Precise large deviation results for the total claim
amount under subexponential claim sizes
Aleksandras Baltrūnas, Remigijus Leipus, Jonas Šiaulys
To cite this version:
Aleksandras Baltrūnas, Remigijus Leipus, Jonas Šiaulys. Precise large deviation results for
the total claim amount under subexponential claim sizes. Statistics and Probability Letters,
Elsevier, 2010, 78 (10), pp.1206. .
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Accepted Manuscript
Precise large deviation results for the total claim amount under
subexponential claim sizes
Aleksandras Baltrūnas, Remigijus Leipus, Jonas Šiaulys
PII:
DOI:
Reference:
S0167-7152(07)00394-X
10.1016/j.spl.2007.11.016
STAPRO 4826
To appear in:
Statistics and Probability Letters
Received date: 26 July 2005
Revised date: 5 September 2007
Accepted date: 6 November 2007
Please cite this article as: Baltrūnas, A., Leipus, R., Šiaulys, J., Precise large deviation results
for the total claim amount under subexponential claim sizes. Statistics and Probability Letters
(2007), doi:10.1016/j.spl.2007.11.016
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Precise large deviation results for the total claim amount under
subexponential claim sizes
1
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Aleksandras Baltrūnas2 , Remigijus Leipus1,2 and Jonas Šiaulys1,2
Department of Mathematics and Informatics, Vilnius University,
Naugarduko 24, Vilnius LT-03225, Lithuania
2
Institute of Mathematics and Informatics, Akademijos 4, Vilnius LT-08663, Lithuania
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September 5, 2007
Abstract
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The paper deals with the renewal risk model. A precise large deviation result in the case
of subexponential claim sizes is proved. As a special case, the example of Pareto distributed
claim sizes and inter-occurence times is investigated.
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Keywords: renewal risk model; subexponential distribution; large deviations
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Corresponding author: Remigijus Leipus, Department of Mathematics and Informatics, Vilnius University, Naugarduko 24, Vilnius LT-03225, Lithuania; email: [email protected]
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Introduction
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In this paper we investigate the precise large deviations for the renewal risk model (total claim
amount process), having the following structure:
Assumption H1 The claim sizes Z1 , Z2 , . . . form a sequence of independent identically distributed (i.i.d.) nonnegative random variables with a common distribution function (d.f.)
B(u) = P(Z1 ≤ u), which has a finite mean a = EZ1 and a finite second moment EZ12 < ∞.
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Assumption H2 The inter-occurrence times θ1 = T1 , θ2 = T2 − T1 , θ3 = T3 − T2 , . . . are i.i.d.
non-negative random variables with mean 1/λ and finite second moment Eθ12 < ∞. In addition,
θ1 , θ2 , . . . are mutually independent of Z1 , Z2 , . . ..
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In the special case, where θ1 , θ2 , . . . have exponential distribution, this model is called a
Poisson model.
Pk
The random variables Tk =
i=1 θi , k = 1, 2, . . . constitute a renewal counting process
N (t) = #{k = 1, 2, . . . : Tk ∈ (0, t]}, t ≥ 0 with a mean function λ(t) = EN (t), for which
P
λ(t) ∼ λt as t → ∞. Define a random walk process Sn = nk=1 Zk , n ≥ 1, S0 = 0.
We are interested in a precise large deviation result for random sums (total claim amount
process) SN (t) under the assumption that claim sizes’ distribution B is heavy-tailed. A natural
class of heavy-tailed distributions is the class of subexponential distributions. Recall that the
d.f. B(x) on [0, ∞) is called subexponential and is denoted B ∈ S ) if the tail B = 1 − B satisfies
equality
lim B ∗ B(u)/B(u) = 2,
u→∞
where B ∗ B denotes the Stieltjes convolution of B with itself. The precise large deviations for
random sums SN (t) in special cases of S were studied in Klüppelberg and Mikosch (1997), Tang
et al. (2001) (the tails B are of extended regular variation), Ng et al. (2004) (B has a consistent
variation). They prove that, under corresponding regularity conditions, it holds
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P(SN (t) > x + aλ(t)) ∼ λB(x),
t→∞
uniformly for x ≥ γλ(t) for every γ > 0, i.e.
