VMSTA Modern Stochastics: Theory and Applications 2351-6054 2351-6046 2351-6046 VTeXMokslininkų g. 2A, 08412 Vilnius, Lithuania VMSTA99 10.15559/18-VMSTA99 Research Article Exponential bounds for the tail probability of the supremum of an inhomogeneous random walk KievinaitėDominykad.kievinaite@gmail.com ŠiaulysJonasjonas.siaulys@mif.vu.lt∗1 Faculty of Mathematics and Informatics, Vilnius University, Naugarduko 24, Vilnius LT-03225, Lithuania Corresponding author.

The second author was supported by grant No S-MIP-17-72 from the Research Council of Lithuania.

 2018 153201852129143 13102017 1022018 1322018 © 2018 The Author(s). Published by VTeX2018 Open access article under the CC BY license.

Let {ξ1,ξ2,…} be a sequence of independent but not necessarily identically distributed random variables. In this paper, the sufficient conditions are found under which the tail probability P(supn⩾0∑i=1nξi>x) x)$]]> can be bounded above by ϱ1exp{−ϱ2x} with some positive constants ϱ1 and ϱ2 . A way to calculate these two constants is presented. The application of the derived bound is discussed and a Lundberg-type inequality is obtained for the ultimate ruin probability in the inhomogeneous renewal risk model satisfying the net profit condition on average.

 Exponential bound supremum of sums tail probability risk model inhomogeneity ruin probability Lundberg’s inequality 62E20 60E15 60G50 91B30 Introduction

Let {ξ1,ξ2,…} be a sequence of independent real-valued random variables (r.v.’s), and let M∞=supn⩾0{ ∑k=1nξk}. Here and subsequently, all empty sums are assumed to be zero.

Sgibnev [30] generalized results by Kiefer and Wolfowitz [21] obtaining the upper bound for the submultiplicative moment E(φ(M∞)) in the case of independent and identically distributed (i.i.d.) r.v.’s. In Theorem 2 of that paper, the following assertion is proved.

Let {ξ1,ξ2,…} be a sequence of i.i.d. r.v.’s with distribution function (d.f.) F, and let φ be a nondecreasing submultiplicative function defined on [0,∞) . Then E(φ(M∞))<∞ under the following conditions: ∙Eξ1<0,∙∫0∞φ(x)F‾(x)dx<∞,∙E(erξ1)<1ifr:=limx→∞logφ(x)x>0. 0.\end{array}\]]]>

Recall that a function φ defined on the interval [0,∞) is said to be submultiplicative if φ(0)=1andφ(x+y)⩽φ(x)φ(y)for allx,y∈[0,∞). Theorem 1 was applied several times to find the asymptotic behavior of the ruin probabilities in the homogeneous renewal risk models.

We say that the insurer’s surplus process R(t) varies according to the homogeneous renewal risk model if R(t)=u+pt− ∑i=1Θ(t)Zi,t⩾0, where:

u⩾0 denotes the initial insurer’s surplus;

p>0 0$]]> denotes a constant premium rate;

the claim sizes {Z1,Z2,…} form a sequence of i.i.d. nonnegative r.v.’s;

Θ(t)=∑n=1∞1{θ1+θ2+⋯+θn⩽t} is the renewal counting process generated by the inter-occurrence times {θ1,θ2,…} , which form another sequence of i.i.d. nonnegative and nondegenerate at 0 r.v.’s;

the sequences {Z1,Z2,…} and {θ1,θ2,…} are mutually independent.

The ultimate ruin probability ψ(u)=P(inft⩾0R(t)<0)=P(supn⩾1 ∑k=1n(Zk−pθk)>u) u\Bigg)\]]]> and the probability of ruin within time T ψ(u,T)=P(inf0⩽t⩽TR(t)<0)=P(sup1⩽n⩽Θ(T) ∑k=1n(Zk−pθk)>u) u\Bigg)\]]]> are the main characteristics of the renewal risk model.

