Lecture Notes on Risk Theory

February 20, 2010

Contents

1 Introduction and basic definitions 1

2 Accumulated claims in a fixed time interval 3

3 Reinsurance 7

4 Risk processes in discrete time 10

5 The Adjustment Coefficient 15

6 Some risk processes in continuous time 176.1 The Sparre Andersen model . . . . . . . . . . . . . . . . . . . . 176.2 The compound Poisson model . . . . . . . . . . . . . . . . . . . 18

i

ii

Chapter 1

Introduction and basic definitions

An insurance company needs to pay claims from time to time, while collectingpremiums from its customers continuously over time. We assume that it startswith an initial (risk) reserve u ≥ 0 and the premium income is linear with someslope c > 0. At times Tn, n ∈ N, a claim occurs. Thus we call Tn the nthclaim time. Naturally, we assume that T1 > 0 and Tn+1 > Tn for n ∈ N. Thesequence of claim times forms a counting process N = (Nt : t ≥ 0), defined byNt := max{n ∈ N : Tn ≤ t}, where we define max ∅ := 0. This process N iscalled the claim process. The nth claim size is denoted by Un, n ∈ N. Then theinsurance company’s risk reserve at any time t is given by is given by

Rt = u+ c · t−Nt∑k=1

Uk (1.1)

where the empty sum is defined as zero, i.e.∑0

k=1 Uk = 0. Since Nt and Uk arerandom, (Rt : t ≥ 0) is a stochastic process. A typical path looks like figure 1.

We see that at time T4 something special happens: The risk reserve RT4 isnegative, implying of course the ruin of the insurance company. Therefore, thestopping time

τ(u) := min{t ≥ 0 : Rt ≤ 0}

is called the time of ruin. The following questions arise:

1. The only random term on the right–hand side of equation 1.1 is the sum∑Ntk=1 Uk representing the accumulated claims until time t. We shall study

this in chapter 2.

1

2. An insurance company wants to insure itself against unusually high amountsof claims in any period. Some general results on this so–called reinsurancewill be presented in chapter 3.

3. An obvious criterion for the viability of an insurance policy is the ruinprobability ψ(u) := P(τ(u) <∞).

2

Chapter 2

Accumulated claims in a fixed timeinterval

Define X :=∑Nt

k=1 Uk for some time t > 0 that shall be fixed throughout thischapter. We assume that the claim sizes Un, n ∈ N, are iid and further that Nt andall Un are independent. Denote

pk := P(Nt = k)

for k ∈ N0 and write

δ0(x) :=

{1, x ≥ 0

0, x < 0

for the distribution function of the Dirac measure on zero. Let Q denote the dis-tribution function of U1. Further denote the distribution function of the sum of afixed number of claims by

Rk(x) := P(U1 + . . .+ Uk ≤ x)

for k ∈ N and x ≥ 0. In order to obtain Rk in terms of Q, we define

Q∗0 := δ0, Q∗1 := Q, and Q∗n+1(x) :=

∫ x

0

Q∗n(x− y)dQ(y)

for n ∈ N. The distribution function Q∗n is called the nth convolutional powerof Q.

Lemma 2.1 Rk = Q∗k for all k ∈ N.

3

Proof: For k = 1 the statement holds by definition. The induction step from k tok + 1 is straightforward:

Rk+1(x) = P(U1 + . . .+ Uk+1 ≤ x)

=

∫ x

0

P(U1 + . . .+ Uk ≤ x− y)dQ(y) by conditioning on Uk+1 = y

=

∫ x

0

Q∗k(x− y)dQ(y) by induction hypothesis

= Q∗k+1(x) by definition

�

Theorem 2.1 P(X ≤ x) =∑∞

k=0 pkQ∗k(x)

Proof:

P(X ≤ x) =∞∑k=0

P(X ≤ x,Nt = k) =∞∑k=0

pkP(X ≤ x|Nt = k)

