1) = ∞ n=0 P(S n = k), k ∈ AE,
with the convention U (-1) = 0. We use the basic fact that S n has Fourier transform g(iθ) n and that the probabilities P(S n = k)'s are linked to the Fourier transform by

P(S n = k) = 1 2π π -π e -ikθ g(iθ) n dθ, n ∈ AE, k ∈ AE,
which can be verified by writing g(iθ) n = (e iθSn ) and using Fubini. Hence, by Lebesgue's dominated convergence and Fubini's theorems,

u k = lim r→1 - ∞ n=0 r n P(S n = k) = lim r→1 - ∞ n=0 r n 1 2π π -π e -ikθ g(iθ) n dθ = lim r→1 - 1 2π π -π e -ikθ 1 1 -rg(iθ)
dθ.

Note that ℑ(u k ) = 0. We deduce, thanks to Lemma 18,

u k = ℜ(u k ) = 1 2µ + 1 2π π -π ℜ e -ikθ 1 1 -g(iθ)
dθ.

We apply the same argument to the i.i.d. r.v. (X ′ n ) n∈AE with distribution X ′ n ∼ δ 1 . This yields, for all k ∈ AE,

1 = 1 2 + 1 2π π -π ℜ e -ikθ 1 1 -e iθ dθ,
whence we deduce

u k - 1 µ = 1 2π π -π ℜ e -ikθ 1 1 -g(iθ) - 1 µ 1 1 -e iθ dθ = 1 2π π -π e -ikθ 1 1 -g(iθ) - 1 µ 1 1 -e iθ dθ, k ∈ AE, (46)
the last line justified by the fact that the integral is convergent. The integrand function

z → e -kz 1 1 -g(z) - 1 µ 1 1 -e z is meromorphic on the domain {z ∈ C; -π ≤ ℑ(z) ≤ π, 0 ≤ ℜ(z) ≤ r}, r < R 0 .
For R 0 -r small enough, the poles inside this domain are exactly z 1 , . . . , z N (the pole at z 0 = 0 has been removed). Cauchy's residue Theorem with contour given in the left panel of Figure 3 implies

u k - 1 µ = - N j=1
Res e -kz 1 -g(z)

; Using this, we obtain finally

z j + 1 2π π -π e -k(r+iθ) 1 1 -g(r + iθ) - 1 µ 1 1 -e r+iθ dθ.
v(k) = ∞ m=0 [v(k + m) -v(k + m + 1)] = ∞ m=0 [1/µ -u(k + m + 1)] = ∞ m=0 N j=1 Res e -(k+m+1)z 1 -g(z) ; z j - 1 2π ∞ m=0 π -π e -(k+m+1)(r+iθ) 1 1 -g(r + iθ) - 1 µ 1 1 -e r+iθ dθ = N j=1 Res e -kz (e z -1)(1 -g(z)) ; z j - e -kr 2π π -π e -ikθ e (r+iθ) -1 1 1 -g(r + iθ) - 1 µ 1 1 -e r+iθ dθ. (48)
Here we have used that all residues Res e -(k+m+1)z 1-g(z) ; z j are obtained by integrating e -(k+m+1)z

1-g(z)

on a compact contour around z j , so that exchanging ∞ m=0 and Res(.) is indeed justified by Fubini's theorem. By the Lebesgue lemma, the last term in (48) is o(e -rk ) and this proves Equation (5).

Proof of Proposition 2

We prove that if F is absolutely continuous, then for all R 0 < R,

lim θ→±∞ sup 0≤r≤R0 |g(r + iθ)| = 0. ( 49
)
Clearly, Equation (49) implies Proposition 2. It is worth noting that this is a uniform version of the Riemann-Lebesgue Lemma. Let us set for all θ ∈ Ê, f θ (z) := g(z + iθ), z ∈ . Since F admits a density, the Riemann-Lebesgue Lemma implies that f θ (z) converges pointwise to 0 as θ → ∞ when z

∈ Ω := {z ∈ | 0 ≤ ℜ(z) ≤ R 0 }.
Let us now note that we have the uniform bound

|f θ (z)| = [e (z+iθ)X ] ≤ [e R0X ] < +∞, ∀z ∈ Ω.
By Theorem 1.6.4 p.26 of [START_REF] Schiff | Normal families[END_REF], f θ (z) converges uniformly towards 0 as θ → ∞ when z lies in any compact subset K ⊂ Ω. Picking in particular

K := {z ∈ | ℜ(z) ∈ [0, R 0 ] and ℑ(z) = 0} yields sup z∈K |f θ (z)| = sup r∈[0,R0] |g(r + iθ)| -→ 0, θ → ∞,
which we were set to prove.

Proof of Theorem 3

The proof of Theorem 3 follows the same lines as the proof of the main Theorem in Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF].

For the sake of clarity, we divide the proof into several steps.

Step 1: We prove that Condition [START_REF] Blanchet | Uniform renewal theory with applications to expansions of random geometric sums[END_REF] implies that the equation g(z) = 1 has only a finite number of solutions in S R0 . Condition (6) entails the existence of M > 0 such that the equation g(r +iθ) = 1 has no solution with 0 ≤ r ≤ R 0 and |θ| > M . Since furthermore the obvious bound |g(z)| < 1 if ℜ(z) < 0 exclude solutions in the half-plane ℜ(z) < 0, the only possible solutions of g(z) = 1 in S R0 belong to the compact set

K = {z ∈ C; 0 ≤ ℜ(z) ≤ R 0 , -M ≤ ℑ(z) ≤ M }.
The function g(z) -1 being holomorphic, its zeros are isolated. Hence the equation g(z) = 1 has finitely many solution in the compact set K.

