Algorithmic Analysis of the Sparre Andersen Model in Discrete Time

25 Pages·2007·0.5 MB·English

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ALGORITHMIC ANALYSIS OF THE SPARRE ANDERSEN MODEL
IN DISCRETE TIME
BY
ATTAHIRU SULE ALFA AND STEVE DREKIC
ABSTRACT
In this paper,we show that the delayed Sparre Andersen insurance risk model
in discrete time can be analyzed as a doubly infinite Markov chain. We then
describe how matrix analytic methods can be used to establish a computa-
tional procedure for calculating the probability distributions associated with
fundamental ruin-related quantities of interest, such as the time of ruin, the
surplus immediately prior to ruin, and the deficit at ruin. Special cases of
the model, namely the ordinary and stationary Sparre Andersen models, are
considered in several numerical examples.
KEYWORDS
Sparre Andersen model, discrete time, matrix analytic methods, time of ruin,
surplus immediately prior to ruin, deficit at ruin.
1. INTRODUCTION
In this paper, we consider the delayed Sparre Andersen insurance risk model
in discrete time. In particular, we assume that the number of claims process
{N : t=0,1,2,…} is a modified discrete-time renewal process with indepen-
t
dent positive interclaim times {W,W,W,…},where W is the duration from
1 2 3 1
time 0 until the first claim occurs and W,i=2,3,4,…,is the time between the
i
(i–1)-th and i-th claims.For i=2,3,4,…,let W have probability mass function
i
(pmf) a =Pr{W =j}, j=1,2,3,…,n ,and corresponding survival function
j i a
A =Pr{W >j}=1- !j a , where we adopt the usual convention that
j i k=1 k
!0 f^kh = 0for an arbitrary function f. We assume in this paper that n <3
k=1 a
(i.e., the interclaim time distribution of W, i=2,3,4,…, has finite support).
i
In the ordinarySparre Andersen model,it is assumed that a claim has occurred
at time 0,so that W has the same distribution as the ordinary interclaim times
1
{W,W,W,…}.If W is not a “full”interclaim time,however,it is well known
2 3 4 1
from standard renewal theory (e.g.,see Karlin and Taylor (1975,pp.192-193))
that,asymptotically in time,the limiting distribution of this forward recurrence
Astin Bulletin37(2),293-317.doi:10.2143/AST.37.2.2024068 © 2007 by Astin Bulletin.All rights reserved.
0345-07_Astin37/2_05 28-11-2007 14:56 Pagina 294
294 A.S.ALFA AND S.DREKIC
n
time is defined by the pmf a =A /! a A , j=1,2,3,…,n . As a result,
j j–1 k=1 k-1 a
in the stationary Sparre Andersen model, W has pmf a rather than a. Since
1 j j
the motivation for the use of a is asymptotic in nature, it is not evident how
j
appropriate this assumption is for finite time.To accommodate other possible
alternatives,we assume more generally that W has pmf r =Pr{W =j}, j=1,
1 j 1
2,3,…,n , where n <3. By appropriate choice of r, it is obvious that both
r r j
the ordinary and stationary Sparre Andersen models are special cases of this
more general risk model.
We further assume that the individual claim amounts {Y,Y,Y ,…} form
1 2 3
an independent and identically distributed (iid) sequence of positive random
variables with common pmf a, j=1,2,3,…,m , and corresponding survival
j a
function L =1- !j a .We remark that unlike the interclaim time distribu-
j k=1 k
tions defined above, the claim amount distribution can be of finite or infinite
support (i.e.,ma≤3).Premiums are collected at the rate of c!(cid:2)+per unit time.
Beginning with an initial reserve of u!{0,1,2,…}, the insurer’s surplus at
time t is given by U =u+ct– !Nt Y, t=0,1,2,…. At any given time point,
t i=1 i
we adopt the usual convention that premiums are collected first before any
claims are paid.Let T=min{t!(cid:2)+:U <0} be defined as the time of ruin with
t
T=3 if U ≥ 0 ∀t ! (cid:2)+. If ruin does occur, we also define |U | as the deficit
t T
at ruin and U =U +c as the surplus immediately prior to ruin. Clearly,
T– T–1
T=3if ma ≤ c.If ma>c,however,then |UT| ! {1,2,3,…,ma–c} and UT– !