=∞∑k=0

pkQ∗k(x)

by definition of X =∑Nt

k=1 Uk and lemma 2.1.�

With p := 1− p0 and

Q :=∞∑k=1

pkpQ∗k

we can writeP(X ≤ x) = (1− p)δ0(x) + pQ(x)

for x ≥ 0. The problem remains of course to determine a simple expression forQ(x). This is possible in the following

Example 2.1 Assume that Nt has a geometric distribution, i.e. pk = (1−p)pk forall k ∈ N0, and Q = Exp(µ). Then Q∗k = Erlk(µ), with density function

fk(x) = µ(µx)k−1

(k − 1)!e−µx

4

for x ≥ 0 (proof as an exercise). Further, Q has the density function

f(x) =∞∑k=1

pkpµ

(µx)k−1

(k − 1)!e−µx =

∞∑k=1

(1− p)pk−1µ(µx)k−1

(k − 1)!e−µx

= (1− p)µe−µx∞∑k=1

(pµx)k−1

(k − 1)!= (1− p)µe−µxepµx

= (1− p)µe−(1−p)µx

for all x ≥ 0. Hence Q = Exp((1− p)µ) and

P(X ≤ x) = (1− p)δ0(x) + p(1− e−(1−p)µx) = 1− pe−(1−p)µx

for x ≥ 0.

Theorem 2.2 E(X) = E(Nt) · E(U1)

Proof:

E(X) = E

(Nt∑k=1

Uk

)= p0 · 0 +

∞∑k=1

pk · E(U1 + . . .+ Uk) =∞∑k=1

pk · kµ

= E(Nt) · E(U1)

�

Theorem 2.3

Var(X) = E(Nt) · Var(U1) + Var(Nt) · E(U1)2

≤ E(U21 ) ·max(E(Nt),Var(Nt))

Proof:

Var(X) = E(X2)− E(X)2 =∞∑k=1

pk · E

( k∑n=1

Un

)2− E(Nt)

2 · E(U1)2

according to theorem 2.2. For the expectation in the first term, we obtain

E

( k∑n=1

Un

)2 = E

(k∑

n=1

U2n +

∑i 6=j

UiUj

)=

k∑n=1

E(U2n) +

∑i 6=j

E(UiUj)

5

Since all Un are iid, we arrive at

E

( k∑n=1

Un

)2 = k · E(U2

1 ) + k(k − 1)E(U1)2

Hence

Var(X) =∞∑k=1

pk · k ·(E(U2

1 )− E(U1)2)

+∞∑k=1

pk · k2 · E(U1)2

− E(Nt)2 · E(U1)

2

= E(Nt) · Var(U1) + Var(Nt) · E(U1)2

The inequality follows from here via E(U21 ) = Var(U1) + E(U1)

2.�

Example 2.2 If N is a Poisson process, then E(Nt) = Var(Nt) and Var(X) =E(Nt) · E(U2

1 ).

6

Chapter 3

Reinsurance

LetX denote the random variable of the accumulated claims after some fixed timet > 0 (e.g. after one year). X is also called the risk at time t. Let F denote thedistribution function of X .

A reinsurance policy is a function R : R+ → R+ such that 0 ≤ R(x) ≤ xfor all x ∈ R+. In the event {X = x} the insurance company pays R(x) of theclaims and the resinsurer pays the rest, x−R(x).

We assume first that the reinsurance premium is the fair net premium

P = P (R) = E(X −R(X))

without safety loading. A special reinsurance policy is the stop loss reinsuranceat price P , defined as

R∗(x) =

{x, x ≤M

M, x > M

where M is the unique solution of

P =

∫ ∞M

(x−M)dF (x)

M is called the deductible.