(1.1)
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¯ P(S
¯
N (t) > x + µλ(t))
¯
¯
− 1¯ = 0.
lim sup ¯
t→∞ x≥γλ(t)
λ(t)B(x)
For applications of the precise deviation results in insurance and finance see, e.g., Klüppelberg
and Mikosch (1997), Mikosch and Nagaev (1998), among others.
In our paper we prove that, under mild additional assumptions on the subexponential d.f.
of the claim size Z1 and on the d.f. of the inter-occurrence time θ1 , for every nonnegative µ ≥ 0,
the relation
P(SN (t) > x + (a + µ)λt) ∼ λtB(x + µλt), x → ∞
(1.2)
holds uniformly for all t ∈ [f (x), γx/Q(x)] in case µ > 0, and for all t ∈ [f1 (x), o(x/Q(x))]
in case µ = 0, where f (x), f1 (x) are arbitrary infinitely increasing functions and γ > 0 is an
arbitrary positive constant. Note that, in general, the zones of uniform convergence in (1.1) and
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(1.2) are different. The relation in (1.2) with respect to x → ∞ is more natural for studying
the asymptotics of the finite time ruin probabilities, where the initial capital of an insurance
company, x, tends to infinity.
The rest of the paper is organized as follows. Additional assumptions on the distribution B
and on the renewal process N (t), together with the main theorem are formulated in Section 2.
The proof of the theorem is given in Section 3. An example of the Pareto distributed claim sizes
and inter-occurence times is presented in Section 4.
Additional assumptions and main result
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To formulate our main result we need to introduce some additional notations and assumptions.
Let Q(u) = − log B(u), u ∈ R+ be the hazard function of distribution B. We assume also
Ru
that there exists a non-negative function q : R+ → R+ such that Q(u) = 0 q(v)dv, u ∈ R+ .
The function q is called the hazard rate of d.f. B. Denote by
r := lim sup uq(u)/Q(u)
u→∞
(2.1)
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a hazard ratio index.
The next two assumptions A and B are essential for our purposes.
Assumption A The distribution B is subexponential and satisfies the following conditions:
(
2
if r = 0,
r < 1/2; lim inf uq(u) ≥
u→∞
4/(1 − r) if r 6= 0.
Assumption B For any positive δ > 0 there exists a positive ² > 0 such that
X
P(N (t) ≥ k)(1 + ²)k → 0 as t → ∞.
(2.2)
k>(1+δ)λt
X
P(N (t) ≥ k)(1 + ²)k
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k>(1+δ)λt
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Remark 2.1 Condition (2.2) is satisfied for the Poisson model. Indeed, if N (t) is the homogeneous Poisson process with intensity λ, then
≤
≤
¿
X
k>(1+δ)λt
´
1
1
(λ(1 + ²)t)k ³
1+
+
+
.
.
.
k!
1 + δ (1 + δ)2
´
1 + δ (λ(1 + ²)t)[(1+δ)λt]+1 ³
1 + ² ³ 1 + ² ´2
1+
+
+ ...
δ
([(1 + δ)λt] + 1)!
1+δ
1+δ
½³
¾
´
2
(1 + δ)
1+²
1
exp
δ + (1 + δ) ln
λt p
,
δ(δ − ²)
1+δ
λt(1 + δ)
if ² < δ. In general case, the verification of Assumption B is more complicated. As conjectured
by Klüppelberg and Mikosch (1997) (see Lemma 2.3 therein) Assumption B can be satisfied if
the following stochastic ordering relation holds:
P(θ1 ≤ u) ≤ P(E1 ≤ u),
3
u ∈ R+ ,
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for some exponential random variable E1 , implying
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P(N (t) ≥ k) = P(θ1 + . . . + θk ≤ t) ≤ P(E1 + . . . + Ek ≤ t) = P(Ñ (t) ≥ k),
(2.3)
Obviously,
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where Ñ (t), t ≥ 0 is a homogenous Poisson process. However, the last inequality is not sufficient
for Assumption B to hold. For example, take i.i.d. Pareto(2) random variables θ1 , θ2 , . . ., i.e.
