The Compound Binomial Risk Model with Randomly Charging Premiums and Paying Dividends to Shareholders

Abstract

Based on characteristics of the nonlife joint-stock insurance company, this paper presents a compound binomial risk model that randomizes the premium income on unit time and sets the threshold 𝑥 for paying dividends to shareholders. In this model, the insurance company obtains the insurance policy in unit time with probability 𝑝₀ and pays dividends to shareholders with probability 𝑝₁ when the surplus is no less than 𝑥. We then derive the recursive formulas of the expected discounted penalty function and the asymptotic estimate for it. And we will derive the recursive formulas and asymptotic estimates for the ruin probability and the distribution function of the deficit at ruin. The numerical examples have been shown to illustrate the accuracy of the asymptotic estimations.

1. Introduction

The compound binomial risk model is one of the classical actuarial models that have been studied extensively. The classic literatures about the compound risk model primarily include [1–9]. With the emergence and development of dividend insurance, compound binomial risk models considering the case of paying dividends to policyholder are attracting more and more attention of actuarial scholars; see [10–14]. Reference [10] builds the compound binomial risk model with randomly dividends payment and derives the ruin problem by recursive algorithm for cases where the company pays dividends to its policyholders with a certain probability. Reference [11] obtains and solves two defective renewal equations for the Gerber-Shiu penalty function under the compound binomial model proposed in [10]. Reference [12] generalizes the model of [10] and derives the discounted penalty function under the compound binomial model with a multithreshold dividend structure and randomized dividend payments. Considering the fact that the joint-stock company may pay dividends to the policyholders and shareholders, [13] builds the compound binomial risk model with random dividends payment to the policyholders and shareholders and studies the ruin problem with the model. furthermore, [14] derives the arbitrary moments of discounted dividend payments under the compound binomial risk model with interest on the surplus and periodically paying dividend.

The previously mentioned models are compound binomial risk models with a constant premium rate suited for depicting the surplus of the life insurance companies which collect installment premiums. However, nonlife insurance companies (e.g., property insurance companies) charge premiums immediately, and insurance policies are obtained randomly in unit time. Thus the model with a constant premium rate cannot reasonably describe the surplus of the nonlife insurance companies. Moreover, joint-stock nonlife insurance companies need to pay dividends to the shareholders randomly (see [10]). However, these characteristics have not been considered in [10] together. Thus, [10] cannot be suitable for describing the surplus of the joint-stock nonlife insurance companies. In order to address the deficiencies of the models in [10], a compound binomial risk model has been developed with random premiums and dividends payment to shareholders, and it derives the recursive formulas and asymptotic estimates of the ruin probability and the distribution of the deficit at ruin.

This paper is organized as follows. In Section 2, we build the compound binomial risk model with randomly charging premiums and paying dividends to shareholders. In Section 3, we derive the recursive formulas of discounted penalty function. In Section 4, we derive the asymptotic estimates for the discounted penalty function. In Section 5, we obtain recursive formulas, and asymptotic estimates of the ruin probability and the distribution function of deficit at ruin are obtained. Finally, the conclusion is shown in Section 6.

2. The Model and Preliminaries

Consider the compound binomial model with randomized decisions on paying dividends, which is described by𝑈(𝑡)=𝑢+𝑡−𝑆(𝑡)−𝐷(𝑡),𝑡=1,2,…, (1) with the initial surplus of the insurance 𝑢 (≥0). 𝑆_𝑡 is the aggregate claim up to time 𝑡; that is, 𝑆(𝑡)=^𝑡_𝑘=1𝜃_𝑘𝑋_𝑘,𝑡=1,2,…, (2) where 𝜃_𝑡 denotes whether the claim occurs or not in (𝑡−1,𝑡]; the event in which the claim occurs is denoted by 𝜃_𝑡+1=1; the event in which no claim occurs is denoted by 𝜃_𝑡+1=0. The probability of a claim is 𝑝 and the probability of no claim is 𝑞=1−𝑝 in any period (𝑡,𝑡+1]. 𝜃={𝜃_𝑡,𝑡=1,2,…} is independent and identically distributed random series. 𝑋={𝑋_𝑡,𝑡=1,2,…} is independent and identically distributed as 𝐹={𝑝(𝑘)=Pr(𝑋=𝑘);𝑘=1,2,…}. 𝐷(𝑡) is the aggregate dividends payment; that is, 𝐷(𝑡)=^𝑡_𝑘=1𝜖_𝑘𝐼(𝑈(𝑘−1)≥𝑥),𝑡=1,2,…, (3) where 𝑥 (>0) is the threshold such that the insurance company may pay dividends to the shareholders, and 𝐼(𝐵) is the indicator function of a set 𝐵. 𝜖_𝑡 denotes whether dividend is paid or not in (𝑡−1,𝑡]. When the surplus is no less than 𝑥, the company pays one dividend to the shareholders with probability 𝑝₁ (denoted by 𝜖_𝑡=1) and not with the probability 𝑞₁=1−𝑝₁ (denoted by 𝜖_𝑡=0); 𝜖={𝜖_𝑡,𝑡=1,2,…} is independent and identically distributed as 𝐵(𝑝₁) (0<𝑝₁<1).

According to the feature of the variety of the surplus in joint-stock nonlife insurance companies, we further assume that premium is charged randomly in unit period (𝑡−1,𝑡]. Then the aggregate premium is 𝑀(𝑡)=^𝑡_𝑘=1𝐼_𝑘,𝑡=1,2,…, (4) where 𝐼_𝑡 is variable with distribution 𝐵(𝑝₀) (0<𝑝₀≤1), 𝑞₀=1−𝑝₀, and denotes obtaining an insurance policy by 𝐼_𝑡=1 in (𝑡−1,𝑡], as well as denotes not obtaining an insurance policy by 𝐼_𝑡=0. 𝐼={𝐼_𝑡,𝑡=1,2,…} is independent and identically distributed. Furthermore, the random series 𝜃,𝑋,𝜖,𝐼 are assumed to be mutually independent. Then, the surplus of the nonlife insurance joint-stock company is 𝑈(0)𝑈(𝑡)(𝑡)(𝑡)(𝑡)=𝑢,=𝑢+𝑀−𝑆−𝐷,𝑡=1,2,…. (5)(6) The model is called as the compound binomial risk model with random premiums and dividends payment to shareholders.

