Probabilistic approach to risk processes with level-dependent premium rate

We study risk processes with level dependent premium rate. Assuming that the premium rate converges, as the risk reserve increases, to the critical value in the net-profit condition, we obtain upper and lower bounds for the ruin probability. In contrast to existing in the literature results, our approach is purely probabilistic and based on the analysis of Markov chains with asymptotically zero drift.


Introduction
In context of the collective theory of risk, the classical Cramér-Lundberg (Sparre Andersen) model is defined as follows.An insurance company receives the constant inflow of premium at rate v, that is, the premium income is assumed to be linear in time with rate v.It is also assumed that the claims incurred by the insurance company arrive according to a homogeneous renewal process N (t) with intensity λ and the sizes (amounts) ξ n ≥ 0 of the claims are independent copies of a random variable ξ with finite mean.The ξ's are assumed independent of the process N (t).The company has an initial risk reserve x = R(0) ≥ 0. Then the risk reserve R(t) at time t is equal to The probability P{R(t) ≥ 0 for all t ≥ 0} = P min t≥0 R(t) ≥ 0 is the probability of ultimate survival and ψ(x) := P{R(t) < 0 for some t ≥ 0} = P min t≥0 R(t) < 0 is the probability of ruin.We have Since v > 0, the ruin can only occur at a claim epoch.Therefore, where T n is the nth claim epoch, so that T n = τ 1 + . . .+ τ n where the τ k 's are independent copies of a random variable τ with finite mean 1/λ, so that N (t) := max{n ≥ 1 : T n ≤ t}.Denote X i := ξ i − vτ i and S n := X 1 + . . .+ X n , then This relation represents the ruin probability problem as the tail probability problem for the maximum of the associated random walk {S n }.Let the net-profit condition hold, thus {S n } has a negative drift: ES 1 = Eξ 1 − vEτ < 0. Hence by the strong law of large numbers S n → −∞ a.s., so ψ(x) ↓ 0 as x → ∞.
The most classical case is when the distribution of X 1 satisfies the following well-known Cramér condition: there exists a β > 0 such that Ee βX 1 = 1. ( Under this condition, the sequence e βSn is a martingale and, by the Doob maximal inequality, the following Lundberg's inequality ψ(x) = P sup n≥1 e βSn > e βx ≤ e −βx , x > 0. ( If we additionally assume that EX 1 e βX 1 < ∞ and the distribution of X 1 is non-lattice, then the Cramér-Lundberg approximation holds, that is, there exists a constant c 0 ∈ (0, 1) such that ψ(x) ∼ c 0 e −βx as x → ∞, see e.g.Theorem VI.3.2 in Asmussen and Albrecher [1]; in the lattice case x must be taken as a multiple of the lattice step.The most important feature of these results is the fact that the upper bound (3) and the asymptotic relation (4) depend on the distribution of X 1 only via the parameter β.If the moment condition (2) on the distribution of X 1 does not hold then the tail asymptotics for ψ(x) are typically determined by the tail of the claim size ξ.The most prominent situation is when the distribution of ξ is of subexponential type.We discuss this case in more detail later.
The risk models with non-constant premium rates have also become rather popular in the collective risk literature.There are two main approaches, one of them leads to a Markovian model when the premium rate is a function of the current level of the risk reserve R(t), see e.g.Asmussen and Albrecher [1, Chapter VIII], Albrecher et al. [2], Boxma and Mandjes [3], Czarna et al. [4], Marciniak and Palmowski et al. [12].The second approach considers the premium rate that depends on the whole claims history, see e.g.Li et al. [11].
