64
4.
4. THE CRAMÉR–LUNDBERG MODEL
The Cramér–Lundberg Model
4.1. Definition of the Cramér–Lundberg Process
We have seen that the compound Poisson model has nice properties. For instance
it can be derived as a limit of individual models. This was the reason for Filip
Lundberg [63] to postulate a continuous time risk model where the aggregate claims
in any interval have a compound Poisson distribution. Moreover, the premium
income should be modelled. In a portfolio of insurance contracts the premium
payments will be spread all over the year. Thus he assumed that that the premium
income is continuous over time and that the premium income in any time interval is
proportional to the interval length. This leads to the following model for the surplus
of an insurance portfolio
Nt
X
Ct = u + ct −
Yi .
i=1
u is the initial capital, c is the premium rate. The number of claims in (0, t] is a Poisson process {Nt } with rate λ. The claim sizes {Yi } are a sequence of iid. positive random variables independent of {Nt }. This model is called the Cramér–Lundberg
process or classical risk process.
We denote the distribution function of the claims by G, its moments by µn =
IIE[Y1n ] and its moment generating function by MY (r) = IIE[exp{rY1 }]. Let µ = µ1 .
We assume that µ < ∞. Otherwise, no insurance company would insure such a risk.
Note that G(x) = 0 for x < 0. We will see later that it is no restriction to assume
that G(0) = 0.
For an insurance company it is important that {Ct } stays above a certain level.
This sovency level is given by legal restrictions. By adjusting the initial capital it is
no loss of generality to assume this level to be 0. We define the ruin time
(inf ∅ = ∞) .
τ = inf{t > 0 : Ct < 0} ,
We will mostly be interested in the probability of ruin in a time interval (0, t]
ψ(u, t) = IIP[τ ≤ t | C0 = u] = IIP[ inf Cs < 0 | C0 = u]
0<s≤t
and the probability of ultimate ruin
ψ(u) = lim ψ(u, t) = IIP[inf Ct < 0 | C0 = u] .
t→∞
t>0
4. THE CRAMÉR–LUNDBERG MODEL
65
It is easy to see that ψ(u, t) is decreasing in u and increasing in t.
We denote the claim times by T1 , T2 , . . . and by convention T0 = 0. Let Xi =
c(Ti − Ti−1 ) − Yi . If we only consider the process at the claim times we can see that
CTn = u +
n
X
Xi
i=1
is a random walk. Note that ψ(u) = IIP[inf n∈IIN CTn < 0]. From the theory of random
walks we can see that ruin occurs a.s. iff IIE[Xi ] ≤ 0 (compare with Lemma E.1 or
[41, p.396]). Hence we will assume in the sequel that
1
IIE[Xi ] > 0 ⇐⇒ c − µ > 0 ⇐⇒ c > λµ ⇐⇒ IIE[Ct − u] > 0 .
λ
Recall that
IIE
Nt
hX
i
Yi = λtµ .
i=1
The condition can be interpreted that the average income is strictly larger than the
average outflow. Therefore the condition is also called the net profit condition.
If the net profit condition is fulfilled then CTn tends to infinity as n → ∞. Hence
inf{Ct − u : t > 0} = inf{CTn − u : n ≥ 1}
is a.s. finite. So we can conclude that
lim ψ(u) = 0 .
u→∞
4.2. A Note on the Model and Reality
In reality the average number of claims in an interval will not be the same all the
time. There will be a claim rate λ(t) which may be periodic in time. Moreover,
the number of individual contracts in the portfolio may vary with time. Let a(t) be
the volume of the portfolio at time t. Then the claim number process {Nt } is an
inhomogeneous Poisson process with rate a(t)λ(t). Let
Z t
Λ(t) =
a(s)λ(s) ds
0
and Λ−1 (t) be its inverse function. Then Ñt = NΛ−1 (t) is a Poisson process with rate
1. Let now the premium rate vary with t such that ct = ca(t)λ(t) for some constant
c. This assumption is natural for changes in the risk volume. It is artificial for the
66
4. THE CRAMÉR–LUNDBERG MODEL
changes in the intensity. For instance we assume that the company gets more new
customers at times with a higher intensity. This effect may arise because customers
withdraw from their old insurance contracts and write new contracts with another
company after claims occurred because they were not satisfied by the handling of
claims by their old companies.
The premium income in the interval (0, t] is cΛ(t). Let C̃t = CΛ−1 (t) . Then
NΛ−1 (t)
−1
C̃t = u + cΛ(Λ (t)) −
X
Yi = u + ct −
i=1
Ñt
X
Yi
i=1
is a Cramér–Lundberg process. Thus we should not consider time to be the real
time but operational time.
The event of ruin does almost never occur in practice. If an insurance company
observes that their surplus is decreasing they will immediately increase their premia.
On the other hand an insurance company is built up on different portfolios. Ruin
in one portfolio does not mean bankruptcy. Therefore ruin is only a technical term.
The surplus will also be a technical term in practice. If the business is going well
then the share holders will decide to get a higher dividend. To model this we would
have to assume a premium rate dependent on the surplus. But then it would be
hard to obtain any useful results. For some references see for instance [43] and [11].
The probability of ruin should be regarded as a measure for the risk and is used
for decision taking; for example, the premium calculation or the computation of
reinsurance retention levels. One therefore freezes the present portfolio and looks
how the surplus would behave in the future. For an actuary it is important to be
able to take a good decision in reasonable time. Therefore it is fine to simplify the
model in order to be able to check whether a decision has the desired effect or not.
In Section 1 about risk models we saw that a negative binomial distribution for
the number of claims in a certain time interval would be preferable. We will later
consider such models. But often actuaries use the Cramér–Lundberg model for their
calculations, even though there is a certain lack of reality. The reason is that this
model is well understood, and the basic properties for a risk also are present in this
simple model.
4.3. A Differential Equation for the Ruin Probability
We first prove that {Ct } is a strong Markov process.
4. THE CRAMÉR–LUNDBERG MODEL
67
Lemma 4.1. Let {Ct } be a Cramér–Lundberg process and T be a finite stopping
time. Then the stochastic process {CT +t − CT : t ≥ 0} is a Cramér–Lundberg process
with initial capital 0 and independent of FT .
Proof.
We can write CT +t − CT as
NT +t
X
ct −
Yi .
i=NT +1
Because the claim amounts are iid. and independent of {Nt } we only have to prove
that {NT +t −NT } is a Poisson process independent of FT . Because {Nt } is a renewal
process it is enough to show that TNT +1 − T is Exp(λ) distributed and independent
of FT . Condition on TNT and T . Then
IIP[TNT +1 − T > x | TNT , T ]
= IIP[TNT +1 − TNT > x + T − TNT | TNT , T, TNT +1 − TNT > T − TNT ] = e−λx
by the lack of memory property of the exponential distribution. The assertion
follows because TNT +1 depends on FT via TNT and T only. The latter, even though
intuitively clear, follows readily noting that FT is generated by sets of the form
{NT = n, An } where n ∈ IIN and An ∈ σ(Y1 , . . . , Yn , T1 , . . . , Tn ).
Let h be small. If ruin does not occur in the interval (0, T1 ∧ h] then a new
Cramér–Lundberg process starts at time T1 ∧ h with new initial capital CT1 ∧h . Let
δ(u) = 1 − ψ(u) denote the survival probability. Using that the interarrival times
are exponentially distributed we have the density λe−λt of the distribution of T1 and
IIP[T1 > h] = e−λh . We obtain noting that δ(x) = 0 for x < 0
δ(u) = e
−λh
Z hZ
u+ct
δ(u + ct − y) dG(y)λe−λt dt .
δ(u + ch) +
0
0
Letting h tending to 0 shows that δ(u) is right continuous. Rearranging the terms
and dividing by h yields
Z Z
1 − e−λh
1 h u+ct
δ(u + ch) − δ(u)
=
δ(u + ch) −
δ(u + ct − y) dG(y)λe−λt dt .
c
ch
h
h 0 0
Letting h tend to 0 shows that δ(u) is differentiable from the right and
Z u
h
i
0
cδ (u) = λ δ(u) −
δ(u − y) dG(y) .
0
(4.1)
68
4. THE CRAMÉR–LUNDBERG MODEL
Replacing u by u − ch gives
−λh
δ(u − ch) = e
Z hZ
u−c(h−t)
δ(u − (c − h)t − y) dG(y)λe−λt dt .
δ(u) +
0
0
We conclude that δ(u) is continuous. Rearranging the terms, dividing by h and
letting h ↓ 0 yields the equation
Z u−
h
i
0
cδ (u) = λ δ(u) −
δ(u − y) dG(y) .
0
We see that δ(u) is differentiable at all points u where G(u) is continuous. If G(u)
has a jump then the derivatives from the left and from the right do not coincide.
But we have shown that δ(u) is absolutely continuous, because G(u) can have at
most countably many jumps. If we now let δ 0 (u) be the derivative from the right we
have that δ 0 (u) is a density of δ(u).
The difficulty with equation (4.1) is that it contains both the derivative of δ(u)
and an integral. Let us try to get rid of the derivative. We find
Z uZ x
Z u
Z
1 u 0
c
(δ(u) − δ(0)) =
δ(x − y) dG(y) dx
δ(x) dx −
cδ (x) dx =
λ
λ 0
0 0
0
Z uZ u
Z u
δ(x − y) dx dG(y)
δ(x) dx −
=
0
0
y
Z uZ
u
Z
u−y
δ(x) dx −
=
0
Z uZ
u
Z
δ(x) dx dG(y)
0
0
u−x
δ(x) dx −
=
Z0 u
=
dG(y) δ(x) dx
Z u
δ(u − x)(1 − G(x)) dx
δ(x)(1 − G(u − x)) dx =
0
0
0
0
R∞
Note that 0 (1−G(x)) dx = µ. Letting u → ∞ we can by the bounded convergence
theorem interchange limit and integral and get
Z ∞
c(1 − δ(0)) = λ
(1 − G(x)) dx = λµ
0
where we used that δ(u) → 1. It follows that
λµ
λµ
,
ψ(0) =
.
c
c
Replacing δ(u) by 1 − ψ(u) we obtain
Z u
cψ(u) = λµ − λ
(1 − ψ(u − x))(1 − G(x)) dx
0
Z ∞
Z u
=λ
(1 − G(x)) dx +
ψ(u − x)(1 − G(x)) dx .
