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

Authors: Yujuan Huang; Wenguang Yu
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

In insurance mathematics, the classical risk model has been the center of focus for decades [1–3]. The surplus U(t) in the classical model at time t can be expressed as
(1)U(t)=u+ct-∑i=1N1(t)‍Yi,
where u=U(0)>0 is the initial capital, c>0 is the constant rate of premium, and {N1(t),t≥0} is a Poisson process, with Poisson rate λ1>0 denoting the number of claims up to time t. The individual claim sizes Y1,Y2,…, independent of {N1(t),t≥0}, are independent and identically distributed nonnegative random variables with common distribution function F(y) with mean μY, variance σY2, and moment generating function MY(r)=E[erY].But in the Poisson process, the expectation and variance are equal. This is obviously not consistent with actual situation. 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 exponential 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 time t>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. In Section2, the risk model is introduced. In Section 3, we obtain the adjustment coefficient equation and the formula of ruin probability. Then we present the effect of the related parameters on the adjustment coefficient. In Section 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
(2)exp{λ(t-1)1-ρt},
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 parameters λ, ρ if it satisfies (1)
N(0)=0;(2)
{N(t),t≥0} has stationary and independent increments;(3)
forallt>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) is MN(t)(r)=exp[λt(er-1)/(1-ρer)].
Then the double Poisson-Geometric risk model with interference is defined as(3)U(t)=u+cN2(t)-∑k=1N3(t)‍Yk+σW(t),
where N2(t) is the number of premium up to time t and follows a Poisson-Geometric distribution with parameters λ2 and ρ2; N3(t) is the number of claims up to time t and follows a Poisson-Geometric distribution with parameters λ3 and ρ3. W(t) is the standard Brownian motion and σ is a constant, representing the diffusion volatility parameters. Throughout this paper, we assume that N2(t), N3(t), W(t), and {Yk} are mutually independent.
In order to ensure the insurance company's stable operation, we assume(4)E[cN2(t)-∑k=1N3(t)‍Yk+σW(t)]>0,
which implies
(5)λ2c1-ρ2-λ3μY1-ρ3>0.
Let(6)λ2c1-ρ2=(1+θ)λ3μY1-ρ3.
Then θ>0 is the relative security loading factor.
For the risk model (3), the time to ruin, denoted by T, is defined as
(7)T=inf{t≥0∣U(t)<0}.
And define the ruin probability with an initial surplus u>0 by ψ(u), namely,
(8)ψ(u)=Pr(T<∞∣U(0)=u).


