1.Introduction YujuanHuang andWenguangYu StudiesonaDoublePoisson-GeometricInsuranceRiskModelwithInterference ResearchArticle

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Volume 2013, Article ID 128796,8pages http://dx.doi.org/10.1155/2013/128796

Research Article

Studies on a Double Poisson-Geometric Insurance Risk Model with Interference

Yujuan Huang

¹

and Wenguang Yu

²

1School of Science, Shandong Jiaotong University, Jinan 250023, China

2School of Insurance, Shandong University of Finance and Economics, Jinan 250014, China

Correspondence should be addressed to Yujuan Huang; [email protected] Received 11 January 2013; Accepted 5 March 2013

Academic Editor: Hua Su

Copyright © 2013 Y. Huang and W. Yu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper mainly studies a generalized double Poisson-Geometric insurance risk model. By martingale and stopping time approach, we obtain adjustment coefficient equation, the Lundberg inequality, and the formula for the ruin probability. Also the Laplace transformation of the time when the surplus reaches a given level for the first time is discussed, and the expectation and its variance are obtained. Finally, we give the numerical examples.

1. Introduction

In insurance mathematics, the classical risk model has been the center of focus for decades [1–3]. The surplus𝑈(𝑡)in the classical model at time𝑡can be expressed as

𝑈 (𝑡) = 𝑢 + 𝑐𝑡 −^𝑁∑¹^(𝑡)

𝑖=1

𝑌_𝑖, (1)

where𝑢 = 𝑈(0) > 0is the initial capital,𝑐 > 0is the constant rate of premium, and{𝑁₁(𝑡), 𝑡 ≥ 0}is a Poisson process, with Poisson rate𝜆₁ > 0 denoting the number of claims up to time𝑡. The individual claim sizes𝑌₁, 𝑌₂, . . ., independent of {𝑁₁(𝑡), 𝑡 ≥ 0}, are independent and identically distributed nonnegative random variables with common distribution function 𝐹(𝑦) with mean 𝜇_𝑌, variance 𝜎_𝑌², and moment generating function𝑀_𝑌(𝑟) = 𝐸[𝑒^𝑟𝑌].

But in the Poisson process, the expectation and variance are equal. This is obviously not consistent with actual situa- tion. So recently there is a huge amount of literature devoted to the generalization of the classical model in different ways.

Lu and Li [4] consider a Markov-modulated risk model in which the claim interarrivals, claim sizes, and premiums are influenced by an external Markovian environment process.

Tan and Yang [5] discuss the compound binomial risk model with an interest on the surplus under a constant dividend barrier and periodically paying dividends. Vellaisamy and

Upadhye [6] study the convolution of compound negative binomial distributions with arbitrary parameters. The exact expression and also a random parameter representation are obtained. Cossette et al. [7] present a compound Markov binomial model, which is an extension of the compound binomial model. The compound Markov binomial model is based on the Markov Bernoulli process which introduces dependency between claim occurrences. Recursive formulas are provided for the computation of the ruin probabilities over finite- and infinite-time horizons. A Lundberg exponen- tial bound is derived for the ruin probability, and numerical examples are also provided. Yang and Zhang [8] investigate a Sparre Andersen risk model in which the inter-claim times are generalized Erlang(n) distributed. Czarna and Palmowski [9] focus on a general spectrally negative Levy insurance risk process. For this class of processes, they analyze the so- called Parisian ruin probability, which arises when the surplus process stays below 0 longer than a fixed amount of time𝑡 > 0.

In this paper, we will consider a double Poisson- Geometric risk model with diffusion in which the arrival of policies is a Poisson-Geometric process and the claims process follows the compound Poisson-Geometric process.

For more details and new developments on the Poisson- Geometric risk model, the interested readers can refer to [10–

13].

The rest of the paper is organized as follows. InSection 2, the risk model is introduced. In Section 3, we obtain the

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ability. Then we present the effect of the related parameters on the adjustment coefficient. InSection 4, using the martingale method, the time when the surplus reaches a level firstly is considered, and the expectation and its variance are obtained.

Numerical illustrations are also given.