δ(0) = 1 −
u
0
In Section 4.8 we shall obtain a natural interpretation of (4.2).
(4.2)
4. THE CRAMÉR–LUNDBERG MODEL
69
Example 4.2. Let the claims be Exp(α) distributed. Then the equation (4.1) can
be written as
Z u
i
h
αy
0
−αu
δ(y)αe dy .
cδ (u) = λ δ(u) − e
0
Differentiating yields
Z
h
0
−αu
cδ (u) = λ δ (u) + αe
00
u
i
δ(y)αeαy dy − αδ(u) = λδ 0 (u) − αcδ 0 (u) .
0
The solution to this differential equation is
δ(u) = A + Be−(α−λ/c)u .
Because δ(u) → 1 as u → ∞ we get A = 1. Because δ(0) = 1 − λ/(αc) the solution
is
λ
δ(u) = 1 − e−(α−λ/c)u
αc
or
λ −(α−λ/c)u
e
.
ψ(u) =
αc
4.4. The Adjustment Coefficient
Let
θ(r) = λ(MY (r) − 1) − cr
(4.3)
provided MY (r) exists. Then we find the following martingale.
Lemma 4.3. Let r ∈ IR such that MY (r) < ∞. Then the stochastic process
{exp{−rCt − θ(r)t}}
(4.4)
is a martingale.
Proof.
By the Markov property we have for s < t
IIE[e−rCt −θ(r)t | Fs ] = IIE[e−r(Ct −Cs ) ]e−rCs −(λ(MY (r)−1)−cr)t
= IIE[er
PNt
Ns +1
Yi
]e−rCs −λ(MY (r)−1)t+crs
= e−rCs −λ(MY (r)−1)s+crs = e−rCs −θ(r)s .
70
4. THE CRAMÉR–LUNDBERG MODEL
ΘHrL
0.8
0.6
0.4
0.2
R
-1.0
0.5
-0.5
1.0
1.5
r
Figure 4.1: The function θ(r) and the adjustment coefficient R
It would be nice to have a martingale that only depends on the surplus and not
explicitly on time. Let us therefore consider the equation θ(r) = 0. This equation
has obviously the solution r = 0. We differentiate θ(r).
θ0 (r) = λMY0 (r) − c ,
θ00 (r) = λMY00 (r) = λIIE[Y 2 erY ] > 0 .
The function θ(r) is strictly convex. For r = 0
θ0 (0) = λMY0 (0) − c = λµ − c < 0
by the net profit condition. There might be at most one additional solution R to
the equation θ(R) = 0 and R > 0. If this solution exists we call it the adjustment
coefficient or the Lundberg exponent. The Lundberg exponent will play an
important rôle in the estimation of the ruin probabilities.
Example 4.2 (continued). For Exp(α) distributed claims we have to solve
α
λr
λ
− 1 − cr =
− cr = 0 .
α−r
α−r
For r 6= 0 we find R = α − λ/c. Note that
ψ(u) =
λ −Ru
e
.
αc
4. THE CRAMÉR–LUNDBERG MODEL
71
In general the adjustment coefficient is hard to calculate. So we try to find some
bounds for the adjustment coefficient. Note that the second moment µ2 exists if the
Lundberg exponent exists. Consider the function θ(r) for r > 0.
θ00 (r) = λIIE[Y 2 erY ] > λIIE[Y 2 ] = λµ2 ,
Z r
0
0
θ00 (s) ds > −(c − λµ) + λµ2 r ,
θ (r) = θ (0) +
Z 0r
r2
θ0 (s) ds > λµ2 − (c − λµ)r .
θ(r) = θ(0) +
2
0
The last inequality yields for r = R
R
0 = θ(R) > R λµ2 − (c − λµ)
2
from which an upper bound of R
R<
2(c − λµ)
λµ2
follows.
We are not able to find a lower bound in general. But in the case of bounded
claims we get a lower bound for R. Assume that Y1 ≤ M a.s..
Let us first consider the function
x RM
e
− 1 − eRx − 1 .
f (x) =
M
Its second derivative is
f 00 (x) = −R2 eRx < 0 .
f (x) is concave with f (0) = f (M ) = 0. Thus f (x) > 0 for 0 < x < M . The
function h(x) = xex − ex + 1 has a minimum in 0. Thus h(x) > h(0) = 0 for x 6= 0,
in particular
1
eRM − 1 < eRM .
RM
We calculate
Z M
Z M
x RM
µ RM
Rx
e
− 1 dG(x) =
e
−1 .
MY (R) − 1 =
e − 1 dG(x) <
M
M
0
0
From the equation determining R we get
0 = λ(MY (R) − 1) − cR <
λµ RM
e
− 1 − cR < λµReRM − cR .
M
It follows readily
R>
1
c
log
.
M
λµ
72
4. THE CRAMÉR–LUNDBERG MODEL
4.5. Lundberg’s Inequality
We will now connect the adjustment coefficient and the ruin probabilities.
Theorem 4.4. Assume that the adjustment coefficient R exists. Then
ψ(u) < e−Ru .
Proof.
Assume that the theorem does not hold. Let
u0 = inf{u ≥ 0 : ψ(u) ≥ e−Ru } .
Because ψ(u) is continuous we get
ψ(u0 ) = e−Ru0 .
Because ψ(0) < 1 we conclude that u0 > 0. Consider the equation (4.2) for u = u0 .
Z u0
hZ ∞
i
cψ(u0 ) = λ
(1 − G(x)) dx +
ψ(u0 − x)(1 − G(x)) dx
u0
Z0 u0
i
hZ ∞
e−R(u0 −x) (1 − G(x)) dx
<λ
(1 − G(x)) dx +
0
Z u∞0
Z ∞Z ∞
−R(u0 −x)
−Ru0
e
(1 − G(x)) dx = λe
eRx dG(y) dx
≤λ
0
0
x
Z ∞Z y
Z ∞
1
(eRy − 1) dG(y)
= λe−Ru0
eRx dx dG(y) = λe−Ru0
R
0
0
0
λ −Ru0
= e
(MY (R) − 1) = ce−Ru0
R
which is a contradiction. This proves the theorem.
We now give an alternative proof of the theorem. We will use the positive
martingale (4.4) for r = R. By the stopping theorem
e−Ru = e−RC0 = IIE[e−RCτ ∧t ] ≥ IIE[e−RCτ ∧t ; τ ≤ t] = IIE[e−RCτ ; τ ≤ t] .
Letting t tend to infinity yields by monotone convergence
e−Ru ≥ IIE[e−RCτ ; τ < ∞] > IIP[τ < ∞] = ψ(u)
(4.5)
because Cτ < 0. We have got an upper bound for the ruin probability. We would
like to know whether R is the best possible exponent in an exponential upper bound.
This question will be answered in the next section.
4. THE CRAMÉR–LUNDBERG MODEL
73
Example 4.5. Let G(x) = 1 − pe−αx − (1 − p)e−βx where 0 < α < β and 0 < p ≤
and thus
β(β − α)−1 . Note that the mean value is αp + 1−p
β
c>
λp λ(1 − p)
+
.
α
β
For r < α
Z
MY (r) =
0
∞
αp
β(1 − p)
erx αpe−αx + β(1 − p)e−βx dx =
+
.
α−r
β−r
Thus, we have to solve
pr
αp
β(1 − p)
(1 − p)r +
− 1 − cr = λ
+
− cr = 0 .
λ
α−r
β−r
α−r
β−r
We find the obvious solution r = 0. If r 6= 0 then
λp(β − r) + λ(1 − p)(α − r) = c(α − r)(β − r)
or equivalently
cr2 − ((α + β)c − λ)r + αβc − λ((1 − p)α + pβ) = 0 .
The solution is
1
r=
α+β−
2
1
=
α+β−
2
(4.6)
r
λ 2
λ
λ
α+β−
− 4 αβ − pβ − (1 − p)α
c
c
c
r
λ
λ 2
λ
±
β−α−
+ 4p (β − α) .
c
c
c
λ
±
c
Now we got three solutions. But there should only be two. Note that
p 1 − p λ
λ
αβ c−λ
+
>0
αβ − pβ − (1 − p)α =
c
c
c
α
β
and thus both solutions are positive. The larger of the solutions can be written as
r
λ
1
λ
λ 2
+ 4p (β − α)
α+
β−α− +
β−α−
2
c
c
c
and is thus larger than α. But the moment generating function does not exist for
r ≥ α. Thus
r
1
λ
λ 2
λ
R=
α+β− −
+ 4p (β − α) .
β−α−
2
c
c
c
From Lundberg’s inequality it follows that
ψ(u) < e−Ru .
74
4. THE CRAMÉR–LUNDBERG MODEL
4.6. The Cramér–Lundberg Approximation
Consider the equation (4.2). This equation looks almost like a renewal equation,
but
Z ∞
λµ
λ
(1 − G(x)) dx =
<1
c
c
0
is not a probability distribution. Can we manipulate (4.2) to get a renewal equation?
We try for some measurable function h and h̄
Z ∞
Z u
λ
λ
ψ(u−x)h(u−x)h̄(x) (1−G(x)) dx (4.7)
ψ(u)h(u) = h(u)
(1−G(x)) dx+
c
c
u
0
where h(u) = h(u−x)h̄(x). We can assume that h(0) = 1. Setting x = u shows that
h̄(u) = h(u). From a general theorem it follows that h(u) = eru where r = log h(1).
In order that (4.7) is a renewal equation we need
Z ∞
Z Z
λ ∞ ∞ rx
λ(MY (r) − 1)
rx λ
1=
e
(1 − G(x)) dx =
e dG(y) dx =
.
c
c 0 x
cr
0
The only solution to the latter equation is r = R. Let us assume that R exists.
Moreover, we need that
Z ∞
λ
xeRx (1 − G(x)) dx < ∞
(4.8)
c
0
which is equivalent to MY0 (R) < ∞, see below. The equation is now
Z ∞
Z u
λ
λ
Ru
Ru
ψ(u)e = e
(1 − G(x)) dx +
ψ(u − x)eR(u−x) eRx (1 − G(x)) dx .
c
c
u
0
It can be shown that
Z
∞
λ
(1 − G(x)) dx
c
u
is directly Riemann integrable. Thus we get from the key renewal theorem
e
Ru
Theorem 4.6. Assume that the Lundberg exponent exists and that (4.8) is fulfilled.
