D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267 255
3. Algebraic formulae for selected non-Poisson claims processes
In this section, two simple extensions of the Poisson claim number process are considered. The Poisson process has exponential inter-event times whose variance equals the square of the mean. The first extension we consider is
the Erlang-α considered in Section 2 as an example of inter-claim times which are less variable than the exponential. The latter case considered is that of a mixture of exponentials, in order to consider a simple example of inter-claim
times that are more variable than the exponential distribution. For the former, we will see that the methods of Stanford and Stroi´nski 1994 extend readily with a minor increase in complexity over the Poisson case. For the
latter, we find that the extension becomes significantly more complicated even in its simplest case, although it is still tractable. In Section 4, we show that a pairing of our underlying recursion and the De Vylder–Goovaerts method
can be used in these situations.
In each case considered, we first determine a difference equation involving the Laplace transforms L
n
s, and then obtain an explicit expression for L
n
s via mathematical induction. Finally, corresponding expressions for the probability of ruin on the nth claim are found.
3.1. Recursions for Erlang-α inter-claim revenues When the inter-claim revenue follows an Erlang-α distribution and the claim sizes have a mixture of N exponential
distributions, a set of general formulae for L
n
s and P n can be determined as follows.
Theorem 3.1. For Erlang-α distributed inter-claim revenue and a mixture of N exponentials claim sizes, L
n
s, n ≥ 1, is given by
L
n
s =
N
X
i=1
p
i
µ
i
µ
i
− s
λ λ + s
α
L
n−1
s − λ
λ + µ
i α
L
n−1
µ
i
, 3.1
and P n =
N
X
i=1
p
i
λ λ + µ
i α
L
n−1
µ
i
.
Proof. Under the stated assumptions
gy =
N
X
i=1
p
i
λ λ + µ
i α
µ
i
e
µ
i
y
, y 0,
Gs = λ
λ + s
α N
X
i=1
p
i
µ
i
µ
i
− s
. After substituting these expressions into 2.5 and after some manipulation one obtains the result 3.1.
Theorem 3.2. The general form of L
n
s, n ≥ 1, when the inter-claim revenue distribution is Erlangian-α-distributed and the claim sizes follow a mixture of two exponential distributions, is given by
L
n
s =
nα
X
j =1
c
n j
λ λ + s
j
+ λ
λ + s
nα
A
n
s, 3.2
where A
n
s =
∞
X
k=n
u
k−n
k
2
X
m=1
p
m
×
D
n m
µ
m
− s
k−n
+
n−2
X
j =0
f
n j m
k−n
X
ℓ=0
k − n − ℓ + j j
µ
m
− s
ℓ
µ
m
− µ
3−m k−n−ℓ
,
256 D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267
and where the coefficients c
n j
, D
n m
and f
n j m
, n ≥ 2, satisfy the recursions
c
n j
=
2
X
i=1
p
i
µ
i
λ + µ
i
n−1α
X
k=max1,j −α
c
n−1 k
λ λ + µ
i k+α−j
+ λ
λ + µ
i nα−j
A
n−1 i
,
3.3
D
n m
= D
n−1 m
h
m
, 3.4
f
n j m
=
h
m
f
n−1 0m
+ h
3−m
D
n−1 m
, j = 0,
h
m
f
n−1 j m
+ h
3−m
f
n−1 j −1,m
, j = 1, 2, . . . , n − 3,
h
3−m
f
n−1 n−3,m
, j = n − 2,
3.5
where h
m
= p
m
uaµ
m
and A
n−1 i
≡ A
n−1
µ
i
. The coefficients are initialised as follows:
c
1 j
=
2
X
i=1
p
i
µ
i
λ + µ
i
e
− uµ
i
λ λ + µ
i α−j
, j = 1, . . . , α,
D
1 i
= e
− uµ
i
uµ
i
, i = 1, 2
f
2 0m
= h
3−m
D
1 m
, m = 1, 2.
Proof. The proof uses mathematical induction. When n = 1, 3.1 already satisfies the general form of L
n
s in the theorem. We establish the remainder of the proof through the addition and subtraction of A
n−1 i
λλ + s
nα
to facilitate the factoring of the root µ
i
− s. In particular, we obtain
λ λ + s
α
L
n−1
s − λ
λ + µ
i α
L
n−1
µ
i
= µ
i
− s
λ + µ
i
n−1α
X
j =1
c
n−1 j
j +α−1
X
l=0
λ λ + s
l+1
λ λ + µ
i j +α−l−1
+ A
n−1 i
nα−1
X
l=0
λ λ + s
l+1
λ λ + µ
i nα−l−1
+ λ
λ + s
nα
h A
n−1
s − A
n−1 i
i .