In this paper we follow the first approach and consider a risk process where the premium rate v(y) only depends on the current level of risk reserve R(t) = y, so R(t) satisfies the equality (5) v(y) is assumed to be a measurable non-negative bounded function.The probability of ruin given initial risk reserve x is again denoted by ψ(x).Notice that ψ(x) is no longer a decreasing function of x as it is in the classical case.The ruin probability for such processes with level dependent premium rate is much less studied in the literature than with constant premium rate, and all known results are exact expressions for some particular distributions of τ , ξ and/or for particular choices of the rate function v(y).The first example of the risk process where ψ(x) is explicitly calculable is the case of exponentially distributed τ and ξ, say with parameters λ and µ respectively, so hence v c = λ/µ.In this case, for some c 0 ∈ (0, 1), provided the outer integral is convergent from 0 to infinity, see, e.g.Corollary 1.9 in Albrecher and Asmussen [1, Ch.VIII].Some further examples of solutions in closed form can be found in Albrecher et al [2].The authors of that paper use a purely analytical approach, which works however only in the situations where the Laplace transforms of ξ and τ are rational functions.
The main goal of our paper is to develop a probabilistic method of the asymptotic analysis of risk processes with level-dependent premium rate, which is not based on exact calculations and uses only moment and tail conditions on ξ and τ .The following two qualitatively different cases can be identified: In the first case (7) one could expect that the ruin probability ψ(x) decays similarly to the classical collective risk model with constant premium rate v ∞ .In this paper we concentrate on a more difficult case (8) where the ruin is more likely due to the approaching the critical premium rate.We start by discretising the time; this procedure is standard for risk processes with constant rate.Since the ruin can only occur at a claim epoch, the ruin probability may be reduced to that for the embedded Markov chain So, our main goal is to analyse the down-crossing probabilities for the chain {R n }.In contrast to the constant premium case, we deal with a Markov chain instead of a random walk with independent increments.
As mentioned above, we shall restrict our attention to the case (8) where v(y) approaches the critical value v c at infinity.Then the Markov chain {R n } has asymptotically zero drift, that is, see Theorem 1 below.The study of Markov chains with vanishing drift was initiated by Lamperti in a series of papers [8,9,10].For further development in Lamperti's problem, see Menshikov et al [13,14].We also show in Theorem 1 that under (9) the ruin probability decays slower than any exponential function, that is, for any λ > 0, As well-known, for the classical Cramér-Lundberg model under the net-profit condition (1), the ruin probability decays slower than any exponential function if and only if the claim size tail distribution is so.As just mentioned, risk processes under the condition (8) give rise to heavy-tailed ruin probabilities whatever the distribution of the claim size, even if it is a bounded random variable.So, risk processes with near critical premium rate provide an important example of a stochastic model where light-tailed input produces heavy-tailed output.
We want to investigate how the rate of convergence in (8) is reflected in how slowly the ruin probability ψ(x) decreases.Let us get some intuition on what kind of phenomena we could expect here by considering the case of exponentially distributed ξ and τ .As we have mentioned above, the ruin probability ψ(x) is given in this case by (6).Combining ( 6) and ( 8), we obtain If the premium rate v(z) ≥ v c approaches v c at the rate of θ/z, θ > 0, more precisely, if where p(z) > 0 is an integrable at infinity decreasing function, then we get and consequently A similar asymptotic equivalence can be obtained also in the case where the Laplace transforms of variables ξ 1 and τ 1 are rational functions, see Albrecher et al. [2].
If the premium rate v(z) approaches v c at a slower rate of θ/z α , θ > 0 and α ∈ (0, 1), more precisely, if where p(z) > 0 is an integrable at infinity decreasing function, then we get where p 1 (z) = p(z) + z −γα is integrable at infinity.Consequently, if 1/α is not integer, then where c 3 is a finite real because p 1 (x) is integrable.In the case of integer 1/α one has Let, for example, α ∈ (1/2, 1).Then Therefore, for C 1 := c 0 e c 3 /θµ > 0 and We are going to extend these results to not necessarily exponential distributions where there are no closed form expressions like (6) available for ψ(x).In that case we can only derive lower and upper bounds for ψ(x) which have the same decay rate at infinity.

Heavy-tailedness of the ruin probability in the critical case
Denote the jumps of the chain {R n = R(T n )} by ξ(x), that is, for all Borel sets B. The dynamics of the risk reserve between two consequent claims is governed by the differential equation R ′ (t) = v(R(t)).Let V x (t) denote its solution with initial value x, so then Therefore, where = st stands for the equality in distribution.
