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Finite time ruin problems for the Erlang(2) risk model
Insurance: Mathematics and Economics 46 (2010) 12–18
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Finite time ruin problems for the Erlang(2) risk model David C.M. Dickson ∗ , Shuanming Li Centre for Actuarial Studies, Department of Economics, University of Melbourne, Victoria 3010, Australia
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Article history: Received August 2008 Received in revised form March 2009 Accepted 6 May 2009
abstract We consider the Erlang(2) risk model and derive expressions for the density of the time to ruin and the joint density of the time to ruin and the deficit at ruin when the individual claim amount distribution is (i) an exponential distribution and (ii) an Erlang(2) distribution. We also consider the special case when the initial surplus is zero. © 2009 Elsevier B.V. All rights reserved.
1. Introduction In this paper we use results given by Dickson and Hipp (2001) and ideas given in Cheung et al. (2008) and Dickson (2008) to study finite time ruin problems for the Erlang(2) risk model. In particular we aim to find formulae for finite time ruin probabilities and for joint densities of the time to ruin and deficit at ruin. In the literature there are very few exact formulae for finite time ruin probabilities. For the classical risk model, Dickson and Willmot (2005) give a formula for the finite time ruin probability in the case when the individual claims have a distribution that is an infinite mixture of Erlang distributions. Willmot and Woo (2007) explain why this formula covers a range of individual claim amount distributions and covers all cases for which formulae for the finite time ruin probability exist. See Drekic and Willmot (2003) and Garcia (2005). In the case of Sparre Andersen risk models formulae for finite time ruin probabilities exist only in the case of exponential claims — see Dickson et al. (2005) and Borovkov and Dickson (2008). Dickson (2008) found formulae for the joint density of the time to ruin and the deficit at ruin in the classical risk model when the distribution of individual claims was either Erlang(2) or a mixture of two exponential distributions. However, no such results presently exist for Sparre Andersen risk models. Here we apply the methodology in Dickson (2008) to derive such results for the Erlang(2) risk model. The outline of this paper is as follows. In Section 2 we set out the mathematical preliminaries and give transform relationships that are central to our derivations in subsequent sections. In Section 3 we discuss the special case when u = 0 and find that the derivation of exact results is somewhat complicated in general, but less so when the individual claim amount distribution has a
particular form. The cases of individual claim amounts having (i) an exponential distribution and (ii) an Erlang(2) distribution are discussed in Sections 4 and 5 respectively. We make some concluding remarks in Section 6. 2. Preliminaries We adopt the model of Dickson and Hipp (2001). Thus, we are dealing with a Sparre Andersen risk model under which claim inter-arrival times are distributed as Erlang(2) with scale parameter β . We denote by p the density function of individual claim amounts, and denote the kth moment as mk . Further, the Laplace transform of p is denoted by p˜ where p˜ (s) =
Corresponding author. E-mail addresses: [email protected] (D.C.M. Dickson), [email protected] (S. Li). 0167-6687/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.insmatheco.2009.05.001
e−sx p(x)dx. 0
Let P = 1 − P¯ denote the distribution function. We denote by c the insurer’s premium income per unit time and assume that 2c /β > m1 . Let {U (t )}t ≥0 denote the surplus process, let T denote the time of ruin, and let Y = | − U (T )| denote the deficit at ruin. Define
φ(u) = E e−δT −sY I (T < ∞)|U (0) = u to be the bivariate Laplace transform of T and Y . Let w(u, t ) denote the (defective) density of T , and let w(u, y, t ) denote the (defective) joint density of T and Y so that
φ(u) =
∞
Z 0
∞
Z
e−δ t −sy w(u, y, t )dydt . 0
For this risk model, Lundberg’s fundamental equation is s2 − 2
∗
∞
Z
β +δ c
s+
β +δ c
2 =
β2 c2
p˜ (s)
and Dickson and Hipp (2001) show that this equation has two solutions r1 and r2 such that r1 < (β +δ)/c < r2 . We easily deduce
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
that r1 = r2 =
β +δ c
β +δ c
−
βp
+
βp
c c
p˜ (r1 ),
(1)
p˜ (r2 ).
(2)
For the remainder of this paper we assume that p˜ (s) = q˜ (s)2 for some density function q. An easy way of thinking of this assumption is that an Erlang(2) risk process is a classical risk process modified so that the first two claims are paid together at the time of the second claim, claims three and four are paid together at the time of the fourth claim and so on. Indeed, such a transform relationship holds if the individual claim amount distribution is infinitely divisible. Rewriting expressions (1) and (2) as
β + δ − cr1 = β q˜ (r1 ), β + δ − cr2 = −β q˜ (r2 ),
(3)
we see that Eq. (3) is identical in form to Lundberg’s fundamental equation for the classical risk model, λ + δ − c ρ = λ˜p(ρ) (see Gerber and Shiu (1998)). Thus we can apply the ideas of Dickson and Willmot (2005, page 49) to obtain the following transform relationship. If ∞
Z
f˜ (r1 ) =
e−r1 t f (t )dt = g˜ (δ) = 0
computing w(0, t ) for specific individual claim amount distributions. However, for certain individual claim amount distributions, we can take a different approach to find solutions both for w(0, t ) and w(0, y, t ), as follows. Let us now assume that the individual claim amount density is such that p(x + y) =
P¯ (x + y) =
n=1
β n t n−1 e−β t n!
f˜ (r2 ) =
∞
e−r2 t f (t )dt = h˜ (δ) = 0
yqn∗ (ct − y)f (y)dy
(4)
0
m X
ηj (x)T¯j (y)
e−δ t h(t )dt
where Tj = 1 − T¯j is the distribution function associated with τj . From Dickson and Hipp (2001) we know that ∞
Z
e−δ t w(0, t )dt = 0
∞
Z
e
0
−δ t
w(0, t )dt =
0
h(t ) = ce−β t f (ct ) +
∞ X (−β)n t n−1 e−β t
=
n!
n=1 ct
Z
yqn∗ (ct − y)f (y)dy.
×
(5)
0
In the following sections we obtain Laplace transforms with transform parameters r1 and r2 . Our approach is to invert these Laplace transforms using relationships (4) and (5). 3. The case u = 0 We consider first the density of the time of ruin. From Dickson and Hipp (2001) (see also Li and Garrido (2004)) we know that ∞
e−δ t w(0, t )dt =
c 2 r1 r2 − (β + δ)2 + β 2
and if we use Eqs. (1) and (2) for r1 and r2 we obtain ∞
e−δ t w(0, t )dt 0
=
β 2 [1 − q˜ (r1 ) + q˜ (r2 ) − q˜ (r1 )˜q(r2 )] − βδ[˜q(r1 ) − q˜ (r2 )]
=
β2
c 2 r1 r2 c2
1 − q˜ (r1 ) r1
1 + q˜ (r2 ) r2
−
βδ[˜q(r1 ) − q˜ (r2 )] c 2 r1 r2
m Z β2 X
.
(6)
Although formula (6) can be inverted, the resulting expression for w(0, t ) does not seem particularly attractive in terms of
c2
∞
Z
∞
Z
0
e−r1 x e−r2 y P¯ (x + y)dydx, 0
c2
j =1
β2
∞
Z
c2
e−r1 x ηj (x)dx 0
∞
Z
0
∞
e− r 1 x e− r 2 y 0
m X
ηj (x)T¯j (y)dydx
j =1
∞
Z
e−r2 y T¯j (y)dy.
(8)
0
The significance of Eq. (8) is that each of the integrals is a Laplace transform — with respect to either r1 or r2 , and hence by the transform relationships (4) and (5) they are Laplace transforms with respect to δ . By inverting each transform, we can obtain the inverse of the product of two transforms by finding the convolution of the two inverses. For example, the simplest case to satisfy Eq. (7) is p(x) = α e−α x giving η1 (x) = e−α x and τ1 (y) = α e−α y , with T¯1 (y) = e−α y . Then Eq. (8) gives the already known equation ∞
Z
e−δ t w(0, t )dt = 0
c 2 r1 r2
0
β2
so if p satisfies Eq. (7) we obtain
then
Z
ηj (x)τj (v)dv
j =1
j=1
ct
∞
Z
m ∞X y
where qn∗ denotes the n-fold convolution of q. The derivation is exactly as in Dickson and Willmot (2005) — all that is different is the notation. The same arguments give a further relationship involving r2 . If
Z
p(x + v)dv
Z
=
g (t ) = ce−β t f (ct ) +
∞
Z
=
e−δ t g (t )dt
Z
(7)
y
0
∞ X
ηj (x)τj (y)
for some functions {ηj , τj }m j=1 . This factorisation was introduced by Willmot (2007), and he shows that if the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter then this factorisation applies. Results in Willmot and Woo (2007) show that this infinite mixture of Erlangs contains many well-known distributions as special cases. We make the further assumption that the functions {τj } are in fact density functions (as is the case when p is an infinite mixture of Erlang densities). Then
then
Z
m X j =1
∞
Z
13
c2
β2 , (α + r1 )(α + r2 )
(see Dickson and Hipp (2001)), and we show how to invert this Laplace transform in the next section. The same techniques apply to other individual claim amount distributions whose densities satisfy Eq. (7). This approach can be extended if the functions {ηj } satisfy the same type of factorisation as Eq. (7). As Willmot (2007) shows, this is the case when the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter. Specifically, let us suppose that
ηj (x + y) =
n X i =1
ξij (x)ζij (y).