Step 2: Following Stone [START_REF] Stone | On moment generating functions and renewal theory[END_REF], we introduce for positive a and h,

V (x, h, a) := [U (x + h -aZ) -U (x -aZ)] = E[U (x -aZ, h)] with U (x, h) = U (x + h) -U (x),
and Z a random variable with standard Gaussian distribution. Using Fourier analysis, one can show (similarly to Equation ( 7) in [START_REF] Stone | On moment generating functions and renewal theory[END_REF])

V (x, h, a) = h 2µ + h 2π +∞ -∞ ℜ e -ixθ 1 -e -ihθ ihθ e -a 2 θ 2 /2 1 1 -g(iθ) dθ = h 2µ + I(x, h, a). ( 50
)
Setting T (h, z) := 1-e -hz hz and ϕ(z) := 1 1-g(z) + 1 µz , we then split the integral I(x, h, a) into

I(x, h, a) = I 1 (x, h, a) -I 2 (x, h, a), (51)
I 1 (x, h, a) = h 2π +∞ -∞ ℜ e -ixθ T (h, iθ)e -a 2 θ 2 /2 ϕ(iθ) dθ, I 2 (x, h, a) = h 2π +∞ -∞ ℜ e -ixθ T (h, iθ)e -a 2 θ 2 /2 1 iµθ dθ.
The analysis of the first term I 1 (x, h, a) relies on Cauchy's residue theorem. In the domain S R0 , the function ϕ(.) is meromorphic with poles z 1 , . . . , z N (note that the pole at z 0 = 0 has been removed). We apply Cauchy's residue theorem with the contour given in the right panel of Figure 3 and we let the parameter L tend to +∞. Thanks to condition ( 6), the contribution of the horizontal parts of the contour vanishes as L → +∞ and we get, for all r ∈ (ℜ(z N ), R 0 ),

1 2π +∞ -∞ e -ixθ T (h, iθ)e -a 2 θ 2 /2 ϕ(iθ)dθ = - N j=1
Res e -xz T (h, z)e a 2 z 2 /2 ϕ(z); z j + ℜ Ĩ1 (x, h, a, r) (52

) with Ĩ1 (x, h, a, r) = 1 2π +∞ -∞ e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 ϕ(r + iθ)dθ. ( 53
)
Note that the sum of residues is a real number because for conjugate poles z and z, the residues at z and z are also conjugate so that the sum is real. Similarly for the second term I 2 (x, h, a), Cauchy's residue theorem yields

1 2π +∞ -∞ e -ixθ T (h, iθ)e -a 2 θ 2 /2 1 iµθ dθ = - 1 2µ + Ĩ2 (x, h, a, r) (54) with Ĩ2 (x, h, a, r) = 1 2π +∞ -∞ e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 1 µ(r + iθ) dθ. (55)
Equations ( 50), ( 51), ( 52) and (54) together yield

V (x, h, a) = h µ -h N j=1
Res e -xz T (h, z)e a 2 z 2 /2 ϕ(z); z j + hℜ{ Ĩ1 (x, h, a, r)} -hℜ{ Ĩ2 (x, h, a, r)}. (56)

Note that the Rieman-Lebesgue Lemma implies that the second term I 2 (x, h, a, r) satisfies, for all r > 0, Ĩ2 (x, h, a, r) = o(e -rx ) uniformly for a and h in bounded sets.

(57)

Step 3: We analyze here the term Ĩ1 (x, h, a, r) and, similarly as Equation ( 8) from [START_REF] Stone | On moment generating functions and renewal theory[END_REF], we show that Ĩ1 (x, h, a, r) = o(e -rx (1 + | ln(a)|)) uniformly for a and h in bounded sets.

(58) The argument for this statement is given almost without proof in [START_REF] Stone | On moment generating functions and renewal theory[END_REF] but can be adapted from a similar argument in [START_REF] Stone | On characteristic functions and renewal theory[END_REF] in the following way. One notices that

ϕ(z) = g(z) -1 -µz µ 2 z 2 + (g(z) -1 -µz) 2 µ 2 z 2 (1 -g(z)
) , so that we have, for a <

= J 1 + J 2 + J 3 + J 4 , 1, Ĩ1 (x, h, a, r)
with

J 1 = 1 2π +∞ -∞ e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 g(r + iθ) -1 -µ(r + iθ) µ 2 (r + iθ) 2 dθ, J 2 = 1 2π |θ|≤1 e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 (g(r + iθ) -1 -µ(r + iθ)) 2 µ 2 (r + iθ) 2 (1 -g(r + iθ)) dθ. J 3 = 1 2π 1<|θ|<a -2 e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 (g(r + iθ) -1 -µ(r + iθ)) 2 µ 2 (r + iθ) 2 (1 -g(r + iθ)) dθ. J 4 = 1 2π |θ|≥a -2 e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 (g(r + iθ) -1 -µ(r + iθ)) 2 µ 2 (r + iθ) 2 (1 -g(r + iθ)) dθ. Since g(z)-1-µz µ 2 z 2
is analytic on S R (the pole at z = 0 has been removed), Cauchy's residue theorem and Lebesgue's Lemma yield

J 1 = o(e -rx
) uniformly for a and h in bounded sets.