To avoid trivial case where ψ(x) = 0 for all sufficiently large x, we assume that ψ(x) > 0 for all x.As the function ψ(x) is not necessarily decreasing, we need in the sequel a stronger condition: for all x 0 , inf A sufficient condition for that is that, for all x 0 , there exists an ε = ε(x 0 ) > 0 such that P{ξ(x) ≤ −ε} ≥ ε for all x ≤ x 0 .
In its turn, it is sufficient to assume that the random variable ξ is unbounded, due to the inequality ξ(y) ≤ vτ − ξ valid for all y, where v := sup z>0 v(z).
Theorem 1.Let v c = Eξ/Eτ and let v(x) → v c as x → ∞.Then the chain {R n } has asymptotically zero drift, that is, (9) holds true.
Let λ > 0. Consider a bounded decreasing function U λ (x) := min(e −λx , 1).For all x > 0, Due to (16), there exists a sufficiently large x λ > 0 such that Therefore, the process {U λ (R n∧τ B λ )} is a bounded submartingale, where B λ := (−∞, x λ ] and τ B = min{n : R n ∈ B}.Hence by the optional stopping theorem, for z > x λ and x ∈ (x λ , z), Letting z → ∞ we conclude that On the other hand, since U λ is bounded by 1, This allows us to deduce the lower bound and hence the conclusion (ii) follows, because by the Markov property, for all λ > 0 and x > 0, where , owing to the condition ( 14).

Transience of the underlying Markov chain
In this section we find conditions on the rate function v(z) which ensure that the ruin probability is strictly less than one.
Theorem 2. Let, for some θ > 0, Let both Eτ 2 1 and Eξ 2 1 be finite.If then the underlying Markov chain As we see from the convergence to a Γ-distribution, in the case (19) the chain R n escapes to infinity in probability at rate √ n in quite specific way as there is now law of large numbers.In the case where v(z) − v c ∼ c/z α with α ∈ (0, 1), the chain R n is transient too, however as follows from Lamperti [9, Theorem 7.1], it follows a law of large numbers, R 1+α n /n → c(1 + α) as n → ∞.Below we prove Theorem 2 via Lyapunov (test) functions approach, so we start with moment computations for the jumps of {R n }.Denote by m k (x) the kth moment of the jump ξ(x) of the chain {R n } from state x, that is, m k (x) = Eξ k (x).Lemma 3. If both Eτ 2 and Eξ 2 are finite, then, under the rate of convergence (10), for some decreasing integrable at infinity function Proof.By (10), On the other hand, again by (10), where the second inequality follows from the upper bound (26).Therefore, 26) and ( 27) that Recalling that v c = Eξ/Eτ , we get By the inequality log Therefore, the relation (21) follows.From that expression we have
Proof of Theorem 2. Let us consider the function v θ (z) := min(v(z), v c + θ/z).The dynamics of the risk reserve between two consequent claims with premium rate denote its solution with the initial value x, so then z for all sufficiently large z, Lemma 3 applies.As a result we have By the condition on θ, there exists an ε > 0 such that Further, again by Lemma 3 with γ 0 = 0, for any fixed δ > 0, for some decreasing integrable at infinity function p(x), due to Eξ 2 < ∞.
The bounds (32) and (33) show that the conditions ( 11) and ( 13) from Theorem 3 in [6] hold true.In addition, the Markov chain {R θ,n }-the embedded Markov chain for the ruin process with premium rate v θ (z)-dominates a similar Markov chain generated by a risk process with constant premium rate v c which represents a zerodrift random walk.The latter is null-recurrent and hence satisfying the condition (12) from Theorem 3 in [6], thus lim sup n→∞ R θ,n = ∞.Therefore, Theorem 3 from [6] applies and we conclude that the chain {R θ,n } is transient.Then the original chain {R n } is transient too, due to the domination property (31).
