Differential equations for ruin probability in a special risk model with FGM copula for the claim size and the inter-claim time

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Research

Abstract

In this paper, we consider a risk process in which the distribution of the inter-claim time is the sum of two independent exponential random variables. We introduce a dependence structure between the claim size and the inter-claim time. The structure is based on FGM copula. An integro-differential equation for the expected discounted penalty function is derived and an explicit expression for the Laplace transform of ruin probability is given for exponential claim size.

Keywords

Risk Model Risk Process Claim Size Surplus Process Initial Surplus 

1 Introduction

In classical risk models, the surplus process usually relies on the assumption of independence between the claim size and the inter-claim time. However, in many applications this assumption is too restrictive and unrealistic. Actually, we know that the greater the claim size is, the longer the inter-claim time is. The requirement for generalization has led to some papers on the modeling of dependence. Among them, Albrecher and Teugels [1] consider general dependence structure based on a copula for the claim size and the inter-claim time, they derive asymptotic results for both the finite and infinite time ruin probabilities. Boudreault et al. [2] consider a particular dependence structure among the inter-claim time and the claim size and derive the defective renewal equation satisfied by expected discounted penalty function. Cossette et al. [3] consider the compound Poisson risk model in which the claim size and the inter-claim time are dependent, and the dependence structure is based on Farlie-Gumbel-Morgenstern (FGM) copula. They derive the Laplace transform of the expected discounted penalty function, and give explicit expression for the Laplace transform of the time of ruin for exponential claim sizes. Barges and Cossette [4] investigate the computation of the moments of the compound Poisson risk model with FGM copula.

In this article, we consider a Sparre Andersen risk process where the claim size and the inter-claim time are dependent with FGM copula, and the distribution of the inter-claim time is the sum of two independent exponential random variables. In ruin theory, a common approach is to obtain an integro-differential equation for the expected discounted penalty function and apply it to derive the Laplace transform of the function. Dickson and Hipp [5] investigate ruin probability for Erlang ( 2 ) Open image in new window risk process, Li and Garrido [6] consider this problem of Erlang ( n ) Open image in new window risk model, Gerber and Shui [7] also do some relative works.

The paper is arranged as follows. In Section 2, we present the risk model and give some notations. An integro-differential equation of the expected discounted penalty function is formulated and the main results are presented in Section 3. In Section 4, we apply the integro-differential equation to derive the Laplace transform of the expected discounted penalty function. The special case where the claim size is exponentially distributed is considered in the final section.

2 The risk model

The surplus process is defined as U ( t ) = u + c t i = 1 N ( t ) X i Open image in new window, where U ( 0 ) = u Open image in new window is the initial surplus, c is the premium rate, { N ( t ) , t R + } Open image in new window is a renewal process, and X i Open image in new window ( i = 1 , 2 , Open image in new window) is the random variable (r.v.) corresponding to the amount of the i th claim. The time between the ( i 1 ) Open image in new windowth and i th claim is defined by the r.v. V i Open image in new window with V 1 Open image in new window being the time of the first claim.

The claim amounts { X i , i N + } Open image in new window form a sequence of independent identically distributed (i.i.d.) random variables (r.v.s) as the r.v. X with probability density function (p.d.f.) f X Open image in new window, cumulative distribution function (c.d.f.) F X Open image in new window. The inter-claim times { V i , i N + } Open image in new window form a sequence of independent r.v.’s identically distributed as the canonical r.v. V, V = W 1 + W 2 Open image in new window, where { W j } Open image in new window are two independent exponentially distributed r.v.s with parameters λ j Open image in new window, j = 1 , 2 Open image in new window, V has p.d.f. f V Open image in new window, and c.d.f. F V Open image in new window. Note that the Erlang ( 2 ) Open image in new window model is the special case where λ 1 = λ 2 Open image in new window. { ( X i , V i ) , i N + } Open image in new window form a sequence of i.i.d. random vectors distributed as the canonical random vector ( X , V ) Open image in new window. The joint p.d.f. of ( X , V ) Open image in new window is denoted by f X , V ( x , t ) Open image in new window with x , t R + Open image in new window.

The joint distribution of ( X , V ) Open image in new window is based on the classical FGM copula, which is defined by
C ( u , v ) = u v + θ u ( 1 u ) v ( 1 v ) Open image in new window
(1)

for every ( u , v ) Open image in new window in [ 0 , 1 ] 2 Open image in new window and the dependence parameter θ takes value in [ 1 , 1 ] Open image in new window.