For the other terms, we use the fact that the function

F (θ) := T (h, r + iθ)e a 2 (r+iθ) 2 /2 (g(r + iθ) -1 -µ(r + iθ)) 2 µ 2 (r + iθ) 2 (1 -g(r + iθ))
is bounded for |θ| ≤ 1 and satisfies

F ′ (θ) = O 1 θ as |θ| ≥ 1 and F (θ) = O e -a 2 θ 2 /2 as |θ| ≥ 1.
This justifies the following estimates

|J 2 | = 1 2π |θ|≤1 e -x(r+iθ) F (θ)dθ = O(e -rx ), |J 3 | = 1 2π 1<|θ|<a -2 e -x(r+iθ) F (θ)dθ = O(e -rx | log(a)|), |J 4 | = 1 2π |θ|≥a -2 e -x(r+iθ) F (θ)dθ = O(e -rx ).
Gathering the above inequalities in (59), we obtain (58). The term O(e -rx | log(a)|) can be replaced by o(e -rx | log(a)|) because r ∈ (ℜ(z N ), R 0 ) is arbitrary. Equations (56), ( 57) and (58) together yield

V (x, h, a) - h µ = -h N j=1 Res e -xz T (h, z)e a 2 z 2 /2 ϕ(z); z j + o(e -rx (1 + | ln(a)|)). (60)
Step 4: We prove the following inequality:

V (x + e -rx , 1 -2e -rx , e -rx /x) -M e -rx ≤ U (x, 1) ≤ (1 -e -rx ) -1 V (x -e -rx , 1 + 2e -rx , e -rx /x). (61) Recall the definition V (x, h, a) = [U (x-aZ, h)] with U (x, h) = U (x+h)-U (x)
and Z a standard Gaussian random variable. Using the fact that U (.) is non decreasing, we check, for |y| ≤ e -rx ,

U (x + e -rx -y, 1 -2e -rx ) ≤ U (x, 1) ≤ U (x -e -rx -y, 1 + 2e -rx ).
Let x 0 > 0 be such that, for all x ≥ x 0 , È(|Z| ≥ x) ≤ e -rx ≤ 1/2. Besides, the fact that renewal function x → U (x) is sub-additive implies that there exists some constant M independent from h and x such that 0 ≤ U (x, h) ≤ M, for all x > 0 and h in a bounded set.

(62)

By the definition V (x, h, a) = [U (x -aZ, h)], both sides of Equation ( 61) are obtained by splitting

V (x±e -rx , h, a) = [U (x±e -rx -aZ, h)½ {|Z|≤x} ]+ [U (x±e -rx -aZ, h)½ {|Z|>x} ]
with the values a = a(x) = e -rx /x and h = h(x) = 1 ± 2e -rx .

Step 5: Let us now prove the following estimate

U (x, 1) - 1 µ = - N j=1 Res e -xz T (1, z)ϕ(z); z j + o(e -rx ). (63)
This is roughly obtained from (60) by setting

a = a(x) = e -rx /x and h = h ± (x) = 1 ± 2e -rx .
This can be made rigorous thanks to Equation (61). We have, uniformly in the neighborhood of z j , j = 1, . . . , N ,

e -(x±e -rx )z T (h(x), z)e -a(x) 2 z 2 /2 = e -xz [1 + o(e -rx )] . T (1, z) + o(e -rx ) [1 + o(e -rx )] = e -xz T (1, z) + o(e -rx ) so that Res e -(x±e -rx )z T (h, z)e a 2 z 2 /2 ϕ(z); z j = Res e -xz T (1, z)ϕ(z); z j + o(e -rx ).
(64) Besides, left hand side of Equation (61) and Equation (64) entail

U (x, 1) - 1 µ + N j=1 Res e -xz T (1, z)ϕ(z); z j ≥ V (x + e -rx , 1 -2e -rx , e -rx /x) -M e -rx - 1 µ + N j=1 Res e -xz T (1, z)ϕ(z); z j = V (x + e -rx , h(x), a(x)) -M e -rx - 1 µ + N j=1
Res e -(x±e -rx )z T (h, z)e a 2 z 2 /2 ϕ(z); z j + o(e -rx ) (65)

Now from (60), we have

V (x+e -rx , h(x), a(x))- h(x) µ +h(x) N j=1 Res e -(x±e -rx )z T (h, z)e a 2 z 2 /2 ϕ(z); z j = o(xe -rx ).
Together with (65) and since h(x) = 1 + O(e -rx ) , this yields

U (x, 1) - 1 µ + N j=1 Res e -xz T (1, z)ϕ(z); z j ≥ o(xe -rx ), . (66)
A similar upper bound is proved in the same way, using the right hand side of Equation (61). We deduce

U (x, 1) - 1 µ + N j=1
Res e -xz T (1, z)ϕ(z); z j = o(xe -rx ).

We can replace o(xe -rx ) by o(e -rx ) because r is arbitrary in (ℜ(z N ), R 0 ). This proves Eq. (63).