(9)
14
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
4. Exponential claims
From Eq. (4) of Sun (2005), we know that ∞
Z
∞
Z
In this√section we consider the case p(x) = α e−α x , x > 0, so that q(x) = α/(xπ )e−αx . Although formulae for w(u, t ) are known for this case (see Dickson et al. (2005) and Borovkov and Dickson (2008)), it is interesting to explore this case for two reasons. First, it introduces the techniques needed for the more complicated case of Erlang(2) claims discussed in the next section, and, second, we end up with a new formula for w(u, t ) which parallels the formula for the density of the time to ruin in a classical risk model with exponential claims obtained by Drekic and Willmot (2003). Dickson (2008) shows that
e−δ t −sy w(0, y, t )dydt 0
0
β2
=
∞
Z
c2
Z
∞
e−r1 x e−r2 y $ (x + y)dydx
0
0
where
$ (u) =
∞
Z
e−sy p(u + y)dy. 0
Applying (7) in the above expression we obtain
$ (u) =
∞
Z
e
−sy
0
m X
ηj (u)τj (y)dy =
j =1
m X
w(u, t ) =
ηj (u)τ˜j (s),
Z tZ 0
j =1
u
w(0, y, τ )w(u − y, t − τ )dydτ
0
∞
Z
w(0, y, t )dy,
+
and if we now apply (9) we obtain
(11)
u
$ (x + y) =
m X n X
and it is well known that by the memoryless property of the exponential distribution
ξij (x)ζij (y)τ˜j (s),
j =1 i =1
w(u, y, t ) = w(u, t )α e−αy .
so that ∞
Z 0
Following the approach in Section 3 of Dickson (2008), we insert this expression in Eq. (11) then take the Laplace transform of the resulting equation. Defining
∞
Z
e−δ t −sy w(0, y, t )dydt 0
n m X β2 X
=
c2
ξ˜ij (r1 )ζ˜ij (r2 )τ˜j (s).
(10)
˜˜ s, δ) = w(
j =1 i =1
η1 (x) = α xe , τ2 (y) = α 2 ye−αy ,
τ1 (y) = α e
−α y
,
η2 ( x) = e
−α x
,
w( ˜ 0, δ) =
e−δ t w(0, t )dt , 0
we obtain
˜˜ s, δ) = w(
1 w( ˜ 0, δ) α+ s
α 1 − w( ˜ 0, δ) α+ s
η1 (x + y) = ξ11 (x)ζ11 (y) + ξ21 (x)ζ21 (y), η2 (x + y) = ξ12 (x)ζ12 (y),
w(u, t ) =
−α y
Then
0
=
∞
Z
w( ˜ 0, δ) =
e−δ t −sy w(0, y, t )dydt 0
β2
α α + 2 2 c (α + r1 ) (α + r2 ) (α + r1 )(α + r2 )2 2 β2 1 α + 2 . c (α + r1 )(α + r2 ) α + s
wn∗ (0, t )
=
∞ 1X
α
˜ 0, δ))n (w(
n =1
α α+s
(α u)n−1 e−αu . Γ ( n)
n
(12)
We remark that the derivation of this result does not depend on the claim inter-arrival distribution being Erlang(2) — formula (12) is a general result. We saw in Section 3 that when the claim inter-arrival distribution is Erlang(2) with scale parameter β ,
ξ11 (x) = α xe , ζ11 (y) = ζ12 (y) = e , −α x ξ21 (x) = ξ12 (x) = e , ζ21 (y) = α ye−αy . ∞
∞ X n=1
where
and
∞
Z
which inverts to
Z
e−su−δ t w(u, t )dtdu
0
and we can write
−α x
∞
Z
0
Thus, the bivariate Laplace transform factorises in terms of transforms with transform parameters r1 , r2 and s. For example, when p(x) = α 2 xe−α x , x > 0, we get −α x
∞
Z
α α+s
As r1 and r2 are both functions of δ , the above equation tells us that
w(0, y, t ) = f1 (t )α e−αy + f2 (t )α 2 ye−αy , and we show in Section 5 how the functions f1 and f2 can be identified. We remark that this example easily extends to the general case when p is the Erlang(n, α ) density with n > 2 since we know from Willmot (2007) that the factorisation (9) applies in this case and that the functions ξij and ζij in (9) are just scaled Erlang density functions. Thus when we apply (10), we obtain terms which are multiples of (α + r1 )−n (α + r2 )−m where n and m are positive integers. The inversion of such terms is described in Section 5.
c2
β2 (α + r1 )(α + r2 )
so that w n∗ (0, t ) can be found by inverting
β 2n . c 2n (α + r1 )n (α + r2 )n Now let V˜ n (δ) = (α + r1 )−n (where r1 depends on δ ) be the Laplace ˜ n (δ) = (α + r2 )−n . transform of a function Vn (t ), and similarly let W Then inversion of V˜ n (δ) with respect to r1 yields t n−1 e−α t /Γ (n) and by (4) inversion with respect to δ yields ∞ (ct )n−1 e−αct X β m t m−1 e−β t + Γ (n) m! m =1 Z ct yn−1 e−α y × yqm∗ (ct − y) dy. Γ (n) 0
ce−β t
We find that ct
Z
yqm∗ (ct − y) 0
yn−1 e−α y
Γ (n)
dy =
nα m/2 e−α ct (ct )n+m/2
Γ (m/2 + n + 1)
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
giving
where
∞ X α m/2 β m c n+m/2 t n+3m/2−1 Vn (t ) = ne−(β+α c )t . m! Γ (m/2 + n + 1) m=0
(B1 , B2 , . . . , Bp , C1 , C2 , . . . , Cq ; Z ) ∞ X (B1 )m (B2 )m · · · (Bp )m Z m = (C1 )m (C2 )m · · · (Cq )m m! m=0
p Fq
Similarly, by applying (5) we obtain Wn (t ) = ne−(β+α c )t
is the generalised hypergeometric function and (a)n = Γ (a + n)/Γ (a) is Pochhammer’s symbol. This follows by looking at the ratio of terms in the sum in (13). (See, for example, Graham et al. (1994, pp. 207–8).) Thus we have
∞ X α m/2 (−1)m β m c n+m/2 t n+3m/2−1 . m! Γ (m/2 + n + 1) m=0
Thus,
β2
wn∗ (0, t ) =
c2
w(u, t ) = e−αu−(β+αc )t
Vn ∗ Wn (t ).
Vn ( t ) =
∞ X
am
t
n+3m/2−1 −(β+α c )t
e
where nα m/2 β m c n+m/2 Γ (n + 3m/2)
Γ (m/2 + n + 1)
m!
,
and so we can write Wn (t ) =
∞ X
(−1)m am
t n+3m/2−1 e−(β+α c )t
Γ (n + 3m/2)
m=0
.
If we take the convolution of the ith term in Vn with the jth term in Wn we get
(−1)j ai aj
e
4
5. Erlang(2) claims
Γ (2n + 32 (i + j))
w(u, y, t ) = h(u, t )α 2 ye−αy + k(u, t )α e−αy
t
β
2n
∞ X
c 2n r =0
kr
e
t
Γ (2n + 23 r )
5.1. Main results
r X (−1)r −i ai ar −i . i=0
By symmetry, kr = 0 if r is odd, so we obtain
wn∗ (0, t ) =
∞ β 2n t 2n−1 e−(β+αc )t X
c 2n
r =0
k2r
t 3r
Γ (2n + 3r )
.
(13)
Tidying up gives k2r = α β c
2r 2n+r
As the form of the solution for w(u, y, t ) is exactly the same as in the classical risk model with Erlang(2, α ) claims, we can apply the approach given in Dickson (2008, Section 3). The reason we can do this is that we are simply inserting formula (14) into Eq. (1) of Dickson (2008), and that equation applies to any Sparre Andersen model. Let
σ (2r , n)
∞
Z
˜
h˜ (s, δ) =
0
∞
Z
e−su−δ t h(u, t )dtdu 0
and let
where 2r X σ (2r , n) = n2 (−1)i i =0
Γ (n + 3i/2) i!Γ (i/2 + n + 1)
Γ (n + 3(2r − i)/2) × . (2r − i)!Γ ((2r − i)/2 + n + 1) We now state a result, which we prove in the Appendix. For n = 1, 2, 3, . . . and r = 0, 1, 2, . . .,
σ (2r , n) =
(14)
where h(u, t ) and k(u, t ) are functions that we will identify. For the sake of brevity we omit the details of why we know w(u, y, t ) is of this form. The approach to showing this is essentially that given by Cheung et al. (2008), but adapted to our risk model.
