In this paper we analyze the time of ruin in a risk process with the interclaim times being Erlang(n) distributed and a constant dividend barrier. We obtain an integro-differential equation for the Laplace Transform of the time of ruin. Explicit solutions for the moments of the time of ruin are presented when the individual claim amounts have a distribution with rational Laplace transform. Finally, some numerical results and a compare son with the classical risk model, with interclaim times following an exponential distribution, are given.risk theory, constant dividend barrier, laplace transform, time of ruin, generalized erlang (n) distribution

M. Mercedes Claramunt Bielsa

Maite Teresa Marmol Jimenez

English

Research Papers in Economics

DOCUMENTS DE TREBALL 
  DE LA FACULTAT DE CIÈNCIES                          
ECONÒMIQUES I EMPRESARIALS 
 
 
 
 
 
Col·lecció d’Economia 
 
 
Time of ruin in a risk model with generalized Erlang(n) interclaim 
times and a constant dividend barrier 
 
 
 
 
 
Maite Mármol and M. Mercè Claramunt 
 
 
 
 
Adreça correspondència: 
Departament de Matemàtica Econòmica, Financera i Actuarial 
Facultat de Ciències Econòmiques i Empresarials 
Universitat de Barcelona 
Avda. Diagonal, 690 
08034 Barcelona 
Tfn. 93 403 5744 
Fax  93 403 7272 
e-mail: mmarmol@ub.edu, mmclaramunt@ub.edu 
  
 
Abstract 
 
In this paper we analyze the time of ruin in a risk process with the 
interclaim times being Erlang(n) distributed and a constant dividend 
barrier. We obtain an integro-differential equation for the Laplace 
Transform of the time of ruin. Explicit solutions for the moments of the 
time of ruin are presented when the individual claim amounts have a 
distribution with rational Laplace transform. Finally, some numerical 
results and a compare son with the classical risk model, with interclaim 
times following an exponential distribution, are given 
 
 
Resumen: 
 
En este artículo analizamos el momento de ruina en un proceso del riesgo 
donde el tiempo de ocurrencia entre los siniestros se distribuye según una 
Erlang(n) y con una barrera de dividendos constate. Obtenemos una 
ecuación integro diferencial para la Transformada de Laplace del momento 
de ruina. 
 
Presentamos soluciones explicitas para el momento de ruina cuando la 
cuantía individual de un siniestro cumple que la Transformada de Laplace 
de su función distribución es racional. Finalmente, se muestran resultados 
numéricos y una comparación con el modelo clásico (con tiempos de 
interocurrencia exponencial) 
 
 
Keywords: Risk theory, Generalized Erlang (n) distribution, constant 
dividend barrier, time of ruin, Laplace Transform 
 