Step 6: We finally prove Equation [START_REF] Breiman | Probability[END_REF]. Since lim x→∞ v(x) = 0, we have

v(x) = ∞ k=0 [v(x + k) -v(x + k + 1)] = ∞ k=0 [-U (x + k, 1) + 1/µ]. (67)
Using Equation (63), we deduce, for all r ∈ (ℜ(z N ), R 0 ),

v(x) = ∞ k=0 N j=1 Res e -(x+k)z T (1, z)ϕ(z); z j + o(e -r(x+k) ) = N j=1 Res e -xz 1 -e -z T (1, z)ϕ(z); z j + o(e -rx ) = N j=1 Res e -xz z(1 -g(z)) ; z j + o(e -rx ). (68)
Equation ( 7) follows easily.

Proof of Corollaries 4 and 6

Proof of Corollary 4. We consider first the lattice case. According to Condition [START_REF] Doering | Non) differentiability and asymptotics for potential densities and subordinators[END_REF], one can consider (r n ) n≥1 an increasing sequence such that r n → +∞ and

lim n→+∞ sup -π≤θ≤π 1 e rn (1 -g(r n + iθ)) = 0.
According to Equation (48),

v(k) = N (rn) j=1 Res e -kz (e z -1)(1 -g(z)) ; z j - 1 2π π -π e -k(rn+iθ) e rn+iθ -1 1 1 -g(r n + iθ) - 1 µ 1 1 -e rn+iθ dθ
with N (r n ) the number of solutions of the equation g(z) = 1 in S f rn . Lebesgue's dominated convergence Theorem implies, for k ≥ 1,

lim n→+∞ π -π e -k(rn+iθ) e rn+iθ -1 1 1 -g(r n + iθ) - 1 µ 1 1 -e rn+iθ dθ = 0.
This yields the result [START_REF] Feller | An introduction to probability theory and its applications[END_REF].

We now consider the non-lattice case. Thanks to Assumption [START_REF] Gerber | Mathematical fun with compound binomial process[END_REF], let (r n ) n≥1 be such that r n → +∞ and

sup θ∈R 1 r n (1 -g(r n + iθ)) -→ 0, n → ∞.
We will prove below that setting h = 1, r = r n and a = a n (x) = e -rnx /x in Equation ( 56) and letting n → +∞, we obtain

U (x, 1) = 1 µ - ∞ j=1 Res e -xz T (1, z)ϕ(z); z j = 1 µ - ∞ j=1
e -xzj (e -zj -1)

z j g ′ (z j ) , (69)
remembering that roots (z j ) j∈AE are simple. This is justified as follows:

-Condition (62) together with Lebesgue's dominated convergence Theorem implies

V (x, h, a) = V (x, 1, e -rnx /x) → U (x, h) as n → +∞. ( 70
)
-Recalling that Ĩ1 (x, h, a, r) and Ĩ2 (x, h, a, r) are respectively given by ( 53) and ( 55), the integral term Ĩ1 (x, h, a, r) -Ĩ2 (x, h, a, r) with h = 1, r = r n and a = e -rnx /x converge to 0 as n → +∞. Indeed, condition [START_REF] Doering | Non) differentiability and asymptotics for potential densities and subordinators[END_REF] and the upper bound

|T (h, r + iθ)| ≤ 2/r imply | Ĩ1 (x, h, a, r) -Ĩ2 (x, h, a, r)| = 1 2π +∞ -∞ e -x(r+iθ) T (h, r + iθ)e a 2 (r+iθ) 2 /2 1 1 -g(r + iθ) dθ ≤ e -xr π +∞ -∞ 1 r(1 -g(r + iθ)) e a 2 r 2 /2-a 2 θ 2 /2 dθ ≤ sup θ∈R 1 r(1 -g(r + iθ)) × e -xr π e a 2 r 2 /2 √ 2π a ≤ sup θ∈R 1 r n (1 -g(r n + iθ)) × √ 2x √ π exp r 2 n e -2rnx 2x
The following lemma can be verified easily. Proof of Theorem 10. It is well known that ψ(x) satisfies the following defective renewal equation

ψ(x) = L(x) + x 0 ψ(x -y)dL(y), (73)
where

L(x) := x 0 α c Ḡ(y)dy, x ≥ 0, (74)
can be seen as the cdf of a defective distribution (because L(+∞) = α c m < 1), see e.g. Corollary 3.3 p.79 of Asmussen and Albrecher [START_REF] Asmussen | Ruin probabilities[END_REF] or Eq. (7.2) p.377 of Feller [START_REF] Feller | An introduction to probability theory and its applications[END_REF].

