Research Article | Open Access
On the Expected Discounted Penalty Function for the Classical Risk Model with Potentially Delayed Claims and Random Incomes
We focus on the expected discounted penalty function of a compound Poisson risk model with random incomes and potentially delayed claims. It is assumed that each main claim will produce a byclaim with a certain probability and the occurrence of the byclaim may be delayed depending on associated main claim amount. In addition, the premium number process is assumed as a Poisson process. We derive the integral equation satisfied by the expected discounted penalty function. Given that the premium size is exponentially distributed, the explicit expression for the Laplace transform of the expected discounted penalty function is derived. Finally, for the exponential claim sizes, we present the explicit formula for the expected discounted penalty function.
In the classical risk theory, assumption of independence among claims is an important condition to the study of risk models. However, in many practical situations, the assumption is often inconsistent with the operation of insurance companies. In reality, claims may be time-correlated for various reasons, and it is important to study risk model which is able to depict this phenomenon. Since the work by Waters and Papatriandafylou , many researchers have studied various kinds of dependencies among claim amounts and claim numbers, such as Gerber , Shiu , Dickson , Willmot , and Ambagaspitiya [6, 7]. Among others, in the case of the compound binomial model, Yuen and Guo  consider a specific dependence structure between the claim sizes and interclaim times. Under their assumption, each claim causes a byclaim but the occurrence of the byclaim may be delayed. Further, based on the same model, Xiao and Guo  investigate the joint distribution of the surplus immediately prior to ruin and the deficit at ruin.
Note that the risk model referred above is based on the assumption that the probability of delay of each byclaim is constant and independent of claim amounts. Albrecher and Boxma  consider a generalization of the classical risk model to a dependent setting where the distribution of the time between two claim occurrences depends on the previous claim size. Motivated by the idea, Zou and Xie  introduce a risk model with an interesting dependence structure between the amount of main claim and the occurrence of byclaim. It is a natural extension for the delayed claims model due to the fact that the bigger the claim amount for main claim (such as car damage) is, the greater odds of the byclaim (such as injury) would be delayed in the actual practice with insurer. Based on the structure, we consider an improved payment mode named potentially delayed claims where the main claim induces a byclaim with a certain probability. The improvement is inspired from a series of examples similar to the case referred above, in which main claim does not induce byclaim with probability 1.
Because the insurance company may have lump sums of income, we apply potentially delayed claims to the compound Poisson risk model in the presence of random incomes. Since Boucherie et al.  described the random incomes by adding a compound Poisson process with positive jumps to the classical risk model, many authors have studied similar topics. Boikov  studies ruin problem of a risk model with stochastic premium process. Bao  considers a risk model, in which the premium is a Poisson process instead of a linear function of time. Labbé et al.  consider a risk model where the stochastic incomes follow a compound Poisson process and research the case when the premiums have Erlang distributions in more depth. Hao and Yang  analyze the expected discounted penalty function of a compound Poisson risk model with random incomes and delayed claims. Yu  also studies the expected discounted penalty function in a Markov regime-switching risk model with random income.
In this paper, we aim at the expected discounted penalty function of an extensive risk model with random incomes and potentially delayed claims. This paper generalizes the model of Hao and Yang . Based on the extensive model, we obtain explicit expression of the expected discounted penalty function, while  derives defective renewal equations of it only. When the main claim induces a byclaim with probability 1 and the byclaim is delayed with a constant probability, the results in this paper will reduce to them in . So  can be seen as a special case of this paper. In addition, Zou and Xie  derive the probability of ruin in the risk model with delayed claims, but this paper obtains the expected discounted penalty function which contains the probability of ruin. If we define the expected discounted penalty function with the same expression as ruin probability and assume that the premium is a linear function of time, we can get the same results as .
The rest of this paper is structured as follows. In Section 2, we introduce the compound Poisson risk model with random incomes and potentially delayed claims. In Section 3, we derive an integral equation for the expected discounted penalty function and obtain explicit expression of its Laplace transform when the premium income is exponentially distributed. The defective renewal equation satisfied by the expected discounted penalty function is studied in Section 4. Section 5 obtains explicit result for the expected discounted penalty function with positive initial surplus when the claim amounts from both classes are exponentially distributed. Section 6 concludes the paper.
Now, we can show the extensive risk model with random incomes and potentially delayed claims in mathematics. On the one hand, we denote the aggregate premium incomes at time by which is a compound Poisson process, and is the corresponding Poisson income number process with parameter . The premium incomes amounts are assumed to be independent and identically distributed (i.i.d.) positive random variables with common distribution , probability function , and mean . So we get . On the other hand, we consider a continuous time model which involves two types of insurance claims, namely, the main claims and the byclaims. Let the aggregate main claims process be a compound Poisson process and let be the corresponding Poisson claim number process with parameter . Its jump times are denoted by with . The main claim amounts and the byclaim amounts are assumed to be independent and identically distributed (i.i.d.) positive random variables with common distribution and , respectively. Moreover, they are independent and their means are denoted by and . Then the surplus process of the risk model is defined as where is the initial capital and is the sum of all byclaims that occurred before time . We assume that and are mutually independent.
With the assumption of potentially delayed claims, the claim occurrence process is to be of the following type: there will be a main claim at every epoch of the Poisson process and the main claim will induce a byclaim with probability . If the main claim amount induces a byclaim and the main claim amount is less than a threshold , the byclaim and its associated main claim occur simultaneously; otherwise, the occurrence of the byclaim is delayed to and main claim occurs simultaneously. From Zou and Xie , we know that
Therefore, we further assume that This assumption ensures that the safety loading is positive.
