The expected time to ruin in a risk process with constant
barrier via martingales
Esther Frostig
Department of Statistics
University of Haifa, 31905, Israel
Email: [email protected]
July 17, 2004
Abstract
Two risk models with a constant dividend barrier are considered. In the two models
claims arrive according to a Poisson process. In the first model the claim size has a phase
type distribution. In the second model the claim size is exponentially distributed, but
the arrival rate, the mean claim size, and the premium rate are governed by a random
environment, which changes according to a Markov process. Kella and Whitt (1992)
martingale is applied in the first model. Asmussen and Kella (2000) multidimensional
martingale is applied in the second model. The expected time to ruin and the amount
of dividends paid until ruin occurs are obtained for both models.
Key-words: martingales, time to ruin, exponential distribution, phase type distribition, Markov additive process, Lévy process, reflected process, Laplace transform.
1
1
Introduction
Let U (t), t ≥ 0 be a risk process, i.e. U (t) = u − X(t), where u is the initial surplus,
N (t)
X(t) =
X
Vj − ct.
(1)
j=0
N (t) is the number of claims arriving up to time t , Vj is the claim size of the jth claim,
and c is the premium rate. In the classical compound Poisson risk process, N (t) is a
Poisson process at rate λ, and Vj , j = 1, 2, · · · are independent identically distributed
(i.i.d.) random variables with distribution function (df) G, moment generating function
MV (α), and mean µ. Ruin occurs the first time that the process becomes negative. Let
b > u. When the surplus reaches the barrier b, all the premium income is paid as dividends.
That is, dividends are paid continuously at rate c until a new claim occurs. Let Ub (t) be
the surplus process under the barrier strategy b. Let τ be the time that ruin occurs,
i.e. τ = inf t≥0 {Ub (t) < 0}. In this model ruin occurs with probability 1. Thus, we are
interested in the distribution of the time until ruin occurs, the moments of this time, and
the distribution of the deficit at ruin. Another interesting quantity is the total amount of
dividends paid until ruin occurs. Two models are considered:
1. Model 1: Claims arrive according to a Poisson process and the claim size has a phase
type distribution.
2. Model 2: The risk process is governed by a random environment. The environment
state at time t, {J(t), t = 0} is a jump Markov process with n states. When the
environment state is j, j = 1, · · · , n, the claims arrive according to a Poisson process
at rate λj , the claim size is exponentially distributed with rate δj , and the premium
rate is cj .
The main tools to find E[τ ] are Kella and Whitt (1992) martingale and the multivariate
version of this martingale introduced in Asmussen and Kella (2000). These tools were exploited by Asmussen et. al. (2002) to find the expected time until the number of customers
in a Markov modulating queueing system exceeds the level n. The same tools were also
exploited by Kella et. al. (2003) to analyze a stochastic clearing system with Lévy components. Perry and Stadje (2000) applied similar methods to study the behavior of the time
to bankruptcy for a stochastic cash management system where the cash flow is modelled
as a Lévy process.
The novelty in this paper is the derivation of the expected time to ruin, the distribution of the time to ruin, and the total dividends paid until ruin occurs in the case of
2
horizontal barrier, via Kella and Whitt (1992) martingale or Asmussen and Kella (2000)
multidimensional martingale.