−(β+α c )t 2n+ 32 r −1
where
2n 2n + 3r
2n + 3r
r
.
Using this result we can write
β 2n t 2n−1 e−(β+αc )t 1 αc β 2 t 3 w (0, t ) = , n + 1; 0 F2 n + Γ (2n) 2 4 n∗
2
We now consider the situation when p(x) = α 2 xe−α x , x > 0, so that q(x) = α e−α x , x > 0. In this case we have
wn∗ (0, t ) =
r
× 0 F2
−(β+α c )t 2n+ 32 (i+j)−1
so that
kr =
Γ (n)Γ (2n) 1 αc β 2 t 3 n + , n + 1; .
Whilst this expression is easily computed, it is not as compact as solutions for w(u, t ) presented in Dickson et al. (2005) and Borovkov and Dickson (2008). However, it is the counterpart of Drekic and Willmot’s (2003) formula for the density of the time to ruin in the classical risk model with exponentially distributed claims. Their formula can also be obtained by applying formula (12) to their model, and their formula, which is expressed in terms of Bessel functions, can be written in terms of 0 F1 functions using the relationship between these functions and Bessel functions.
Γ (n + 3m/2)
m =0
∞ X (α u)n−1 β 2n t 2n−1 n =1
We can evaluate the convolution Vn ∗ Wn (t ) by taking a term by term approach. We write
am =
15
h˜ (0, δ) =
∞
Z
e−δ t h(0, t )dt
and k˜ (0, δ) =
0
Z
∞
e−δ t k(0, t )dt .
0
Exactly as in Dickson (2008) we have
n+r +1 ∞ n α ˜h˜ (s, δ) = 1 X X n h˜ (0, δ)r +1 k˜ (0, δ)n−r α n =0 r =0 r α+s and — this is where we differ from Dickson (2008) – using results from Section 3, h˜ (0, δ)r +1 k˜ (0, δ)n−r =
×
β2 c 2 (α + r1 )(α + r2 )
r +1
β 2α β 2α + c 2 (α + r1 )2 (α + r2 ) c 2 (α + r1 )(α + r2 )2
n−r
16
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
r +1
β2 β 2α 2 2 c (α + r1 )(α + r2 ) c (α + r1 )(α + r2 ) n−r 1 1 + × α + r1 α + r2 n +1 β 2n+2 α n−r 1 = c 2n+2 (α + r1 )(α + r2 ) n − r X n−r 1 1 × x (α + r1 ) (α + r2 )n−r −x x x=0 n −r β 2n+2 α n−r X n−r = 2n+2
n −r
To find Cn,x,r let us write
=
c
×
An,x (t ) =
1
1
(α + r1 )n+1+x (α + r2 )2n+1−r −x
am =
and let us write Bn,x,r (t ) =
bk =
.
Γ (2k + 2n − r − x + 1) c 2n−r −x+1 (−αβ c )k . (2n − r + 2 − x)k k! (2n − r − x)!
(β + α c + s)2m+3n−r +2k+2 (15)
which inverts to am bk t 2m+3n−r +2k+1 e−(β+α c )t
. Γ (2m + 3n − r + 2k + 2) Then for l = 0, 1, 2, . . . the coefficient of
α m (ct − y)m−1 e−α(ct −y) yk e−αy dy Γ (m) Γ (k + 1) 0 (k + 1)α m e−αct (ct )m+k+1 = Γ (k + 2 + m)
t 3n−r +2l+1 e−(β+α c )t
Γ (3n − r + 2l + 2) is τl where
and expression (15) becomes
τl =
∞ X (ct )k e−αct β m t m−1 e−β t α m e−αct (ct )m+k+1 ce−β t + (k + 1) Γ (k + 1) m! Γ (k + 2 + m) m=1
=
∞ X β m t m e−β t α m e−αct (ct )m+k (ct )k e−αct + c (k + 1) Γ (k + 1) m! Γ (k + 2 + m) m=1 m ∞ αβ ct 2 c (ct )k e−(β+α c )t X 1 = k! (k + 2)m m! m=0
=
say.
0 F1
Γ (2i + n + x + 1) Γ (2(l − i) + 2n − r − x + 1) (n + 2 + x)i (2n − r + 2 − x)l−i
(αβ c )l σl,n,r ,x (n + x)!(2n − r − x)! l! c 3n−r +2
σl,n,r ,x =
(n + 2; −αβ ct 2 )
c (ct )2n−r −x e−(β+α c )t
(2n − r − x)! say.
0 F1
l X
×
2
c2
(2n − r − x + 2; −αβ ct 2 ) = Bn,x,r (t ),
Cn,x,r (t ) =
∞ X
×
∞ n X (αβ)2n X n
un+r
(n + r )!
c 2n n−r
X x =0
r =0
n−r x
=
r Cn,x,r (t ).
l
i
Γ (2i + n + x + 1) ( n + 2 + x) i
Thus,
l =0
n=0
Γ (2(l − i) + 2n − r − x + 1) × . (2n − r + 2 − x)l−i
Thus, if we define Cn,x,r to be the convolution of An,x and Bn,x,r , then −α u β
(−1)l−i
i=0
and hence 1/(α + r2 )2n+1−r −x inverts to
h(u, t ) = e
(n + x)!
and
n −(β+α c )t
n!
(n + 2 + x)i i!
Γ (2(l − i) + 2n − r − x + 1) c 2n−r −x+1 (−αβ c )l−i (2n − r + 2 − x)l−i (l − i)! (2n − r − x)! l 3n−r +2 l X c (αβ c ) l = (−1)l−i (n + x)!(2n − r − x)!l! i=0 i
×
Next, by the symmetry of (4) and (5), 1/(α + r2 )n+1 inverts to c (ct ) e
l X Γ (2i + n + x + 1) c n+x+1 (αβ c )i
×
(k + 2; αβ ct 2 ).
(n + 2 + x; αβ ct 2 ) = An,x (t ),
ai b l − i
i=0
Hence, 1/(α + r1 )n+1+x inverts to 0 F1
l X i=0
= ce−β t
(n + x)!
Γ (2k + 2n − r − x + 1)
am b k
y
c (ct )n+x e−(β+α c )t
t 2k+2n−r −x e−(β+α c )t
We recall that n and r are fixed. The convolution of the mth term of An,x with the kth term of Bn,x,r has Laplace transform with transform parameter s
ct
0 F1
bk
where
As qm∗ is the Erlang(m, α ) density, the integral becomes
k!
∞ X k=0
∞ β m t m−1 e−β t (ct )k e−αct X + Γ (k + 1) m! m=1 Z ct yk e−α y × yqm∗ (ct − y) dy. Γ (k + 1) 0
c (ct )k e−(β+α c )t
Γ (2m + n + x + 1)
Γ (2m + n + x + 1) c n+x+1 (αβ c )m , ( n + 2 + x) m m ! (n + x)!
ce−β t
=
t 2m+n+x e−(β+α c )t
where
Now by Eq. (4), 1/(α + r1 )k+1 inverts to
Z
am
m=0
x
x =0
∞ X
(16)
(αβ c )l σl,n,r ,x (n + x)!(2n − r − x)! l! c 3n−r +2
t 3n−r +2l+1 e−(β+α c )t
Γ (3n − r + 2l + 2)
c (ct )3n−r +1 e−(β+α c )t
(n + x)!(2n − r − x)! ∞ X (αβ ct 2 )l σl,n,x,r × l ! Γ ( 3n − r + 2l + 2) l =0
(17)
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
and h(u, t ) = e
−α u−(β+α c )t
∞ n β2 X (αβ)2n X n
c
×
n−r X
n−r
×
∞ X l =0
c 2n
n=0
x
x=0
h(0, t ) = e−(β+α c )t β 2 t 2 0 F3
r =0
un + r
(ct )3n−r +1 (n + x)!(2n − r − x)!
k(0, t ) = e
×
n−r X n−r x
x =0
∞ n X (αβ)2n X n
c 2n
n=0
r =0
r
un+r +1
(n + r + 1)!
(ct )3n−r +1 (n + x)!(2n − r − x)!
σl,n,x,r Γ ( 3n − r + 2l + 2) l =0 n ∞ X (αβ)2n X n un + r + αβ 2 ct 2 e−αu 2n c r (n + r )! r =0 n=0 nX −r +1 n−r +1 (ct )3n−r e−(β+αc )t × x (n + x)!(2n − r − x + 1)! x =0 ×
×
∞ X (αβ ct 2 )l
l!