 
JEL Classification: G22 
 1 Introduction
In the classical model of risk theory, the insurer’s surplus process at a given
time t,R(t), is given by
R(t)=u + c · t −
N(t) X
i=1
Zi ,t ∈ [0,∞)
with u = R(0) being the insurer’s initial surplus. N (t), the number of
claims occurred until time t, follows a Poisson process with parameter λ,
and Zi i st h ea m o u n to ft h ei-th claim and has density function f (z). The
instantaneous premium rate, c,i sc = E [N] · E [Z] · (1 + ρ), where ρ, called
the security loading, is a positive constant.
Figure 1: Sample path for the classical risk process
Deﬁne τ to be the time of ruin so that τ =i n f {t : R(t) < 0}, with
τ = ∞ if R(t) ≥ 0 for all t>0. We denote the ultimate ruin probability
from initial surplus u by ψ(u),s ot h a tψ(u)=P [τ<∞]. T h et i m et or u i n
in the classical risk model is considered in Gerber and Shiu (1998), Lin and
Willmot (2000), Dickson and Waters (2002), Drekic and Willmot (2003), or
Ren (2005) where a perturbed model is analyzed.
3In this paper the Poisson number process of the classical risk model is
replaced with a more general renewal risk process with inter-occurrence times
of a General Erlang(n) type. The time of ruin in an Erlang risk process is
considered in Albrecher et al. (2005), Dickson and Hipp (2001) and Dickson
et al. (2003) and Li and Garrido (2004) where an integro-diﬀerential equation
for the Gerber-Shiu function is derived.
In what follows we shall use the modiﬁed model with a constant divi-
dend barrier b , 0 ≤ u ≤ b, so that when the surplus reaches the level b,
premium income is paid out as dividend to shareholders and the modiﬁed
surplus process remains at level b until the occurrence of the next claim. In
this model the probability of ruin is equal to one, so we can assure that τ
< ∞.
Figure 2: Modiﬁed model with a constant dividend barrier
The layout of this paper is as follows. In the next section we obtain the
integro-diﬀerential equation for the Laplace transform of the time of ruin,
φ(u), in a model modiﬁed with a constant dividend barrier in a Sparre
Andersen model with generalized Erlang(n) interclaim times. Using rational
Laplace Transforms, we solve the corresponding diﬀerential equation.
Section 3 presents the results for the particular case when the interclaim
time follows an Erlang(2,λ) process. In 3.1 we obtain the boundary condition
4when the individual claim amount is distributed as an exponential distrib-
ution, and in 3.2 when the individual claim amount follows an Erlang(2,β)
distribution.
In section 4 some ideas about the interpretation of the Laplace transform
are given, and by diﬀerentiating φ(u) we will obtain the n-moments of the
time of ruin and some numerical results are presented. Finally a comparison
with the classical risk model is presented.
2 Laplace Transform of the time of ruin
We start by obtaining the integro-diﬀerential equation for the Laplace Trans-
form of the time of ruin. We deﬁne,
φ(u)=E
£
e
−δτ¤
.
We assume that the interocurrence times Ti,i=1 ,2,..follow a generalized
Erlang(n) distributed, i.e. each Ti is a sum of n independent exponential
random variables with possibly diﬀerent parameters λ1,...,λn each causing
a "sub-claim" of size 0 and at the time of the nth subclaim a claim with
distribution F occurs. We consider n states of the risk process (starting at
time 0 in state 1) , where every subclaim causes a transition to the next state
and in the time of occurrence of the nth subclaim the risk process jumps into
state 1 again (see Albrecher at al. (2005)).
Let φ
j (u) denote the Laplace transform of the time of ruin if the risk
process is in state j =1 ,...,n.
Theorem 1 The integro-diﬀerential equation for φ(u) is,
Ã
n Y
j=1
µ
δ + λj − c
∂·
∂u
¶!
φ(u)−
n Y
j=1
λj
Z u
0
φ(u − z)dF (z)−
n Y
j=1
λj [1 − F (u)] = 0,
(1)
5with boundary condition,
k−1 Y
j=1
Ã
δ + λj − c ∂·
∂u
λj
!
∂φ(u)