De Finetti (1940) was the first to suggest the risk surplus process under the assumption
that dividends are paid when the surplus process reaches a given barrier b. Bühlmann
(1970) discusses various dividend policies for the surplus process in discrete time. For the
continuous time he presents an integro-differential equation for the expected discounted
dividends paid until ruin for the fixed barrier strategy. He arrives at an explicit solution to
this equation for the case where the claim size is exponentially distributed. Gerber (1979)
also discusses an integro-differential equation for the cumulative expected discounted dividends pay-out. By presenting a certain martingale he got an expression for the Laplace
transform of the time to ruin. Explicit results are presented only for the exponentially
distributed claim size. Segerdahl (1970) discusses the ruin probability in the case of fixed
absorbing barrier, i.e. the risk process is stopped once reaching the barrier. He found an
explicit solution for the case of exponentially distributed claim size. Dickson and Gray
(1984a) extended Segerdahl (1970) results to the cases of gamma and hyper-exponential
claim size distributions. In another paper, Dickson and Gray (1984b) developed an alternative method to approximate the ruin probability in the Segerdahl (1970) model. Lately,
the barrier strategy has been studied by several researchers. In some papers the focus is
on the optimality of the barrier strategy for different optimality criteria. Assuming that
the company reduces risk by reinsurance, Asmussen et. al. (2000) studied the optimal
retention level and the optimal dividend policy which maximize the cumulative expected
discounted dividends pay-out. Højgaard (2002) discusses the optimal premium policy in
order to maximize the expected total discounted pay-out of dividends. The other stream
of research is devoted to the analysis of probabilistic properties and the calculations of
various quantities as the time to ruin, and the deficit at the time to ruin, for various barrier
models. Wang and Politis (2002) studied some properties of the classical risk model with
a constant barrier. Let Aj be the inter-arrival time between the j − 1 and the jth claim,
P
and let Yj = cAj − Vj , j = 1, 2, · · · . For n ≥ 1 define Sn = nj=1 Yj . Wang and Politis
(2002) discuss the distribution of ρb = inf {n : Sn > b}. Applying the theory of a general
random walk, they find the expected time until the random walk Sn crosses the level b.
Finally, applying a zero-mean martingale, they arrive at an approximation and an upper
bound for the expected time that the surplus process reaches level 0 or level b, and for the
expected time that the process reaches the level b, given that level b is reached before level
0. However, these results are obtained assuming that the surplus can be negative, i.e. the
risk process is not stopped when ruin occurs. Irbäck (2003) discusses the time to ruin in
the classical model, assuming that the barrier b is asymptotically infinity. When u is close
3
to the barrier he proved that the time to ruin is asymptotically exponentially distributed.
He also obtained the Laplace transform of the time to the first visit to the dividend barrier,
given the process reaches the barrier, and the Laplace transform of the time between two
visits to the barrier. When the initial reserve u is very small compared to the barrier, he
got the asymptotic distribution of the time to ruin, an approximation for the probability
of ruin before reaching the barrier, and the conditional probability of ruin before reaching
the barrier. Irbäck (2003) also obtained an upper bound for the expected time to reach
the barrier before ruin, given this event occurs. Lin Willmot and Drekic (2003), apply the
well-known discounted penalty function defined by Gerber and Shiu (1998). They obtained
an integral equation for the expected penalty function. The moments of the time to ruin
are given by mathematical expression that are functions of the ruin probabilities in the case
without barriers. Simplified expression are given for the cases where the claim size is either
exponentially distributed or is a mixture of two exponentials.
2
Model 1: Phase type claim size distribution
In this section we assume that the claim size distribution G is phase type. That is, Vi
is distributed as the time to absorption in a terminal state in a continuous time Markov
process with n states, initial probability vector θ, and a transition rate matrix T . Thus
1 − G(x) = θeT x 1, 1 is a column vector of 1’s. Let T o = −T 1 be a column vector of the
transition rates to the absorbing state. Note that E[Vi ] = θT −1 1.
Let Ub (t) be the surplus process under the barrier strategy b, and let Z(t) = b − Ub (t).
Let
L(t) = − inf {b − u + X(s)}− ,
0≤s≤t
(2)
where x− = min {x, 0}. The process L(t) is known as the local time. Note that the total
dividends paid until ruin is L(τ ). It is not difficult to see that
Z(t) = b − u + X(t) + L(t).
(3)
X(t) is a Lévy process with an exponent ϕ(α),
ϕ(α) = log[EeαX(1) ] = −cα + λMV (α) − λ,
(4)
Z(t) is called a reflected Lévy process. Note that the time to ruin is τ = inf t>0 {Z(t) > b}.
Remark 1 Note that L(t) increases at t, if and only if, Z(t) = 0.
