Some results in risk theory with Erlang(2)

M. M. Claramunt1, M. Mármol1 and R. Lacayo2mmclaramunt@ub.edu mmarmol@ub.edu rlacayo@mat.uab.es

AbstractIn this paper the process of aggregated claims in a non-life insurance portfo-

lio as defined in the classical model of risk theory is modified. The CompoundPoisson process is replaced with a more general renewal risk process with inte-roccurrence times of Erlangian type. In this model we focus our analysis on theprobability that the process of surplus reaches a certain level before ruin occurs,χ (u, b).Our main contribution is the generalization obtained in the computation

of χ (u, b) for the case of interoccurrence time between claims distributed asErlang(2,β) and the individual claim amount as Erlang(n,γ).

KeywordsErlang distribution, upper barrier, ordinary differential equation, boundary

conditions

1 Departament de Matemàtica Econòmica, Financera i Actuarial.Universitat de BarcelonaAvda. Diagonal, 69008034 BarcelonaTlf: 93.4035744

2 Departament de Matemàtiques.Universitat Autònoma de Barcelona. Edifici C08193 Bellaterra (Barcelona)Tlf. 93.5812534

1

1 IntroductionRuin theory is concerned basically with the study of the insurer’s solvencythrough the analysis of the level of reserves as a function of time and otherimportant aspects such as the probability of ruin, the time of ruin and theseverity of ruin. The source of risk in the insurance business is the randomcharacter of claim occurrence.Much of the literature on ruin theory is concerned with deriving results for

the classical risk model. In this model, the surplus at a given time is a stochasticprocess calculated as the sum of the insurer’s initial surplus and the premiumincome minus the claims occurred. The aggregate claim amount (the sum of allclaims up to a given moment) is assumed to follow a compound Poisson processin which the number of claims is Poisson and the individual claim amount isarbitrary with a given density function.One of the most important probabilities related with ruin is the probability

that the process of surplus reaches a certain level before ruin occurs. The aimof this paper is the study of this probability, χ (u, b) .In the classical risk model this probability has been extensively analyzed

(see, for example, Dickson and Gray (1984), Dickson (1992) and Dickson andEgidio dos Reis (1994)). It has been used to calculate ruin probabilities in thepresence of an absorbing upper barrier and plays also an important role in themodel with dividend pay-outs.In this paper the Poisson number process of the classical risk model is re-

placed with a more general renewal risk process with interoccurrence times ofErlangian type (see, e.g., Dickson and Hipp (1998), Dickson (1998) and Sunand Yang (2004)). Dickson (1998) analyzed χ (u, b) for the particular case inwhich the interoccurrence times between claims are distributed as Erlang(2,2)and the individual claim amount has also an Erlang(2,2) distribution. Our maincontribution in this paper is the generalization obtained in the computation ofχ (u, b) for the case of interoccurrence time distributed as Erlang(2,β) and theindividual claim amount as Erlang(n,γ), both β and γ being arbitrary.The organization of the paper is as follows. In Section 2 we summarize the

main results related to χ (u, b) in the classical risk model. In Section 3, weobtain an integro-differential equation for χ (u, b) in an ordinary Erlangian(2,β)model, i.e. with interoccurrence time Erlang(2,β). In 3.1 we obtain and solve thecorresponding differential equation for χ (u, b) assuming a general Erlang(n,γ)distribution for the individual claim amount. In 3.2 we provide numerical resultsfor the particular case when the individual claim amount is distributed as anErlang(2, γ) , and in 3.3 for the case of an Erlang (1,γ) , i.e. Exponential(γ)distribution. In 3.4 we analyze the influence of the individual claim amountdistribution on χ (u, b) by comparing the numerical results.Finally, in Section 4, we calculate χ (u, b) assuming that claims occur as an

equilibrium Erlangian model associated to an ordinary Erlangian model.

2

2 Classical modelIn the classical model of risk theory, the surplus, R(t), at a given time t ∈ [0,∞)is defined as R (t) = u + ct − S (t) , with u = R (0) being the insurer’s initialsurplus, S (t) the aggregate claims and c the rate at which the premiums arereceived.

