Exact non-ruin probabilities in infinite time
Using the Wiener-Hopf method, for the fundamental equation of the risk theory it is obtained an exact solution in terms of the Fourier transforms and factorization.
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| Цитувати: | Exact non-ruin probabilities in infinite time / V. Chernecky // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 20–25. — Бібліогр.: 3 назв.— англ. |
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irk-123456789-44542009-11-12T12:00:31Z Exact non-ruin probabilities in infinite time Chernecky, V. Using the Wiener-Hopf method, for the fundamental equation of the risk theory it is obtained an exact solution in terms of the Fourier transforms and factorization. 2006 Article Exact non-ruin probabilities in infinite time / V. Chernecky // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 20–25. — Бібліогр.: 3 назв.— англ. 0321-3900 http://dspace.nbuv.gov.ua/handle/123456789/4454 en Інститут математики НАН України |
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Using the Wiener-Hopf method, for the fundamental equation of the risk theory it is obtained an exact solution in terms of the Fourier transforms and factorization. |
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Chernecky, V. |
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Chernecky, V. Exact non-ruin probabilities in infinite time |
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Chernecky, V. |
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Chernecky, V. |
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Exact non-ruin probabilities in infinite time |
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Exact non-ruin probabilities in infinite time |
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Exact non-ruin probabilities in infinite time |
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Exact non-ruin probabilities in infinite time |
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Exact non-ruin probabilities in infinite time |
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exact non-ruin probabilities in infinite time |
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Інститут математики НАН України |
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2006 |
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Exact non-ruin probabilities in infinite time / V. Chernecky // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 20–25. — Бібліогр.: 3 назв.— англ. |
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AT cherneckyv exactnonruinprobabilitiesininfinitetime |
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2025-07-02T07:41:43Z |
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Theory of Stochastic Processes
Vol. 12 (28), no. 3–4, 2006, pp. 20–25
VASILY CHERNECKY
EXACT NON-RUIN PROBABILITIES IN INFINITE TIME
Using the Wiener-Hopf method, for the fundamental equation of the risk
theory it is obtained an exact solution in terms of the Fourier transforms
and factorization.
1. Introduction
Let F (v) be the distribution of claims Zk, EZk = μ, K(v) be the distri-
bution of waiting time Tk, ETk = 1/α, k ∈ Z+, random variables Zk and
Tk be independent, and c > αμ be the gross premium rate. Probability of
solvency of an insurance company, ϕ(u), with initial capital u, in ordinary
renewal process, satisfies the Feller integral equation, [1,2]:
ϕ(u) =
∫ ∞
0
dK(v)
∫ u+cv
0
ϕ(u + cv − z) dF (z). (1)
We are interested by a bounded, monotonically nondecreasing solution of
(1), for which
lim
u→+∞
ϕ(u) = 1. (2)
The example from the classical model, when T and Z have the exponential
distributions, [1], p. 5-6, shows, that the condition (2) is unsufficient for
determination of the interesting us solution. Therefore, in general case,
the solution of equation (1) should satisfy one more additional condition,
for example, a condition in zero or some equivalent one. Note that in the
classical model such additional condition is
lim
u→+0
ϕ(u) = 1 − αμ/c. (3)
But, unfortunately, in general case, the condition in zero for ϕ(u) is not
known, however this condition is known for the solution ϕ0(u) in the case
of accompanying stationary renewal process, calculated by the formula:
ϕ0(u) = α
∫ ∞
0
[1 − K(v)]
{∫ u+cv
0
ϕ(u + cv − z) dF (z)
}
dv, (4)
for which
lim
u→+0
ϕ0(u) = 1 − αμ/c ⇐⇒ lim
u→+∞
ϕ0(u) = 1. (5)
2000 Mathematics Subject Classification. Primary 60G35, 45E10, 62P05.
Key words and phrases. Integral equation of the risk theory, classical risk model,
ordinary renewal process, stationary renewal process, one-sided Wiener-Hopf integral
equation, Wiener-Hopf factorization method.
20
EXACT NON-RUIN PROBABILITIES 21
Therefore hereinafter we shall search for the solution of equation (1) such
that for function ϕ(u) two initial conditions (2) and (5) are fulfilled.
At the derivation of the equations (1) and (4) nothing does not prevent
us to consider u to be negative. Really, the right side of the equation (1) in
this case means the probability for the company, being in the state of ruin,
to be found in the state of non-ruin at the epoch of the first claim and later
on, [1], p. 4, [2], p. 149. Note also that in the case u + cv ≤ z ≤ 0∫ u+cv
0
ϕ(u + cv − z) dF (z) = 0,
since F (z) = 0 for z < 0, and∫ u+cv
0
ϕ(u + cv − z) dF (z) = P{0 < Z1 ≤ u + cv},
when 0 ≤ z ≤ u + cv. Therefore hereinafter we shall search for the solution
ϕ(u) of the problem (1) for u on all numerical axis, u ∈ R.