4
2.1
The basic martingale
Let X(t) be a Lévy process with no negative jumps, and a Lévy exponent ϕ(α). Let
{Y (t), t ≥ 0} be an adapted process, and let Z(t) = Z(0) + X(t) + Y (t). Theorem 2 in Kella
and Whitt (1992) yields that under certain conditions on {Y (t), t ≥ 0}, the process
Z
M (α, t) = ϕ(α)
t
eαZ(s) ds + eαZ(0) − eαZ(t) + α
0
Z
t
eαZ(s) dY (s)
(5)
0
is martingale. Let X(t) and L(t) be as in Equations (1), and (2), respectively. Substitute
Y (t) = L(t), and Z(0) = b − u, in Equation (5). The process {L(t), t ≥ 0} is continuous
with finite expected variation, thus Theorem 2 in Kella and Whitt (1992) holds. Applying
Remark 1 yields that:
t
Z
eαZ(s) ds + eαZ(0) − eαZ(t) + αL(t).
M (α, t) = ϕ(α)
(6)
0
Since L(0) = 0, we conclude that M (α, t) is a zero-mean martingale. Applying (6), and
the optional sampling theorem, Asmussen and Kella (2001) proved that (See Lemma 3 p.
52):
E[Z(τ )] = Z(0) + ϕ0 (0)E[τ ] + E[L(τ )].
(7)
To apply this equation to our problem we have to find E[L(τ )] and E[Z(τ )]. Note that L(τ )
is the amount of dividends paid until ruin occurs. We first present an intuitive method to
find the distribution of L(τ ). Next, we derive a method to calculate E[Z(τ )] and E[L(τ )]
by solving a set of linear equations.
2.2
The distribution of L(τ )-method 1
Each time that the surplus process hits the barrier b, or equivalently the process Z(t) hits
0, L(·) increases at rate c for a time interval which is exponentially distributed with rate λ.
Let Nb (u) be the number of times that the surplus process hits b, and let Wj , j = 1, 2, · · ·
be the interval length where L(t) increases at the j − th time, i.e., Wj , j = 1, 2, · · · are i.i.d
exponentially distributed with rate λ.
Nb (u)
L(τ ) = c
X
Wj .
j=1
Let qu be the probability of hitting the barrier before ruin and let pu = 1 − qu . Let q be
the probability of hitting the barrier before ruin, given that the initial reserve is b, and let
5
p = 1 − q. It is shown in Irbäck (2003), Dickson and Gray (1984) that
qu =
1 − ψ(u)
,
1 − ψ(b)
where ψ(u) is the ruin probability, when the initial capital is u. The ruin probability in
the compound Poisson process, when the claim size has a phase type distribution (See,
Asmussen (2000), p. 227) is:
o
ψ(u) = θ + e(T +T θ+ )u 1,
where
λ
θ + = − θT −1 ,
c
and T o θ+ is a n × n matrix whose i, j element is Tio θj . Thus,
Z b
o
1 − θ+ e(T +T θ+ )(b−x) 1 θeT x T o dx
0
Z b
o
T
b
= 1 − θe 1 −
θ+ e(T +T θ+ )(b−x) 1θeT x T o dx.
q(1 − ψ(b)) =
(8)
0
Similar to the derivation of the convolution of two phase type densities it can be shown
that the integral in (8) is equal to:
(θ+ , 0n )eAb
0>
n
To
0
= (θ, 0n )eA b
0>
n
1
,
where, 0n (0>
n ) is a row (column) vector with n zero’s,
T + T o θ+ 1θ
A=
0
T
and,
A0
=
T
0
T o θ+
T + T o θ+
.
The total sum of dividends paid until ruin is 0 with probability pu , and with probability qu
it is exponentially distributed with parameter pλ/c. The expected sum of dividends paid
until ruin occurs is qu c/pλ.
2.3
The distribution of Z(τ )
Clearly Z(τ ) is equal to b + ξ, where ξ is the deficit at ruin and has a phase type distribution with presentation (π, T ), where, π = (π1 , · · · , πn ) is an n dimensional row probability
vector, where πj is the probability that at the time of ruin the claim phase is j, or equivP
alently, the process Z(t) hits b at phase j. Note that n1 πj = 1. We will use the fluid
6
description of the risk process. Thus, if there is an upwards jump of size v in the process
Z(t), the fluid process increases at rate 1 for v time units. This description enables us
to apply the multi-dimensional martingale for Markov additive process introduced in Asmussen and Kella (2000), to derive equations for the probabilities πj , j = 1, · · · , n, and for
E[L(τ )]. Denote by X 0 (t) the fluid model corresponding to X(t). The process X 0 (t) is a
Markov additive process, where the modulating states are 0, 1, · · · , n. When in state 0 the
process decreases at slope c, and when in states 1, . . . , n the process increases at slope 1.