The convergence to a Γ-distribution follows from Theorem 4 in [6].
Remark 4. It is worth mentioning that the condition (19) is close to be minimal one for ψ(x) < 1.More precisely, one can show that if v(z) ≤ v c + θ/z for all sufficiently large z with some This statement follows by similar arguments applied to a dominating Markov chain with premium rate v θ (z) := max(v(z), v c + θ/z) that satisfies, for some ε > 0, 2zm θ,1 (z) ≤ (1 − ε)m θ,2 (z) for all sufficiently large z, and hence the classical Lamperti criterion (see, e.g.Lamperti [8]) for recurrence of Markov chains applies.4 Approaching critical premium rate at rate of θ/x Theorem 5. Assume (14) and the rate of convergence (10) with some θ satisfying (19), that is, If both Eτ 2 log(1 + τ ) and Eξ ρ+2 are finite then there exist positive constants c 1 and c 2 such that These bounds are quite similar to the classical estimates (3) and (4).Indeed, they are universal and only depend on a single parameter ρ of the distribution of (ξ, τ ).In contrast to the classical Cramer case, the crucial parameter ρ is very easy to compute.A further advantage of the bounds in Theorem 5 is the fact that they are applicable to a wide class of claim size distributions: the only restriction is that the moment of order ρ + 2 should be finite; otherwise, the probability of ruin is higher, see Section 6.
By the condition on θ, ρ > 0. Define q(x) := (ρ + 1) min(1, 1/x) and hereinafter we define Q(x) = 0 for x < 0. The increasing function Q(x) is concave on the positive half line because q(x) is decreasing.We have, for c = ρ + 1, so the function e −Q(x) is integrable at infinity, due to ρ > 0. It allows us to define the following bounded decreasing function which plays the most important rôle in our analysis of the ruin probabilities: and U (x) = U (0) for x ≤ 0. For all x ≥ 1 we have Let us also define the following auxiliary decreasing functions needed for our analysis.Without loss of generality we assume that p 1 (x) ≤ p(x) ≤ q(x) for all x, where p 1 (x) is given by Lemma 3; otherwise we can always consider the function max(p 1 (x), p(x)) instead of p(x).Consider the functions q + (x) := q(x) + p(x) and q − (x) := q(x) − p(x) and let and we have Therefore, Since p(x) is decreasing and integrable, p(x)x → 0 as x → ∞.We also assume that It follows from Lemma 24 that the condition on p ′ (x) is always satisfied for a properly chosen function p. Lemma 6.As x → ∞, and Proof.We start with the following decomposition: The third term on the right hand side is negative because U ± decreases and it may be bounded below as follows: due to the upper bound (25) which implies P{ξ(x) > x/2} = o(p 1 (x)/x), and due to the relations (38) and (39).Further, the first term on the right hand side of ( 43) is positive and possesses the following upper bound: due to the upper bound (23) and due to the relations (36) and (39).
To estimate the second term on the right hand side of (43), we make use of Taylor's expansion: where 0 ≤ θ = θ(x, ξ(x)) ≤ 1.By the construction of U ± , Then it follows that by Lemma 3 which yields It follows from ( 25) and (47) that Finally, let us estimate the last term in (46).Notice that by the condition (40) on the derivative of p(x), hence, as x → ∞ uniformly for |y| ≤ x/2 which implies Then, in view of ( 24), Substituting ( 48)-( 50) into (46), we obtain that Substituting (44)-or (45)-and ( 51) into (43) and recalling that p 1 (x) ≤ p(x), we finally come to the desired conclusions.
Lemma 6 implies the following result.
Corollary 7.There exists an x such that, for all x > x, Proof of Theorem 5.The process U − (R n ) is bounded above by U − (0).Let x be any level guaranteed by the last corollary, B = (−∞, x] and τ B = min{n ≥ 1 : X n ∈ B}.By Corollary 7, U − (R n∧τ B ) is a bounded supermartingale.Hence by the optional stopping theorem, for z > x and x ∈ ( x, z), Letting z → ∞ we conclude that On the other hand, since U − is decreasing, Therefore, which implies, by (39), that, for some constant c 2 < ∞, P x {R n ≤ x for some n} ≤ c 2 U (x) for all x > x.
