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

18 Pages·2003·0.27 MB·English

by  David Dickson

The density of the time to ruin for a Sparre
Andersen process with Erlang arrivals and
exponential claims
David C M Dickson, Barry D Hughes and Zhang Lianzeng
Abstract
We derive expressions for the density of the time to ruin given that
ruinoccursinaSparreAndersenmodelinwhichindividualclaimamounts
are exponentially distributed and inter-arrival times are distributed as
Erlang(n,β). We provide numerical illustrations of finite time ruin prob-
abilities, as well as illustrating features of the density functions.
1 Introduction
The Sparre Andersen surplus process U(t) is defined by
t 0
{ }≥
N(t)
U(t)=u+ct X ,
i
−
i=1
X
where u is the initial surplus, c is the rate of premium income per unit time,
X isasequenceofindependentandidenticallydistributed(i.i.d.) random
{ i}∞i=1
variables, where X represents the amount of the ith claim, and N(t)
i t 0
is a counting process with N(t) denoting the number of claims up{to tim}e≥t.
The sequence of i.i.d. random variables W represents the claim inter-
{ i}∞i=1
arrival times, with W being the time until the first claim, and we assume
1
that claim amounts are independent of claim inter-arrival times. Further, we
assume that cE(W ) > E(X ), and define the premium loading factor (cid:39) by
1 1
cE(W )=(1+(cid:39))E(X ).
1 1
We denote by p the density function of X , and throughout this paper we
1
let p(x) = αexp αx , x > 0. We also assume that W has an Erlang(n,β)
1
{− }
distribution, so that
n−1(βt)j
Pr(W t)=1 e βt
1 ≤ − − j!
j=0
X
with density function βntn 1e βt/(n 1)!, and hence the premium loading
− −
−
factor becomes (cid:39)=cαn/β 1.
−
Let T denote the time of ruin, and be given by
inf t: U(t)<0
T = { }
if U(t) 0 for all t>0
½ ∞ ≥
and let T =T T < . The probability of ultimate ruin from initial surplus u
c
| ∞
is defined as
ψ(u)=Pr[T < U(0)=u].
∞|
Similarly, the probability of ruin by time t, t>0, is defined as
ψ(u,t)=Pr[T tU(0)=u].
≤ |
We define the function φ by
φ(u)=E e δTI(T < )U(0)=u
−
∞ |
where δ is a non-negative para£meter and I is the indi¤cator function. From
Gerber and Shiu (1998) we have
∂
φ(u)= ∞e δt ψ(u,t)dt.
−
∂t
Z0
Inthecaseoftheclassicalriskmodel,i.e. whenW hasanexponentialdistri-
1
bution, it is well known that an explicit solution for ψ(u,t) exists when X has
1
an exponential distribution. See Seal (1978) and references therein. More re-
cently, attention has focussed on finding the density of T by using φ as a
c
Laplace transform and inverting that transform. Garcia (2002) uses the com-
plex inversion formula to produce explicit solutions for ψ(u,t) when X has an
1
exponential distribution and when it has an Erlang(2) distribution. Drekic and
Willmot(2003)derivethedensityofT inthecasewhenX hasanexponential
c 1
distribution by direct inversion of φ.
Westartwithsomepreliminaryresultsinthenextsection. Then,inSection
3, we apply the ideas contained in Garcia (2002) to derive the density of T
c
whenX hasanexponentialdistributionand W is distributedas Erlang(n,β).
1 1
In particular, we derive a more readily computable form of the density of T
c
than the Drekic—Willmot solution when n = 1. Garcia’s basic approach, while
producing correct answers, involves unjustified steps which are here made com-
pletely rigorous. Finally, the results of Section 3 are illustrated in Section 4.