After factoring the term µ
i
− s from the first bracketed term of the last expression, and after reversing the order of
summation, it is readily seen that the coefficients c
n j
are related to the previous coefficients by 3.3. To complete the proof, we need to show that
A
n
s =
2
X
i=1
p
i
µ
i
µ
i
− s
[A
n−1
s − A
n−1 i
]. We observe that
D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267 257
A
n−1
s − A
n−1 i
= µ
i
− s
∞
X
k=n
u
k+1−n
k ×
2
X
m=1
p
m
D
n−1 m
k−n
X
p=0
µ
m
− s
p
µ
m
− µ
i k−n−p
+
n−3
X
j =0
f
n−1 j m
k−n
X
l=0
k − n−l + j j
×
l
X
p=0
µ
m
− s
p
µ
m
− µ
i l−p
µ
m
− µ
3−m k−n−l
. After expanding this about indices m = i and m = 3 − i, using 3.4, and employing the identity
k−n
X
l=p
k − n − l + j j
= k − n − p + j + 1
j + 1 ,
we obtain
2
X
i=1
p
i
µ
i
µ
i
− s
[A
n−1
s − A
n−1 i
] =
∞
X
k=n
u
k−n
k
2
X
i=1
p
i
D
n i
µ
i
− s
k−n
+
k−n
X
l=0
µ
i
− s
l
µ
i
− µ
3−i k−n−l
f
n 0i
+ k − l − 2
n − 2 f
n n−2,i
+
n−3
X
j =1
k − n − l + j j
h
i
f
n−1 j i
+ h
3−i
f
n−1 j −1,i
.
The term multiplying k − n − l + j
j is f
n j i
from 3.5. Therefore the final expression can be recognised as A
n
s, which completes the proof. For exponential claim sizes we can state the following algorithm. We state this result without proof as the derivation
is straightforward but tedious. L
n
s = e
− uµ
µ λ + s
8
A
µ
n nα−1
X
j =0
c
n j
λ + µ λ + s
j
+ 8
A
s
n
uµ
n ∞
X
k=n
uµ − s
k−n
k
,
where 8
A
s = λλ + s
α
and
c
n j
=
uµ
n−1
n − 1 +
µ λ + µ
P
n−1α−1 l=0
c
n−1 l
, 0 ≤ j ≤ α − 1,
uµ
n−1
n − 1 +
µ λ + µ
P
n−1α−1 l=j −α
c
n−1 l
, α ≤ j ≤ nα − 1.
3.6
The algorithm is initialised by setting c
1 j
= 1, j = 0, . . . , α − 1.
In summary, the algorithms for the computation of the coefficients c
n j
in the case of Erlang-α revenues are only slightly more complex than their Poisson counterparts. We note in particular that with the exception of the
A
n
µ
i
term the algorithm consists of multiplications and additions of purely positive coefficients between 0 and 1. Numerical examples based on this algorithm are presented in Section 5.
258 D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267
3.2. Recursions for mixtures of exponentially distributed inter-claim revenues
Theorem 3.3. For inter-claim revenue having a mixture of M exponential distributions with weights r
1
, . . . , r
M
and rates λ
1
, . . . , λ
M
, and claim sizes having a phase-type distribution with representation γ γ
γ , T T
T , where ttt =
− Te
Te Te, L
n
s, n ≥ 1, is given by L
n
s =
M
X
i=1
r
i
v v
v
i
Z
∞ x=0
p
n−1
x expT T
T x dx − γ γ
γ λ
i
λ
i
+ s
L
n−1
s sIII + T
T T
− 1
ttt ,
3.7 where v
v v
i
= λ
i
γ γ
γ λ
i
III − T T
T
− 1
.
Proof. For the stated inter-claim revenue and claim size distributions, the density of the increment becomes
gy =
P
M i=1
r
i
λ
i
e
− λ
i
y
8
B
λ
i
, y ≥ 0,
P
M i=1
r
i
λ
i
R
∞
e
− λ
i
t
bt − y dt, y 0.