2. The Risk Model

Definition 1(see [10]). A distribution is said to be Poisson- Geometric distributed, denoted by𝑃𝐺(𝜆, 𝜌), if its generating function is

exp{𝜆 (𝑡 − 1)

1 − 𝜌𝑡 } , (2)

where 𝜆 > 0, 0 ≤ 𝜌 < 1. Note that if 𝜌 = 0, then the Poisson-Geometric distribution degenerates into Poisson distribution.

Definition 2(see [10]). Let𝜆 > 0 and 0 ≤ 𝜌 < 1, then {𝑁(𝑡), 𝑡 ≥ 0}is said to be a Poisson-Geometric process with parameters𝜆,𝜌if it satisfies

(1)𝑁(0) = 0;

(2){𝑁(𝑡), 𝑡 ≥ 0}has stationary and independent incre- ments;

(3) for all𝑡 > 0,𝑁(𝑡)is a Poisson-Geometric distributed with parameters 𝜆, 𝜌, and 𝐸[𝑁(𝑡)] = 𝜆𝑡/(1 − 𝜌), Var[𝑁(𝑡)] = 𝜆𝑡(1 + 𝜌)/(1 − 𝜌)².

The corresponding moment generating function of𝑁(𝑡) is𝑀_𝑁(𝑡)(𝑟) =exp[𝜆𝑡(𝑒^𝑟− 1)/(1 − 𝜌𝑒^𝑟)].

Then the double Poisson-Geometric risk model with interference is defined as

𝑈 (𝑡) = 𝑢 + 𝑐𝑁₂(𝑡) −^𝑁∑³^(𝑡)

𝑘=1

𝑌_𝑘+ 𝜎𝑊 (𝑡) , (3) where 𝑁₂(𝑡) is the number of premium up to time 𝑡 and follows a Poisson-Geometric distribution with parameters𝜆₂ and𝜌₂;𝑁₃(𝑡)is the number of claims up to time𝑡and follows a Poisson-Geometric distribution with parameters𝜆₃and𝜌₃. 𝑊(𝑡)is the standard Brownian motion and 𝜎is a constant, representing the diffusion volatility parameters. Throughout this paper, we assume that𝑁₂(𝑡),𝑁₃(𝑡),𝑊(𝑡), and{𝑌_𝑘}are mutually independent.

In order to ensure the insurance company’s stable opera- tion, we assume

𝐸 [𝑐𝑁₂(𝑡) −^𝑁∑³^(𝑡)

𝑘=1

𝑌_𝑘+ 𝜎𝑊 (𝑡)] > 0, (4) which implies

𝜆₂𝑐

1 − 𝜌₂ − 𝜆₃𝜇_𝑌

1 − 𝜌₃ > 0. (5)

Let

𝜆₂𝑐

1 − 𝜌₂ = (1 + 𝜃) 𝜆₃𝜇_𝑌

1 − 𝜌₃. (6)

Then𝜃 > 0is the relative security loading factor.

defined as

𝑇 =inf{𝑡 ≥ 0 | 𝑈 (𝑡) < 0} . (7) And define the ruin probability with an initial surplus𝑢 > 0 by𝜓(𝑢), namely,

𝜓 (𝑢) =Pr(𝑇 < ∞ | 𝑈 (0) = 𝑢) . (8)

3. The Ruin Probability

Define the profits process by{𝑆(𝑡); 𝑡 ≥ 0}; that is, 𝑆 (𝑡) = 𝑐𝑁₂(𝑡) −^𝑁∑³^(𝑡)

𝑘=1

𝑌_𝑘+ 𝜎𝑊 (𝑡) . (9)

Obviously we have

𝐸 [𝑆 (𝑡)] = [ 𝜆₂𝑐

1 − 𝜌₂ − 𝜆₃𝜇_𝑌 1 − 𝜌₃] 𝑡,

Var[𝑆 (𝑡)] = 𝑐²Var[𝑁₂(𝑡)] +Var[𝑁₃(𝑡)] ⋅ 𝐸²[𝑌_𝑘] + 𝐸 [𝑁₃(𝑡)] ⋅Var[𝑌_𝑘] + 𝜎²Var[𝑊 (𝑡)]

= [𝜆₂𝑐²(1 + 𝜌₂)

(1 − 𝜌₂)² +𝜆₃(1 + 𝜌₃) 𝜇_𝑌² (1 − 𝜌₃)² +𝜆₃𝜎_𝑌²

1 − 𝜌₃+ 𝜎²] 𝑡.