³ 1 ´2
, u ≥ 0.
P (θ1 > u) =
1+u
P (θ1 ≤ u) = 1 − e−2 log(1+u) ≤ 1 − e−2u = P (E1 ≤ u),
where E1 ∼ Exp(2). Then inequality (2.3) reads as
´
(2t)k+1
+ ... .
(k + 1)!
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³ (2t)k
u ≥ 0,
P(N (t) ≥ k) ≤ P(Ñ (t) ≥ k) = e−2t
k!
+
(2.4)
Since λ = (Eθ1 )−1 = 1, for fixed 0 < δ ≤ 1/2 and every ² > 0 we have
X
P(Ñ (t) ≥ k)(1 + ²)k = e−2t
X
³ (2t)k
k!
k>(1+δ)t
´
(2t)k+1
+ . . . (1 + ²)k
(k + 1)!
(2t)[2t]
(2t)[2t]
p
(1 + ²)[2t] ∼ e−2t
(1 + ²)[2t] → ∞
[2t]!
[2t][2t] e−[2t] [2t]
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k>(1+δ)t
+
> e−2t
by Stirling formula, so that (2.2) does not hold when N (t) is replaced by Ñ (t). This difficulty
was also noted in Tang et al. (2001, p. 92). The reason for this discrepancy lies in the ”shifted”
set of summation indices. In this example, instead of summation with respect to k > (1 + δ)2t
(because EÑ (t) = 2t) as in (2.2), the set of summation indices is k > (1 + δ)t. More precise
verification of (2.2) in the Pareto(β) (β > 2) case is given in Section 4.
The main result of our paper is the following theorem.
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Theorem 2.1 Let assumptions H1 , H2 , A and B be satisfied. Then, for every positive µ > 0,
it holds
P(SN (t) > x + (a + µ)λt) ∼ λtB(x + µλt)
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as x → ∞ uniformly for all t ∈ [f (x), γx/Q(x)], where f (x) is an arbitrary infinitely increasing
function and γ > 0 is arbitrary positive constant. (Note that x/Q(x) is nondecreasing function
for sufficiently large x, see Lemma 3.3 (i).)
The proof of the theorem is given in Section 3. In case µ = 0 we have:
Corollary 2.1 Under assumptions H1 , H2 , A and B it holds
P(SN (t) > x + aλt) ∼ λtB(x),
as x → ∞ uniformly for all t ∈ [f1 (x), f2 (x)], where f1 (x) is an arbitrary infinitely increasing
function and f2 (x) = o(x/Q(x)).
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P(SN (t) > x + aλt) ≥ P(SN (t) > x + (a + µ)λt)
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Proof. It follows from Theorem 2.1 and Lemma 3.4 that with some fixed positive µ
B(x + µλt)
B(x)
½ Z x+µλt
¾
= (1 − o(1))λtB(x) exp −
q(u)du
≥ (1 − o(1))λtB(x)
x
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x+µλt
µ
¶
Z
uq(u) Q(u)
≥ (1 − o(1))λtB(x) 1 −
du
Q(u) u
x
µ
µ
¶¶
tQ(x)
≥ (1 − o(1))λtB(x) 1 − O
x
≥ (1 − o(1))λtB(x)
as x → ∞ and t ∈ [f1 (x), f2 (x)].
¡
¢
¡
¢
On the other hand, for such fixed positive µ, P SN (t) > x + aλt ≤ P SN (t) > x∗ + (a + µ)λt ,
where x∗ = x − µλf2 (x) = x − o(1)x/Q(x) = x(1 − o(1)). Hence, similarly we can obtain that
for sufficiently large x
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P(SN (t) > x + aλt) ≤ (1 + o(1))λtB(x∗ + µλt)
B(x∗ ) B(x∗ + µλt)
B(x)
B(x∗ )
½
¾
Q(x∗ )
∗
≤ (1 + o(1))λtB(x) exp
(x − x + µλt)
x∗
≤ (1 + o(1))λtB(x)
≤ (1 + o(1))λtB(x)
uniformly for t from interval [f1 (x), f2 (x)].