Example 3.1 Assume that X ∼ F := (1 − p) · δ0 + p · Exp((1 − p)µ) as inexample 2.1. Set β := (1− p)µ. Then dF (x) = p · βe−βxdx for x > 0 and

P =

∫ ∞M

(x−M)dF (x) = p ·(∫ ∞

M

xβe−βxdx−M∫ ∞M

βe−βxdx

)7

Integration by parts yields for the first integral[x · (−1) · e−βx

]∞M

+

∫ ∞M

e−βxdx = M · e−βM +1

βe−βM

As the second integral equals e−βM , we obtain

P = p · 1

βe−βM

from which the deductible is determined as

M = − 1

βln

(P · β

p

)Thus the deductible has a meaningful (i.e. positive) value if and only if the condi-tion

P < p · 1

β= E(X)

holds. This is reasonable, as for larger premiums P > E(X) the insurance com-pany would do better to cover the risk itself.

Theorem 3.1 For a fixed fair net premium P the stop loss reinsurance yields thesmallest variance Var(R(X)) of all reinsurance policies.

Proof: Let R denote any reinsurance policy with fair net premium P . DefineP1 := E(X)− P = E(R(X)) and note that P1 = E(R∗(X)), too. Then

Var(R(X)) + P 21 =

∫ ∞0

R2(x)dF (x)

=

∫ ∞0

(R(x)−M)2dF (x) + 2MP1 −M2

≥∫ M

0

(R(x)−M)2dF (x) + 2MP1 −M2

Over the range 0 < x < M we obtain for the integrand

(R(x)−M)2 = (M −R(x))2 ≥ (M − x)2 = (M −R∗(x))2 = (R∗(x)−M)2

As R∗(x) = M for x > M , we further get

Var(R(X))+P 21 ≥

∫ ∞0

(R∗(x)−M)2dF (x)+2MP1−M2 = Var(R∗(X))+P 21

8

which implies the statement.�

Now assume that the reinsurance premium is of the form

P ′(R) = E(X −R(X)) + f(Var(X −R(X)))

where f is some positive increasing function. The term f(Var(X − R(X))) iscalled the safety loading.

Theorem 3.2 Given a fixed value V > 0 and the condition Var(R(X)) = V , thecheapest reinsurance policy R∗ for the premium P ′ is given by

R∗(x) =

√V

Var(X)· x

for all x ≥ 0.

Remark 3.1 The term ”cheapest” is to be understood in the following sense: Af-ter the time period [0, t], the insurance company will have to pay the premiumP ′(R) to the reinsurer and R(X) as its own contribution to cover the accumulatedclaims. Altogether, the expected amount of this is

E(X−R(X))+f(Var(X−R(X)))+E(R(X)) = E(X)+f(Var(X−R(X)))

The goal is to minimise this expectation.

Proof: By the above remark it suffices to minimise f(Var(X−R(X))) and henceVar(X −R(X)), as f is increasing. For this we can write

Var(X −R(X)) = Var(X) + Var(R(X))− 2 · Cov(X,R(X))

which shows that we need to maximise the covariance Cov(X,R(X)). Thisis achieved by the form R(X) = a · X for a constant a. Now the conditionVar(R(X)) = V yields

V = a2 · Var(X), i.e. a =

√V

Var(X)

which is the statement.�

9

Chapter 4

Risk processes in discrete time

Let Xn denote the accumulated claims in the time interval ]n− 1, n], n ∈ N (e.g.the nth year). We assume that the random variables Xn, n ∈ N, are iid. As inchapter 1, the initial reserve and the rate of premium income are denoted by u ≥ 0and c > 0. The random variable

Kn := u+ c · n−n∑k=1

Xk

is called the risk reserve at time n ∈ N. Then the survival probability until timem (with initial reserve u) is defined as

ψ(u,m) := P(Kn ≥ 0 ∀n ≤ m)

Remark 4.1 Let X1 ∼ F . Conditioning on {X1 = x} yields the recursive for-mula

ψ(u,m+ 1) =

∫ u+c

0

ψ(u+ c− x,m)dF (x)

for all u > 0 and m ∈ N. However, this is in general numerically intractable.