The underlying modulating process is a continuous time Markov process with transition
rate matrix
Q=
Let
−λ λθ
To T
F (α) = Q +
.
−cα
0
0
αIn
,
(9)
where In is an n dimensional identity matrix. Note that the diagonal elements of the
second matrix in (9) are the Lévy exponent of the process X 0 (t) on J(t) = j. Let 1j be
an n + 1 dimensional row vector with all components 0, except the j-th which is 1. Let
L0 (t) = − inf 0≤s≤t {X 0 (t) + b − u}− , and let Z 0 (t) = b − u + X 0 (t) + L0 (t). By Theorem 2.1
of Asmussen and Kella (2000),
Z t
Z t
αZ 0 (t)
αZ 0 (0)
αZ 0 (t)
e
1J(s) dsF (α) + e
1J(0) − e
1J(t) + α
1J(s) dL0 (s)
0
(10)
0
is row vector zero mean martingale. The determinant of F (α) is a polynomial of degree
n + 1. Let α0 , · · · , αn be the roots of det F (α) = 0, and let h(αj ) be the eigenvectors
corresponding to the eigenvalue 0 of F (αj ), i.e. F (αj )h(αj ) = 0, j = 0, · · · , n. Clearly one
root, say α0 , is zero, and its corresponding eigenvector h(αo ) is 1.
Multiplying (10) by h(αj ) yields that
αj Z 0 (0)
e
αj Z 0 (t)
hJ(0) (αj ) − e
Z
hJ(t) (αj ) + αj
t
hJ(s) (αj )dL0 (s)
(11)
0
is a zero-mean martingale. Note that dL0 (s) > 0 if, and only if, J(s) = 0. Applying the
optional stopping theorem yields that:
eαj (b−u) h0 (αj ) − eαj b
m
X
πi hi (αj ) + αj h0 (αj )E[L(τ )] = 0.
(12)
i=1
Thus, we obtain n + 1 equations for π1 , · · · , πn and E[L(τ )].
Corollary 2
1. The deficit at the time of ruin has a phase type distribution with repre-
sentation (π, T ), and expectation −πT −1 1.
7
2. E[τ ] =
b+(−πT −1 1)−(b−u)−E[L(τ )]
.
−c+λ/δ
Proof. Statement 1 is straightforward. Statement 2 follows from (7) and statement 1.
3
Example: The exponential claim size distribution
In this section we analyze the case where the claim size is exponentially distributed with
rate δ. Assuming that cδ − λ > 1, The ruin probability is:
ψ(u) =
λ −(δ− λ )u
c
e
.
cδ
Thus,
δc−λ
λ − c u
1 − cδ
e
1 − ψ(u)
qu =
=
.
λ − δc−λ
1 − ψ(b)
1 − cδ
e c b
(8) yields that,
Z
b
q(1 − ψ(b)) =
(1 −
o
λ − δc−λ (b−x) −δx
e c
)δe dx.
cδ
After some calculation we arrive at
q=
1 − e−
1−
b
λ − δc−λ
c
cδ e
and
p=1−q =
δc−λ
b
c
e−
δc−λ
b
c
1−
,
λ
δc )
,
b
λ − δc−λ
c
e
cδ
(1 −
thus,
E[Lτ ] =
u
λ − δc−λ
c
c
cδ e
.
− δc−λ
b
λ
λ
e c (1 − δc )
1−
(13)
An alternative method to compute E[Lτ ] is via the fluid model. Consider a two state
Markov additive process (MAP) (Xt , Jt ). Jt is a Markov jump process with states 1 and
2, where the sojourn time at 1 is exponential with rate λ and the sojourn time at 2 is
exponential with rate δ. While in state 1 X is linear with slope −c, and while in state 2,
X is linear with slope 1. The matrix F (α) is as follows:
−λ
λ
−cα 0
−(cα) + λ
λ
F (α) =
+
=
.
