62 T. Siegl, R.F. Tichy Insurance: Mathematics and Economics 26 2000 59–73
3. Analytic solution
The analytic solution will solve the case where F y = 1 − exp−βy is the exponential distribution. Without
loss of generality we can restrict ourselves to the parameter of the exponential distribution β = 1. As we will solve
the equation by a series type solution we will group together certain parts of the equation and try to eliminate each group. This solution procedure is very convenient for this type of problem, but looks unusual for everyone trying to
solve the equation as a whole. Eq. 2 is an integro-differential equation with deviating argument. For the theory of differential equations with
deviating argument see, e.g. El’sgol’ts and Norkin 1973. One way to obtain a solution for 2 is to convert this equation by differentiating it with respect to x and solving the resulting differential equation with deviating argument.
However, there is no obvious gain by this, and thus we will solve the equation directly by using a solution of the form:
U x =
X
i,j ∈I
U
i,j
x =
X
i,j ∈I
A
i,j
e
R
i,j
x
, 3
for a suitable index set I = {0, 0} ∪ {1, 2} × N
. We have used two indices for notational convenience. Inserting this into 2 yields:
X
i,j ∈I
e
R
i,j
x
A
i,j
cR
i,j
− λ + γ + λ 1
R
i,j
+ 1 ,
4 +
X
i,j ∈I
e
R
i,j
αx
e
R
i,j
K
γ A
i,j
5 −
X
i,j ∈I
λ e
−x
A
i,j
R
i,j
+ 1 = 0.
6 For R
0,0
= 0 we obtain a solution of 4+5 = 0 with arbitrary A
0,0
. For the further steps of the solution we will forget Eq. 5 for a moment,
X
i,j ∈I
e
R
i,j
x
A
i,j
cR
i,j
− λ + γ + λ 1
R
i,j
+ 1 − λ e
−x
A
i,j
R
i,j
+ 1 = 0,
and concentrate on a given set of indices i, j ∈ {1, 0, 2, 0}. To obtain a non-trivial solution we eliminate the
coefficient of expR
i,j
x by setting R
1,0
and R
2,0
equal to the two solutions of the quadratic equation resulting from Eq. 4:
cR − λ + γ + λ
1 R
+ 1 = 0.
7 Next, we will eliminate the terms resulting from 6 by a restriction to the coefficients
X
i,j ∈I
A
i,j
R
i,j
+ 1 = 0.
8 What remains is to eliminate the terms that were newly introduced by U
1,0
and U
2,0
in the part with deviating argument Eq. 5. To do so let us rewrite Eqs. 4
+5+6 once more, with R
i,j
= αR
i,j −1
. Furthermore, in writing down the equation we rewrite the indices using expR
i,j
x = expαR
i,j −1
x ,
X
i,j ∈I
e
R
i,j
x
A
i,j
cR
i,j
− λ + γ + λ 1
R
i,j
+ 1 ,
9
T. Siegl, R.F. Tichy Insurance: Mathematics and Economics 26 2000 59–73 63
+ X
i,j ∈I,
j 1
e
R
i,j −1
αx
e
R
i,j −1
K
γ A
i,j −1
10
− X
i,j ∈I
λ e
−x
A
i,j
R
i,j
+ 1 = 0,
11 and change the summation index to avoid references to the index j
−1. Finally we insert expR
i,j
x = expαR
i,j −1
x and obtain
X
i,j ∈{0,0,1,0,2,0}
e
R
i,j
x
A
i,j
cR
i,j
− λ + γ + λ 1
R
i,j
+ 1 12
+ X
i,j ∈I,
j 1
e
R
i,j
x
A
i,j
cR
i,j
− λ + γ + λ 1
R
i,j
+ 1 13
+ X
i,j ∈I,
j 1
e
R
i,j
x
e
R
i,j −1
K
γ A
i,j −1
14
− X
i,j ∈I
λ e
−x
A
i,j
R
i,j
+ 1 = 0.