{c,c+1,c+2,…,m –1}.
a
It is of considerable interest to risk practitioners to be able to calculate the
joint probability distribution (as well as associated marginal distributions) of
T, U , and |U |. Surprisingly, however, there appear to be few results in the
T– T
literature for computing the joint probability distribution of these fundamental
ruin-related quantities of interest in the delayed Sparre Andersen model described
above. Most of the results in the literature tend to be specialized and mainly
concern the computation of ruin probabilities over finite- and infinite-time
horizons for the well-known compound binomialmodel — a discrete analogue
of the classical compound Poisson risk model in which interclaim times are geo-
metrically distributed and the ordinary and stationary variants are identical
(e.g.,see Gerber (1988),Michel (1989),Shiu (1989),Dickson and Waters (1991),
Willmot (1993),Dickson (1994),De Vylder and Marceau (1996),Cheng et al.
(2000),Cardoso and Egidio dos Reis (2002),and Li and Garrido (2002)).For
the compound binomial model,however,Dickson et al.(1995) developed recur-
sive numerical procedures for calculating the joint and marginal probability
distributions of the surplus immediately prior to ruin and the deficit at ruin.
While recent papers by Pavlova and Willmot (2004), Li (2005a, 2005b), and
Cossette et al.(2006) have analyzed variants of the discrete-time Sparre Ander-
sen model described above,the emphasis in these papers has been primarily the-
oretical, focussing on bounds on ruin probabilities as well as distributional
properties and mathematical connections to other related renewal risk models
of interest via the methodology of the Gerber-Shiu discounted penalty func-
tion (e.g., see Gerber and Shiu (1998)).
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ALGORITHMIC ANALYSIS OF THE SPARRE ANDERSEN MODEL 295
The thrust of this paper is more computational in flavour. To begin with,
we will show how the delayed Sparre Andersen model described above can be
set up as a doubly infinite Markov chain with finite blocks. Using this set-up,
we will then show how matrix analytic methods can be employed as a means
of computing the bivariate joint probabilities
wn,i(u)=Pr{T=n,UT–=i |U0=u}, n ! (cid:2)+; i=c,c+1,c+2,…,ma–1
and
ƒn,j(u) = Pr{T=n, |UT|=j |U0=u}, n ! (cid:2)+; j=1,2,3,…,ma–c,
as well as the trivariate joint probability
c (u) = Pr{T=n,U =i, |U |=j |U =u}, n ! (cid:2)+;
n,i,j T– T 0
i=c,c+1,c+2,…,m –1;
a
j=1,2,3,…,m –i.
a
Finally, we will demonstrate how these discrete-time probabilities can be
employed to approximate analogous quantities in the continuous-time version
of this model, where considerably more attention has been devoted in recent
years to characterizing the probability distributions of the time of ruin,the sur-
plus immediately prior to ruin, and the deficit at ruin.
2. FORMULATION OF THE MODEL
We begin by considering the interclaim time distribution defined by the pmf a.
j
Letting W denote an arbitrary W, i=2,3,4,…, we introduce t =Pr{W>j|
i j
W>j–1}=A /A , j=1,2,3,…,n . We immediately note that t =A and
j j–1 a 1 1
t =0. If we now define the n ≈ n probability transition matrix
na a a
J0 t 0 g 0 N
K 1 O
K0 0 t1 j 0 O
S = K h h j j h O, (2.1)
K O
KK0 0 g 0 tna-1OO
0 0 g 0 0
L P
and the 1≈n row vector e =(1,0,0,…,0),Alfa and Neuts (1995) have shown
a 1
that the random variable Wcan be modelled as a discrete phase-type random
variable with representation (e ,S) of order n .Alfa and Neuts (1995) refer to
1 a
this as the elapsed time representation of the discrete random variable W. We
remark that one may interpret {1,2,3,…,n } as the transient states of the
a
underlying Markov chain having probability transition matrix S among these
states,whereas state n +1 is the absorbingstate (i.e.,the state which marks the
a
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296 A.S.ALFA AND S.DREKIC
end of an interclaim time and the subsequent occurrence of a claim).If we then
define the n ≈1 column vector of absorption probabilities
a
J 1- t N
K 1 O
1- t
K 2 O
s = K h O,
K1- t O
KK na-1OO
1
L P
it can be shown that
a = e Sj–1s, j =1,2,3,…,n .
j 1 a
Further details concerning this result can be found in Alfa and Neuts (1995)
as well as in Alfa (2004).