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Let

𝛼 = 𝜆₂𝑐

1 − 𝜌₂ − 𝜆₃𝜇_𝑌 1 − 𝜌₃, 𝛽 =𝜆₂𝑐²(1 + 𝜌₂)

(1 − 𝜌₂)² +𝜆₃(1 + 𝜌₃) 𝜇_𝑌²

(1 − 𝜌₃)² + 𝜆₃𝜎_𝑌² 1 − 𝜌₃ + 𝜎².

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Then

𝐸 [𝑆 (𝑡)] = 𝛼𝑡,

Var[𝑆 (𝑡)] = 𝛽𝑡. (12)

Lemma 3. The profits process{𝑆(𝑡); 𝑡 ≥ 0}has the following properties:

(1)𝑆(0) = 0;

(2){𝑆(𝑡); 𝑡 ≥ 0}has stationary and independent incre- ments.

Theorem 4. For the profits process {𝑆(𝑡); 𝑡 ≥ 0}, there is a function𝑔(𝑟)such that

𝐸 [𝑒^{−𝑟𝑆(𝑡)}] = 𝑒^{𝑡𝑔(𝑟)}. (13)

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0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.154

0.156 0.158 0.16 0.162 0.164

𝑐 𝑅

Figure 1: The impact of𝑐on𝑅.

Proof. Consider

𝐸 [𝑒^{−𝑟𝑆(𝑡)}] = 𝐸 {exp[−𝑟𝑐𝑁₂(𝑡)]} ⋅ 𝐸

× {exp[𝑟^𝑁∑³^(𝑡)

𝑘=1

𝑌_𝑘]} ⋅ 𝐸 {exp[−𝑟𝜎𝑊 (𝑡)]}

= exp{𝑡 [𝜆₂(𝑒^−𝑟𝑐− 1) 1 − 𝜌₂𝑒^−𝑟𝑐 +𝜆₃ 𝑀_𝑌(𝑟) − 1

1 − 𝜌₃𝑀_𝑌(𝑟)+1

2𝜎²𝑟²] } . (14)

Let

𝑔 (𝑟) = 𝜆₂(𝑒^−𝑟𝑐− 1)

1 − 𝜌₂𝑒^−𝑟𝑐 + 𝜆₃ 𝑀_𝑌(𝑟) − 1 1 − 𝜌₃𝑀_𝑌(𝑟)+1

2𝜎²𝑟². (15) Then we obtain (13).

Theorem 5. Equation

𝑔 (𝑟) = 0 (16)

has a unique positive solution𝑟 = 𝑅 > 0, and(16)is said to be an adjustment coefficient equation of the risk model(3)and 𝑅 > 0is said to be an adjustment coefficient.

Proof. From (15), we have𝑔(0) = 0, and since 𝑔^󸀠(𝑟) = 𝑐𝜆₂𝑒^−𝑟𝑐(𝜌₂− 1)

(1 − 𝜌₂𝑒^−𝑟𝑐)² +𝜆₃(1 − 𝜌₃) 𝐸 [𝑌𝑒^𝑟𝑌] (1 − 𝜌₃𝑀_𝑌(𝑟))² + 𝜎²𝑟, 𝑔^󸀠󸀠(𝑟) = 𝑐²𝜆₂𝑒^−𝑟𝑐(1 − 𝜌₂) (1 + 𝜌₂𝑒^−𝑟𝑐)

(1 − 𝜌₂𝑒^−𝑟𝑐)³ +𝜆₃(1 − 𝜌₃) [1 − 𝜌₃𝑀_𝑌(𝑟)]

[1 − 𝜌₃𝑀_𝑌(𝑟)]⁴

× {(1− 𝜌₃𝑀_𝑌(𝑟)) 𝐸 [𝑌²𝑒^𝑟𝑌]+ 2𝜌[𝐸 (𝑌𝑒^𝑟𝑌)]²}+𝜎², (17)

0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0.82

0.825 0.83 0.835 0.84 0.845 0.85 0.855

𝜆₂ 𝑅

Figure 2: The impact of𝜆₂on𝑅.