S (t) is modeled as a compound Poisson process

S (t) =

N(t)Xi=1

Xi

where N (t) , the number of claims occurred until time t, follows a Poissonprocess with parameter λ, and Xi is the amount of the i-th claim and hasdensity function f (x) with mean µ.The instantaneous premium rate, c, is proportional to the product of the

mean number of claims, λ, and the mean value of the claim’s amount, µ. Inother words, c = λµ (1 + ρ) , where ρ, called the security loading, is a positiveconstant.In this model, and in the more general ordinary renewal model, the inte-

roccurrence time between claims, Ti, i = 1, 2, ... is modeled as a sequence ofindependent and identically distributed random variables. T1 denotes the timeuntil the first claim and, in general, Ti denotes the time between the i − 1-thand i-th claims.Note that in a Poisson process with parameter λ, Ti, i ≥ 1 hasan exponential distribution with mean

1

λ.

Given that the time of the first claim, T1, follows an exponential distributionwith density function fT1(t) = λe−λt, the probability χ (u, b) that the surplusprocess reaches the level b > u before the time until ruin, defined as τ =min {t | R (t) < 0}, can be obtained as

χ (u, b) =

Z t0

0

λe−λtZ u+ct

0

χ (u+ ct− x, b) f (x) dxdt+

Z ∞t0

λe−λtdt, (1)

where u+ct0 = b, so that the surplus process will reach b at time t0 if no claimsoccur by time t0 (Dickson and Gray (1984)).The function χ (u, b) has also been related with ruin probabilities. The

probability of ultimate ruin is defined as

ψ (u) = P [R (t) < 0 | t <∞]

and δ (u) = 1 − ψ (u) denotes the survival probability. It can be proved that(Dickson and Gray (1984)),

δ (u) = χ (u, b) δ (b) (2)

It is clear then that χ (u, b) can be computed as this ratio of survival prob-abilities as an alternative to using expression (1) .

3

In a model with upper absorbing barrier b, such that when the reserve levelreaches this barrier the process is finished, the quantity 1 − χ (u, b) is also, bydefinition, the probability of ruin given that the initial reserve is u.

χ (u, b) plays also an important role in the model with a constant dividendbarrier. In this model whenever the surplus reaches the level b, dividends arepaid out in such amount that surplus stays at the barrier until the next claim.The present value of the dividends paid out, D (u, b) , can then be calculated.Obviously D (u, b) is a random variable that has a non-null probability at zero.This is the probability that dividends paid out are zero (Mármol, Claramuntand Alegre (2003)), i.e.,

P [D (u, b) = 0] = 1− χ (u, b)

and then,

P [D (u, b) > 0] = χ (u, b)

3 Ordinary renewal model with Ti ∼ Erlang (2, β)

The classical Poisson risk model is an ordinary renewal process, with Ti ∼Erlang (1, λ).In this section we assume that the number of claims is an ordinary renewal

process in which the Ti are iid Erlang(2, β) with density function,

k (t) = β2te−βt, t > 0, (3)

and distribution function

K (t) = 1− e−βt (βt+ 1) for t ≥ 0Then, as in expression(1) , we obtain

χ (u, b) =

Z t0

0

k (t)

Z u+ct

0

χ (u+ ct− x, b) f (x) dxdt+

Z ∞t0

k (t) dt (4)

Substituting s = u+ ct in (4) and differentiating twice with respect to u, weget

c2χ00 (u, b)− 2βcχ0 (u, b) + β2χ (u, b) = β2uZ0

χ(u− x, b)f (x) dx (5)

Notice that equation (2), which in the classical model relates the survivalprobability with χ (u, b), is not true in the Ordinary Erlangian (2,β) modelbecause the lack of memory property is exclusive of the Exponencial and doesnot hold for the general Erlang distribution. As a result, χ (u, b) cannot be

4

obtained as a ratio of survival probabilities, and for its calculation expression(5) must be used.

From (5) we obtain and solve the differential equation assuming that theindividual claim amount is Erlang(n, γ)

3.1 Individual claim amount Erlang(n,γ)

In this section we assume that the individual claim amount follows an Erlang(n, γ)distribution with pdf

f (x) =γnxn−1e−γx

(n− 1)! (6)

To solve (5) let us define

h (u) = β2uZ0

χ(x, b)f (u− x) dx (7)

Substituting (6) in (7) yields

h (u) =β2γne−γu

(n− 1)!

uZ0

χ(x, b) (u− x)n−1

eγxdx.