Consider now the delayed Sparre Andersen model described in Section 1.
Suppose we fix the value of W to be k,where k!{1,2,3,…,n}.Assuming that
1 r
W =k, define L as the “elapsed time”(in the sense described above) at time
1 t
t,t=k, k+1, k+2,…,since the occurrence of the last claim.For t=k, k+1,
k+2,…,consider the bivariate stochastic process (U,L ) which possesses the
t t
following Markovian relationship, namely
⎧
(U +c,L +1) if there is no claim at time t+1
(U ,L ) = ⎨ t t (2.2)
t+1 t+1 ⎩ (U +c–Y,1) if there is a claim Y at time t+1.
t
Figure 1 depicts a simple transition scheme of the process assuming that u=6,
c=1, k=3, and U =5. An important observation is that L =1, since we are
3 k
assuming that a claim has occurred at time k. Let D ={(U,L ) :U ! (cid:2); L =
t t t t
1,2,3,…,n } denote the state space for this Markov chain. We refer to the U
a t
component as the level of the process and the L component as the phase of
t
the process. Since the delayed process reverts to the ordinary process (having
interclaim time distribution defined by the pmf a ) upon occurrence of the first
j
claim, the probability transition matrix P associated with this Markov chain
for t=k, k+1, k+2,… is given as
f –3 –2 –1 0 1 2 f c c+1 c+2 c+3 f
h ⎧ j j j j ⎫
⎪ ⎪
–3 j B B B B
⎪ c c–1 c–2 c–3 ⎪
–2 j B B B B B
⎪ c+1 c c–1 c–2 c–3 ⎪
–1 j B B B B B B
⎪ c+2 c+1 c c–1 c–2 c–3 ⎪
P = 0 j B B B B B B f B ,
⎪ c+3 c+2 c+1 c c–1 c–2 0 ⎪
1 j B B B B B B f B B
⎪ c+4 c+3 c+2 c+1 c c–1 1 0 ⎪
2 j B B B B B B f B B B
⎪ c+5 c+4 c+3 c+2 c+1 c 2 1 0 ⎪
3 j B B B B B B f B B B B
c+6 c+5 c+4 c+3 c+2 c+1 3 2 1 0
h ⎩ j j j j j j j j j j j j j ⎭
0345-07_Astin37/2_05 28-11-2007 14:56 Pagina 297
ALGORITHMIC ANALYSIS OF THE SPARRE ANDERSEN MODEL 297
ess.
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0345-07_Astin37/2_05 28-11-2007 14:56 Pagina 298
298 A.S.ALFA AND S.DREKIC
where
⎧
S if i=0,
B = ⎨ (2.3)
i ⎩ (se )a if i=1,2,3,….
1 i
For c=1, we note that this is a doubly infinite, right skip-free Markov chain,
with finite blocks of size n . Moreover, since a =0 ∀ i >m , B becomes an
a i a i
n ≈ n matrix of zeros if i >m .
a a a
We now partition the state space D into two state spaces, namely
D = {(i,j) : i=0,1,2, …; j =1,2,…,n }
1 a
and
D = {(i,j) : i = –1,–2,–3,…; j =1,2,…,n }.