2 Preliminaries
We start by establishing Lundberg’s fundamental equation for our model. The
arguments that follow are essentially those of Gerber and Shiu (1998), who
consider the classical risk model, but modified to our circumstances. Define
m
V(m)=u+ (cW X )
i i
−
i=1
X
for m=1,2,3,... with V(0)=u, so that V(m) gives the surplus at the time of
the mth claim, and let
m
τ(m)= W
i
i=1
X
denote the time of the mth claim, with τ(0)=0. Then by standard arguments
the process
{exp{−δτ(m)+ξV(m)}}∞m=0
2
is a martingale if
E[exp (δ cξ)W ]E[exp ξX ]=1,
1 1
{− − } {− }
which becomes
β n α
=1 (2.1)
β+δ cξ α+ξ
µ − ¶
forourchoiceofdistributionsforW andX . Further,thereisauniquenegative
1 1
value ξ = R where R > 0 which satisfies equation (2.1), which is Lundberg’s
−
fundamental equation for our model.
By applying the optional stopping theorem we find that
R
φ(u)= 1 e Ru
−
− α
µ ¶
a result which is derived by this argument by Gerber and Shiu (1998) when
the Erlang parameter n is 1, and by a different argument by Dickson and Hipp
(2001) when the Erlang parameter n is 2.
When δ =0, we get the well known result
R
ψ(u)= 1 0 e−R0u
− α
µ ¶
where R is the adjustment coefficient. See, for example, Grandell (1991).
0
3 Main Results
Applying results from the previous two sections to our model we have
∂ R
∞e δt ψ(u,t)dt= 1 e Ru, (3.1)
− −
∂t − α
Z0 µ ¶
where
β n α
=1. (3.2)
β+δ+cR α R
µ ¶ −
As R is a function of δ, the right hand side of equation (3.1) is a function of δ
and hence we can find (∂/∂t)ψ(u,t) if we can invert the Laplace transform.
DrekicandWillmot(2002)havefoundtheinverseLaplacetransformforthe
case n = 1 by first expanding the right-hand side of equation (3.1) in a series
in powers of 1 R/α and then inverting the Laplace transform term-by-term.
−
Briefly, if one notes that for n = 1 the unique solution of equation (3.2) with
0<R<α corresponds to
R β+δ+αc (β+δ+αc)2 4αβc
1 = − − ,
− α 2αc
p
each term in the expansion can be inverted in terms of the modified Bessel
function I (Abramowitz and Stegun 1965, Chapter 9) and one finds from this
ν
that1
∂ ψ(u,t)= e−αu−(αc+β)t ∞ (n+1)[u(αβ/c)1/2]n+1I ((4αβc)1/2t).
∂t αut n! n+1
n=0
X
1ToconverttothenotationofDrekicandWillmot,setαu=µx/(1+θ),β=λ(1+θ)and
αc=λ.
3
This result, though readily computable, is not especially convenient and the
technique does not generalise to n>1, where the solution for 1 R/α is either
−
too unpleasant to permit term-by-term inversion (n = 2, 3, 4) or unable to be
found analytically (n 5).
≥
Garcia (2002) has noted that the Laplace transform encountered in the
present context may also be evaluated using the complex inversion formula for
n = 1, and we shall follow this approach here for arbitrary positive integer n.
Ascertainsubtletiesareencounteredwhenusingcomplexvariablemethods,we
shallfirstbrieflysummarisetheimportantrelevantaspectsofthecomplexvari-
able theory of the Laplace transform (Sneddon 1972, Chapter 3; Henrici 1977,
Chapter 10).
Let f(t) be a function defined for t>0 with at most finitely many points of
discontinuity. If the Laplace transform integral
∞e stf(t)dt=f(s)
−
Z0
exists for s = s , then the integral exists alsobat least for all complex s with
0
Re s > Re s , and the resulting function f(s) is holomorphic in the half-
0
{ } { }
plane in which Re s >Re s . For any fixed real number κ with κ>Re s ,
0 0
{ } { } { }
the function f(t) can be recovered from f(s) bby a contour integral taken along
the ‘Bromwich contour’, which is the line Re s =κ, traversed in the direction
{ }
of increasing Im s : b
{ }
1 κ+i
f(t)= ∞estf(s)ds (t>0).
2πi
Zκ−i∞
b
There are mild regularity conditions on f(t) for the inversion formula to hold
as stated, and at points of discontinuity of f, the contour integral returns
[lim f(τ)+lim f(τ)]/2 rather than f(t). Also, if the Bromwich contour
τ t τ t
integ↓ral fails to co↑nverge in the classical sense, it is to be interpreted as the
limit as L of the contour integral from κ iL to κ+iL.