Then
c − λµ
lim ψ(u)eRu =
.
u→∞
λMY0 (R) − c
Proof. It only remains to compute the limit.
Z ∞
Z ∞
Z Z
λ
λ ∞ x Ru
Ru
e
(1 − G(x)) dx du =
e du (1 − G(x)) dx
c
c 0 0
0
u
Z ∞
λ
1
λµ
1
=
(eRx − 1)(1 − G(x)) dx = −
=
(c − λµ) .
cR 0
R cR
cR
4. THE CRAMÉR–LUNDBERG MODEL
75
The mean value of the distribution is
Z ∞
Z Z
λ ∞ ∞ Rx
Rx λ
xe
(1 − G(x)) dx =
xe dG(y) dx
c
c 0 x
0
Z ∞Z y
Z ∞
λ
λ
Rx
xe dx dG(y) =
(RyeRy − eRy + 1) dG(y)
=
c 0 0
cR2 0
λ(RMY0 (R) − MY (R) + 1)
λRMY0 (R) − cR
λMY0 (R) − c
=
.
=
=
cR2
cR2
cR
The limit value follows readily.
Remark.
The theorem shows that it is not possible to obtain an exponential
upper bound for the ruin probability with an exponent strictly larger than R. The theorem can be written in the following form
c − λµ
ψ(u) ∼
e−Ru .
0
λMY (R) − c
Thus for large u we get an approximation to ψ(u). This approximation is called the
Cramér–Lundberg approximation.
Example 4.2 (continued).
From MY (r) we get that
α
MY0 (r) =
(α − r)2
and thus
lim ψ(u)eRu =
c−
λα
(α−R)2
λ
α
c−
λα
λ
α
=
2 − c
( λc )
Hence the Cramér–Lundberg approximation
λ −(α−λ/c)u
ψ(u) ∼
e
αc
becomes exact in this case.
u→∞
−c
=
λ αc − λ
λ
=
.
αc αc − λ
αc
Example 4.7. Let c = λ = 1 and G(x) = 1 − 13 (e−x + e−2x + e−3x ). The mean
value of claim sizes is µ = 0.611111, i.e. the net profit condition is fulfilled. One can
show that
ψ(u) = 0.550790 e−0.485131u + 0.0436979 e−1.72235u + 0.0166231 e−2.79252u .
From Theorem 4.6 it follows that app(u) = 0.550790 e−0.485131u is the Cramér–
Lundberg approximation to ψ(u). Table 4.1 below shows the ruin function ψ(u), its
Cramér–Lundberg approximation app(u) and the relative error (app(u)−ψ(u))/ψ(u)
multiplied by 100 (Er). Note that the the relative error is below 1% for u ≥
1.71358 = 2.8µ.
76
4. THE CRAMÉR–LUNDBERG MODEL
u
ψ(u)
app(u)
Er
u
ψ(u)
app(u)
Er
0
0.25
0.5
0.75
1
0.6111 0.5246 0.4547 0.3969 0.3479
0.5508 0.4879 0.4322 0.3828 0.3391
-9.87
-6.99
-4.97
-3.54
-2.54
1.25
1.5
1.75
2
2.25
0.3059 0.2696 0.2379 0.2102 0.1858
0.3003 0.2660 0.2357 0.2087 0.1849
-1.82
-1.32
-0.95
-0.69
-0.50
Table 4.1: Cramér–Lundberg approximation to ruin probabilities
4.7. Reinsurance and Ruin
4.7.1.
Proportional Reinsurance
Recall that for proportional insurance the insurer covers YiI = αYi of each claim,
the reinsurer covers YiR = (1 − α)Yi . Denote by cI the insurer’s premium rate. The
insurer’s adjustment coefficient is obtained from the equation
λ(MαY (r) − 1) − cI r = 0 .
The new moment generating function is
MαY (r) = IIE[erαYi ] = MY (αr) .
Assume that both insurer and reinsurer use an expected value premium principle
with the same safety loading. Then cI = αc and we have to solve
λ(MY (αr) − 1) − cαr = 0 .
This is almost the original equation, hence R = αRI where RI is the adjustment
coefficient under reinsurance. The new adjustment coefficient is larger, hence the
risk has become smaller.
We assume now that there is r∞ such that MY (r) < ∞ if r < r∞ and MY (r∞ ) =
∞. Write c = (1 + ξ)λµ for some ξ > 0. Suppose the reinsurer charges the premium
(1 + ϑ)(1 − α)λµ. We now want to choose the retention level α, such that the
corresponding adjustment coefficient R(α) becomes maximal. In some sense, this
means that the ruin probability is minimised for large initial capital. In order that
the solution is not trivial, we assume ϑ > ξ. R(α) is determined by the equation
λ[MY (αR(α)) − 1] − [α(1 + ϑ) − (ϑ − ξ)]R(α)λµ = 0 .
4. THE CRAMÉR–LUNDBERG MODEL
77
In order that the net profit condition is fulfilled, we need α(1 + ϑ) − (ϑ − ξ) > α, or
equivalently, α > 1 − ξ/ϑ. By the implicit function theorem, we have
R0 (α)[αMY0 (αR(α)) − {α(1 + ϑ) − (ϑ − ξ)}µ] + R(α)MY0 (αR(α)) − (1 + ϑ)R(α)µ = 0 .
Thus, R0 (α∗ ) = 0 implies MY0 (α∗ R(α∗ )) = (1 + ϑ)µ. In particular, MY0 (α∗ R(α∗ )) −
{α∗ (1 + ϑ) − (ϑ − ξ)}µ = {(1 − α∗ )(1 + ϑ) + (ϑ − ξ)}µ > 0. Differentiating a second
time yields
R00 (α)[αMY0 (αR(α)) − {α(1 + ϑ) − (ϑ − ξ)}µ] + R0 (α)h(α) + R2 (α)MY00 (αR(α)) = 0
for some function h(α). Because R2 (α)MY00 (αR(α)) > 0, we get that R00 (α∗ ) < 0.
We see that R(α) has a unique maximum, and R(α) is increasing on [1 − ξ/ϑ, α∗ ∧ 1]
and decreasing on [α∗ ∧ 1, 1].
We now write r(α) = αR(α). Then we have to solve
MY (r(α)) − 1 − (1 + ϑ)r(α)µ + (ϑ − ξ)r(α)µ/α = 0 .
We have seen above that r∗ = r(α∗ ) is the unique value for which MY0 (r∗ ) = (1+ϑ)µ.
Note that such a r∗ < r∞ exists under our assumptions.
r(α) is increasing on [0, α0 ] where α0 > α∗ . We can therefore invert r(α), write
α(r). We get
(ϑ − ξ)rµ
α(r) =
.
(1 + ϑ)rµ − (MY (r) − 1)
Thus the optimal retention level is
o
n
(ϑ − ξ)r∗ µ
∗
,1 .
α = min
(1 + ϑ)r∗ µ − (MY (r∗ ) − 1)
If α(r∗ ) ≤ 1 then R(α∗ ) = r∗ /α(r∗ ).
Example 4.8. Suppose that the claims are Γ(γ, α) distributed. Then µ = γ/α,
MY (r) = (α/(α − r))γ and MY0 (r) = γ(α/(α − r))γ+1 /α = (1 + ϑ)γ/α. Thus
r∗ = α(1 − (1 + ϑ)−1/(γ+1) ). Thus, after some simplifications,
n γ(ϑ − ξ)[1 − (1 + ϑ)−1/(γ+1) ]
o
α∗ = min
,
1
.
γϑ + (γ + 1)(1 − (1 + ϑ)γ/(γ+1) )
If α∗ < 1, then
αγ(ϑ − ξ)
.
γϑ + (γ + 1)(1 − (1 + ϑ)γ/(γ+1) )
Note that in general there is no closed formula for the adjustment coefficient for
Gamma distributed claim sizes. However, for the maximal adjustment coefficient
we obtain an explicit formula.
R∗ =
78
4. THE CRAMÉR–LUNDBERG MODEL
4.7.2.
Excess of Loss Reinsurance
Under excess of loss reinsurance with retention level M the insurer has to pay
YiI = min{Yi , M } of each claim. The adjustment coefficient is the strictly positive
solution to the equation
Z M
rx
rM
λ
e dG(x) + e (1 − G(M )) − 1 − cI r = 0 .
0
There is no possibility to find the solution from the problem without reinsurance.
We have to solve the equation for every M separately. But note that RI exists
in any case. Especially for heavy tailed distributions this shows that the risk has
become much smaller. By the Cramér–Lundberg approximation the ruin probability
decreases exponentially as the initial capital increases.
We will now show that for the insurer the excess of loss reinsurance is optimal.
We assume that both insurer and reinsurer use an expected value principle.
Proposition 4.9. Let all premia be computed via the expected value principle.
Under all reinsurance forms acting on individual claims with premium rates cI and
cR fixed the excess of loss reinsurance maximizes the insurer’s adjustment coefficient.
Proof. Let h(x) be an increasing function with 0 ≤ h(x) ≤ x for x ≥ 0. We
assume that the insurer pays YiI = h(Yi ). Let h∗ (x) = min{x, U } be the excess of
loss reinsurance. Because cI is fixed U can be determined from
Z ∞
Z U
∗
h(y) dG(y) .
y dG(y) + U (1 − G(U )) = IIE[h (Yi )] = IIE[h(Yi )] =
0
0
Because ez ≥ 1 + z we obtain
∗ (y))
er(h(y)−h
≥ 1 + r(h(y) − h∗ (y))
and
erh(y) ≥ erh
∗ (y)
(1 + r(h(y) − h∗ (y))) .
Thus
Z
∞
Z
∞
∗
dG(y) ≥
erh (y) (1 + r(h(y) − h∗ (y))) dG(y)
0
Z ∞ 0
∗
= Mh∗ (Y ) (r) + r
(h(y) − h∗ (y)) erh (y) dG(y) .
Mh(Y ) (r) =
rh(y)
e
0
4. THE CRAMÉR–LUNDBERG MODEL
79
For y ≤ U we have h(y) ≤ y = h∗ (y). We obtain for r > 0
Z ∞
∗
(h(y) − h∗ (y))erh (y) dG(y)
0
Z ∞
Z U
∗
∗
rh∗ (y)
(h(y) − h (y))e
dG(y) +
(h(y) − h∗ (y))erh (y) dG(y)
=
0
U
Z ∞
Z U
(h(y) − h∗ (y))erU dG(y) +
(h(y) − h∗ (y))erU dG(y)
≥
0
U
Z ∞
(h(y) − h∗ (y)) dG(y)
= erU
0
= erU (IIE[h(Y )] − IIE[h∗ (Y )]) = 0 .
