1. Ruin Theoretical Comparisons
Vesa Pekkanen
Aalto University School of Science
Master’s Thesis
Espoo, April 28, 2016
Thesis advisor: Lasse Leskel¨a
Supervisor: Lasse Leskel¨a
2. Aalto University
School of Science
PL 11000, 00076 Aalto
http://www.aalto.fi
Abstract of master’s thesis
Author: Vesa Pekkanen
Title of thesis: Ruin Theoretical Comparisons
Master’s programme: Applied physics and mathematics
Major: mathematics Code: F3006
Supervisor: Prof. Lasse Leskel¨a
Advisor: Prof. Lasse Leskel¨a
This thesis compares ruin probabilities given by the Cram´er–Lundberg model and two of its
extensions, a diffusion perturbed model and a model with risky investments. For each model, ruin
probabilities are calculated for degenerately, uniformly and exponentially distributed claim sizes
as well as for Pareto-distributed claim sizes. These ruin probabilities are then compared with one
another. For the extended models, the thesis also analyzes a pure diffusion risk reserve process.
For every model, this work demonstrates that the key distinction is whether the claim sizes
follow a heavy-tailed probability distribution or not. Specifically, models with heavy-tailed claims
sizes do not permit constructing exponentially decreasing bounds for the ruin probability.
If the insurer invests part of his surplus in an asset that follows a geometric Brownian mo-
tion, it is shown that the ruin probability, at best, exhibits a power law decay. Nevertheless, under
certain circumstances the insurer benefits from investing in a risky asset even when he only aims
to minimize the ruin probability. Under other circumstances, there is a trade-off between higher
expected returns and a lower ruin probability.
Date: April 28, 2016 Language: English Number of pages: 68
Key words: Ruin theory, ruin probability, Cram´er–Lundberg model, degenerate distribution, uni-
form distribution, exponential distribution, Pareto distribution, heavy tails
4. Notations and abbreviations
Abbreviations
a.s. almost surely
i.i.d. independently and identically distributed
p. page(s)
SD standard deviation
General stochastic notation
E expectation operator
P probability
Var variance operator
MF moment generating function of a probability distribution with cumulative distribution
function F
FI integrated tail distribution function with F as the underlying distribution function
⊥ left and right side are independent of one another
D
= both sides have the same probability distribution
F filtration
Deg(β) a degenerate distribution with parameter β
Exp(β) an exponential distribution with rate β
N(µ, σ2) a normal distribution with expectation µ and variance σ2
Par(α, M) Pareto distribution with scale parameter α and minimum value parameter M
U(a, b) continuous uniform distribution with parameters a < b
Nt a Poisson process with parameter λ
Wt a standard Wiener process
W
(µ,σ)
t a Wiener process with drift µ and volatility σ
Ruin theory specific notation
ψ(u) ruin probability
φ(u) survival probability
ψd(u) probability of ruin due to diffusion
ψc(u) probability of ruin due to a claim
τu time to ruin with initial reserve u
4
5. Ti the occurrence of i:th claim
u initial reserve
c premium rate
Rt risk reserve process
Yt surplus process
λ the Poisson arrival intensity of the claims
B(u) cumulative distribution function of the claim size distribution
η(n) n:th moment of the claim size distribution
ρ := λη
c
Jt a compound Poisson process representing total claims
θ(s) Lundberg equation
γ the adjustment coefficient
Vt a standard Wiener process representing the diffusion component of Rt
ς volatility of the diffusion component
St a geometric Brownian motion representing the risky asset
µ drift of St
σ volatility of St
u∗ such initial reserve that ψ(u∗) = 0.2
Other notation
R set of real numbers, real axis
N set of natural numbers, zero not included
an arbitrarily small positive number
⇔ left and right side are equivalent
⇒ left side follows from right side
x→y
−−−→ left side tends to right side as x tends to y
f(x) ∼ g(x) limx→c
f(x)
g(x) = 1. By default, c = ∞.
left-side limit
right-side limit
Xt− := lims t Xs
F(x) := 1 − F(x)
5
6. f Laplace transform of function f
L Laplace transform operator
x floor function, gives largest integer not larger than x
F ∗ G Stieltjes convolution of functions F and G
F∗n n:th convolution power of function F
y ∨ x := max{x, y}
y ∧ x := min{x, y}
1{} indicator function
:= left side is defined to be the right side
6
7. Introduction
The classic approach to modelling net cash flows of an insurance company is to construct a
stochastic process which includes some non-negative initial value that represents the initial re-
serve of the insurance company, a linear time dependent function that represents premiums paid
by the insured and a compound Poisson process that represents the total value of the insurance
claims. This stochastic process is referred to as the risk reserve process. Both the number of
the claims and the size of each claim is random. The problem to solve involves finding the
probability with which this process will become negative for a given time horizon which may be
infinite. This probability is called the ruin probability and is typically expressed as a function
of the initial reserve. This work extends this basic model by a adding a diffusion component to
the risk reserve process and by compounding the process with a geometric Brownian motion.
This thesis first calculates or, more commonly, approximates the ruin probability for the classic
model and its extensions, and then explores how this ruin probability depends on the choice of
the claim size distribution. The plain vanilla model and its two extensions are also compared
with one another
The degenerate distribution, uniform distribution, exponential distribution and Pareto distri-
bution comprise the four different claim size distributions chosen for comparisons. Degenerate
claims correspond to a situation where an insurance company faces equal-sized claims. Assuming
uniform claims is justifiable for claims that are evenly distributed between some minimum and
maximum value. The exponential distribution emerges as a natural choice for claims that may
in principle be as large as possible but tend not to vary too much in size. If the insurer provides
insurance against both fairly minor tragedies, like car burglaries, and major catastrophes, like
hurricanes, then assuming Pareto-distributed claim sizes might be appropriate.
Swedish mathematicians Filip Lundberg and Harald Cram´er laid the groundwork for ruin the-
ory, and the classical risk model, the Cram´er–Lundberg model, derives its name from these
two pioneers in actuarial mathematics. Of the more contemporary insurance mathematicians,
one should mention Søren Asmussen and Hansj¨org Albrecher whose book Ruin probabilities
[4] is an absolute gem. Naturally, this thesis draws heavily on [4]. Others worthy of mention-
ing include Fran¸cois Dufresne and Hans Gerber who introduced a diffusion component to the
Cram´er–Lundberg model in their crisp paper Risk theory for the compound Poisson process that
is perturbed by diffusion [7]. Section 3 owes a considerable amount to [7]. As for ruin theory
when investment income is included, a prominent figure here is Jostein Paulsen. Not only has
he contributed substantially to this subfield of ruin theory, but he has also concisely summed
the contribution’s of others in his paper Ruin models with investment income [9].
The first section introduces and briefly explains some key tools and definitions which are repeat-
edly featured in the later sections. The second section presents the Cram´er–Lundberg model and
compares the ruin probabilities given by different claim size distributions. In the third section,
a diffusion component is added to the risk reserve process. Similar comparisons as in the end
of Section 2 and comparing the diffused and non-diffused models conclude the section. In the
fourth section, the insurer’s surplus is invested in an asset that follows a geometric Brownian
motion. The section explores how this influences the ruin probability. Internal comparisons of
the model as well as comparing the model to the model in Section 3 again complete the section.
7
8. 1 Some key tools and definitions
This section briefly introduces some concepts and example values of parameters that are re-
peatedly featured in this work.
1.1 Example values of parameters
Unless otherwise specified, these values are used in the examples, tables and figures throughout
the thesis.
• premium rate: c = 1
• intensity of the compound Poisson process: λ = 100
• expectation of the claim size distribution: η = 1/120
• degenerate distribution Deg(η): η = 1/120
• uniform distribution U(a, b): a = 0, b = 2η = 1/60
• exponential distribution Exp(β): β = η−1 = 120
• Pareto distribution Par(M, α) (M is the minimum value parameter and α is the scale
parameter): M = 1/360, α = 3/2
• volatility of diffusion: ς = cη = 1/120
• drift of diffusion in a pure diffusion model: ν = −λη = −5/6
• drift of the compounding geometric Brownian motion: µ = ln(6/5)
• volatility of the compounding geometric Brownian motion: σ = 8 ln(5/4)
11
In the examples and figures, the time t is in years. Note that with the given parameters, the
expectation of the degenerate, uniform, exponential and Pareto distributions are all the same:
η = 1/120.
1.2 Tail function
The tail function of a probability distribution is defined F(x) := 1 − F(x), where F is the
cumulative distribution function. The tail function delivers the probability P(X > x). In ruin
theory, how slowly the tail function of the claim size distribution decays to zero gives some idea
of how dangerously the claims are distributed and therefore how the ruin probability decays
as a function of the insurer’s initial reserve u. Specifically, in the Cram´er-Lundberg model an
exponentially bounded tail function of the claim size distribution typically implies an exponen-
tially bounded ruin probability. In the opposite case, exponential bounds do not exist and in
general calculus becomes more complicated. Both cases will be discussed in more detail later.
Example 1.1. (Tail functions of degenerate, uniform, exponential and Pareto distributions) Let
X1 be degenerate at η = 1/120, let X2 be uniform on [0, 1/60], let X3 be exponentially distributed
with rate 120 and let X4 be Pareto-distributed with scale parameter 3/2 and minimum value
parameter 1/360. As these are the example distributions (see Section 1.1), all four distributions
8
9. F
Degenerate
Uniform
Exponential
Pareto
0.005 0.010 0.015 0.020 0.025 0.030
0.2
0.4
0.6
0.8
1.0
x
F
Exponential
Pareto
0.02 0.04 0.06 0.08 0.10 0.12 0.14
0.02
0.04
0.06
0.08
x
Figure 1: Tail functions of degenerate, uniform, exponential and Pareto distributions.
have the same expectation value. The probability that the random variable Xi is ten times bigger
than its expectation value is
P(X1 > 1/12) = P(X2 > 1/12) = 0
P(X3 > 1/12) ≈ 0.000045
P(X4 > 1/12) ≈ 0.0061.
In other words, that Xi is ten times bigger than E[Xi] is 134 times more likely to occur for
Pareto distributed random variables than for exponentially distributed ones. For degenerate and
uniform random variables, such a value for Xi is impossible. This implies that, of these four
distributions, the Pareto distribution has the heaviest tail. Figure 1 supports this implication.
9
10. Changing the parameters of the distributions in Example 1.1 does not change its implication
because a Pareto tail function always exhibits power law decay. In contrast, the tail function of
an exponential distribution, as one might surmise from its name, decreases exponentially. For
uniform and degenerate distributions, F(x) = 0 for a sufficiently large x.
1.3 Integrated tail distribution
For a non-negative random variable X with a mean 0 < η < ∞ and a distribution function F,
it holds that
[0,∞[
F(x)dx =
[0,∞[ [x,∞[
dF(z)dx =
[0,∞[ [0,z]
dxdF(z) =
[0,∞[
zdF(z) = η.
Hence F(x)/η is the density function of a proper probability distribution.
Degenerate F
Degenerate F I
Uniform F
Uniform F I
Pareto F
Pareto F I
0.005 0.010 0.015 0.020
0.2
0.4
0.6
0.8
1.0
x
Figure 2: The tail function of a degenerate, uniform and Pareto distribution and their corre-
sponding integrated tail distribution.
Definition 1.2. (Integrated tail distribution) If a non-negative random variable has a dis-
tribution function F on [0, ∞[ and expectation η > 0, then the corresponding integrated tail
distribution is
FI(x) =
1
η [0,x]
F(y)dy
If X is exponentially distributed with rate η−1, then FI(x) = F(x). Example 1.3 explores the
relationship between the integrated tail distribution and its underlying distribution for a few
other probability distributions.
Example 1.3. Assume that X1 is degenerate at η, X2 is uniform on [0, 2η] and X3 is Pareto-
distributed with minimum value parameter M > 0 and scale parameter α > 0. Let XI
i , i = 1, 2, 3
be a random variable whose distribution function FI is based on the distribution function of Xi.
Table 1 gives the tail function, expectation and variance of Xi and XI
i .
10
11. X1 XI
1
F 1{u<η}
η−u
η 1{u<η}
E η η
2
Var 0 η2
12
X2 XI
2
F 2η−u
2η 1{u<2η}
u
2η 2 − u
2η 1{u<2η}
E η 2η
3
Var η2
3
2η2
9
X3 XI
3
F 1{u<α−1
α
η} + α−1
αu η
α
1{u≥α−1
α
η}
η−u
η 1{u<α−1
α
η} + 1
α
α−1
αu η
α−1
1{u≥α−1
α
η}
E η
α−1
α
2 αη
2(α−2) if α > 2
∞ if α ≤ 2
Var
η2
α(α−2) if α > 2
∞ if α ≤ 2
α−1
α
3 α3−4α2+7α
12(α−3)(α−2)2 η2 if α > 3
∞ if α ≤ 3
Table 1: The tail function, expectation and variance of a degenerate, uniform and Pareto dis-
tribution and their corresponding integrated tail distributions.
From Table 1 one may infer that the integrated tail distribution of a degenerate distribution
with parameter η is a uniform distribution on [0, η]. The integrated tail distribution of a uni-
form distribution resembles a special case of the beta distribution. Finally, the integrated tail
distribution of a Pareto distribution with a shape parameter α is very nearly a Pareto distri-
bution with a shape parameter α − 1. The tail function only differs from a Pareto tail function
in the vicinity of zero where it behaves more like a uniform distribution. Figure 2 depicts the
tail functions of the distributions in Example 1.3 (example values have been chosen for the
parameters).
1.4 Stieltjes convolution
Suppose that X and Y are independent non-negative continuous random variables with dis-
tribution functions F and G. Then, via the law of total probability, the sum X + Y has a
distribution function
P(X + Y ≤ x) =
[0,∞[
P X ≤ x − y Y = y dG(y) =
[0,∞[
F(x − y)dG(y).
This result motivates the following definition.
Definition 1.4. (Stieltjes convolution) The Stieltjes convolution of functions F on R and G
on [0, ∞[ is
(F ∗ G) (x) :=
[0,∞[
F(x − y)dG(y).
As demonstrated, a Stieltjes convolution represents the distribution function of the sum of
two independent random variables with distributions F and G. The convolution of a function
F with itself shall be denoted by F∗2(x). Rather obviously, F∗2(x) describes the distribution
function of the sum of two independently and identically distributed (i.i.d.) random variables.
The distribution function of n i.i.d. random variables is given by the n:th convolution of F :
F∗n
(u) := F∗(n−1)
∗ F (x), n ∈ N.
11
12. The 0:th convolution has a special definition F∗0(x) := 1{x≥0}. For non-negative i.i.d. random
variables Xi, undoubtedly P(X1 + X2 > x) ≥ P(X1 > x) so
F∗n(x) ≤ F∗m(x) for all n ≥ m, n, m ∈ N ∪ {0}. (1)
F
F * 2
F * 3
0.01 0.02 0.03 0.04 0.05 0.06
0.2
0.4
0.6
0.8
1.0
x
Figure 3: The tail function of an exponential distribution and its second and third convolution.