→∞ −
In most cases where Laplace transforms are used, the Laplace transform
f(s) has an analytic continuation to the rest of the complex plane, apart from
certain singular points (singularities), and subject to modest requirements on
tbhe decay of f(s) as s , one may complete the integration contour and
| | | | → ∞
then usually either
(a) evaluate thbe integral using the residue theorem, giving
f(t)= Residue estf(s) ,
{ }
singularities
X
b
this is the case that applies if all singularities of the analytic continuation of
f(s) are isolated (that is, poles or isolated essential singularities); or
(b) convert the contour integral to a more tractable equivalent real integral —
tbhis is the case when the analytic continuation of f(s) has one or two branch
points.
We shall scale the problem by writing b
∂ δ cα
ψ(u,t)=βf(βt), s= , T =βt and (cid:18)= .
∂t β β
4
We shall also set
1 R/α=zn,
−
where from the context of the original problem we have 0<R<α, so that we
have z real, with z (0,1) when s is a positive real number. We now have
∈
f(s)=znexp( αu+αuzn),
−
where s is to be determined from
b
z[1+s+(cid:18)(1 zn)]=1. (3.3)
−
We remark that if we define
P (z)=(cid:18)zn+1 (1+s+(cid:18))z+1,
s
−
the relation between s and z is equivalent to P (z) = 0. For real, positive s
s
and real z, we may observe that P (z) is real, P (0) = 1, P (1) = s < 0,
s s s
−
and P (z) as z , so that P (z) = 0 has real solutions in each of the
s s
→ ∞ → ∞
intervals (0,1) and (1, ). Moreover, P (0)<0 and P (z)>0 for z >0. This
∞ s0 s00
ensures that for positive real s, equation (3.3) has precisely one real solution
in the interval (0,1), and precisely one real solution in the interval (1, ). All
∞
other solutions are either complex, or are real negative numbers. The Mapping
Theorem discussed by Marsden and Hoffman (1999, pp. 399—400) ensures that
for each real s (0, ), there exists ∆(s ) > 0 such that there is a unique
0 0
∈ ∞
solution z = Z(s) of P (z) = 0 in s s < ∆(s ). The initial analytic
s 0 0
| − |
continuation of f(s) beyond the positive real axis is therefore unambiguously
defined using the root z = Z(s) of equation (3.3) that converges as s s to
0
→
the real root z =bZ(s0)∈(0,1) of Ps0(z)=0.
The case n = 1: s-plane analysis
Consider as an illustration the case n = 1, for which we have as the solutions
for real positive s of P (z)=0 the choices
s
1+s+(cid:18) (1+s+(cid:18))2 4(cid:18)
z = ± − .
2(cid:18)
p
It is easily shown that the negative sign is to be selected to obtain the root in
(0,1), and so we have
1
Z(s)= 1+s+(cid:18) (1+s+(cid:18) 2(cid:18)1/2)1/2(1+s+(cid:18)+2(cid:18)1/2)1/2 . (3.4)
2(cid:18) − −
n o
To obtain a valid analytic continuation of Z(s) we assign each of the square
roots their principal values. Then if s=σ+iτ we have
iθ
(1+s+(cid:18) 2(cid:18)1/2)1/2 =[(1+σ+(cid:18) 2(cid:18)1/2)2+τ2]1/4exp ± ,
± ± 2
µ ¶
where θ ( π,π] with
± ∈ −
τ
sinθ = ,
± [(1+σ+(cid:18) 2(cid:18)1/2)2+τ2]1/2
±
1+σ+(cid:18) 2(cid:18)1/2
cosθ = ± .