It follows that Mh(Y ) (r) ≥ Mh∗ (Y ) (r) for r > 0. Since
0 = θ(RI ) = λ(Mh(Y ) (RI ) − 1) − cI RI ≥ λ(Mh∗ (Y ) (RI ) − 1) − cI RI = θ∗ (RI )
and the assertion follows from the convexity of θ∗ (r).
Suppose now that we want to maximise the adjustment coefficient. Suppose
again that the safety loading of the insurer is ξ and the safety loading of the reinsurer
is ϑ > ξ. Note that
Z M
r(Y ∧M )
IIE[e
− 1] =
rery (1 − G(y)) dy
0
and
Z
∞
(1 − G(y)) dy .
IIE[max{Y − M, 0}] =
M
The net profit condition becomes, after dividing by λ,
Z M
Z ∞
Z
(1 + ξ)
(1 − G(y)) dy − (ϑ − ξ)
(1 − G(y)) dy >
0
M
or equivalently
Z
M
(1 − G(y)) dy ,
0
∞
(1 − G(y)) dy .
ξµ > ϑ
M
For the ajustment coefficient we have to find the solution to
Z M
Z ∞
ry
λ
r(e − (1 + ξ))(1 − G(y)) dy + (ϑ − ξ)λr
(1 − G(y)) dy = 0 .
0
M
Since R 6= 0, we look for the solution to
Z M
Z
ry
(e − (1 + ξ))(1 − G(y)) dy + (ϑ − ξ)
0
∞
M
(1 − G(y)) dy = 0 .
80
4. THE CRAMÉR–LUNDBERG MODEL
Taking the derivative, we get from the implicit function theorem
Z M
0
yeR(M )y (1 − G(y)) dy + [eR(M )M − (1 + ϑ)](1 − G(M )) = 0 .
R (M )
0
If R0 (M ) = 0, then eR(M )M = 1 + ϑ. We conclude that it is not optimal not to
take reinsurance. Indeed, as M → ∞ the adjustment coefficient R(M ) tends to the
adjustemt coefficient of the model without reinsurance, and therefore R(M )M → ∞.
If the adjustment coefficient does not exist but exponential moments exist, then the
ruin probability decreases exponentially at the rate r∞ , and also r∞ M → ∞ as
M → ∞. Finally, if no exponential moments exist, the adjustment coefficient exists
after reinsurance.
Suppose for the moment that G(y) is differentiable. Then the second derivative
gives
Z M
00
yeR(M )y (1 − G(y)) dy + R0 (M )h(M )
R (M )
0
+ R(M )eR(M )M (1 − G(M )) − [eR(M )M − (1 + ϑ)]G0 (M ) = 0
for some function h(M ). At a point where R0 (M ) = 0, we have
Z M
00
R (M )
yeR(M )y (1 − G(y)) dy + R(M )eR(M )M (1 − G(M )) = 0 ,
0
where we used that eR(M )M = 1 + ϑ. We conclude that R00 (M ) < 0. This shows
that R(M ) has a unique local maximum, and hence is unimodal. For M < M ∗ , the
optimal retention level, we have R0 (M ) > 0, and thus eR(M )M < 1+ϑ. For M > M ∗ ,
we have R0 (M ) < 0, and therefore eR(M )M > 1 + ϑ. The solution is thus found by
calculating R(M )M , and find the unique value for which R(M )M = log(1 + ϑ).
We consider now the portfolio of the reinsurer. What is the claim number process
of the claims the reinsurer is involved? Because the claim amounts are independent
of the claim arrival process we delete independently points from the Poisson process
with probability G(M ) and do not delete them with probability 1 − G(M ). By
Proposition C.3 this process is a Poisson process with rate λ(1 − G(M )). Because
the claim sizes are iid. and independent of the claim arrival process the surplus of
the reinsurer is a Cramér–Lundberg process with intensity λ(1 − G(M )) and claim
size distribution
G̃(x) = IIP[Yi − M ≤ x | Yi > M ] =
G(M + x) − G(M )
.
1 − G(M )
4. THE CRAMÉR–LUNDBERG MODEL
81
4.8. The Severity of Ruin, the Capital Prior to Ruin and the Distribution of inf{Ct : t ≥ 0}
For an insurance company ruin is not so dramatic if −Cτ is small, but it could ruin
the whole company if −Cτ is very large. So we are interested in the distribution of
−Cτ if ruin occurs. The random variable −Cτ is called the severity of ruin. We
further want to know how ruin happens. Thus, we also consider the capital prior to
ruin Cτ − . Let
ψx,y (u) = IIP[τ < ∞, Cτ < −x, Cτ − > y] .
We proceed as in Section 4.3. For h small
Z h hZ u+ct
−λh
ψx,y (u) = e ψx,y (u + ch) +
ψx,y (u + ct − z) dG(z)
0
0
i
+ 1Iu+ct>y (1 − G(u + ct + x)) λe−λt dt .
It follows that ψx,y (u) is right continuous. We obtain the integro-differential equation
Z u
h
i
0
cψx (u) = λ ψx,y (u) −
ψx,y (u − z) dG(z) − 1Iu≥y (1 − G(u + x)) .
0
Replacing u by u − ch gives the derivative from the left. But ψx,y (u) is not differentiable at points where G(u) jumps and at u = y. Integration of the integrodifferential equation yields
Z u
Z u
c
(ψx,y (u) − ψx,y (0)) =
ψx,y (u − t)(1 − G(t)) dt −
(1 − G(t + x)) dt
λ
0
u∧y
Z x+u
Z u
(1 − G(t)) dt .
ψx,y (u − t)(1 − G(t)) dt −
=
0
(x+u)∧(x+y)
Because ψx,y (u) ≤ ψ(u) we can see that ψx,y (u) → 0 as u → ∞. By the bounded
convergence theorem we obtain
Z ∞
c
− ψx,y (0) = −
(1 − G(t)) dt
λ
x+y
and
IIP[Cτ < −x, Cτ − > y | τ < ∞, C0 = 0] =
λ
c
1
=
µ
Z
∞
(1 − G(t)) dt
x+y
Z
∞
λµ
c
(1 − G(t)) dt .
(4.9)
x+y
If we choose y = u = 0, we get that −Cτ has the conditional distribution B(x) =
Rx
µ−1 0 (1 − G(t)) dt given that ruin occurs.
82
4. THE CRAMÉR–LUNDBERG MODEL
Remark. In order to get an equation for the ruin probability we can consider the
first time point τ1 where the surplus is below the initial capital. At this point, by
the strong Markov property, a new Cramér–Lundberg process starts. We get three
possibilities:
• The process gets never below the initial capital.
• τ1 < ∞ but Cτ1 ≥ 0.
• Ruin occurs at τ1 .
Thus we get
Z
Z
λ ∞
λ u
λµ ψ(u) = 1 −
ψ(u − y)(1 − G(y)) dy +
1 (1 − G(y)) dy .
0+
c
c 0
c u
This is equation (4.2). Thus we have now a natural interpretation of (4.2).
Let τ0 = 0 and τi = inf{t > τi−1 : Ct < Cτi−1 }, called the ladder times,
and define Li = Cτi−1 − Cτi , called the ladder heights. Note that Li only is
defined if τi < ∞. By Lemma 4.1 we find IIP[τi < ∞ | τi−1 < ∞] = λµ/c. Let
K = sup{i ∈ IIN : τi < ∞} be the number of ladder epochs. We have just seen that
K ∼ NB(1, 1 − λµ/c) and that, given K, the random variables (Li : i ≤ K) are
iid. and absolutely continuous with density (1 − G(x))/µ. We only have to condition
on K because Li is not defined for i > K. If we assume that all (Li : i ≥ 1) have the
same distribution, then we can drop the conditioning on K and (Li ) is independent
of K. Then
K
X
inf{Ct : t ≥ 0} = u −
Li
i=1
and
K
hX
i
IIP[τ < ∞] = IIP[inf{Ct : t ≥ 0} < 0] = IIP
Li > u .
i=1
We can use Panjer recursion to approximate ψ(u) by using an appropriate discretization.
A formula that is useful for theoretical considerations, but too messy to use for
the computation of ψ(u) is the Pollaczek–Khintchine formula.
K
∞
n
hX
i X
hX
i
ψ(u) = IIP
Li > u =
IIP
Li > u IIP[K = n]
n=1
i=1
= 1−
∞
λµ X λµ n
c
n=1
c
i=1
(1 − B ∗n (u)) .
(4.10)
4. THE CRAMÉR–LUNDBERG MODEL
83
4.9. The Laplace Transform of ψ
Definition 4.10. Let f be a real function on [0, ∞). The transform
Z ∞
ˆ
e−sx f (x) dx
(s ∈ IR)
f (s) :=
0
is called the Laplace transform of f .
If X is an absolutely continuous positive random variable with density f then fˆ(s) =
MX (−s). The Laplace transform has the following properties.
Lemma 4.11.
i) If f (x) ≥ 0 a.e.
fˆ(s1 ) ≤ fˆ(s2 )
⇐⇒
s1 ≥ s2 .
c|(s1 ) < ∞ =⇒ |f
c|(s) < ∞ for all s ≥ s1 .
|f
Z ∞
0
fb (s) =
e−sx f 0 (x) dx = sfˆ(s) − f (0)
ii)
iii)
0
c|(s) < ∞.
provided f 0 (x) exists a.e. and |f
lim sfˆ(s) = lim f (x)
iv)
s→∞
x→0
0
c|(s) < ∞ for an s large
provided f (x) exists a.e., limx→0 f (x) exists and |f
enough.
lim sfˆ(s) = lim f (x)
v)
x→∞
s↓0
c|(s) < ∞ for some s > 0.
provided f 0 (x) exists a.e., limx→∞ f (x) exists and |f
vi)
fˆ(s) = ĝ(s) on (s0 , s1 )
=⇒
f (x) = g(x) Lebesgue a.e.
∀x ∈ [0, ∞).
We want to find the Laplace transform of δ(u). We multiply (4.1) with e−su and
then integrate over u. Let s > 0.