We choose this class of copula since it provides an easy manner to construct bivariate models with a variety of dependence structures. Even if the FGM copula introduces only light dependence, it admits positive as well as negative dependence between a set of random variables and includes the independence copula when θ = 0 Open image in new window (see Nelsen [8]).

The p.d.f. associated to (1) is given by
c ( u , v ) = 1 + θ ( 1 2 u ) ( 1 2 v ) . Open image in new window
(2)
Given (1), the joint c.d.f. F X , V Open image in new window is defined by
F X , V ( x , t ) = C ( F X ( x ) , F V ( t ) ) = F X ( x ) F V ( t ) + θ F X ( x ) ( 1 F X ( x ) ) F V ( t ) ( 1 F V ( t ) ) . Open image in new window
With (2), the joint p.d.f. f X , V Open image in new window is
f X , V ( x , t ) = f X ( x ) f V ( t ) c ( F X ( x ) , F V ( t ) ) = f X ( x ) f V ( t ) + θ f X ( x ) f V ( t ) ( 1 2 F X ( x ) ) ( 1 2 F V ( t ) ) . Open image in new window
(3)
For simplicity, we define the following functions:
k X ( x ) = f X ( x ) ( 1 2 F X ( x ) ) Open image in new window
(4)
and
k V ( t ) = f V ( t ) ( 1 2 F V ( t ) ) . Open image in new window
(5)
Let T denote the time to ruin, so that
T = inf { t | U ( t ) < 0 ,  for  t > 0 } . Open image in new window
Then the probability of ultimate ruin from initial surplus u is defined as
ψ ( u ) = P ( T < | U ( 0 ) = u ) . Open image in new window
To ensure that ruin will not occur almost surely, we assume that
E ( c V X ) > 0 . Open image in new window
(6)
Besides the ruin probability, other important ruin quantities in ruin theory include the Laplace transform of ruin time; the surplus immediately before ruin denoted by U ( T ) Open image in new window; the deficit at ruin denoted by | U ( T ) | Open image in new window, etc. A unified approach to study these ruin quantities is to consider the so-called expected discounted penalty function introduced by Gerber and Shiu [9]; the function is given by
m δ ( u ) = E [ e δ T ω ( U ( T ) , | U ( T ) | ) I ( T < ) | U ( 0 ) = u ] , u > 0 , Open image in new window
(7)

where ω ( x , y ) Open image in new window, for all x , y 0 Open image in new window, is the penalty function at the time of ruin for the surplus prior to ruin and the deficit at ruin, I ( ) Open image in new window is the indicator function, and δ is a nonnegative parameter. We can think of δ either as being a force of interest or as a dummy variable in the context of the Laplace transform. A special case of the expected discounted penalty function with ω ( x , y ) = 1 Open image in new window is the Laplace transform of the time of ruin. The cases δ = 0 Open image in new window and ω ( x , y ) = 1 Open image in new window correspond to the infinite-time ruin probability.

3 Integro-differential equation

In this section, we derive an integro-differential equation for the expected discounted penalty function m δ ( u ) Open image in new window. By conditioning on the time and the amount of the first claim, we have
m δ ( u ) = 0 0 u + c t e δ t m δ ( u + c t x ) f X , V ( x , t ) d x d t + 0 u + c t e δ t ω ( u + c t , x u c t ) f X , V ( x , t ) d x d t . Open image in new window
(8)
With (3), (4) and (5), (8) becomes
m δ ( u ) = 0 0 u + c t e δ t m δ ( u + c t x ) f X ( x ) f V ( t ) d x d t + θ 0 0 u + c t e δ t m δ ( u + c t x ) k X ( x ) k V ( t ) d x d t + 0 u + c t e δ t ω ( u + c t , x u c t ) f X ( x ) f V ( t ) d x d t + θ 0 u + c t e δ t ω ( u + c t , x u c t ) k X ( x ) k V ( t ) d x d t . Open image in new window
(9)
We define two functions:
σ 1 ( y ) = 0 y m δ ( y x ) f X ( x ) d x + ω 1 ( y ) Open image in new window
(10)
and
σ 2 ( y ) = 0 y m δ ( y x ) k X ( x ) d x + ω 2 ( y ) , Open image in new window
(11)
Given (10) and (11), (9) becomes
m δ ( u ) = 0 e δ t f V ( t ) σ 1 ( u + c t ) d t + θ 0 e δ t k V ( t ) σ 2 ( u + c t ) d t . Open image in new window
(12)
Substituting s = u + c t Open image in new window into (12), we have
Theorem 1 In the risk model introduced in Section 2, the excepted discounted penalty function m δ ( u ) Open image in new window satisfies the following integro-differential equation:
α ( D ) β ( D ) m δ ( u ) = α ( D ) ( σ 1 ( u ) θ σ 2 ( u ) ) + θ β ( D ) γ ( D ) σ 2 ( u ) , Open image in new window
(14)

I and D are the identity and differential operators.