The asymptotic behavior of ψ(u,T) was considered by Tang [31] when random claims in the homogeneous model have d.f. with consistently varying tail. The author of the paper uses the assertion of Theorem 1 for function φ(x)=(1+x)q with q>0 0$]]> to get the main term of the asymptotics for the probability ψ(u,T) . Leipus and Šiaulys considered the asymptotic behavior of ψ(u,T) in [22, 23] but for subexponentially distributed r.v.’s {Z1,Z2,…} . In their proofs, the assertion of Theorem 1 was used for function φ(x)=exp(ρx) with some ρ>0 0$]]> (see Lemma 3.3 in [22] and Lemma 2.1 in [23]). In the case of exponential function, Theorem 1 implies the following assertion.

Let {ξ1,ξ2,…} be a sequence of i.i.d. r.v.’s. If Eξ1<0 and Eehξ1<∞ for some positive h, then there exists a positive constant ϱ such that eϱxP(M∞>x)→x→∞0. x)\underset{x\to \infty }{\to }0.\]]]>

The Sgibnev’s proof of Theorem 1 is substantially related to the techniques of Banach algebras, while Corollary 1 can be derived using only the probabilistic approach. Wang et al. (see Lemma 4.4 in [32]) demonstrated such a probabilistic way to obtain the assertion of Corollary 1 supposing, in addition, that r.v.’s {ξ1,ξ2,…} follow some dependence structure. Corollary 1 can be applied not only as auxiliary assertion in the consideration of the asymptotic behavior of ψ(u,T) . The assertion of Corollary 1 is closely related to the following statement on the upper bound for ψ(u) in the homogeneous renewal risk model.

Let the claim sizes {Z1,Z2,…} and the inter-occurrence times {θ1,θ2,…} form a homogeneous renewal risk model. Let, in addition, the net profit condition EZ1−pEθ1<0 hold and EehZ1<∞ for some positive h. Then, there exists a positive H such that ψ(u)⩽e−Hu,u⩾0. If EeR(Z1−pθ1)=1 for some positive R, then we can take H=R in (3).

The above assertion is the well-known Lundberg inequality. There exist a lot of different proofs of this inequality. For example, some of the proofs can be found in [4], [14], [15], [16], [25]. The existing proofs of Lundberg’s inequality are essentially based on the renewal idea. However, the classical methods used for consideration of the ruin probability in the homogeneous renewal risk model are not applicable in the case of the inhomogeneous model because at any time moment distribution of the future is completely new.

Another way to derive the Lundberg inequality is related to Theorem 1. Namely, the first part of Theorem 2 follows from Theorem 1 and the additional inequality ψ(0)<1 . We use this approach to get the inequality similar to the Lundberg inequality but for the inhomogeneous renewal risk model with not necessarily identically distributed claim sizes {Z1,Z2,…} and the inter-occurrence times {θ1,θ2,…} .

In this paper, we consider a sequence of independent but not necessarily identically distributed r.v.’s {ξ1,ξ2,…} . We obtain an assertion similar to that in Corollary 1. We present an algorithm to get the numerical values of the two positive constants in the exponential bound for P(M∞>x) x)$]]> in the case of not necessarily identically distributed r.v.’s {ξ1,ξ2,…} . We apply the obtained estimate to derive two Lundberg-type inequalities similar to that in Theorem 2 but for the inhomogeneous renewal risk model with possibly nonidentically distributed random claim amounts Z1,Z2,… . Corollaries 2 and 3 below show that the exponential bound for the ruin probability in the inhomogeneous renewal risk model holds if the model satisfies the net profit condition on average. This means that the quantity 1n ∑k=1nE(Zk−pθk) is negative for all sufficiently large n.

The results of the present paper are complementary to those obtained by Albrecher et al. [1], Ambagaspitiya [2], Bernackaitė and Šiaulys [5, 6], Castañer et al. [7], Cojocaru [9], Constantinescu et al. [10], Czarna and Palmowski [11], Damarackas and Šiaulys [12], Danilenko et al. [13], Grigutis et al. [17, 18], Jordanova and Stehlík [20], Mishura et al. [24], Răducan et al. [26–28], Ragulina [29], Zhang et al. [33], Zhang et al. [34] and other authors who dealt with different inhomogeneous risk models.