Remark. (1) Model (5) is the general form of the model in [10] and is the model in [10] if 𝑝₀=1. And also, model (5) can be regarded as general form of the classic risk model and exactly the classic risk model if 𝑝₀=1, 𝑝₁=0.

(2) In this model, the initial time 𝑡=0 is some time in the past at which point we begin studying the surplus of the joint-stock nonlife insurance company, but not the time when the company is created.

Define ruin time with 𝑇=inf{𝑡≥0∣𝑈(𝑡)<0}. The ultimate ruin probability is defined by 𝜓(𝑢)=Pr(𝑇<+∞∣𝑈(0)=𝑢). Define the (expected discounted) penalty function by 𝜙_𝑟(𝑢)=𝐸𝑓𝑈_𝑇−1,𝑈_𝑇𝐼(𝑇<+∞)𝑟^−𝑇∣𝑈(0)=𝑢, (7) where 𝑓(𝑥,𝑦) (𝑥≥0,𝑦≥0) is the non-negative bounded function, 0<𝑟≤1. In this paper, the fact that ∑⁻¹_𝑘=0𝑚_𝑘=0 is adopted.

Let 𝑃(𝑛)=∑^𝑛_𝑘=1𝑝(𝑘), 𝑃(𝑛)=1−𝑃(𝑛). We always assume that 𝜇=∑^+∞_𝑘=1𝑘𝑝(𝑘)=∑^+∞_𝑘=0𝑃(𝑛)<∞, and the 𝐸[𝐼_𝑘−𝜃_𝑘𝑋_𝑘−𝜖_𝑘]=𝑝₀−𝑝𝜇−𝑝₁>0, which leads to the positive security loading. Denote that the security loading 𝛿∶𝛿=(𝑝₀−𝑝𝜇−𝑝₁)/𝑝𝜇>0. Let 𝜙(𝑢)=𝜙₁(𝑢).

3. The Recursive Formulas of the Penalty Function

Theorem 1. Let 𝐽(𝑥)=𝑞₀𝑝₁𝑝(𝑥)+(𝑞₀𝑞₁+𝑝₀𝑝₁)𝑝(𝑥+1)+𝑝₀𝑞₁𝑝(𝑥+2), 𝐽(𝑥)=𝑞₀𝑝₁𝑃(𝑥)+(𝑞₀𝑞₁+𝑝₀𝑝₁)𝑃(𝑥+1)+𝑝₀𝑞₁𝑃(𝑥=2), 𝑇(𝑥)=𝑝₀𝑝(𝑥+1)+𝑞₀𝑝(𝑥), 𝑇(𝑥)=𝑝₀𝑃(𝑥+1)+𝑞₀𝑃(𝑥). Then(1)𝜙(0),𝜙(1),…,𝜙(𝑥) satisfy the following linear equations: 𝑝₀𝑞𝜙(0)−𝑝₁𝜙()𝑝𝑥−1=𝛼,₀𝑞𝜙(𝑢+1)+𝑞𝑞₀+𝑝𝑝₀𝑝(1)−1𝜙(𝑢)+𝑝^𝑢−1_𝑘=0𝜙(𝑘)𝑇(𝑢−𝑘)=𝛽,𝑢=0,1,2,…,𝑥−1, (8)(9)where 𝛼=𝑝𝑞₁^𝑥−1_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝑇(𝑖)+𝑝𝑝₁^𝑥−2_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝑇(𝑖)+𝑝^+∞_𝑘=𝑥^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1),𝛽=−𝑝^∞_𝑘=𝑢+1𝑓(𝑘,𝑘−𝑢)𝐽(𝑘); (10)(2)for 𝑢≥𝑥, the penalty functions 𝜙(𝑢) satisfy 𝜙()=𝑞𝑢+11−𝑞₀𝑞₁+𝑝₀𝑝₁𝑞𝑝₀𝑞₁𝜙(𝑢)−𝑞₀𝑝₁𝑝₀𝑞₁𝜙()−𝑝𝑢−1𝑞𝑝₀𝑞₁^𝑢_𝑘=0𝜙(𝑘)𝐽()−𝑝𝑢−𝑘−1𝑞𝑝₀𝑞₁^+∞_𝑘=𝑢+1𝑓()𝐽(),𝜙()=𝑞𝑢,𝑘−𝑢𝑘−1𝑢+1⁰𝑝₁𝑝₀𝑞₁𝜙(𝑢)+𝑝𝑞𝑝₀𝑞₁^𝑢_𝑘=0𝜙(𝑘)𝐽()+𝑝𝑥−𝑘−1𝑞𝑝₀𝑞₁^+∞_𝑘=𝑢+1^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1). (11)(12)