Later on we will need an expression for the n-th derivative of the functionabove in terms of the lower order derivatives. This result is the essence of thefollowing

Lemma 1 The n-th derivative of the function h (u) is given by

h(n) (u) = −n−1Xi=0

µn

i

¶h(i) (u) γn−i + β2γnχ(u, b) (8)

(See the proof in Appendix A)

After rewriting equation (5) in the form

c2χ00 (u, b)− 2βcχ0 (u, b) + β2χ (u, b) = h (u) ,

it is clear that differentiating

c2χ(i+2) (u, b)− 2βcχ(i+1) (u, b) + β2χ(i) (u, b) = h(i) (u) (9)

c2χ(n+2) (u, b)− 2βcχ(n+1) (u, b) + β2χ(n) (u, b) = h(n) (u) (10)

Substitution in (10) of the value of h(n) (u) found in (8) and the value ofh(i) (u) from (9) yields the following ordinary differential equation of order(n+ 2) for χ (u, b):

5

an+2χ(n+2) (u, b)+ an+1χ

(n+1) (u, b)+ anχ(n) (u, b)−

n−1Xj=1

ajχ(j) (u, b) = 0 (11)

The value of the constant coefficients is given by

an+2 = c2

an+1 = c2γn− 2βcan = β2 − 2βcγn+ ¡ n

n−2¢c2γ2

aj = −c2¡

nj−2¢γn+2−j +

¡n

j−1¢2βcγn+1−j − β2

¡nj

¢γn−j , j = 1, ..., n− 1.

If all the roots of the characteristic equation of (11) , {ri}n+1i=0 , are different,it is a trivial matter to write down the solution for the ODE above, to wit,

χ (u, b) =n+1Xi=0

αieriu (12)

where {ri}n+1i=0 are functions of γ, β and c, while the {αi}n+1i=0 depend additionallyon b. To obtain the values of {αi}n+1i=0 we need (n+ 2) equations.The first of them is obtained from the boundary condition χ(b, b) = 1. Then,

n+1Xi=0

αierib = 1 (13)

From (12) we know that,

χ0(u, b) =n+1Pi=0

riαieriu

χ00(u, b) =n+1Pi=0

r2iαieriu,

(14)

and their substitution in (5) , after rearranging terms, yields

n+1Xi=0

αieriu¡c2r2i − 2βcri + β2

¢= β2γn

n+1Xi=0

αieriu

1

(n− 1)!

uZ0

xn−1e−(ri+γ)xdx

The integral above can be found by parts1, thus

1 Zxn−1eaxdx = eax

a

³xn−1 − n−1

axn−2 + ... (−1)

n−1(n−1)!an−1

´

6

n+1Pi=0

αieriu³c2r2i − 2βcri + β2 + β2γn (−1)n−1

(−(ri+γ))n´=

= β2γnn+1Pi=0

αi1

(n−1)!e−γu−(ri+γ)

³un−1 − n−1

−(ri+γ)un−2 + ... (−1)

n−1(n−1)!(−(ri+γ))n−1

´ (15)

It is not difficult to show that the term c2r2i − 2βcri + β2 + β2γn (−1)n−1(−(ri+γ))n

is equal to the characteristic equation of (11) , whence (15) becomes

β2γnn+1Xi=0

αi1

(n−1)!e−γu−(ri+γ)

³un−1 − n−1

−(ri+γ)un−2 + ... (−1)

n−1(n−1)!(−(ri+γ))n−1

´= 0

whence we obtain n equations, namely,

n+1Xi=0

αi(ri + γ)

s = 0 , s = 0, 1, ..., n− 1 (16)

From (4) , differentiating with respect to u, one easily obtains

cχ0 − βχ = −1c

bZu

β2e−βc (s−u)

sZ0

f (x)χ(s− x, b)dxds+

∞Zb

β2e−βc (s−u)ds

(17)

substituting (12) and (14) in the left side, and solving the integral term , it isobtained

n+1Xi=0

αieriu (cri − β) = −

n+1Xi=0

αiβ2γn

(ri + γ)n1

(cri − β)

¡eribe−βt0 − eriu

¢− βe−βt0

(18)then,

1 = α0 +1

β

3Xi=1

αi (β − cri) erib (19)

Consequently, after the combination of (13) , (16) and (19), we obtain the re-quired set of (n+ 2) equations from which to calculate the coefficients {αi}n+1i=0 .They are

n+1Pi=0

αierib = 1

n+1Pi=0

αi(ri + γ)s

= 0 , s = 1, 2, ..., n

1β

n+1Pi=0

αi (β − cri) erib = 1

(20)

7

3.2 Ti ∼ Erlang (2, β) and X ∼ Erlang (2, γ)

In this case we study the case n = 2. The ODE can be obtained directly from(11)as

c2χ0000 (u, b) +¡2γc2 − 2βc¢χ000 (u, b)+¡

β2 − 4γβc+ γ2c2¢χ00 (u, b) +

¡2γβ2 − 2γ2βc¢χ0 (u, b) = 0 (21)