The rest of the paper is organized as follows. In Section 2, we present our main result together with its proof. In Section 3, we recall the concept of the inhomogeneous renewal risk model and we present two corollaries from the main theorem, which yield exponential bounds for the ruin probability in this model. Finally, in Section 4, we present some examples which show the applicability of the theorem and corollaries.

 Main result

In this section, we formulate and prove our main result. The assertion below is a generalization of Lemma 1 by Andrulytė et al. [3]. In that lemma, the exponential bound for P(M∞>x) x)$]]> was established under more restrictive conditions. In addition, the assertion below provides an algorithm to calculate two positive constants establishing this exponential bound. For this reason, the conditions of the main theorem are formulated in an explicit form in contrast to the conditions of Lemma 1 in [3]. It should be noted that the presented proof of the main result has some similarities with the classical approach by Chernoff [8] and Hoeffding [19].

Let {ξ1,ξ2,…} be independent r.v.’s such that: (i)1n ∑i=1nEξi⩽−aifn⩾b,(ii)supn⩾b1n ∑i=1nE(|ξi|1{ξi⩽−c})⩽ε,(iii)supn⩾b1n ∑i=1n(P(ξi⩽0)+E(ehξi1{ξi>0}))⩽d1,(iv)max1⩽n⩽b−11n ∑i=1n(P(ξi⩽0)+E(ehξi1{ξi>0}))⩽d2, 0\}}}\big)\big)\leqslant {d_{1}},\\{} & \displaystyle (\text{iv})& \displaystyle \underset{1\leqslant n\leqslant b-1}{\max }\hspace{0.1667em}\frac{1}{n}{\sum \limits_{i=1}^{n}}\big(\mathbb{P}({\xi _{i}}\leqslant 0)+\mathbb{E}\big({\mathrm{e}}^{h{\xi _{i}}}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\big)\big)\leqslant {d_{2}},\hspace{2.5pt}\end{array}\]]]> for some a>0 0$]]>, b∈N , c>0 0$]]>, ε⩾0 , h>0 0$]]>, d1⩾1 and d2⩾1 .

If −Δ:=ε+δhd1max{c22,2h2}−a<0, with some δ∈(0,1/2] , then P(supn⩾0 ∑i=1nξi>x)⩽min{1,c1e−δhx},x⩾0, x\Bigg)\leqslant \min \big\{1,\hspace{0.1667em}{c_{1}}{\mathrm{e}}^{-\delta hx}\big\},\hspace{1em}x\geqslant 0,\]]]> where c1=(S(b,d2)+exp{−δhΔb}1−exp{−δhΔ}), with S(b,d2)=d2d2b−1−1d2−1ifd2>1,b−1ifd2=1. 1,\hspace{1em}\\{} b-1\hspace{1em}\textit{if}\hspace{2.5pt}{d_{2}}=1.\hspace{1em}\end{array}\right.\]]]>

We observe that, for all x⩾0 , P(x):=P(supn⩾0 ∑i=1nξi>x)=P( ⋃n=1∞{ ∑i=1nξi>x})⩽ ∑n=1∞P( ∑i=1nξi>x). x\Bigg)=\mathbb{P}\Bigg({\bigcup \limits_{n=1}^{\infty }}\Bigg\{{\sum \limits_{i=1}^{n}}{\xi _{i}}>x\Bigg\}\Bigg)\\{} & \leqslant {\sum \limits_{n=1}^{\infty }}\mathbb{P}\Bigg({\sum \limits_{i=1}^{n}}{\xi _{i}}>x\Bigg).\end{aligned}\]]]>

Since r.v.’s {ξ1,ξ2,…} are independent, by the exponential Chebyshev inequality, we get P(x)⩽e−yx ∑n=1∞ ∏i=1nEeyξi=e−yx ∑n=1b−1 ∏i=1nEeyξi+e−yx ∑n=b∞ ∏i=1nEeyξi:=P1(x)+P2(x), for all x⩾0 and 0<y⩽h .