∞ X l =0
σl∗,n,r ,x
(αβ ct ) l! Γ (2l + 3n − r + 3) 2 l
where
σl∗,n,r ,x =
l X l Γ (2i + n + x + 1) (−1)l−i (n + 2 + x)i i i=0 Γ (2(l − i) + 2n − r − x + 2) × . (2n − r + 3 − x)l−i
Finally, we remark that the density of the time to ruin is obtained as w(u, t ) = h(u, t ) + k(u, t ). 5.2. Hypergeometric solutions From a computational point of view, it would be useful if we could write the above solutions for h(u, t ) and k(u, t ) in terms of hypergeometric functions. Whilst we believe this may be possible, we are unable to do this in a systematic way. The route to finding such solutions is to find closed form solutions for σl,n,x,r and σl∗,n,r ,x , and we can do this using techniques described in Graham et al. (1994, Chapter 5). However, we are unable to find general expressions in terms of n, x and r, and the solutions we obtained for specific values of n, x and r do not point to general expressions. An exception is the case when u = 0. We find that
σl,0,0,0 = l!
l X
(−1)
l−i
2i + 1
×
2(l − i) + 1
1 2i + 1
i
i=0
is zero when l is odd (by symmetry), and equals l!4l+1
Γ ( l+21 )
Γ( )
2Γ ( + 2)
1 2
l 2
−
Γ (l + 23 ) Γ (l + 3)
when l is even, leading to
42
4
−(β+α c )t
αc β t
2 2
0 F3
3 3
, ,2 ;
2 2
α2β 2 c 2t 4 42
.
6.1. Other Erlang densities From the results in the previous section and the comments at the end of Section 3 we see that we can find an explicit solution for w(0, y, t ) when p is an Erlang(n, α ) density where n > 2. Finding a formula for w(u, y, t ) in this more general case is not as straightforward as in the case n = 2. For example, in the case n = 3, it can be shown that the solution for w(u, y, t ) is of the form
w(u, y, t ) =
3 X
hi (u, t )
i=1
β i yi−1 e−β y Γ (i)
and the bivariate transform
˜
h˜ i (s, δ) =
∞
Z 0
∞
Z
e−su−δ t hi (u, t )dtdu 0
can be written in terms of products of the univariate transforms h˜ i (0, δ). As stated at the end of Section 3 we know that all such products are of the form (α + r1 )−n (α + r2 )−m where n and m are positive integers, and can thus be inverted. The workings are messy, but no new ideas are involved. 6.2. Other quantities of interest We note that the solutions for h(u, t ) and k(u, t ) in Section 5.1 can be written as infinite sums including Erlang densities. Thus, we can express the finite time ruin probability as an infinite mixture of Erlang distributions. Similarly, we can use this observation to obtain expressions for the moments of the time to ruin. For the sake of brevity we do not include these expressions here. Similar comments apply in the case of exponential claims from Section 4. 6.3. A dual risk model According to Mazza and Rullière (2004), since we know the density of the time to ruin in a Sparre Andersen model with Erlang(2, β ) claim inter-arrival times, Erlang(2, α ) claim amounts and premium rate c per unit time, we can compute the density of the time to ruin in a dual Sparre Andersen model with Erlang(2, α ) gain inter-arrival times, Erlang(2, β ) gains and expenses rate E = 1/c. Our formulae for h(u, t ) and k(u, t ) in Section 5.1 allow us to apply Mazza and Rullière’s results to obtain a formula for the density of the time to ruin in this dual model. We omit the details, again for the sake of brevity.
1 2(l − i) + 1
l−i
2 2 4
6. Concluding remarks
Similarly, we can show that 2
2 2
α β c t 2
Similarly, we can show that
(αβ ct 2 )l σl,n,x,r . l! Γ (3n − r + 2l + 2)
αβ k(u, t ) = e−α u−(β+α c )t c
3 5
, ,2 ; 4 4 2 2 2 4 1 3 α β c t − 0 F3 , ,2 ; . 2
(n + r )!
r
17
!
Acknowledgements The authors are grateful to two referees whose comments have improved this paper. Appendix In Section 4 we stated that
18
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18 2r X (−1)i
n2
i=0
=
Γ (n + 3i/2)
Γ (n + 3(2r − i)/2)
for r = 0, 2, 4,. .3r., and is 0 otherwise. Next, +1 1 , [z r ]B3/2 (z ) = 2 3r r +1 2
i!Γ (i/2 + n + 1) (2r − i)!Γ ((2r − i)/2 + n + 1)
2n
2n + 3r
2n + 3r
r
.
(18)
We do not claim that this result is new, but as we have been unable to find it in the mathematics literature, we give a proof which draws heavily on ideas presented in Graham et al. (1994). Formula (18) is true if
B3/2 (z )n B3/2 (−z )n = B3 (z 2 )2n
Bt (z ) =
k=0
[z ]z B3 (z ) = r
2
2 3
3r 2
r 2
1 r 2
−1
+2
r 2
zk
+
r 2
2 3r 2
3r 2
tk + 1
k
3r 2
for r = 2, 4, 6, . . ., and is 0 otherwise. Then we find that [z 0 ]h(z ) = 0, and for r = 1, 3, 5, . . ., [z r ]h(z ) = 0. Finally, for r = 2, 4, 6, . . . ,
[ z r ] h( z ) =
where
∞ X tk + 1
and
−1
+2 1 r 2
3r −
2
+1 r
2 3r 2
+1
,
is what Graham et al. (1994) call the generalised binomial series, with
and the right-hand side simplifies to 0, which gives the result.
∞ X tk + n nz k . Bt (z ) = tk + n k k=0
References
n
(19)
Thus, to show (18), it suffices to show that
B3/2 (z )B3/2 (−z ) = B3 (z 2 )2 .
(20)
From the first formula of (5.59) of Graham et al. (1994) we know that
B3/2 (z ) − 1 = z B3/2 (z )3/2 so that
B3/2 (z )B3/2 (−z ) − B3/2 (z ) − B3/2 (−z ) + 1
3/2 = −z 2 B3/2 (z )B3/2 (−z ) . P∞ n If a series A(z ) = n=0 an z satisfies A(z ) − B3/2 (z ) − B3/2 (−z ) + 1 = −z 2 A(z )3/2 , then A(z ) = B3/2 (z )B3/2 (−z ). We now show that B3 (z 2 )2 is such a series and this establishes that formula (20) holds. Let h(z ) = B3 (z 2 )2 − B3/2 (z ) − B3/2 (−z ) + 1 + z 2 B3 (z 2 )3 and consider [z r ]h(z ) (i.e. the coefficient of z r in h(z )). Then using (19),
[z ]B 3 ( z ) = r
2 2
3r 2
+2 r 2
2 3r 2
+2
Borovkov, K.A., Dickson, D.C.M., 2008. On the ruin time distribution for a Sparre Andersen process with exponential claim sizes. Insurance: Mathematics & Economics 42, 1104–1108. Cheung, E.C.K., Dickson, D.C.M., Drekic, S., 2008. Moments of discounted dividends for a threshold strategy in the compound Poisson risk model. North American Actuarial Journal 12 (3), 299–318. Dickson, D.C.M., 2008. Some explicit solutions for the joint density of the time of ruin and the deficit at ruin. ASTIN Bulletin 38, 259–276. Dickson, D.C.M., Hipp, C., 2001. On the time to ruin for Erlang(2) risk processes. Insurance: Mathematics & Economics 29, 333–344. Dickson, D.C.M., Willmot, G.E., 2005. The density of the time to ruin in the classical Poisson risk model. ASTIN Bulletin 35, 45–60. Dickson, D.C.M., Hughes, B.D., Lianzeng, Z., 2005. The density of the time to ruin for a Sparre Andersen process with Erlang arrivals and exponential claims. Scandinavian Actuarial Journal 2005 (5), 358–376. Drekic, S., Willmot, G.E., 2003. On the density and moments of the time to ruin with exponential claims. ASTIN Bulletin 33, 11–21. Garcia, J.M.A., 2005. Explicit solutions for survival probabilities in the classical risk model. ASTIN Bulletin 35, 113–130. Gerber, H.U., Shiu, E.S.W., 1998. On the time value of ruin. North American Actuarial Journal 2 (1), 48–78. Graham, R.L., Knuth, D.E., Patashnik, O., 1994. Concrete Mathematics, 2nd ed.. Addison-Wesley, Upper Saddle River, NJ. Li, S., Garrido, J., 2004. On ruin for the Erlang(n) risk process. Insurance: Mathematics & Economics 34, 391–408. Mazza, C., Rullière, C., 2004. A link between wave governed random motions and ruin processes. Insurance: Mathematics & Economics 35, 205–222. Sun, L.-J., 2005. The expected discounted penalty at ruin in the Erlang(2) risk process. Statistics & Probability Letters 72, 205–217. Willmot, G.E., 2007. On the discounted penalty function in the renewal risk model with general interclaim times. Insurance: Mathematics & Economics 41, 17–31. Willmot, G.E., Woo, J.K., 2007. On the class of Erlang mixtures with risk theoretic applications. North American Actuarial Journal 11 (2), 99–115.