Z ∞
Z ∞
Z ∞Z u
0
−su
−su
c
δ (u)e
du = λ
δ(u)e
du − λ
δ(u − y) dG(y)e−su du.
0
0
0
0
We have to determine the last integral.
Z ∞Z u
Z ∞Z ∞
−su
δ(u − y) dG(y)e
du =
δ(u − y)e−su du dG(y)
0
0
Z0 ∞Zy ∞
=
δ(u)e−s(u+y) du dG(y) = δ̂(s)MY (−s) .
0
0
84
4. THE CRAMÉR–LUNDBERG MODEL
Thus we get the equation
c(sδ̂(s) − δ(0)) = λδ̂(s)(1 − MY (−s))
which has the solution
δ̂(s) =
c − λµ
cδ(0)
=
.
cs − λ(1 − MY (−s))
cs − λ(1 − MY (−s))
The Laplace transform of ψ can easily be found as
Z ∞
1
(1 − δ(u))e−su du = − δ̂(s) .
ψ̂(s) =
s
0
Example 4.2 (continued).
δ̂(s) =
For exponentially distributed claims we obtain
c − λ/α
c − λ/α
(c − λ/α)(α + s)
=
=
.
cs − λ(1 − α/(α + s))
s(c − λ/(α + s))
s(c(α + s) − λ)
and
1 (c − λ/α)(α + s)
c(α + s) − λ − (c − λ/α)(α + s)
−
=
s
s(c(α + s) − λ)
s(c(α + s) − λ)
λ
1
λ
1
=
=
.
α c(α + s) − λ
αc α − λ/c + s
ψ̂(s) =
By comparison with the moment generating function of the exponential distribution
we recognize that
λ −(α−λ/c)u
ψ(u) =
e
.
αc
Example 4.5 (continued).
For the Laplace transform of δ we get
c − λp/α − λ(1 − p)/β
cs − λ(1 − αp/(α + s) − β(1 − p)/(β + s))
c − λp/α − λ(1 − p)/β
=
s(c − λp/(α + s) − λ(1 − p)/(β + s))
δ̂(s) =
and for the Laplace transform of ψ
c − λp/(α + s) − λ(1 − p)/(β + s) − c + λp/α + λ(1 − p)/β
s(c − λp/(α + s) − λ(1 − p)/(β + s))
sp/(α(α + s)) + s(1 − p)/(β(β + s))
=λ
s(c − λp/(α + s) − λ(1 − p)/(β + s))
p(β + s)/α + (1 − p)(α + s)/β
=λ
c(α + s)(β + s) − λp(β + s) − λ(1 − p)(α + s)
p(β + s)/α + (1 − p)(α + s)/β
=λ 2
.
cs + ((α + β)c − λ)s + αβc − λ((1 − p)α + pβ)
ψ̂(s) =
4. THE CRAMÉR–LUNDBERG MODEL
85
The denominator can be written as c(s + R)(s + R̄) where R and R̄ are the two
solutions to (4.6) with R < R̄. The Laplace transform of ψ can be written in the
form
A
B
+
ψ̂(s) =
R + s R̄ + s
for some constants A and B. Hence
ψ(u) = Ae−Ru + Be−R̄u .
A must be the constant appearing in the Cramér–Lundberg approximation. The
constant B can be found from ψ(0). We can see that in this case the Cramér–
Lundberg approximation is not exact.
Recall that τ1 is the first ladder epoch. On the set {τ1 < ∞} the random variable
sup{u − Ct : t ≥ 0} is absolutely continuous with distribution function
IIP[sup{u − Ct : t ≥ 0} ≤ x | τ1 < ∞] = 1 −
c
ψ(x) .
λµ
Let Z be a random variable with the above distribution. Its moment generating
function is
Z ∞
c
c c − λµ
λµ MZ (r) =
eru δ 0 (u) du =
−r
− 1−
λµ
λµ
−cr − λ(1 − MY (r))
c
0
c(c − λµ)
c
r
+
.
=1−
λµ
λµ
cr − λ(MY (r) − 1)
We will later need the first two moments of the above distribution function. Assume
that µ2 < ∞. The first derivative of the moment generating function is
c(c − λµ) cr − λ(MY (r) − 1) − r(c − λMY0 (r))
λµ
(cr − λ(MY (r) − 1))2
c(c − λµ) rMY0 (r) − (MY (r) − 1)
=
.
µ
(cr − λ(MY (r) − 1))2
MZ0 (r) =
Note that
1
lim (cr − λ(MY (r) − 1)) = c − λµ .
r→0 r
We find
rMY0 (r) − (MY (r) − 1)
MY0 (r) + rMY00 (r) − MY0 (r)
µ2
=
lim
=
r→0
r→0
r2
2r
2
lim
and thus
IIE[Z] =
c(c − λµ)
µ2
cµ2
=
.
2
µ
2(c − λµ)
2µ(c − λµ)
(4.11)
86
4. THE CRAMÉR–LUNDBERG MODEL
Assume now that µ3 < ∞. The second derivative of MZ (r) is
c(c − λµ)
1
00
MZ (r) =
rMY00 (r)(cr − λ(MY (r) − 1))
3
µ
(cr − λ(MY (r) − 1))
− 2(rMY0 (r) − (MY (r) − 1))(c − λMY0 (r)) .
For the limit to 0 we find
1 lim 3 rMY00 (r)(cr − λ(MY (r) − 1))
r→0 r
− 2(rMY0 (r) − (MY (r) − 1))(c − λMY0 (r))
1 00
= lim 2 (MY (r) + rMY000 (r))(cr − λ(MY (r) − 1))
r→0 3r
+ rMY00 (r)(c − λMY0 (r)) − 2rMY00 (r)(c − λMY0 (r))
+ 2(rMY0 (r) − (MY (r) − 1))λMY00 (r)
rMY0 (r) − (MY (r) − 1)
µ3
(c − λµ) + λµ2 lim
r→0
3
r2
µ3
λ
= (c − λµ) + µ22 .
3
2
Thus the second moment of Z becomes
c(c − λµ) µ3 (c − λµ)/3 + λµ22 /2
c
µ3
λµ22
2
IIE[Z ] =
=
+
.
µ
(c − λµ)3
µ 3(c − λµ) 2(c − λµ)2
(4.12)
=
4.10. Approximations to ψ
4.10.1.
Diffusion Approximations
Diffusion approximations are based on the following
(n)
Proposition 4.12. Let {Ct } be a sequence of Cramér–Lundberg processes with
initial capital u(n) = u, claim arrival intensities λ(n) = λn, claim size distributions
√
G(n) (x) = G(x n) and premium rates
√
c − λµ (n) (n)
(n)
√ λ µ = c + ( n − 1)λµ .
c = 1+
λµ n
R∞
R∞
Let µ = 0 y dG(y) and assume that µ2 = 0 y 2 dG(y) < ∞. Then
(n)
d
{Ct } → {u + Wt }
in distribution in the topology of uniform convergence on finite intervals where {Wt }
is a (c − λµ, λµ2 )-Brownian motion.
Proof.
See [57] or [46].
4. THE CRAMÉR–LUNDBERG MODEL
87
Intuitively we let the number of claims in a unit time interval go to infinity and
(n)
make the claim sizes smaller in such a way that the distribution of C1 − u tends
(n)
to a normal distribution and IIE[C1 − u] = c − λµ. Let τ (n) denote the ruin time
(n)
of {Ct } and τ = inf{t ≥ 0 : u + Wt < 0} the ruin probability of the Brownian
motion. Then
(n)
Proposition 4.13. Let (Ct ) and (Wt ) be as above. Then
lim IIP[τ (n) ≤ t] = IIP[τ ≤ t]
n→∞
and
lim IIP[τ (n) < ∞] = IIP[τ < ∞] .
n→∞
Proof. The result for a finite time horizon is a special case of [89, Thm.9], see
also [57] or [46]. The result for the infinite time horizon can be found in [72].
The idea of the diffusion approximation is to approximate IIP[τ (1) ≤ t] by IIP[τ ≤ t]
and IIP[τ (1) < ∞] by IIP[τ < ∞]. Thus we need the ruin probabilities of the Brownian
motion.
Lemma 4.14. Let {Wt } be a (m, η 2 )-Brownian motion with m > 0 and τ =
inf{t ≥ 0 : u + Wt < 0}. Then
IIP[τ < ∞] = e−2um/η
and
Proof.
2
mt − u mt + u −2um/η 2
√
√
+e
Φ
.
IIP[τ ≤ t] = 1 − Φ
η t
η t
By Lemma D.3 the process
n 2m(u + W ) o
t
exp −
η2
is a martingale. By the stopping theorem
n 2m(u + W ) o
τ ∧t
exp −
η2
is a positive bounded martingale. Thus, because limt→∞ Wt = ∞,
n 2um o
h n 2m(u + W ) oi
τ
exp − 2
= IIE exp −
= IIP[τ < ∞] .
η
η2
88
4. THE CRAMÉR–LUNDBERG MODEL
u
ψ(u)
DA
Er
u
ψ(u)
DA
Er
0
0.25
0.5
0.75
1
0.6111 0.5246 0.4547 0.3969 0.3479
1.0000 0.8071 0.6514 0.5258 0.4244
63.64 53.87 43.26 32.49 21.98
1.25
1.5
1.75
2
2.25
0.3059 0.2696 0.2379 0.2102 0.1858
0.3425 0.2765 0.2231 0.1801 0.1454
11.96
2.54
-6.22 -14.32 -21.78
Table 4.2: Diffusion approximation to ruin probabilities
It is easy to see that {sW1/s − m} is a (0, η 2 )-Brownian motion (see [60, p.351]) and
thus {s(u + W1/s ) − m} is (u, η 2 )-Brownian motion. Denote the latter process by
{W̃s }. Then
IIP[τ ≤ t] = IIP[inf{u + Ws : 0 < s ≤ t} < 0]
= IIP[inf{s(u + W1/s ) : s ≥ 1/t} < 0]
= IIE[IIP[inf{m + W̃s : s ≥ 1/t} < 0 | W̃1/t ]]
Z ∞ 2u(y+m)
Z −m
(y−u/t)2
(y−u/t)2
1
1
−
−
−
2
2
2η
/t
η
p
p
e
dy +
e
e 2η2 /t dy
=
2
2
2πη /t
2πη /t
−m
−∞
Z ∞
2
−m − u/t (y+u/t)
1
−
− 2um
√
p
+ e η2
=Φ
e 2η2 /t dy
η/ t
2πη 2 /t
−m
mt + u mt − u − 2um
√
√
+ e η2 Φ
.