For all i∈N , we have Eeyξi=1+yEξi+E((eyξi−1)1{ξi⩽−c})−yE(ξi1{ξi⩽−c})+E((eyξi−1−yξi)1{−c<ξi⩽0})+E((eyξi−1−yξi)1{ξi>0}). 0\}}}\big).\end{aligned}\]]]>

It is obvious that ev−1⩽0ifv⩽0,ev−v−1⩽v22ifv⩽0,ev−v−1⩽v22evifv⩾0, and v2⩽ev for nonnegative v. Using these inequalities we get Eeyξi⩽1+yEξi+yE(|ξi|1{ξi⩽−c})+y22E(ξi21{−c<ξi⩽0})+y22E(ξi2eyξi1{ξi>0})⩽1+yEξi+yE(|ξi|1{ξi⩽−c})+y2c22P(ξi⩽0)+2y2h2E(ehξi1{ξi>0}), 0\}}}\big)\\{} & \leqslant 1+y\mathbb{E}{\xi _{i}}+y\hspace{0.1667em}\mathbb{E}\big(|{\xi _{i}}|{\mathbb{1}_{\{{\xi _{i}}\leqslant -c\}}}\big)+\frac{{y}^{2}{c}^{2}}{2}\hspace{0.1667em}\mathbb{P}({\xi _{i}}\leqslant 0)\\{} & \hspace{1em}+\frac{2{y}^{2}}{{h}^{2}}\hspace{0.1667em}\mathbb{E}\big({\mathrm{e}}^{h{\xi _{i}}}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\big),\end{aligned}\]]]> if 0<y⩽h/2 , because E((hξi2)2eyξi1{ξi>0})⩽E((hξi2)2ehξi/21{ξi>0})⩽E(ehξi1{ξi>0}), 0\}}}\bigg)\leqslant \mathbb{E}\bigg({\bigg(\frac{h{\xi _{i}}}{2}\bigg)}^{2}{\mathrm{e}}^{h{\xi _{i}}/2}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\bigg)\leqslant \mathbb{E}\big({\mathrm{e}}^{h{\xi _{i}}}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\big),\]]]> in this case.

If n⩾b , then conditions (ii), (iii) and relation (6) together with the inequality 1+u⩽eu , u∈R , imply that ∏i=1nEeyξi⩽ ∏i=1n(1+yEξi+yE(|ξi|1{ξi⩽−c})+y2c22P(ξi⩽0)+2y2h2E(ehξi1{ξi>0}))⩽exp{y ∑i=1nEξi+nysupn⩾b1n ∑i=1nE(|ξi|1{ξi⩽−c})+ny2max{c22,2h2}supn⩾b1n ∑i=1n(P(ξi⩽0)+E(ehξi1{ξi>0}))}⩽exp{y ∑i=1nEξi+nyε+ny2d1max{c22,2h2}}. 0\}}}\big)\bigg)\\{} & \leqslant \exp \Bigg\{y{\sum \limits_{i=1}^{n}}\mathbb{E}{\xi _{i}}+ny\hspace{0.1667em}\underset{n\geqslant b}{\sup }\frac{1}{n}{\sum \limits_{i=1}^{n}}\mathbb{E}\big(|{\xi _{i}}|{\mathbb{1}_{\{{\xi _{i}}\leqslant -c\}}}\big)\\{} & \hspace{1em}+n{y}^{2}\max \bigg\{\frac{{c}^{2}}{2},\frac{2}{{h}^{2}}\bigg\}\underset{n\geqslant b}{\sup }\frac{1}{n}{\sum \limits_{i=1}^{n}}\big(\mathbb{P}({\xi _{i}}\leqslant 0)+\mathbb{E}\big({\mathrm{e}}^{h{\xi _{i}}}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\big)\big)\Bigg\}\\{} & \leqslant \exp \Bigg\{y{\sum \limits_{i=1}^{n}}\mathbb{E}{\xi _{i}}+ny\varepsilon +n{y}^{2}{d_{1}}\max \bigg\{\frac{{c}^{2}}{2},\frac{2}{{h}^{2}}\bigg\}\Bigg\}.\end{aligned}\]]]>

Hence, by condition (i) P2(x)⩽e−yx ∑n=b∞exp{ny(−a+ε+yd1max{c22,2h2})}, for all x⩾0 and 0<y⩽h/2 .