Example 1.5. Suppose that Xi, i = 1, 2, 3, are i.i.d. and follow an exponential distribution with
rate 1/η. Then the tail function, expectation and variance of X1, X1 + X2 and X1 + X2 + X3
are the ones given in Table 2.
X1 X1 + X2 X1 + X2 + X3
F e−x/η e−x/η(1 + x/η) e−x/η 1 + x/η + x2
2η2
E η 2η 3η
Var η2 2η2 3η2
Table 2: The tail function, expectation and variance of an exponential distribution and its second
and third convolution.
Example 1.5 evinces that the tail function of an exponential distribution changes very system-
atically as the distribution is convoluted. Of course, the same is true for the expectation value
and variance because both E and Var (for uncorrelated random variables) are linear operators.
In fact, one can prove that the sum of i.i.d. exponential random variables follows an Erlang
distribution. Figure 3 depicts the three tail functions in Example 1.5.
1.5 Heavy-tailed distributions
Heavy-tailed distributions comprise probability distributions with at least one non-exponentially
bounded tail. For modelling claim sizes, it suffices to analyze distributions with heavy right tails.
12
13. Definition 1.6. A probability distribution with a distribution function F has a heavy right tail
if
lim
x→∞
eωx
F(x) = ∞ for all ω > 0 (2)
or, equivalently, if the moment generating function, M(s), does not exist (is not finite) for any
s > 0.
Proof of the equivalence in Definition 1.6: Suppose that lim
x→∞
eωxF(x) = ∞ for all ω >
0. Then, assuming that x > 0 and s > 0,
MF (s) =
[0,∞[
esy
dF(y) ≥
[x,∞[
esy
dF(y) ≥ esx
[x,∞[
dF(y) = esx
F(x)
x→∞
−−−→ ∞.
As for the proof to the other direction, suppose that M(s) = ∞ for all s > 0 and that
X = sup
x∈R
eˆsxF(x) < ∞ for some ˆs > 0. Let us approximate the Lebesgue–Stieltjes
integral [0,∞[ esydF(y) from above by dividing the interval [0, ∞[ into subintervals with
length one. Now, for all s < ˆs,
[0,∞[
esy
dF(y) ≤
∞
k=0
es(k+1)
≤F(k)
(F(k + 1) − F(k))
≤
∞
k=0
es(k+1)−ˆsk
≤X
eˆsk
F(k)
≤ Xes
∞
k=0
e(s−ˆs)k
=
Xes
1 − es−ˆs
< ∞.
But this cannot be, as M(s) = [0,∞[ esydF(y) = ∞ for all s > 0, and therefore X = ∞.
Since F(x) is finite for all x ∈ R ∪ {∞} and and esx is finite for all x ∈ R, clearly X
obtains its supremum as x approaches infinity.
Many typical heavy-tailed distributions fall in the class of subexponential distributions.
Definition 1.7. A distribution F on [0, ∞[ is subexponential if
F∗n(x)
F(x)
x→∞
−−−→ n, for all n ∈ N ∪ {0}. (3)
Subexponential random variables satisfy the so-called principle of a single big jump.
Proposition 1.8. (The principle of a single big jump) Let Xi be i.i.d. subexponential random
variables. Then, for any n ∈ N,
P max
1≤i≤n
{Xi} > x ∼ nF(x) as x → ∞.
13
14. Here f(x) ∼ g(x) means that limx→c
f(x)
g(x) = 1. Unless otherwise explicitly indicated, henceforth
it is assumed that c = ∞.
Proof : The independence of the random variables and the binomial theorem yield
1 − P max
1≤i≤n
{Xi} ≤ x = 1 −
n
i=1
P(Xi ≤ x)
= 1 − 1 − F(x)
n
= 1 −
n
i=0
n
i
−F(x)
i
= nF(x) −
n
i=2
n
i
−F(x)
i
∼ nF(x).
So, if Xi are i.i.d. subexponential random variables, then Definition 1.7 and Proposition 1.8
imply that for large x
P
n
i=1
Xi > x ∼ P max
1≤i≤n
{Xi} > x ,
that is, the cumulative sum n
i=1 Xi, given that it is big, mostly consists of its maximum com-
ponent. This principle of a single big jump can be thought of as a more extreme version of the
Pareto principle, which states that about 80 % of the effects come from 20 % of the causes. As
one might guess, Pareto distributions belong to the subexponential family.
All subexponential distributions on [0, ∞[ are also long-tailed, the definition of which is given
in Proposition 1.9. Furthermore, all long-tailed distributions are heavy-tailed.
Proposition 1.9. If F on [0, ∞[ is subexponential, then F is also long-tailed, that is, F(x−y)
F(x)
x→∞
−−−→
1 for any fixed y ∈ R.
Proof : The proof is similar to the proof of Proposition 1.5 in [4]. Suppose that 0 ≤ y < x.
Integration by parts leads to the following inequality:
F∗2(x)
F(x)
=
F(x) + F(x) − [0,x] F(x − z)dF(z)
F(x)
= 1 +
[0,x] F(x − z)dF(z)
F(x)
= 1 +
[0,y] F(x − z)dF(z)
F(x)
+
]y,x] F(x − z)dF(z)
F(x)
≥ 1 +
F(x) [0,y] dF(z)
F(x)
+ F(x − y)
]y,x] dF(z)
F(x)
= 1 + F(y) +
F(x − y)
F(x)
[F(x) − F(y)].
14
15. If lim sup
x→∞
F(x−y)
F(x)
> 1, then that and the inequality above imply that lim sup
x→∞
F
∗2
(x)
F(x)
> 2.
This violates the subexponentiality of F and hence, for y ≥ 0 it must hold that
lim sup
x→∞
F(x−y)
F(x)
≤ 1. On the other hand, if y ≥ 0, then, as F is a decreasing func-
tion, lim inf
x→∞
F(x−y)
F(x)
≥ 1, which concludes the proof of the proposition for y ≥ 0. If
F(x−y)
F(x)
x→∞
−−−→ 1 then obviously also F(x)
F(x−y)
x→∞
−−−→ 1 so the proposition holds for y < 0
as well.
Corollary 1.10. Any long-tailed distribution, and hence any subexponential distribution, is
also heavy-tailed.
Proof : Let be an arbitrarily small positive number. From Proposition 1.9 it follows that
F(n)/F(n − 1) ∼ 1 and thus for a sufficiently large n one gets F(n)/F(n − 1) ≥ e− /2.
Choose n0 ∈ N to be sufficiently large and assume that n > n0. Then
F(n) ≥ e− /2
F(n − 1) ≥ e−
F(n − 2) ≥ · · · ≥ e−(n−n0) /2
F(n0) ≥ e−n /2
F(n0).
On the other hand, F(n) ≥ F(n0) ≥ e−n /2F(n0) for all n ≤ n0. So there exists some
positive constant C such that F(x) ≥ Ce−x /2 for all x ≥ 0 and thus
F(x)e x
≥ Cex /2 x→∞
−−−→ ∞
1.6 Geometric Brownian Motion
A geometric Brownian motion St satisfies the stochastic differential equation
dSt = µStdt + σStdWt, (4)
where µ is the drift, σ is volatility and Wt is a standard Wiener process (or Brownian motion).
A geometric Brownian motion is an Itˆo drift-diffusion process so Itˆo’s lemma yields
df(t, St) =
∂
∂t
f(t, St) + µSt
∂
∂St
f(t, St) +
σ2S2
t
2
∂2
∂S2
t
f(t, St) dt + σSt
∂
∂St
f(t, St)dWt (5)
where f is a twice differentiable function. Applying (5) to the function ln(St) and then inte-
grating over [0, t] gives
St = S0e
µ−σ2
2
t+σWt
. (6)
As Wt follows a normal distribution with zero mean and volatility t, one can easily calculate
the expectation and variance for a geometric Brownian motion:
E[St] = S0eµt
; (7)
Var[St] = S2
0e2µt
(eσ2t
− 1). (8)
15
16. 2 The Cram´er–Lundberg model
In the Cram´er–Lundberg model, the risk reserve process Rt and the surplus process Yt are
defined by
Rt := u + ct −
Nt
k=0
Uk; (9)
Yt := u − Rt, (10)
where
• u is the initial capital,
• c > 0 is the premium rate,
• Nt is a homogeneous Poisson process with parameter λ > 0 and time t ≥ 0,
• Ui are i.i.d. non-negative random variables with a distribution function B and, if they
exist, moments η(n). These random variables are independent of Nt and represent the
claim sizes.
Rt
0.00 0.02 0.04 0.06 0.08 0.10
0.05
0.06
0.07
0.08
0.09
t
Figure 4: A sample path of a risk reserve process with exponential claims.
As Figure 4 indicates, in the Cram´er–Lundberg model both the size and the number of claims
are random. In between the claims, the risk reserve process increases linearly. Ruin occurs the
moment that Yt > u and thus the time to ruin with infinite time horizon may be expressed as
τu := inf
t≥0
{t|Yt > u}. (11)
The corresponding ruin probability has the following equivalent definitions:
16
17. ψ(u) := P inf
t≥0
Rt < 0 = P sup
t≥0
Yt > u = P(τu < ∞). (12)
With infinite time horizon, obviously the risk reserve process cannot exhibit negative drift if
certain ruin is to be avoided. Even a zero drift, as later will be proven, does not suffice, no
matter how large the initial reserve is.
The drift and the other moments of Rt can be calculated from the moment generating function
of Yt via the identity
E[Y n
t ] = M
(n)
Yt
(0),
where MYt is the characteristic function of Yt.
Proposition 2.1. Provided that MB(s) < ∞, the moment generating function of the surplus
process Yt is
MYt (s) = eλt(MB(s)−1)−cts
. (13)
Proof : Applying the law of total expectation and recalling that the claim sizes are i.i.d yield
MYt (s) := E[eYts
] = e−cts
E
E
e
s
Nt
k=1
Uk
Nt
= e−cts
E
Nt
k=1
E eUks
= e−cts
E MNt
B (s)
= e−cts−λt
∞
n=0
[λtMB(s)]n
n!
= eλt[MB(s)−1]−cts
.
The expectation and variance of Yt follow directly from Proposition 2.1 and (13):
E[Yt] = MYt
(0) = λtη − ct. (14)
Var[Yt] = MYt
(0) − E2
[Yt] = λtη(2)
. (15)
Let us denote ρ := λη
c . The ratio ρ describes the infinitesimal increase in expected liabilities
divided by the infinitesimal increase in assets. Expectation (14) of Yt suggests that the drift and
therefore the asymptotic behaviour of Yt may be inferred from the the value of ρ.
Proposition 2.2.
- If ρ > 1, then Yt
t→∞
−−−→
a.s.
∞
17
18. - If ρ < 1, then Yt
t→∞
−−−→
a.s.
− ∞
- If ρ = 1, then lim inf
t→∞
= −∞ and lim sup
t→∞
= ∞.
Proof : The proof closely follows the proof of Proposition 1.2 in [4]. Suppose that nα ≤ t ≤
(n + 1)α, where n ∈ N and α is some fixed positive constant. Then
• Yt = Ynα − (t − nα)c + Nt
k=Nαn+1 Uk ≥ Ynα − (t − nα)c
• Yt = Y(n+1)α + ((n + 1)α − t)c −
N(n+1)α
k=Nt+1 Uk ≤ Y(n+1)α + ((n + 1)α − t)c.
Now, Yαn =
n
k=1
Yαk − Yα(k−1) . Furthermore, because disjoint intervals of the Poisson
process are independent, and because the distribution of each interval only depends on
its length, it follows that
Yαk − Yα(k−1) = −αc +
Nαk
j=Nα(k−1)+1
Uj
D
= −αc +
Nα
j=1
Uj
for all k ∈ N. This implies that Yαk − Yα(k−1) are i.i.d. Equation (14) gives the expec-
tation E Yαk − Yα(k−1) = (λη − c) α = c(ρ − 1) so applying the strong law of large
numbers to Yα/n leads to
Yαn
n
n→∞
−−−→
a.s.
Yαn
n + 1
n→∞
−−−→
a.s.
Yα(n+1)
n
n→∞
−−−→
a.s.
E [Yα] = c (ρ − 1) α.
Hence
lim inf
t→∞
Yt
t
≥ lim inf
n→∞
inf
nα≤t≤(n+1)α
Yt
t
≥ lim inf
n→∞
inf
nα≤t≤(n+1)α
Ynα − (t − nα)c
t
≥ lim inf
n→∞
Ynα + nαc
(n + 1)α
− c
≥ lim inf
n→∞
Ynα
(n + 1)α
n→∞
−−−→
a.s.
c(ρ − 1).
Similarly,
lim sup
t→∞
Yt
t
≤ lim sup
n→∞
sup
nα≤t≤(n+1)α
Yt
t
≤ lim sup
n→∞
sup
nα≤t≤(n+1)α
Y(n+1)α + ((n + 1)α − t)c
t
≤ lim sup
n→∞
Y(n+1)α + (n + 1)αc
nα
− c
≤ lim sup
n→∞
Y(n+1)α
nα
n→∞
−−−→
a.s.
c(ρ − 1).
18
19. This means that Yt
t→∞
−−−→
a.s.
∞ if ρ > 1 and Yt
t→∞
−−−→
a.s.
−∞ if ρ < 1.
The case ρ = 1 calls for a slightly different kind of approach. Denote Y +
∞ := lim sup
n→∞
Ynα,
Y −
∞ := lim inf
n→∞
Ynα and let Y± be the event that Y ±
∞ is finite. Now the value of
Yαn = n
k=1 Yαj − Yα(j−1) does not depend on in which order the terms Yαj −Yα(j−1)
are summed so the Hewitt-Savage -zero-one law states that event Y± occurs with either
probability one or probability zero (hence the name of the law... ). Suppose that Y ±
∞ is
finite. Then, as lim sup
n→∞
(Yαn − U1) = Y +
∞ and lim inf
n→∞
(Yαn − U1) = Y −
∞, this would imply
that Y +
∞ − U1
D
= Y +
∞ and Y −
∞ − U1
D
= Y −
∞, which is impossible. Therefore Y +
∞ = ∞ and
Y −
∞ = −∞
Proposition 2.2 clearly implies that, with infinite time horizon, the ruin will occur almost surely
whenever ρ ≥ 1 no matter how large the initial capital u is. In contrast, if ρ < 1 (that is, the
surplus process Yt drifts negatively), the ruin probability, as later shall be proven, will be less
than one even when u = 0. For negative u, ruin of course occurs immediately, regardless of the
value of ρ. Henceforth it is assumed that ρ < 1 unless specified otherwise.
Rt
200000 400000 600000 800000 1 × 10 6
- 50
50
100
t
Figure 5: For ρ = 1, the risk reserve process oscillates around the initial reserve.