± [(1+σ+(cid:18) 2(cid:18)1/2)2+τ2]1/2
±
5
Im s Im s
(a) (b)
cut Re s Re s
Figure3.1: Completionandcontractioninthes-planeoftheBromwichcontour
Re s =κ. (a)ThefiniteBromwichcontourwith L Im s Liscompleted
{ } − ≤ { }≤
with a circular arc that encloses completely the branch cut. (b) The resulting
closed contour is contracted to circles of radius η centred on the branch points,
andstraightlinesalongthetopandbottomedgesofthecut. Thecontributions
from the three circular arcs vanish in the limits η 0 and L .
→ →∞
We need to evaluate
1 κ+i
f(T)= ∞esTZ(s)exp( αu+αuZ(s))ds (T >0).
2πi −
Zκ−i∞
The integrand has branch point singularities at s= a and s= b, where
− −
a=1+(cid:18) 2(cid:18)1/2 =(1 (cid:18)1/2)2, b=1+(cid:18)+2(cid:18)1/2 =(1+(cid:18)1/2)2.
− −
If we now consider the limiting behaviour of Z(σ+iτ) as τ 0 we find that
→
Z(σ+iτ) is continuous across τ =0 except for b σ a. We therefore cut
− ≤ ≤−
thes-planealongthestraightlinejoiningaandb,andnotethatfor b<σ < a
− −
1
Z(σ+iτ) Z (σ)= 1+(cid:18)+σ ib+σ 1/2 a+σ 1/2 as τ 0 . (3.5)
± ±
→ 2(cid:18){ ∓ | | | | |} →
We complete the contour in the standard manner with a circular arc in the
left half-plane and contract the contour down to a ‘dumb-bell’ shaped contour
consistingoftheupperandlowersurfacesofthebranchcutbetweens= b+η
−
ands= a η andcirclesofradiusηcentredons= aands= b(seeFigure
− − − −
3.1). In the limit η 0, the contributions from the circular arcs vanish (the
→
integrand is bounded, and the contour length vanishes in the limit). Making
careful use of equation (3.5) and writing x= σ, we find that
−
1 b
f(T)= e−xT−αu Z−( x)eαuZ−(−x) Z+( x)eαuZ+(−x) dx.
2πi { − − − }
Za
6
If we make the change of variable x=a+(b a)ξ =1+(cid:18) 2(cid:18)1/2+4(cid:18)1/2ξ, we
− −
find that
Z ( x)=(cid:18) 1/2 1 2ξ 2i[ξ(1 ξ)]1/2 ,
± −
− { − ∓ − }
giving
f(T) = 4eαu((cid:18)−1/2−1)−(1+(cid:18)−2(cid:18)1/2)T 1e−4(cid:18)1/2Tξ−2αu(cid:18)−1/2ξ
π
Z0
2αu
(1 2ξ)sin [ξ(1 ξ)]1/2
− (cid:18)1/2 −
· ½ ¾
2αu
+2[ξ(1 ξ)]1/2cos [ξ(1 ξ)]1/2 dξ.
− (cid:18)1/2 −
½ ¾¸
Re-expressing this result in the original variables, we find that
∂ 4βeu(αβ/c)1/2 αu [β+αc 2(αβc)1/2]t 1
ψ(u,t) = − − − e 4(αβc)1/2[t+u/(2c)]ξ
−
∂t π
Z0
2u(αβ)1/2
(1 2ξ)sin [ξ(1 ξ)]1/2
− c1/2 −
· ½ ¾
2u(αβ)1/2
+2[ξ(1 ξ)]1/2cos [ξ(1 ξ)]1/2 dξ.
− c1/2 −
½ ¾¸
(3.6)
WecanuseWatson’sLemma(Olver1974,Chapter3;Henrici1977,Chapter11)
to deduce the asymptotic expansion for this integral as 4(αβc)1/2[t+u/(2c)]
∞. Watson’s Lemma states that if φ(ξ) ∼ ∞n=0φnξλn−1 as ξ → 0 (wi→th
0<λ <λ <λ < ), then as Z ,
0 1 2
··· →∞ P
∞e Zξφ(ξ)dξ ∞ Γ(λm)φm,
Z0 − ∼m=0 Zλm
X
where Γ(λ)= 0∞e−ξξλ−1dξ is the usual gamma function. The expansion is in
general a divergent asymptotic expansion, rather than a convergent series.