The survival probability over infinite time is defined as

ψ(u) := limm→∞

ψ(u,m)

Clearly, ψ(u,m) is decreasing in m and bounded below by zero, such that thelimit ψ(u) does exist.

If we define Yn := Xn − c as the net claim in ]n − 1, n] and Sn :=∑n

k=1 Ykfor all n ∈ N0, then the Yn are iid (because the Xn are so and c is constant) and

10

thus S := (Sn : n ∈ N0) is a random walk. Yn is called the nth increment ofthe random walk. For technical reasons (needed in the third statement of 4.1), weexclude the trivial case by assuming that P(Xn = c) < 1, i.e. P(Yn = 0) < 1, forall n ∈ N.

Theorem 4.1 Let (Sn : n ∈ N0) be a random walk with increments Yn. Then

E(Y1) > 0 ⇒ limn→∞

Sn =∞

E(Y1) < 0 ⇒ limn→∞

Sn = −∞

E(Y1) = 0 ⇒ lim supn→∞

Sn =∞, lim infn→∞

Sn = −∞

where the implications hold almost certainly, i.e. with probability one.

Proof: The first two statements are immediate consequences of the strong law oflarge numbers. For a proof of the third statement, see Rolski et al. [3], p.234f.�

We say that the random walk has a positive (resp. negative) drift if E(Y1) > 0(resp. E(Y1) < 0), and that it is oscillating if E(Y1) = 0. For the remainder ofthis chapter we assume a negative drift, i.e. the net profit condition E(X1) < c.Define

M := sup{Sn : n ∈ N0}

and note that theorem 4.1 states that M < ∞ almost certainly. The crucial con-nection to our risk process is now the observation

ψ(u) = P(M ≤ u)

Define the (first strong) ascending ladder epoch by

ν+ := min{n ∈ N : Sn > 0}

with the understanding that min ∅ := ∞. The random variable ν+ is a stoppingtime with respect to the sequence Y = (Yn : n ∈ N), i.e.

P(ν+ = n|Y) = P(ν+ = n|Y1, . . . , Yn)

for all n ∈ N. This means we do not need any future information to determine theevent {ν+ = n}.

11

The corresponding random variable

L+ :=

{Sν+ , ν+ <∞∞ otherwise

is called the (first strong) ascending ladder height. Denote the distribution func-tion of L+ by G+(x) := P(L+ ≤ x) and note that

G+(∞) := limx→∞

G+(x) = P(L+ <∞)

may be strictly less than one.

Theorem 4.2 If E(Y1) < 0, then E(ν+) =∞.

Proof: Assume that E(ν+) < ∞. Then Wald’s lemma (see lemma 4.1 below)applies to

E(L+) = E

(ν+∑n=1

Yn

)= E(ν+) · E(Y1)

But E(L+) > 0, while E(ν+) > 0 and E(Y1) < 0, which leads to a contradiction.Hence E(ν+) =∞.�

Lemma 4.1 Wald’s Lemma Let Y = (Yn : n ∈ N) be a sequence of iid randomvariables and assume that E(|Y1|) <∞. Let S be a stopping time for the sequenceY with E(S) <∞. Then

E

(S∑n=1

Yn

)= E(S) · E(Y1)

Proof: For all n ∈ N define the random variables In := 1 on the set {S ≥ n} andIn := 0 else. Then

E

(S∑n=1

Yn

)≤ E

(S∑n=1

|Yn|

)= E

(∞∑n=1

In|Yn|

)=∞∑n=1

E(In|Yn|)

by monotone convergence, as In|Yn| ≥ 0 for all n ∈ N. Since S is a stopping timefor Y , we obtain P(S ≥ 1) = 1 and further

P(S ≥ n|Y) = 1− P(S ≤ n− 1|Y) = 1− P(S ≤ n− 1|Y1, . . . , Yn−1)