Substituting the expressions for phase-distributed claim sizes, and letting v v
v
i
= λ
i
γ γ
γ λ
i
III − T T
T
− 1
, the following is obtained:
gy =
P
M i=1
r
i
v v
v
i
e
− λ
i
y
ttt ,
y ≥ 0, P
M i=1
r
i
v v
v
i
exp−T T
T yttt ,
y 0. 3.8
Therefore Z
∞ y=x
e
sy
g−y dy =
M
X
i=1
r
i
v v
v
i
Z
∞ y=x
e
sy
expT T
T y dyttt =
M
X
i=1
r
i
v v
v
i
Z
∞ y=x
expsIII + Ty dyttt = −
e
sx M
X
i=1
r
i
v v
v
i
expT T
T xsI + T T
T
− 1
ttt .
3.9 Substitution of 3.9 into the second term of 2.5 completes the proof. Evaluating 3.7 at s = 0, and recalling Eq.
2.10 of Stanford and Stroi´nski 1994 one obtains the probability of ruin on the nth claim.
P n =
M
X
i=1
r
i
v v
v
i
Z
∞ x=0
p
n−1
xexpT T
T x dxeee. 3.10
Theorem 3.4. The following general expressions for P n and L
n
s are obtained for mixtures of M exponential inter-claim revenues and Erlang-α claim sizes:
P n =
M
X
i=1 α−1
X
k=0
r
i
L
k n−1
µ −µ
k
k 1 −
µ λ
i
+ µ
α−k
, 3.11
D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267 259
L
n
s =
M
X
i=1
r
i
L
n−1
s λ
i
λ
i
+ s
, 0, . . . , 0 −
λ
i
λ
i
+ µ
L
n−1
µ, λ
i
λ
i
+ µ
−µL
′ n−1
µ + λ
i
λ
i
+ µ
µ λ
i
+ µ
L
n−1
µ, . . . ,
α−1
X
ℓ=0
λ
i
λ
i
+ µ
µ λ
i
+ µ
ℓ
−µ
α−1−ℓ
α − 1 − ℓ L
α−1−ℓ n−1
µ
µ
µ − s
α
µ µ − s
α−1
.. .
µ µ − s
. 3.12
Proof. The proof is nearly identical to that presented in Section 2.3 of Stanford and Stroi´nski 1994, and is contained in Lee et al. 1994.
Since 3.12 involves the term µµ − s
α
it stands to reason that µ − s
α
must be factored from the bracketed terms so that these factors will cancel when finding L
n
s. Therefore we hypothesise the following general form for L
n
s when claim sizes follow an Erlang-α distribution: L
n
s = e
− uµ
n
X
j =1 j
X
l=0
c
n j l
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
+ 8
A
s
n
uµ
αn ∞
X
k=αn
uµ − s
k−αn
k
.
3.13 The proof of the correctness of this hypothesis when α = 1, together with the recursion for the coefficients, is
given in the following theorem. The interested reader is directed to Lee et al. 1994 for the derivations and proofs for a comparable form when α = 2. The same report includes numerical examples for both α = 1 and α = 2.
Theorem 3.5 α = 1: Exponential claim sizes. When the inter-claim revenue follows a mixture of two exponential distributions, and for exponentially distributed claim sizes, the Laplace transform of the distribution of the reserve
following the nth claim, L
n
s, n ≥ 1, is given by 3.13 where the coefficients c
n j l
at the nth stage are related to those at the n − 1 th stage by the following expressions for j = 1, 2, . . . , n:
c
n j l
=
µ λ
2
+ µ
uµ
n−1
n − 1 8
A
µ
n−j
+ c
n−1 j −1,0
+ 8
A
µ P
n−j −1 p=0
P
p q=0
c
n−1 p+j,q
r
1
λ
1
λ
1
+ µ
q
r
2
λ
2
λ
2
+ µ
p−q
, l = 0,
uµ
n−1
n − 1 8
A
µ
n−j
µ λ
1
+ µ
j − 1 l − 1
+ µ
λ
2
+ µ
j − 1 l
+ µ
λ
1
+ µ
c
n−1 j −1,l−1
+ µ
λ
2
+ µ
c
n−1 j −1,l
+ µ
λ
1
+ µ
8
A
µ P
n−j −1 p=0
c
n−1 p+j,p+l
r
1
λ
1
λ
1
+ µ
p
, 0 l j, µ
λ
1
+ µ
uµ
n−1
n − 1 8
A
µ
n−j
+ c
n−1 j −1,j −1
+ 8
A
µ P
n−j −1 p=0
c
n−1 p+j,p+j
r
1
λ
1
λ
1
+ µ
p
, l = j, 3.14
where it is understood that sums of the form P
-
1
. equal 0, and that c
n−1 0l
= 0, for all l and n. The algorithm is
initialised by setting c
1 10
= µλ
2
+ µ and c
1 11
= µλ
1
+ µ.