Let be the time of ruin with if for all . The ruin probability is defined by , . The expected discounted penalty function is of the following form: where is a constant and is a nonnegative measurable function defined on . is the indicator function of event . is the deficit at ruin and is the surplus immediately prior to ruin.
3. The Expected Discounted Penalty Function of the Exponential Premium Income
To handle the surplus process (1), we consider a slight change in the risk model. Instead of having one main claim and a byclaim with probability at the first epoch , another byclaim is added at the first epoch ; that is, byclaim and main claim occur at simultaneously. Hence, the corresponding surplus process of this auxiliary risk model is defined as where denotes the other byclaim amount, . Assume that and are i.i.d. positive random variables.
The expected discounted penalty function for this auxiliary risk model is denoted by which is useful to derive .
Obviously there will be a main claim at the first epoch . Let be the time for the first premium. The first claim can be or cannot be earlier than the first premium. If it is, there are three situations.(1)The main claim does not induce a byclaim; then the surplus process gets renewed except for the initial value. The probability of this event is .(2)The main claim induces a byclaim and the main claim size ; then the byclaim also occurs at the first epoch ; the surplus process will renew itself with different initial reserve. The probability of this event is .(3)The main claim induces a byclaim and the main claim size ; then the occurrence of the byclaim will be delayed to ; that is, the delayed byclaim and the main claim occur simultaneously. In this case, will not renew itself but transfer to the auxiliary model described in the paragraph above. The probability of this event is .
Conditioning on the time of the first event, we have where Similarly, for the expected discounted penalty function of the auxiliary risk model, we have where
In the following, we will give the Laplace transforms of the and .
Let and . For , we define
Now we introduce the Dickson-Hipp operator studied in Dickson and Hipp . Define where is a real-valued function and is a complex number. As in Li and Garrido , we find . For distinct and , If ,
Suppose that the premium incomes are exponentially distributed; that is,
According to the definition and properties of the Dickson-Hipp operator, we take the Laplace transform of and ; then
Proof. To prove Lemma 1, it is equal to show that (22) has exactly two roots in the right half complex plan. Firstly, (22) can be rewritten as
For , , it is easy to check that and . And which implies is a strictly decreasing function of . So has exactly one positive real root, say, . Obviously, is also one positive real root of (22). Note that is another positive real root of (22), say, and . That means , so we conclude that (22) has exactly two distinct positive real roots, say and .
Now, we prove that is the exactly one positive real root of equation on the right half of the complex plane. When is on the half-circle, and on the complex plane, for sufficiently large, while for on the imaginary axis, , the last inequality is true as well. That is to say, on the boundary of the contour enclosed by the half-circle and the imaginary axis, Then we conclude, by Rouché’s theorem, that on the right half of the complex plane, the number of roots of the equation equals the number of roots of the equation . Furthermore, the latter has exactly one root on the right half of the complex plane. It follows that has exactly one positive real root, say, , on the right half of the complex plane. It is easy to see that is the exactly one positive real root of equation on the right half of the complex plane. It follows from all of the above that (22) has exactly two distinct positive real roots and on the right half of the complex plane. Hence, the lemma is proved.
Remark 2. From Klimenok , we know that . Thereafter, we denote them by , for simplicity.
Since is finite for all with , we know , , should also be zeros of the numerator in (19); that is,
By solving these linear equations, we get
4. The Defective Renewal Equation for the Expected Discounted Penalty Function
In this section, we study the defective renewal equation satisfied by the expected discounted penalty function. Note that (19) can be rewritten as where
Lemma 3. The Laplace transform of the expected discounted penalty function satisfies
Proof. Since is analytic for all with , we know , , are zeros of the numerator in (30). It means , . Because is a polynomial of degree 1, using Lagrange interpolating theorem, we obtain
The denominator of (30) can be dealt with in a similar way. From Lemma 1, we know that , . Because is a polynomial of degree 2, using Lagrange interpolating theorem, we obtain
Then using Property 6 of the Dickson-Hipp operator given in Li and Garrido , we have
It is easy to check that which makes (36) become
Invoking (34) and (37) into (30), we could obtain which leads to (32). This completes the proof.
Now, we are ready to derive the defective renewal equation for .
Theorem 4. satisfies the following integral equation:
Proof. Equation (39) follows easily from the inverse Laplace transform in (32). We could like to point out that (39) is also a defective renewal equation. This can be verified by showing that
For , putting in (36), it follows that
Now we consider the case . Setting in the denominators of (19) and (20), we have
Differentiating both sides of this equation with respect to and then setting , we have
Then taking the limit in (41) and using L’Hôspital’s rule, we obtain
Thus, (39) is a defective renewal equation, and the proof is complete.
Remark 5. When and , then , . In the case, each main claim induces a byclaim, and its associated byclaim occurs simultaneously with probability , or the occurrence of the byclaim may be delayed with probability . Actually, the risk model given by (1) will be the compound Poisson risk model with delayed claims and random incomes studied by Hao and Yang . Then, by some simple calculations, we can find that (39) in Theorem 4 is consistent with in .
5. Explicit Results for Exponential Claim Size Distributions
We now consider the case where both claim sizes are exponentially distributed, that is, distribution functions and , where and . Then we have
For the special case , we obtain So we have
Let , ; then can be written as So we can derive where
From (32) and (37), we know that It turns out that (51) can be transformed to another expression by multiplying both denominator and numerator by : The common denominator of (52), denoted by , is a polynomial of degree 5 with the leading coefficient , given by Obviously, has five roots on the complex plane and all the complex roots are in conjugate pairs. Noting that , , and are three roots, we have
Denote . Furthermore, if , , and are distinct, we obtain, by partial fractions, that where Then (52) can be simplified to where
Taking the inverse Laplace transforms, we can derive explicit expressions for :