Thus, P x {R n ≤ 0 for some n} ≤ P x {R n ≤ x for some n} ≤ c 2 U (x) for all x > x.
This gives the desired upper bound.
On the other hand, the process {U + (R n∧τ B )} is a bounded submartingale due to the lower bound provided by Corollary 7. Hence again by the optional stopping theorem, for x > x 0 , On the other hand, since U + is bounded by U + (0), This allows us to deduce a lower bound which completes the proof of the lower bound, for some constant c 1 > 0, due to (39).To complete the proof of the lower bound it remains to refer to the arguments in (17).
5 Approaching critical premium rate at rate of θ/x α In this section we consider the case (12) with some α ∈ (0, 1).In order to understand the asymptotic behaviour of the ruin probability under this rate of approaching the critical value v c , we first derive asymptotic estimates for the moments of V x (τ ) − x.Define γ := min{k ≥ 1 : αk > 1}.
Lemma 8. Let Eτ γ < ∞ and there exists an where both v − (x) and v + (x) are decreasing functions on [x 0 , ∞).Then, for all k ≤ γ, If, in addition, Eτ γ+1−α < ∞ and (12) holds true, then there exists an integrable decreasing function p 1 (x) such that, for all k ≤ γ, Proof.Fix some x ≥ x 0 .Due to (53), v(z) ≤ v + (x) for all z ≥ x.Hence, and the inequality on the right hand side of (54) follows.It follows from the left hand side inequality in (53) and from the last upper bound for V x (t) that and the left hand side bound in (54) is proven.
Owing to (12), v(z) is sandwiched between the two eventually decreasing functions v ± (z) := v c + θ/z α ± p(z).Therefore, applying the right hand side bound in (54) we get From the lower bound in (54) we deduce that, for all k ≤ γ, By the inequality 1/(1 + y) α ≥ 1 − αy ∧ 1, we infer that, for c 2 = αv, Therefore, for all k ≤ γ, for some c 3 < ∞.Then, due to the integrability of p(x), in order to prove that for some decreasing integrable function p 1 (x), it suffices to show that x −α E{τ γ ; τ > x} and x −j(α+1) E{τ γ+j ; τ ≤ x} are bounded by decreasing integrable at infinity functions.Indeed, the integral of the first function-which decreases itself-is finite due to the finiteness of the (γ + 1 − α) moment of τ .Concerning the second function, first notice that The right hand side is bounded by a decreasing integrable at infinity function due to the moment condition on τ and Lemma 18.So, ( 60) is proven which together with (58) completes the proof.
Proposition 9. Assume the rate of convergence (12).If both Eτ 1+γ and Eξ 1+γ are finite, then, for all k ≤ γ, where p 2 (x) is a decreasing integrable at infinity function and In addition, Proof.It follows from the definition of ξ(x) that Applying Lemma 8, we then obtain where By the definition of the γ, αγ > 1.The function 1/x αγ is integrable at infinity.The same arguments work for ξ, so the value of x −α(k−1) E{|ξ k (x)|; |ξ(x)| > x α } is bounded by a decreasing integrable at infinity function, and the proof is complete.Now we state the main result in this section.
We can determine all these numbers recursively.Indeed, as proven in Proposition 9, For r 1 defined defined in (62), for any choice of r 2 , r 3 , . . ., r γ−1 .Then we can choose r 2 such that the coefficient of x −2α is also zero, and so on.It is clear that the numbers r 1 , r 2 , . . ., r γ−1 do not depend on the parameter b.Therefore, we can take b so large that the function q(x) is decreasing on [0, ∞).