2.1 Integro-differential representations of the ruin probability and the Pol-
laczek–Khinchin formula
Let us denote φ(u) := 1−ψ(u). The function φ(u) is called the survival probability. Calculating
with φ rather than ψ in many instances facilitates analysis. For example, φ and B allow chang-
ing their convolution order but ψ and B do not because ψ(u) = 1 for all u < 0.
The survival functions satisfy the following integro-differential equations.
19
20. Theorem 2.3. Suppose that ρ ≤ 1. Then
I. φ (u) =
λ
c
φ(u) − (φ ∗ B)(u) ;
II. φ(0) = 1 − ρ;
III. φ(u) = φ(0) + ρ(φ ∗ BI)(u).
Proof : The proof follows closely the proof of Proposition 1.1.1 in [11]. Let B be the event ”no
ruin for infinite time horizon with initial capital u”. The time to the first claim, T1, is
exponentially distributed with rate λ. Conditioning with respect to the time and size
of the first claim, T1 and U1, and applying the transformation ˜t = u + ct yield
φ(u) =
[0,∞[
λe−λt
[0,∞[
P(B|T1 = t, U1 = x)dB(x)dt
=
[0,∞[
λe−λt
[0,u+ct]
φ(u + ct − x)dB(x)dt
=
λ
c
e
λ
c
u
[u,∞[
e−λ
c
˜t
(φ ∗ B)(˜t)d˜t.
Clearly φ(u) differentiates, and differentiating leads to
φ (u) =
λ
c
φ(u) −
λ
c
(φ ∗ B)(u),
which proves I. Integrating both sides over [0, u], applying a change of variable ˜x = u−x,
integrating by parts and changing the integration order give
c
λ
[φ(u) − φ(0)] =
[0,u]
φ(x)dx −
[0,u] [0,x]
φ(x − z)dB(z)dx
=
[0,u]
φ(u − ˜x)d˜x −
[0,u]
φ(0)B(x)dx −
[0,u] [0,x]
φ (x − z)B(z)dzdx
=
[0,u]
φ(u − ˜x)d˜x −
[0,u]
φ(0)B(x)dx −
[0,u]
B(z)[φ(u − z) − φ(0)]dz
=
[0,u]
φ(u − z)B(z)dz.
Now B(z)dz = ηdBI(z), so
φ(u) = φ(0) +
λη
c [0,u]
φ(u − z)dBI(z)
= φ(0) + ρ(φ ∗ BI)(u),
which proves III. This is called the defective renewal equation of φ.
Evidently, only finitely many claims can occur before time t < ∞ and so sup0≤t<∞ Yt <
20
21. ∞ Moreover, Proposition 2.2 suggests that Yt/t
t→∞
−−−→
a.s.
c(1 − ρ), which again is finite.
Therefore supt≥0 Yt < ∞. This leads to the limit
lim
u→∞
φ(u) = lim
u→∞
P sup
t≥0
Yt ≤ u = 1.
Monotone convergence gives
1 = lim
u→∞
φ(u) = φ(0) + ρ
[0,∞[
=1
lim
u→∞
φ(u − z) dBI(z)
= φ(0) + ρ,
which proves II.
Perhaps not so obviously, Theorem 2.3 states that ruin can be avoided even with zero initial
reserve. Even less obviously, the theorem asserts that for u = 0 the ruin probability, in addition
to the Poisson rate and the premium rate, only depends on the expectation of the claim size
distribution. The claim size’s variance, skewness, etc. are irrelevant. Moreover, Theorem 2.3
reduces solving the ruin probability into a purely analytical problem. Solving ψ from III is pos-
sible, but generally the solution involves an infinite sum of convolutions of BI. In some special
cases, such as the exponential claims case 1 a more satisfying solution to ψ can be obtained
from the theorem.
Aside from exponential claims, fairly concise solutions to I in Theorem 2.3 exist for degen-
erate claims and whenever the minimum claim size exceeds the initial capital, that is, whenever
B(u) = 0.
Corollary 2.4. If B(u) = 0 and ρ ≤ 1, then ψ(u) = 1 − (1 − ρ)e
λ
c
u
.
Proof : When B(u) = 0, I in Theorem 2.3 reduces to φ (u) = λ
c φ(u) and solving this with the
initial condition φ(0) = 1 − ρ proves the corollary.
Corollary 2.5. If the claim sizes Ui are degenerate at η, then
ψ(u) = 1 − (1 − ρ)
u/η
k=0
[ρ(k − u/η)]k
k!
e−ρ(k−u/η)
(16)
Proof : Ignoring some minor, mostly notational, alterations, the proof is directly from [4] (proof
of Corollary 3.7).
If u < η, then B(u) = 0 and one easily sees that the ruin probability in (16) equals that
in Corollary 2.4. As for the case u ≥ η, since Ui are degenerate, then B(y) = 1{y≤η}.
Recall that dBI(y) = B(y)dy/η. Integral equation III in Theorem 2.3 and a change of
variable ˜y = u − y lead to the following representation of the survival probability:
1
see, for instance, [11] p. 6
21
22. φ(u) = 1 − ρ +
ρ
η [0,u]
φ(u − y)1{y≤η}dy
= 1 − ρ +
ρ
η [0,η]
φ(u − y)dy
= 1 − ρ +
ρ
η [u−η,u]
φ(˜y)d˜y.
Differentiating this produces φ (u) = ρ
η [φ(u) − φ(u − η)]. The corollary can be proved
using induction. Assume that (16) is shown for (n − 1)η ≤ u < nη, n ∈ N (for n=1
(16) was already proven in Corollary 2.4). For nη ≤ u < (n + 1)η, one needs to show
that φ (u) = ρ
η [φ(u) − φ(u − η)]. From (16),
φ(u) = (1 − ρ)
n
k=0
[ρ(k − u/η)]k
k!
e−ρ(k−u/η)
;
φ (u) =
ρ
η
φ(u) − (1 − ρ)
n
k=1
[ρ(k − u/η)]k−1
(k − 1)!
e−ρ(k−u/η)
=
ρ
η
φ(u) − (1 − ρ)
n−1
j=0
[ρ(j − u−η
η )]j
j!
e
−ρ(j−u−η
η
)
=
ρ
η
[φ(u) − φ(u − η)].
Pollaczek–Khinchin formula gives the general solution to the ruin probability.
Proposition 2.6. (Pollaczek–Khinchin formula) For u ≥ 0,
ψ(u) = (1 − ρ)
∞
k=0
ρk
B∗k
I (u). (17)
Proof : Recoursing III in Theorem 2.3 n times produces
φ(u) = (1 − ρ)
n−1
k=0
ρk
B∗k
I (u) + ρn
φ ∗ B∗n
I (u).
As was discussed in Section 1.4, the distribution function of the sum n
i=1 UI
i , where
UI
i are i.i.d. and have a distribution function BI, is given by the convolution power B∗n
I .
Since B∗n
I is a distribution function, it must hold that 0 ≤ B∗n
I ≤ 1 for all n ∈ N and
u ∈ R. What is more, since φ(u) ≤ 1 for all u ∈ R, it holds that [0,u] φ(u−y)dB∗k
I (y) ≤
B∗k
I (u). Therefore
ρn
φ ∗ B∗n
I ≤ ρn
B∗n
I ≤ ρn n→∞
−−−→ 0
22
23. and letting n → ∞ produces φ(u) = (1 − ρ) ∞
k=0 ρkB∗k
I (u). As ρ < 0, it holds that
1 = (1 − ρ) ∞
k=0 ρk and so
ψ(u) = 1 − φ(u) = (1 − ρ)
∞
k=0
ρk
B∗k
I (u).
2.2 Ruin for light-tailed claim sizes
Suppose that the moment generating function MB(s) := E[esU ] of the claim size distribution
exists for at least some s > 0. Proposition (2.1) yields
MYt (s) = eθ(s)t
, (18)
where θ(s) := λ(MB(s) − 1) − cs is the Lundberg function. If MB(s0) < ∞ for some s0 > 0,
then θ(s) is smooth on ] − ∞, s0[. Furthermore, the function θ(s) is convex on ] − ∞, s0[ as
θ (s) = λMB(s) = λE U2esU > 0. Figure 6 depicts these properties.
θ (s)
Degenerate
Uniform
Exponential
10 20 30 40 50
- 2
2
4
6
8
10
s
Figure 6: The Lundberg function in degenerate, uniform and exponential cases.
That MYt (s) < ∞ for some s > 0 bodes well because if one can find a unique γ > 0 such that
θ(γ) = 0, then one can also calculate uniform decreasing upper and lower bounds for the ruin
probability. The upper bound always decreases exponentially and, quite frequently, so does the
lower bound. The following lemma 2 states conditions for the existence of γ. Theorem 2.8 3 then
provides the bounds.
2
Lemma 3.1 (p. 31) in [12]
3
Theorem 3.2 (p. 31) in [12]
23
24. Lemma 2.7. If there exists s∞ ∈ R ∪ {∞} such that MB(s) < ∞ for all s < s∞ and
MB(s s∞) = ∞, then a unique positive solution γ to the equation θ(s) = 0 exists.
Proof : Since θ(0) = 0, θ (0) = λη − c = c(ρ − 1) < 0 and θ is convex and approaches infinity
when s s∞, θ intersects the positive real axis once. When s∞ < ∞, clearly θ
approaches infinity. On the other hand, when s∞ = ∞, acknowledge that MB(s) grows
supralinearly because both the first and second derivatives of MB(s) are positive.
The constant γ is referred to as the adjustment coefficient. The adjustment coefficient satisfies
the following tremendously useful identity:
ρ
[0,∞[
eγy
dBI(y) = 1. (19)
Proof of (19): Because θ(γ) = 0, then MB(γ) − 1 = cγ
λ . Reshaping the term dBI(y) and
changing the integration order lead to
[0,∞[
eγy
dBI(y) =
1
η [0,∞[
eγy
d
[0,y]
B(z)dz
=
1
η [0,∞[
eγy
B(y)dy
=
1
η [0,∞[
eγy
[y,∞[
dB(z)dy
=
1
η [0,∞[ [0,z]
eγy
dydB(z)
=
1
ηγ
(MB(γ) − 1)
=
1
ρ
.
Theorem 2.8. Suppose that γ > 0 exists. Then for all u ≥ 0
a−e−γu
≤ ψ(u) ≤ a+e−γu
,
where
a− := inf
x∈[0,b[
eγxBI(x)
]x,∞[ eγydBI(y)
, a+ := sup
x∈[0,b[
eγxBI(x)
]x,∞[ eγydBI(y)
and
b := sup{z|BI(z) < 1}.
24
25. Proof : Define h0(u) := ae−γu for u ≥ 0 and h0(u) = 1 for u < 0, where a ∈]0, 1]. Further
define hk(u) := ρBI(u) + ρ [0,u] hk−1(u − y)dBI(y) for u ≥ 0 and hk(u) = 1 for u < 0,
where k ∈ N. Evidently 0 ≤ h0(u) ≤ 1 for all u ∈ R. As
hk(u) = ρBI(u) + ρ
[0,u]
hk−1(u − y)dBI(y)
the tail function of hk may be represented
hk+1(u) = 1 − ρ + ρ hk ∗ BI (u).
Following similar steps as in the proof of Proposition 2.6 results in
hk→∞(u) = (1 − ρ)
∞
k=0
ρk
BI(u) = φ(u),
or, equivalently, hk→∞(u) = ψ(u).
Suppose that hk(u) ≤ hk−1(u). Then
hk+1(u) = ρBI(u) + ρ
[0,u]
hk(u − x)dBI(x) ≤ ρBI(u) + ρ
[0,u]
hk−1(u − x)dBI(x) = hk.
Therefore, if h1(u) ≤ h0(u), induction proves that hk(u) ≤ hk−1(u) for all k ∈ N and
hence ψ(u) ≤ h0(u). Proving the upper bound requires finding a constant a such that
h1(u) ≤ h0(u). If u ≥ b, then BI(u) = 1 and (19) gives that
h1(u) = aρ
[0,∞[
e−γ(u−y)
dBI(y) = ae−γu
= h0(u).
As for 0 ≤ u < b, via (19), the function h0 may be expressed
h0(u) = ae−γu
ρ
[0,∞[
eγy
dBI(y) = aρ e−γu
]u,∞[
eγy
dBI(y) +
[0,u]
e−γ(u−y)
dBI(u) .
On the other hand,
h1(u) = ρ BI(u) + a
[0,u]
e−γ(u−y)
dBI(u)
so for h0 ≥ h1 to hold, it suffices to find constant a such that
ρ ae−γu
]u,∞[
eγy
dBI(y) +
$$$$$$$$$$$
a
[0,u]
e−γ(u−y)
dBI(u) ≥ ρ BI(u) +
$$$$$$$$$$$
a
[0,u]
e−γ(u−y)
dBI(u) .
So, that h0 ≥ h1 holds requires that a ≥ eγuBI (u)
]u,∞[ eγydBI (y)
holds. Since this inequality
must hold for all u ∈ [0, b[, a+ must be the supremum of the term eγuBI (u)
]u,∞[ eγydBI (y)
. This
concludes the proof for the upper bound. The proof for the lower bound is analogous.
Corollary 2.9. 0 ≤ a− ≤ ρ and ρ ≤ a+ ≤ 1.
25
26. Proof : Obviously a− ≥ 0. Additionally, a+ ≤ sup
x∈[0,b[
]x,∞[ eγydBI (y)
]x,∞[ eγydBI (y)
= 1 as eγxBI(x) ≤ ]x,∞[ eγydBI(y)
for all x ≥ 0.
As for a+ ≥ ρ, choose x = 0 and recall identity (19) to obtain
a+ ≥
BI(0)
]u,∞[ eγydBI(y)
=
1
1/ρ
= ρ.
Same approach also gives a− ≤ ρ.
The constant a− might equal zero because nothing prohibits the ruin probability from decreas-
ing non-exponentially. For instance, as was proven in 2.5, when B(u) = 0 the ruin probability
decreases more quickly than exponentially.
Figure 6 illustrates the function θ(s) for degenerate, uniform and exponential claims sizes. The
adjustment coefficient γ appears to be, unsurprisingly, the smallest for exponential claims which
is equivalent with exponential claims having the most slowly decreasing bounds. As discussed in
Section 1.2, of these three distributions the heaviest tail belongs to the exponential distribution.
For exponentially distributed claim sizes, the bounds in Theorem 2.8 collapse, resulting in
an analytical representation of the ruin probability.
Corollary 2.10. If the claim sizes are exponentially distributed with rate 1/η, then
ψ(u) = ρe
−1−ρ
η
u
.
Proof : The moment generating function of the exponential distribution is
MExp(s) =
1
η [0,∞[
e(s−1/η)x
dx =
1
1 − ηs
for s < 1/η. Therefore
θ(s) = λ
1
1 − ηs
− 1 − cs = s
λη
1 − ηs
− c .