Proof of main theorem
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3
2
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The statement of Theorem 2.1 follows from Theorem 3.1, Lemma 3.3 below and the estimate
½ Z x+µλt
¾
½
¾
B(x + µλt(1 − ∆))
q(u)u Q(u)
∆µλtQ(x)
= exp
du ≤ exp
u
x
B(x + µλt)
x+µλt(1−∆) Q(u)
provided 0 < ∆ < 1 and x is sufficiently large.
Theorem 3.1 Under assumptions H1 , H2 , A and B, for every positive µ > 0, ∆1 , ∆2 (∆2 < 1)
and any domain D ⊂ [f1 (x), ∞), where f1 (x) is an infinitely increasing function, it holds
lim sup sup
x→∞ t∈D
and
lim inf inf
x→∞ t∈D
P(SN (t) > x + (a + µ)λt)
λtB(x + µλt)
P(SN (t) > x + (a + µ)λt)
λtB(x + µλt)
≤ 1 + ∆1 lim sup sup
x→∞ t∈D
≥ 1 − ∆1 lim inf inf
x→∞ t∈D
5
B(x + µλt(1 − ∆2 ))
B(x + µλt)
B(x + µλt(1 − ∆2 ))
.
B(x + µλt)
The statement of this theorem follows from lemmas 3.1 and 3.4 below.
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Lemma 3.1 Let assumptions H1 , H2 , A and B be satisfied and let ²(t) be a monotonically
vanishing function satisfying limt→∞ t²2 (t) = ∞. Then for every µ > 0 and 0 < ∆ < 1
X
P(N (t) = k)P(Sk > x + λ(µ + a)t) = o(1)tB(x + µλt(1 − ∆))
|k−λt|>²(t)λt
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as x → ∞ uniformly for t ∈ [f3 (x), ∞), where f3 (x) is an infinitely increasing function.
Proof. Write
X
P(N (t) = k)P(Sk > x + λ(µ + a)t)
|k−λt|>ε(t)λt
k∈D1
+
X ¶
P(N (t) = k)P(Sk > x + λ(µ + a)t) =: I1 + I2 + I3 ,
+
X
k∈D2
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=
µ X
k∈D3
(3.1)
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where D1 = [0, (1 − ²(t))λt), D2 = ((1 + ²(t))λt, (1 + δ)λt], D3 = ((1 + δ)λt, ∞).
First we estimate the therm I1 . For this we use the next large deviation result (see Theorem
4.1 of Baltrūnas et al. (2004)) and auxiliary Lemma 3.3.
Lemma 3.2 Assume that random variables Z1 , Z2 , . . . satisfy assumptions H1 and A. Then
P(Sn − ESn > τ ) ∼ nB(τ )
as n → ∞ uniformly for τ ≥ τn , where τn is any infinitely increasing sequence satisfying
√
Q(τ )
n sup
= 0.
n→∞
τ
τ ≥τn
lim
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Lemma 3.3 Suppose that the hazard function Q(u) satisfies condition r < 1. Then:
(i) Q(u)/u does not increase for sufficiently large u;
(ii) for every ² > 0 there exists positive u² and c² , such that
Q(u) ≤ c² ur+²
for u > u² .
(3.2)
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Proof of Lemma 3.3. (i) The proof easily follows from the proof of Lemma 2.1 in Baltrūnas
(2005).
(ii) We have
(log Q(u))0 =
q(u)
uq(u) 1
1
=
≤ (r + ²)
Q(u)
Q(u) u
u
for u > u² . Hence,
Z
u
(log Q(v))0 dv
Z u
1
u
≤ (r + ²)
dv = (r + ²) log ,
u²
u² v
log Q(u) − log Q(u² ) =
u²
6
implying Q(u) ≤
Q(u² )
ur+²
²
ur+² = c² ur+² for large u.