If n⩽b−1 and 0<y⩽h then, due to the condition (iv), we have 1n ∑i=1nEeyξi=1n ∑i=1n(E(eyξi1{ξi⩽0})+E(eyξi1{ξi>0}))⩽d2. 0\}}}\big)\big)\leqslant {d_{2}}.\]]]> Therefore, P1(x)⩽e−yx ∑n=1b−1d2n, because ∏i=1nEeyξi⩽(1n ∑i=1nEeyξi)n, by the inequality of arithmetic and geometric means.

Equality (5) and inequalities (7), (8) imply that P(x)⩽e−yx(S(b,d2)+ ∑n=b∞(exp{y(−a+ε+yd1max{c22,2h2})})n), for all x⩾0 and 0<y⩽h/2 .

Let now y=δh with some δ∈(0,1/2] satisfying condition (4). For such y, we obtain from (9) that P(x)⩽e−δhx(S(b,d2)+exp{−δhΔb}1−exp{−δhΔ}).

This is the desired inequality. The theorem is proved.  □

 Lundberg-type inequalities

In this section, we present two corollaries from Theorem 3, which yield the Lundberg-type inequalities for the inhomogeneous renewal risk model.

We say that the insurer’s surplus process R(t) varies according to the inhomogeneous renewal risk model if equality (1) holds for all t⩾0 with the initial insurer’s surplus u⩾0 , a constant premium rate p>0 0$]]>, a sequence of independent nonnegative and not necessarily identically distributed claim amounts {Z1,Z2,…} and with the renewal counting process Θ(t) generated by the inter-occurrence times {θ1,θ2,…} , which form a sequence of independent nonnegative nondegenerate at zero and possibly not identically distributed r.v.’s. In addition, sequences {Z1,Z2,…} and {θ1,θ2,…} are supposed to be independent.

It is obvious that definitions and expressions of the ruin probabilities ψ(u) and ψ(u,T) for the inhomogeneous renewal risk model remain the same as those given in Section 1.

The main requirement to get the Lundberg-type bounds for ψ(u) is the net profit condition. In both assertions below, it is supposed that this condition holds on average. Our first corollary follows immediately from Theorem 3 and representation (2).

Let us consider the inhomogeneous renewal risk model such that r.v.’s ξk=Zk−pθk , k∈N , satisfy conditions (i)–(iv) of Theorem 3. Then the ruin probability in the model satisfies the following inequality ψ(u)⩽min{1,c1e−δhu},u⩾0, where constants h>0 0$]]>, δ∈(0,1/2] and c1>0 0$]]> are the same as in Theorem 3 for the sequence {ξ1=Z1−pθ1,ξ2=Z2−pθ2,…} .

Our second corollary is more convenient to use because the requirements are formulated separately for r.v.’s {Z1,Z2,…} and {θ1,θ2,…} in it. We present the corollary below together with a short proof.

Let the inhomogeneous renewal risk model with a sequence of random claim amounts {Z1,Z2,…} , a sequence of random inter-occurrence times {θ1,θ2,…} and premium rate p satisfy the following additional requirements (i)1n ∑i=1n(EZi−pEθi)⩽−αifn⩾β,(ii)supn⩾β1n ∑i=1nE(θi1{θi⩾ϰp})⩽ϵ,(iii)supn⩾β1n ∑i=1nEeγZi⩽ν1,(iv)max1⩽n⩽β−11n ∑i=1nEeγZi⩽ν2, for some α>0 0$]]>, β∈N , ϰ>0 0$]]>, ϵ⩾0 , γ>0 0$]]>, ν1⩾1 and ν2⩾1 .

If −Δˆ:=pϵ+δγ(1+ν1)max{ϰ22,2γ2}−α<0, for some δ∈(0,1/2] , then ψ(u)⩽min{1,c2e−δγu},u⩾0, with the positive constant c2=(1+ν2ν2((1+ν2)b−1−1)+exp{−δγΔˆβ}1−exp{−δγΔˆ}).

Let ξi=Zi−pθi for all i∈N . Then obviously 1n ∑i=1nEξi⩽−αifn⩾β, by condition (i) of the corollary.