The function θ has roots s1 = 0 and s2 = 1−λη/c
η = 1−ρ
η . Clearly γ = s1 as s1 is not
positive. For s2 to be the adjustment coefficient, it must be positive but also smaller
than 1/η because otherwise MExp(s) would not exist. But since 0 < ρ < 1, evidently
0 < s2 = 1−ρ
η < 1/η and thus γ = s2 = 1−ρ
η .
The exponential distribution has expectation MExp(0) = η(1−η ×0)−2 = η so BI(u) =
1
η [0,u] e−x/ηdx = 1 − e−x/η. Recall that dBI(y) = 1
η B(y)dy = 1
η e−y/ηdy, and compute
a+ = sup
x∈[0,b[
eγxBI(x)
]x,∞[ eγydBI(y)
= sup
x∈[0,b[
e¡¡
x
η
−ρx
η
&
&&
e
−x
η
1
η ]x,∞[ e¡¡
y
η
−ρy
η
&
&&
e
− y
η dy
= sup
x∈[0,b[
e
−ρx
η
1
ρe
−ρx
η
= ρ.
The proof that a− = ρ follows similarly.
26
27. In addition to the bounds, the existence of the adjustment coefficient provides a limit for the ruin
probability. Proving this requires the following well-known lemma, the so-called Key Renewal
Theorem 4.
Lemma 2.11. (Key Renewal Theorem) Suppose that function L(u) can be expressed through
the renewal equation
L(u) = v(u) +
[0,u]
L(u − z)dF(z),
where function L on [0, ∞[ is unknown, function v is known and function F on [0, ∞[ is the
distribution of some known random variable X. Then, if
• F is non-lattice, that is, there exists no α ≥ 0 such that ∞
k=1 P (X = kα) = 1,
• v is directly Riemann-integrable 5,
there exits a limit
L(u)
u→∞
−−−→
[0,∞[ v(z)dz
[0,∞[ zdF(z)
.
Theorem 2.12. (Cram´er–Lundberg approximation) If the adjustment coefficient γ exists, then
ψ(u)eγu u→∞
−−−→
1 − ρ
λ
c MB(γ) − 1
where 1−ρ
λ
c
MB(γ)−1
> 0.
Proof : With some modifications, the proof is from [4] (Alternative proof of Cram´er–Lundberg’s
approximation, p. 90). Recalling that ηdBI(y) = B(y)dy, III in Theorem 2.3 gives
ψ(u)eγu
= ρeγu
BI(u) +
λ
c [0,u]
eγ(u−y)
ψ(u − y)eγy
B(y)dy. (20)
Identity (19) attests that λ
c eγuB(u) is the density function of a proper probability
distribution, rendering (20) a proper renewal equation. Applying the Key Renewal
Theorem and choosing L(u) = eγuψ(u), v(u) = ρeγuBI(u) and dF(u) = λ
c eγyB(u)du
produce
ψ(u)eγu u→∞
−−−→
ρ [0,∞[ eγyBI(y)dy
λ
c [0,∞[ yeγyB(y)dy
.
For the numerator, changing the integration order gives
ρ
[0,∞[
eγy
BI(y)dy =
λ
c [0,∞[ ]y,∞[
eγy
B(z)dzdy =
λ
cγ [0,∞[
(eγy
− 1)B(z)dz
Again, recall that ηdBI(y) = B(y)dy and then apply identity (19) to compute
λ
c [0,∞[
eγy
B(y)dy = ρ
[0,∞[
eγy
dBI(y) = 1.
4
Proposition A1.1, in [4]
5
for an explanation of direct Riemann-integrability, see [15]
27
28. Thus, as [0,∞[ B(z)dz = η, it follows that
ρ
[0,∞[
eγy
BI(y)dy =
1 − λ
c [0,∞[ B(z)dz
γ
=
1 − ρ
γ
.
As for the denominator, note that MB(γ) − 1 = cγ
λ and [0,∞[ zeγzdB(z) = MB(γ),
change the integration order and integrate by parts to get
λ
c [0,∞[
yeγy
]y,∞]
dB(z)dy =
λ
c [0,∞[ [0,z[
y
d [eγy]
γ
dB(z)
=
λ
cγ [0,∞[
zeγz
−
[0,z[
eγy
dy dB(z)
=
λ
cγ [0,∞[
zeγz
−
eγz − 1
γ
dB(z)
=
λ
cγ
MB(γ) −
1
γ
(MB(γ) − 1)
=
1
γ
λ
c
MB(γ) − 1 .
To prove that 1−ρ
λ
c
MB(γ)−1
> 0, note that θ (s) = λMB(s)−c and that θ (γ) > 0 because γ
is a unique positive root to θ(s) = 0, θ(0) = 0, θ (0+) < 0 and θ is convex on [0, ∞[
Lundberg bounds
Cramér–Lundberg approximation
Figure 7: Lundberg bounds and the Cram´er–Lundberg approximation when the claim sizes
follow a uniform distribution.
Example 2.13. Choose example values (see Section 1.1) for the parameters. Suppose that U
in uniform on [0, 1/60]. Then γ ≈ 31.4163. Theorems 2.8 and 2.12 yield 5
6e−γu ≤ ψ(u) ≤ e−γu
and ψ(u)eγu u→∞
−−−→ 0.8759.
In Example 2.13, note that 5/6 < 0.8759 < 1. So in this case Theorem 2.12 provides a reasonably
accurate approximation of the ruin probability even for small u because the approximation stays
within the the Lundberg bounds and the Lundberg bounds themselves are not too far apart.
Figure 7 illustrates Example 2.13 graphically.
28
29. 2.3 Heavy-tailed claim sizes
As finite moment generating functions do not exist for heavy-tailed distributions, the adjust-
ment coefficient cannot be calculated whenever the claims sizes follow one of these distributions.
This means that all the previously calculated bounds and asymptotes vanish for heavy-tailed
claims. However, as discussed earlier, many heavy-tailed distributions fall into the family of
subexponential distributions and claim sizes belonging to this family typically permit deriving
an alternative asymptote.
Assuming subexponential claims, producing a limit for the ruin probability requires two lemmas.
Lemma 2.14. Let F be the distribution function of a subexponential distribution on [0, ∞[.
Then, for every > 0 there exists a constant C such that F∗n(x)
F(x)
≤ C (1 + )n for x ≥ 0 and
n ≥ 1.
Proof : The proof is from [4] (proof of Lemma 1.8). Define αn := supx≥0
F∗n(x)
F(x)
and choose x
such that
x
0
F(x−y)
F(x)
dF(y) ≤ 1 + for all x ≥ x (note that F(x−y)
F(x)
1 as x → ∞).
Then
αn+1 = sup
x≥0
F(x) + F(x) − F∗(n+1)(x)
F(x)
= 1 + sup
x≥0
x
0
F∗n(x − y)
F(x)
dF(y)
≤ 1 + sup
0≤x≤x
1
F(x)
≤1
x
0
F∗n(x − y)dF(y) + sup
x>x
x
0
≤αn
F∗n(x − y)
F(x − y)
F(x − y)
F(x)
dF(y)
≤ 1 +
1
F(x )
+ αn sup
x>x
x
0
F(x − y)
F(x)
dF(y)
≤ 1 +
1
F(x )
+ αn(1 + ).
Recoursing all the way to α1 = 1 results in
αn+1 ≤ (1 + )n
+ 1 +
1
F(x )
n−1
k=0
(1 + )k
= (1 + )n
+ 1 +
1
F(x )
(1 + )n − 1
= (1 + )n+1 1
1 +
1 +
1 + 1
F(x )
1 −
1
(1 + )n
≤ (1 + )n+1 1
1 +
1 +
1 + 1
F(x )
.
Set C = 1
1+ 1 +
1+ 1
F (x )
.
29
30. Lemma 2.15. Let Xi be i.i.d. non-negative subexponential random variables with a distribution
function F and let J ∈ N ∪ {0} be an independent random variable that satisfies E[zJ ] < ∞ for
some z > 1. Then
P
J
i=1
Xi > x
F(x)
x→∞
−−−→ E[J]. (21)
Proof : The proof is from [4] (proof of Lemma 2.2). Applying the definition of subexponentiality,
the law of total probability and the dominated convergence theorem give
P
J
i=1
Xi > x
F(x)
=
∞
n=0
P
J
i=1
Xi > x|J = n P (J = n)
F(x)
=
∞
n=0
x→∞
−−−→n
F∗n(x)
F(x)
P(J = n)
x→∞
−−−→ E[J].
Lemma 2.14 permits using dominated convergence as
∞
n=0
F∗n(x)
F(x)
P(J = n) ≤ C E (1 + )J
< ∞.
Theorem 2.16. If BI is subexponential, then ψ(u) ∼ ρ
1−ρBI(u).
Proof : Reformulate the The Pollaczek–Khinchin formula (17) into
ψ(u) =
∞
n=0
P(J = n)P
n
i=1
UI
i > u = P
J
i=1
UI
i > u , (22)
where P(J = n) = (1−ρ)ρn is the probability mass function of a geometric probability
distribution with parameter 1 − ρ and where UI
i are i.i.d. random variables with a
distribution function BI. The random variable J has expectation
E[J] = (1 − ρ)
∞
k=0
kρk
= (1 − ρ)ρ
∞
k=0
kρk−1
= (1 − ρ)ρ
d
dρ
∞
k=0
kρk
= (1 − ρ)ρ
d
dρ
1
1 − ρ
=
ρ
1 − ρ
and E[zJ ] < ∞ when ρz < 1 so applying Lemma 2.15 to (22) proves the theorem.
Generally, it does not hold that the integrated tail distribution of a subexponential distribution
is subexponential too or vice versa. For the most common subexponential distributions, this
equivalence does hold however. Furthermore, the tail function of a subexponential distribution’s
integrated tail distribution is heavier than the underlying distribution’s.
30
31. Proposition 2.17. If B is subexponential, then limu→∞
BI (u)
B(u)
= ∞.
Proof : The proof is from [4] (proof of Proposition 2.3). Proposition 1.9 indicates that
B(u + y) ∼ B(u) for y ∈ R so
lim inf
u→∞
∞
u B(z)dz
ηB(u)
≥ lim inf
u→∞
B(u + y)
u+y
u dz
ηB(u)
≥ lim inf
u→∞
u+y
u dz
η
=
y
η
y→∞
−−−→ ∞.
Rt
0.2 0.4 0.6 0.8 1.0
- 0.10
- 0.05
0.05
0.10
0.15
0.20
Degenerate Uniform Exponential Pareto
t
Rt 20 40 60 80 100
- 20
- 10
10
Degenerate Uniform Exponential Pareto
t
Figure 8: Sample paths of the risk reserve process for a one-year and a 100-year time horizon.
31
32. 2.4 Comparisons and considerations
Let us compare Deg(η)-distributed, U(0, 2η) -distributed, Exp(1/η)-distributed and Par(α−1
α η, α)-
distributed claims. Choose example values (c = 1, λ = 100, η = 1/120, α = 3/2) for the param-
eters
Figure 8 depicts the simulated risk-reserve processes for all four distributions. Sample paths
for both a one-year and a 100-year time horizon are simulated. The risk reserve process with
Pareto-distributed stands out, especially for the 100-year time horizon. In contrast to the other
three cases, the Pareto case seems to portray a certain degree of scale invariance.
ψ
Degenerate
Uniform
Exponential
0.02 0.04 0.06 0.08 0.10
0.2
0.4
0.6
0.8
u
ψ
Pareto
Exponential
0.0 0.5 1.0 1.5 2.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
u
Figure 9: Ruin probability as a function of the initial reserve for degenerate, uniform, exponential
and Pareto claims.
Let initial capital u∗ be such that the ruin probability equals 0.2. Table 3 illustrates the one-
year standard deviation (SD) of the risk reserve process, the adjustment coefficient and the
initial reserve u∗ in all four cases. The standard deviations are calculated from (15). The initial
reserve u∗ is computed from Corollary 2.5 for degenerate claims and from Corollary 2.10 for
exponential cases. For uniform claims, u∗ has been approximated from Theorem (2.12) and for
Pareto claims from Theorem 2.16. The (approximated) ruin probabilities, as a function of the
32
33. initial reserve u, are depicted in Figure 9.
SD γ u∗
Degenerate 0.0833 42.5 0.0351
Uniform 0.0962 31.4 0.0470
Exponential 0.118 20.0 0.0714
Pareto ∞ - 0.772
Table 3: Results.
Predictably, degenerate claims are safest whereas Pareto claims are, by quite a margin, the
most dangerous for an insurer. Compared to degenerate claims, u∗ is 33 % larger in the uniform
case, 103 % larger in the exponential case and a whopping 2200 % larger in the Pareto case.
Although u∗ is only approximated in the Pareto case, surely this result alludes that heavy-tailed
claim sizes complicates things for an insurer. Achieving a fairly low ruin probability suggest that
a very large initial reserve is required. Holding a large initial reserve, either in cash or some
low-yielding safe asset, of course has its disadvantages.
33
34. 3 Diffusion perturbed model
Now let us extend the Cram´er–Lundberg model by adding a diffusion term to the risk reserve
process (9):
Rt = u + ct + ςVt −
Nt
k=1
Uk (23)
Yt = u − Rt. (24)
Here Vt is a standard Wiener process and ς > 0 is the volatility of diffusion. The term Vt is
assumed to be independent of claim sizes and the number of claims. Figure 10 illustrates the
dynamics of a diffused and a non-diffused risk reserve processes.
Rt
0.02 0.04 0.06 0.08 0.10
0.01
0.02
0.03
0.04
Non - diffused
Diffused
Pure diffusion
t
Figure 10: Sample paths of the risk reserve process.
Proposition 2.1 suggests that the analogous moment generating function of the surplus process
Yt is
MYt (s) = eλt(MB(s)−1)−cts+ς2ts2/2
and so the expectation and the variance of Yt are
E [Yt] = (λη − c) t;
Var(Yt) = λtη(2)
+ ς2
t. (25)
Expectedly, including of a zero-drift diffusion term does not alter the expectation of the surplus
process but is does increase its variance. The inequality ρ < 1 still needs to hold to avoid certain
ruin.
With the introduction of a diffusion term, either a claim or the surplus process Yt diffusing
above u may cause ruin. This motivates decomposing the ruin probability as follows:
ψ(u) = ψc(u) + ψd(u).
Here ψc(u) represents the probability of claim-induced ruin and ψd represents the probability
of diffusion-induced ruin. The moment that Yt = u, ruin occurs almost surely due to the path
34
35. properties of Vt near zero. Therefore τu = inf t Yt ≥ u gives the moment of ruin and the two
ruin probabilities have the representations:
ψd(u) = P (τu < ∞, Yτu = u) ; (26)
ψc(u) = P (τu < ∞, Yτu > u) . (27)
From equation (26) one easily infers that ψ(0) = ψd(0) = 1 and so ψc(0) = 0. As opposed to
the non-diffused case, where ψ(0) = ρ, ruin occurs almost surely with zero initial reserve if the
surplus process is diffused. Moreover, (27) yields ψ(u) = ψc(u) = 1 and ψd(u) = 0 for u < 0.