2 a
Note that D comprises the “non-ruined”states of the system whereas D com-
1 2
prises the “ruined” states of the system, so that D=D (cid:2) D . Moreover, we
1 2
define two matrices Cand Dsuch that C:D " D and D:D " D .Therefore,
1 1 1 2
the matrix Chas block elements C containing transition probabilities which
v,w
map U =v ! {0,1,2,…} into U =w ! {0,1,2,…}. Clearly, the non-zero
t t+1
block elements are given as C =B ,w=0,1,…, v+c.In a similar fash-
v,w v–w+c
ion,the matrix Dhas block elements D =B ,which contain the transi-
v,w v–w+c
tion probabilities mapping U =v!{0,1,2,…} into U =w!{–1,–2,–3,…}.
t t+1
As a result, the two matrices C and D look as follows:
0 1 2 f c c+1 c+2 f
0 ⎧ B B B f B ⎫
⎪ c c–1 c–2 0 ⎪
1 ⎪ B B B f B B ⎪
C = ⎪ c+1 c c–1 1 0 ⎪ (2.4)
2 B B B f B B B
c+2 c+1 c 2 1 0
h ⎩ h j j j j j j j ⎭
and
–1 –2 –3 f
0 ⎧ B B B f ⎫
⎪ c+1 c+2 c+3 ⎪
1 ⎪ B B B f⎪.
D = ⎪ c+2 c+3 c+4 ⎪ (2.5)
2 B B B f
c+3 c+4 c+5
h ⎩ h h h ⎭
We remark that the matrix Cis the lower right quadrant of Pand the matrix D
is the lower left quadrant of P horizontally reversed.
Assuming that W =k, it readily follows that
1
Pr{U =v|W =k,U =u}=a , v=u+ck–1, u+ck–2, u+ck–3,….
k 1 0 u+ck–v
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ALGORITHMIC ANALYSIS OF THE SPARRE ANDERSEN MODEL 299
Therefore,the initial(i.e.,starting at time k) probability row vector correspon-
ding to the states in D is given by
1
b(k) = (a e ,a e ,a e ,…, a e ,a e ,0,0,…), (2.6)
u+ck 1 u+ck–1 1 u+ck–2 1 2 1 1 1
where 0denotes the 1≈n row vector of zeros.Note that the i-th level of b(k)is
a
given by the 1≈n row vector a e for each i!W ={0,1,2,…,u+ck–1}.
a u+ck–i 1 k
Moreover, b(k) contains zeros from level u+ck onward.
We now define two additional row vectors, namely
g(k) = (g(k), g(k), g(k),…) = b(k)Cn, n = 0,1,2,… (2.7)
n n,0 n,1 n,2
and
h(k) = (h(k) , h(k) , h(k) ,…) = g(k) D = b(k)Cn–1D, n = 1,2,3,…. (2.8)
n n,–1 n,–2 n,–3 n–1
Note that g(k) contains the probabilities of being in the various “non-ruined”
n
states at time k+nwithout having visited a “ruined”state during the previous
n–1 transitions, given that U ! W according to the probability vector b(k).
k k
In a similar fashion, h(k) contains the probabilities of being in the various
n
“ruined”states for the first timeat time k+n,given that U ! W according to
k k
the probability vector b(k).Since ruin can only occur at claim instants (thereby
resulting in the phase component being reset to 1 with probability 1 at such
instants), it immediately follows that the structure of the 1 ≈ n row vector
a
h(k) is simply given by h(k) =(ƒ(k)(u),0,0,…,0),where ƒ(k)(u)=Pr{T=k+n,
n,–j n,–j n,j n,j
|U |= j|U ! W }. Hence, we simply obtain
T k k
ƒ(k)(u) = h(k) e(cid:2), (2.9)
n,j n,–j 1
where e(cid:2) denotes the transpose of e .
1 1
We can also apply similar probabilistic reasoning to obtain a representation
for c(k) (u)=Pr{T=k+n, U =i,|U |=j |U ! W }. In order for ruin to
n,i,j T– T k k
occur at time k+n with a surplus prior to ruin equal to i, we observe that:
(i) none of the previous n–1 transitions must have included a visit to any state
in D ,and (ii) the surplus at time k+n–1 must be equal to i–c.The quantity
2
corresponding to points (i) and (ii) is the 1 ≈ n row vector g(k) . At the
a n–1,i–c
next time unit (i.e., time k+n), a claim must necessarily occur but not before
a premium of c is first collected, thereby raising the surplus level to i. Since s
contains the absorption (to claim occurrence) probabilities from the n possi-
a
ble phase states, and the claim causing ruin must be of size i+j in order to
ensure that the deficit at ruin is equal to j, it immediately follows that
c(k) (u) = (g(k) s)a . (2.10)
n,i,j n–1,i–c i+j
Our analysis up to this point has assumed that W =k,where k!{1,2,3,…,n}.