Further, by conditions (iii) and (iv), we have supn⩾β1n ∑i=1nE(eγξi1{ξi>0})⩽supn⩾β1n ∑i=1nEeγZi⩽ν1, 0\}}}\big)& \leqslant \underset{n\geqslant \beta }{\sup }\frac{1}{n}{\sum \limits_{i=1}^{n}}\mathbb{E}{\mathrm{e}}^{\gamma {Z_{i}}}\leqslant {\nu _{1}},\end{aligned}\]]]> max1⩽n⩽β−11n ∑i=1nE(eγξi1{ξi>0})⩽ν2, 0\}}}\big)& \leqslant {\nu _{2}},\end{aligned}\]]]> because of the nonnegativity of the inter-occurrence times θi,i∈N .

For the use of Theorem 3, it remains to estimate supn⩾β1n ∑i=1nE(|ξi|1{ξi⩽−c}), for some suitable positive c.

Choosing c=ϰ we get supn⩾β1n ∑i=1nE(|ξi|1{ξi⩽−ϰ})⩽psupn⩾β1n ∑i=1nE(θi1{Zi−pθi⩽−ϰ})=psupn⩾11n ∑i=1nE(θi1{θi⩾1p(Zi+ϰ)})⩽psupn⩾11n ∑i=1nE(θi1{θi⩾ϰp})⩽pϵ.

The obtained inequalities (10), (11), (12) and (13) imply that r.v.’s {ξ1,ξ2,…} satisfy conditions (i)–(iv) of Theorem 3 with a=α,b=β,c=ϰ,h=γ,d1=1+ν1,d2=1+ν2andε=pϵ. The assertion of the corollary follows now from Theorem 3.  □

 Examples

In this section, we present four examples. The first two examples show the applicability of Theorem 3. The third example demonstrates how to get the exponential bound for the ruin probability applying Corollary 3. The last example shows that the Lundberg-type inequality of the form ψ(u)⩽ϱ1e−ϱ2u , u⩾0 , with ϱ1=1 and a positive constant ϱ2 is impossible if the inhomogeneous renewal risk model is “good” only on average.

Suppose that {ξ1,ξ2,…} are independent r.v.’s such that:

ξi are uniformly distributed on interval [0,2] for all i≡1mod3 ;

ξi are uniformly distributed on interval [−2,0] for all i≡2mod3 ;

F‾ξi(x)=1(−∞,−2)(x)+e−x−21[−2,∞)(x) if i≡0mod3 .

We can see that the presented sequence {ξ1,ξ2,…} has three subsequences. Two of them generate random walks with negative drifts, and one subsequence generates random walk with a positive drift. Fortunately, sequence {ξ1,ξ2,…} has a negative drift on average. Therefore, we can use Theorem 3 to get the exponential bound for P(M∞>x) x)$]]>.

It is evident that Eξi=1ifi≡1mod3,−1ifi≡2mod3ori≡0mod3. Therefore, after some calculations, we get 1n ∑i=1nEξi⩽−17forn⩾7. Additionally, supn⩾11n ∑i=1nE(|ξi|1{ξi⩽−2})=0 and E(e45ξi1{ξi>0})=58(e8/5−1)<2.48ifi≡1mod3,0ifi≡2mod3,5/e2<0.68ifi≡0mod3. 0\}}}\big)=\left\{\begin{array}{l@{\hskip10.0pt}l}\frac{5}{8}({\text{e}}^{8/5}-1)<2.48\hspace{1em}& \text{if}\hspace{2.5pt}\hspace{2.5pt}i\hspace{0.1667em}\equiv \hspace{0.1667em}1\hspace{0.1667em}\text{mod}\hspace{0.1667em}3,\\{} 0\hspace{1em}& \text{if}\hspace{2.5pt}\hspace{2.5pt}i\hspace{0.1667em}\equiv \hspace{0.1667em}2\hspace{0.1667em}\text{mod}\hspace{0.1667em}3,\\{} 5/{\text{e}}^{2}<0.68\hspace{1em}& \text{if}\hspace{2.5pt}\hspace{2.5pt}i\hspace{0.1667em}\equiv \hspace{0.1667em}0\hspace{0.1667em}\text{mod}\hspace{0.1667em}3.\end{array}\right.\]]]> The last expression implies supn⩾71n ∑i=1n(P(ξi⩽0)+E(e45ξi1{ξi>0}))<1.79, 0\}}}\big)\big)<1.79,\end{aligned}\]]]> max1⩽n⩽61n ∑i=1n(P(ξi⩽0)+E(e45ξi1{ξi>0}))<2.48. 0\}}}\big)\big)<2.48.\end{aligned}\]]]>