This equation generalizes the one obtained by Dickson (1998) for the particularcase c = 1.1, β = 2 and γ = 2.The solution of (21) gives

χ(u, b) =3X

i=0

αieriu

where {ri}3i=0 are the roots of the characteristic equation of (21) .From (20) , the system of equations required to find {αi}3i=0 is,

3Pi=0

αierib = 1

3Pi=0

αi(ri + γ)

= 0

3Pi=0

αi

(ri + γ)2 = 0

1

β

3Pi=0

αi (β − cri) erib = 1

(22)

For γ = 2, β = 2, and c = 1.1, solving (22) and finding the roots of the char-acteristic equation (21) , we obtain the following results for χ (u, b) for differentvalues of u and b,

u\b 0 0.5 1 2 3 4 50 1 0.7963 0.5802 0.3694 0.2805 0.2335 0.2049

0.5 1 0.8542 0.5679 0.4324 0.3599 0.3157

1 1 0.7600 0.5828 0.4854 0.4258

2 1 0.8472 0.7096 0.6228

3 1 0.8939 0.7875

4 1 0.9224

5 1

Graphically we can represent the evolution of χ(u, b) with respect to b fordifferent values of the initial surplus u. In the Figure 1 χ(u, b) is plotted foru = 0, 0.5, 1, 2,

8

5 10 15 20 b

0.2

0.4

0.6

0.8

1χHu,bL 8u=0,u=0.5,u=1,u=2<

Figure 1 : χ (u, b) assuming f (x) = 4xe−2t

For a given value of b, the probability of the reserves reaching that valuebefore ruin, χ(u, b), is increasing in u. On the other hand, for a given value of u,the probability χ(u, b) is decreasing in b, and for each value of u tends towarda limiting value, as shown in the table below.

u 0 0.5 1 2 3 4 5

limb→∞

χ(u, b) 0.1268 0.1954 0.2636 0.3855 0.4876 0.572793 0.643815

Obviuosly, as b tends to infinity, the probability χ(u, b) includes only thosetrajectories of the reserve process which do not lead to ruin. In other words,

limb→∞

χ(u, b) = δ (u)

Consequently, the limiting values for χ(u, b) just obtained are the values ofthe survival probability for the corresponding initial reserves u (they can befound in the Discussion section written by De Vyler and Goovaerts in Dickson(1998)).

3.3 Ti ∼ Erlang (2, β) and X ∼ exp (γ)Here we study the case n = 1. The corresponding ODE, from (11) is

c2χ000(u, b) +¡γc2 − 2βc¢χ00(u, b) + ¡β2 − 2βγc¢χ0(u, b) = 0,

with solution

χ(u, b) = α0 +2Xi=1

αieriu

where r1 and r2 are the roots of

c2r2 +¡γc2 − 2βc¢ r + ¡β2 − 2βγc¢ = 0.

9

In order to obtain {αi}2i=0 , we put n = 1 in (20) ,

2Pi=0

αierib = 1

2Pi=0

αi(ri + γ)

= 0

1β

2Xi=0

(β − cri)αierib = 1

For γ = 1, β = 2, and c = 1.1, we get the following results of χ (u, b)

u\b 0 0.5 1 2 3 4 5 · · · ∞0 1 0.8237 0.6363 0.4318 0.3339 0.2779 0.2419 · · · 0.1199

0.5 1 0.8772 0.6151 0.4765 0.3966 0.3452 · · · 0.1711

1 1 0.7838 0.6106 0.5083 0.4425 · · · 0.2194

2 1 0.8518 0.7125 0.6204 · · · 0.3076

3 1 0.8906 0.7781 · · · 0.3858

4 1 0.9155 · · · 0.4552

5 1 · · · 0.5168

The behavior of this probability in this case turns out to be the same as inSection 3.1 where the Erlang distribution was assumed.