0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.79

0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89

𝑅

𝜆₃

Figure 3: The impact of𝜆₃on𝑅.

which imply

𝑔^󸀠(0) = − 𝜆₂𝑐

1 − 𝜌₂ + 𝜆₃𝜇_𝑌

1 − 𝜌₃ = −𝜃𝜆₃𝜇_𝑌

1 − 𝜌₃ < 0. (18) It is easy to see that the moment generating function 𝑀_𝑌(𝑟)is an increasing function. Due to0 < 𝜌₃ < 1, there exists an𝑟₁such that𝑀_𝑌(𝑟₁) = 1/𝜌₃; that is,1 − 𝜌₃𝑀_𝑌(𝑟) > 0 when0 < 𝑟 < 𝑟₁. So when 0 < 𝑟 < 𝑟₁,𝑔^󸀠󸀠(𝑟) > 0and 𝑔(𝑟)is a convex function with lim_{𝑟 → +∞}𝑔(𝑟) = +∞. Then it can be shown that𝑔(𝑟)has a unique positive solution on (0, +∞).

Example 6. Suppose𝑐 = 0.5,𝜆₂= 0.4,𝜆₃= 0.2,𝜌₂= 0.9, and 𝜌₃ = 0.6,𝛼 = 0.9,𝜎 = 1.4. By (16), we obtain the adjustment coefficient𝑅 = 0.158. Moreover, we give the effect of related parameters on adjustment coefficient𝑅; see Figures1,2,3,4, 5,6, and7.

For the profits process{𝑆(𝑡); 𝑡 ≥ 0}, let𝐹_𝑡^𝑠= 𝜎{𝑆 (V);V≤ 𝑡}.

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0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.83

0.835 0.84 0.845 0.85 0.855

𝑅

𝜌₂

Figure 4: The impact of𝜌₂on𝑅.

0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.82

0.825 0.83 0.835 0.84 0.845 0.85 0.855 0.86

𝑅

𝜌3

Figure 5: The impact of𝜌₃on𝑅.

Theorem 7. {𝐻_𝑢(𝑡); 𝐹_𝑡^𝑠; 𝑡 ≥ 0}is a martingale, where𝐻_𝑢(𝑡) = 𝑒^{−𝑟(𝑢+𝑆(𝑡))}/𝑒^{𝑡𝑔(𝑟)}.

Proof. Consider

𝐸 [𝐻_𝑢(𝑡) | 𝐹_V^𝑠] = 𝐸 [𝑒^{−𝑟(𝑢+𝑆(𝑡))} 𝑒^{𝑡𝑔(𝑟)} | 𝐹_V^𝑠]

= 𝐸 [𝑒^{−𝑟(𝑢+𝑆(V))} 𝑒^V𝑔(𝑟)

𝑒−𝑟(𝑆(𝑡)−𝑆(V))

𝑒^{(𝑡−V)𝑔(𝑟)} | 𝐹_V^𝑠]

= 𝐻_𝑢(V) 𝐸 [𝑒−𝑟(𝑆(𝑡)−𝑆(V))

𝑒^{(𝑡−V)𝑔(𝑟)} | 𝐹_V^𝑠]

= 𝐻_𝑢(V) .

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Theorem 8. If𝑟and𝑠satisfy the equation𝑔(𝑟) = 𝑠, then the surplus{𝑒^{−𝑟𝑆(𝑡)−𝑡𝑠}; 𝑡 ≥ 0}is a martingale.

0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.83

0.832 0.834 0.836 0.838 0.84 0.842 0.844 0.846

𝑅

𝛼

Figure 6: The impact of𝛼on𝑅.

1.36 1.37 1.38 1.39 1.4 1.41 1.42 1.43 1.44 1.45 0.81

0.82 0.83 0.84 0.85 0.86 0.87

𝑅

𝜎

Figure 7: The impact of𝜎on𝑅.