By (14)–(17), we conclude that conditions of Theorem 3 hold with a=1/7 , b=7 , c=2 , ε=0 , h=4/5 , d1=1.8 and d2=2.5 . Therefore, −Δ=ε+δhd1max{c22,2h2}−a=92δ−17=−114, if δ=1/63 . It follows now from Theorem 3 that P(M∞>x)⩽min{1,(d2d26−1d2−1+exp{−δhΔb}1−exp{−δhΔ})e−δhx}⩽min{1,1502exp{−0.01269x}}, x)& \leqslant \min \bigg\{1,\hspace{0.1667em}\bigg({d_{2}}\frac{{d_{2}^{\hspace{0.1667em}6}}-1}{{d_{2}}-1}+\frac{\exp \{-\delta h\varDelta b\}}{1-\exp \{-\delta h\varDelta \}}\bigg)\hspace{0.1667em}{\mathrm{e}}^{-\delta hx}\bigg\}\\{} & \leqslant \min \big\{1,\hspace{0.1667em}1502\exp \{-0.01269x\}\big\},\end{aligned}\]]]> for all positive x.

The last inequality works if x⩾578 . Though the obtained bound is not as good as one would prefer, it is still exponential. Its weakest point is the large constant before the main term. By Theorem 3, the value of this constant is closely related to the behavior of the first elements of the sequence {ξ1,ξ2,…} . In our case, the first elements of {ξ1,ξ2,…} increase this constant because the subsequence {ξ1,ξ4,ξ7,…} has a positive drift. The second example shows that the better exponential bound can be obtained from Theorem 3 in the case when only some of the r.v.’s {ξ1,ξ2,…} drag the model to the positive side.

Suppose that {ξ1,ξ2,…} are independent r.v.’s such that: P(ξi=−1)=1−1i+1andP(ξi=1)=1i+1fori∈{1,2,…}.

For all i⩾1 , we have Eξi=2i+1−1andE(eξi1{ξi>0})=ei+1. 0\}}}\big)=\frac{\mathrm{e}}{i+1}.\]]]> Consequently, 1n ∑i=1nEξi⩽−518ifn⩾3,supn⩾31n ∑i=1n(P(ξi⩽0)+E(eξi1{ξi>0}))=13e+2336<1.621,max1⩽n⩽61n ∑i=1n(P(ξi⩽0)+E(eξi1{ξi>0}))=e+12<1.86. 0\}}}\big)\big)& =\frac{13\mathrm{e}+23}{36}<1.621,\\{} \underset{1\leqslant n\leqslant 6}{\max }\frac{1}{n}{\sum \limits_{i=1}^{n}}\big(\mathbb{P}({\xi _{i}}\leqslant 0)+\mathbb{E}\big({\text{e}}^{{\xi _{i}}}{\mathbb{1}_{\{{\xi _{i}}>0\}}}\big)\big)& =\frac{\mathrm{e}+1}{2}<1.86.\end{aligned}\]]]>

Due to the derived bounds conditions of Theorem 3 hold with a=5/18,b=3,c=11/10,ε=0,h=1,d1=1.625andd2=e+12. In this case, we have −Δ=ε+δhd1max{c22,2h2}−a=134δ−518<0, for δ<10/117 .

If we chose δ=1/20 , then by Theorem 3, we have P(M∞>x)⩽min{1,178exp{−x/20}},x⩾0. x)\leqslant \min \big\{1,\hspace{0.1667em}178\exp \{-x/20\}\big\},\hspace{1em}x\geqslant 0.\]]]>

As was stated before, the next example shows the possibility of the exponential bound for the ruin probability in the case when the inhomogeneous renewal risk model satisfies net profit condition on average.