2. One-sided Wiener-Hopf equation
Reducing the equation (1) by the difference method to the system of
linear algebraic equations, we observe that the received system is a discrete
system of Wiener-Hopf type. This makes us think that the equation (1) is
the homogeneous one-sided integral equation of Wiener-Hopf type
ϕ(u) =
∫ ∞
0
ϕ(v) dk(u − v), −∞ < u < +∞, (6)
with some kernel k(v) defined by K(v) and F (v). Using the Wiener-Hopf
method presented in the monographs [3], pp. 55-56, 103, for the symbol
A(x) of the equation (1) we receive the formula
A(x) = 1 − FT (cx) · FZ(x), −∞ < x < ∞,
where F·(·) denotes a characteristic function of a corresponding distribution
(Fourier transform of density), and the line denotes the complex conjuga-
tion. Returning to the equation (6), we obtain that k(u) can be found with
the help of the inverse Fourier transform of FT (cx) · FZ(x) and integration.
The symbol A(x) is differentiable on the numerical line and has always
the zero of the first order at x = 0. The first follows from the existence
of EZk and ETk, and the latter follows from the condition c > αμ. There
is yet the possibility of existence of infinite number of other zeros xk of
A(x) on the numerical line, which represent not the interest for us since
they generate nonmonotonic solutions. Therefore the equation (1) is of
nonnormal (nonelliptic) type, for the solution of which we can also apply
the Wiener-Hopf factorization method [3].
There exists a factorization
A(x) = x · A+(x) · A−(x) ·
∏
k
(x − xk), −∞ < xk < +∞,
22 VASILY CHERNECKY
where the function A+(x) (A−(x)) is assumed to be analytically extendable
in the domains D+ = {z : Im z > 0}, (D− = {z : Im z < 0}) continuous in
D+ (D−), and A+(z) �= 0 in D+ (1/A−(z) can have singularity at z = ∞).
For the simplicity of the presentation, we some digress from the definition
of factorization given in the monograph [3] since for us it is sufficient to
know only the function A+(x). We also leave aside the question about the
index of the operator corresponding to the equation (1).
Following the Wiener-Hopf method, the solution of the problem (1), equal
to zero on the interval (−∞, 0), is represented in the form of a linear com-
bination:
ϕ+(u) = C1 F−1
[
1
x+
· 1
A+(x)
]
+ C2 F−1
[
1
A+(x)
]
= C1 ϕ1(u) + C2 ϕ2(u), (7)
where
1
x+
=
i
x
+ πδ(x)
is the boundary value of the function 1/z+ analytic in the upper half-plane,
obtained from the factorization of the function 1
x
[3], δ(x) is the Dirac
function, and F−1[·] denotes the inverse Fourier transform in the sense of
L. Schwartz. Index ‘+’ at ϕ means that the function ϕ+(u) is equal to 0
for u ∈ (−∞, 0). Here ϕ1(u) is a monotonically increasing and bounded
function, and ϕ2(u) is a monotonically decreasing function tending to zero
when u → +∞. The constants C1 and C2 are uniquely determined by the
conditions (2) and (5).
Calculating
ϕ(u) = F−1
[
FT (cx) · FZ(x) · Fϕ+(x)
]
,
we obtain the solution of the problem (1) on the axis (−∞, +∞), non equal
0 on the negative semiaxis, which represents the probability of exit from
the state of ruin when u < 0, with ϕ(u) = ϕ+(u) for u > 0. Note that
Fϕ(x) = FT (cx) · FZ(x) · Fϕ(x),
so that the corresponding ϕ(u) satisfies the problem (1) also, and ϕ(u) =
ϕ+(u) for u ∈ (0, +∞). Note that in the literature, in general, the meaning
is not attached to the solution of the Wiener-Hopf equation arising in the
case when u < 0, [3], p. 56.
The solution ϕ0(u) in (2) can be also calculated in the terms of the Fourier
transform
ϕ0(u) = F−1
[
Fα(1−K)(cx) · FZ(x) · Fϕ+(x)
]
, u ∈ (−∞, +∞)
or
ϕ0(u) = F−1
[
Fα(1−K)(cx) · FZ(x) · Fϕ(x)
]
, u ∈ (−∞, +∞).
Since the characteristic function for Gamma distribution with integer
exponent (Erlang distribution) is a rational function, the solution for the
EXACT NON-RUIN PROBABILITIES 23
problem (1) in the case when Tk and Zk have such distributions can be
found in explicit form.
3. Examples
Example 1. Classical model, exponentially distributed claim. Consider
the case when the inter-occurrence times and claims have the exponential
distributions with α = 2, μ = 1/2, respectively, and c = 3. Then
FT (x) =
2
2 + ix
, FZ(x) =
2
2 − ix
, A(x) =
x(3x + 4i)
(3x − 2i)(x + 2i)
,
A+(x) =
3x + 4i
x + 2i
, A−(x) = 3x − 2i,
ϕ1(u) = F−1
[
1
x+
· 1
A+(x)
]
= F−1
[(
1
x
+ πδ(x)
)
· x + 2i
3x + 4i
]
=
=
1
2
· H(u) ·
(
1 − 1
3
e−
4
3
u
)
,
ϕ2(u) = F−1
[
1
A+(x)
]
= F−1
[
x + 2i
3x + 4i
]
=
1
3
· δ(u) +
2
9
H(u)e−
4
3
u,
where H(u) is the Heaviside function (nondetermined at u = 0). From the
condition (2) we obtain C1 = 2. The constant C2 = 0 since
ϕ+(u) = ϕ0+(u) = 2 · ϕ1(u) = H(u) ·
(
1 − 1
3
e−
4
3
u
)
satisfies the condition (3). The obtained solution coincides with solution
given by the well known formula for exponentially distributed claims in
classical model, [1,2], obtained by reduction (1) to a differential equation.