Proof. Consider

𝐸 [𝑒^{−𝑟𝑆(𝑡)−𝑡𝑠}| 𝐹_V^𝑠] = 𝐸 [𝑒−𝑟𝑆(𝑡)−𝑡𝑔(𝑟)| 𝐹_V^𝑠]

= 𝐸 [𝑒−𝑟𝑆(V)−𝑡𝑔(𝑟)−𝑟[𝑆(𝑡)−𝑆(V)]−(𝑡−V)𝑔(𝑟)| 𝐹_V^𝑠]

= 𝑒−𝑟𝑆(V)−𝑡𝑔(𝑟)⋅ 𝐸 [𝑒−𝑟[𝑆(𝑡)−𝑆(V)]−(𝑡−V)𝑔(𝑟)| 𝐹_V^𝑠]

= 𝑒^{−𝑟𝑆(V)−𝑡𝑠}.

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Lemma 9. The ruin time𝑇is the stopping time of𝐹_𝑡^𝑠. Theorem 10. For for all𝑟, the ultimate ruin probability satisfies

𝜓 (𝑢) ≤ 𝑒^−𝑟𝑢𝐵 (𝑟) , (21) where𝐵(𝑟) = 𝐸[sup_𝑡≥0{exp[𝑡𝑔(𝑟)]}].

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0 2 4 6 8 10 0

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

𝑢 𝑅 = 0.2

𝑅 = 0.3 𝑅 = 0.4

𝜓(𝑢)

Figure 8: The impact of𝑅on ruin probability𝜓(𝑢).

0.65 0.7 0.75 0.8 0.85 0.9

100 200 300 400 500 600 700 800 900

𝜌2

𝐸[𝜏]

Figure 9: The impact of𝜌₂on𝐸[𝜏].

Proof. For a fixed time𝑡₀,𝑡₀∧ 𝑇is a bounded stopping time;

using the theorem of martingale and stopping time, we have

𝑒^−𝑟𝑢= 𝐸 [𝐻_𝑢(0)] = 𝐸 [𝐻_𝑢(𝑇 ∧ 𝑡₀)]

= 𝐸 [𝐻_𝑢(𝑇) | 𝑇 ≤ 𝑡₀]Pr(𝑇 ≤ 𝑡₀) + 𝐸 [𝐻_𝑢(𝑡) | 𝑇 > 𝑡₀]Pr(𝑇 > 𝑡₀)

≥ 𝐸 [𝐻_𝑢(𝑇) | 𝑇 ≤ 𝑡₀]Pr(𝑇 ≤ 𝑡₀) ,

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0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75

380 400 420 440 460 480 500

𝜌3

𝐸[𝜏]

Figure 10: The impact of𝜌₃on𝐸[𝜏].

0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 300

400 500 600 700 800 900

𝜆₂

𝐸[𝜏]

Figure 11: The impact of𝜆₂on𝐸[𝜏].

which implies

Pr(𝑇 ≤ 𝑡₀) = 𝑒^−𝑟𝑢

𝐸 [𝐻_𝑢(𝑇) | 𝑇 < 𝑡₀]≤ 𝑒^−𝑟𝑢

inf_{0≤𝑡≤𝑡}₀exp[−𝑡𝑔 (𝑟)]

= 𝑒^−𝑟𝑢sup

0≤𝑡≤𝑡0

{exp[𝑡𝑔 (𝑟)]} ,

(23) by expectation on both sides of (23), and letting𝑡₀ → +∞, we can obtain (21).

Theorem 11. The probability of the risk model(3)is

𝜓 (𝑢) = 𝑒^−𝑅𝑢

𝐸 [𝑒^{−𝑅𝑈(𝑇)}| 𝑇 < ∞]. (24)

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0 0.2 0.4 0.6 0.8 1 300

400 500 600 700 800 900

𝜆₃

𝐸[𝜏]

Figure 12: The impact of𝜆₃on𝐸[𝜏].

0 0.2 0.4 0.6 0.8 1

300 400 500 600 700 800 900 1000

𝐸[𝜏]

𝜇𝑌

Figure 13: The impact of𝜇_𝑌on𝐸[𝜏].

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

200 300 400 500 600 700 800 900 1000

𝑐

𝐸[𝜏]

Figure 14: The impact of𝑐on𝐸[𝜏].

0.65 0.7 0.75 0.8 0.85 0.9

0 2000 4000 6000 8000 10000

𝜌2

Var [𝜏]

Figure 15: The impact of𝜌₂on Var[𝜏].