15 The terms in Eq. 12 were eliminated by restriction Eq. 7 to the exponents R
i,j =0
, and Eq. 15 was already eliminated by restriction Eq. 8 to the coefficients A
i,j
. The remaining terms in the equation can be eliminated by a further restriction to the coefficients A
i,j
. This can be achieved by a recursion of the following type. We assume A
i,j −1
and R
i,j −1
are known and set, R
i,j
= αR
i,j −1
, A
i,j
= −γ e
R
i,j −1
K
cR
i,j
− λ + γ + λR
i,j
+ 1 A
i,j −1
, or equivalently,
R
i,j
= α
j
R
i,
, A
i,j
=
j
Y
k =1
−γ e
R
i,
α
k −1
K
cα
k
R
i,
− λ + γ + λα
k
R
i,
+ 1 A
i,
. 16
In other words, we introduce a new remainder for every part of Eq. 5 that has not been canceled out already. Of course this will generate further terms that have to be canceled out later. As long as this remainder converges to 0
for j → ∞ the series will solve the equation at the limit. As α 1, we have R
i,j
→ 0 i = 1, 2 for j → ∞, and thus we require that the coefficients converge as well:
A
1,j
+ A
2,j
= 0 for j → ∞. 17
As A
i,j
is not arbitrary, this is actually a further restriction to A
i,
i = 1, 2. We add this condition as:
lim
j →∞
X
i =1,2
j
Y
k =1
−γ e
R
i,
α
k −1
K
cα
k
R
i,
− λ + γ + λα
k
R
i,
+ 1 A
i,
= 0.
64 T. Siegl, R.F. Tichy Insurance: Mathematics and Economics 26 2000 59–73
Table 1 Parameters for the numerical experiments
Parameter c
λ γ
α x
max
K Value
1110 1
110 910
10 1
Due to Eq. 17 and as R
i,j
→ 0 for j → ∞, the maximum norm of the remaining defect goes to 0 as j → ∞. Finally, we want to fit the solution to the boundary value and get the restriction
U x
max
= 1. 18
We have three arbitrary coefficients A
0,0
, A
1,0
and A
2,0
and three equations 18, 17 and 8. Thus, our solution is the only solution of the form 3. What remains is to show that the solution is unique.
The theory of differential equations with deviating arguments gives us uniqueness of the solution under the condition that the corresponding integral operator is contracting. We have chosen to prove existence and uniqueness
in this way as the integral operator for Eq. 2 is of interest itself for numerical simulation. The operator is defined by the integral equation U
= IU, where we define I as the expectation over time over
all events at time t, claims of size y and devaluations of the expected probability of survival in each case.
I U x
= Z
T
e
−λ+γ t
λ Z
x +ct
U x + ct − y dF y + γ Uαx + ct + K
dt + e
−λ+γ T
, 19
where T = x − x
max
c is the remaining time until the barrier is hit and the additive term corresponds to the case
where the barrier is reached before an event occurs. We prove that the operator is strictly contracting by using the maximum norm of the difference of two solutions U and V
kIU − IV k
∞
= max
≤xx
max
|IUx − IV x|.
We write U x − V x as U − V x and obtain:
kIU − IV k
∞
= max
≤xx
max
Z
T
e
−λ+γ t
λ Z
x +ct
U − V x + ct − y dF y + γ U − V αx + ct + K
dt ≤ max
0x
max
Z
T
e
−λ+γ t
λ Z
x +ct
dF y + γ
kU − V k
∞
dt ≤ max
≤xx
max
Z
T
e
−λ+γ t
λ + γ dtkU − V k
∞
≤ 1kU − V k
∞
, where 1
= 1 − exp−λ + γ x
max
c 1. Therefore, we have proved contraction with an explicit contraction
constant 1. Depending on the given cumulative distribution function F the contraction constant can in general be improved. In our analytic solution, we have the exponential distribution with parameter 1. This leads to a maximal
contraction constant of ≈ 0.991 for the parameters in Table 1.
Furthermore, due to the additive term exp −λ + γ T , we cannot have the trivial solution U ≡ 0 if x
max
is finite. By using Banach’s fixed point theorem on L
∞
R
+
we have proved that the solution of Eq. 3 is the only measurable bounded solution of Eq. 2 with Eq. 1.
4. Model with random time horizon