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k∈D1
X
∼ (1 − ²(t))λtB(x + µλt + λta²(t))
P(N (t) = k)
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k∈D1
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Since Q(u)/u does not increase for sufficiently large u, it follows from lemmas 3.2–3.3 and
Assumption A that for t = t(x) ∈ [f2 (x), ∞),
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P(N (t) = k)
I1 ≤ P(S[(1−ε(t))λt] > x + (µ + a)λt)
= o(1)tB(x + µλt)
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≤ λtB(x + µλt) P(N (t) < (1 − ²(t))λt)
s
¾
½
N (t) − λt
t²2 (t)
√
= λtB(x + µλt) P
<−
λVar θ1
λ3 tVar θ1
(3.3)
by assumption t²2 (t) → ∞ and the central limit theorem for the renewal process (see, e.g.,
Theorem 2.5.13 in Embrechts et al. (1997)):
N (t) − λt d
√
−→ N(0, 1).
λ3 tVar θ1
(3.4)
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Consider the term I2 . We can apply the large deviation Lemma 3.2 once again for term I2
to obtain, by the dominated convergence theorem, that for t ∈ [f3 (x), ∞)
X
I2 =
P (N (t) = k)P(Sk − ESk > x − ka + (µ + a)λt)
k∈D2
X
∼
P(N (t) = k)kB(x − (k − λt)a + µλt)
k∈D2
≤ B(x − δaλt + µλt)
X
P(N (t) = k)k.
k∈D2
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Let the positive δ > 0 be such that δa < ∆µ. Similarly as in the estimate of term I1 we have
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I2 ≤ (1 + o(1))(1 + δ)λtB(x + µλt(1 − ∆))P(N (t) > (1 + ²(t))λt)
r
½
¾
N (t) − λt
t²2 (t)
>
≤ (1 + o(1))(1 + δ)P √
λtB(x + µλt(1 − ∆))
λ Var θ
λ3 tVar θ
= o(1)tB(x + µλt(1 − ∆)).
(3.5)
Using the property of subexponential distribution B (see, e.g., Lemma 1.3.5 in Embrechts
et al. (1997)) we have that for each ² > 0 there exists a constant K(²) such that
P(Sn > x) ≤ K(²)(1 + ²)n B(x),
x ≥ 0.
Therefore, applying Assumption B and taking into account the equivalence of (2.2) to (see
Remark (ii) on p. 296 in Klüppelberg and Mikosch (1997))
X
P(N (t) = k)(1 + ²)k → 0,
k>(1+δ)λt
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P(N (t) = k)(1 + ²)k B(x + λ(µ + a)t)
k∈D3
≤ K(²)B(x + µλt)
X
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X
I3 ≤ K(²)
T
we obtain
P(N (t) = k)(1 + ²)k
k∈D3
= o(1)B(x + µλt).
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The proof of the lemma follows from (3.1) and the estimates (3.3), (3.5), (3.6).
(3.6)
2
Lemma 3.4 Assume that assumptions H1 , H2 and A are satisfied. Then for every positive
µ>0
X
P(N (t) = k)P(Sk > x + λ(µ + a)t) ∼ λtB(x + µλt)
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|k−λt|≤²(t)λt
as x → ∞ uniformly for t > f4 (x), where f4 (x) is arbitrary infinitely increasing function,
√
²(t) = c1 log t/ t and c1 > 0 is any positive constant.
Proof. Rewrite
|k−λt|≤²(t)λt
P(N (t) = k)P(Sk > x + λ(µ + a)t)
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X
X
= B(x + µλt)
P(N (t) = k)ϕ(x, t)
|k−λt|≤²(t)λt
P(Sk > x + λ(µ + a)t)
,
B(x − (k − λt)a + µλt)
where ϕ(x, t) = B(x−(k−λt)a+µλt)
.
B(x+µλt)
Using Lemma 3.2 we obtain
P(Sk > x + λ(µ + a)t) = P(Sk − ESk > x − (k − λt)a + µλt)
∼ kB(x − (k − λt)a + µλt)
(3.7)
lim ϕ(x, t) = 1
(3.8)
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uniformly for |k − λt| ≤ ²(t)λt.
We will show that
x→∞
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uniformly for t ∈ [f4 (x), ∞).