As in the non-diffused case, the ruin probability satisfies certain integro-differential equations.
First, define FE(u) := 1−e−2cu/ς2
, which is the distribution function of a Exp(2c/ς2)-distribution.
Proposition 3.1. If the survival probability φ(u) differentiates twice and η is finite, then
I.
ς2
2
φ (u) + cφ (u) + λ (φ ∗ B) (u) − λφ(u) = 0 for u = 0;
II. φ (0+
) =
2c(1 − ρ)
ς2
;
III. φ(u) = (1 − ρ)FE(u) + ρ (φ ∗ FE ∗ BI) (u).
Proof : The proof is inspired by the proof of Theorem 2.4 in [14]. Let Zt := u + ct + ςVt and
T1 be the time when the first claim occurs. Furthermore, let A := {inft≥0 Rt > 0} be
the event ”ruin never occurs” and define conditional probability Mt := E[1A|Ft] and
stopping time Tt := inf{v|Zv ≤ 0}∧t. Clearly φ(u) = M0. Moreover, Mt is martingale:
E[Mt|Fs] = E E[1A|Ft] Fs = E[1A|Ft ∩ Fs] = E[1A|Fs] = Ms.
Hence optional stopping theorem gives φ(u) = E[M0] = E[MTt∧T1 ]. The process RTt∧T1
also satisfies the strong Markov property so
φ(u) = E [MTt∧T1 ] = E [E [MTt∧T1 |FTt∧T1 ]] = E[φ(RTt∧T1 )].
Recall that RT1
D
= ZT1 − U1 and Z ⊥ T1. Expanding the expectation above produces
φ(u) = E φ(ZTt )1{t<T1} + E φ(RTt∧T1 )1{T1≤t}
= e−λt
E [φ(ZTt )] + E φ(ZTt )1{Tt<T1}1{T1≤t} + E φ(ZT1 − U1)1{Tt≥T1}1{T1≤t} .
Conditioning the two latter terms with respect to T1 and the third term further with
respect to U1, and then applying Itˆo’s Lemma to the first term yield
φ(u) = e−λt
E[φ(ZTt )] +
[0,t]
λe−λs
E φ(ZTs )1{Ts<s} + (φ ∗ B) (Zs)1{Ts≥s} ds
= e−λt
E φ(u) +
[0,t]
cφ (ZTs ) +
ς2
2
φ (ZTs ) ds + e−λt
=0
E
[0,t]
ςφ (ZTs )dWs
+
[0,t]
λe−λs
E φ(ZTs )1{Ts<s} + (φ ∗ B) (Zs)1{Ts≥s} ds
35
36. This rearranges into
1 − e−λt
t
φ(u) =
1
teλt
E
[0,t]
cφ (ZTs ) +
ς2
2
φ (ZTs ) ds
+
1
t [0,t]
λe−λs
E φ(ZTs )1{Ts<s} + (φ ∗ B) (Zs)1{Ts≥s} ds.
Let t 0 and note that Tt
t 0
−−→ 0. L’Hospital’s rule then gives
λφ(u) = cφ (u) +
ς2
2
φ (u) + λ (φ ∗ B) (u),
which completes the proof of I.
For II, integrate φ(x) over [0, u], integrate by parts and change the integration order
to obtain
ς2
2
[φ (u) − φ(0+
)] = c
=0
φ(0) −cφ(u) + λ
[0,u]
φ(x)dx + λ
[0,u] [0,x]
φ(x − y)dB(y)dx
= −cφ(u) +
¨¨
¨¨¨
¨¨
λ
[0,u]
φ(x)dx + λ
[0,u]
φ(0)B(u) −
&
&
&
&
&
φ(x)
=1
B(0)
dx
−λ
[0,u] [0,x]
B(y)dφ(x − y)dx
= −cφ(u) + λ
[0,u]
B(y)
[y,u]
φ (x − y)dxdy
= −cφ(u) + λ
[0,u]
B(y)φ(u − y)dy. (28)
Since η is finite, then [0,∞[ B(y)dy = η. Also, φ (u → ∞) = 0 and φ(u → ∞) = 1 so
letting u → ∞ gives
−
ς2
2
φ (0+
) = −c + λη,
which concludes the proof for II. For III, rearrange (28) and then multiply both sides
by e2c/ς2x to obtain
e
2c
ς2 x
φ (x) +
2c
ς2
φ(x) =
2c(1 − ρ)
ς2
e
2c
ς2 x
+
λη
c [0,x]
2c
ς2
e
2c
ς2 x
φ(x − y)dBI(y).
Integrating over [0, u] produces
e
2c
ς2 u
φ(u) = (1 − ρ) e
2c
ς2 u
− 1 + ρ
[0,u] [0,x]
2c
ς2
e
2c
ς2 x
φ(x − y)dBI(y)dx.
As FE(x) = 1−e
− 2c
ς2 x
, then FE(u−x) = 2c
ς2 e
− 2c
ς2 (u−x)
. Now dividing both sides by e
2c
ς2 u
,
36
37. changing the integration order and applying a change of variable ˜x := u − x yield
φ(u) = (1 − ρ)FE(u) + ρ
[0,u] [0,x]
FE(u − x)φ(x − y)dBI(y)dx
= (1 − ρ)FE(u) + ρ
[0,u] [y,u]
FE(u − x)φ(x − y)dxdBI(y)
= (1 − ρ)FE(u) + ρ
[0,u] [0,u−y]
φ(u − y − ˜x)FE(˜x)d˜xdBI(y)
= (1 − ρ)FE(u) + ρ (φ ∗ FE ∗ BI) (u).
For the case ς = 0, Theorem 2.3 in [14] establishes conditions for the differentiability of ψ. For
the case ς > 0, [6] points out that conditions for the differentiability ψ are yet to be determined.
Henceforth, this thesis will simply assume that ψ is twice continuously differentiable.
Proposition (3.1) induces several corollaries.
Corollary 3.2. If the minimum claim size exceeds the initial capital (B(u) = 0), then
φ(u) =
c(1 − ρ)
√
c2 + 2λς2
e
√
c2+2λς2−c
ς2 u
− e
−c+
√
c2+2λς2
ς2 u
Proof : If B(u) = 0, then I in Proposition 3.1 simplifies to
ς2
2
φ (u) + cφ (u) − λφ(u) = 0.
Solving this differential equation with initial conditions φ(0) = 0 and
φ (0+) = 2c(1−ρ)
ς2 completes the proof.
Corollary 3.3. (Ruin probability of a pure diffusion process) Suppose that the risk reserve
process is given by
Rt = u + ct + V
(−ν,ς)
t , (29)
where V
(−ν,ς)
t is a Wiener process with drift −ν and volatility ς. Moreover, the drift satisfies the
inequality ν < c. Then the ruin probability is
ψ(u) = P sup
t≥0
Yt ≥ u = e
−
2(c−ν)
ς2 u
. (30)
Proof : It holds that
Rt
D
= u + (c − ν)t + ςVt
so Rt, in distribution, equals the process (23) where λ = 0 and the premium rate is
c − ν. Hence Proposition 3.1 may be applied. As λ = 0, then ρ = 0 and thus from III
in Proposition 3.1 it follows that
φ(u) = FE(u) = 1 − e
−
2(c−ν)
ς2
.
37
38. In a pure diffusion model, the surplus process Yt is a Brownian motion with drift ν − c and
volatility ς. Therefore Corollary 3.3 reinforces the well-known result that the supremum of a
Brownian motion is exponentially distributed with rate 2(ν − c)/ς2.
Like φ, the components of the ruin probability, ψd and ψc, also have integro-differential repre-
sentations.
Proposition 3.4. For u > 0, it holds that
I.
ς2
2
ψc (u) + cφc(u) + λ
[0,u]
ψc(u − x)dB(x) − λψc(u) + λB(u) = 0;
II. ψc(0+
) =
2cρ
ς2
;
III.
ς2
2
ψd(u) + cψd(u) + λ
[0,u]
ψd(u − x)dB(x) − λψd(u) = 0;
IV. ψd(0+
) = −
2c
ς2
;
V. ψ (0+
) = −
2c(1 − ρ)
ς2
.
Proof : Following similar steps as in the proof of I in Proposition 3.1 leads to
λψc(u) = cψc(u) +
ς2
2
ψc (u) + λ (ψc ∗ B) (u).
As ψc(u) = 1 for negative u,
(ψc ∗ B) (u) =
[0,∞[
ψc(u − x)dB(x) =
[0,u]
ψc(u − x)dB(x) + B(u),
which proves I. The same approach proves III too, except now ψd(u) = 0 for u < 0.
As for II, rearrange I and then integrate both sides over [0, u], apply a change of variable
˜x = u − x, integrate by parts and finally change the integration order to obtain
ψc(u) − ψc(0+
) = −
2c
ς2
ψc(u) − ψc(0)
+
2λ
ς2
[0,u]
ψc(x)dx −
[0,u] [0,x]
ψc(x − y)dB(y)dx −
[0,u]
B(x)dx
= −
2c
ς2
ψc(u) +
2λ
ς2
[0,u]
ψc(u − ˜x)d˜x −
[0,u]
ψc(0)B(x)dx
−
[0,u] [0,x]
B(y)ψc(x − y)dydx −
[0,u]
B(x)dx
= −
2c
ς2
ψc(u) +
2λ
ς2
[0,u]
ψc(u − ˜x)d˜x
$$$$$$$$$$
−
[0,u]
ψc(0)B(x)dx
−
[0,u]
B(y) [ψc(u − y)$$$$−ψc(0)] dy −
[0,u]
B(x)dx
= −
2c
ς2
ψc(u) +
2λ
ς2
[0,u]
B(x) ψc(u − x) − 1 dx. (31)
38
39. Let u → ∞. Because ψc(u → ∞) = ψc(u → ∞) = 0 and [0,∞[ B(x)dx = η,
ψc(0+
) =
2λη
ς2
=
2cρ
ς2
so now II is proven. To prove IV, similar computations for III as for IV (recall that
now ψd(0) = 1) produce
ψd(u) − ψd(0+
) = −
2c
ς2
ψd(u) − 1 +
2λ
ς2
[0,u]
B(x)ψc(u − x)dx. (32)
As ψd(u → ∞) = ψd(u → ∞) = 0, clearly ψd(0+) = −2c
ς2 . Finally, V follows directly
from II and IV.
3.1 Diffusion and light-tailed claim sizes
So long as MB(s) < ∞ for some s > 0, the adjustment coefficient exists for the diffused model
as well. Now, from (3) it follows that
MYt (s) = eθ(s)t
,
where θ(s) = λ (MB(s) − 1) − cs + ς2s2/2. In the diffused model, the term ς2s2/2 gets added to
θ(s). As with the non-diffused case, the adjustment coefficient γ must satisfy θ(γ) = 0 and γ > 0.
It holds that θ(0) = 0, θ (0) = λη − c > 0 and θ (s) = E[U2esU ] + ς2 > 0. Thus, Lemma
2.7, applies in the diffusion perturbed model. An identity corresponding to (19) exists too:
ρ
1 − ς2γ
2c [0,∞[
eγx
dBIx = 1. (33)
Proof of (33): Equation θ(γ) = 0, implies that MB(γ)−1 = cγ−ς2γ2/2
λ . Following similar steps
as in the proof of (19) produces
[0,∞[
eγx
dBI(x) =
1
ηγ
MB(γ) − 1 =
c − ς2γ/2
λη
.
That the integral [0,∞[ eγxdBI(x) is positive suggests that c−ς2γ/2
λη > 0, which is equivalent with
γ <
2c
ς2
. (34)
Similarly to the classical ruin model, this adjustment coefficient provides an upper bound and
an approximation for the ruin probability.
Theorem 3.5. (Generalized Lundberg bounds) If γ > 0 exists, then the ruin probability is
exponentially bounded:
ψ(u) ≤ e−γu
.
39
40. Proof : The proof is from [7]. For s < t, Rt
D
= Rs+Rt−s−u. Furthermore, because λ (MB(γ) − 1)−
cγ + ς2γ2/2 = 0, the term e−γRt is a martingale:
E e−γRt
Fs = e−γRs+γu
E e−γu+(t−s) λ(MB(γ)−1)−cγ+ς2γ2/2
= e−γRs
.
Optional stopping theorem gives
e−γu
= E e−γR0
= E e−γRτu ,
where τu is the time of ruin. The probability of ruin due to oscillation is ψd(u) =
P τu < ∞, Rτu = 0 and the probability that ruin is caused by a claim is ψc(u) =
P τu < ∞, Rτu < 0 . If τu = ∞, then Rτu = ∞, since Rt exhibits positive drift. Hence
E e−γRτu = E e−γRτu τu < ∞ P τu < 0 + E e−γRτu τu = ∞ P τu = 0
= E e−γRτ(u) τu < ∞, Rτu = 0 ψd(u) + E e−γRτu τu < ∞, Rτu < 0 ψc(u)
≥ ψd(u) + ψc(u) = ψ(u).
In contrast to the Lundberg bounds in the non-diffused ruin model, the upper bound in The-
orem 3.5 cannot be refined by any constant other than one because ψ(0) = 1. An analogous
Cram´er–Lundberg approximation does exists however.
Theorem 3.6. Cram´er–Lundberg approximation for a diffusion perturbed model.
I. ψc(u)eγu u→∞
−−−→
c 1 − ρ − γς2 2
λMB(γ) + γς2 − c
;
II. ψd(u)eγu u→∞
−−−→
γς2/2
λMB(γ) + γς2 − c
;
III. ψ(u)eγu u→∞
−−−→
c 1 − ρ
λMB(γ) + γς2 − c
;
Proof : For I, rearranging (31), multiplying both sides with e
2c
ς2 x
, integrating over [0, u] and
changing the integration order yields
[0,u]
e
2c
ς2 u
ψc(x) +
2c
ς2
ψc(u) dx =
ς2
2c
ψc(0+
)
[0,u]
2c
ς2
e
2c
ς2 x
dx
+
λ
c [0,u]
2c
ς2
e
2c
ς2 x
[0,x]
B(y) ψc(x − y) − 1 dydx
[0,u]
d ψc(x)e
2c
ς2 x
= ρ e
2c
ς2 u
− 1
+
λ
c [0,u] [y,u]
2c
ς2
e
2c
ς2 x
B(y) ψc(x − y) − 1 dxdy.