R
From the first term of the Watson’s Lemma expansion, on recalling that
Γ(3/2)=π1/2/2 we deduce that if 4(αβc)1/2[t+u/(2c)] 1,
À
∂ 4β[1+u(αβ/c)1/2]eu(αβ/c)1/2 αu [β+αc 2(αβc)1/2]t
ψ(u,t) − − − . (3.7)
∂t ≈ π1/2 4(αβc)1/2[t+u/(2c)] 3/2
{ }
This result has the status of a long-time limit, but for a given large t, the
differencebetweentheapproximationandthetruevaluecanbesensitivetothe
values of parameters α, β, c and u. Further terms in the expansion are easily
derived, but for parameter regimes where the first term alone is inadequate,
an alternative expression discussed below is readily computable to very high
precision. We remark that the change of variable ξ = sin2θ (0 θ π/2)
≤ ≤
enables the solution (3.6) to be rewritten as
∂ 4βeu(αβ/c)1/2 αu [β+αc 2(αβc)1/2]t π/2
ψ(u,t) = − − − e 4(αβc)1/2[t+u/(2c)]sin2θ
−
∂t π
Z0
u(αβ)1/2
sin 2sinθcosθ+2θ 2sinθcosθdθ.
× c1/2
½ ¾
7
As the integrand is even, the integration interval can be extended to ( π,π)
−
and a subsequent change of variable φ=2θ gives
∂ βe αu (β+αc)t π
ψ(u,t) = − − e2(αβc)1/2[t+u/(2c)]cosφ
∂t π
Z−π
u(αβ)1/2
sin sinφ+φ sinφdφ. (3.8)
× c1/2
½ ¾
The case n = 1: z-plane analysis
Garcia (2002) avoids the issue of determining Z(s) by making a change of vari-
able in the contour integral, producing an integral over a new complex variable
z. In our notation the change of variable is
s=z 1+(cid:18)z 1 (cid:18).
−
− −
However,asthechangeofvariablealsoinvolvesachangeofintegrationcontour,
and effectively runs the contour through an isolated essential singularity, the
argumentrequiresmorecare. Asbefore,wewrites=σ+iτ. OntheBromwich
contour, we have Re s =σ =κ. Observe that as τ for fixed σ = κ we
{ } | | →∞
have
sinθ sgn(τ), cosθ =O(τ 1),
−
± → ±
so that θ sgn(τ)π/2. In more detail,
± →
(1+κ+(cid:18) 2(cid:18)1/2)2
sinθ± =sgn(τ) 1− 2τ±2 +O(τ−4) ,
· ¸
and so on setting θ =sgn(τ)[π/2 η] and examining the small-η behaviour, we
−
deduce that
π 1+κ+(cid:18) (cid:18)1/2
θ =sgn(τ) ± +O(τ−2) ,
± 2 − τ
· | | ¸
giving
1 π 1+κ+(cid:18)
(θ +θ )=sgn(τ) +O(τ 2) .
2 − + 2 − τ −
· | | ¸
Also,
[(1+κ+(cid:18) 2(cid:18)1/2)2+τ2]1/4 = τ 1/2[1+(1+κ+(cid:18) 2(cid:18)1/2)2/τ2]1/4
± | | ±
= τ 1/2[1+O(τ 2)]
−
| |
anditfollows,afteralittlealgebrainwhichmanycancellationsoccurinequation
(3.4), that
i 1+κ+(cid:18)
Z(κ+iτ)= + +O(τ 3) as τ .
−τ τ2 − →∞
The image in the z-plane of the Bromwich contour is illustrated in Figure 3.2.
Interpretedasaclosedcontour,itistraversedclockwiseandpassesthroughthe
point z =0. However z =0 is an isolated essential singularity of
eδt =esT =e(z−1+(cid:18)z 1 (cid:18))T.
− −
8
0.2
Im z τ = −1
0.1
τ = −10
0
0.1 0.2 0.3 0.4 Re z
τ = 10
−0.1
τ = 1
−0.2
Figure 3.2: The image in the z-plane of the Bromwich contour Re s = κ, for
{ }
the case n = 1, κ = 1, (cid:18) = 1. The discs represent points corresponding to
integer values of τ =Im s with 10 τ 10.