As in the previous section, we define For x < 0 we set U (x) = U (0).It is immediate from the definition of q(x) that We define also We further assume that, for all 1 ≤ k ≤ γ − 1, and If q(x) ∼ c/x α where γα < 2, then it follows from Lemma 24 that the condition on the derivatives of p(x) is always satisfied for a properly chosen function p, so the condition (66) on the derivatives of p does not restrict generality under this specific choice of r(x).
It is clear that Noting that and applying the L'Hôpital rule, we conclude that, as x → ∞, Lemma 11.As x → ∞, we have the following estimates: Proof.We start with the following decomposition: The third term on the right hand side is negative because U ± decreases and it may be bounded below as follows: due to the upper bound (61) which implies P{ξ(x) > x α } = o(p 1 (x)/x α ), and due to the relation (68).
Further, the first term on the right hand side of (71) is positive.To obtain an upper bound for that expectation we first notice that, due to the fact that Q(z) is monotone increasing, Since q(x) is chosen to be decreasing, Q(z) is concave and, consequently, Using this inequality we obtain and The moment assumption on ξ implies that the decreasing function E{ξe Q(ξ) ; ξ > x α } is integrable at infinity.As a result we have To estimate the second term on the right hand side of (71), we make use of Taylor's expansion with γ + 1 terms: where 0 ≤ θ = θ(x, ξ(x)) ≤ 1.By the construction of U ± , and, for k = 3, . . ., γ + 1, as x → ∞, where the remainder terms in the parentheses on the right are of order o(p(x)) by the conditions (66) and (67).By the definition of q ± (x), From these equalities we get x) , Combining this with (61), we obtain It follows from the equalities (75) and (76) that by the equality (65).Owing to the condition (66) on the derivatives of r(x) and (67), Then, the last term in (74) possesses the following bound: by the condition (67).Therefore, it follows from (74), ( 77) and (78) that Together with (72), (73), and (71) this completes the proof.
The remaining part of the proof repeats literally the final part of the proof of Theorem 5 and we omit it.

Heavy-tailed claim sizes
In this section we study the case where the distribution of the claim size is so heavy that the moment conditions in the theorems proved above are not met.
We assume that v(x) converges towards v c at rate θ/x and that the distribution of ξ is regularly varying at infinity with index −(β + 2) for some β ∈ (0, ρ).Then Eξ ρ+2 is infinite and, consequently, Theorem 5 does not apply.
Under the condition (79), by Karamata's theorem, Therefore, the claim of Theorem 12 can be reformulated in the following way: Notice that, for the classical ruin process with constant premium rate and with claim size of subexponential type, ψ(x) is asymptotically equivalent to the integral ∞ x P{ξ > y}dy (see e.g.[7,Section 5.11]).So, the main difference between our case and the classical one is that the probability of ruin is higher in our case owing to the additional weight y in the integral, which is not surprising and reflects the fact that our system is close to a critical one, v(y) → v c as y → ∞.
Notice that the condition (14) follows by (79).The proof of Theorem 12 is split into two parts, where we derive the upper and lower bounds.For both, we need the following result on the left tail distribution of the jumps of the chain {R n }.Proof.Using the equality ξ(x) = V x (τ ) − x − ξ, we get the following upper bound For a lower bound, let us notice that, for any fixed u, due to the long-tailedness of the distribution of ξ.Also, by the stochastic boundedness of the family of random variables {V which implies the following lower bound, uniformly for all x ≥ 0, hence the desired result.

Proof of the upper bound
As in the previous sections, we analyse the behaviour of the chain R n = R(T n ), n ≥ 0. In order to understand the impact of large claim sizes on ruin probabilities from the point of view of an upper bound, we introduce an auxiliary chain with jumps truncated below.For every x ≥ 0 we define jump ξ(x) as follows: Let { R n } be a Markov chain with jumps ξ(x).The connection between { R n } and {R n } is described in the next lemma.
Proof.We use the induction in n.If n = 1, then For the induction step n − 1 → n it suffices to apply the Markov property: which completes the proof.