∂u
¯ ¯ ¯ ¯
u=b
=0 for k =1 ,...,n (2)
Proof. For 0 ≤ u<b ,by diﬀerential argument
φ
j (u)=( 1− λjdt)φ
j (u + cdt)e−δdt+
λjdtφ
j+1 (u + cdt)e−δdt + θ(dt)
for j =1 ,...,n− 1 (3)
beeing θ(dt) the probability of more than one claim occurs in dt.
φ
n (u)=( 1− λndt)φ
n (u + cdt)e−δdt+
λndt
R u+cdt
0 φ
1 (u + cdt − z)dF (z)e−δdt+
λndte−δdtR ∞
u+cdt dF (z)+θ(dt)
for j = n
(4)
Then by linear approximation, dividing by dt,a n dt a k i n gdt → 0, from
(3) and (4) we obtain,
c
∂φ
j (u)
∂u
− (δ + λj)φ
j (u)+λjφ
j+1 (u)=0 , for j =1 ,...,n − 1. (5)
c
∂φn(u)
∂u − (δ + λn)φ
n (u)+
λn
R u
0 φ
1 (u − z)dF (z)+λn [1 − F (u)] = 0.
for j = n (6)
From (5),
φ
j+1 (u)=
Ã
(δ + λj) − c ∂·
∂u
λj
!
φ
j (u),j =1 ,...,n − 1,
6and following a recursive process,
φ
n (u)=
Ã
n−1 Y
j=1
(δ + λj) − c ∂·
∂u
λj
!
φ
1 (u). (7)
We can write (6) as,
µ
(δ + λn) − c
∂·
∂u
¶
φ
n (u)−λn
Z u
0
φ
1 (u − z)dF (z)−λn [1 − F (u)] = 0. (8)
Substituting (7) in (8),(1) is obtained.
For u = b using an analogous process for j =1 ,...,n − 1
φ
j (b)=( 1− λjdt)φ
j (b)e
−δdt + λjdtφ
j+1 (b)e
−δdt + θ(dt)
we obtain
λjφ
j+1 (b) − (δ + λj)φ
j (b)=0
that comparing with (5),
∂φ
j (u)
∂u
¯ ¯ ¯ ¯
u=b
=0 (9)
And for j = n,
φ
n (b)=( 1− λndt)φ
n (b)e−δdt + λndt
R b
0 φ
1 (b − z)dF (z)e−δdt+
λndte−δdtR ∞
b dF (z)+θ(dt)
then,
λn
Z b
0
φ(b − z)dF (z)+λn
Z ∞
b
dF (z) − (δ + λn)φ
n (b)=0
and comparing with (6)
∂φ
n (u)
∂u
¯ ¯ ¯ ¯
u=b
=0 (10)
From (9), (10) and (7) the boundary condition (2) is obtained.
7Using an alternative method, Li and Garrido (2004) obtained an equiv-
alent integro-diﬀerential equation and boundary conditions for the Gerber-
Shiu function.
Applying Laplace transforms to (1) we obtain,
ϑ(s)e φ(s)+Ln−1 (s) −
n Y
j=1
λj
Ã
e φ(s) e f (s) −
1 − e f (s)
s
!
=0 , (11)
being e φ(s) the Laplace transform of φ(u) and e f (s) the Laplace transform
of the claim density function f (z).
Ln−1 (s) represents the n − 1 degree polynomial, whose coeﬃcients in-
volve the quantities
∂φj)(u)
∂u
¯ ¯ ¯
u=0
,j=0 ,...,n − 1,a n dϑ(s) is the n degree
polynomial ϑ(s)=
n Y
j=1
(δ + λj − cs).
From (11) we obtain e φ(s),
e φ(s)=
n Y
j=1
(λj)
³
1−h f(s)
s
´
− Ln−1 (s)
ϑ(s) −
n Y
j=1
λj e f (s)
. (12)
Example 2 Assuming the classical model of risk theory with interocurrence
claims Erlang(1,λ), expression (12) is,
e φ(s)=
λ
³
1−h f(s)
s
´
− cφ(0)
(δ + λ − cs) − λe f (s)
. (13)
Let us now restrict the further analysis to the case of claim size distribu-
tion with rational Laplace transform
e f (s)=
Qr−1 (s)
Pr (s)
,
8where Pr (s) and Qr−1 (s) denote polynomials of degree r and (at most) r−1
respectively with no common zeros. Moreover Pr (s) has leading coeﬃcient
1, no roots in the positive half plane and Pr (0) = Qr−1 (0).For this claim size
distribution, the denominator of (12) has n+r zeroes, which are s1,...,sn+r.,
and we assume that the roots are real and distinct. From the above we
conclude that r of these zeroes are located in the negative halfplane. Then
using partial fractions, it is obtained,
φ(u)=
n+r X
i=1
αi (b)e
siu. (14)
Note that αi (b) are functions of b, but for brevity we write αi. Now, we
need to ﬁnd n + r equations satisﬁed by αi,i=1 ,...,n + r. The ﬁrst n are
obtained from (2).
3 Laplace transform of Time of ruin in Erlang(2,λ)
risk model
Assuming the case of a Sparre Andersen model with Erlang(2,λ) interclaim
times (n =2and λ1 = λ2 = λ) the integro-diﬀerential equation for φ(u) is
µ
δ + λ − c
∂·
∂u
¶2
φ(u) − λ
2
Z u
0
φ(u − z)dF (z) − λ
2 [1 − F (u)] = 0, (15)
From (12),i ti se a s yt oo b t a i n
e φ(s)=
λ
2
³
1−h f(s)
s
´
+ c2φ
0 (0) + (c2s − 2c(δ + λ))φ(0)
(δ + λ − cs)
2 − λ
2 e f (s)
, (16)
and assuming a claim distribution with rational Laplace transform, it is ob-
9tained
φ(u)=
2+r X
i=1
αi (b)e
siu. (17)
3.1 Claim distribution z ∼ exp(γ)
If we assume that f (z)=γe−γz,t h e ne f (s)=
γ
γ+s. We obtain the roots
si,i=1 ,2,3, from
(δ + λ − cs)
2 (γ + s) − λ
2γ =0 . (18)