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Finite time ruin problems for the Erlang(2) risk model David C.M. Dickson ∗ , Shuanming Li Centre for Actuarial Studies, Department of Economics, University of Melbourne, Victoria 3010, Australia
article
info
Article history: Received August 2008 Received in revised form March 2009 Accepted 6 May 2009
abstract We consider the Erlang(2) risk model and derive expressions for the density of the time to ruin and the joint density of the time to ruin and the deficit at ruin when the individual claim amount distribution is (i) an exponential distribution and (ii) an Erlang(2) distribution. We also consider the special case when the initial surplus is zero. © 2009 Elsevier B.V. All rights reserved.
1. Introduction In this paper we use results given by Dickson and Hipp (2001) and ideas given in Cheung et al. (2008) and Dickson (2008) to study finite time ruin problems for the Erlang(2) risk model. In particular we aim to find formulae for finite time ruin probabilities and for joint densities of the time to ruin and deficit at ruin. In the literature there are very few exact formulae for finite time ruin probabilities. For the classical risk model, Dickson and Willmot (2005) give a formula for the finite time ruin probability in the case when the individual claims have a distribution that is an infinite mixture of Erlang distributions. Willmot and Woo (2007) explain why this formula covers a range of individual claim amount distributions and covers all cases for which formulae for the finite time ruin probability exist. See Drekic and Willmot (2003) and Garcia (2005). In the case of Sparre Andersen risk models formulae for finite time ruin probabilities exist only in the case of exponential claims — see Dickson et al. (2005) and Borovkov and Dickson (2008). Dickson (2008) found formulae for the joint density of the time to ruin and the deficit at ruin in the classical risk model when the distribution of individual claims was either Erlang(2) or a mixture of two exponential distributions. However, no such results presently exist for Sparre Andersen risk models. Here we apply the methodology in Dickson (2008) to derive such results for the Erlang(2) risk model. The outline of this paper is as follows. In Section 2 we set out the mathematical preliminaries and give transform relationships that are central to our derivations in subsequent sections. In Section 3 we discuss the special case when u = 0 and find that the derivation of exact results is somewhat complicated in general, but less so when the individual claim amount distribution has a
particular form. The cases of individual claim amounts having (i) an exponential distribution and (ii) an Erlang(2) distribution are discussed in Sections 4 and 5 respectively. We make some concluding remarks in Section 6. 2. Preliminaries We adopt the model of Dickson and Hipp (2001). Thus, we are dealing with a Sparre Andersen risk model under which claim inter-arrival times are distributed as Erlang(2) with scale parameter β . We denote by p the density function of individual claim amounts, and denote the kth moment as mk . Further, the Laplace transform of p is denoted by p˜ where p˜ (s) =
Corresponding author. E-mail addresses: [email protected] (D.C.M. Dickson), [email protected] (S. Li). 0167-6687/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.insmatheco.2009.05.001
e−sx p(x)dx. 0
Let P = 1 − P¯ denote the distribution function. We denote by c the insurer’s premium income per unit time and assume that 2c /β > m1 . Let {U (t )}t ≥0 denote the surplus process, let T denote the time of ruin, and let Y = | − U (T )| denote the deficit at ruin. Define
φ(u) = E e−δT −sY I (T < ∞)|U (0) = u to be the bivariate Laplace transform of T and Y . Let w(u, t ) denote the (defective) density of T , and let w(u, y, t ) denote the (defective) joint density of T and Y so that
φ(u) =
∞
Z 0
∞
Z
e−δ t −sy w(u, y, t )dydt . 0
For this risk model, Lundberg’s fundamental equation is s2 − 2
∗
∞
Z
β +δ c
s+
β +δ c
2 =
β2 c2
p˜ (s)
and Dickson and Hipp (2001) show that this equation has two solutions r1 and r2 such that r1 < (β +δ)/c < r2 . We easily deduce
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
that r1 = r2 =
β +δ c
β +δ c
−
βp
+
βp
c c
p˜ (r1 ),
(1)
p˜ (r2 ).
(2)
For the remainder of this paper we assume that p˜ (s) = q˜ (s)2 for some density function q. An easy way of thinking of this assumption is that an Erlang(2) risk process is a classical risk process modified so that the first two claims are paid together at the time of the second claim, claims three and four are paid together at the time of the fourth claim and so on. Indeed, such a transform relationship holds if the individual claim amount distribution is infinitely divisible. Rewriting expressions (1) and (2) as
β + δ − cr1 = β q˜ (r1 ), β + δ − cr2 = −β q˜ (r2 ),
(3)
we see that Eq. (3) is identical in form to Lundberg’s fundamental equation for the classical risk model, λ + δ − c ρ = λ˜p(ρ) (see Gerber and Shiu (1998)). Thus we can apply the ideas of Dickson and Willmot (2005, page 49) to obtain the following transform relationship. If ∞
Z
f˜ (r1 ) =
e−r1 t f (t )dt = g˜ (δ) = 0
computing w(0, t ) for specific individual claim amount distributions. However, for certain individual claim amount distributions, we can take a different approach to find solutions both for w(0, t ) and w(0, y, t ), as follows. Let us now assume that the individual claim amount density is such that p(x + y) =
P¯ (x + y) =
n=1
β n t n−1 e−β t n!
f˜ (r2 ) =
∞
e−r2 t f (t )dt = h˜ (δ) = 0
yqn∗ (ct − y)f (y)dy
(4)
0
m X
ηj (x)T¯j (y)
e−δ t h(t )dt
where Tj = 1 − T¯j is the distribution function associated with τj . From Dickson and Hipp (2001) we know that ∞
Z
e−δ t w(0, t )dt = 0
∞
Z
e
0
−δ t
w(0, t )dt =
0
h(t ) = ce−β t f (ct ) +
∞ X (−β)n t n−1 e−β t
=
n!
n=1 ct
Z
yqn∗ (ct − y)f (y)dy.
×
(5)
0
In the following sections we obtain Laplace transforms with transform parameters r1 and r2 . Our approach is to invert these Laplace transforms using relationships (4) and (5). 3. The case u = 0 We consider first the density of the time of ruin. From Dickson and Hipp (2001) (see also Li and Garrido (2004)) we know that ∞
e−δ t w(0, t )dt =
c 2 r1 r2 − (β + δ)2 + β 2
and if we use Eqs. (1) and (2) for r1 and r2 we obtain ∞
e−δ t w(0, t )dt 0
=
β 2 [1 − q˜ (r1 ) + q˜ (r2 ) − q˜ (r1 )˜q(r2 )] − βδ[˜q(r1 ) − q˜ (r2 )]
=
β2
c 2 r1 r2 c2
1 − q˜ (r1 ) r1
1 + q˜ (r2 ) r2
−
βδ[˜q(r1 ) − q˜ (r2 )] c 2 r1 r2
m Z β2 X
.
(6)
Although formula (6) can be inverted, the resulting expression for w(0, t ) does not seem particularly attractive in terms of
c2
∞
Z
∞
Z
0
e−r1 x e−r2 y P¯ (x + y)dydx, 0
c2
j =1
β2
∞
Z
c2
e−r1 x ηj (x)dx 0
∞
Z
0
∞
e− r 1 x e− r 2 y 0
m X
ηj (x)T¯j (y)dydx
j =1
∞
Z
e−r2 y T¯j (y)dy.
(8)
0
The significance of Eq. (8) is that each of the integrals is a Laplace transform — with respect to either r1 or r2 , and hence by the transform relationships (4) and (5) they are Laplace transforms with respect to δ . By inverting each transform, we can obtain the inverse of the product of two transforms by finding the convolution of the two inverses. For example, the simplest case to satisfy Eq. (7) is p(x) = α e−α x giving η1 (x) = e−α x and τ1 (y) = α e−α y , with T¯1 (y) = e−α y . Then Eq. (8) gives the already known equation ∞
Z
e−δ t w(0, t )dt = 0
c 2 r1 r2
0
β2
so if p satisfies Eq. (7) we obtain
then
Z
ηj (x)τj (v)dv
j =1
j=1
ct
∞
Z
m ∞X y
where qn∗ denotes the n-fold convolution of q. The derivation is exactly as in Dickson and Willmot (2005) — all that is different is the notation. The same arguments give a further relationship involving r2 . If
Z
p(x + v)dv
Z
=
g (t ) = ce−β t f (ct ) +
∞
Z
=
e−δ t g (t )dt
Z
(7)
y
0
∞ X
ηj (x)τj (y)
for some functions {ηj , τj }m j=1 . This factorisation was introduced by Willmot (2007), and he shows that if the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter then this factorisation applies. Results in Willmot and Woo (2007) show that this infinite mixture of Erlangs contains many well-known distributions as special cases. We make the further assumption that the functions {τj } are in fact density functions (as is the case when p is an infinite mixture of Erlang densities). Then
then
Z
m X j =1
∞
Z
13
c2
β2 , (α + r1 )(α + r2 )
(see Dickson and Hipp (2001)), and we show how to invert this Laplace transform in the next section. The same techniques apply to other individual claim amount distributions whose densities satisfy Eq. (7). This approach can be extended if the functions {ηj } satisfy the same type of factorisation as Eq. (7). As Willmot (2007) shows, this is the case when the individual claim amount distribution is an infinite mixture of Erlang distributions with the same scale parameter. Specifically, let us suppose that
ηj (x + y) =
n X i =1
ξij (x)ζij (y).