The solution of the problem on (−∞, +∞) is given by the formula
ϕ(u) = ϕ0(u) = F−1
[
FT (cx) · FZ(x) · Fϕ+(x)
]
= 4 · F−1
[
4πδ(x)x − 3iπδ(x)x2 + 4i + 2x
(2 + 3ix)(−2 + ix)x(−4 + 3ix)
]
=
2
3
H(−u)e
2
3
u + H(u)(1 − 1
3
e−
4
3
u).
The value of ϕ(u) = ϕ0(u) at u = 0 is determined by the continuity
ϕ(0) = lim
u→0
ϕ(u) =
2
3
.
Example 2. Classical model, Γ-distributed claims. Consider the case when
the inter-occurrence times have the exponential distributions with α = 2,
c = 3, and the claims density is given by
fZ(t) = te−2t for t > 0
24 VASILY CHERNECKY
so that μ = 1. Then
FT =
2
2 − ix
, FZ =
4
(−2 + ix)2
, A(x) =
x(3x2 + 10ix − 4)
(3x − 2i)(x + 2i)2
,
A+(x) =
3x2 + 10ix − 4
(x + 2i)2
, A−(x) =
1
3x − 2i
.
Omitting obvious intermediate calculations, we obtain the solution on
(0, +∞)
ϕ+(u) = ϕ0+(u) =
H(u) ·
[
1 − 1
39
(
(13 − 4
√
13)e−
5+
√
13
3
u + (13 + 4
√
13)e−
5−√
13
3
u
)]
which coincides with solution obtained by using the Laplace transform, [1],
p. 13,
ϕ+(u) = H(u)
[
1 −
(
8
39
√
13 sinh
(√
13
3
u
)
− 2
3
cosh
(√
13
3
u
))
e−
5
3
u
]
.
The solution on (−∞,∞) is
ϕ(u) = ϕ0(u) = H(u)×[
1 − 1
39
(
(13 − 4
√
13)e−
5+
√
13
3
u + (13 + 4
√
13)e−
5−√
13
3
u
)]
+
1
3
H(−u)e
2
3
u,
ϕ(0) = ϕ0(0) =
1
3
.
Example 3. Γ-distributed inter-occurrence times, Γ-distributed claims.
Consider the case when the inter-occurrence times and claims have the Γ-
distributions
fT (t) = te−t, fZ(t) = 4te−2t for t > 0,
with α = 1/2, μ = 1, respectively, and c = 1. Then
FT =
1
(1 + ix)2
, FZ =
4
(−2 + ix)2
, A(x) =
x(x2 + ix + 4)(x + i)
(x + 2i)2(x − i)2
,
A+(x) =
(2x + (1 +
√
17)i)(x + i)
(x + 2i)2
, A−(x) =
2x + (1 −√
17)i
4(x − i)2
.
After obvious calculations, we obtain the solutions of (1) on (0, +∞)
ϕ+(u) = H(u) ·
[
1 − 1
32
(
(19 − 5
√
17)e−
1+
√
17
2
u + (9 +
√
17)e−u
)]
,
and on (−∞,∞)
ϕ(u) = H(u) ·
[
1 − 1
32
(
(19 − 5
√
17)e−
1+
√
17
2
u + (9 +
√
17)e−u
)]
+
1
32
H(−u)
(
(−28 + 4
√
17)u + 4(1 +
√
17)
)
eu,
EXACT NON-RUIN PROBABILITIES 25
ϕ(0) = lim
u→0
ϕ(u) =
1 +
√
17
8
.
The solution of (4) on (−∞,∞) is as follows
ϕ0(u) = H(u)
[
1 − 1
64
(
32 − (27 + 3
√
17)e−u − (5 − 3
√
17)e−
−1+
√
17
2
u
)]
+
1
64
H(−u)
(
−(28 − 4
√
17)u + 32
)
eu, ϕ0(0) = lim
u→0
ϕ0(u) =
1
2
.
Note that in all these three examples C2 = 0 so that there is not necessity
to calculate the function ϕ2(u).
References
1. Grandell J., Aspects of Risk Theory, Springer, 1991.
2. Bühlmann H., Mathematical Methods in Risk Theory, Springer, 1970.
3. Gakhov F. D. and Cherski Yu. I., Equations of convolution type, M.: Nauka,
1978.
Department of Mathematical Analysis, Mechnikov Odessa National Uni-
versity, 65026, Odessa, Ukraine
E-mail address: chern.va@paco.net
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