3.4 Numerical comparison

To study the influence of the claim amount distribution on the probabilityχ(u, b), we first define the difference d(u, b) as

d(u, b) = χexp(u, b)− χerl(u, b),

and plot it for the values u = 0, 0.5, 1, 2

5 10 15 20 b

-0.06

-0.04

-0.02

0.02

0.04

0.06dHu,bL 8u=0,u=0.5,u=1,u=2<

Figure 2 : d (u, b)

10

For a given value of b, the probability of the reserves reaching that valuebefore ruin, χ(u, b), is increasing in u. On the other hand, for a given value of u,the probability χ(u, b) is decreasing in b, and for each value of u tends towarda limiting value, as shown in the table below.

u 0 0.5 1 2 3 4 5

limb→∞

χ(u, b) 0.1268 0.1954 0.2636 0.3855 0.4876 0.572793 0.643815

The difference d(u, b) for each value of u tends toward a limiting value.This value indicates the difference between survival probabilities assuming anErlang(2,2) and an exponential(1) distribution for the individual claim amount,i.e. two distributions with the same mean

u 0 0.5 1 2 3 4 5

limb→∞

d(u, b) -0.0068 -0.0242 -0.0441 -0.0778 -0.1017 -0.1175 -0.1269

We can observe that the survival probability obtained assuming an Erlang(2,2)distribution is higher than assuming an exponential(1) distribution.

4 Equilibrium Renewal ProcessIn an equilibrium renewal process, the distribution of the time between the i-

1 th and i-th claims is K (t) for i ≥ 2, with finite mean 1α, and the distribution

of the time until first claim K1 (t) ,

K1 (t) = α

Z t

0

(1−K (s)) ds

being the density function, k1 (t)

k1 (t) = α (1−K (t))

Assuming that K (t) is an Erlang(2, β) , it is obtained

k1 (t) =β

2e−βt +

β2

2te−βt (23)

In this section we find an expression for χe (u, b) when claims occur as anequilibrium renewal process rather than as an ordinary renewal process.By considering the time and the amount of the first claim, we have,

χe (u, b) =

Z t0

0

k1 (t)

Z u+ct

0

χ (u+ ct− x, b) f (x) dxdt+

Z ∞t0

k1 (t) dt (24)

11

Substituting (23)in (24) we get,

χe (u, b) =R t00

β2 e−βt R u+ct

0χ (u+ ct− x, b) f (x) dxdt+

+R t00

β2

2 te−βt R u+ct

0χ (u+ ct− x, b) f (x) dxdt+

+R∞t0

β2 e−βtdt+

+R∞t0

β2

2 te−βtdt

(25)

And from (1)

χe (u, b) =12χ (u, b)+

+R t00

β2 e−βt R u+ct

0χ (u+ ct− x, b) f (x) dxdt+

+R∞t0

β2 e−βtdt

(26)

or

χe (u, b) =1

2χ (u, b)+

+1

2β

hR t00

β2e−βtR u+ct0

χ (u+ ct− x, b) f (x) dxdt+R∞t0

β2e−βtdti(27)

And substituting (17) in (27) we get

χe (u, b) =1

2χ (u, b)− 1

2β(cχ0 (u, b)− βχ (u, b)) =

= χ (u, b)− c

2βχ0 (u, b)

4.1 Numerical results

Assuming Erlang, for γ = 2, β = 2, and c = 1.1, the numerical results forχe (u, b) obtained are,

u\b 0 0.5 1 2 3 4 5

0 1 0.6211 0.4250 0.2652 0.2012 0.1674 0.1469

0.5 1 0.7215 0.4580 0.3477 0.2894 0.2539

1 1 0.6606 0.5027 0.4184 0.3671

2 1 0.7835 0.6533 0.5732

3 1 0.8497 0.7463

4 1 0.8900

5 1

and numerically

u 0 0.5 1 2 3

limb→∞

χe(u, b) 0.0909 0.1571 0.2272 0.3548 0.4620

12

Assuming Exponential distribution the result obtained , for γ = 2, β = 2, and

c = 1.1, are,

u\b 0 0.5 1 2 3 4 5

0 1 0.6627 0.4890 0.3276 0.2531 0.2106 0.1833

0.5 1 0.7651 0.5179 0.4004 0.3332 0.2901

1 1 0.6962 0.5391 0.4487 0.3906

2 1 0.7918 0.6598 0.5744

3 1 0.8463 0.7374

4 1 0.8813

5 1

and numerically

u 0 0.5 1 2 3 4 5

limb→∞

χe(u, b) 0.0909 0.1438 0.1936 0.2848 0.3656

13

Appendix AProof of Lema 1.Since the function h (u) depends explicitly on n, for notational convenience

we rewrite it as

hn (u) =β2γne−γu

(n− 1)!

uZ0

χ(x, b) (u− x)n−1 eγxdx

= Ane−γu

uZ0

B (u− x)n−1

dx, (28)

where, An =β2γn

(n− 1)! and B = χ(x, b)eγx.