(9)
14
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
4. Exponential claims
From Eq. (4) of Sun (2005), we know that ∞
Z
∞
Z
In this√section we consider the case p(x) = α e−α x , x > 0, so that q(x) = α/(xπ )e−αx . Although formulae for w(u, t ) are known for this case (see Dickson et al. (2005) and Borovkov and Dickson (2008)), it is interesting to explore this case for two reasons. First, it introduces the techniques needed for the more complicated case of Erlang(2) claims discussed in the next section, and, second, we end up with a new formula for w(u, t ) which parallels the formula for the density of the time to ruin in a classical risk model with exponential claims obtained by Drekic and Willmot (2003). Dickson (2008) shows that
e−δ t −sy w(0, y, t )dydt 0
0
β2
=
∞
Z
c2
Z
∞
e−r1 x e−r2 y $ (x + y)dydx
0
0
where
$ (u) =
∞
Z
e−sy p(u + y)dy. 0
Applying (7) in the above expression we obtain
$ (u) =
∞
Z
e
−sy
0
m X
ηj (u)τj (y)dy =
j =1
m X
w(u, t ) =
ηj (u)τ˜j (s),
Z tZ 0
j =1
u
w(0, y, τ )w(u − y, t − τ )dydτ
0
∞
Z
w(0, y, t )dy,
+
and if we now apply (9) we obtain
(11)
u
$ (x + y) =
m X n X
and it is well known that by the memoryless property of the exponential distribution
ξij (x)ζij (y)τ˜j (s),
j =1 i =1
w(u, y, t ) = w(u, t )α e−αy .
so that ∞
Z 0
Following the approach in Section 3 of Dickson (2008), we insert this expression in Eq. (11) then take the Laplace transform of the resulting equation. Defining
∞
Z
e−δ t −sy w(0, y, t )dydt 0
n m X β2 X
=
c2
ξ˜ij (r1 )ζ˜ij (r2 )τ˜j (s).
(10)
˜˜ s, δ) = w(
j =1 i =1
η1 (x) = α xe , τ2 (y) = α 2 ye−αy ,
τ1 (y) = α e
−α y
,
η2 ( x) = e
−α x
,
w( ˜ 0, δ) =
e−δ t w(0, t )dt , 0
we obtain
˜˜ s, δ) = w(
1 w( ˜ 0, δ) α+ s
α 1 − w( ˜ 0, δ) α+ s
η1 (x + y) = ξ11 (x)ζ11 (y) + ξ21 (x)ζ21 (y), η2 (x + y) = ξ12 (x)ζ12 (y),
w(u, t ) =
−α y
Then
0
=
∞
Z
w( ˜ 0, δ) =
e−δ t −sy w(0, y, t )dydt 0
β2
α α + 2 2 c (α + r1 ) (α + r2 ) (α + r1 )(α + r2 )2 2 β2 1 α + 2 . c (α + r1 )(α + r2 ) α + s
wn∗ (0, t )
=
∞ 1X
α
˜ 0, δ))n (w(
n =1
α α+s
(α u)n−1 e−αu . Γ ( n)
n
(12)
We remark that the derivation of this result does not depend on the claim inter-arrival distribution being Erlang(2) — formula (12) is a general result. We saw in Section 3 that when the claim inter-arrival distribution is Erlang(2) with scale parameter β ,
ξ11 (x) = α xe , ζ11 (y) = ζ12 (y) = e , −α x ξ21 (x) = ξ12 (x) = e , ζ21 (y) = α ye−αy . ∞
∞ X n=1
where
and
∞
Z
which inverts to
Z
e−su−δ t w(u, t )dtdu
0
and we can write
−α x
∞
Z
0
Thus, the bivariate Laplace transform factorises in terms of transforms with transform parameters r1 , r2 and s. For example, when p(x) = α 2 xe−α x , x > 0, we get −α x
∞
Z
α α+s
As r1 and r2 are both functions of δ , the above equation tells us that
w(0, y, t ) = f1 (t )α e−αy + f2 (t )α 2 ye−αy , and we show in Section 5 how the functions f1 and f2 can be identified. We remark that this example easily extends to the general case when p is the Erlang(n, α ) density with n > 2 since we know from Willmot (2007) that the factorisation (9) applies in this case and that the functions ξij and ζij in (9) are just scaled Erlang density functions. Thus when we apply (10), we obtain terms which are multiples of (α + r1 )−n (α + r2 )−m where n and m are positive integers. The inversion of such terms is described in Section 5.
c2
β2 (α + r1 )(α + r2 )
so that w n∗ (0, t ) can be found by inverting
β 2n . c 2n (α + r1 )n (α + r2 )n Now let V˜ n (δ) = (α + r1 )−n (where r1 depends on δ ) be the Laplace ˜ n (δ) = (α + r2 )−n . transform of a function Vn (t ), and similarly let W Then inversion of V˜ n (δ) with respect to r1 yields t n−1 e−α t /Γ (n) and by (4) inversion with respect to δ yields ∞ (ct )n−1 e−αct X β m t m−1 e−β t + Γ (n) m! m =1 Z ct yn−1 e−α y × yqm∗ (ct − y) dy. Γ (n) 0
ce−β t
We find that ct
Z
yqm∗ (ct − y) 0
yn−1 e−α y
Γ (n)
dy =
nα m/2 e−α ct (ct )n+m/2
Γ (m/2 + n + 1)
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
giving
where
∞ X α m/2 β m c n+m/2 t n+3m/2−1 Vn (t ) = ne−(β+α c )t . m! Γ (m/2 + n + 1) m=0
(B1 , B2 , . . . , Bp , C1 , C2 , . . . , Cq ; Z ) ∞ X (B1 )m (B2 )m · · · (Bp )m Z m = (C1 )m (C2 )m · · · (Cq )m m! m=0
p Fq
Similarly, by applying (5) we obtain Wn (t ) = ne−(β+α c )t
is the generalised hypergeometric function and (a)n = Γ (a + n)/Γ (a) is Pochhammer’s symbol. This follows by looking at the ratio of terms in the sum in (13). (See, for example, Graham et al. (1994, pp. 207–8).) Thus we have
∞ X α m/2 (−1)m β m c n+m/2 t n+3m/2−1 . m! Γ (m/2 + n + 1) m=0
Thus,
β2
wn∗ (0, t ) =
c2
w(u, t ) = e−αu−(β+αc )t
Vn ∗ Wn (t ).
Vn ( t ) =
∞ X
am
t
n+3m/2−1 −(β+α c )t
e
where nα m/2 β m c n+m/2 Γ (n + 3m/2)
Γ (m/2 + n + 1)
m!
,
and so we can write Wn (t ) =
∞ X
(−1)m am
t n+3m/2−1 e−(β+α c )t
Γ (n + 3m/2)
m=0
.
If we take the convolution of the ith term in Vn with the jth term in Wn we get
(−1)j ai aj
e
4
5. Erlang(2) claims
Γ (2n + 32 (i + j))
w(u, y, t ) = h(u, t )α 2 ye−αy + k(u, t )α e−αy
t
β
2n
∞ X
c 2n r =0
kr
e
t
Γ (2n + 23 r )
5.1. Main results
r X (−1)r −i ai ar −i . i=0
By symmetry, kr = 0 if r is odd, so we obtain
wn∗ (0, t ) =
∞ β 2n t 2n−1 e−(β+αc )t X
c 2n
r =0
k2r
t 3r
Γ (2n + 3r )
.
(13)
Tidying up gives k2r = α β c
2r 2n+r
As the form of the solution for w(u, y, t ) is exactly the same as in the classical risk model with Erlang(2, α ) claims, we can apply the approach given in Dickson (2008, Section 3). The reason we can do this is that we are simply inserting formula (14) into Eq. (1) of Dickson (2008), and that equation applies to any Sparre Andersen model. Let
σ (2r , n)
∞
Z
˜
h˜ (s, δ) =
0
∞
Z
e−su−δ t h(u, t )dtdu 0
and let
where 2r X σ (2r , n) = n2 (−1)i i =0
Γ (n + 3i/2) i!Γ (i/2 + n + 1)
Γ (n + 3(2r − i)/2) × . (2r − i)!Γ ((2r − i)/2 + n + 1) We now state a result, which we prove in the Appendix. For n = 1, 2, 3, . . . and r = 0, 1, 2, . . .,
σ (2r , n) =
(14)
where h(u, t ) and k(u, t ) are functions that we will identify. For the sake of brevity we omit the details of why we know w(u, y, t ) is of this form. The approach to showing this is essentially that given by Cheung et al. (2008), but adapted to our risk model.