Proof Case 1. λ 1 λ 2 Open image in new window, we obtain the p.d.f. and c.d.f. of r.v. V,
hence
k V ( t ) = f V ( t ) ( 1 2 F V ( t ) ) = λ 1 λ 2 λ 2 λ 1 ( e λ 1 t e λ 2 t ) ( 2 λ 1 λ 1 λ 2 e λ 2 t 2 λ 2 λ 1 λ 2 e λ 1 t 1 ) . Open image in new window
(16)
Let
λ = λ 1 λ 2 λ 2 λ 1 , r 1 = 2 λ 1 λ 1 λ 2 , r 2 = 2 λ 2 λ 1 λ 2 , Open image in new window
then (13) becomes
m δ ( u ) = λ c u ( e ( δ + λ 1 ) s u c e ( δ + λ 2 ) s u c ) ( σ 1 ( s ) θ σ 2 ( s ) ) d s + θ λ c u ( e ( δ + λ 1 ) s u c e ( δ + λ 2 ) s u c ) ( r 1 e λ 2 s u c r 2 e λ 1 s u c ) σ 2 ( s ) d s = λ c ξ ( u ) + θ λ c η ( u ) , Open image in new window
(17)
where
ξ ( u ) = u ( e ( δ + λ 1 ) s u c e ( δ + λ 2 ) s u c ) ( σ 1 ( s ) θ σ 2 ( s ) ) d s Open image in new window
(18)
and
η ( u ) = u ( e ( δ + λ 1 ) s u c e ( δ + λ 2 ) s u c ) ( r 1 e λ 2 s u c r 2 e λ 1 s u c ) σ 2 ( s ) d s = u ( ( r 1 + r 2 ) e ( λ 1 + λ 2 + δ ) s u c r 1 e ( 2 λ 2 + δ ) s u c r 2 e ( 2 λ 1 + δ ) s u c ) σ 2 ( s ) d s . Open image in new window
(19)

In order to get (14), we firstly apply the operator β ( D ) Open image in new window to both sides of (18) with respect to (w.r.t.) u.

Given (18), we have
D ξ ( u ) = u ( λ 1 + δ c e ( δ + λ 1 ) s u c λ 2 + δ c e ( δ + λ 2 ) s u c ) ( σ 1 ( s ) θ σ 2 ( s ) ) d s Open image in new window
and
D 2 ξ ( u ) = u ( ( λ 1 + δ c ) 2 e ( δ + λ 1 ) s u c ( λ 2 + δ c ) 2 e ( δ + λ 2 ) s u c ) ( σ 1 ( s ) θ σ 2 ( s ) ) d s λ 1 λ 2 c ( σ 1 ( u ) θ σ 2 ( u ) ) . Open image in new window
These allow us to derive the following result:
β ( D ) ξ ( u ) = 1 λ 1 λ 2 ( ( λ 1 + δ ) ( λ 2 + δ ) I c ( λ 1 + λ 2 + 2 δ ) D + c 2 D 2 ) = c λ ( σ 1 ( u ) θ σ 2 ( u ) ) . Open image in new window
(20)

Secondly, we take the operator α ( D ) Open image in new window to both sides of (19) w.r.t. u.