Proof. Considering the insurance policy and dividend and claim number in the first time period (0,1], there are twelve cases as follows:(1)no insurance policy is obtained, no claim occurs, and no dividend is paid in (0,1]; (2)an insurance policy is obtained, no claim occurs, and no dividend is paid in (0,1];(3)no insurance policy is obtained, a claim occurs, no dividend is paid, and 𝑈(1)≥0 in (0,1];(4)an insurance policy is obtained, a claim occurs, no dividend is paid, and 𝑈(1)≥0 in (0,1];(5)no insurance policy is obtained, a claim occurs, no dividend is paid, and 𝑈(1)<0 in (0,1];(6)an insurance policy is obtained, a claim occurs, no dividend is paid, and 𝑈(1)<0 in (0,1];(7)no insurance policy is obtained, no claim occurs, and a dividend is paid in (0,1] (if 𝑈(0)<𝑥, the case does not exist); (8)an insurance policy is obtained, no claim occurs, and a dividend is paid in (0,1] (if 𝑈(0)<𝑥, the case does not exist);(9)no insurance policy is obtained, a claim occurs, and a dividend is paid in (0,1] (if 𝑈(0)<𝑥, the case does not exist);(10)an insurance policy is obtained, a claim occurs, a dividend is paid, and 𝑈(1)≥0 in (0,1] (if 𝑈(0)<𝑥, the case does not exist);(11)no insurance policy is obtained, a claim occurs, a dividend is paid, and 𝑈(1)≥0 in (0,1] (if 𝑈(0)<𝑥, the case does not exist);(12)an insurance policy is obtained, a claim occurs, a dividend is paid, and 𝑈(1)<0 in (0,1] (if 𝑈(0)<𝑥, the case does not exist).Using the total probability formula, when 0≤𝑢<𝑥, 𝜙(𝑢)=𝑞₀𝑞𝜙(𝑢)+𝑝₀𝑞𝜙(𝑢+1)+^𝑢_𝑘=0𝜙(𝑘)𝑇(𝑢−𝑘)+𝑝^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝑇(𝑘). (13) Equation (13) is equivalent to (9).
When 𝑢≥𝑥, 𝜙(𝑢)=𝑞𝑞₀𝑝₁𝜙(𝑢−1)+𝑞₀𝑞₁+𝑝₀𝑝₁𝜙(𝑢)+𝑝₀𝑞₁𝜙(𝑢+1)+𝑝^𝑢_𝑘=0𝜙(𝑘)𝐽(𝑢−𝑘−1)+𝑝^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (14) Equation (11) comes from (14). Equation (14) is equivalent to 1−𝑞𝑞₀𝑞₁+𝑝₀𝑝₁𝜙(𝑢)=𝑞𝑞₀𝑝₁𝜙(𝑢−1)+𝑞𝑝₀𝑞₁𝜙(𝑢+1)+𝑝^𝑢_𝑘=0𝜙(𝑘)𝐽(𝑢−𝑘−1)+𝑝^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (15) Subtracting 𝑞(𝑞₀𝑝₁+𝑝₀𝑞₁)𝜙(𝑢) from (15), we obtain 𝑝𝜙(𝑢)=𝑞𝑞₀𝑝₁𝜙(𝑢−1)−𝜙(𝑢)+𝑞𝑝₀𝑞₁𝜙(𝑢+1)−𝜙(𝑢)+𝑝^𝑢_𝑘=0𝜙(𝑘)𝐽(𝑢−𝑘−1)+𝑝^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (16) When 𝑡≥𝑥, summing (16) over 𝑢 from 𝑥 to 𝑡 yields 𝑝^𝑡_𝑢=𝑥𝜙(𝑢)=𝑞𝑞₀𝑝₁𝜙(𝑥−1)−𝜙(𝑡)+𝑞𝑝₀𝑞₁𝜙(𝑡+1)−𝜙(𝑥)+𝑝^𝑡𝑢_{𝑢=𝑥𝑘=0}𝜙(𝑘)𝐽(𝑢−𝑘−1)+𝑝^𝑡_𝑢=𝑥^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (17) Interchanging the summing order of the third term in the right-hand side of (17), we get 𝑝^𝑡𝑢_{𝑢=𝑥𝑘=0}𝜙(𝑘)𝐽(𝑢−𝑘−1)=𝑝^𝑡_𝑘=0𝜙(𝑘)1−𝐽(𝑡−𝑘−1)−𝑝^𝑥−1_𝑘=0𝜙(𝑘)1−𝐽(𝑥−𝑘−2). (18) Adding (18) to (17), we obtain 𝑞𝑞₀𝑝₁𝜙(𝑥−1)−𝜙(𝑡)+𝑞𝑝₀𝑞₁𝜙(𝑡+1)−𝜙(𝑥)=𝑝∑^𝑡_𝑘=0𝜙(𝑘)𝐽(𝑡−𝑘−1)−𝑝^𝑥−1_𝑘=0𝜙(𝑘)𝐽(𝑥−𝑘−2)−𝑝^𝑡_𝑢=𝑥^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (19)
The security loading 𝛿>0 leads to lim_𝑢→+∞𝜓(𝑢)=0. 𝑓(𝑥,𝑦) is bounded function, 𝜙(𝑢)≤𝜓(𝑢)‖𝑓‖, where ‖𝑓‖=sup{𝑓(𝑥,𝑦)∣𝑥∈𝑁,𝑦∈𝑁}. Therefore, lim_𝑢→+∞𝜙(𝑢)=0. By the Dominated Convergence Theorem, we can obtain 𝑝^𝑡_𝑘=0𝜙(𝑘)𝐽(𝑡−𝑘−1)≤𝑝^𝑡+1_𝑘=1𝜙(𝑡+1−𝑘)𝐽(𝑘−2)≤𝑝^∞_𝑘=1𝜙(𝑡+1−𝑘)𝐽(𝑘−2)⟶0(𝑡⟶+∞). (20) Let 𝑡→+∞ in (19), and we get 𝑞𝑞₀𝑝₁𝜙(𝑥−1)−𝑞𝑝₀𝑞₁𝜙(𝑥)=−𝑝^𝑥−1_𝑘=0𝜙(𝑘)𝐽(𝑥−𝑘−2)−𝑝^+∞_𝑢=𝑥^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘−1). (21) Subtracting (21) from (19) −𝑞𝑞₀𝑝₁𝜙(𝑡)+𝑞𝑝₀𝑞₁𝜙(𝑡+1)=𝑝^𝑡_𝑘=0𝜙(𝑘)𝐽(𝑥−𝑘−1)+𝑝^+∞_𝑘=𝑡+1^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1). (22) Equation (12) is equivalent to (22).
Subtracting 𝑝₀𝑞𝜙(𝑢+1) from both sides of (13), we obtain 𝑝𝜙(𝑢)=𝑝₀𝑞𝜙(𝑢+1)−𝜙(𝑢)+𝑝^𝑢_𝑘=0𝜙(𝑘)𝑇(𝑢−𝑘)+𝑝^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝑇(𝑘). (23)
When 𝑥≥1, summing (23) over 𝑢 from 0 to 𝑥−1, we get 𝑝^𝑥−1_𝑢=0𝜙(𝑢)=𝑝₀𝑞𝜙(𝑥)−𝜙(0)+𝑝^𝑥−1𝑢_𝑢=0𝑘=0𝜙(𝑘)𝑇(𝑢−𝑘)+𝑝^𝑥−1_𝑢=0^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝑇(𝑘). (24) Interchanging the summation order of the second term on the right-hand side of (24), (24) is equivalent to 𝑝₀𝑞𝜙(𝑥)−𝜙(0)=𝑝^𝑥−1_𝑘=0𝜙(𝑘)𝐽(𝑥−1−𝑘)−𝑝^𝑥−1_𝑢=0^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝑇(𝑘). (25)
Replacing 𝑥 by 𝑥−1 and adding 𝑝𝜙(𝑥−1) to both sides of (25), we yield 𝑝₀𝑞𝜙(𝑥)−𝜙(0)+𝑝𝜙(𝑥−1)=𝑝^𝑥−1_𝑘=0𝜙(𝑘)𝐽(𝑥−1−𝑘)−𝑝^𝑥−2_𝑢=0^+∞_𝑘=𝑢+1𝑓(𝑢,𝑘−𝑢)𝐽(𝑘). (26) From 𝑞₁× (25) + 𝑝₁× (26), we obtain 𝑞𝑝₀𝑞₁𝜙(𝑥)+𝑞𝑝₀𝑝₁+𝑝𝑝₁𝜙(𝑥−1)−𝑝₀𝑞𝜙(0)=𝑝^𝑥−1_𝑘=0𝜙(𝑘)𝐽(𝑡−𝑘−2)−𝑝𝑞₁^𝑥−1_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖)−𝑝𝑝₁^𝑥−2_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖). (27) Equation (27) minus (21) is (9). The theorem has been proved.