−(β+α c )t 2n+ 32 r −1
where
2n 2n + 3r
2n + 3r
r
.
Using this result we can write
β 2n t 2n−1 e−(β+αc )t 1 αc β 2 t 3 w (0, t ) = , n + 1; 0 F2 n + Γ (2n) 2 4 n∗
2
We now consider the situation when p(x) = α 2 xe−α x , x > 0, so that q(x) = α e−α x , x > 0. In this case we have
wn∗ (0, t ) =
r
× 0 F2
−(β+α c )t 2n+ 32 (i+j)−1
so that
kr =
Γ (n)Γ (2n) 1 αc β 2 t 3 n + , n + 1; .
Whilst this expression is easily computed, it is not as compact as solutions for w(u, t ) presented in Dickson et al. (2005) and Borovkov and Dickson (2008). However, it is the counterpart of Drekic and Willmot’s (2003) formula for the density of the time to ruin in the classical risk model with exponentially distributed claims. Their formula can also be obtained by applying formula (12) to their model, and their formula, which is expressed in terms of Bessel functions, can be written in terms of 0 F1 functions using the relationship between these functions and Bessel functions.
Γ (n + 3m/2)
m =0
∞ X (α u)n−1 β 2n t 2n−1 n =1
We can evaluate the convolution Vn ∗ Wn (t ) by taking a term by term approach. We write
am =
15
h˜ (0, δ) =
∞
Z
e−δ t h(0, t )dt
and k˜ (0, δ) =
0
Z
∞
e−δ t k(0, t )dt .
0
Exactly as in Dickson (2008) we have
n+r +1 ∞ n α ˜h˜ (s, δ) = 1 X X n h˜ (0, δ)r +1 k˜ (0, δ)n−r α n =0 r =0 r α+s and — this is where we differ from Dickson (2008) – using results from Section 3, h˜ (0, δ)r +1 k˜ (0, δ)n−r =
×
β2 c 2 (α + r1 )(α + r2 )
r +1
β 2α β 2α + c 2 (α + r1 )2 (α + r2 ) c 2 (α + r1 )(α + r2 )2
n−r
16
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
r +1
β2 β 2α 2 2 c (α + r1 )(α + r2 ) c (α + r1 )(α + r2 ) n−r 1 1 + × α + r1 α + r2 n +1 β 2n+2 α n−r 1 = c 2n+2 (α + r1 )(α + r2 ) n − r X n−r 1 1 × x (α + r1 ) (α + r2 )n−r −x x x=0 n −r β 2n+2 α n−r X n−r = 2n+2
n −r
To find Cn,x,r let us write
=
c
×
An,x (t ) =
1
1
(α + r1 )n+1+x (α + r2 )2n+1−r −x
am =
and let us write Bn,x,r (t ) =
bk =
.
Γ (2k + 2n − r − x + 1) c 2n−r −x+1 (−αβ c )k . (2n − r + 2 − x)k k! (2n − r − x)!
(β + α c + s)2m+3n−r +2k+2 (15)
which inverts to am bk t 2m+3n−r +2k+1 e−(β+α c )t
. Γ (2m + 3n − r + 2k + 2) Then for l = 0, 1, 2, . . . the coefficient of
α m (ct − y)m−1 e−α(ct −y) yk e−αy dy Γ (m) Γ (k + 1) 0 (k + 1)α m e−αct (ct )m+k+1 = Γ (k + 2 + m)
t 3n−r +2l+1 e−(β+α c )t
Γ (3n − r + 2l + 2) is τl where
and expression (15) becomes
τl =
∞ X (ct )k e−αct β m t m−1 e−β t α m e−αct (ct )m+k+1 ce−β t + (k + 1) Γ (k + 1) m! Γ (k + 2 + m) m=1
=
∞ X β m t m e−β t α m e−αct (ct )m+k (ct )k e−αct + c (k + 1) Γ (k + 1) m! Γ (k + 2 + m) m=1 m ∞ αβ ct 2 c (ct )k e−(β+α c )t X 1 = k! (k + 2)m m! m=0
=
say.
0 F1
Γ (2i + n + x + 1) Γ (2(l − i) + 2n − r − x + 1) (n + 2 + x)i (2n − r + 2 − x)l−i
(αβ c )l σl,n,r ,x (n + x)!(2n − r − x)! l! c 3n−r +2
σl,n,r ,x =
(n + 2; −αβ ct 2 )
c (ct )2n−r −x e−(β+α c )t
(2n − r − x)! say.
0 F1
l X
×
2
c2
(2n − r − x + 2; −αβ ct 2 ) = Bn,x,r (t ),
Cn,x,r (t ) =
∞ X
×
∞ n X (αβ)2n X n
un+r
(n + r )!
c 2n n−r
X x =0
r =0
n−r x
=
r Cn,x,r (t ).
l
i
Γ (2i + n + x + 1) ( n + 2 + x) i
Thus,
l =0
n=0
Γ (2(l − i) + 2n − r − x + 1) × . (2n − r + 2 − x)l−i
Thus, if we define Cn,x,r to be the convolution of An,x and Bn,x,r , then −α u β
(−1)l−i
i=0
and hence 1/(α + r2 )2n+1−r −x inverts to
h(u, t ) = e
(n + x)!
and
n −(β+α c )t
n!
(n + 2 + x)i i!
Γ (2(l − i) + 2n − r − x + 1) c 2n−r −x+1 (−αβ c )l−i (2n − r + 2 − x)l−i (l − i)! (2n − r − x)! l 3n−r +2 l X c (αβ c ) l = (−1)l−i (n + x)!(2n − r − x)!l! i=0 i
×
Next, by the symmetry of (4) and (5), 1/(α + r2 )n+1 inverts to c (ct ) e
l X Γ (2i + n + x + 1) c n+x+1 (αβ c )i
×
(k + 2; αβ ct 2 ).
(n + 2 + x; αβ ct 2 ) = An,x (t ),
ai b l − i
i=0
Hence, 1/(α + r1 )n+1+x inverts to 0 F1
l X i=0
= ce−β t
(n + x)!
Γ (2k + 2n − r − x + 1)
am b k
y
c (ct )n+x e−(β+α c )t
t 2k+2n−r −x e−(β+α c )t
We recall that n and r are fixed. The convolution of the mth term of An,x with the kth term of Bn,x,r has Laplace transform with transform parameter s
ct
0 F1
bk
where
As qm∗ is the Erlang(m, α ) density, the integral becomes
k!
∞ X k=0
∞ β m t m−1 e−β t (ct )k e−αct X + Γ (k + 1) m! m=1 Z ct yk e−α y × yqm∗ (ct − y) dy. Γ (k + 1) 0
c (ct )k e−(β+α c )t
Γ (2m + n + x + 1)
Γ (2m + n + x + 1) c n+x+1 (αβ c )m , ( n + 2 + x) m m ! (n + x)!
ce−β t
=
t 2m+n+x e−(β+α c )t
where
Now by Eq. (4), 1/(α + r1 )k+1 inverts to
Z
am
m=0
x
x =0
∞ X
(16)
(αβ c )l σl,n,r ,x (n + x)!(2n − r − x)! l! c 3n−r +2
t 3n−r +2l+1 e−(β+α c )t
Γ (3n − r + 2l + 2)
c (ct )3n−r +1 e−(β+α c )t
(n + x)!(2n − r − x)! ∞ X (αβ ct 2 )l σl,n,x,r × l ! Γ ( 3n − r + 2l + 2) l =0
(17)
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18
and h(u, t ) = e
−α u−(β+α c )t
∞ n β2 X (αβ)2n X n
c
×
n−r X
n−r
×
∞ X l =0
c 2n
n=0
x
x=0
h(0, t ) = e−(β+α c )t β 2 t 2 0 F3
r =0
un + r
(ct )3n−r +1 (n + x)!(2n − r − x)!
k(0, t ) = e
×
n−r X n−r x
x =0
∞ n X (αβ)2n X n
c 2n
n=0
r =0
r
un+r +1
(n + r + 1)!
(ct )3n−r +1 (n + x)!(2n − r − x)!
σl,n,x,r Γ ( 3n − r + 2l + 2) l =0 n ∞ X (αβ)2n X n un + r + αβ 2 ct 2 e−αu 2n c r (n + r )! r =0 n=0 nX −r +1 n−r +1 (ct )3n−r e−(β+αc )t × x (n + x)!(2n − r − x + 1)! x =0 ×
×
∞ X (αβ ct 2 )l
l!