## 3. The Ruin Probability

Define the profits process by{S(t);t≥0}; that is,
(9)S(t)=cN2(t)-∑k=1N3(t)‍Yk+σW(t).
Obviously we have
(10)E[S(t)]=[λ2c1-ρ2-λ3μY1-ρ3]t,Var[S(t)]=c2Var[N2(t)]+Var[N3(t)]·E2[Yk]+E[N3(t)]·Var[Yk]+σ2Var[W(t)]=[λ2c2(1+ρ2)(1-ρ2)2+λ3(1+ρ3)μY2(1-ρ3)2+λ3σY21-ρ3+σ2]t.Let(11)α=λ2c1-ρ2-λ3μY1-ρ3,β=λ2c2(1+ρ2)(1-ρ2)2+λ3(1+ρ3)μY2(1-ρ3)2+λ3σY21-ρ3+σ2.
Then
(12)E[S(t)]=αt,Var[S(t)]=βt.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 increments.Theorem 4.
For the profits process{S(t);t≥0}, there is a function g(r) such that
(13)E[e-rS(t)]=etg(r).Proof.
Consider(14)E[e-rS(t)]=E{exp[-rcN2(t)]}·E×{exp[r∑k=1N3(t)‍Yk]}·E{exp[-rσW(t)]}=exp{t[λ2(e-rc-1)1-ρ2e-rc+λ3MY(r)-11-ρ3MY(r)+12σ2r2λ2(e-rc-1)1-ρ2e-rc]λ2(e-rc-1)1-ρ2e-rc}.
Let
(15)g(r)=λ2(e-rc-1)1-ρ2e-rc+λ3MY(r)-11-ρ3MY(r)+12σ2r2.
Then we obtain (13).Theorem 5.
Equation(16)g(r)=0
has a unique positive solution r=R>0, and (16) is said to be an adjustment coefficient equation of the risk model (3) and R>0 is said to be an adjustment coefficient.Proof.
From (15), we have g(0)=0, and since
(17)g′(r)=cλ2e-rc(ρ2-1)(1-ρ2e-rc)2+λ3(1-ρ3)E[YerY](1-ρ3MY(r))2+σ2r,g′′(r)=c2λ2e-rc(1-ρ2)(1+ρ2e-rc)(1-ρ2e-rc)3+λ3(1-ρ3)[1-ρ3MY(r)][1-ρ3MY(r)]4×{(1-ρ3MY(r))E[Y2erY]+2ρ[E(YerY)]2}+σ2,
which imply
(18)g′(0)=-λ2c1-ρ2+λ3μY1-ρ3=-θλ3μY1-ρ3<0.
It is easy to see that the moment generating functionMY(r) is an increasing function. Due to 0<ρ3<1, there exists an r1 such that MY(r1)=1/ρ3; that is, 1-ρ3MY(r)>0 when 0<r<r1. So when 0<r<r1, g′′(r)>0 and g(r) is a convex function with limr→+∞g(r)=+∞. Then it can be shown that g(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 coefficient R=0.158. Moreover, we give the effect of related parameters on adjustment coefficient R; see Figures 1, 2, 3, 4, 5, 6, and 7.
For the profits process{S(t);t≥0}, let Fts=σ{S(v);v≤t}.Figure 1
The impact ofc on R.Figure 2
The impact ofλ2 on R.Figure 3
The impact ofλ3 on R.Figure 4
The impact ofρ2 on R.Figure 5
The impact ofρ3 on R.Figure 6
The impact ofα on R.Figure 7
The impact ofσ on R.Theorem 7.
{
H
u
(
t
)
;
F
t
s
;
t
≥
0
} is a martingale, where Hu(t)=e-r(u+S(t))/etg(r).Proof.
Consider(19)E[Hu(t)∣Fvs]=E[e-r(u+S(t))etg(r)∣Fvs]=E[e-r(u+S(v))evg(r)e-r(S(t)-S(v))e(t-v)g(r)∣Fvs]=Hu(v)E[e-r(S(t)-S(v))e(t-v)g(r)∣Fvs]=Hu(v).Theorem 8.
Ifr and s satisfy the equation g(r)=s, then the surplus {e-rS(t)-ts;t≥0} is a martingale.Proof.
Consider(20)E[e-rS(t)-ts∣Fvs]=E[e-rS(t)-tg(r)∣Fvs]=E[e-rS(v)-tg(r)-r[S(t)-S(v)]-(t-v)g(r)∣Fvs]=e-rS(v)-tg(r)·E[e-r[S(t)-S(v)]-(t-v)g(r)∣Fvs]=e-rS(v)-ts.Lemma 9.
The ruin timeT is the stopping time of Fts.Theorem 10.
For for allr, the ultimate ruin probability satisfies
(21)ψ(u)≤e-ruB(r),
where B(r)=E[supt≥0{exp[tg(r)]}].Proof.
For a fixed timet0, t0∧T is a bounded stopping time; using the theorem of martingale and stopping time, we have
(22)e-ru=E[Hu(0)]=E[Hu(T∧t0)]=E[Hu(T)∣T≤t0]Pr(T≤t0)+E[Hu(t)∣T>t0]Pr(T>t0)≥E[Hu(T)∣T≤t0]Pr(T≤t0),
which implies
(23)Pr(T≤t0)=e-ruE[Hu(T)∣T<t0]≤e-ruinf0≤t≤t0exp[-tg(r)]=e-rusup0≤t≤t0{exp[tg(r)]},
by expectation on both sides of (23), and letting t0→+∞, we can obtain (21).Theorem 11.
The probability of the risk model (3) is
(24)ψ(u)=e-RuE[e-RU(T)∣T<∞].Proof.
T is a ruin time and for a fixed time t0, Tu∧t0 is a bounded stopping time. Using the theorem of martingale and stopping time, we have
(25)e-ru=Hu(0)=E[Hu(T∧t0)]=E[Hu(T∧t0)∣T≤t0]Pr(T≤t0)+E[Hu(T∧t0)∣T>t0]Pr(T>t0).
Let r=R, we have
(26)e-Ru=E[e-RU(T)∣T≤t0]Pr(T≤t0)+E[e-RU(T)∣T>t0]Pr(T>t0).
IfI(A) is an indicator function of the event A, we get
(27)0≤E[e-RU(T)∣T>t0]Pr(T>t0)=E[e-RU(T)I(T>t0)]≤E[e-RU(t0)I(U(t0)≥0)].
Since(28)0≤e-RU(t0)I(U(t0)≥0)≤1,
by the law of large numbers, when t0→∞, U(t0)→∞ (a.s.). By dominated convergence theorem, we have
(29)limt0→∞E[e-RU(T)∣T>t0]Pr(T>t0)=0,(a.s.).
Then when t0→∞ in (26), we can obtain (24).Corollary 12.
Consider(30)ψ(u)≤e-Ru.Example 13.
SupposeR=0.2, R=0.3, and R=0.4. By (30), we give the effect of adjustment coefficient R on the upper bound of the ruin probability; see Figure 8.Figure 8
The impact ofR on ruin probability ψ(u).