According to Theorem 1, 𝜙(0),𝜙(1),…,𝜙(𝑥−1) can be obtained by solving the linear equations (8) and (9) when 𝑥≥1. And we can obtain 𝜙(𝑥+1),𝜙(𝑥+2),… by (11). The following problem that needs to be solved is whether there is a unique solution to the set of linear equations (8) and (9).

Definition 2. Assume that the matrix 𝐴=(𝑎_𝑖𝑗)∈𝐶^𝑛×𝑛, and it satisfies 𝑎_𝑖𝑖>_𝑗≠𝑖𝑎_𝑖𝑗. (28) Then 𝐴 is called a (row) strictly diagonally dominant matrix.

Lemma 3. If 𝐴 is a strictly diagonally dominant matrix, then 𝐴 is a nonsingular matrix.

Proof. For the proof, see [6].

Theorem 4. Under the assumption that the security loading 𝛿>0, the set of linear equations (8) and (9) have a unique solution.

Proof. Let 𝜙=(𝜙(0),𝜙(1),𝜙(2),…,𝜙(𝑥)), Δ=(𝛼,𝛽(0),𝛽(1),𝛽(2),…,𝛽(𝑥)), 𝜒=𝑞𝑞₀+𝑝𝑝₀𝑝(1)−1, 𝜒^′=𝑝𝑝₀𝑝(1)−𝑝, 𝑝𝐵=⁰𝑞000⋯0−𝑝₁0𝜒𝑝₀(1)𝑞00⋯000𝑝𝑇𝜒𝑝₀(2)(1)𝑞0⋯000𝑝𝑇𝑝𝑇𝜒𝑝₀()()()()(1)𝑞⋯000⋯⋯⋯⋯⋯⋯⋯⋯𝑝𝑇𝑥−1𝑝𝑇𝑥−2𝑝𝑇𝑥−3𝑝𝑇𝑥−4⋯𝑝𝑇𝜒𝑝₀𝑞; (29)then, the set of linear equations (8) and (9) can be rewritten as 𝐵𝜙=Δ. (30) The coefficient matrix 𝐵 will be carried out by a series of elementary row operations as follows: the (𝑥+1) row is replaced by itself plus the first 𝑥 rows; the 𝑥 row is replaced by itself plus the first 𝑥−1 rows, and so on; the second row is replaced by itself plus the first row. The matrix 𝐵 is changed into𝑝₀𝑞000⋯00−𝑝₁0𝜒^′𝑝₀𝑞00⋯00−𝑝₁0𝑝𝑇(1)𝜒^′𝑝₀𝑞0⋯00−𝑝₁0𝑝𝑇(2)𝑝𝑇(1)𝜒^′𝑝₀𝑞⋯00−𝑝₁0𝑝⋯⋯⋯⋯⋯⋯⋯⋯⋯𝑇(𝑥−2)𝑝𝑇(𝑥−3)𝑝𝑇(𝑥−4)𝑝𝑇(𝑥−5)⋯𝑝𝑇(1)𝜒^′𝑝₀𝑞−𝑝₁0𝑝𝑇(𝑥−1)𝑝𝑇(𝑥−2)𝑝𝑇(𝑥−3)𝑝𝑇(𝑥−4)⋯𝑝𝑇(2)𝑝𝑇(1)𝜒^′−𝑝₁𝑝₀𝑞. (31) For the first row, because 𝛿>0 and 𝜇=∑^+∞_𝑛=0𝑃(𝑛)>1, 𝑝₀𝑞−𝑝₁=𝑝₀1−𝑝−𝑝₁>𝑝₀−𝑝₁−𝑝𝜇>0. (32) For the second row, because 𝑞₀𝑞+𝑝𝑝₀𝑝(1)−1<0, then 𝑝₀𝑞−𝜒^′−𝑝₁=𝑝₀−𝑝𝑝₀𝑃(1)−𝑝−𝑝₁>𝑝₀−𝑝𝑃(1)+𝑃(0)−𝑝₁>𝑝₀−𝑝𝜇−𝑝₁>0. (33) For the 𝑖 row (2<𝑖≤𝑥−1), 𝑝₀𝑞−𝑝^𝑖−2_𝑘=1𝑇(𝑘)+𝑝𝑝₀𝑝(1)−𝑝−𝑝₁=𝑝₀−𝑝𝑝₀^𝑖−1_𝑘=1𝑃(𝑘)−𝑝𝑞₀^𝑖−2_𝑘=1𝑃(𝑘)−𝑝−𝑝₀≥𝑝₀−𝑝𝑝₀𝜇−1−𝑝𝑞₀𝜇−1−𝑝−𝑝₀=𝑝₀−𝑝𝜇−𝑝₁>0. (34) For the 𝑥 row, owing to 𝑝₀𝑞−𝑝₁=𝑝₀−𝑝₁−𝑝𝑝₀>0, 𝑝₀𝑞−𝑝₁−𝜒^′−𝑝^𝑥−2_𝑘=1𝑇(𝑘)=𝑝₀−𝑝₀𝑝^𝑥−2_𝑘=1𝑃(𝑘)−𝑞₀𝑝^𝑥−1_𝑘=1𝑃(𝑘)≥𝑝₀−𝑝𝑝₀𝜇−1−𝑝𝑞₀𝜇−1−𝑝−𝑝₀=𝑝₀−𝑝𝜇−𝑝₁>0 (35) For 𝑥−1 row, owing to 𝑝𝑝₀𝑝(1)−𝑝−𝑝₁<0, 𝑝₀𝑞−𝜒^′−𝑝₁−𝑝^𝑥−1_𝑘=1𝑇(𝑘)=𝑝₀−𝑝−𝑝₁−𝑝𝑝₀^𝑥_𝑘=1𝑃(𝑘)−𝑝𝑞₀^𝑥−1_𝑘=1𝑃(𝑘)≥𝑝₀−𝑝𝑝₀𝜇−1−𝑝𝑞₀𝜇−1−𝑝₁−𝑝=𝑝₀−𝑝𝜇−𝑝₀>0. (36) Thus, the matrix 𝐵 is a strictly diagonally dominant matrix. According to Lemma 3, matrix 𝐵 is a nonsingular matrix. So the set of linear equations have a unique solution. The theorem has been proved.