∞ X l =0
σl∗,n,r ,x
(αβ ct ) l! Γ (2l + 3n − r + 3) 2 l
where
σl∗,n,r ,x =
l X l Γ (2i + n + x + 1) (−1)l−i (n + 2 + x)i i i=0 Γ (2(l − i) + 2n − r − x + 2) × . (2n − r + 3 − x)l−i
Finally, we remark that the density of the time to ruin is obtained as w(u, t ) = h(u, t ) + k(u, t ). 5.2. Hypergeometric solutions From a computational point of view, it would be useful if we could write the above solutions for h(u, t ) and k(u, t ) in terms of hypergeometric functions. Whilst we believe this may be possible, we are unable to do this in a systematic way. The route to finding such solutions is to find closed form solutions for σl,n,x,r and σl∗,n,r ,x , and we can do this using techniques described in Graham et al. (1994, Chapter 5). However, we are unable to find general expressions in terms of n, x and r, and the solutions we obtained for specific values of n, x and r do not point to general expressions. An exception is the case when u = 0. We find that
σl,0,0,0 = l!
l X
(−1)
l−i
2i + 1
×
2(l − i) + 1
1 2i + 1
i
i=0
is zero when l is odd (by symmetry), and equals l!4l+1
Γ ( l+21 )
Γ( )
2Γ ( + 2)
1 2
l 2
−
Γ (l + 23 ) Γ (l + 3)
when l is even, leading to
42
4
−(β+α c )t
αc β t
2 2
0 F3
3 3
, ,2 ;
2 2
α2β 2 c 2t 4 42
.
6.1. Other Erlang densities From the results in the previous section and the comments at the end of Section 3 we see that we can find an explicit solution for w(0, y, t ) when p is an Erlang(n, α ) density where n > 2. Finding a formula for w(u, y, t ) in this more general case is not as straightforward as in the case n = 2. For example, in the case n = 3, it can be shown that the solution for w(u, y, t ) is of the form
w(u, y, t ) =
3 X
hi (u, t )
i=1
β i yi−1 e−β y Γ (i)
and the bivariate transform
˜
h˜ i (s, δ) =
∞
Z 0
∞
Z
e−su−δ t hi (u, t )dtdu 0
can be written in terms of products of the univariate transforms h˜ i (0, δ). As stated at the end of Section 3 we know that all such products are of the form (α + r1 )−n (α + r2 )−m where n and m are positive integers, and can thus be inverted. The workings are messy, but no new ideas are involved. 6.2. Other quantities of interest We note that the solutions for h(u, t ) and k(u, t ) in Section 5.1 can be written as infinite sums including Erlang densities. Thus, we can express the finite time ruin probability as an infinite mixture of Erlang distributions. Similarly, we can use this observation to obtain expressions for the moments of the time to ruin. For the sake of brevity we do not include these expressions here. Similar comments apply in the case of exponential claims from Section 4. 6.3. A dual risk model According to Mazza and Rullière (2004), since we know the density of the time to ruin in a Sparre Andersen model with Erlang(2, β ) claim inter-arrival times, Erlang(2, α ) claim amounts and premium rate c per unit time, we can compute the density of the time to ruin in a dual Sparre Andersen model with Erlang(2, α ) gain inter-arrival times, Erlang(2, β ) gains and expenses rate E = 1/c. Our formulae for h(u, t ) and k(u, t ) in Section 5.1 allow us to apply Mazza and Rullière’s results to obtain a formula for the density of the time to ruin in this dual model. We omit the details, again for the sake of brevity.
1 2(l − i) + 1
l−i
2 2 4
6. Concluding remarks
Similarly, we can show that 2
2 2
α β c t 2
Similarly, we can show that
(αβ ct 2 )l σl,n,x,r . l! Γ (3n − r + 2l + 2)
αβ k(u, t ) = e−α u−(β+α c )t c
3 5
, ,2 ; 4 4 2 2 2 4 1 3 α β c t − 0 F3 , ,2 ; . 2
(n + r )!
r
17
!
Acknowledgements The authors are grateful to two referees whose comments have improved this paper. Appendix In Section 4 we stated that
18
D.C.M. Dickson, S. Li / Insurance: Mathematics and Economics 46 (2010) 12–18 2r X (−1)i
n2
i=0
=
Γ (n + 3i/2)
Γ (n + 3(2r − i)/2)
for r = 0, 2, 4,. .3r., and is 0 otherwise. Next, +1 1 , [z r ]B3/2 (z ) = 2 3r r +1 2
i!Γ (i/2 + n + 1) (2r − i)!Γ ((2r − i)/2 + n + 1)
2n
2n + 3r
2n + 3r
r
.
(18)
We do not claim that this result is new, but as we have been unable to find it in the mathematics literature, we give a proof which draws heavily on ideas presented in Graham et al. (1994). Formula (18) is true if
B3/2 (z )n B3/2 (−z )n = B3 (z 2 )2n
Bt (z ) =
k=0
[z ]z B3 (z ) = r
2
2 3
3r 2
r 2
1 r 2
−1
+2
r 2
zk
+
r 2
2 3r 2
3r 2
tk + 1
k
3r 2
for r = 2, 4, 6, . . ., and is 0 otherwise. Then we find that [z 0 ]h(z ) = 0, and for r = 1, 3, 5, . . ., [z r ]h(z ) = 0. Finally, for r = 2, 4, 6, . . . ,
[ z r ] h( z ) =
where
∞ X tk + 1
and
−1
+2 1 r 2
3r −
2
+1 r
2 3r 2
+1
,
is what Graham et al. (1994) call the generalised binomial series, with
and the right-hand side simplifies to 0, which gives the result.
∞ X tk + n nz k . Bt (z ) = tk + n k k=0
References
n
(19)
Thus, to show (18), it suffices to show that
B3/2 (z )B3/2 (−z ) = B3 (z 2 )2 .
(20)
From the first formula of (5.59) of Graham et al. (1994) we know that
B3/2 (z ) − 1 = z B3/2 (z )3/2 so that
B3/2 (z )B3/2 (−z ) − B3/2 (z ) − B3/2 (−z ) + 1
3/2 = −z 2 B3/2 (z )B3/2 (−z ) . P∞ n If a series A(z ) = n=0 an z satisfies A(z ) − B3/2 (z ) − B3/2 (−z ) + 1 = −z 2 A(z )3/2 , then A(z ) = B3/2 (z )B3/2 (−z ). We now show that B3 (z 2 )2 is such a series and this establishes that formula (20) holds. Let h(z ) = B3 (z 2 )2 − B3/2 (z ) − B3/2 (−z ) + 1 + z 2 B3 (z 2 )3 and consider [z r ]h(z ) (i.e. the coefficient of z r in h(z )). Then using (19),
[z ]B 3 ( z ) = r
2 2
3r 2
+2 r 2
2 3r 2
+2
Borovkov, K.A., Dickson, D.C.M., 2008. On the ruin time distribution for a Sparre Andersen process with exponential claim sizes. Insurance: Mathematics & Economics 42, 1104–1108. Cheung, E.C.K., Dickson, D.C.M., Drekic, S., 2008. Moments of discounted dividends for a threshold strategy in the compound Poisson risk model. North American Actuarial Journal 12 (3), 299–318. Dickson, D.C.M., 2008. Some explicit solutions for the joint density of the time of ruin and the deficit at ruin. ASTIN Bulletin 38, 259–276. Dickson, D.C.M., Hipp, C., 2001. On the time to ruin for Erlang(2) risk processes. Insurance: Mathematics & Economics 29, 333–344. Dickson, D.C.M., Willmot, G.E., 2005. The density of the time to ruin in the classical Poisson risk model. ASTIN Bulletin 35, 45–60. Dickson, D.C.M., Hughes, B.D., Lianzeng, Z., 2005. The density of the time to ruin for a Sparre Andersen process with Erlang arrivals and exponential claims. Scandinavian Actuarial Journal 2005 (5), 358–376. Drekic, S., Willmot, G.E., 2003. On the density and moments of the time to ruin with exponential claims. ASTIN Bulletin 33, 11–21. Garcia, J.M.A., 2005. Explicit solutions for survival probabilities in the classical risk model. ASTIN Bulletin 35, 113–130. Gerber, H.U., Shiu, E.S.W., 1998. On the time value of ruin. North American Actuarial Journal 2 (1), 48–78. Graham, R.L., Knuth, D.E., Patashnik, O., 1994. Concrete Mathematics, 2nd ed.. Addison-Wesley, Upper Saddle River, NJ. Li, S., Garrido, J., 2004. On ruin for the Erlang(n) risk process. Insurance: Mathematics & Economics 34, 391–408. Mazza, C., Rullière, C., 2004. A link between wave governed random motions and ruin processes. Insurance: Mathematics & Economics 35, 205–222. Sun, L.-J., 2005. The expected discounted penalty at ruin in the Erlang(2) risk process. Statistics & Probability Letters 72, 205–217. Willmot, G.E., 2007. On the discounted penalty function in the renewal risk model with general interclaim times. Insurance: Mathematics & Economics 41, 17–31. Willmot, G.E., Woo, J.K., 2007. On the class of Erlang mixtures with risk theoretic applications. North American Actuarial Journal 11 (2), 99–115.