We introduce the probability distribution dF (x) = e κx dL(x) and use the notation

g(z) := ∞ 0 e zx dF (x) = α c 1 z + κ E e (z+κ)Z -1 , µ := ∞ 0 xdF (x) and µ 2 := ∞ 0 x 2 dF (x) (75)
for the moment generating function and the first and second moments of F . Elementary calculation yields

g ′ (z j ) = 1 z j + κ α c E Ze (zj+κ)Z -1 . (76)
Thanks to the assumption Ḡ ∈ D and Lemma 19, the tail function of F belongs also to D whence the moment generating function g(z) is defined for all z ∈ . Setting ψ κ (x) := e κx ψ(x), (73) entails the non-defective renewal equation

ψ κ (x) = z(x) + x 0 ψ κ (x -y)dF (y), z(x) := e κx L(x), (77)
see Feller [START_REF] Feller | An introduction to probability theory and its applications[END_REF] Eq. (6.12) p.376. The distribution F is non-lattice so that Smith's renewal theorem entails the classical Cramer-Lundberg asymptotics:

ψ κ (x) → C := ∞ 0 z(y)dy ∞ 0 ydF (y) = c -αm c -αE (Z 1 e κZ1 ) as x → +∞, (78)
the last equality being obtained by direct computation, see e.g. Theorem 5.3 p.86 of [START_REF] Asmussen | Ruin probabilities[END_REF]. We now wish to provide an expansion of ψ κ (x) and provide extra terms in (78). Let us introduce

U (x) = ∞ n=0 F * (n) (x), v(x) = U (x) - x µ - µ 2 2µ 2 , x ≥ 0.
The solution to (77) is given by

ψ κ (x) = x 0 z(x -y)dU (y) = x 0 z(x -y) dy µ + x 0 z(x -y)dv(y). (79)
Using the fact that x 0 z(x -y)dy =

x 0 z(y)dy, an integration by parts and the definition of C, we obtain

ψ κ (x) -C = - ∞ x z(y) dy µ + x 0 z ′ (x -y)v(y)dy -v(0-)z(x) + α c mv(x) := I 1 (x) + I 2 (x) + I 3 (x) + I 4 (x). ( 80
)
We recall that x → Ḡ(x) belongs to D defined in (72), so that one easily sees that z(x) defined in (77) belongs to D, and that I 1 (x) and I 3 (x) are o(e -rx ) from Lemma 19. Thus we only need to study asymptotics of the terms I 2 (x) and I 4 (x) as x → ∞ in (80). We start with I 4 (x). Let us note that, as g(z) is defined for all z ∈ , R defined in ( 2) is equal to +∞, as well as R 0 in Theorem 3. Expansion [START_REF] Breiman | Probability[END_REF] in Theorem 3 yields

I 4 (x) = N j=1 α c mRes e -xz z(1 -g(z)) ; z j + o(e -rx ) (81)
Now turning to I 2 (x). By a change of variable one has that I 2 (x) =

x 0 z ′ (y)v(xy)dy, so that, using Expansion [START_REF] Breiman | Probability[END_REF] in Theorem 3 for v(x -y) yields

I 2 (x) = N j=1 x 0 z ′ (y)Res e -(x-y)z z(1 -g(z))
; z j dy + x 0 z ′ (y)e -r(x-y) ε(x -y)dy (82) for some function ε(.) vanishing at +∞. Furthermore, it is easy to check that

x → z ′ (x) ∈ D. One then verifies that ∞ x z ′ (y)Res e -(x-y)z
z(1-g(z)) ; z j dy is an o(e -rx ), by writing the residue as an integral over a fixed contour including z j then using Lemma 19. One also checks that the last term on the right hand side of (82) is o(e -rx ) by a dominated convergence argument. One may thus write in (82)

I 2 (x) = N j=1 ∞ 0 z ′ (y)Res e -(x-y)z z(1 -g(z)) ; z j dy + o(e -rx ). (83)
Deriving z(y) = e κy L(y) with L(y) given in (74), expressing the residue as an integral over a fixed contour and using Fubini, thus yields the following expression for the terms in the summation on the right hand side of (83) :

∞ 0 z ′ (y)Res e -(x-y)z z(1 -g(z)) ; z j dy = Res e -xz z(1 -g(z)) ∞ 0 z ′ (y)e yz dy; z j = Res e -xz z(1 -g(z)) ∞ 0 κe (z+κ)y ∞ y α c Ḡ(v)dvdy - ∞ 0 e (z+κ)y α c Ḡ(y)dy ; z j = Res e -xz z(1 -g(z)) α c E ∞ 0 ∞ 0 κe (z+κ)y ½ [Z≥v≥y] dydv - α c E ∞ 0 e (z+κ)y ½ [Z≥y] dy ; z j = Res e -xz z(1 -g(z)) α c κ z + κ E e (z+κ)Z -1 z + κ -Z - α c 1 z + κ E e (z+κ)Z -1 ; z j ,
which, since z j satisfies g(z j ) = 1 with g(z) given by (75), yields

∞ 0 z ′ (y)Res e -(x-y)z z(1 -g(z)) ; z j dy = Res e -xz z(1 -g(z)) κ z + κ - α c κ z + κ m -1 ; z j = Res e -xz z(1 -g(z)) [-z -α c κm] z + κ ; z j . (84)
Plugging (84) in (83) and summing I 2 (x) and Expression (81) of I 4 (x) then yields from (80)

ψ κ (x) -C = N j=1 Res e -xz 1 -g(z) [-1 + α c m] z + κ ; z j + o(e -rx ).
Let us now note that in the case where g ′ (z j ) = 0, the residue in the jth term in the above can be explicitly given and, thanks to (76), is equal to (27). Let us also note that expression of C provided in (78) is but -

-1+ α c m α c E(Ze (z 0 +κ)Z )-1
with z 0 = 0 and that g ′ (z 0 ) = 0, so that the above yields Expansion (39).