12

for all n ≥ 2. Since the Yn are independent, the equality above implies that In and|Yn| are independent, too. This yields

E(In|Yn|) = E(In) · E(|Yn|) = P(S ≥ n) · E(|Y1|)

for all n ∈ N. Now the relation∑∞

n=1 P(S ≥ n) = E(S) yields

E

(S∑n=1

Yn

)≤

∞∑n=1

P(S ≥ n) · E(|Y1|) = E(S) · E(|Y1|) <∞

Now we can use dominated convergence to obtain

E

(S∑n=1

Yn

)= E

(∞∑n=1

InYn

)=∞∑n=1

E(InYn) =∞∑n=1

E(In)E(Yn)

=∞∑n=1

P(S ≥ n) · E(Y1) = E(S) · E(Y1)

�In the case ν+ <∞ of a finite ladder epoch, we can define a new random walk

starting at (ν+, L+) by S ′ = ((Sν++n − L+) : n ∈ N0). As the increments areiid, S ′ is independent from (S1, . . . , Sν+) and the random walks S and S ′ have thesame distribution. Doing this iteratively, we arrive at the following definitions:

Let ν+0 := 0 and ν+

n+1 := min{k > ν+n : Sk > Sν+

n} for all n ∈ N0, where

min ∅ := ∞. The stopping time ν+n is called the nth (strong ascending) ladder

epoch. Correspondingly, we define the nth (strong ascending) ladder height as

L+n :=

{Sν+

n− Sν+

n−1, ν+

n <∞∞ otherwise

for n ∈ N. By construction the ladder heights are iid. Defining now N :=max{n ∈ N : ν+

n <∞} we obtain the representation

M =N∑k=1

L+k

Remark 4.2 The net profit condition E(Y1) < 0 implies that M < ∞ almostcertainly, see theorem 4.1. Since the (L+

k : k ∈ N) are iid, this entails N < ∞almost certainly. Hence we can conclude that G+(∞) < 1.

13

Theorem 4.3 If E(Y1) < 0, then

P(M ≤ u) = (1− p) ·∞∑k=0

(G+)∗k(u) =∞∑k=0

(1− p) · pkG∗k0 (u)

with p := G+(∞) and

G0(u) :=1

pG+(u) = P(L+ ≤ u|ν+ <∞)

for all u ≥ 0.

Proof: By definition, p = P(ν+ < ∞). The construction of the ladder epochsyields

P(N = k) = P(ν+n <∞ ∀ n ≤ k, ν+

k+1 =∞) = pk · (1− p)

Hence

P(M ≤ u) = P

(N∑n=1

L+n ≤ u

)=∞∑k=0

(1− p) · pk · P

(k∑

n=1

L+n ≤ u

∣∣∣∣∣ ν+k <∞

)by conditioning on {N = k}. The statement follows with the observation that

G∗k0 (u) = P

(k∑

n=1

L+n ≤ u

∣∣∣∣∣ ν+k <∞

)

�Likewise, the ruin probability is given as ψ(u) = P(M > u) and thus

Corollary 4.1 If the net profit condition E(Y1) < 0 holds, then

ψ(u) =∞∑k=1

(1− p) · pkG∗k0 (u)

where G∗k0 (u) = 1−G∗k0 (u).

This corollary is known as Beekman’s formula in risk theory and as thePollaczek–Khinchin formula in queueing theory.

Exercise 4.1 Show that ψ(0) = p and limu→∞ ψ(u) = 0.

14

Chapter 5

The Adjustment Coefficient

Let P denote a probability measure with E(P ) :=∫xdP (x) ∈]−∞, 0[. Assume

that the moment generating functionmP (α) :=∫eαxdP (x) satisfiesmP (α) <∞

for some α > 0 and limα→s−mP (α) ≥ 1 for s := sup{α > 0 : mP (α) < ∞}.Then the solution γ > 0 of mP (γ) = 1 is called the adjustment coefficient of P .