Let us consider the inhomogeneous risk model which is generated by uniformly distributed on [1,3] inter-occurrence times θ1,θ2,… , constant premium rate p=2 and a sequence of the claim amounts {Z1,Z2,…} such that Z1=Z2=0 , Z3=Z4=4 with probability 1 and F‾Zi(x)=1(−∞,0)(x)+e−x(1+xi)1[0,∞)(x),i⩾5.

In this case, we have EZi=1+1i,i⩾5,EeγZi=i−1i11−γ+1i1(1−γ)2,i⩾5,γ∈(0,1). Consequently, 1n ∑i=1n(EZi−pEθi)⩽−2ifn⩾1 and supn⩾11n ∑i=1nE(eZi/3)⩽2.4.

The obtained inequalities imply conditions of Corollary 3 with α=2,β=1,ϰ=6,ϵ=0,γ=13,ν1=2.4,ν2=1. Therefore, for the described model, −Δˆ=1025δ−2<0, if δ<10/102 .

If we choose δ=9/102 , then Δˆ=1/5 , and the assertion of Corollary 3 implies the following Lundberg-type inequality for the model ψ(u)⩽min{1,170e−0.029u},u⩾0.

If we choose δ=5/102 , then Δˆ=1 , and Corollary 3 implies that ψ(u)⩽min{1,61e−0.016u},u⩾0.

It is clear that we can get a lot of different Lundberg-type inequalities for the same model because there exist infinitely many collections of constants {α,β,ϰ,ϵ,γ,ν1,ν2,δ} satisfying conditions of Corollary 3. It follows from the construction of the bound in Corollary 3 that we get the better bound for the smaller constants β,ϰ,ϵ,ν1,ν2 and for larger constants α,γ . If the collection of the constants {α,β,ϰ,ϵ,γ,ν1,ν2} is quite “unfriendly”, then we can still get an exponential bound for the ruin probability but with unsatisfiably small δ. All possible exponential bounds for the ruin probability have the form ϱ1exp{−ϱ2u} with some positive constants ϱ1 and ϱ2 . Theorem 2 shows that ϱ1=1 in the case of the homogeneous renewal risk model satisfying the net profit condition. If the net profit condition holds on average (see condition (i) of Corollaries 2 or 3), then it is impossible to get the exponential bound for ψ(u) with ϱ1=1 in general. The following simple example confirms this.

Let us consider the inhomogeneous risk model with p=1 such that Z1=Z2=10 , Zi=0 for i⩾3 , and θi=1 for i⩾1 almost surely.

The model under consideration is inhomogeneous but satisfies the net profit condition on average and all other conditions of, for instance, Corollary 3. On the other hand, the model is degenerate. It is not difficult to obtain the exact values of the ruin probability. Namely, expression (2) implies that ψ(u)=1if0⩽u<18,ψ(u)=0ifu⩾18. We can see the graph of the function ψ in Figure 1. All the best possible exponential bounds for ψ(u) must go through the point A(18,1) (see colored curves in Figure 1). Therefore, the upper bounds of the form ϱ1exp{−ϱ2u} should satisfy the condition ϱ1e−18ϱ2⩾1.

Hence, it is evident that ϱ1⩾e18ϱ2>1. 1.\]]]>

Ruin probability in the model of Example 4

 Concluding remarks

In the paper, the problem of the estimating of the ruin probability for the inhomogeneous renewal risk models is considered. It is evident that this problem is closely related to the bounds for the tail probability of the supremum of an inhomogeneous random walk. The upper bound of the exponential type ϱ1e−ϱ2u is derived for the renewal risk models satisfying the net profit condition on average. The positive constants ϱ1 and ϱ2 depend on the constants describing the model. Unfortunately, the obtained estimates are not sharp enough. We guess that it is possible to get sharper exponential bounds for the ruin probabilities but for narrower class of the inhomogeneous renewal risk models.

 Acknowledgments

We would like to thank three anonymous referees for the detailed and helpful comments on the previous versions of the manuscript.

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