Divide both sides by e
2c
ς2 u
and recall that FE(u) = 1 − e
− 2c
ς2 u
. Then, via changes of
40
41. variables ˜x = x − y and ˆx = u − ˜x and a few changes of integration order, one gets
ψc(u) = ρFE(u) +
λ
c [0,u] [y,u]
FE(u − x) ψc(x − y) − 1 dxB(y)dy
= ρFE(u) +
λ
c [0,u] [0,u−y]
FE(u − y − ˜x) ψc(˜x) − 1 d˜xB(y)dy
= ρFE(u) + ρ
[0,u]
ψc(˜x) − 1
[0,u−˜x]
FE(u − ˜x − y)dBI(y)d˜x
= ρFE(u) + ρ
[0,u]
ψc(u − ˆx) − 1
[0,ˆx]
FE(ˆx − y)dBI(y)dˆx.
Multiplying by eγu gives
ψc(u)eγu
= ρeγu
FE(u) + ρ
[0,u]
ψc(u − x) − 1 eγ(u−x)
[0,x]
FE(x − y)eγx
dBI(y)dx.
Now, change the integration order, recall that γ < 2c/ς2 see (34 and apply identity
(33) to compute
ρ
[0,∞[
x
0
FE(x − y)eγx
dBI(y)dx = ρ
[0,∞[ [y,∞[
2c
ς2
e γ−2c/ς2 x
dxe2cy/ς2y
dBI(y)
=
ρ
1 − γς2 2c [0,∞[
eγy
dBI(y)
= 1
so [0,x] FE(x−y)eγxdBI(y) is the density function of a proper probability distribution.
Therefore Key Renewal Theorem (2.11) provides a limit
eγu
ψc(u)
u→∞
−−−→
ρ [0,∞[ eγx FE(x) − [0,x] [0,y] FE(y − z)dBI(z)dy dx
ρ [0,∞[ [0,u] xFE(x − y)eγxdBI(y)dx
. (35)
Recall that MB(γ) − 1 = cγ−ς2γ2/2
λ . After changing the integration order over and over
41
42. again, the numerator of the approximation given by Key Renewal Theory simplifies to
[0,∞[
ρeγx
1 − e−2cx/ς2
−
[0,x] [z,x]
FE(y − z)dydBI(z) dx
= −
ρ
2c ς2 − γ
+ ρ
[0,∞[
eγx
1 −
[0,u]
FE(x − z)dBI(z) dx
= −
ρ
2c ς2 − γ
+ ρ
[0,∞[
eγx
BI(x) +
[0,x]
e γ−2c/ς2 x
e2cz/ς2
dBI(z) dx
= −
ρ
2c ς2 − γ
+ ρ
[0,∞[ [x,∞[
eγx
dBI(y)dx +
1
η [0,∞[ [z,∞[
e γ−2c/ς2 x
dxe2cz/ς2
[z,∞[
dB(y)dz
= −
ρ
2c ς2 − γ
+
λ
c [0,∞[
eγy − 1
γ [y,∞[
dB(z)dy +
1
η [0,∞[ [0,y]
eγz
2c ς2 − γ
dzdB(y)
= −
ρ
2c ς2 − γ
+
λ
cγ [0,∞[ [0,z]
eγy
− 1 dydB(z) +
λ
cγ [0,∞[
eγy − 1
2c ς2 − γ
dB(y)
= −
ρ
2c ς2 − γ
+
λ
cγ2
MB(γ) − 1 −
λ
cγ
η +
λ
cγ
1
2c ς2 − γ
MB(γ) − 1
= −
ρ
2c ς2 − γ
+
1 − ς2γ 2c
γ
−
ρ
γ
+
1 − ς2γ 2c
2c ς2 − γ
=
$$$$
−γς2ρ + 1 − γς2 2c 2c −&
&γς2 − ρ(2c −&
&γς2) +@@@@@@@@
γς2 1 − γς2 2c
2c − γς2 γ
=
c(1 − ρ) − γς2/2
c − γς2/2 γ
.
Integrating by parts, repetitively changing the integration order and observing that
MB(s) = [0,∞[ xesxdB(x) simplify the denominator to
ρ
[0,∞[ [y,∞[
x
2c
ς2
e γ−2c/ς2 x
dxe2cy/ς2
dBI(y)
=
λ
c
2c/ς2
2c/ς2 − γ [0,∞[
yeγy
+ e2cy/ς2
[y,∞[
e γ−2c/ς2 x
dx
[y,∞[
dB(z)dy
=
λ
c − γς2/2 [0,∞[ [0,z]
yeγy
+
eγy
2c/ς2 − γ
dydB(z)
=
γ
MB(γ) − 1 [0,∞[
zeγz
γ
−
1
γ [0,z]
eγy
dy +
eγz − 1
γ 2c/ς2 − γ
dB(z)
=
1
MB(γ) − 1
MB(γ) −
MB(γ) − 1
γ
+
MB(γ) − 1
2c/ς2 − γ
dB(z)
=
λMB(γ)
γ c − γς2/2
+
2γ − 2c/ς2
γ 2c/ς2 − γ
=
λMB(γ) + γς2 − c
γ c − γς2/2
Substituting the nominator and denominator into (35) results in
ψd(u)eγu u→∞
−−−→
c(1 − ρ) − γς2/2 c − γς2/2 γ
λMB(γ) + γς2 − c γ c − γς2/2
=
c(1 − ρ) − γς2/2
λMB(γ) + γς2 − c
,
42
43. which proves I.
To prove II, manipulating (32) the same way as (31) produces
[0,u]
d ψd(x)e
2c
ς2 x
=
[0,u] ¡
¡
¡2c
ς2
+$$$$
ψd(0+
) dx +
λ
c [0,u] [0,x]
2c
ς2
e
2c
ς2 x
B(y)ψd(x − y)dydx.
Recall that ψd(0) = 1. Similar steps as in the proof of I yield
ψd(u) = e
− 2c
ς2 u
+
λ
c [0,u] [y,u]
FE(u − x)B(y)ψd(x − y)dxdy
= FE(u) + ρ
[0,u] [0,x]
ψd(u − x)FE(x − y)dBI(y)dx
and so
ψd(u)eγu
= FE(u)eγu
+ ρ
[0,u] [0,x]
ψd(u − x)eγ(u−x)
FE(x − y)eγx
dBI(y)dx.
Again, Key Renewal Theorem gives
ψd(u)eγu u→∞
−−−→
[0,∞[ eγxFE(x)dx
ρ [0,∞[ [0,x] xFE(x − y)eγxdBI(y)dx
.
The numerator simplifies to
[0,∞[
eγx
FE(x)dx =
[0,∞[
e γ−2c ς2 x
dx =
1
2c ς2 − γ
=
ς2/2
c − γς2/2
and the denominator is the same as in I so
ψd(u)eγu u→∞
−−−→
ς2/2 c − γς2/2
λMB(γ) + γς2 − c γ c − γς2/2
=
γς2/2
λMB(γ) + γς2 − c
.
Finally, III follows simply by summing I and II.
As ς 0, the contribution of diffusion approaches zero in Theorem 3.6 while the contribution
of the claim-induced ruin probability approaches the approximation given in Theorem 2.12, as
it should. The main contribution of Theorem 3.6 is, however, that it provides an exponentially
decreasing asymptote for the ruin probability whenever the the adjustment coefficient γ exists.
Let us finally prove that the the constants of the approximations are all positive.
Proposition 3.7.
c 1 − ρ − γς2 2
λMB(γ) + γς2 − c
> 0;
γς2/2
λMB(γ) + γς2 − c
> 0;
c 1 − ρ
λMB(γ) + γς2 − c
> 0.
43
44. Proof : That c(1−ρ) > 0 and γς2/2 > 0 is immediately clear. For the denominators, recall that
θ (s) = λMB(s)−c+ς2s. As θ(0) = 0 and θ(s) is convex on R, it follows that θ (γ) > 0,
which proves the positivity of the second and third constant. For the numerator in the
first constant, remember that γς2
2 = c − λ
γ MB(γ) − 1 and eγz > 1 + γz for z > 0, and
then compute
c(1 − ρ) −
γς2
2
=
λ
γ
MB(γ) − 1 − λη = λ
[0,∞[
eγz − 1
γ
dB(z) − λη
> λ
[0,∞[
zdB(z) − λη
= 0.
Corollary 3.8. If the adjustment coefficient γ exists, then
γ <
2c(1 − ρ)
ς2
.
Proof : The inequality is equivalent with c(1 − ρ) − γς2
2 > 0, and this inequality was shown to
hold in the proof of Proposition 3.7.
θ
Degenerate
Uniform
Exponential
Pure diffusion
10 20 30 40 50
5
10
15
s
Figure 11: The Lundberg functions in the degenerate, uniform, exponential and pure diffusion
cases. Dashed curves indicate the corresponding Lundberg functions for the non-diffused case.
Figure 11 illustrates the Lundberg functions for degenerate, uniform, exponential and pure-
diffusion cases. The corresponding functions for the non-diffused case are also included. Note
that in Section 1.1 drift ν is defined so that the drift of the risk reserve process in the pure
diffusion case is c − λη, the same as in the other cases.
Anticipatedly, the pure diffusion case has the largest adjustment coefficient while for the other
44
45. three the order is the same as in the non-diffused case. For these three cases the adjustment co-
efficient is also smaller than the corresponding adjustment coefficient in the non-diffused model,
which, again, is to be expected.
3.2 Laplace transforms and exponential claim sizes
Often, Laplace transforms offer a convenient way to approximate the ruin probability numeri-
cally. The following proposition gives the Laplace transforms of the ruin and survival probability.
Proposition 3.9. Denote the Laplace transform of function f by f(s) := [0,∞[ f(x)e−sxdx and
the Laplace–Stieltjes transform of the claim size distribution by MB(−s). The Laplace transforms
of the ruin and survival probabilities are
φ(s) =
c(1 − ρ)
ς2s2
2 + cs − λ [1 − MB(−s)]
;
ψ(s) =
ς2s
2 + λη − λB(s)
ς2s2
2 + cs − λ [1 − MB(−s))]
;
ψc(s) =
λη − λB(s)
ς2s2
2 + cs − λ [1 − MB(−s)]
;
ψd(s) =
ς2s
2
ς2s2
2 + cs − λ [1 − MB(−s)]
.
Proof : For the Laplace transform of φ (u), integration by parts gives
φ (s) = sφ(s) − φ(0)
φ (s) = sφ (s) − φ (0+
) = s2
φ(s) − sφ(0) − φ (0+
).
Changing the integration order and using a change of variable ˜u = u − x yields the
Laplace transform of a Stieltjes convolution:
(φ ∗ B)(s) =
[0,∞[ [0,u]
e−su
φ(u − x)dB(x)du
=
[0,∞[ [x,∞[
e−(u−x)s
φ(u − x)e−sx
dudB(x)
=
[0,∞[ [0,∞[
e−˜us
φ(˜u)d˜ue−sx
dB(x)
= φ(s)MB(−s).
Laplace-transforming both sides of I in Theorem 2.3 produces
sφ(s) − φ(0) =
λ
c
φ(s) [1 − MB(−s))] .
Keeping these identities in mind while Laplace transforming the appropriate integro-
differential equations in Propositions 3.1 and 3.4 and then solving with respect to the
transformed ruin and survival probabilities proves the proposition.
45
46. Aside from numerical considerations, Laplace transforming integro-differential representations
of ψ provides a method to analyze the asymptotic behavior of ψ. This shall be explored in
more detail when investment income is incorporated in the model. Moreover, for exponential
claims, Laplace transforming ψ, rearranging and inverse Laplace transforming back result in an
analytical representation of the ruin probability. When ς > 0, this representation is not concise
and it shall not be stated here. For ς = 0, however, this approach provides an alternative way
to prove Corollary 2.10.
ψ
ψ
ψ c
ψ d
0.02 0.04 0.06 0.08 0.10 0.12 0.14
0.2
0.4
0.6
0.8
1.0
u
Figure 12: Components of the ruin probability for exponential claims.
Figure 12 portrays the ruin probability and its components computed from Proposition 3.9 for
exponential claims. As can be seen, ψd dominates only for a very small initial reserve. This
reflects the fact that for small u oscillation fairly likely induces ruin before the first claim even
occurs. As u increases, from Theorem 3.6 one gets the asymptotic ratio
ψc(u)
ψd(u)
u→∞
−−−→
2c(1 − ρ)
γς2
− 1
so the contribution of ψd does not approach zero (of course, this line of reasoning only applies
when the adjustment coefficient γ exists). The reason for this is that claims can bring the risk
reserve process very close to zero after which oscillation can cause ruin with a non-negligible
probability. With example values and exponential claims, the probability ratio ψc/ψd is about
1.88, that is, for a large initial reserve a claim-induced ruin occurs about 88% more likely than
a diffusion-induced ruin.
3.3 Diffusion and heavy-tailed claim sizes
Because an exponentially bounded ruin probability is pertained whenever claims are light-
tailed, it would seem that the contribution of the diffusion term Vt to the ruin probability is
exponentially bounded. Therefore one might surmise that for subexponential claims Theorem
2.16 continues to hold in a diffused environment as well. This intuitive conclusion is indeed
correct, and proving it requires a generalized form of the Pollaczek–Khinchin formula (2.6).
46
47. Proposition 3.10. (Generalized Pollaczek–Khinchin formula)
ψ(u) = 1 − ρ
∞
k=0
ρk
1 − F
∗(k+1)
E ∗ B∗k
I (u)
Proof : Recourse IV in Proposition (3.1) n times to obtain
φ(u) = (1 − ρ)
n−1
k=0
ρk
F
∗(k+1)
E ∗ B∗k
I (u) + ρn
φ ∗ FE ∗ BI (u).
As with the proof of the non-diffused Pollaczek–Khinchin formula, the last convolution
term converges to zero as n → ∞.
Theorem 3.11. If BI is subexponential, then ψ(u) ∼ ρ
1−ρBI(u).
Proof : It suffices to show that
F
∗(n+1)
E ∗ B∗n
I (u)
B∗n
I (u)
u→∞
−−−→ 1,
that is, the generalized Pollaczek–Khinchin formula converges to the corresponding
classical formula. Expand
1 − F
∗(n+1)
E ∗ B∗n
I (u)
BI(u)
=
1 − 1 − F
∗(n+1)
E ∗ 1 − B∗n
I (u)
BI(u)
=
B∗n
I (u) + F
∗(n+1)
E (u) − B∗n
I ∗ F
∗(n+1)
E (u)
BI(u)
.
Let us consider each of the three terms in the nominator separately. F
∗(n+1)
E is the
distribution function of the sum of n+1 i.i.d. exponentially distributed random variables
with rate 2c ς2. That this sum is Erlang-distributed with shape parameter n + 1 and
rate 2c ς2 is well-known and so
F
∗(n+1)
E (u) = e
− 2c
ς2 u
n
k=0
2cu ς2 k
k!
.