Remark 5.1 The adjustment coefficient does exist under the given assumptions,because mP (0) = 1, m′P (0) < 0, and mP (α) is strictly convex as m′′P (α) > 0for all α < s. The number γ is the only positive number such that eγxdP (x) is aprobability measure.

Theorem 5.1 Lundberg inequalityLet (Yn : n ∈ N) denote a sequence of independent random variables with identi-cal distribution P , and define

ψ(u) := P

(sup

{m∑i=1

Yi : m ∈ N

}> u

)for u ≥ 0. If E(Y1) < 0 and γ is the adjustment coefficient of P , then

ψ(u) ≤ e−γu

for all u ≥ 0.

Proof: Define

ψn(u) :=

{P (sup {

∑mi=1 Yi : m ≤ n} > u) , u ≥ 0

1, u < 0

15

for all n ∈ N. We will show that ψn(u) ≤ e−γu for all u ∈ R by induction on n.Since ψ(u) = limn→∞ ψn(u), this will yield the statement.

For u ≤ 0 and all n ∈ N, the bound holds by definition. Let n = 1 and u > 0.Then (see exercise below)

ψ1(u) = P(Y1 > u) = P(γY1 > γu)

≤ e−γuE(eγY1) = e−γu

Independence of the Yn, n ∈ N, yields further

ψn+1(u) = 1−∫ u

−∞1− ψn(u− y) dP (y)

= 1− P (]−∞, u]) +

∫ u

−∞ψn(u− y) dP (y)

=

∫ ∞−∞

ψn(u− y) dP (y)

≤∫ ∞−∞

e−γ(u−y) dP (y) = e−γu

where the third equality holds by definition of ψn, the inequality by inductionhypothesis, and the last equality by the fact that E(eγY1) = 1.�

Exercise 5.1 Show that for any real–valued random variable X , a real numbera ∈ R, and a positive and increasing function f the inequality

P(X > a) ≤ E(f(X))

f(a)

holds.

Remark 5.2 A distribution P is called heavy–tailed if its moment generatingfunction does not exist, i.e. if

∫eαxdP (x) does not converge for any α > 0. If

P is heavy–tailed, then it does not have an adjustment coefficient and Lundberg’sinequality does not apply.

16

Chapter 6

Some risk processes in continuoustime

6.1 The Sparre Andersen modelOur basic assumptions for this chapter are:

1. The claim sizes Un, n ∈ N, are iid with common distribution function F .

2. The claim arrival process N is an ordinary renewal process, which meansthat the inter–claim times Wn := Tn − Tn−1, n ∈ N, with T0 := 0, are iidwith common distribution function G.

3. Claim sizes and inter–claim times are independent.

As in chapter 1, we denote the rate of premium income by c > 0. Define

Yn := Un − cWn and Sn :=n∑k=1

Yk

for n ∈ N. Then the process S = (Sn : n ∈ N) is a random walk (cf. chapter 4).Since c > 0, ruin can occur only at claim epochs Tn, whence we obtain for theruin time

τ(u) = min{t > 0 : Rt < 0} = min{Tn : RTn = u− Sn < 0}

This implies for the ruin probability

ψ(u) = P(M := sup

n∈NSn > u

)17

meaning that all results from chapter 4 apply to the Sparre Andersen model. Fur-ther results using a random walk analysis can be found in [2].

In order to determine the drift of the random walk S, we observe that

E(Y1) = E(U1)− c · E(W1) < 0

⇔ ρ :=E(U1)

c · E(W1)< 1

⇔ η :=1

ρ− 1 =

1

E(U1)· (c · E(W1)− E(U1)) > 0

The number η is called the relative safety loading, while ρ is called the systemload, a name that originated in queueing theory. The inequality

E(U1)− c · E(W1) < 0

is called the net profit condition. Clearly, the random walk S has a negative driftif and only if the net profit condition holds.