Proof of Theorem 15

We let b(k) := P(Z 1 = k), k ∈ AE, the probability mass distribution of Z 1 and denote its survival function by l(k)

:= P[Z 1 ≥ k + 1] = ∞ j=k+1 b(k).
Let us in particular note that b(0) can be positive. In other words, there is a possibility that there is no claim (or equivalently, a claim of size 0) at time n. We define

D d := {f : AE -→ | f (k) = o(e -sk ) ∀s > 0}
and suppose here that the Z k 's verify that k → l(k) ∈ D d . we let κ > 0 its unique real positive solution.

The following lemma is the discrete analog of Lemma 19:

Lemma 20. Let f ∈ D d , then 1. k ∈ AE → e zk f (k) belongs to D d for all z ∈ , 2. k ∈ AE → ∞ j=k f (j) belongs to D d .
Similarly to (74), ( 23) and (75) we define the following discrete measures F (.), associated probability mass function f (.) and complex valued function g(.) which we recall here:

F (x) := x k=0 f (k), x ∈ AE, f (k) := e κk l(k) = e κk P(Z ≥ k + 1), k ∈ AE, (85) g
(z) := ∞ 0 e zx dF (x) = ∞ k=0 e (z+κ)k P(Z ≥ k + 1) = 1 -E(e (z+κ)Z ) 1 -e z+κ , z ∈ . (86)
g(z) is thus defined for all z ∈ and is equal to E(e zX ) for some integer valued r.v. X with probability mass function f (.), and with expectation and second moment E(X) := µ and E(X 2 ) := µ 2 .

Proof of Theorem 15. We proceed along the lines of Proof of Proposition 10. Setting L(x) := ∞ y=x l(y), x ∈ AE, the discrete time analog of (73) is

ψ(x) = x k=0 ψ(x -k)l(k) + L(x), x ∈ AE, (87)
see e.g. Proposition 1.2 p.488 in [START_REF] Asmussen | Ruin probabilities[END_REF]. Letting ψ κ (x) := e κx ψ(x) and z(x) := e κx L(x), and by the definition of f (k) for all k in (85), we obtain from (87) the renewal equation

ψ κ (x) = z(x) + x k=0 ψ κ (x + 1 -k)f (k). ( 88
)
Since X is lattice, the corresponding Smith's renewal theorem implies that

ψ κ (x) -→ C := 1 µ ∞ k=0 z(k) = m -e κ e κ -E(Ze κZ ) , x → ∞, (89)
the last equality can be verified by direct computation. Let us note in passing that (89) provides the Cramer Lundberg asymptotics ψ(x) ∼

m-e κ e κ -E(Ze κZ ) e -κx for x → ∞, x ∈ AE, in the discrete case. Defining now

U (j) := ∞ n=0 F * n (j), u j := U (j)-U (j-1), v(j) := U (j)- j µ - µ 2 + µ 2 2µ 2 , j ∈ AE
with U (-1) = 0, solution to (88) is given by

ψ κ (x) = x k=0 z(x -k)u k = x k=0 z(x -k) 1 µ + v(k) -v(k -1) , = 1 µ x k=0 z(x -k) + x k=0 z(x -k)v(k) - x-1 k=-1 z(x -k -1)v(k), = 1 µ x k=0 z(k) + x-1 k=0 [z(x -k) -z(x -k -1)]v(k) + z(0)v(x) -z(x)v(-1).
Substracting in the above constant C given by (89) yields

ψ κ (x)-C = - 1 µ ∞ k=x+1 z(k)+ x-1 k=0 [z(x-k)-z(x-k-1)]v(k)+z(0)v(x)-z(x)v( -1).
(90) The first and last term on the right hand side of (90) are o(e -rk ) by Lemma 20, hence we are interested in the following quantities ; z j + ε x-k e -r(x-k)

for some (ε n ) n∈AE vanishing at ∞, one obtains,

I 1 (x) = N j=1
Res 1 (e z -1)(1 -g(z)

x k=1

[1 -e -κ ]e κk L(k)e -(x-k)z ; z j (93)

- N j=1
Res 1 (e z -1)(1 -g(z)

x k=1 e κ(k-1) l(k -1)e -(x-k)z ; z j (94)

+ x k=1
[z(k) -z(k -1)]ε x-k e -r(x-k) .

(95)

Using Lemma 20 as well as a dominated convergence theorem, it is not hard to see that ( 95) is an o(e -rx ) as x → ∞, x ∈ AE. We then study (93) and (94). One verifies that x ∈ AE → Res 1 (e z -1)(1-g(z) ∞ k=x+1 e κ(k-1) l(k -1)e kz ; z j ∈ D d by writing the residue as an integral over a fixed contour including z j then using Lemma 20, so that in (94) we compute

One verifies thanks to Lemma 20 and writing the residue as a contour around z j that x ∈ AE → Res m -g(z)e κ+z 1 -e κ+z e -xz ; z j + o(e -rx ).( 99)

Plugging (99) into (90) yields (39), provided that we prove that residue of -1

1-g(z)

m-g(z)e κ+z 1-e κ+z e -xz at z = z 0 = 0 equals Ce -κx , C given by (89). This computation is in fact included in the case where the g ′ (z j )'s are non zero, in which case jth term of (99) is equal to (40), which we proceed to consider now. Direct computation from (86) coupled to the fact that g(z j ) = 1 gives g ′ (z j ) = -E Ze (κ+zj)Z [1 -e κ+zj ] + [1 -E(e (κ+zj )Z )]e κ+zj (1 -e κ+zj ) 2