{ } − ≤ ≤
Garcia (2002) has not addressed this issue, and has in effect evaluated the
contour integral in the z-plane by assuming that the contour encircles z = 0
rather than passing through it. Fortunately, Garcia’s analysis can be set on a
firm foundation by the following construction.
Let L > 0 be given and consider L L (where ultimately we shall let
0 0
≥
L ). Then the image of the segment L Im s L of the Bromwich
→ ∞ − ≤ { } ≤
contour Re s =κ in the s-plane corresponds in the z-plane to a contour with
{ }
Re z χ(L), where
{ }≥
Z(κ iL) = χ(L) iλ(L),
± ∓
1+κ+(cid:18)
χ(L) = +O(L 3),
L2 −
1
λ(L) = +O(L 3).
−
L
We consider
1 1
IL = 2πi e(z−1+(cid:18)z−1−(cid:18))Tze−αu+αuz −z2 +(cid:18) dz,
ZC · ¸
where the contour C, traversed clockwise, consists of four smooth arcs (see
Figure 3.3):
C : theimageinthez-planeofthesegment L Im s LoftheBromwich
1
− ≤ { }≤
contour Re s =κ;
{ }
9
C : the straight line joining z =χ(L)+iλ(L) to z =iλ(L);
2
C : the semicircular arc to the left of the origin joining z = iλ(L) to z =
3
iλ(L);
−
C : the straight line joining z = iλ(L) to z =χ(L) iλ(L).
4
− −
Fromtheresiduetheorem, rememberingthat thecontouristraversed clock-
wise, we have
1
IL = −Residue e(z−1+(cid:18)z−1−(cid:18))Tze−αu+αuz −z2 +(cid:18)
z =0 ½ · ¸¾
1
= e−αu−(1+(cid:18))T Residue ez−1Te((cid:18)T+αu)z (cid:18)z
z −
z =0 ½ · ¸¾
∞ Tr((cid:18)T +αu)r ∞ Tr+2((cid:18)T +αu)r
= e αu (1+(cid:18))T (cid:18)
− − ( (r!)2 − r!(r+2)! )
r=0 r=0
X X
(cid:18)T2
= e αu (1+(cid:18))T F (1;T((cid:18)T +αu)) F (3;T((cid:18)T +αu)) ,
− − 0 1 − 2 0 1
½ ¾
where
∞ Γ(C)Zm
F (C;Z)=
0 1 Γ(C+m)m!
m=0
X
is a special caseof thehypergeometricseries. Notethat these series can also be
recognised as modified Bessel functions, since
Z ν ∞ (Z2/4)m 1 Z ν
I (Z)= = F (ν+1;Z2/4)
ν 2 m!Γ(m+ν+1) Γ(ν+1) 2 0 1
µ ¶ m=0 µ ¶
X
and so
(cid:18)T
=e αu (1+(cid:18))T I 4T((cid:18)T +αu) I 4T((cid:18)T +αu) .
IL − − 0 − (cid:18)T +αu 2
½ ³p ´ ³p ´¾
We now prove that the contributions from the contour integrals along the arcs
C , C and C all vanish in the limit L .
2 3 4
→∞
OnthesemicirculararcC ,wemaywritez =λ(L)eiφ withπ/2 φ 3π/2.
3
≤ ≤
Then
e(z−1+(cid:18)z−1−(cid:18))T =e[λ(L)−1cosφ+(cid:18)λ(L)cosφ−1−(cid:18)]T.
| |
All terms in the argument of the exponential on the right are non-positive and
except near the imaginary axis, the exponential vanishes rapidly as L .
→ ∞
With the constant K depending on α, u, (cid:18), κ, T and L , but independent of L,
0
we have
1 3π/2
e(z−1+(cid:18)z−1−(cid:18))Tze−αu+(cid:18)αuz −z2 +(cid:18) dz ≤K eTλ(L)−1cosφdφ.
¯ZC3 · ¸ ¯ Zπ/2
¯ ¯
¯ ¯
¯ ¯
10