6.2 The compound Poisson modelThis is a special case of the Sparre Andersen model with G(t) := 1 − e−λt fort ≥ 0, where λ > 0. Thus the claim arrival process N is a Poisson process withintensity λ.

Remark 6.1 It can be shown (see Doob [1]) that N is necessarily a Poisson pro-cess if we postulate the following properties:

1. N0 = 0 with probability one.

2. N has independent increments, i.e. Nt −Ns and Nv −Nu are independentfor all 0 ≤ s < t ≤ u < v.

3. N has stationary increments, i.e. Ns+t −Nt and Nt have the same distribu-tion for all s > 0.

4. There are no double events. More exactly, we postulate that

P(Nh ≥ 2) = o(h), i.e.1

hP(Nh ≥ 2)→ 0 as h ↓ 0

18

Now we derive a (defective) renewal equation for the ruin probability ψ(u).Define the complementary distribution function F by F (x) := 1 − F (x) for allx ≥ 0.

Theorem 6.1 For any initial reserve u ≥ 0,

ψ(u) =λ

c

∫ ∞u

F (x) dx+λ

c

∫ u

0

ψ(u− x)F (x) dx

Proof: We condition on the time T1 = t of the first claim. As T1 ∼ Exp(λ), weobtain for the survival probability ψ(u) = 1− ψ(u)

ψ(u) =

∫ ∞0

λe−λt∫ u+ct

0

ψ(u+ ct− s)dF (s)dt

=λ

c

∫ ∞u

e−λc(y−u)

∫ y

0

ψ(y − s)dF (s)dy

after the substitution y = u + ct which entails dy = c dt and t = 0 ⇔ y = u.This shows that the function

ψ(x) =λ

ceλcx

∫ ∞x

e−λcy

∫ y

0

ψ(y − s)dF (s)dy (6.1)

is differentiable on x ∈]0,∞[. Denote g(x) :=∫∞xe−

λcy∫ y

0ψ(y − s)dF (s)dy.

Then the derivative ψ′(x) is given by

ψ′(x) =d

dx

(λ

ceλcxg(x)

)=λ

c

(λ

ceλcxg(x) + e

λcxg′(x)

)As g′(x) = −e−λc x

∫ x0ψ(x− s)dF (s), we obtain further

ψ′(x) =λ

c

(λ

ceλcx

∫ ∞x

e−λcy

∫ y

0

ψ(y − s)dF (s)dy

−eλcxe−

λcx

∫ x

0

ψ(x− s)dF (s)

)=λ

c

(ψ(x)−

∫ x

0

ψ(x− s)dF (s)

)

19

using the representation (6.1). We will determine ψ(u) via its derivative as

ψ(u)− ψ(0) =

∫ u

0

ψ′(x)dx

=λ

c

(∫ u

0

ψ(x)dx−∫ u

0

∫ x

0

ψ(x− s)dF (s)dx

)The second integral transforms as∫ u

x=0

∫ x

s=0

ψ(x− s) dF (s) dx =

∫ u

s=0

∫ u

x=s

ψ(x− s) dx dF (s)

=

∫ u

s=0

∫ u−s

x=0

ψ(x) dx dF (s)

=

∫ u

x=0

∫ u−x

s=0

dF (s) ψ(x) dx

=

∫ u

x=0

F (u− x)ψ(x) dx

Thus we can write

ψ(u) = ψ(0) +λ

c

∫ u

0

(1− F (u− x))ψ(x) dx

= ψ(0) +λ

c

∫ u

0

ψ(u− x)F (x) dx

We can determine ψ(0) if we let u→∞. Since ψ(u− x)→ 0 and F (x)→ 0 asx→∞, we obtain

ψ(0) = 1− λ

c

∫ ∞0

F (x) dx = 1− λ

cE(U1)