Then the solution of φ(u) is,
φ(u)=
3 X
i=1
αie
siu. (19)
The ﬁrst two equations obtained from (2) are
3 X
i=1
siαiesib =0and
3 X
i=1
s2
iαiesib =
0.
To ﬁnd the third equation, we substitute (19) and f (z)=γe−γz in the
integro-diﬀerential equation (15), and resolving the integral,
3 X
i=1
αie
siu
∙
(λ + δ)
2 − 2c(λ + δ)si + c
2s
2
i −
λ
2γ
(si + γ)
¸
+
λ
2γ
3 X
i=1
αie−γu
(si + γ)
− λ
2e
−γu =0 .
From (18) the coeﬃcient of esiu is equal to zero, then we obtain the third
equation,
3 X
i=1
αi
(si+γ) = 1
γ.
10We thus have a system of three equations from which we can easily solve
for αi,i=1 ,2,3 using Mathematica.
3.2 Claim distribution z ∼ Erlang(2,β)
Consider a risk process in which claims occur as an Erlang(2,β) distribution,
f (z)=β
2ze−βz with Laplace transform e f (s)=
β2
(β+s)2. Then the fourth roots
are obtained from
(δ + λ − cs)
2 (β + s)
2 − λ
2β
2 =0 (20)
T h es o l u t i o ni nt h i sc a s ei s ,
φ(u)=
4 X
i=1
αie
siu. (21)
We need to ﬁnd four equations to calculate the unknowns αi,i=1 ,2,3,4.
Two equations are obtained from (2):
4 X
i=1
siαiesib =0and
4 X
i=1
s2
iαiesib =
0.
From (15), (21) and knowing that f (z)=β
2ze−βz,w eﬁnd the other two,
4 X
i=1
αi
(si + β)
2 =
1
β
2 and
4 X
i=1
αi
(si + β)
=
1
β
.
4 Getting information from Laplace transform
o ft h et i m eo fr u i n
From φ(u) we can get two kind of diﬀerent informations about the time of
ruin.
First φ(u)=E
£
e−δτ¤
can be interpreted as the expected present value
of one monetary unit that was paid at the time of ruin. Then the parameter
of the Laplace transform δ can be interpreted as the interest rate used to
11obtain the present value.
But also φ(u) allows us to ﬁnd the the n-moments of the random variable
τ, i.e. E [τn], that can be obtained diﬀerentiating φ(u) with respect to δ,
∂nφ(u)
∂δ
n =
∂n
∂δ
nE
£
e
−δτ¤
= E
£
(−τ)
n e
−δτ¤
.
and evaluating these at δ =0 ,
E [τ
n]=( −1)
n ∂φ
n (u)
∂δ
n
¯ ¯ ¯ ¯
δ=0
.
In what follows, some numerical results are obtained.
If we assume that f (z)=γe−γz, for γ =1 ,λ=1 ,b=1 0and c =0 .6, the
results for E [τ],σ[τ] and the variation coeﬃcient of τ deﬁned by cv [τ]=
100.σ[τ]
E[τ] are summarized in Table 1,
E [τ] σ[τ] cv [τ]
u =0 93.9577 217.63 231.625
u =1 157.031 265.267 168.926
u =2 205.805 288.281 140.075
u =3 243.077 299.847 123.355
u =4 271.099 305.565 112.714
u =5 291.68 308.253 105.682
u =6 306.277 309.411 101.023
u =7 316.06 309.845 98.0336
u =8 321.974 309.973 96.2727
u =9 324.794 309.996 95.4437
u =1 0 325.372 309.997 95.2744
Table 1: E [τ],σ[τ] and cv [τ]
for f (z)=e−z,λ=1 ,b=1 0and c =0 .6
12Graphically,
2 4 6 8 10
u
100
150
200
250
300
E@τD
Figure 3: E[τ] assuming Ti ∼ Erlang(2,1)
and Z ∼ exp(1)
2 4 6 8 10
u
240
260
280
300
α@τD
Figure 4: α[τ] assuming Ti ∼ Erlang(2,1)
and Z ∼ exp(1
2 4 6 8 10
u 100
120
140
160
180
200
220
cv@τD
Figure 5: cv[τ] assuming Ti ∼ Erlang(2,1)
and Z ∼ exp(1)
134.1 Comparison with the classical model
In this section we are going to compare the time of ruin with a constant
barrier in an Erlang risk model with the corresponding in the classical risk
model.
In the classical risk model, claims occur as a Poisson process. Note that in
a Poisson process with parameter λ, Ti,i≥ 1 has an exponential distribution
with mean 1
λ, i.e. with an Erlang(1,λ) distribution. So the time of ruin in the
classical risk model is included in Theorem 1 as a particular case. From (1),
for n =1 ,w eg e tt h ei n t e g r o - d i ﬀerential equation for the Laplace transform
of the time of ruin in the classical risk model and the boundary condition,
(δ + λ)φ(u) − cφ
0 (u) − λ
Z u
0
φ(u − z)dF (z) − λ[1 − F (u)] = 0 (22)
∂φ(u)
∂u
¯ ¯ ¯ ¯
u=b
=0 .
This equation can be obtained too from equation (2.5) in Lin et al (2003)
for the Gerber-Shiu function in the classical risk model. (See too Dickson
and Waters (2004))
To solve this model, from (13), assuming that f (z)=γe−γz, the roots