δ −δ
0 α
δ
−δ + α
The equation det F (α) = 0 has two roots: α1 = 0 and α2 =
8
δc−λ
c .
The right eigenvectors for the eigenvalue 0 for F (αj ), j = 1, 2 are:
h(0) = (1, 1)>
h(
(14)
δc − λ
δc
) = (1, )> .
c
λ
Applying equations (11) and (12) and taking expectation, yields that
e
δc−λ
(b−u)
c
−e
δc−λ
b
c
δc δc − λ δc
+
E[Lτ ] = 0.
λ
c
λ
This equation yields an expression for E[Lτ ]. Applying (7), and noting that ϕ0 (0) =
−c + λ/δ, we arrive at
E[τ ] =
b + 1/δ − (b − u) − E[Lτ ]
.
−c + λδ
Next we will find the moment generating function of the time to ruin. Let Yt = Lt + β/αt.
Equation (5) yields that:
Z
t
α(Zs +β/α)s
M̃ (β, α, t) = (ϕ(α) + β)
e
αZ0
ds + e
αZt +βt
−e
Z
t
+α
0
eα(Zs +β/αs) dLs
(15)
0
is a zero mean martingale. Solving the equation
ϕ(α) + β = 0
yields two solutions: α1 (β) and α2 (β), where
α1,2 (β) =
cδ − λ + β ±
p
(cδ − λ + β)2 − 4δβc
2c
(16)
Substituting α1,2 (β) in Equation (15) yields that
αj (β)Z0
M̃ (β, α, t) = e
αj (β)Zt +βt
−e
Z
+ αj (β)
t
eβs dLs
(17)
0
is a zero martingale. Applying the optional sampling theorem yields that
Z τ
αj (β)(b−u)
αj (β)Zτ
βτ
e
− E[e
]E[e ] + αj (β)E[
eβs) dLs ] = 0.
(18)
0
Note that
E[eαj (β)Zτ ] = eαj (β)b
δ
.
δ − αj (β)
Rτ
Therefore we can solve for E[eβτ ], and E[ 0 eβs dLs ]. Similarly, we can derive the Laplace
transform of the time to ruin.
9
We can exploit equations (16) and (17) to find the mean and the variance of τ . This
is accomplished by taking first and second derivatives of (17),at β = 0. Let f (t) be a
real valued function. In the sequel we denote by f 0 (t) and f 00 (t) the first and the second
derivative of f at t. Note that
cδ − λ
c
α2 (0) = 0
α1 (0) =
−λ
c(cδ − λ)
δ
cδ − λ
α10 (0) =
α20 (0) =
2δλ
(cδ − λ)3
2δλ
.
cδ − λ
α100 (0) = −
α200 (0) =
Taking first derivative of (18) with respect to β, yields:
αj0 (β)E[(b − u)eαj (β)(b−u) ] − αj0 (β)E[Zτ eαj (β)Zτ +βτ ] − E[τ eαj (β)Zτ +βτ ]
Z τ
Z τ
0
βs
+αj (β)E[
e dLs ] + αj (β)E[
seβs dLs ] = 0.
0
(19)
0
Note that:
eαj (0)b δ
δ − αj (0)
1
δ − αj (0)
δ
E[τ eαj (0)Zτ ] = E[τ ]eαj (0)b
δ − αj (0)
Z τ
E[
dLs ] = E[Lτ ].
E[Zτ eαj (0)Zτ ] =
b+
0
Thus, substituting in (19) β = 0, j = 1 (the non-zero root), E[τ ], E[L(τ ] yields an
Rτ
expression for E[ 0 sdLs ]. Taking the second derivative of (18) with respect to β, yields:
αj00 (β)E[Z0 eαj (β)Z0 ] + (αj0 (β))2 E[(Z0 )2 eαj (β)Z0 ] − (αj0 (β))2 E[(Zτ )2 eαj (β)Zτ +βτ ]
−αj00 (β)E[Zτ eαj (β)Zτ +βτ ] − 2αj0 (β))E[τ Zτ eαj (β)Zτ +βτ ] − E[τ 2 eαj (β)Zτ ]
Z τ
Z τ
Z τ
+αj00 (β)E[
eβs dLs ] + 2αj0 (β)E[
seβs dLs ] + αj (β)E[
s2 eβs dLs ] = 0.