0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75

1000 1500 2000 2500 3000 3500

𝜌3

Var [𝜏]

Figure 16: The impact of𝜌₃on Var[𝜏].

0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 0

2000 4000 6000 8000 10000 12000

Var [𝜏]

𝜆2

Figure 17: The impact of𝜆₂on Var[𝜏].

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0 0.2 0.4 0.6 0.8 1 0

0.5 1 1.5 2 2.5

Var [𝜏]

×10⁴

Figure 18: The impact of𝜆₃on Var[𝜏].

0 0.5 1 1.5 2 2.5 3

0 0.2 0.4 0.6 0.8 1

Var [𝜏]

×10⁴

𝜇𝑌

Figure 19: The impact of𝜇_𝑌on Var[𝜏].

Proof. 𝑇is a ruin time and for a fixed time𝑡₀,𝑇_𝑢 ∧ 𝑡₀ is a bounded stopping time. Using the theorem of martingale and stopping time, we have

𝑒^−𝑟𝑢= 𝐻_𝑢(0) = 𝐸 [𝐻_𝑢(𝑇 ∧ 𝑡₀)]

= 𝐸 [𝐻_𝑢(𝑇 ∧ 𝑡₀) | 𝑇 ≤ 𝑡₀]Pr(𝑇 ≤ 𝑡₀) + 𝐸 [𝐻_𝑢(𝑇 ∧ 𝑡₀) | 𝑇 > 𝑡₀]Pr(𝑇 > 𝑡₀) .

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Let𝑟 = 𝑅, we have

𝑒^−𝑅𝑢= 𝐸 [𝑒^{−𝑅𝑈(𝑇)}| 𝑇 ≤ 𝑡₀]Pr(𝑇 ≤ 𝑡₀)

+ 𝐸 [𝑒^{−𝑅𝑈(𝑇)}| 𝑇 > 𝑡₀]Pr(𝑇 > 𝑡₀) . (26) If𝐼(𝐴)is an indicator function of the event𝐴, we get

0 2000 4000 6000 8000 10000 12000

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

𝑐

Var [𝜏]

Figure 20: The impact of𝑐on Var[𝜏].

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

3100 3200 3300 3400 3500 3600 3700

𝜎

Var (𝜏)

Figure 21: The impact of𝜎on Var[𝜏].

0 ≤ 𝐸 [𝑒^{−𝑅𝑈(𝑇)}| 𝑇 > 𝑡₀]Pr(𝑇 > 𝑡₀)

= 𝐸 [𝑒^{−𝑅𝑈(𝑇)}𝐼 (𝑇 > 𝑡₀)] ≤ 𝐸 [𝑒^{−𝑅𝑈(𝑡}⁰⁾𝐼 (𝑈 (𝑡₀) ≥ 0)] . (27) Since

0 ≤ 𝑒^{−𝑅𝑈(𝑡}⁰⁾𝐼 (𝑈 (𝑡₀) ≥ 0) ≤ 1, (28) by the law of large numbers, when𝑡₀ → ∞,𝑈(𝑡₀) → ∞ (a.s.). By dominated convergence theorem, we have

𝑡0lim→ ∞𝐸 [𝑒^{−𝑅𝑈(𝑇)}| 𝑇 > 𝑡₀]Pr(𝑇 > 𝑡₀) = 0, (a.s.) . (29) Then when𝑡₀ → ∞in (26), we can obtain (24).

Corollary 12. Consider

𝜓 (𝑢) ≤ 𝑒^−𝑅𝑢. (30)

Example 13. Suppose𝑅 = 0.2,𝑅 = 0.3, and𝑅 = 0.4. By (30), we give the effect of adjustment coefficient𝑅on the upper bound of the ruin probability; seeFigure 8.

(8)

Let

𝜏 =inf{𝑡 ≥ 0 | 𝑈 (𝑡) = 𝑥} . (31) Then 𝜏 is the time when the surplus reaches a given level firstly.