are obtained from −c2s2 +( δ + λ − cγ)s + δγ =0 .T h e nφ(u)=
2 X
i=1
αiesiu.
To calculate αi, substituting φ(u)=
2 X
i=1
αiesiu and f (z)=γe−γz in (22) we
get
2 X
i=1
αi
(si + γ)
=
1
γ
.
In order to analyze the inﬂuence that the distribution of the interclaim
times has in the time of ruin, the two models have to be comparable, i.e.
E [Nt] when t →∞must be assimptotically the same in the Erlangian model
and in the classical model.
14Following Cox (1962), in an ordinary renewal process the renewal func-
tion, E [Nt], is related with Ti
e L(s)=
1
s
e gTi (s)
1 − e gTi (s)
, for e gTi (s) < 1
being e L(s) the Laplace transform of E [Nt] and e gTi (s) the Laplace transform
of the density function of the interrocurrence times.
For an Erlangian model, i.e. Ti ∼ Erlang(n,λ) with e gTi (s)=
¡
λ
λ+s
¢n,
then
e L(s)=
λ
n
s((λ + s)
n − λ
n)
. (23)
From (23),i fTi ∼ Erlang(2,λ),E[Nt] is easily obtained inverting e L(s)=
λ
2
s
¡
(λ + s)
2 − λ
2¢, then E [Nt]=
1
E [Ti]
t −
1
4
+
1
4
e
−
4
E [Ti]
t
.
From (23),i fTi ∼ Erlang(1,λ), i.e. for the classical risk model with
E [Ti]=
1
λ
, e L(s)=
λ
s2 and inverting, we have E [Nt]=
t
E [Ti]
.
So, if E [Ti] is the same, the renewal function E [Nt] has the same behavior
when t →∞ . I ng e n e r a li na no r d i n a r yr e n e w a lp r o c e s slim
t→∞
E[Nt]
t = 1
E[Ti] (see
Parzen (1972))
It´s easy to see also that the security loading included in the premium
is diﬀerent in the two models, and that if E [Ti] is the same, asymptot-
ically the security loading in the two models coincides: It´s known that
the total premium income until time t is ct = E [Z]E [Nt](1+ρt), then
ρt =
ct
E [Z]E [Nt]
− 1, and assuming E [Z]=1 , for the Erlangian model
ρt =
ct
1
E [Ti]
t −
1
4
+
1
4
e
−
4
E [Ti]
t
− 1
15and for the classical model
ρt = cE [Ti] − 1
Then, in order to compare the time of ruin in the Erlang(2,1) model with
E [Ti]=2with the classical model, where the interocurrence time follows an
exponential distribution, the parameter of the exponential must be 0.5, i.e.
Ti ∼ exp(0.5) with E [Ti]=2 .
The diﬀerence between E [τ] in the Erlang(2,1) model and in the classical
model with Ti ∼ Exp(0.5) is represented in Figure 6. In the Figures 7 and
Figure 8 the diﬀerence for σ[τ] and the variation coeﬃcient of τ is repre-
sented,
2 4 6 8 10
u
50
60
70
80
90
100
110
Difference E@τD
Figure 6: EErlang[τ]− EExp[τ]
2 4 6 8 10
u
90
95
100
105
110
Difference α@τD
Figure 7: σErlang[τ] − σExp[τ]
162 4 6 8 10
u
-15
-12.5
-10
-7.5
-5
-2.5
Difference cv@τD
Figure 8: cvErlang[τ]− cvExp[τ]
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18

On a class of renewal risk models with a constant dividend barrier&quot;

On the density and moments of t h et i m et or u i nw i t he x p o n e n t i a lc l a i m s &quot;ASTIN

On the distribution of dividend payments in a Sparre Andersen model with generalized Erlang(n) interclaim time&quot;

On the time to ruin for Erlang(2) risk processes&quot;

On the time value of ruin&quot;

Procesos Estocásticos&quot; Paraninfo.

Renewal theory&quot;

The classical risk model with a constant dividend barrier: analysis of the Gerber—Shiu discounted penalty function&quot;

The density of the time to ruin for a Sparre Andersen process with Erlang arrivals and exponential claims&quot;

The distribution of the time to ruin in the classical risk model&quot;

The expected value of the time of ruin and the moments of the discounted deﬁcit at ruin in the perturbed classical risk process&quot;

The moments of time of ruin, the surplus before ruin, and the deﬁcit at ruin&quot;

Time of ruin in a risk model with generalized Erlang (n) interclaim times and a constant dividend barrier

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Time of ruin in a risk model with generalized Erlang (n) interclaim times and a constant dividend barrier

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Research Papers in Economics