0
0
0
10
(20)
Note that,
E[(Z0 )2 ] = (b − u)2
2
b
E[Zτ2 ] = b2 + 2 + 2
δ δ 1
E[τ Zτ ] = E[τ ] b +
.
δ
Rτ
Thus, substituting β = 0, the zero root α2 (0) = 0, the expression for E[ 0 sdLs ] in (20)
yields an expression for E[τ 2 ].
4
A Markov modulating risk process with constant barrier
The risk process is governed by a random environment. The environmental state at time
t is J(t), where J(t), t ≥ 0 is a Markov jump process with n states and a transition rate
matrix Q. When the environmental state is j, j = 1, · · · , n, the claims arrive according to
a Poisson process at rate λj , the claim size is exponentially distributed with rate δj and
the premium rate is cj . When the surplus reaches level b all the premium income is paid
as dividends until a new claim occurs.
4.1
Some results for X(t)
Let X(t) =
PN (t)
j=1
Vj −
Pm R t
j=1 0
cj I(J(s) = j)ds, where N (t) is the number of the claim
arrivals up to time t, and I is the indicator function. Note that on {J(t) = j}, X(t) is a
Lévy process with an exponent:
ϕj (α) = −αcj + λj Mj (α) − λj = −αcj + λj
δj
− λj .
λj − α
Let Ft (α) be a matrix where its i, j entry is Ei [eαXt ; Jt = j]. It is shown in Chapter XI p.
311 of Asmussen (2003) that Ft (α) = etK(α) , where
K(α) = Q + ∆(ϕ(α)) = Q + (−α∆(c) + ∆(λ
δ
) − ∆(λ)),
δ−α
(21)
where ∆(a) is a diagonal matrix with the elements (a1 , · · · , an ) on the diagonal. The
matrix K(α) has a real eigenvalue κ(α) with maximal real part (cf. chapter XI, p. 312
Asmussen (2003)). The corresponding left and right eigenvectors ν α and hα may be chosen
with strict positive components. Moreover, with out loss of generality it can be assumed
that ν α hα = 1, and that πhα = 1, where π = ν 0 is the stationary probability vector for
Q, and h0 is a column vector of 1’s, (See Section 2c, p. 312 Asmussen (2003)). Proposition
2.4, p. 312 in Asmussen (2003), implies that:
11
1. Ei [eαX(t) hαJt ] = eκ(α)t hαi .
2. eαX(t)−tκ(α) hαJt is a martingale.
Let k = (k1 , · · · , kn ) be the derivative of hα at α = 0.
Differentiating the expression in 2 at α = 0 yields that (See Corollaries 2.6 , Chapter
XI p. 313 Asmussen (2003)):
E[X(τ )] = mE[τ ] + E[kJ(0) ] − E[kJ(τ ) ],
(22)
where, by Corollaries 2.7 and 2.8 , Chapter XI, p. 313 Asmussen (2003)):
0
m = κ (0) =
n
X
πj (λj /δj − cj ).
(23)
j=1
To find the vector k we apply the same technique as in Asmussen and Kella (2000, p. 385),
and Asmussen et. al. (2002, p. 68 ). Note that
κ(α)hα = K(α)hα .
(24)
Taking the derivative of Equation (24) at α = 0, yields that
λ
κ0 (0)h0 + κ(0)k = (−∆(c) + ∆( ))h0 + Qk.
δ
(25)
Substituting κ(0) = 0, and (23), in (25), and then subtracting from the two sides 1πk = 0,
yields that
λ
m1 = (−∆(c) + ∆( ))1 + (Q − 1π)k.
δ
Thus
λ
k = (Q − 1π)−1 (m + ∆(c) − ∆( ).