From (19), we get
and
D 3 η ( u ) = u ( ( r 1 + r 2 ) ( λ 1 + λ 2 + δ c ) 3 e ( λ 1 + λ 2 + δ ) s u c r 1 ( 2 λ 2 + δ c ) 3 e ( 2 λ 2 + δ ) s u c r 2 ( 2 λ 1 + δ c ) 3 e ( 2 λ 1 + δ ) s u c ) σ 2 ( s ) d s ( ( r 1 + r 2 ) ( λ 1 + λ 2 + δ c ) 2 r 1 ( 2 λ 2 + δ c ) 2 r 2 ( 2 λ 1 + δ c ) 2 ) σ 2 ( u ) ( ( r 1 + r 2 ) ( λ 1 + λ 2 + δ c ) r 1 ( 2 λ 2 + δ c ) r 2 ( 2 λ 1 + δ c ) ) D σ 2 ( u ) . Open image in new window
Hence
α ( D ) η ( u ) = 1 2 λ 1 λ 2 { ( λ 1 + λ 2 + δ ) ( 2 λ 1 + δ ) ( 2 λ 2 + δ ) η ( u ) c ( 2 ( λ 1 + λ 2 + δ ) 2 + ( 2 λ 1 + δ ) ( 2 λ 2 + δ ) ) D η ( u ) + 3 c 2 ( λ 1 + λ 2 + δ ) D 2 η ( u ) c 3 D 3 η ( u ) } = c λ ( ( 2 λ 1 + 2 λ 2 + δ ) σ 2 ( u ) c D σ 2 ( u ) ) = c λ γ ( D ) σ 2 ( u ) . Open image in new window
(21)
Applying the operator α ( D ) β ( D ) Open image in new window to both sides of (17), we can obtain
α ( D ) β ( D ) m δ ( u ) = α ( D ) β ( D ) ( λ c ξ ( u ) + θ λ c η ( u ) ) = α ( D ) ( σ 1 ( u ) θ σ 2 ( u ) ) + θ β ( D ) γ ( D ) σ 2 ( u ) . Open image in new window
Case 2. λ 1 = λ 2 = λ Open image in new window, we know
then (13) becomes
m δ ( u ) = λ 2 c D ξ ( u ) + θ λ 2 c D η ( u ) , Open image in new window
(24)
where
D ξ ( u ) = u s u c e ( δ + λ ) s u c ( σ 1 ( s ) θ σ 2 ( s ) ) d s Open image in new window
and
D η ( u ) = u ( λ ( s u c ) 2 + s u c ) e ( δ + 2 λ ) s u c σ 2 ( s ) d s . Open image in new window
When repeating a similar procedure to Case 1, the following formulae can be obtained:
β ( D ) D ξ ( u ) = c λ 2 ( σ 1 ( u ) θ σ 2 ( u ) ) , Open image in new window
(25)
α ( D ) D η ( u ) = c λ 2 ( ( δ + 4 λ ) σ 2 ( u ) c D σ 2 ( u ) ) = c λ 2 γ ( D ) σ 2 ( u ) . Open image in new window
(26)
Given (25) and (26), applying the operator α ( D ) β ( D ) Open image in new window to both sides of (24), we can derive
α ( D ) β ( D ) m δ ( u ) = α ( D ) β ( D ) D ( λ 2 c ξ ( u ) + θ λ 2 c η ( u ) ) = α ( D ) ( σ 1 ( u ) θ σ 2 ( u ) ) + θ β ( D ) γ ( D ) σ 2 ( u ) . Open image in new window

Hence, the integro-differential equation is true. □

4 The Laplace transform of m δ ( u ) Open image in new window for δ = 0 Open image in new window

Throughout this paper we denote the Laplace transform of a function f ( x ) Open image in new window by
f ( s ) = 0 e s x f ( x ) d x . Open image in new window

Before deriving the Laplace transform of Gerber-Shiu function, one important step is to develop Lundberg’s fundamental equation and examine its properties.

To derive Lundberg’s fundamental equation, we consider the process
{ exp ( δ i = 1 k V i + s U k ) , k = 0 , 1 , } , for  s > 0 . Open image in new window
It is a martingale if and only if
E ( e δ V e s ( c V X ) ) = 1 , Open image in new window
(27)
which corresponds to Lundberg’s fundamental equation, see Gerber and Shui [9]. Due to (3), (4) and (5), (27) can be written as
E ( e δ V e s ( c V X ) ) = 0 0 e t ( c s δ ) e s x f X , V ( x , t ) d x d t + θ 0 0 e t ( c s δ ) e s x k X ( x ) k V ( t ) d x d t = f X ( s ) f V ( δ c s ) + θ k X ( s ) k V ( δ c s ) = 1 . Open image in new window
(28)
From (15), (16) and (22), (23), we conclude that f V ( δ c s ) Open image in new window and k V ( δ c s ) Open image in new window have the same form in Case 1 or Case 2,
f V ( δ c s ) = ( λ 1 + δ λ 1 c λ 1 s ) 1 ( λ 2 + δ λ 2 c λ 2 s ) 1 = 1 β ( s ) Open image in new window
and
k V ( δ c s ) = ( 2 λ 1 + 2 λ 2 + δ ) c s α ( s ) 1 β ( s ) = γ ( s ) α ( s ) 1 β ( s ) . Open image in new window
Thereby, (28) is equivalent to
α ( s ) β ( s ) α ( s ) f X ( s ) θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) = 0 . Open image in new window
(29)

To derive the expression of m δ ( s ) Open image in new window, we need to know the number of roots in the right-half-plane of Lundberg’s fundamental equation (28). For δ = 0 Open image in new window and θ 0 Open image in new window, by Theorem 1 of Klimenok [10], we can determine the number of roots to (28) with a positive real part. However, for δ > 0 Open image in new window and θ 0 Open image in new window we do not reach the conclusion about it.