1 r
Therefore,conditioning on the value of W yields,by the Law of Total Probability,
1
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300 A.S.ALFA AND S.DREKIC
n
ƒ (u) = !r r ƒ(k) (u), n=n +1, n +2, n +3, … (2.11)
n,j k n–k,j r r r
k=1
and
n
c (u) = !r r c(k) (u), n=n +1, n +2, n +3, …. (2.12)
n,i,j k n–k,i,j r r r
k=1
However,in order for T=nwhere n=1,2,3,…,n,it must be that either:(i) time
r
nis the time of the first claim which happens to cause ruin,or (ii) the first claim
occurs at some time k!{1,2,3,…,n–1} which does not cause ruin, but ruin
subsequently occurs n–k time units later. Mathematically, this translates to
n-1
ƒ (u) = !r ƒ(k) (u)+r a , n=1,2,3,…,n (2.13)
n,j k n–k,j n u+cn+j r
k=1
and
n-1
c (u) = !r c(k) (u)+d r a , n=1,2,3,…,n, (2.14)
n,i,j k n–k,i,j i,u+cn n u+cn+j r
k=1
where d denotes the Kronecker delta of iand u+cn.Since r =0∀n>n ,
i,u+cn n r
equations (2.11) and (2.13) (as well as equations (2.12) and (2.14)) can be com-
bined to give
min!n-1,nr+
ƒ (u) = ! r ƒ(k) (u)+r a , n ! (cid:2)+ (2.15)
n,j k n–k,j n u+cn+j
k=1
and
min!n-1,nr+
c (u) = ! r c(k) (u)+d r a , n ! (cid:2)+. (2.16)
n,i,j k n–k,i,j i,u+cn n u+cn+j
k=1
For fixed j!{1,2,3,…,ma–c},we simply note that summing (2.16) from i=c
to ma–j yields (2.15), as required. Similarly, for fixed i !{c,c+1,c+2,…,
m –1}, summing (2.16) from j=1 to m –i yields
a a
w ^uh = min!!n-1,nr+r m!a-ic]kg ^uh+ d r m!a-ai
n,i k n-k,i,j i,u+cn n u+cn+j
k=1 j=1 j=1
= L min!!n-1,nr+r cg]kg sm + d r _L - L i,
i k n-k-1,i-c i,u+cn n u+cn u+cn+ma-i
k=1
n ! (cid:2)+, (2.17)
where (2.10) was used to establish the last equality.Clearly,L = 0 in
u+cn+ma–i
(2.17) if ma=3.
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ALGORITHMIC ANALYSIS OF THE SPARRE ANDERSEN MODEL 301
3. COMPUTATIONAL PROCEDURE
In order to calculate the probability distributions of the ruin-related quantities
of interest,it is clear from (2.9) and (2.10) that we need to be able to compute
the row vectors g(k)and h(k),quantities which essentially characterize a matrix-
n n
based means of sample path enumeration via (2.7) and (2.8) respectively.At first
glance, however, this is not entirely straightforward since C and D are both
infinite-dimensional matrices defined by (2.4) and (2.5) respectively. In what
follows,we capitalize on the structure of these matrices to develop an efficient
and stable computational procedure which ultimately enables one to calculate
quantities (2.15), (2.16), and (2.17).
We begin by observing that in the computation of g(k) given by (2.7), we
1
pre-multiply C by the row vector b(k) given by (2.6). Since b(k) contains zeros
from level u+ck onward, we obtain that g(k) is of the form
1
g(k) = (g(k), g(k), g(k), …, g(k) , g(k) , 0,0,…),
1 1,0 1,1 1,2 1,u+c(k+1)–2 1,u+c(k+1)–1
where
u+ck-1
g(k) = !a e B , i =0,1,2,…, c–1,
1,i u+ck-j 1 j–i+c
j=0
and
u+ck-1
g(k) = !a e B , i =c, c+1, c+2,…,u+c(k+1)–1.