Since all subexponential distributions are also heavy-tailed, then
e u
BI(u) =
∞
k=0
u)kBI(u)
k!
u→∞
−−−→ ∞
for all > 0. This means that there exists m ∈ N such that BI(u)uk u→∞
−−−→ ∞ for all
k ≥ m. Hence
BI(u)
F
∗(n+1)
E (u)
=
e
2c
ς2 u
BI(u)
n
k=0
2cu ς2
k
k!
= BI(u) +
∞
k=n+1
BI(u)
2cu ς2 k
k!
u→∞
−−−→ ∞
or, equivalently,
F
∗(n+1)
E (u)
BI(u)
x→∞
−−−→ 0.
47
48. Now, as B∗n
I ≤ 1,
B∗n
I ∗ F
∗(n+1)
E (u)
BI(u)
≤
F
∗(n+1)
E (u)
BI(u)
u→∞
−−−→ 0.
The definition of a subexponential distribution yields
B∗n
I (u)
BI (u)
u→∞
−−−→ n and so
F
∗(n+1)
E ∗ B∗n
I (u)
BI(u)
u→∞
−−−→
B∗n
I (u)
BI(u)
⇒
F
∗(n+1)
E ∗ B∗n
I (u)
B∗n
I (u)
u→∞
−−−→ 1.
For large u and subexponential claims, the diffusion term Vt contributes very little to the ruin
probability. The diffusion term inconveniences an insurer primarily by rendering gambling with
zero initial reserve a rather untenable strategy.
3.4 Comparisons and cogitations
Similarly to Section 2.4, let us compare ruin probabilities for differently distributed claims. A
pure diffusion case shall be included. Moreover, let us compare the ruin probabilities given by
the diffused and non-diffused model. Recall that u∗ solves ψ(u∗) = 0.2. Corollary 3.3 gives the
pure diffusion ruin probability and Proposition 3.9 provides the ruin probability for exponential
claims. The ratio ψc/ψc, as well as the ruin probability for degenerate and uniform claims, is
computed from Theorem 3.6. One-year standard deviations are computed from (25). For Pareto
claims, as was demonstrated in Section 3.3, the approximated ruin probability equals the cor-
responding approximation in the non-diffused case. Table 4 collects the results and Figure 13
illustrates the ruin probabilities graphically.
SD γ u∗ ψc/ψd u∗, %-change
Pure diffusion 0.0913 40.0 0.0402 - -
Degenerate 0.124 21.2 0.0745 0.885 112 %
Uniform 0.133 18.2 0.0862 1.20 83.4 %
Exponential 0.149 13.9 0.110 1.88 54.4 %
Pareto ∞ - 0.772 - 0 %
Table 4: Results for the diffused model.
Predictably, the pure diffusion case is the safest for an insurer while the Pareto case is the
most dangerous, again with some margin. For a large initial reserve, how diffusion-induced and
claim-induced ruins compare depends on the claim size distribution. For degenerate claims, ψd
somewhat dominates ψc. For exponential claims ruin is 88 % more likely caused by a claim. For
Pareto claims, this percentage would presumably be higher still. One may plausibly infer that
the ratio ψc/ψd depends positively on the variance of the claim size distribution
The extent to which u∗ increases compared to the non-diffused case (the last column in Table
48
49. ψ
Degenerate
Uniform
Exponential
Pure diffusion
0.02 0.04 0.06 0.08 0.10 0.12 0.14
0.2
0.4
0.6
0.8
1.0
u
Figure 13: Ruin probabilities in degenerate, uniform, exponential and pure diffusion cases.
Dashed curves (except the horizontal one) indicate corresponding ruin probabilities in the non-
diffused case.
4) also depends on the claim size distribution. For degenerate claims, u∗ more than doubles
whilst for exponential claims u∗ merely increases by 50 %. This again suggests that increases
in u∗ hinge on the variance of the claim size distribution: the less volatile the claim sizes, the
more prominent the effect of the diffusion component.
Rt
0.2 0.4 0.6 0.8 1.0
- 0.1
0.1
0.2
Degenerate
Uniform
Exponential
Pareto
Pure diffusion
t
Figure 14: Sample paths of the risk reserve process.
For subexponential claims and a large initial reserve, adding a diffusion component makes little
difference. For a sufficiently small u, the inclusion of a diffusion component does, however, lead
49
50. to non-negligible increase in the ruin probability.
Finally, for diffused risk reserve processes, the main separation is not between jump-diffusion
processes and pure diffusion processes, but rather between processes with heavy-tailed jumps
and those with either light-tailed jumps or no jumps at all. As the results in Table 4 indicate,
the degenerate, uniform and exponential cases resemble the pure diffusion case far more than
the Pareto case. Figure 14, where the process with Pareto jumps distinctly separates from the
others, further confirms this conclusion. Therefore, one might justifiably approximate Rt with
a pure diffusion process (which would render the ruin probability easily solvable) if Rt includes
light-tailed jumps. For Rt with heavy-tailed jumps, such an approximation could be considered
overly audacious.
50
51. 4 Ruin probability and risky investments
Let us now suppose that the insurer invests its surplus into an asset that follows a geometric
Brownian motion St = e(µ−σ2/2)t+σWt , where Wt is a standard Wiener process, µ is the drift
and σ is the volatility. Assume that Wt, Vt, Nt, and Ui are all independent of each other. Fur-
thermore, denote Jt :=
Nt
i=1
Ui and κ := µ − σ2
2 . For the premium rate, still assume that c > λη,
although including investment income does permit constructing processes with a positive drift
for smaller or even negative c (such models, however, could be contested on economic grounds).
Rt
0.00 0.02 0.04 0.06 0.08 0.10
0.95
1.00
1.05
1.10
1.15
Non - diffused
Diffused
Diffused +Investment
t
Figure 15: Sample paths of the risk reserve process.
Let δ ∈ [0, 1]. Suppose that the insurer continuously invests fraction δRt of the reserve surplus
in the risky asset. Then the reserve process satisfies the stochastic differential equation
dRt = (µδRt− + c)dt + σδRt−dWt + ςdVt − dJt (36)
where Rt− := lim
s t
Rs. Clearly, investing a constant fraction δ is equivalent (in distribution)
with investing everything in asset that follows a geometric Brownian motion with drift δµ
and volatility δσ and thus including δ does not fundamentally change anything. Therefore the
fraction parameter, without loss of generality, shall be set δ = 1. Recalling that R0 = u, solving
(36) yields
Rt = eκt+σWt
u +
[0,t]
e−κs−σWs
(cds − dJs + ςdVs) , (37)
which can also be expressed in terms of the risky asset St:
Rt = uSt +
[0,t]
St−s (cds − dJs + ςdVs) . (38)
51
52. The independence of the random variables, Fubini’s theorem and the expectation of geometric
Brownian motion (7) produce
E [Rt] = ueµt
+
c − λη
µ
(eµt
− 1). (39)
Expectation (39) suggests that the risk reserve process has a positive drift if u + c−λη
µ > 0. As
ρ < 1, this inequality holds for all u ≥ 0. However, Rt having a positive drift does not suffice to
prevent almost sure ruin, as Theorem 4.3 will demonstrate. The proof requires two lemmas.
Lemma 4.1. Suppose that a Markov chain (Xn)n∈N∪{0} on Rd satisfies the random coefficient
autoregressive model
Xn = PnXn−1 + Qn, (40)
where Pn > 0, Qn ∈ Rd and the random variable pairs (Pn, Qn)n≥1 are i.d.d, and also indepen-
dent of X0. Furthermore, assume that
I. P(P1x + Q1 = x) < 1 for all x ∈ Rd
;
II. E (ln(P1) + ln(|Q1| ∨ 1)2+
< ∞ for some > 0;
III. E[ln(P1)] = 0 and P(P1 = 1) < 1.
Then Xn has an unbounded invariant Radon measure υ, which is unique up to a constant factor.
The measure υ is sigma-finite on B(Rd), where B(Rd) is the sigma-field of the subsets of Rd.
That υ is an invariant Radon measure means that
υ(A) =
Rd
P(P1x + Q1 ∈ A)υ(dx),
where P(P1x + Q1 ∈ A) is the probability that the process Xn with current value x will belong
to the set A after one step. 6
For the proof and further discussion on the topic, see [5].
Lemma 4.2. For all x ∈ R and p > 0
lnp
(x ∨ 1) ≤ cp|x|, where cp :=
p
e
p
. (41)
Proof : . For x ≤ 1, the inequality is obvious. For x > 1, simply maximize lnp
(x)/x to attain
cp.
Theorem 4.3. If µ ≤ σ2/2, then ψ(u) = 1 for all u ∈ R.
Proof : With some modifications, the proof is from that of Theorem 3.1 in [10]. Rearrange (37)
into
Rn = S∆nSn−1 u +
[0,n−1[
S−s cds − dJs + ςdVs +
[n−1,n]
Sn−s(cds − dJs + ςdVs),
D
= PnRn−1 + Qn,
6
This definition of an invariant radon measure is from [8], p. 234.
52
53. where Pn
D
= P1 = eκ+σW1 and Qn
D
= Q1 =
1
0 eκs+σWs (cds − dJs + ςdVs).
For the case κ < 0, note that |Q1| ≤
1
0 Ss(cds + dJs + d|Vs|) and
dE [|Vs|] = d
1
√
2πs [0,∞[
xe−x2 2s
dx −
]−∞,0[
xe−x2 2s
dx
= d
2s
π
=
s−1/2
√
2π
ds.
Then, apply Fubini’s theorem and the independence of the processes Jt, Wt and Vt to
get
E[|Q1|] ≤
[0,1]
E [Ss(cds + dJs + d|Vs|)] =
[0,1]
eµs
c + λη +
s−1/2
√
2π
ds < ∞.
If κ < 0, then E[ln(P1)] < 0. Condition Rn−1 = r > 0 and rearrange Rn = Pnr 1 + Qn
rPn
,
to obtain (assume that ln(0) = −∞)
ln(Rn ∨ 0) = ln(Pn) + ln(r) + ln 0 ∨ 1 +
Qn
Pnr
≤ ln(Pn) + ln(r) + ln 1 +
|Qn|
Pnr
⇒ ln
Rn ∨ 0
r
≤ ln Pn 1 +
|Qn|
Pnr
.
The inequality above and Jensens’s inequality give
E ln
Rn ∨ 0
r
Rn−1 = r ≤ E ln(Pn) + ln 1 +
1
r
|Qn|
Pn
(42)
≤ E[ln(Pn)] + ln 1 +
1
r
E
|Qn|
Pn
.
Now,
E
|Qn|
Pn
≤ E
[0,1]
Ss−1 cds + dJs + d|Vs| ≤ E
[0,1]
eµ(s−1)
c + λη +
s−1/2
√
2π
ds < ∞
so E[ln(Pn)] + ln 1 + 1
r E[|Qn|/Pn] ≤ 0 for a sufficiently large r (say r0). The process
Rt is an irreducible Markov chain (that is, all states communicate with each other).
Expectation E ln Rn∨0
r0
Rn−1 = r0 ≤ 0 implies that Rn ∨0 visits the interval ]0, r0]
infinitely often (see Proposition 5.3 (p. 21) in [3]). For any sensible ruin model, ψ(u) > 0
53
54. where u ∈]0, r0] so a geometric trial argument suggests that ψ(u) = 1 for all u ∈ R.
For κ = 0, E[ln(P1)] = 0 and clearly P1Rn−1 + Q1 is not degenerate, so conditions
I and III in Lemma 4.1 are met. Set q = 2 + and apply Minkowski’s inequality and
Lemma 4.2 to get
E ln(P1) + ln(|Q1| ∨ 1)
q
≤ E ln(P1) + ln(|Q1| ∨ 1)
q 1/q
≤ E[|κ + σW1|q
]1/q
+ E ln(|Q1| ∨ 1)
q 1/q
≤ E[(κ + σ|W1|)q
]1/q
+ E[cq|Q1|]1/q
.
To check that condition II is satisfied, it suffices to prove that E[|W1|q] and E[|Q1|] are
finite for q > 2. That E[|Q1|] < ∞ was already shown and because W1 is a standard
normal random variable, it straightforwardly follows that
E[|W1|q
] =
2q/2Γ(q+1
2 )
√
π
< ∞ for all q > 0.
Thus Rt has an unbounded invariant Radon measure υ, which is unique up to a constant
factor. Following Corollary 4.2 in [5] Rt visits any open set of positive υ measure
infinitely often. Since P1 can become arbitrarily close to zero, the interval ]0, ] will
be visited infinitely often. Q1 can get negative values so a geometric trial argument
concludes the proof. For a more general proof, see [10].
It, rather surprisingly, follows from Theorem 4.3 that if κ ≤ 0 (high volatility case), investing
the surplus into the risky asset will eventually force the risk reserve process Rt below zero, no
matter how high the initial reserve or the premium rate are.
As with the ruin model with no investment income, one may construct (under certain circum-
stances) an integro-differential representation of the ruin probability. This will enable applying
many analytical techniques to determine the ruin probability or its asymptotes.
Proposition 4.4. Assume that φ(u) is twice continuously differentiable. Then
ς2 + σ2u2
2
φ (u) + (c + µu)φ (u) − λφ(u) + λ(φ ∗ B)(u) = 0 (43)
or, equivalently,
ς2 + σ2u2
2
ψ (u) + (c + µu)ψ (u) − λψ(u) + λ
[0,u]
ψ(u − x)dB(x) + λB(u) = 0. (44)
Proof : Set Zt = eκt+σWt u +
t
0 e−κs−σWs cds + ςdVs and Tt := inf{v|Zv ≤ 0}∧t. Following
54
55. similar steps as in the proof of Propostion (3.1) results in
φ(u) = e−λt
E[φ(ZTt )] +
[0,t]
λe−λs
E φ(ZTs )1{Ts<s} + (φ ∗ B) (Zs)1{Ts≥s} ds
= e−λt
E φ(u) +
[0,t]
(c + µZt)φ (ZTs ) +
ς2 + σ2Z2
Ts
2
φ (ZTs ) ds
+e−λt
=0
E
[0,t]
(ς + σZTs )φ (ZTs ) dWs + dVs
+
[0,t]
λe−λs
E φ(ZTs )1{Ts<s} + (φ ∗ B) (Zs)1{Ts≥s} ds
Rearrange this into
1 − e−λt
t
φ(u) =
1
teλt
E
[0,t]
(c + µZt)φ (ZTs ) +
ς2 + σ2Z2
Ts
2
φ (ZTs ) ds
+
1
t [0,t]
λe−λs
E φ(ZTs )I{Ts<s} + (φ ∗ B) (Zs)I{Ts≥s} ds.
Let t 0 and note that Tt
t 0
−−→ 0. L’Hospital’s rule then produces
λφ(u) = (c + µu)φ (u) +
ς2 + σ2u2
2
φ (u) + λ (φ ∗ B) (u).
4.1 Degenerate, uniform and pure diffusion cases
For a pure diffusion risk reserve process, even when it is compounded by a geometric Brownian
motion, a closed form solution of ψ(u) exists.