4. The Asymptotic Estimate of the Penalty Function

Let 𝐷=𝜃₁𝑋₁+𝜖₁−𝐼₁ denote the generating function of 𝐷; then 𝐺_𝐷(𝑟)=𝑝𝐺_𝑋+𝑞𝑝₁𝑟+𝑞₁𝑝₀𝑟+𝑞₀, (37) where 𝐺_𝑋 is the generating function of 𝑋.

Assumption 5. There exists a 𝑟_∞ such that 𝐺_𝑋(𝑟)→+∞ (𝑟→𝑟_∞) (𝑟_∞ is possibly +∞).
This assumption is similar to the one in [11].
Let 𝐺_𝐷(𝑟)=1, and then 𝑝𝐺_𝑋(𝑟)+𝑞𝑝₁𝑟+𝑞₁𝑝₀+𝑞₀𝑟=𝑟. (38) Let 𝐻(𝑟)=(𝑝𝐺_𝑋(𝑟)+𝑞)(𝑝₁𝑟+𝑞₁)(𝑝₀+𝑞₀𝑟). 𝐻(0)=𝑞𝑞₁𝑝₀, 𝐻(1)=1, 𝐻(𝑟) is a convex and increasing function in [0,𝑟_∞), and thus (38) has two real nonnegative roots at most, and one of them is 1. Because 𝛿>0, 𝐻^′(1)=1−(𝑝₀−𝑝𝜇−𝑝₁)<1. Because 𝐻^′′(𝑟)>0 in [0,𝑟_∞), 𝐻(𝑟) is strictly convex in [0,𝑟_∞). Therefore, there exist two real roots in (38). Denote the other root by 𝑅, and then 𝑅>1.
Note that if 𝑞₀=0, (38) becomes 𝑝𝐺_𝑋(𝑟)+𝑞𝑝₁𝑟+𝑞₁=𝑟, (39) which is just the adjustment coefficient equation of the compound binomial model with randomized decisions on dividends payment (see [10]). The following lemma will be used to derive the asymptotic estimates of 𝜙(𝑢).

Lemma 6. 𝑍 is a set of integers, {𝑎_𝑘,𝑘∈𝑍}, {𝑏_𝑘,𝑘∈𝑍}, {𝑢_𝑘,𝑘∈𝑍} are sequences that satisfy 𝑎_𝑘≥0, ∑^+∞_−∞𝑎_𝑘=1, ∑^+∞_−∞|𝑘|𝑎_𝑘<+∞, ∑^+∞_−∞𝑘𝑎_𝑘>0, ∑^+∞_−∞|𝑏_𝑘|<+∞, the greatest common divisor of the integers 𝑘 for which 𝑎_𝑘>0 is 1, and the bounded series 𝑢_𝑘 satisfies the following renewal equation: 𝑢_𝑛=^+∞_𝑘=−∞𝑎_𝑛−𝑘𝑢_𝑘+𝑏_𝑛,𝑛=0,±1,±2,…; (40) then lim_𝑛→∞𝑢_𝑛 and lim_{𝑛→−∞}𝑢_𝑛 exist. Furthermore, if lim_{𝑛→−∞}𝑢_𝑛=0, then lim_𝑛→∞𝑢_𝑛=∑^∞_𝑘=−∞𝑏_𝑘∑^∞_𝑘=−∞𝑘𝑎_𝑘. (41)

Proof. The proof can be seen in Karlin and Taylor [5, Chapter 3].

Theorem 7. The asymptotic estimate for the penalty 𝜙(𝑢) is 𝜙(𝑢)∼𝐾𝑅^−𝑢, (42) where 𝐾=𝑞𝑝₀𝑝₁(𝑅−1)+𝑝₀𝑞𝑅𝑅^𝑥𝜙(0)+()−1𝑅−1𝑞𝑞₀𝑝₁−𝑞𝑝₀𝑝₁−𝑝𝑝₁^𝑥_𝑚=1𝑅^𝑚𝜙(𝑚−1)+𝐾₁(𝑅−1)×(𝑅−1)𝐾₂𝑅+𝑝^+∞_𝑘=2𝐽(𝑘−2)⁻¹,𝐾₁=−𝑝𝑞₁𝐾₃−𝑝𝑝₁𝐾₄+𝐾₅,𝐾₂=𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑝𝑞₀𝑝₁+𝑝𝑝₀𝑝₁+𝑞₀𝑞₁,𝐾₃=^𝑥_𝑚=1𝑅^𝑚𝑚−1_𝑘=0^+∞_𝑖=𝑘+1𝑓()𝑇(𝑖),𝐾𝑘,𝑖−𝑘₄=^𝑥_𝑚=1𝑅^𝑚𝑚−2_𝑘=0^+∞_𝑖=𝑘+1𝑓()𝑇(𝑖),𝐾𝑘,𝑖−𝑘₅=^+∞_𝑚=𝑥+1𝑅^𝑚+∞_𝑘=𝑚+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1). (43)