Lemma 1 For δ = 0 Open image in new window and θ 0 Open image in new window, Lundberg’s fundamental equation (28) has exactly 4 roots denoted by { ρ i , i = 1 , , 4 } Open image in new window with Re ( ρ i ) > 0 Open image in new window and a 5th root ρ 5 = 0 Open image in new window.

Proof Let z ( s ) = k s k Open image in new window, C k = { s : | z ( s ) | = 1 } Open image in new window.

Firstly, α ( s ) β ( s ) Open image in new window and α ( s ) f X ( s ) + θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) Open image in new window are analytic inside the unit contour C k Open image in new window and continuous on C k Open image in new window. Let k Open image in new window and denote by C the limiting contour. We want to show
| α ( s ) β ( s ) | > | α ( s ) f X ( s ) θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) | , Open image in new window
which is equivalent to
| f X ( s ) | 1 | β ( s ) | + | θ k X ( s ) | | β ( s ) γ ( s ) α ( s ) | | α ( s ) | | β ( s ) | < 1 , Open image in new window
1 | β ( s ) | Open image in new window and | β ( s ) γ ( s ) α ( s ) | | α ( s ) | | β ( s ) | Open image in new window are the ratios of polynomials with a strictly higher degree at the denominator. From the definitions, we have | f X ( s ) | 1 Open image in new window and | θ k X ( s ) | 1 Open image in new window. Hence
| f X ( s ) | 1 | β ( s ) | + | θ k X ( s ) | | β ( s ) γ ( s ) α ( s ) | | α ( s ) | | β ( s ) | 0 Open image in new window
on C (excluding s = 0 Open image in new window). Moreover, due to (4),

Because α ( s ) β ( s ) Open image in new window have 5 positive roots, by Theorem 1 of Klimenok [10], we can conclude that the number of solutions to (28) inside C is equal to 4. Finally, it is clear that the 5th root to (28) is ρ 5 = 0 Open image in new window with δ = 0 Open image in new window. Hence, the conclusion is true. □

In the following sections, we only consider the case that the roots { ρ i , i = 1 , , 5 } Open image in new window are distinct.

Theorem 2 In the risk model introduced in Section 2, the Laplace transform of m δ ( u ) Open image in new window for δ = 0 Open image in new window denoted by m 0 ( s ) Open image in new window is
m 0 ( s ) = ω ˜ ( s ) q ˜ ( s ) f ˜ ( s ) , Open image in new window
(30)
and
f ˜ ( s ) = α ( s ) β ( s ) α ( s ) f X ( s ) θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) . Open image in new window
Proof First, from the properties of the Laplace transform, we have
( f ( k ) ) ( s ) = s k f ( s ) l = 0 k 1 s k l 1 D l f ( 0 ) , for  k N + Open image in new window
and
( 0 x f 1 ( y ) f 2 ( x y ) d y ) ( s ) = f 1 ( s ) f 2 ( s ) . Open image in new window

In order to get (30), we must take the Laplace transform of both sides of (14).

The Laplace transform of α ( D ) β ( D ) m 0 ( u ) Open image in new window is
α ( s ) β ( s ) m 0 ( s ) + q ( s ) , Open image in new window

where q ( s ) Open image in new window is a polynomial of degree four or less, with coefficients in terms of c, λ 1 Open image in new window, λ 2 Open image in new window, and the value of m 0 ( 0 ) Open image in new window and its first 4 derivatives at u = 0 Open image in new window.

The Laplace transform of α ( D ) ( σ 1 ( u ) θ σ 2 ( u ) ) + θ β ( D ) γ ( D ) σ 2 ( u ) Open image in new window is

where q 1 ( s ) Open image in new window and q 2 ( s ) Open image in new window are polynomials of degree three or less, with coefficients in terms of c, λ 1 Open image in new window, λ 2 Open image in new window.