Theorem 4.5. Assume that µ > σ2/2. Then, in a pure diffusion model with investment, the
ruin probability has the following solution and asymptote:
ψ(u) = Cd
[u,∞[
(x2
σ2
+ ς2
)− µ
σ2 exp −
2c
σς
arctan
xσ
ς
dx ∼
σ−2µ/σ2
e−cπ/σς
2µ/σ2 − 1
u1−2µ/σ2
, (45)
where
C−1
d =
[0,∞[
(x2
σ2
+ ς2
)− µ
σ2 exp −
2c
σς
arctan
xσ
ς
dx.
Proof : The proof is from [6]. Now λ = 0 and thus the integro-differential equation (43) sim-
plifies to
ς2 + σ2u2
2
ψ (u) + (c + µu)ψ (u) = 0.
The solution is
ψ(u) = Cd
[u,∞[
(x2
σ2
+ ς2
)− µ
σ2 exp −
2c
σς
arctan
xσ
ς
dx.
55
56. Evidently, this solution satisfies the boundary conditions ψ(0) = 1 and
ψ(u → ∞) = 0. Acknowledge that arctan xσ
ς
x→∞
−−−→ π/2. L’Hospital’s rule gives
ψ(u)
u1−2µ/σ2 ∼ Cd
(u2σ2 + ς2)− µ
σ2 e−cπ
σς
2µ/σ2 − 1 u−2µ/σ2 ∼ Cd
(uσ)− 2µ
σ2 e−cπ
σς
2µ/σ2 − 1 u−2µ/σ2 .
Since obviously ψ(u) ≥ ψd(u), Theorem 4.13 provides a lower bound for ψ(u). The theorem also
helps determine approximations for degenerate and uniform claims.
Appproximation 4.6. If the claims Ui are degenerately distributed at η, or if the claims are
uniformly distributed with parameters b > a > 0, then, assuming that µ > σ2/2,
ψ(u) ≈ C
σ−2µ/σ2
e−(c−λη)π/σς
2µ/σ2 − 1
u1−2µ/σ2
as u → ∞ (46)
where C ≥ Cde−ληπ
ςσ
Let us prove Approximation 4.6 heuristically. For degenerate claims, B(u) = 1{u≥η}. If u ≥ η,
then integration by parts yields
[0,u]
ψ(u − x)dB(x) = ψ(0) −
[0,u]
1{u≥η}dψ(u − x) = ψ(u − η).
Thus, for u ≥ η, the integro-differential equation (4.4) becomes
ς2 + u2σ2
2
ψ (u) + (c + µu)ψ (u) − λη
ψ(u) − ψ(u − η)
η
= 0.
Note that B(u) = 0 for u ≥ η and apply approximation ψ (u) ≈ ψ(u)−ψ(u−η)
η . Since ψ(u) is twice
continuously differentiable, monotonically decreasing and gets values on [0, 1], this seems like a
reasonable approximation for large u. Hence, for large u, the solution to the integro-differential
equation above is approximately the same as the solution to the differential equation
ς2 + u2σ2
2
ψ (u) + (c − λη + µu)ψ (u) = 0.
This type of differential equation was already solved in the proof of Theorem 4.5. Simply re-
place c with c − λη. Of course, now the initial condition ψ(0) = 1 is not applicable. However,
as ψd(u) ≤ ψ(u), clearly C ≥ Cde−ληπ
ςσ
As for uniform claims, B(u) = 0 for u ≥ b. Furthermore, for u ≥ b, approximate
[0,u]
ψ(u − x)dB(x) =
[a,b]
ψ(u − x)
b − a
=
[u−b,u−a]
ψ(˜x)d˜x ≈ ψ(u − a) +
1
2
ψ(u − b) − ψ(u − a) .
Thus
ψ(u) −
[0,u]
ψ(u − x)dB(x) ≈
b
2
ψ(u) − ψ(u − b)
b
+
a
2
ψ(u) − ψ(u − a)
a
≈
a + b
2
ψ (u).
Recall that the expectation of the uniform distribution is a+b
2 . For large u, the solution to
the integro-differential equation (4.4) is, again, approximately the same as the solution to the
differential equation
ς2 + u2σ2
2
ψ (u) + (c − λη + µu)ψ (u) = 0,
which justifies 4.6.
56
57. Appproximation 4.7. If ς = 0 and µ > σ2/2, then for degenerate and uniform claims the
ruin probability has an approximation.
ψ(u) ≈ C
2µ
σ2
− 1
[0,1/u]
x2µ/σ2−2
e2cx/σ2
dx ∼ Cu1−2µ/σ2
as u → ∞,
where C > 0.
Let us prove Approximation 4.7 heuristically. The differential equation to solve now simplifies
to
σ2u2
2
ψ (u) + (c + µu)ψ (u) = 0.
Evidently, any given constant satisfies this differential equation. The boundary condition ψ(u →
∞) = 0 however ensures that this constant must be zero.
As for the second solution, since
d
du [0,1/u]
x2µ/σ2−2
e2cx/σ2
dx = −u−2µ/σ2
e2cu−1/σ2
and
d2
du2
[0,1/u]
x2µ/σ2−2
e2cx/σ2
dx = −
2µ
σ2
u−1
+
2c
σ2
u−2
ψ (u),
clearly C 2µ
σ2 − 1 [0,1/u] x2µ/σ2−2e2cx/σ2
dx is a solution. Furthermore, this solution satisfies the
initial condition ψ(u → ∞) = 0. To conclude the proof, apply l’Hospital’s rule to obtain
2µ
σ2 − 1 [0,1/u] x2µ/σ2−2e2cx/σ2
dx
u1−2µ/σ2 ∼
1 − 2µ
σ2 u−2µ/σ2
e2cu−1/σ2
1 − 2µ
σ2 u−2µ/σ2
u→∞
−−−→ 1,
which justifies 4.7.
4.2 Exponential claim sizes
The asymptotic power law decay of order 1 − 2µσ2 that the pure diffusion, degenerate and uni-
form cases exhibit exists for exponential claim sizes as well, although the proof in the exponential
case will be proved quite differently7.
Corollary 4.8. If the claims are exponentially distributed, then ψ(u) satisfies the third order
differential equation
(a1u2
+ a2)ψ (u) + (b1u2
+ b2u + b3)ψ (u) + (c1u + c2)ψ (u) = 0 (47)
with boundary conditions
ψ(u → ∞) = 0
ς2
2 ψ (0+) + cψ (0+) + λ 1 − ψ(0) = 0
ς2
2 ψ (0+) − λ
η 1 − ψ(0) + cψ (0+) + (µ − λ)ψ (0+) = 0,
where a1 = ησ2/2, a2 = ς2/2, b1 = σ2/2, b2 = η(σ2+µ), b3 = ς2+ηc, c1 = µ and c2 = c+ηµ−ηλ.
7
That ψ(u) ∼ Cu1−2µ/σ2
is a well-established result for light-tailed claims. See, for instance, [9], p. 424.
57
58. Proof : Now
[0,u]
ψ(u − y)dB(y) =
1
η [0,u]
ψ(u − y)e−y/η
dy =
1
η
e−u/η
[0,u]
ψ(˜y)e˜y/η
d˜y.
Rearranging (44) yields
−
λ
η
e−u/η
[0,u]
ψ(y)ey/η
dy =
ς2 + σ2u2
2
ψ (u) + (c + µu)ψ (u) − λψ(u) + λe−u/η
.
Differentiating this with respect to u and then multiplying both sides by −η gives
−
λ
η
e−u/η
[0,u]
ψ(y)ey/η
dy = −λψ(u) −
η(ς2 + σ2u2)
2
ψ (u) − η(σ2
u + c + µu)ψ (u)
−η(µ − λ)ψ (u) + λe−u/η
Combining the two equations above enables discarding the integral term and then divid-
ing both sides by η(ς2+σ2u2)
2 gives the desired differential equation. The first boundary
condition is obvious, the second one follows by setting u = 0+ in (44) and the third
one follows by setting u = 0+ in (47) and from the second boundary condition.
Proving the asymptotic behaviour of ψ for the exponential case requires two theorems, a Kara-
mata–Tauberian and a monotone density theorem8.
Theorem 4.9. (Karamata–Tauberian)
Let F be a cumulative distribution function on [0, ∞[. Assume that p ≥ 0 and C > 0. Moreover,
let L(u) be a slowly varying function, that is, L(αu) ∼ L(u) for all α > 0. Then the following
are equivalent:
I. MF (−s) ∼ Cs−p
L(1/s) as s 0;
II. F(u) ∼
C
Γ(p + 1)
up
L(u),
where Γ(p + 1) := [0,∞[ xpe−xdx.
Theorem 4.10. (Monotone Density Theorem)
Let L(u) be a slowly varying function and G(u) = [0,u] g(y)dy where g is monotone on ]w, ∞[
for some w > 0. If
G(u) ∼ Cup
L(u),
for some C > 0 and p ∈ R {0}, then
g(u) ∼ Cpup−1
L(u).
Theorem 4.11. If the claim sizes are exponentially distributed and µ > σ2/2, then
ψ(u) ∼ Cu1−2µ/σ2
where C > 0.
8
both are from [4]
58
59. Proof : The proofs follows similar steps as that of Theorem 3.1 in [2]. Laplace transforms satisfy
ψ(n)(s) = sn
ψ(s) −
n
i=1
sn−i
ψ(i−1)
(0+
); (48)
unψ(s) = (−1)n
ψ(n)
(s) (49)
Keeping properties (48) and (49) in mind, transforming the components of the differ-
ential equation in Corollary 4.8 produces
L a1u2
+ a2 ψ (s) = a1
d2
ds2
+ a2 s3
ψ(s) − s2
ψ(0) − sψ (0+
) − ψ (0+
)
= a1s3
ψ (s) + 6a1s2
ψ (s) + (6a1s + a2s3
)ψ(s)
−2a1ψ(0) − a2 s2
ψ(0) + sψ (0+
) + ψ (0+
) ;
L b1u2
+ b2u + b3 ψ (s) = b1
d2
ds2
− b2
d
ds
+ b3 s2
ψ(s) − sψ(0) − ψ (0+
)
= b1s2
ψ (s) + (4b1s − b2s2
)ψ (s) + (2b1 − 2b2s + b3s2
)ψ(s)
+b2ψ(0) − b3 sψ(0) + ψ (0+
) ;
L c1u + c2 ψ (s) = − c1
d
ds
+ c2 sψ(s) − ψ(0)
= −c1sψ (s) + c2s − c1 ψ(s) − c2ψ(0).
The Laplace transform corresponding to (47) is
v(s)ψ (s) + w(s)ψ (s) + k(s)ψ(s) + l(s) + A = 0 (50)
where
v(s) = (a1s3
+ b1s2
)
w(s) = 6a1s2
− b2s2
+ (4b1 − c1)s
k(s) = a2s3
+ b3s2
+ (6a1 − 2b2 + c2)s + 2b1 − c1
l(s) = −a2ψ(0)s2
− a2ψ (0+
) + b3ψ(0) s
A = b2 − 2a1 − c1 ψ(0) − b3ψ (0+
) − a2ψ (0+
).
Let us apply a Frobenius type approach to solve the homogeneous form of (50) in
the vicinity of zero (for an extensive treatment of the method, see Chapter 4 in [13]).
Suppose that the solution is of form ψH(s) = 2
j=1 Cj
∞
k=0 δj,ksrj+k where Cj and
δj,k are constants. Constants δ1,0 and δ2,0 are arbitrary non-zero constants.
2
j=1
Cj
∞
k=0
δj,k (rj − 1 + k)(rj + k)
v(s)
s2
+ (rj + k)
w(s)
s
+ k(s) srj+k
= 0.
Now, because v(s)/s2 s 0
−−−→ b1, w(s)/s
s 0
−−−→ 4b1 − c1 and k(s)
s 0
−−−→= 2b1 − c1, it must
hold that
b1(rj − 1)rj + (4b1 − c1) + 2b1 − c1 = 0.
This is called the indical equation of the differential equation to be solved. Solving rj
produces r1 = −1 and r2 = c1/d1 −2 = 2µ/σ2 −2. If r2 −r1 ∈ N, the the solutions given
by r1 and r2 may not be linearly independent. Then one of the solutions is of form
59
60. ψ1(s) = C1
∞
k=0 δ1,ksr1+k + C3δ2,k ln(s)sr2+k . But this would lead to a requirement
that
C3 (2r2 + 3 − c1/b1) = 0.
Since µ > σ2/2, this cannot hold unless C3 = 0. Therefore the homogeneous solutions
do not depend on whether r2 − r1 is an integer or not.
Because r1 < r2, ψH(s) ∼ C1s−1 as s 0. Define Ψ(u) := [0,u] ψ(x)dx. Applying
the Karamata–Tauberian theorem yields ΨH(u) ∼ C1δ1,0u and monotone density the-
orem further gives ψH(u) ∼ C1δ1,0. Deduce from the boundary conditions of ψ(u) that
C1 = 0.
The new leading term is sr2 . For r2 < 0, Karamata–Tauberian and monotone den-
sity theorems give
ΨH(u) ∼
C2δ2,0
Γ(1 − r2)
u−r2
⇒ ψH(u) ∼
C2δ2,0
Γ(−r2)
u−(r2+1)
.
Here property Γ(x+1) = xΓ(x) has been applied. If r2 ≥ 0, then lims 0 ψH(s) < ∞ is fi-
nite. Dominated convergence theorem gives [0,∞[ ψ(x)dx = limu→∞ [0,∞[ e−x/uψ(x)dx.
Thus, via monotone density theorem,
ΨH(u) ∼
[0,∞[
e−x/u
ψ(−x/u)dx ∼ C2δ2,0u−r2
;
⇒ ψ(u) ∼ −r2C2δ2,0u−(r2+1)
Denote the solutions to the homogeneous solution with ψ1 for r = −1 and ψ2 for r = r2.
Applying variation of parameters leads to the following expression of the particular
solution:
ψp(s) = ψ2(s)
[ 0,s]
ψ1(x) (−l(x) − A) dx
x2W(x)
− ψ1(s)
[0,s]
ψ2(x) (−l(s) − A) dx
x2W(x)
,
where 0 = I{r2≥0}, is arbitrarily small positive number and
W(s) := ψ1(s)ψ2(s) − ψ1(s)ψ2(s) = sr2−2
∞
k=0
γksk
.
Here γk = k
j=0 δ1,jδ2, k − j r2 + 1 + k − 2j . Note that γ0 = δ1,0δ2,0(r2 + 1) = 0 since
from µ > σ2/2 it follow that r2 + 1 > 0. Furthermore, one has limits
ψ1(s)
s2W(s)
∼
s−(r2+1)
β0(r2 + 1)
, s 0;
ψ2(s)
s2W(s)
s 0
−−−→
1
α0(r2 + 1)
;
l(s)
s 0
−−−→ 0.
60