Proof. When 𝑢>𝑥, from (12), we can obtain 𝜙(𝑢)=𝑞₀+𝑝₀𝑝₁+𝑝𝑝₁𝑞₁𝜙(𝑢)+𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑞₀𝑝₁𝑝+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁×𝜙(𝑢−1)+𝑝^𝑢−2_𝑘=0𝜙(𝑘)𝐽(𝑢−2−𝑘)+𝑝^+∞_𝑘=𝑢^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1). (44) Equation (27) is equivalent to 𝜙(𝑢)=𝑞₀+𝑝₀𝑝₁+𝑝𝑝₁𝑞₁𝜙(𝑢)+𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑞₀𝑝₁𝑝+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁×𝜙(𝑢−1)+𝑝^𝑢−2_𝑘=0𝜙(𝑘)𝐽(𝑢−2−𝑘)+𝑝₀(0)+𝑞𝜙𝑞𝑞₀𝑝₁−𝑝₀𝑝₁𝑞−𝑝𝑝₁𝜙(𝑢−1)−𝑝𝑞₁^𝑢−1_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝑇(𝑖)−𝑝𝑝₁^𝑢−2∞_{𝑘=0𝑖=𝑘+1}𝑓(𝑘,𝑖−𝑘)𝑇(𝑖). (45) Combining (44) and (45), we can obtain the renewal equation 𝜙(𝑢)=𝑞₀+𝑝₀𝑝₁+𝑝𝑝₁𝑞₁𝜙(𝑢)+𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑞₀𝑝₁𝑝+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁𝜙(𝑢−1)+𝑝^𝑢−2_𝑘=0𝜙(𝑘)𝐽()+𝑝𝑢−2−𝑘₀𝑞𝜙(0)+𝑞𝑞₀𝑝₁−𝑝₀𝑝₁𝑞−𝑝𝑝₁𝜙(𝑢−1),−𝑝𝑞∑₁^𝑢−1∑_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝑇(𝑖)−𝑝𝑝∑₁^𝑢−2∑_𝑘=0^+∞_𝑖=𝑘+1𝑓()𝑇(𝑖),𝑝𝑘,𝑖−𝑘0<𝑢≤𝑥,_+∞∑∑_𝑘=𝑢^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1),𝑥<𝑢. (46) Denoting that 𝜙(𝑢)=𝜙(𝑢)𝑅^𝑢, 𝑎_𝑢=𝑞₀+𝑝₀𝑝₁+𝑝𝑝₁𝑞₁𝑅,𝑢=0,𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑞₀𝑝₁𝑝+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁,𝑢=1,𝑝𝑅^𝑢𝐽()𝑏𝑢−2,𝑢≥2,_𝑢=𝑞𝑝₀𝑞₁𝑝,𝑢=0,₀𝑞𝜙(0)+𝑞𝑞₀𝑝₁−𝑝₀𝑝₁𝑞−𝑝𝑝₁𝜙(𝑢−1)−𝑝𝑞∑₁^𝑢−1∑_𝑘=0^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝑇(𝑖)−𝑝𝑝∑₁^𝑢−2_𝑘=0∑_𝑖=𝑘+1()𝑇(𝑖),𝑝+∞𝑓𝑘,𝑖−𝑘0<𝑢≤𝑥,_+∞∑∑_𝑘=𝑢^+∞_𝑖=𝑘+1𝑓()𝐽()𝑏𝑘,𝑖−𝑘𝑖−1,𝑥<𝑢,_𝑢=𝑝𝑅^𝑢+∞_𝑘=𝑢^+∞_𝑖=𝑘+1𝑓(𝑘,𝑖−𝑘)𝐽(𝑖−1),𝑢>𝑥. (47) Multiplying (46) by 𝑅^𝑢, we can obtain 𝜙(𝑢)=^𝑢_𝑘=0𝑎_𝑛−𝑘𝜙(𝑘)+𝑏_𝑢,𝑢=0,1,2,…. (48) We will prove that (48) satisfies the conditions of Lemma 6.
Consider ^+∞_𝑘=0𝑎_𝑘=𝑞₀+𝑝₀𝑝₁+𝑝𝑝₀𝑞₁+𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑞₀𝑝₁𝑝+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁𝑅+^+∞_𝑘=2𝑝𝐽(𝑘−2)𝑅^𝑘=𝑞₀+𝑝₀𝑝₁+𝑞₀𝑝₁𝑅+𝑅²𝑝𝑞₀𝑝₁^+∞_𝑘=0𝑃(𝑘)𝑅^𝑘+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁^+∞_𝑘=0𝑃(𝑘)𝑅^𝑘+𝑝𝑝₀𝑞₁^+∞_𝑘=0𝑃(𝑘)𝑅^𝑘=𝑞₀+𝑝₀𝑝₁+𝑞₀𝑝₁𝑅+𝑅²𝑝𝑞₀𝑝₁+𝑝𝑞₀𝑞₁+𝑝₀𝑝₁+𝑝𝑝₀𝑞₁𝐺_𝑋(𝑅)−1𝑅−1=1, (49) where the last equation is valid because 𝑅 is the root of (38). The following steps will prove that ∑^+∞_𝑘=0|𝑏_𝑘|<∞. For 𝑢>𝑥, 0≤𝑏_𝑢≥𝑝𝑓𝑅^𝑢+∞_𝑘=𝑢𝐽(𝑘−1); (50) then ^+∞_𝑢=𝑥+1𝑏_𝑢≤𝑝𝑓^+∞_𝑢=1𝑅^𝑢+∞_𝑘=𝑢𝐽(),𝑝𝑘−1_+∞𝑢=1𝑅^𝑢+∞_𝑘=𝑢𝐽(𝑘−1)=𝑝^+∞𝑘_𝑘=1𝑢=1𝐽(𝑘−1)𝑅^𝑢=𝑝^+∞_𝑘=1𝐽()𝑅𝑘−1^𝑘−𝑅=𝑝𝑅−1𝑅(𝑅−1)^∞_𝑘=2𝑅^𝑘𝐽()−𝑅𝑘−2𝑅−1^+∞_𝑘=2𝐽()≤𝑝𝑘−2𝑅(𝑅−1)^∞_𝑘=2𝑎_𝑘≤𝑝𝑅().𝑅−1 (51) From (51), we can obtain ∑^+∞_𝑢=𝑥+1𝑏_𝑢<+∞. And because |𝑏_𝑘|<+∞(0≤𝑢≤𝑥), ∑^+∞_𝑢=0𝑏_𝑢<+∞. Further, we can get ^+∞_𝑘=0𝑏_𝑘=𝑞𝑝₀𝑝₁+𝑝₀𝑞𝑅𝑅^𝑥−1()𝜙(0+𝑅−1)𝑞𝑞₀𝑝₁−𝑞𝑝₀𝑝₁−𝑝𝑝₁^𝑥_𝑚=1𝑅^𝑚𝜙(𝑚−1)+𝐾_1+∞𝑘=1𝑘𝑎_𝑘=𝐾₂+𝑝^+∞_𝑘=2𝑘𝑅^𝑘𝐽(𝑘−2). (52) According to Lemma 6, we can derive lim_𝑢→+∞𝜙=𝑞𝑝₀𝑝₁+𝑝₀𝑞𝑅𝑅^𝑥−1𝜙(0)+𝑅−1𝑞𝑞₀𝑝₁−𝑞𝑝₀𝑝₁−𝑝𝑝₁^𝑥_𝑚=1𝑅^𝑚𝜙(𝑚−1)+𝐾₁×𝐾₂𝑅+𝑝^+∞_𝑘=2𝑘𝐽(𝑘−2)⁻¹. (53) Equation (53) is equivalent to (42). The theorem has been proved.