1,i u+ck-j 1 j–i+c
j=i-c
Note that g(k) contains zeros from level u+c(k+1) onward, which is c levels
1
further than that in g(k)=b(k).Since (2.7) infers g(k)=g(k) Cfor n!(cid:2)+,we can
0 n n–1
continue this process inductively to establish that g(k)contains zeros from level
n
u+c(k+n) onward, so that
g(k) = (g(k), g(k), g(k), …, g(k) , g(k) , 0,0,…),
n n,0 n,1 n,2 n,u+c(k+n)–2 n,u+c(k+n)–1
where
g(k) = ⎨⎪⎧ !uj=+0c]k+n-1g-1g]nk-g1,jBj–i+c if i=0,1,2,…,c–1, (3.1)
n,i ⎪⎩ !u+c]k+n-1g-1g]kg B if i=c,c+1,c+2,…,u+c(k+n)–1.
j=i-c n-1,j j–i+c
Substituting (2.3) into (3.1), the following recursive procedure for computing
the 1≈ n row vector g(k), n !(cid:2)+, can then be constructed:
a n
Ju+c]k+n-1g-1 N
g(k) = K ! a g]kg s,0,0,...,0O, i=0,1,2,…, c–1, (3.2)
n,i K j-i+c n-1,j O
L j=0 P
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302 A.S.ALFA AND S.DREKIC
and
Ju+c]k+n-1g-1 N
g(k)= g(k) S + K ! a g]kg s,0,0,...,0O,
n,i n–1,i–c K j-i+c n-1,j O
L j=i-c+1 P (3.3)
i =c, c+1, c+2, …, u+c(k+n)–1,
starting with g(k)= a e ,j=0,1,2,…,u+ck–1.We remark that further
0,j u+ck–j 1
simplifications are possible. In particular, we first note that g(k) =
n,u+c(k+n)–1
g(k) S from (3.3). If we repeatedly apply (3.3) to each resulting
n–1,u+c(k+n–1)–1
equation, we eventually obtain that g(k) = g(k) Sn. However, S is
n,u+c(k+n)–1 0,u+ck–1
of the form (2.1),and one can readily verify that Snbecomes an n ≈ n matrix
a a
of zeros if n≥n .Hence,it immediately follows that g(k) =0if n ≥ n .
a n,u+c(k+n)–1 a
Moreover,it is a straightforward exercise to prove by induction that g(k) =
n,u+c(k+n)–j
0 if n ≥ jn , j ! (cid:2)+. The verification of this result is left to the reader.
a
With the above basic recursive procedure for g(k) in place, other recursive
n
procedures can readily be established. For example, noting that e s=1–t =
1 1
a , (2.10) reduces to give c(k) (u)=a a a , and for n=2,3,4,…,
1 1,i,j 1 i+j u+c(k+1)–i
Z
]a a !u+c]k+n-2g-1a g]kg s
] 1 i+j ,=0 ,-i+2c n-2,,
]kg ] if i =c,c+1,...,min"2c-1, ma- j,,
c ^uh = [
n,i,j ]a g]kg Ss+a a !u+c]k+n-2g-1a g]kg s
] i+j n-2,i-2c 1 i+j ,=i+1-2c ,-i+2c n-2,,
] if i =2c, 2c+1,...,ma- j.
\
In addition, we observe that since g(k) =0 ∀ =cn,cn+1, cn+2,…,it
n,u+ck+
follows from (2.3), (2.5), and (2.8) that
u+c]k+ng-1 Ju+c]k+ng-1 N
h]kg = ! g]kg B = K ! a g]kg s,0,0,...,0O. (3.4)
n,-j n-1,,-c j+, KK j+, n-1,,-c OO
,=c L ,=c P
Hence, substituting (3.4) into (2.9) immediately yields
u+c]k+ng-1
ƒ(k)(u) = ! a g]kg s. (3.5)
n,j j+, n-1,,-c
,=c
4. NUMERICAL EXAMPLES
In this section, we illustrate the application of our proposed algorithm with
several numerical examples.All calculations in this section were carried out on
an IBM Pentium IV clone with a 3 Ghz CPU and 4 GB of RAM.