Let (k − λt)a ≥ 0. Then, applying the mean value theorem, rewrite
ϕ(x, t) = exp{Q(x + µλt) − Q(x − (k − λt)a + µλt)} = exp{(k − λt)aq(ξ)} ≥ 1,
where x + λt(µ − ε(t)) ≤ x + µλt − (k − λt)a ≤ ξ ≤ x + µλt. By (2.1), we have
q(ξ) =
Q(ξ)
ξq(ξ) Q(ξ)
≤ (r + ²)
Q(ξ) ξ
ξ
for every ² > 0 and sufficiently large x. Therefore
n
Q(ξ) o
ϕ(x, t) ≤ exp (r + ²)(k − λt)a
ξ
8
(3.9)
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for sufficiently large x. The function Q(u)/u does not increase for sufficiently large u. Hence,
√
by inequality (k − λt)a ≤ λt²(t) = λc1 t log t, we obtain from (3.9) that
½
¾
½
¾
√
Q(x + λt(µ − ²(t)))
Q(λµt/2)
√
ϕ(x, t) ≤ exp c1 (r + ²) t log t
≤ exp c2 log t
x + λt(µ − ²(t))
t
for sufficiently large x. Since r < 1/2, we obtain for large t
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Q(λµt/2) log t
log t
log t
√
≤ c² (λµt/2)r+² √ = c∗² 1/2−r−² → 0
t
t
t
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by (3.2), choosing ² > 0 such that r + ² < 1/2. If (k − λt)a < 0, the proof of (3.8) is similar.
Hence, by the dominated convergence theorem, we have from (3.7)–(3.8) that
X
P(N (t) = k)P(Sk > x + λ(µ + a)t)
|k−λt|≤²(t)λt
∼ λtB(x + µλt)P(|N (t) − λt| ≤ ²(t)λt) ∼ λtB(x + µλt)
according to relation (3.4). The proof of the lemma is complete.
Example of Pareto distribution
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4
2
In this section we will show that assumptions H1 , H2 , A and B are satisfied in case of claim sizes
and inter-occurence times having heavy-tailed Pareto distribution. Assume that i.i.d. claim sizes
Z1 , Z2 , . . . and inter-occurence times θ1 , θ2 , . . . have the Pareto(α) and Pareto(β) distributions,
respectively:
³
B(u) = P(Z1 > u) =
1 ´α
,
1+u
³
P(θ1 > u) =
1 ´β
,
1+u
u≥0
EP
with α > 2 and β > 2. Obviously, assumptions H1 and H2 are satisfied and a = EZ1 = (α−1)−1 ,
EZ12 = 2(α − 1)−1 (α − 2)−1 , λ = (Eθ1 )−1 = β − 1, Eθ12 = 2(β − 1)−1 (β − 2)−1 .
It is well-known that Pareto distribution is subexponential. Moreover, in our case Q(u) =
α log(1 + u), q(u) = α(1 + u)−1 , implying
r = lim sup
AC
C
u→∞
uq(u)
1
=0<
and lim inf uq(u) = α > 2.
u→∞
Q(u)
2
Hence, Assumption A is satisfied.
The most complicated is the verification of Assumption B. First we will estimate P(N (t) ≥
k) = P(θ1 + . . . + θk ≤ t) showing that for k > (1 + δ)(β − 1)t and every δ > 0 the following
inequality holds:
n
P(θ1 + . . . + θk ≤ t) ≤ Cσ∗ ,δ,β exp
where σ∗ is specified in (4.3) below.
9
−
o
σ∗ δk
,
(1 + δ)(2 + δ)(β − 1)
(4.1)
0
∞
e−us
du.
(1 + u)β+1
RI P
The Laplace–Stieltjes transform of d.f. P (θ1 ≤ u) is
Z ∞
Z
−us
θ̂(s) =
e dP(θ1 ≤ u) = β
T
ACCEPTED MANUSCRIPT
0
Obviously, θ̂(s) is analytic function for Re s > 0. The properties of the Laplace-Stieltjes transform imply
Z ∞
k
θ̂ (s) =
e−us dP(θ1 + . . . + θk ≤ u) ∀s : Re s > 0,
0
For any real s > 0 we have
Z
θ̂(s) =
Z
σ+i∞
σ−i∞
ets − 1 k
θ̂ (s)ds for any σ > 0.
s
(4.2)
AN
U
1
P(θ1 + . . . + θk ≤ t) =
2πi
SC
so that the inverse formula gives
∞
e−us dP(θ1 ≤ u) = 1 − sEθ1 + O(s2 Eθ12 )
n
o
s
s
+ O(s2 ) = exp −
+ O(s2 ) .