For n = 1 we have

h1(u) = β2γe−γυZ u

0

χ(x, b)eγxdx,

and it readily follows through differentiation and substitution that

h01(u) = −γh1(u) + β2γχ(u, b) (29)

For n > 1, differentiating (28) once yields

h0n (u) = Ane−γu (n− 1)

uZ0

B (u− x)n−2 dx− γAne−γu

uZ0

B (u− x)n−1 dx,

which, after dropping the argument u, can be written as the recurrencerelation

h0n = −γhn + γhn−1 , n > 1 (30)

Obviously we can obtain all the required derivatives of hn(u) by successivedifferentiation of (30).From (30) we have

h00n = −γh0n + γh0n−1 = −γh0n + γ (−γhn−1 + γhn−2)= −γh0n − γ (h0n + γhn) + γ2hn−2= −2γh0n − γ2hn + γ2hn−2. (31)

In the first line we used (30) to obtain h0n−1, and again in the second line toobtain γhn−1.In the same manner, from (31) we get

14

h000n = −2γh00n − γ2h0n + γ2h0n−2 = −2γh00n − γ2h0n + γ2 (−γhn−2 + γhn−3)= −2γh00n − γ2h0n − γ

¡h00n + 2γh

0n + γ2hn) + γhn−3

¢= −3γh00n − 3γ2h0n − γ3hn + γ3hn−3. (32)

Here we used in the first line (30) to obtain h0n−2, and in the second (31) toobtain γ2hn−2.It is clear that in the fourth derivative we would have to make use of (30) and

of (32) and, in general, each derivative requires the use of (30) and of the previousone. Moreover, it follows easily from (30), (31) and (32), after transposing, thatthe rightmost term in the derivative of order k can be formally written as thek-th derivative of the product γhn (this follows from Leibniz formula) providedthe derivatives of γ are interpreted as regular powers. In other words, recallingthat h(0)n = hn,

γkhn−k = (γhn)(k)=

kXj=0

µk

j

¶γjh(k−j)n , k < n. (33)

In particular, for k = n− 1, (33) becomes

γn−1h1 = (γhn)(n−1) =

n−1Xj=0

µn− 1j

¶γjh(n−1−j)n . (34)

Note that in (29) h01 is given in terms of h1, that is, in terms of the summa-tion appearing in (34). If we now differentiate (34) and make the appropriatesubstitutions, after transposing we obtain

β2γnχ(u, b) =n−1Xj=0

µn− 1j

¶γjh(n−j)n +

n−1Xj=0

µn− 1j

¶γj+1h(n−1−j)n .

In the first summation above we can safely change the upper limit to nbecause

¡n−1n

¢= 0. In the second summation, by substitution of the dummy

variable j with j − 1 the summation limits are changed from j = 1 to j = n.But since

¡n−1−1¢= 0, we put the lower limit at j = 0. Combining the resulting

summations and given that¡n−1j

¢+¡n−1j−1¢=¡nj

¢we finally obtain

β2γnχ(u, b) =nXj=0

µn

j

¶γjh(n−j)n .

The equivalence of this expression to(8) is evident. The proof of the Lemmais complete.

15

References[1] Dickson, D.C.M. and J.R. Gray (1984). Approximations to ruin probabil-

ity in the presence of an upper absorbing barrier. Scandinavian ActuarialJournal. 105-115.

[2] Dickson, D.C.M. and A.D. Egídio dos Reis (1994). Ruin problems and dualevents. 14. 51-60.

[3] Dickson, D.C.M. (1996). On the distribution of the surplus prior to ruin.Insurance: Mathematics and Economics. 11. 197-207.

[4] Dickson, D.C.M. (1998). On a class of renewal risk process. North AmericanActuarial Journal, Vol 2, no 3

[5] Dickson, D.C.M. and C.Hipp (2001). On the time to ruin for Erlang(2) riskprocess. Insurance: Mathematics and Economics.29, 333-344

[6] Mármol, M., Claramunt, M.M. and A. Alegre (2003). Reparto de dividendosen una cartera de seguros no vida. Obtención de la barrera constante óptimabajo criterios económico-actuariales. Documents de treball de la Divisió deCiències Jurídiques, Econòmiques i Socials. E03/99.

[7] Sun, L. and Yang, H. (2004). On the joint distributions of surplus inmedi-ately before ruin and the deficit at ruin for Erlang(2) risk processes. Insur-ance: Mathematics and Economics.34, 121-125

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