5. The Application of the Penalty Function

We will give some examples of ruin quantities such as the ultimate ruin probability, the distribution of the surplus of the deficit at ruin to illustrate the application of the recursive formulas, and asymptotic estimates for the penalty function 𝜙(𝑢).

5.1. Ruin Quantities

Example 8. Let 𝑓(𝑥,𝑦)=1, 𝜙(𝑢)=Pr(𝑇<+∞∣𝑈(0)=𝑢)=𝜓(𝑢), which is the ultimate ruin probability. By Theorem 1, we can show that(1)𝜓(0),𝜓(1),…,𝜓(𝑥) satisfy the following linear equations: 𝑝₀𝑞𝜓(0)−𝑝₁𝜓()𝑝𝑥−1=𝛼,₀𝑞𝜓(𝑢+1)+𝑞𝑞₀+𝑝𝑝₀𝑝(1)−1𝜓(𝑢)+𝑝^𝑢−1_𝑘=0𝜓(𝑘)𝑇(𝑢−𝑘)=𝛽,𝑢=0,1,2,…,𝑥−1, (54)where 𝛼=𝑝𝑞₁^𝑥−1_𝑘=0𝑇(𝑘)+𝑝𝑝₁^𝑥−2_𝑘=0𝑇(𝑘)+𝑝^+∞_𝑘=𝑥𝐽(𝑢−1),𝛽=−𝑝𝑇(𝑢); (55)(2)for 𝑢≥𝑥, the penalty functions 𝜙(𝑢) satisfy 𝜓()=𝑞𝑢+11−𝑞₀𝑞₁+𝑝₀𝑝₁𝑞𝑝₀𝑞₁𝜓(𝑢)−𝑞₀𝑝₁𝑝₀𝑞₁𝜓()−𝑝𝑢−1𝑞𝑝₀𝑞₁^𝑢_𝑘=0𝜓(𝑘)𝐽()−𝑝𝑢−𝑘−1𝑞𝑝₀𝑞₁𝐽(𝑘−1). (56)By Theorem 7, the asymptotic estimates of the ultimate ruin probability are 𝜓(𝑢)∼𝐾_𝜓𝑅^−𝑢, (57) where 𝐾_𝜓=𝑞𝑝₀𝑝₁(𝑅−1)+𝑝₀𝑞𝑅𝑅^𝑥𝜓(0)+()−1𝑅−1𝑞𝑞₀𝑝₁−𝑞𝑝₀𝑝₁−𝑝𝑝₁×^𝑥_𝑚=1𝑅^𝑚𝜓(𝑚−1)+𝐾₁(×)𝐾𝑅−1)(𝑅−1₂𝑅+𝑝^+∞_𝑘=2𝐽(𝑘−2)⁻¹,𝐾₁=−𝑝𝑞₁𝐾₃−𝑝𝑝₁𝐾₄+𝐾₅,𝐾₂=𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑝𝑞₀𝑝₁+𝑝𝑝₀𝑝₁+𝑞₀𝑞₁,𝐾₃=^𝑥_𝑚=1𝑅^𝑚𝑚−1_𝑘=0𝑇(𝑘),𝐾₄=^𝑥_𝑚=1𝑅^𝑚𝑚−2_𝑘=0𝑇(𝑘),𝐾₅=^+∞_𝑚=𝑥+1𝑅^𝑚𝐽(𝑚−1). (58)