Theorem 14. The Laplace transform of𝜏is

𝐸 [𝑒^−𝑠𝜏] = 𝑒^𝑟𝑥, (32) where𝑟and𝑠satisfy

𝑔 (𝑟) = 𝑠. (33)

Proof. For the surplus process {𝑈(𝑡); 𝑡 ≥ 0}, using the theorem of martingale and stopping time, we see that𝜏is a stopping rime of𝐹_𝑡^𝑠. Let𝑄(𝑡) = 𝑒^{−𝑟𝑈(𝑡)−𝑡𝑠}. ByTheorem 8, the surplus process{𝑄(𝑡); 𝑡 ≥ 0}is a martingale; hence, we have

𝐸 [𝑄 (𝜏)] = 𝐸 [𝑄 (0)] , (34) implying that

𝐸 [𝑒^{−𝑟𝑈(𝑡)−𝑡𝑠}] = 1. (35)

Since𝑈(𝑡) = 𝑥, so we get

𝐸 [𝑒^−𝑠𝜏] = 𝑒^𝑟𝑥. (36)

Theorem 15. The expectation and variance of𝜏satisfy 𝐸 [𝜏] = 𝑥

𝛼, Var[𝜏] = 𝑥𝛽

𝛼³.

(37)

Proof. Let 𝜑(𝑠) = ln𝐸[𝑒^−𝑠𝜏]. Using Theorem 11, we have 𝜑(𝑠) = 𝑟𝑥. Then

𝑑𝜑 (𝑠)

𝑑𝑠 = 𝑑𝜑 (𝑠) 𝑑𝑟 ⋅𝑑𝑟

𝑑𝑠 = 𝑑𝜑 (𝑠)

𝑑𝑟 ⋅ 1

𝑑𝑔 (𝑟) /𝑑𝑟= 𝑥 𝑔^󸀠(𝑟), 𝑑²𝜑 (𝑠)

𝑑𝑠² =𝑑𝜑^󸀠(𝑠)

𝑑𝑠 = 𝑑𝜑^󸀠(𝑠) 𝑑𝑟 ⋅ 𝑑𝑟

𝑑𝑠 =𝑑𝜑^󸀠(𝑠)

𝑑𝑟 ⋅ 1

𝑑𝑔 (𝑟) /𝑑𝑟

= −𝑥𝑔^󸀠󸀠(𝑟) [𝑔^󸀠(𝑟)]² ⋅ 1

𝑔^󸀠(𝑟) = −𝑥𝑔^󸀠󸀠(𝑟) [𝑔^󸀠(𝑟)]³.

(38) Let𝑠 = 𝑟 = 0. We have

𝐸 [𝜏] = −𝑑𝜑 (𝑠) 𝑑𝑠 󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨𝑠=0 = − 𝑥 𝑔^󸀠(0)= 𝑥

𝛼, Var[𝜏] = −𝑑²𝜑 (𝑠)

𝑑𝑠² 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠=𝑟=0= 𝑥𝛽 𝛼³.

(39)

Example 16. Suppose𝜌₂ = 0.75,𝜌₃ = 0.75,𝜆₂ = 0.75,𝜆₃ = 0.5,𝜇_𝑌= 0.5,𝜎_𝑌= 0.5,𝜎 = 1, and𝑐 = 1. By (37), we give the effect of related parameters on𝐸[𝜏]and Var[𝜏]; see Figures9, 10,11,12,13,14,15,16,17,18,19,20, and21.

Y. Huang thanks the three anonymous referees for the thoughtful comments and suggestions that greatly improved the presentation of this paper. This work was supported by the National Natural Science Foundation of China (Grant no. 11171187, Grant no. 10921101), National Basic Research Program of China (973 Program, Grant no. 2007CB814906), Natural Science Foundation of Shandong Province (Grant no. ZR2012AQ013, Grant no. ZR2010GL013), Humanities and Social Sciences Project of the Ministry Education of China (Grant no. 10YJC630092, Grant no. 09YJC910004), and 2013 Major Project Cultivation Plan of Shandong Jiaotong Univer- sity.

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[8] H. Yang and Z. M. Zhang, “Gerber-Shiu discounted penalty function in a Sparre Andersen model with multi-layer dividend strategy,”Insurance: Mathematics and Economics, vol. 42, no. 3, pp. 984–991, 2008.

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23–28, 2007.

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1.Introduction YujuanHuang andWenguangYu StudiesonaDoublePoisson-GeometricInsuranceRiskModelwithInterference ResearchArticle

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Studies on a Double Poisson-Geometric Insurance Risk Model with Interference

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