=1−Φ
η t
η t
Diffusion approximations only work well if c/(λµ) is close to 1. There also exist
corrected diffusion approximations which work much better, see [79] or [7].
Example 4.7 (continued). Let c = λ = 1 and G(x) = 1 − 13 (e−x + e−2x + e−3x ).
We find c − λµ = 7/18 and λµ2 = 49/54. This leads to the diffusion approximation ψ(u) ≈ exp{−6u/7}. Table 4.2 shows exact values (ψ(u)), the diffusion
approximation (DA) and the relative error multiplied by 100 (Er). Here we have
c/(λµ) = 18/11 = 1.63636 is not close to one. This is also indicated by the figures.
4. THE CRAMÉR–LUNDBERG MODEL
4.10.2.
89
The deVylder Approximation
In the case of exponentially distributed claim amounts we know the ruin probabilities
explicitly. The idea of the deVylder approximation is to replace {Ct } by {C̃t } where
{C̃t } has exponentially distributed claim amounts and
IIE[(Ct − u)k ] = IIE[(C̃t − u)k ]
for k = 1, 2, 3.
The first three (centralized) moments are
λ̃ t,
IIE[Ct − u] = (c − λµ)t = c̃ −
α̃
Var[Ct ] = Var[u + ct − Ct ] = λµ2 t =
2λ̃
t
α̃2
and
IIE[(Ct − IIE[Ct ])3 ] = −IIE[(u + ct − Ct − IIE[u + ct − Ct ])3 ] = −λµ3 t = −
6λ̃
t.
α̃3
The parameters of the approximation are
α̃ =
λ̃ =
3µ2
,
µ3
λµ2 α̃2
9µ3
= 22 λ
2
2µ3
and
λ̃
3µ2
= c − λµ + 2 λ .
α̃
2µ3
Thus the approximation to the probability of ultimate ruin is
c̃ = c − λµ +
λ̃ −
ψ(u) ≈
e
α̃c̃
α̃− λ̃c̃ u
.
There
is also a formula for the probability of ruin within finite time. Let η =
q
λ̃/(α̃c̃). Then
Z
λ̃ −(α̃−λ̃/c̃)u 1 π
ψ(u, t) ≈
e
−
f (x) dx
(4.13)
α̃c̃
π 0
where
f (x) = η
exp{2η α̃c̃t cos x − (α̃c̃ + λ̃)t + α̃u(η cos x − 1)}
1 + η 2 − 2η cos x
·(cos(α̃uη sin x) − cos(α̃uη sin x + 2x)) .
A numerical investigation shows that the approximation is quite accurate.
90
4. THE CRAMÉR–LUNDBERG MODEL
u
0
ψ(u)
DV
Er
u
0.25
0.75
1
0.6111 0.5246 0.4547 0.3969 0.3479
0.5774 0.5102 0.4509 0.3984 0.3520
-5.51
-2.73
-0.86
0.38
1.18
1.25
ψ(u)
DV
Er
0.5
1.5
1.75
2
2.25
0.3059 0.2696 0.2379 0.2102 0.1858
0.3110 0.2748 0.2429 0.2146 0.1896
1.67
1.95
2.07
2.09
2.03
Table 4.3: DeVylder approximation to ruin probabilities
Example 4.7 (continued). In addition to the previously calculated values we also
need µ3 = 251/108. The approximation parameters are α̃ = 1.17131, λ̃ = 0.622472
and c̃ = 0.920319. This leads to the approximation ψ(u) ≈ 0.577441e−0.494949u . It
turns out that the approximation works well.
4.10.3.
The Beekman–Bowers Approximation
Recall from (4.11) and (4.12) that
F (u) = 1 −
is a distribution function and that
Z ∞
z dF (z) =
0
and that
c
ψ(u)
λµ
cµ2
2µ(c − λµ)
∞
c
µ3
λµ22
+
.
µ 3(c − λµ) 2(c − λµ)2
0
The idea is to approximate the distribution function F by the distribution function
F̃ (u) of a Γ(γ, α) distributed random variable such that the first two moments
coincide. Thus the parameters γ and α have to fulfil
γ
cµ2
=
,
α
2µ(c − λµ)
γ(γ + 1)
c
µ3
λµ22
=
+
.
α2
µ 3(c − λµ) 2(c − λµ)2
Z
z 2 dF (z) =
The Beekman–Bowers approximation to the ruin probability is
λµ
λµ
ψ(u) =
(1 − F (u)) ≈
(1 − F̃ (u)) .
c
c
Remark. If 2γ ∈ IIN then 2αZ ∼ χ22γ is χ2 distributed.
4. THE CRAMÉR–LUNDBERG MODEL
u
0
0.25
0.5
91
0.75
1
ψ(u)
0.6111 0.5246 0.4547 0.3969 0.3479
BB1
Er
0.6111 0.5227 0.4553 0.3985 0.3498
0.00
-0.35
0.12
0.42
0.54
BB2
Er
0.6111 0.5105 0.4456 0.3914 0.3450
0.00
-2.68
-2.02
-1.38
-0.83
u
1.25
1.5
1.75
2
2.25
ψ(u)
0.3059 0.2696 0.2379 0.2102 0.1858
BB1
Er
0.3076 0.2709 0.2387 0.2106 0.1859
0.54
0.47
0.34
0.19
0.04
BB2
Er
0.3046 0.2693 0.2383 0.2110 0.1869
-0.42
-0.11
0.18
0.40
0.59
Table 4.4: Beekman–Bowers approximation to ruin probabilities
Example 4.7 (continued).
solve the equations
For the Beekman–Bowers approximation we have to
γ
= 1.90909 ,
α
γ(γ + 1)
= 7.71429 ,
α2
which yields the parameters γ = 0.895561 and α = 0.469104. From this the
Beekman–Bowers approximation can be obtained. Here we have 2γ = 1.79112
which is not close to an integer. Anyway, one can interpolate between the χ21 and
the χ22 distribution function to get the approximation
0.20888χ21 (2αu) + 0.79112χ22 (2αu)
to 1 − c/(λµ) ψ(u). Table 4.4 shows the exact values (ψ(u)), the Beekman–Bowers
approximation (BB1) and the approximation obtained by interpolating the χ2 distributions (BB2). The relative errors (Er) are given in percent. One can clearly see
that all the approximations work well.
4.11. Subexponential Claim Size Distributions
Let us now consider subexponential claim size distributions. In this case the Lundberg exponent does not exist (Lemma F.3).
92
4. THE CRAMÉR–LUNDBERG MODEL
Theorem 4.15. Assume that the ladder height distribution
Z
1 x
(1 − G(y)) dy
µ 0
is subexponential. Then
lim R ∞
u→∞
u
ψ(u)
λ
.
=
c − λµ
(1 − G(y)) dy
Remark.
Recall that the probability that ruin occurs at the first ladder time
given there is a first ladder epoch is
Z
1 ∞
(1 − G(y)) dy .
µ u
Hence the ruin probability is asymptotically (c − λµ)−1 λµ times the probability of
ruin at the first ladder time given there is a first ladder epoch. But (c − λµ)−1 λµ is
the expected number of ladder times. Intuitively for u large ruin will occur if one of
the ladder heights is larger than u.
Proof.
Let B(x) denote the distribution function of the first ladder height L1 , i.e.
Z
1 x
B(x) =
(1 − G(y)) dy .
µ 0
Choose ε > 0 such that λµ(1 + ε) < c. By Lemma F.6 there exists D such that
1 − B ∗n (x)
≤ D(1 + ε)n .
1 − B(x)
From the Pollaczek–Khintchine formula (4.10) we obtain
∞
ψ(u)
λµ X λµ n 1 − B ∗n (u)
= 1−
1 − B(u)
c n=1 c
1 − B(u)
∞
λµ X λµ n
(1 + ε)n < ∞ .
≤D 1−
c n=1 c
Thus we can interchange sum and limit. Recall from Lemma F.7 that
1 − B ∗n (u)
= n.
u→∞ 1 − B(u)
lim
Thus
n
∞
∞
ψ(u)
λµ X λµ n λµ X X λµ n
lim
= 1−
n
= 1−
u→∞ 1 − B(u)
c n=1
c
c n=1 m=1 c
∞
∞
∞
λµ X X λµ n X λµ m
λµ/c
λµ
= 1−
=
=
=
.
c m=1 n=m c
c
1 − λµ/c
c − λµ
m=1
4. THE CRAMÉR–LUNDBERG MODEL
93
u
ψ(u)
App
Er
1
2
3
4
5
10
20
30
40
50
0.364
0.150
6.18 · 10−2
2.55 · 10−2
1.05 · 10−2
1.24 · 10−4
1.75 · 10−8
2.50 · 10−12
1.60 · 10−15
1.21 · 10−16
8.79 · 10−3
1.52 · 10−4
8.58 · 10−6
9.22 · 10−7
1.49 · 10−7
3.47 · 10−10
5.40 · 10−13
1.10 · 10−14
6.71 · 10−16
7.56 · 10−17
-97.588
-99.898
-99.986
-99.996
-99.999
-100
-99.997
-99.56
-58.17
-37.69
Table 4.5: Approximations for subexponential claim sizes
Example 4.16. Let G ∼ Pa(α, β). Then (see Example F.5)
1 − B(zx)
1 − G(zx)
= z lim
= z −(α−1) .
x→∞ 1 − B(x)
x→∞ 1 − G(x)
lim
By Lemma F.4 B is a subexponential distribution. Note that we assume that µ < ∞
and thus α > 1. Because
Z ∞
β α
β β α−1
dy =
β+y
α−1 β+x
x
we obtain
ψ(u) ≈
β α−1
λβ/(α − 1) β α−1
λβ
=
.
c − λβ/(α − 1) β + u
c(α − 1) − λβ β + u
Choose now c = 1, λ = 9, α = 11 and β = 1. Table 4.5 gives the exact value
(ψ(u)), the approximation (App) and the relative error in percent (Er). Consider
for instance u = 20. The ruin probability is so small, that it is not interesting for
practical purposes anymore. But the approximation still underestimates the true
value by almost 100%. That means we are still not far out enough in the tail. This
is the problem by using the approximation. It should, however, be remarked, that
for small values of α the approximation works much better. Values α ∈ (1, 2) are
also more interesting from a practical point of view.