= -E Ze (κ+zj)Z 1 -e κ+zj + e κ+zj 1 -e κ+zj so that it is easy to verify that Res

Proof of Proposition 16

Proof of Proposition 16. We recall that two functions f and h satisfy f (x) ≫ h(x) iff h(x)/f (x) tends to 0 as x → ∞. In the following we consider cases where q > 0 is such that functions respectively satisfy C 1 0 e -κ1x ≫ ℜ C 1 1 e -(κ1+z 1 1 )x ≫ C 2 0 e -qκ2x , ℜ C 2 1 e -q(κ2+z 2 1 )x , (100)

C 1 0 e -κ1x ≫ C 2 0 e -qκ2x ≫ ℜ C 1 1 e -(κ1+z 1 1 )x , ℜ C 2 1 e -q(κ2+z 2 1 )x , (101)
all other cases being treated similarly. We will use the following inequalities (an easy and direct consequence of Lemma 2 (i) of [START_REF] Palmowski | Exit problem of a twodimensional risk process from the quadrant: Exact and asymptotic results[END_REF])

ψ 1 (x 1 ) ≤ ψ or (x 1 , x 2 ) ≤ ψ 1 (x 1 ) + ψ 2 (x 2 ). ( 102
)
Let us first consider the case (100). Using expansions (42) with x 1 = x, x 2 = qx, as well as (102) we get ε 1 (x)e -(r+κ1)x ≤ ψ or (x, qx) -C 1 0 e -κ1x -ℜ C 1 1 e -(κ1+z 1 1 )x ≤ C 2 0 e -qκ2x + ℜ C 2 1 e -q(κ2+z 2 1 )x + ε 1 (x)e -(r+κ1)x + ε 2 (qx)e -(r+κ1)qx . (103)

We now note that the terms on the far left and right hand side of (103) divided by e -ℜ(κ1+z 1 1 )x tend to 0 because of Assumption (100), which proves (43). We now consider case (101). Using again (102) we get, similarly to (103),

  defines a meromorphic function on C. The poles are the solutions of e z = 1 -(1 -p)e 2iπ j n p , 0 ≤ j ≤ n -1.

Fig 2 .

 2 Fig 2. Two term asymptotic expansions for the stop loss model.

Fig 3 .

 3 Fig 3. Contours used for the application of Cauchy's residue Theorem in the lattice case (left) and the non-lattice case (right)

. 2 j / 2 - 1 . 2 .

 2212 This last quantity goes to 0 as r n → +∞. -Let us recall inequality |e z -1| ≤ e|z| for all |z| ≤ 1. As roots (z n ) n∈AE are simple and real, one has for all j = 1, ..., N (r n ), and for n large enough,Res e -(x±e -rn x )z T (1, z)e an(x) 2 z 2 /2 ϕ(z); z j -Res e -xz T (1, z)ϕ(z); z j = Res e ±e -rn x z e an(x) 2 z 2 /2 -1 e -xz T (1, z)ϕ(z); z j = e ±e -rnx zj e an(x) 2 z Res e -xz T (1, z)ϕ(z); z j ≤ e ±e -rnx z j + a n (x) 2 z 2 j /Res e -xz T (1, z)ϕ(z); z j ≤ M e -2rnxr 2 n . Res e -xz T (1, z)ϕ(z); z j exponentially fast, i.e. Ḡ is in D with D := f : [0, +∞) -→ ∀r > 0, f (x) = o(e -rx ) as x → ∞ .

Lemma 19 .

 19 Let f ∈ D, then 1. x → e zx f (x) belongs to D for all z ∈ ; 2. x → ∞ x f (y)dy belongs to D.

I 1 (

 1 x -k) -z(x -k -1)]v(k) = x k=1 [z(k) -z(k -1)]v(x -k), I 2 (x) := z(0)v(x) = L(0)v(x) = mv(x).Thanks to Theorem 1, one has thatI 2 (x) = N j=0 Res e -kz (e z -1)(1 -g(z)) ; z j + o(e -rx )(91)and we then turn to I 1 (x). Writingz(k) -z(k -1) = e κk L(k) -e κ(k-1) L(k -1) = [1 -e -κ ]e κk L(k) + e κ(k-1) [ L(k) -L(k -1)] = [1 -e -κ ]e κk L(k) -e κ(k-1) l(k -1) = [1 -e -κ ]z(k) -e κ(k-1) l(k -1),(92)then writing, in view of expansion[START_REF] Biard | Asymptotic multivariate finite-time ruin probabilities with heavy-tailed claim amounts: Impact of dependence and optimal reserve allocation[END_REF],v(x -k) = N j=1Res e -(x-k)z (e z -1)(1 -g(z))

e 1 - 1 -

 11 k-1) l(k -1)e kz ; z j = Res 1 (e z -1)(1 -g(z)) e -xz e z ∞ k=0 e (z+κ)k P(Z ≥ k + 1); z j -Res 1 (e z -1)(1 -g(z)) e -xz∞ k=x+1 e κ(k-1) l(k -1)e kz ; z j = Res 1 (e z -1)(1 -g(z)) e -xz e z g(z); z j + o(e -rx ), j = 1, ..., N. (96) Next, we compute, for all z, κ+z 1 -e (κ+z)n 1 -e κ+z l(n) e κ+z [m -l(0)] -1 e κ+z e κ+z [g(z) -l(0)] = e κ+z 1 -e κ+z [m -g(z)],