For simplicity, we define the following functions:
and
f ˜ ( s ) = α ( s ) β ( s ) α ( s ) f X ( s ) θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) , Open image in new window
then
α ( s ) β ( s ) m 0 ( s ) + q ( s ) = ( α ( s ) f X ( s ) + θ ( β ( s ) γ ( s ) α ( s ) ) k X ( s ) ) m 0 ( s ) + ω ( s ) + q 1 ( s ) + θ q 2 ( s ) . Open image in new window
Hence
m 0 ( s ) = ω ˜ ( s ) q ˜ ( s ) f ˜ ( s ) . Open image in new window
(31)
Because m 0 ( s ) Open image in new window is finite for Re ( s ) 0 Open image in new window, the numerator on the right-hand side of (31) must be zero whenever the denominator is zero. From Lemma 1, it follows that q ˜ ( s ) Open image in new window is the collocation polynomial of function ω ˜ ( s ) Open image in new window with respect to { ρ i , i = 1 , , 5 } Open image in new window. Then, by the Lagrange interpolation formula, we obtain
q ˜ ( s ) = j = 1 5 ω ˜ ( ρ j ) k = 1 , k j 5 ρ k s ρ k ρ j . Open image in new window
(32)

Thereby, we complete the proof. □

5 Exponential claim size

In this section, we assume that the individual claim size follows an exponential distribution with parameter μ.

Let
f X ( x ) = μ e μ x Open image in new window
and
k X ( x ) = f X ( x ) ( 1 2 F X ( x ) ) = 2 μ e 2 μ x μ e μ x , Open image in new window

In order to get the expression of the ruin probability, we consider a special case of the expected penalty function with ω ( x , y ) = 1 Open image in new window for all x , y > 0 Open image in new window.

Corollary 1 In the risk model introduced in Section 2, if X Exp ( μ ) Open image in new window and ω ( x , y ) = 1 Open image in new window, the Laplace transform of ruin probability denoted by ψ ( s ) Open image in new window is
Proof Given ω ( x , y ) = 1 Open image in new window, (31) and (32), we know
and
f ˜ ( s ) = α ( s ) β ( s ) α ( s ) μ μ + s θ ( β ( s ) γ ( s ) α ( s ) ) ( 2 μ 2 μ + s μ μ + s ) . Open image in new window
For simplicity, to invert the Laplace transform of ψ ( s ) Open image in new window, we multiply ( μ + s ) ( 2 μ + s ) Open image in new window on f ˜ ( s ) Open image in new window, ω ˜ ( s ) Open image in new window, q ˜ ( s ) Open image in new window respectively, yielding

 □

Example 1 For the numerical results, we choose λ 1 = 1 Open image in new window, λ 2 = 1 Open image in new window, μ = 1 Open image in new window and the premium rate c = 1 Open image in new window. We can invert the Laplace transform in (35) leading to ψ ( u ) Open image in new window. In Table 1, the analytic expressions of ψ ( u ) Open image in new window are provided for differential dependence parameters: θ = 1 ; 0.5 ; 0.25 ; 0 ; 0.25 ; 0.5  and  1 Open image in new window, respectively.
Table 1

Analytic expressions of ruin probability

θ

Expressions for the ruin probability ψ ( u ) Open image in new window

−1

0.5201exp(−0.5214u)−0.02806exp(−2.2149u)

−0.5

0.4665exp(−0.5661u)−0.01537exp(−2.0653u)

−0.25

0.4372exp(−0.5910u)−0.008086exp(−2.0335u)

0

0.3820exp(−0.6180u)

0.25

0.3557exp(−0.6474u)+0.008384exp(−1.9646u)

0.5

0.3270exp(−0.6794u)+0.01841exp(−1.9271u)

1

0.2600exp(−0.7534u)+0.04565exp(−1.8434u)

From analytic expressions of ruin probability in Table 1, the resulting ruin probabilities are depicted in Figure 1. We can see that for fixed value of initial surplus u and the impact of the dependence parameters range from −1 to 1, the ruin probabilities decrease.
Figure 1

The result of ruin probability.

Author’s contributions

WY and HX work together to complete the composition of this paper. HX finished the draft and WY modified it.

Notes

Acknowledgements

This research was supported by the Preliminary Doctoral Program for Mathematics of Chongqing University of Technology. The authors would like to express their thanks to the referees for helpful suggestions.

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© Yong and Xiang; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.School of Mathematics and Statistics, Chongqing University of TechnologyChongqingP.R. China
  2. 2.Department of Risk Management and InsuranceSchool of Economics, Nankai UniversityTianjinP.R. China

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