Remark.
The conditions of the theorem are also fulfilled for LN(µ, σ 2 ), for
LG(γ, α) and for Wei(α, c) (α < 1) distributed claims, see [35] and [62].
94
4. THE CRAMÉR–LUNDBERG MODEL
4.12. The Time to Ruin
Consider the function
fα (u) = IIE[e−ατ 1I{τ <∞} | C0 = u] .
The function is defined at least for α ≥ 0. We will first find a differential equation
for fα (u).
Lemma 4.17. The function fα (u) is absolutely continuous and fulfils the equation
hZ u
i
0
cfα (u) + λ
fα (u − y) dG(y) + 1 − G(u) − fα (u) − αfα (u) = 0 .
(4.14)
0
Proof.
For h small we get
fα (u) = e−λh e−αh fα (u + ch)
Z h hZ u+ct
i
+
e−αt fα (u + ct − y) dG(y) + e−αt (1 − G(u + ct)) λe−λt dt .
0
0
We see that fα (u) is right continuous. Reordering of the terms and dividing by h
yields
c
fα (u + ch) − fα (u) 1 − e−(λ+α)h
−
fα (u + ch)
ch
h
Z h hZ u+ct
i
1
+
fα (u + ct − y) dG(y) + (1 − G(u + ct)) λe−(λ+α)t dt = 0 .
h 0
0
Letting h → 0 shows that fα (u) is differentiable and that (4.14) for the derivative
from the right holds. Replacing u by u − ch shows the derivative from the left. This differential equation is hard to solve. Let us take the Laplace transform
R∞
with respect to the initial capital. Let fˆα (s) = 0 e−su fα (u) du. For the moment
we assume s > 0. Note that (see Section 4.9)
Z ∞
fα0 (u)e−su du = sfˆα (s) − fα (0) ,
0
Z
0
and
Z
∞Z ∞
∞Z
u
fα (u − y) dG(y)e−su du = fˆα (s)MY (−s)
0
−su
dG(y)e
0
u
Z
∞Z y
du =
0
0
e−su du dG(y) =
1 − MY (−s)
.
s
4. THE CRAMÉR–LUNDBERG MODEL
95
Multiplying (4.14) by e−su and integrating yields
i
h
1 − MY (−s)
ˆ
ˆ
ˆ
− fα (s) − αfˆα (s) = 0 .
c(sfα (s) − fα (0)) + λ fα (s)MY (−s) +
s
Solving for fˆα (s) yields
cfα (0) − λs−1 (1 − MY (−s))
fˆα (s) =
.
cs − λ(1 − MY (−s)) − α
(4.15)
We know that fˆα (s) exists if α > 0 and s > 0 and is positive. The denominator
cs − λ(1 − MY (−s)) − α
is convex, has value −α < 0 at 0 and converges to ∞ as s → ∞. Thus there exists a
strictly positive root s(α) of the denominator. Because fˆα (s) exists also for s = s(α)
the numerator must have a root at s(α) too. Thus
cfα (0) = λs(α)−1 (1 − MY (−s(α))) .
The function s(α) is differentiable by the implicit function theorem
s0 (α)(c − λMY0 (−s(α))) − 1 = 0 .
(4.16)
Because s(0) = 0 we obtain that limα→0 s(α) = 0.
Example 4.18. Let the claims be Exp(β) distributed. We have to solve
λs
cs −
−α=0
β+s
which admits the two solutions
p
−(βc − λ − α) ± (βc − λ − α)2 + 4αβc
s± (α) =
2c
where s− (α) < 0 ≤ s+ (α). Thus
λ/(β + s+ (α)) − λ/(β + s)
λ
fˆα (s) =
=
.
cs − λs/(β + s) − α
c(β + s+ (α))(s − s− (α))
It follows that
λ
fα (u) =
es− (α) u .
c(β + s+ (α))
Noting that
IIE[τ 1I{τ <∞} ] = lim IIE[τ e−ατ 1I{τ <∞} ] = lim −
α↓0
we see
Z ∞
d
IIE[τ 1I{τ <∞} | C0 = u]e
du = lim −
α↓0
dα
0
We can find the following “explicit” formula.
−su
α↓0
Z
0
d
IIE[e−ατ 1I{τ <∞} ]
dα
∞
fα (u)e−su du = lim −
α↓0
d ˆ
fα (s) .
dα
96
4. THE CRAMÉR–LUNDBERG MODEL
Lemma 4.19. Assume µ2 < ∞. Then
1 h λµ2
δ(u) −
IIE[τ 1I{τ <∞} ] =
c − λµ 2(c − λµ)
Proof.
−
Z
u
i
ψ(u − y)δ(y) dy .
(4.17)
0
We get
d ˆ
λs0 (α)
1 − MY (−s(α)) − MY0 (−s(α))s(α)
fα (s) =
dα
cs − λ(1 − MY (−s)) − α
s(α)2
λ((1 − MY (−s(α)))s(α)−1 − (1 − MY (−s))s−1 )
. (4.18)
−
(cs − λ(1 − MY (−s)) − α)2
It follows from (4.16) that
s0 (0) =
1
c − λµ
and we have already seen that
lim
α↓0
1 − MY (−s(α))
1 − MY (−s)
= lim
= lim MY0 (−s) = µ .
s↓0
s↓0
s(α)
s
Moreover,
lim
α↓0
(1 − MY (−s(α))) − MY0 (−s(α))s(α)
(1 − MY (−s)) − MY0 (−s)s
=
lim
s↓0
s(α)2
s2
µ2
sMY00 (−s)
=
.
= lim
s↓0
2s
2
Letting α tend to 0 in (4.18) yields
Z ∞
IIE[τ 1I{τ <∞} | C0 = u] e−su du
0
λµ2
λµ − λs−1 (1 − MY (−s))
−
2(c − λµ)(cs − λ(1 − MY (−s)))
(cs − λ(1 − MY (−s)))2
c − λµ
λµ2
=
2
2(c − λµ) cs − λ(1 − MY (−s))
1
1
c − λµ
c − λµ
−
−
c − λµ cs − λ(1 − MY (−s)) s cs − λ(1 − MY (−s))
h
i
1
λµ2
=
δ̂(s) − δ̂(s)ψ̂(s) .
c − λµ 2(c − λµ)
=
But this is the Laplace transform of the assertion.
4. THE CRAMÉR–LUNDBERG MODEL
97
Corollary 4.20. Let t > 0. Then
λµ2
.
2(c − λµ)2 t
IIP[t < τ < ∞] <
Proof.
Because δ(u) and ψ(u) take values in [0, 1] it is clear from (4.17) that
IIE[τ 1I{τ <∞} ] <
λµ2
.
2(c − λµ)2
By Markov’s inequality
λµ2
1
.
IIP[t < τ < ∞] = IIP[τ 1I{τ <∞} > t] < IIE[τ 1I{τ <∞} ] <
t
2(c − λµ)2 t
From (4.17) it is possible to get an explicit expression for u = 0
IIE[τ 1I{τ <∞} | C0 = 0] =
λµ λµ2
λµ2
1
−
=
2(c − λµ)2
c
2c(c − λµ)
and
IIE[τ | τ < ∞, C0 = 0] =
Example 4.18 (continued).
µ2
.
2µ(c − λµ)
For Exp(β) distributed claims we get for R = β −λ/c
IIE[τ 1I{τ <∞} ]
Z u
i
λ −Ru λ −R(u−y) λ
β h
2λ
1− e
−
e
1 − e−Ry dy
=
cβ − λ 2β(cβ − λ)
βc
βc
0 βc
h
β
λ
λ
λ
c
λ i
=
1 − e−Ru − e−Ru
(eRu − 1) − u
cβ − λ β(cβ − λ)
βc
βc
βc − λ
βc
1
λ
= 2
e−Ru (λu + c) =
ψ(u)(λu + c) .
βc (βc − λ)
c(βc − λ)
The conditional expectation of the time of ruin is linear in u
IIE[τ | τ < ∞] =
λu + c
.
c(βc − λ)
98
4. THE CRAMÉR–LUNDBERG MODEL
4.13. Seal’s Formulae
We consider now the probability of ruin within finite time ψ(u, t). But first we want
to find the conditional finite ruin probability given Ct for some t fixed.
Lemma 4.21. Let t be fixed, u = 0 and 0 < y ≤ ct. Then
IIP[Cs ≥ 0, 0 ≤ s ≤ t | Ct = y] =
Proof.
Consider
Nt
X
Ct − C(t−s)− = cs −
y
.
ct
Yi .
i=N(t−s)− +1
This is also a Cramér–Lundberg model. Thus
IIP[Cs ≥ 0, 0 ≤ s ≤ t | Ct = y] = IIP[Ct − Ct−s ≥ 0, 0 ≤ s ≤ t | Ct − C0 = y]
= IIP[Cs ≤ Ct , 0 ≤ s ≤ t | Ct = y] .
Let Ss = cs − Cs denote the aggregate claims up to time s. Denote by σ the
permutations of {1, 2, . . . , n}. Then
k
1 hX X
IIE[Ss | St = y, Nt = n, Ns = k] = IIE
Yσ(i)
n!
σ i=1
n
k(n − 1)! hX
IIE
Yi
=
n!
i=1
i
St = y, Nt = n, Ns = k
i ky
.
S
=
y,
N
=
n,
N
=
k
=
t
t
s
n
Because, given Nt = n, the claim times are uniformly distributed in [0, t] (Proposition C.2) we obtain
IIE[Ss | St = y, Nt = n] =
=
n X
n s k t − s n−k ky
t
t
n
n − 1 s k t − s n−k
sy
y=
.
k−1
t
t
t
k=0
n X
k=1
k
This is independent of n. Thus
IIE[Ss | St = y] =
sy
t
and
IIE[Cs | Ct = y] = IIE[cs − Ss | St = ct − y] = cs −
s(ct − y)
sy
=
.
t
t
4. THE CRAMÉR–LUNDBERG MODEL
99
Then for v ≤ s < t
i y − C − IIE[(C − C ) | C , C = y]
hy − C v
s
v
v
t
s IIE
Cv , Ct = y =
t−s
t−s
y − Cv − (s − v)(y − Cv )/(t − v)
y − Cv
=
=
t−s
t−v
and the process
y − Cs
Ms =
(0 ≤ s < t)
t−s
is a conditional martingale. Note that lims↑t Ms = c. Let T = inf{s ≥ 0 : Cs = y}.