Example 9. Let 𝑓(𝑥,𝑦)=𝐼(𝑦≤𝑧)(𝑧=1,2,…), and then 𝜙(𝑢)=Pr((|𝑈(𝑇)|≤𝑧,𝑇<+∞∣𝑈(0)=𝑢))=𝐹(𝑢,𝑧), which is the distribution of the surplus of the deficit at ruin. By Theorem 1, we can show that (1)𝐹(0,𝑧), 𝐹(1,𝑧),…,𝐹(𝑥−1,𝑧) satisfy the following linear equations: 𝑝₀𝑞𝐹(0,𝑧)−𝑝₁𝐹()𝑝𝑥−1,𝑧=𝛼,₀𝑞𝐹(𝑢+1,𝑧)+𝑞𝑞₀+𝑝𝑝₀𝑝(1)−1𝐹(𝑢,𝑧)+𝑝^𝑢−1_𝑘=0𝜙(𝑘)𝑇(𝑢−𝑘)=𝛽,𝑢=0,1,2,…,𝑥−1, (59)where 𝛼=𝑝𝑞₁^𝑥−1_𝑘=0𝑇(𝑘)−𝑇(𝑘+𝑧)+𝑝𝑝₁^𝑥−2_𝑘=0𝑇(𝑘)−𝑇(𝑘+𝑧)+𝑝^+∞_𝑘=𝑥𝐽(𝑘−1)−𝐽(𝑘+𝑧−1)𝛽=−𝑝𝑇(𝑢)−𝑇(𝑢+𝑧); (60)(2)for 𝑢≥𝑥, the penalty functions 𝜙(𝑢) satisfy 𝐹()=𝑞𝑢+1,𝑧1−𝑞₀𝑞₁+𝑝₀𝑝₁𝑞𝑝₀𝑞₁𝐹()−𝑞𝑢,𝑧⁰𝑝₁𝑝₀𝑞₁𝐹()−𝑝𝑢−1,𝑧𝑞𝑝₀𝑞₁^𝑢_𝑘=0𝜙(𝑘)𝐽()−𝑝𝑢−𝑘−1𝑞𝑝₀𝑞₁𝐽(𝑢−1)−𝐽(𝑢+𝑧−1). (61)By Theorem 7, we can obtain the asymptotic estimates of the distribution function of deficit at ruin. Consider 𝐹(𝑢,𝑧)∼𝐾_𝐹(𝑧)𝑅^−𝑢, (62) where 𝐾_𝐹(𝑧)=𝑞𝑝₀𝑝₁(𝑅−1)+𝑝₀𝑞𝑅𝑅^𝑥𝜙(0)+()−1𝑅−1𝑞𝑞₀𝑝₁−𝑞𝑝₀𝑝₁−𝑝𝑝₁×^𝑥_𝑚=1𝑅^𝑚𝐹(𝑚−1,𝑧)+𝐾₁()×()𝐾𝑅−1𝑅−1₂𝑅+𝑝^+∞_𝑘=2𝐽(𝑘−2)⁻¹,𝐾₁=−𝑝𝑞₁𝐾₃−𝑝𝑝₁𝐾₄+𝐾₅,𝐾₂=𝑞𝑞₀𝑝₁+𝑝𝑝₀𝑞₁𝑃(1)+𝑝𝑞₀𝑝₁+𝑝𝑝₀𝑝₁+𝑞₀𝑞₁,𝐾₃=^𝑥_𝑚=1𝑅^𝑚𝑚−1_𝑘=0𝑇(𝑘)−𝑇(),𝐾𝑘+𝑧₄=^𝑥_𝑚=1𝑅^𝑚𝑚−2_𝑘=0𝑇(𝑘)−𝑇(),𝐾𝑘+𝑧₅=^+∞_𝑚=𝑥+1𝑅^𝑚𝐽(𝑚−1)−𝐽(𝑚+𝑧−1). (63)

5.2. Numerical Illustration

The initial term 𝜙(0),𝜙(1),…,𝜙(𝑥) can be obtained by solving the set of linear equations (8) and (9). 𝜙(𝑥+1),𝜙(𝑥+2),… can be computed by two approaches, which are using the recursive formulas and asymptotic estimation, respectively. We will compare the asymptotic values for the ruin probability and distribution of the deficit at ruin with the exact values computed by the recursive formulas and analysis on the impact of the randomly paying dividends on the ruin probability and distribution of the deficit at ruin.

The numerical analysis will be performed using the following assumed parameters. The distribution of claim amount 𝑋_𝑖 is a zero-truncated geometric distribution with parameter 𝛼=9/10, and then 𝑓(𝑘)=(1−9/10)(9/10)^𝑖−1, 𝑖=1,2,…; 𝑝=0.05, and the threshold 𝑥=5. The four cases with the probability of paying dividend 𝑃=(𝑝₀,𝑝₁)=(0.9,0.015),(0.75,0.015),(0.75,0.055),(0.65,0.055) will be performed. The relative security loading are larger than zero in the four case, then there is a unique solution which is adjustment coefficient in each case. The adjustment coefficient 𝑅 is computed and shown in Table 1.

Table 1 

Adjustment coefficients.

The exact values calculated by recursive formulas (11)-(12) and the asymptotic values estimated by (42) are shown in Table 2 and Table 3 for the ruin probability and the distribution of the deficit at ruin, respectively. In both of the tables, the E.V means the exact value, and the A.V means the asymptotic value.

Table 2 

Exact values and asymptotic values for the ruin probability.

Table 3 

Exact values and asymptotic values for the distribution function of the deficit at ruin.

From Table 3, we can find that the asymptotic values of ruin probability are generally close to the exact values with the surplus 𝑢 increasing under the cases 𝑃=(𝑝₀,𝑝₁)=(0.9,0.015),(0.75,0.015),(0.75,0.055),(0.65,0.055). It is easy to see that the ruin probability increases with decreasing probability of obtaining an insurance policy 𝑝₀, and increases with raising the probability of paying dividend 𝑝₁.

In Table 3, the exact values and asymptotic values for the distribution of the deficit at ruin when 𝑧=10,15 are shown. It suggests that the asymptotic values are more close to the exact values with the surplus increasing under the cases 𝑃=(0.75,0.015),(0.65,0.055).

6. Conclusions

In order to describe the surplus of the nonlife insurance companies reasonably, the compound binomial risk model with randomly charging premiums and paying dividends to shareholders is proposed in this paper. Further, we derive the recursive formulas and asymptotic estimation of penalty function using classical method. The results about penalty function are applied to obtain the recursive formulas and asymptotic estimations of the ruin probability and the distribution of the deficit at ruin. The numerical examples show that the actual penalty function can be approached by asymptotic estimation. The results about the model are meaningful to analyze the ruin problem about the joint-stock nonlife insurance companies. It may provide the reference for decision-making of the joint-stock nonlife insurance companies about risk management.

Acknowledgments

The authors would like to thank the editor and anonymous referees for their valuable comments and suggestions leading to improvement of this paper. This research is supported by the Natural Science Foundation of China under Grant no. 71203241.