## 4. The Time to Reach a Given Level

Let(31)τ=inf{t≥0∣U(t)=x}.
Then τ is the time when the surplus reaches a given level firstly.Theorem 14.
The Laplace transform ofτ is
(32)E[e-sτ]=erx,
where r and s satisfy
(33)g(r)=s.Proof.
For the surplus process{U(t);t≥0}, using the theorem of martingale and stopping time, we see that τ is a stopping rime of Fts. Let Q(t)=e-rU(t)-ts. By Theorem 8, the surplus process {Q(t);t≥0} is a martingale; hence, we have
(34)E[Q(τ)]=E[Q(0)],
implying that
(35)E[e-rU(t)-ts]=1.
Since U(t)=x, so we get
(36)E[e-sτ]=erx.Theorem 15.
The expectation and variance ofτ satisfy
(37)E[τ]=xα,Var[τ]=xβα3.Proof.
Letφ(s)=lnE[e-sτ]. Using Theorem 11, we have φ(s)=rx. Then
(38)dφ(s)ds=dφ(s)dr·drds=dφ(s)dr·1dg(r)/dr=xg′(r),d2φ(s)ds2=dφ′(s)ds=dφ′(s)dr·drds=dφ′(s)dr·1dg(r)/dr=-xg′′(r)[g′(r)]2·1g′(r)=-xg′′(r)[g′(r)]3.
Let s=r=0. We have
(39)E[τ]=-dφ(s)ds|s=0=-xg′(0)=xα,Var[τ]=-d2φ(s)ds2|s=r=0=xβα3.Example 16.
Supposeρ2=0.75, ρ3=0.75, λ2=0.75, λ3=0.5, μY=0.5, σY=0.5, σ=1, and c=1. By (37), we give the effect of related parameters on E[τ] and Var[τ]; see Figures 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, and 21.Figure 9
The impact ofρ2 on E[τ].Figure 10
The impact ofρ3 on E[τ].Figure 11
The impact ofλ2 on E[τ].Figure 12
The impact ofλ3 on E[τ].Figure 13
The impact ofμY on E[τ].Figure 14
The impact ofc on E[τ].Figure 15
The impact ofρ2 on Var[τ].Figure 16
The impact ofρ3 on Var[τ].Figure 17
The impact ofλ2 on Var[τ].Figure 18
The impact ofλ3 on Var[τ].Figure 19
The impact ofμY on Var[τ].Figure 20
The impact ofc on Var[τ].Figure 21
The impact ofσ on Var[τ].



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