This implies that To bound the second probability term on the right hand side of (81), we firstly apply the total probability law twice and then we apply again Lemma 14 to the probability on the right hand side: Plugging now this bound and (82) into (81), we get the desired upper bound.
In order to get an upper bound for ψ(x) we need upper bounds for both terms on the right hand side of (80).It turns out that ψ(x) can be estimated by the method used in the proof of Theorem 5.
Proof.Let U − (x) be the function defined in the proof of Theorem 5.By the definition of ξ(x), because P{ξ > y} = o(1/y 2 ) as y → ∞ due to Eξ 2 < ∞, and H x (0, y] = O(y 2 ) due to (84).Next, by Lemma 17, owing to the regular variation of the distribution of ξ and Karamata's theorem.Since β < ρ, we conclude that Together with (86) it yields that This estimate and Lemma 16 imply the desired upper bound.

Proof of the lower bound
To star with, we notice that, for all x > 0 and N ≥ 1, Consequently, putting N = δx 2 , we get the following lower bound for every δ > 0. Thus, it only remains to show that we can choose a δ > 0 so small that the probability on the right hand side is bounded away from zero.We start by stating the following decomposition It follows from Lemma 3 that for every ε < ρ there exists an x 0 such that for every β 0 < β, we infer that all the conditions of Lemma 1 in [6] hold true and, consequently, there exists an x 0 such that In particular, To bound the second probability on the right hand side of (87) we introduce a martingale Due to Lemma 3, we may assume that x 0 is so large that ym 1 (y) ≤ 2θEτ for all y ≥ x 0 .This implies that, for R 0 = x, on the event {R k ≥ x/2 for all k ≥ 1}.Consequently, where c 1 := 4θEτ .Applying the Doob inequality to the right hand side and noting that E x M 2 k ≤ c 2 k for all k and x, we obtain Plugging this estimate and (88) into (87), we conclude that Choosing δ > 0 sufficiently small, we can make the right hand side less than 1, hence inf This completes the proof of the lower bound.
Lemma 19.Let p(x) > 0 be a decreasing function which is integrable at infinity.Then there exists a decreasing integrable at infinity function p 1 (x) > 0 such that p 1 (x)/p(x) → ∞ as x → ∞.
Proof.Since p(x) is integrable at infinity, there exists an increasing sequence n k → ∞, k ≥ 0, such that x 0 = 0 and Define a continuous function p 1 (x) as follows: p 1 (x k ) := (k + 1)p(x k ), between x k and x k+1 we define p 1 (x) piece-wise linearly.Then the function p 1 (x) satisfies the condition p 1 (x)/p(x) → ∞ as x → ∞.Since p(x) decreases, the sequence x k may be chosen in such a way that (k + 2)p(x k+1 ) < (k + 1)p(x k ) for all k ≥ 1, which guarantees that the function p 1 (x) is decreasing.In addition, its integral may be bounded as follows: Lemma 20 (Denisov [5]).Let p(x) > 0 be a decreasing function which is integrable at infinity.Then there exists a decreasing integrable at infinity function p 1 (x) > 0 which dominates p(x) and is regularly varying at infinity with index −1.
Lemma 21.Let ξ ≥ 0 be a random variable and let V (x) ≥ 0 be an increasing function such that EV (ξ) < ∞.Let U (x) ≥ 0 be a function such that the function f (x) := V (x)/xU (x) increases and satisfies the condition Then there exists an increasing function s(x) → ∞ of order o(x) such that E{U (ξ); ξ > s(x)} = o(p(x)xU (x)/V (x)) as x → ∞, where p(x) is a decreasing integrable at infinity function which is only determined by ξ and V (x).
where the integral on the right hand side is integrable with respect to x, since Thirdly, dy k−1 . . . . . .
Since p(x) is decreasing and integrable at infinity, p(x) = O(1/x) as x → ∞, so p (k) k (x) = O(1/x 1+k ).Integrating the kth derivative k − j times we get that the jth derivative of p k (x) is not greater than (k − j)th integral of c/x 1+k which is of order O(1/x 1+j ).This completes the proof.