Proof. When α = 1, Eqs. 3.11 and 3.12 reduce to
260 D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267
L
n
s = µ
µ − s [8
A
sL
n−1
s − 8
A
µL
n−1
µ], 3.15
P n = 8
A
µL
n−1
µ, 3.16
where 8
A
s = r
1
λ
1
λ
1
+ s
+ r
2
λ
2
λ
2
+ s
. Assuming an initial reserve of u, P 1 = 8
A
µe
− uµ
. Therefore L
1
s can be written as L
1
s = e
− uµ
µ µ − s
[8
A
se
uµ−s
− 8
A
µ] =
e
− uµ
2
X
i=1
r
i
λ
i
λ
i
+ s
µ λ
i
+ µ
+ 8
A
suµ
∞
X
k=1
uµ − s
k−1
k .
3.17 3.17 satisfies the general form of L
n
s, with c
1 10
= µλ
2
+ µ and c
1 11
= µλ
1
+ µ. Now assume the general
form is valid up to n = N − 1. Substituting into 3.15 for n = N , one finds L
N
s = µ
µ − s e
− uµ
N −1
X
j =1 j
X
l=0
c
N −1 j l
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
8
A
s −
r
1
λ
1
λ
1
+ µ
l
r
2
λ
2
λ
2
+ µ
j −l
8
A
µ +
uµ
N −1
N − 1 8
A
s
N
− 8
A
µ
N
+ 8
A
s
N
uµ
N −1
uµ − s
∞
X
k=N
uµ − s
k−N
k .
After several changes of indices, and simplifying and cancelling the term µ − s one obtains L
N
s = e
− uµ
N
X
j =2
j
X
l=1
c
N −1 j −1,l−1
µ λ
1
+ µ
+
j −1
X
l=0
c
N −1 j −1,l
µ λ
2
+ µ
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
+ 8
A
µ
N −1
X
j =1 j
X
l=1
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
N −1
X
p=j
c
N −1 p,p+l−j
r
1
λ
1
λ
1
+ µ
p−j
µ λ
1
+ µ
+
8
A
µ
N −1
X
j =1
r
2
λ
2
λ
2
+ s
j
N −j −1
X
p=0 p
X
q=0
c
N −1 p+j,q
r
1
λ
1
λ
1
+ µ
q
r
2
λ
2
λ
2
+ µ
p−q
µ λ
2
+ µ
+
uµ
N −1
N − 1 µ
λ
1
+ µ
N
X
j =1 j
X
l=1
j − 1 l − 1
8
A
µ
N −j
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
+ uµ
N −1
N − 1 µ
λ
2
+ µ
N
X
j =1 j −1
X
l=0
j − 1 l
8
A
µ
N −j
r
1
λ
1
λ
1
+ s
l
r
2
λ
2
λ
2
+ s
j −l
+ 8
A
s
N
uµ
N ∞
X
k=N
uµ − s
k−N
k .
D.A. Stanford et al. Insurance: Mathematics and Economics 26 2000 251–267 261
The proof is completed by collecting like terms and expressing the coefficients c
N j l
as in 3.14 for n = N .
Thus, P n, n ≥ 2 can be explicitly determined from 3.16 with coefficients c
n j l
starting from c
1 10
= µλ
2
+ µ
and c
1 11
= µλ
1
+ µ.
Numerical examples for this case are presented in Section 5. By inspection of Eqs. 3.13 and 3.14, we see that the recursion involves no negative numbers. Furthermore, the multiplications involve terms between 0 and 1,
except for two combinatorial terms in the middle line. Thus, while the form of 3.14 may be complicated, it can nonetheless be programmed to provide accurate results.
Theorem 3.5 shows that the recursion for the coefficients, while tractable, is somewhat cumbersome. The situation is even more cumbersome for α = 2. As an alternative to this exact method when the recursions are either too cum-
bersome or intractable, an approximate method by De Vylder and Goovaerts can be adopted. This is presented next.
4. A numerical approach using the De Vylder–Goovaerts method
Kategori