1 (e z - 1 )[ 1 Res 1 -

 111 (1-g(z) ∞ k=x+1 e (κ+z)k L(k); z j ∈ D d , so that inserting (97) in (93) yields Res 1 (e z -1)(1 -g(z))x k=1 -e -κ ]e κk L(k)e -(x-k)z ; z j = Res 1 (e z -1)(1 -g(z)) e -xz e κ+z 1 -e κ+z [m -g(z)]; z j + o(e -rx ). (98)Respectively inserting (96) and (98) in (94) and (93) in I 1 (x), and adding I 2 (x) in (91), yieldsI 1 (x) + I 2 (x) = N j=1 e -κ (e z -1)(1 -g(z)) e κ+z 1 -e κ+z [m -g(z)]e -xz ; z j -N j=1 Res g(z) (e z -1)(1 -g(z)) e z e -xz ; z j + N j=0Res m (e z -1)(1 -g(z)) e -xz ; z j + o(e -rx ),

1 1 -

 1 g(z) m-g(z)e κ+z 1-e κ+z e -xz ; z j is (40) times e -(κ+zj)x .

ε 1

 1 (x)e -(r+κ1)x +ℜ C 1 1 e -(κ1+z 1 1 )x -C 2 0 e -qκ2x ≤ ψ or (x, qx)-C 1 0 e -κ1x -C 2 0 e -qκ2x ≤ ℜ C 1 1 e -(κ1+z 1 1 )x +ℜ C 2 1 e -q(κ2+z 2 1 )x +ε 1 (x)e -(r+κ1)x +ε 2 (qx)e -(r+κ1)qx .(104)Because of Assumption (101), we get from (104) that-1 = lim x→∞ ε 1 (x)e -(r+κ1)x + ℜ C 1 1 e -(κ1+z 1 1 )x -C 2 0 e -qκ2x C 2 0 e -qκ2x ≤ lim inf x→∞ ψ or (x, qx) -C 1 0 e -κ1x -C 2 0 e -qκ2x C 2 0 e -qκ2x ≤ lim sup x→∞ ψ or (x, qx) -C 1 0 e -κ1x -C 2 0 e -qκ2x C 2 0 e -qκ2x ≤ lim x→∞ ℜ C 1 1 e -(κ1+z11 )x + ℜ C 2 1 e -q(κ2+z 2 1 )x + ε 1 (x)e -(r+κ1)x + ε 2 (qx)e -(r+κ1)qx

  0} is the capital of an insurance company with claims (Z 1 n ) n∈AE distributed as min(V, d) where V is exponentially distributed with parameter λ. The second risk process {R 2 t , t ≥ 0} corresponds to the capital of a reinsurance company which covers the excess of claims with priority d, i.e. claims (Z 1 n ) n∈AE are distributed as (V -d) + , as described in Example 11. In that case, the two risk processes R 1

	by
	α
	t and R 2 t are dependent. Because of the memoryless property of the exponential distribution, the Z 2 n 's are exponentially distributed with parameter λ and the intensity of the Poisson process {N 2 t , t ≥ 0} is given

1 1 ) Fig 1. Two term asymptotic expansions on the four different regions. Example 17. Let us consider the Stop Loss contract scenario with priority d > 0. We assume that {R 1 t , t ≥

Acknowledgment. The authors wish to thank an anonymous referee for pointing out the informal proof given at the end of Section 2.1, comments leading to Remark 5, as well as reference [15] leading to the infinite expansion for ψ(x) in Example 11. This work was partially supported by the French National Agency of Research ANR Grant, project AMMSI, number ANR-2011-BS01-021.

where M = M (x) is a constant independent from n and j, but which may depend on x > 0. One thus deduces

Res e -xz T (1, z)ϕ(z); z j (71) which tends to zero as n → ∞ because of Assumption [START_REF] Feller | A renewal theorem[END_REF].

-The fact that Ĩ1 (x, h, a n (x), r n ) -Ĩ2 (x, h, a n (x), r n ) tends to 0 as n → ∞ and convergence (70) (with h = 1) as well as inequality (71) implies ( 69) from (56).

The end of the proof follows easily, as in Step 6 of the proof of Theorem 3: Equations ( 67) and ( 69) imply [START_REF] Hu | On joint ruin probabilities of a two-dimensional risk model with constant interest rate[END_REF].

Proof of Corollary 6. In the lattice case, we deduce from Corollary 4 that

Res e -kz 1 -g(z)

; z j .

Finally, note that z 0 = 0, g(0) = 1, g ′ (0) = µ so that

; z 0 .

In the non-lattice case, we deduce from Corollary 4 that

Res e -kz 1 -g(z)

; z j .

The result follows since 1 µ = Res e -kz 1-g(z) ; z 0 .

5. Proofs for section 3

Proof of Theorem 10

Assumption (20) that the moment generating function of G is finite everywhere is equivalent to the fact that the tail function Ḡ = 1 -G decreases super-