Because
0 ≤ Ms 1I{T >s} ≤ c
on {Ct = y} we have
h
i
lim IIE[Ms 1I{T >s} | Ct = y] = IIE lim Ms 1I{T >s} Ct = y = cIIP[T = t | Ct = y] .
s↑t
s↑t
Note that by the stopping theorem {MT ∧t } is a bounded martingale. Thus because
MT = 0 on {T < t}
y
= M0 = lim IIE[MT ∧s | Ct = y] = lim IIE[Ms 1I{T >s} | Ct = y] = cIIP[T = t | Ct = y] .
s↑t
s↑t
t
Note that IIP[T = t | Ct = y] = IIP[Cs ≤ Ct , 0 ≤ s ≤ t | Ct = y].
Conditioning on Ct is in fact a conditioning on the aggregate claim size St .
Denote by F (x; t) the distribution function of St . For the integration we use dF (·; t)
for an integration with respect to the measure F (·; t). Moreover, let δ(u, t) = 1 −
ψ(u, t) = IIP[τ > t].
Theorem 4.22. For initial capital 0 we have
1
1
δ(0, t) = IIE[Ct ∨ 0] =
ct
ct
Proof.
Z
ct
F (y; t) dy .
0
By Lemma 4.21 it follows readily that
hC ∨ 0i
t
δ(0, t) = IIE[IIP[τ > t | Ct ]] = IIE
.
ct
Moreover,
Z
ct
IIE[Ct ∨ 0] = IIE[(ct − St ) ∨ 0] =
(ct − z) dF (z; t)
0
Z ctZ ct
Z ctZ y
Z
=
dy dF (z; t) =
dF (z; t) dy =
0
z
0
0
ct
F (y; t) dy .
0
100
4. THE CRAMÉR–LUNDBERG MODEL
Let now the initial capital be arbitrary.
Theorem 4.23. With the notation used above we have for u > 0
Z u+ct v − u
v − u F dv;
.
δ 0, t −
δ(u, t) = F (u + ct; t) −
c
c
u
Proof.
Note that
δ(u, t) = IIP[Ct > 0] − IIP[∃0 ≤ s < t : Cs = 0, Cv > 0 for s < v ≤ t]
and IIP[Ct > 0] = F (u + ct; t). Let T = sup{0 ≤ s ≤ t : Cs = 0} and set T = ∞ if
Cs > 0 for all s ∈ [0, t]. Then, noting that IIP[Cs > 0 : 0 < s ≤ t | C0 = 0] = IIP[Cs ≥
0 : 0 < s ≤ t | C0 = 0],
IIP[T ∈ [s, s + ds)] = IIP[Cs ∈ (−c ds, 0], Cv > 0 for v ∈ [s + ds, t]]
= (F (u + c(s + ds); s) − F (u + cs; s))δ(0, t − s − ds)
= δ(0, t − s)c dF (u + cs; s) .
Thus
Z
δ(u, t) = F (u + ct; t) −
t
δ(0, t − s)c dF (u + cs; s)
0
Z
= F (u + ct; t) −
u
u+ct
v − u
v − u
dF v,
.
δ 0, t −
c
c
4.14. Finite Time Lundberg Inequalities
In this section we derive an upper bound for probabilities of the form
IIP[yu < τ ≤ y u]
for 0 ≤ y < y < ∞. Assume that the Lundberg exponent R exists. We will use the
martingale (4.4) and the stopping theorem. For r ≥ 0 such that MY (r) < ∞,
e−ru = IIE[e−rCτ ∧yu −θ(r)(τ ∧yu) ] > IIE[e−rCτ ∧yu −θ(r)(τ ∧yu) ; yu < τ ≤ y u]
= IIE[e−rCτ −θ(r)τ | yu < τ ≤ y u]IIP[yu < τ ≤ y u]
> IIE[e−θ(r)τ | yu < τ ≤ y u]IIP[yu < τ ≤ y u]
> e− max{θ(r)yu,θ(r)yu} IIP[yu < τ ≤ y u] .
4. THE CRAMÉR–LUNDBERG MODEL
101
Thus
IIP[yu < τ ≤ y u | C0 = u] < e−(r−max{θ(r)y,θ(r)y})u .
Choosing the exponent as small as possible we obtain
IIP[yu < τ ≤ y u | C0 = u] < e−R(y,y)u
(4.19)
where
R(y, y ) = sup min{r − θ(r)y, r − θ(r)y } .
r∈IR
The supremum over r ∈ IR makes sense because θ(r) > 0 for r < 0 and θ(r) = ∞
if MY (r) = ∞. Since θ(R) = 0 it follows that R(y, y ) ≥ R. Hence (4.19) could be
useful.
For further investigation of R(y, y ) we consider the function fy (r) = r − yθ(r).
Let r∞ = sup{r ≥ 0 : MY (r) < ∞}. Then fy (r) exists in the interval (−∞, r∞ )
and fy (r) = −∞ for r ∈ (r∞ , ∞). If MY (r∞ ) < ∞ then |fy (r∞ )| < ∞. Since
fy00 (r) = −yMY00 (r) < 0 the function fy (r) is concave. Thus there exists a unique ry
such that fy (ry ) = sup{fy (r) : r ∈ IR}. Because fy (0) = 0 and fy0 (0) = 1 − yθ0 (0) =
1 + y(c − λµ) > 0 we conclude that ry > 0. ry is either the solution to the equation
1 + y(c − λMY0 (ry )) = 0
or ry = r∞ . We assume now that R < r∞ and let
y0 = (λMY0 (R) − c)
−1
.
It follows ry0 = R. We call y0 the critical value.
Because
d
f (r)
dy y
= −θ(r) it follows readily that
d
fy (r) T 0 ⇐⇒ r S R .
dy
We get
fy (r) S fy(r) as r S R .
Because MY0 (r) is an increasing function it follows that
y T y0 ⇐⇒ ry S R .
Thus

R
if y ≤ y0 ≤ y ,


R(y, y ) = fy(ry) if y < y0 ,


fy (ry ) if y > y0 .
102
4. THE CRAMÉR–LUNDBERG MODEL
If y0 ∈ [y, y ] then we got again Lundberg’s inequality. If y < y0 then R(y, y ) does
not depend on y. By choosing y as small as possible we obtain
IIP[0 < τ ≤ yu | C0 = u] < e−R(0,y)u
(4.20)
for y < y0 . Note that R(0, y) > R. Analogously
IIP[yu < τ < ∞ | C0 = u] < e−R(y,∞)u
(4.21)
for y > y0 . The strict inequality follows from
e−ru ≥ IIE[e−rCτ −θ(r)τ ; yu < τ < ∞] > e−θ(r)yu IIP[yu < τ < ∞]
if 0 < r < R, compare also (4.5). Again R(y, ∞) > R.
We see that IIP[τ ∈
/ ((y0 − ε)u, (y0 + ε)u)] goes faster to 0 than IIP[τ ∈ ((y0 −
ε)u, (y0 + ε)u)]. Intuitively ruin should therefore occur close to y0 u.
Theorem 4.24. Assume that R < r∞ . Then
τ
−→ y0
u
in probability on the set {τ < ∞} as u → ∞.
Proof. Choose ε > 0. By the Cramér–Lundberg approximation IIP[τ < ∞ | C0 =
u] exp{Ru} → C for some C > 0. Then
h τ
i
IIP − y0 > ε τ < ∞, C0 = u
u
IIP[τ < (y0 − ε)u | C0 = u] + IIP[(y0 + ε)u < τ < ∞ | C0 = u]
=
IIP[τ < ∞ | C0 = u]
exp{−R(0, y0 − ε)u} + exp{−R(y0 + ε, ∞)u}
≤
IIP[τ < ∞ | C0 = u]
exp{−(R(0, y0 − ε) − R)u} + exp{−(R(y0 + ε, ∞) − R)u}
=
−→ 0
IIP[τ < ∞ | C0 = u]eRu
as u → ∞.
Bibliographical Remarks
The Cramér–Lundberg model was introduced by Filip Lundberg [63]. The basic
results, like Lundberg’s inequality and the Cramér–Lundberg approximation go back
4. THE CRAMÉR–LUNDBERG MODEL
103
to Lundberg [64] and Cramér [24], see also [25], [47] or [68]. The differential equation
for the ruin probability (4.2) and the Laplace transform of the ruin probability can
for instance be found in [25]. The renewal theory proof of the Cramér–Lundberg
approximation is due to Feller [41]. The martingale (4.4) was introduced by Gerber
[42]. The latter paper contains also the martingale proof of Lundberg’s inequality
and the proof of (4.20). The proof of (4.21) goes back to Grandell [48]. Section 4.14
can also be found in [36]. Some similar results are obtained in [4] and [5]. Segerdahl
[78] proved the asymptotic normality of the ruin time conditioned on ruin occurs.
The results on reinsurance and ruin can be found in [88] and [22]. The approach
for proportional reinsurance as well as the example is from [53]. Proposition 4.9 can
be found in [44, p.130].
The term severity of ruin was introduced in [45], the joint distribution of the
severity of ruin and the capital prior to ruin was first investigated in [34]. More
results on the topic are obtained in [32], [68] and [74]. A diffusion approximation,
even though not obtained mathematically correct, was already used by Hadwiger
[52]. The modern approach goes back to Iglehart [57], see also [46], [7] and [72].
The deVylder approximation was introduced in [31]. Formula (4.13) was found by
Asmussen [7]. It is however stated incorrectly in the paper. For the correct formula
see [16] or [68]. The Beekman–Bowers approximation can be found in [17].
Theorem 4.15 was obtain by von Bahr [87] in the special case of Pareto distributed claim sizes and by Thorin and Wikstad [86] in the case of lognormally
distributed claims. The theorem in the present form is basically found in [35], see
also [39].
Section 4.12 is taken from [71]. Parts can also be found in [72] and [16]. (4.15)
was first obtained in [25]. The expected value of the ruin time for exponentially
distributed claims can be found in [44, p.138]. Seal’s formulae were first obtained
by Takács [82, p.56]. Seal [76] and [77] introduced them to risk theory. The present
presentation follows [44].