This leads to

ψ(u) = 1− ψ(u) =λ

c

∫ ∞0

F (x) dx− λ

c

∫ u

0

ψ(u− x)F (x) dx

=λ

c

∫ ∞0

F (x) dx− λ

c

∫ u

0

(1− ψ(u− x))F (x) dx

=λ

c

∫ ∞u

F (x) dx+λ

c

∫ u

0

ψ(u− x)F (x) dx

completing the proof.�

20

Exercise 6.1 Show that

limu→∞

∫ u

0

ψ(u− x)F (x) dx = E(U1)

Remark 6.2 The proof shows further that the system load

p = ψ(0) =λ

cE(U1) = ρ

depends on U1 only by its mean. This is called insensitivity with respect to theshape of F , the distribution of U1.

Remark 6.3 If c depends on λ by c = (1 + η)λE(U1), then ψ(u) is constant in λ.

Remark 6.4 The equation in the above theorem 6.1 is called a (defective) re-newal equation for ψ(u).

Theorem 6.2 The Laplace transform of ψ(u) is given by

Lψ(s) :=

∫ ∞0

e−suψ(u) du =1

s− c− λE(U1)

cs− λ(1− LF (s))

where LF (s) =∫∞

0e−su dF (u) is the Laplace transform of the claim size distri-

bution.

Proof: Using theorem 6.1, we obtain

Lψ(s) =λ

c

(∫ ∞0

e−su∫ ∞u

F (x) dx du+

∫ ∞0

e−su∫ u

0

ψ(u− x)F (x) dx du

)=λ

c

∫ ∞0

F (x)

∫ x

0

e−su du dx

+λ

c

∫ ∞0

e−sxF (x)

∫ ∞x

e−s(u−x)ψ(u− x) du dx

=λ

c

∫ ∞0

F (x)1

s

(1− e−sx

)dx+

λ

c

∫ ∞0

e−sxF (x)Lψ(s) dx

Isolating Lψ(s) yields

Lψ(s) =λ

cs

(∫ ∞0

F (x) dx−∫ ∞

0

e−sxF (x) dx

)·(

1− λ

c

∫ ∞0

e−sxF (x) dx

)−1

21

Integration by parts yields∫ ∞0

e−sxF (x) dx =1

s−∫ ∞

0

e−sxF (x) dx

=1

s+

[1

se−sxF (x)

]∞0

− 1

s

∫ ∞0

e−sxdF (x)

=1

s(1− LF (s))

whence we obtain

Lψ(s) =λ

cs

(E(U1)−

1

s(1− LF (s))

)·(

1− λ

cs(1− LF (s))

)−1

=λE(U1)− λ

s(1− LF (s))

cs− λ(1− LF (s))

=1

s− c− λE(U1)

cs− λ(1− LF (s))

�

Example 6.1 If U1 ∼ Exp(µ), then LF (s) = µ · (µ+ s)−1 and

Lψ(s) =

λµ− λ

s(1− µ

µ+s)

cs− λ(1− µµ+s

)=

λµ− λ

µ+s

cs− λsµ+s

=

λ(µ+s)µ− λ

cs(µ+ s)− λs

=λs

µ(s · (cµ+ cs− λ))−1 =

λ

µc

(µ− λ

c+ s

)−1

=λ

µc

∫ ∞0

e−sue−(µ−λ/c)u du

Now uniqueness of the Laplace transform yields

ψ(u) =λ

µce−(µ−λ/c)u

for u ≥ 0.

22

Bibliography

[1] J. Doob. Stochastic processes. New York: Wiley, 1953.

[2] S. M. Pitts and K. Politis. The joint density of the surplus before and afterruin in the Sparre Andersen model. J. Appl. Prob., 44:695–712, 2007.

[3] T. Rolski, H. Schmidli, V. Schmidt, and J. Teugels. Stochastic Processes forInsurance and Finance. Wiley, Chichester etc., 1999.

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