δ
4.2
(26)
The expected time to ruin
Let Lj (t) be the total time up to time t that dividends are paid while the environment is at
Rτ
P
j, and `j = E[ 0 I(J(s) = j)dL(s)] = E[Lj (τ )], j = 1, · · · , n. Note that E[L(τ )] = n1 `i .
Let pj , be the probability that the process Z(t) hits b (Ub (t) hits 0) when the environment
is at state j state j = 1, · · · , n. Denote by 1j an n dimensional row vector of which the jth
entry is equal to 1 and all the other entries are 0. Let υ(α) and h(α) be an eigenvalue and
12
a right eigenvector of K(α). To apply equation (22) we exploit the following n-dimensional
process:
t
Z
eαZ(s) 1J(s) dsK(α) + eαZ(0) 1J(0) − eαZ(t) 1J(t) + α
M (α, t) =
Z
0
t
1J(s) dL(s).
(27)
0
By Theorem 2.1 of Asmussen and Kella (2000), M (α, t) is an n dimensional zero-mean
martingale. det(K(α)) can be written as a quotient of two poliynoms where the numerator
is of degree 2n. Assume that the numerator has 2n roots, α1 , · · · , α2n . Note that one root
is 0. Let h[j] be a column vector satisfying K(αj )h[j] = 0. By multiplying (27) by h[j] for
j = 1, · · · , 2n, and taking expextation, we arrive at the following 2n equations for pj , and
`j , j = 1, · · · , n.
[j]
E[eαj Z(0) hJ(0) ]
−
n
X
n
αj b
pi e
i=1
X
δi
[j]
[j]
hi + αj
`i hi = 0.
δi − αj
i=1
Since X(τ ) = Z(τ ) − (b − u) − L(τ ),
E[X(τ )] = E[Z(τ )] − (b − u) − E[L(τ )]
n
n
X
X
1
=
pi (b + ) −
`i − (b − u).
δi
i=1
(28)
i=1
Substituting (28) and (26) in (22) yields E[τ ].
5
Numerical examples
In this section we present some numerical computations. First we consider the case where
the initial reserve u = 50, claims arrive according to a Poisson process at rate λ = 1, the
claim size is exponentially distributed with rate δ, and the premium rate is c = 1, i.e. the
loading factor is ρ = cδ/λ − 1 = δ − 1. Figure 1 and Figure 2 present the expected time
to ruin, Figure 3 and Figure 4 exhibit the mean of the total dividends paid until ruin,
Figure 5 and Figure 6 show the squared coefficient of variation (scv) of the time to ruin,
(scv =
V ar(τ )
).
(E[τ ])2 )
13
Expected time to ruin.
Expected dividends until ruin
Squared coefficient of variation of the time to ruin
Figure 7 exhibits F̄τ (t) = P (τ > t) for different loading factors ρ (ρ = 1 − c/λE(V )),
and Figure 8 shows F̄τ (t) = P (τ > t) for different values of barrier levels. To derive F̄τ (t)
apply the numerical inversion of Laplace transform, using EULER method as presented in
Abate and Whitt (1995).
The probability of no-ruin up time t.
In Figures 9 and 10 we describe the expected time to ruin and the expected dividends
paid for three phase type claim size distributions, all with mean 1. We assume that λ = 1,
c = 1.02, and the following claim size distributions: 1. Erlang distribution with two phases,
(scv=1/2). 2. Exponential. 3. hyper-exponential with mean 1 and scv=1.5.
Expected time to ruin and expected dividends paid for risk process with a
phase type claim size.
Figures 11 and 12 show the expected time to ruin and the expected dividend paid for a
risk process governed by a random environment. We assume that the environment can be
either in state 1 or 2 according to the following transition rate matrix:
−q
q
Q=
q −q
and initial probabilities equal to 0.5. Premium rates are c1 = c2 = 1, arrival rates are
λ1 = 1, λ2 = 1.5, the claim size is exponentially distributed with rates δ1 = 1.5 and
δ2 = 1.8, initial reserve is u = 50, and the barrier is b = 100.
Expected time to ruin and expected dividends paid for risk process with a
random environment.
14
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