= 1−
β−1
β−1
0
Since
2
2+δ
TE
DM
Therefore, for any real positive s
´o
n
n
o
n s ³ 2
cβ s2 o
s
2s
+
− 1 + cβ s .
θ̂(s) ≤ exp −
= exp −
exp
β−1 β−1
(2 + δ)(β − 1)
β−1 2+δ
δ
− 1 + cβ s → − 2+δ
< 0, s → 0, there exists a small number σ∗ > 0, such that
n
o
2σ∗
θ̂(σ∗ ) ≤ exp −
.
(2 + δ)(β − 1)
(4.3)
From (4.2), for such σ∗ we obtain
AC
C
EP
P(θ1 + . . . + θk ≤ t) ≤
≤
≤
≤
1
2π
Z
∞
−∞
Z
|et(σ∗ +iy) − 1|
p
|θ̂(σ∗ + iy)|k dy
2
2
σ∗ + y
∞
|et(σ∗ +iy) − 1| k−1
p
θ̂ (σ∗ )|θ̂(σ∗ + iy)|dy
σ∗2 + y 2
−∞
Z ∞
1
|θ̂(σ∗ + iy)|
p
(etσ∗ + 1)θ̂k−1 (σ∗ )
dy
2π
σ∗2 + y 2
−∞
Z ∞
2σ∗ (k−1)
1 tσ∗ − (δ+2)(β−1)
|θ̂(σ∗ + iy)|
p
e
dy.
π
σ∗2 + y 2
−∞
1
2π
(4.4)
To estimate the integral in (4.4), note that equality (here Re s > 0)
Z
e−us
β β(β + 1) ∞
du
θ̂(s) = −
s
s
(1 + u)β+2
0
implies |θ̂(σ∗ + iy)| ≤
Cβ
|σ∗ +iy|
Z
∞
−∞
C
= √ 2β
σ∗ +y 2
and therefore
|θ̂(σ∗ + iy)|
p
dy ≤ Cβ
σ∗2 + y 2
10
Z
∞
2
−∞ σ∗
Cβ π
dy
=
.
2
+y
σ∗
(4.5)
ACCEPTED MANUSCRIPT
(1 + ²)k exp
k>(1+δ)(β−1)t
AN
U
k>(1+δ)(β−1)t
X
SC
which is needed inequality (4.1).
Now we have
X
P(N (t) ≥ k)(1 + ²)k ≤ Cσ∗ ,δ,β
RI P
T
For k > (1 + δ)(β − 1)t, from (4.4) and (4.5) we obtain
n
Cβ
σ∗ k
2σ∗ (k − 1) o
P(θ1 + . . . + θk ≤ t) ≤
exp
−
π
(1 + δ)(β − 1) (2 + δ)(β − 1)
n
n
o
Cβ
σ∗ k ³ 1
2 ´o
2σ∗
=
exp
−
exp
π
β−1 1+δ 2+δ
(2 + δ)(β − 1)
n
o
σ∗ δk
= Cσ∗ ,δ,β exp −
,
(1 + δ)(2 + δ)(β − 1)
as t → ∞, since
n
−k
σ∗ δ
σ∗ δ
− log(1 + ²) →
> 0,
(1 + δ)(2 + δ)(β − 1)
(1 + δ)(2 + δ)(β − 1)
o
σ∗ δ
→0
(1 + δ)(2 + δ)(β − 1)
when ² ↓ 0.
TE
DM
Therefore, Assumption B is satisfied.
To conclude the section, we recall that the main theorem of the paper says that, for every
positive µ,
µ
µ
¶ ¶
β−1
(β − 1)t
´α , x → ∞,
P SN (t) > x +
+µ t ∼ ³
α−1
1+x+ µ t
α−1
uniformly for t ∈ [f (x), γx/ ln x] with arbitrary infinitely increasing function f (x) and arbitrary
positive constant γ.
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