Seven Introductory Lectures

on Actuarial Mathematics

Mogens Steffensen

31. marts 2005

Indhold

1 Compound Interest 3

1.1 Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Particular Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Life Insurance Risk: Mortality 9

2.1 The lifetime stochastic variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Expected values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Non-life Insurance Risk: Frequency and Severity 13

3.1 Compound risk models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.2 Frequency risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.3 Severity risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Valuation principles 19

4.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.2 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5 Life Insurance: Products and Valuation 23

5.1 Single payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.2 Series of payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.3 Premium calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6 Non-life Insurance: The Aggregate Claim Distribution 30

6.1 The Panjer Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6.2 Normal Power Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.3 Ruin Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

7 Finance and Reinsurance: Products and Valuation 38

7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2

1 Compound Interest

1.1 Time Value of Money

Consider an account with the amount b at time 0. The present value at time t (> 0) is the amount

into which b has turned at time t with compound interest. The present values at time 0, 1, t, . . . , n,

including the interest ij in period j, equal

U (0) = b,

U (1) = b (1 + i1) , . . . ,

U (t) = b∏t

s=1(1 + is) = b (1 + i1) (1 + i2) · · · (1 + it) , . . . ,

U (n) = b∏n

t=1(1 + it) = b (1 + i1) (1 + i2) · · · (1 + in) .

Consider an account with the amount b at time n. The present value at time t (< n) is the

amount which has to be saved at time t such that it turns into b at time n with compound interest.

The present values at time n, n− 1, t, . . . , 0, including the interest ij in period j, equal

V (n) = b,

V (n− 1) = 11 + in

b, . . . ,

V (t) = b∏n

s=t+1

1

1 + is=

1

1 + in

1

1 + in−1· · · 1

1 + it+1b, . . . ,

V (0) = b∏n

t=1

1

1 + it=

1

1 + in

1

1 + in−1· · · 1

1 + i1b.

If in particular it = i,

U (1) = b (1 + i) , . . . ,

U (t) = b (1 + i)t , . . . ,

U (n) = b (1 + i)n

,

V (n− 1) = 11 + i

b, . . . ,

V (t) =

(

1

1 + i

)n−t

b, . . . ,

V (0) =

(

1

1 + i

)n

b.

If the interest is not accumulated with the same frequency as it is announced, one must di-

stinguish carefully between effective interest and nominal interest. A nominal interest of i(m) means

that the interest i(m)/m is accumulated m times per period. Then the value after one period of 1

unit at time 0 equals(

1 +i(m)

m

)m

.

Equating

1 + i =

(

1 +i(m)

m

)m

,

3

we get that a nominal interest i(m) corresponds to the effective interest per period

i =

(

1 +i(m)

m

)m

− 1,

and the effective interest per period i corresponds to the nominal interest

i(m) = m(

(1 + i)1/m − 1

)

.

Note that only if m = 1, the nominal interest and the effective interest coincide.

A special situation arises if we let m tend to infinity. In the limit we say that interest is

accumulated continuously. We realize that

limm→∞

i(m) = limm→∞

m(

(1 + i)1/m − 1)

= limm→∞

(1 + i)1/m − (1 + i)0

1/m

= limh→0

(1 + i)0+h − (1 + i)0h

=d

dt(1 + i)

t

∣

∣

∣

∣

t=0

=d

dtet log(1+i)

∣

∣

∣

∣

t=0

= log (1 + i) et log(1+i)∣

∣

∣

t=0

= log (1 + i) .

We now introduce the following three quantities

δ = log (1 + i) ,

v =1

1 + i

d = 1− v = i1 + i

• The accumulation factor from time s (< t) to time t measures at time t the value of 1 unitat time s,

(1 + i)t−s

= v−(t−s) = eδ(t−s).

• The yearly accumulation factor measures at time t + 1 the value of 1 unit at time t,

1 + i = v−1 = eδ.

• The discount factor from time s (> t) to time t measures at time t the value of 1 unit at times,

(

1

1 + i

)s−t

= vs−t = e−δ(s−t).

• The yearly discount factor measures at time t the value of 1 unit at time t + 1,1

1 + i= v = e−δ.

4

If we replace the payment b by a discrete payment stream b = (b (0) , . . . , b (n)), the present

value adds up the present values of each and every payment, such that this becomes a sum of

payments which are accumulated respectively discounted to the time point of measurement. We

get the present value

PV (t) =∑n

s=0b (s) (1 + i)

t−s=∑n

s=0b (s) v−(t−s) =

∑n

s=0b (s) eδ(t−s).

In particular,

PV (0) =∑n

t=0b (t) (1 + i)

−t=∑n

t=0b (t) vt =

∑n

t=0b (t) e−δt,

PV (n) =∑n

t=0b (t) (1 + i)

n−t=∑n

t=0b (t) v−(n−t) =

∑n

t=0b (t) eδ(n−t).

If we replace the payment b by a discrete payment stream b =(

b (0) , b(

1m

)

, b(

2m

)

, . . . , b(

nmm

))

we get the present value

PV (t) =∑mn

s=0b (s) (1 + i)

t−s/m=∑mn

s=0b (s) v−(t−s/m) =

∑mn

s=0b (s) eδ(t−s/m).

In particular,

PV (0) =∑mn

t=0b (t) (1 + i)

−t/m=∑mn

t=0b (t) vt/m =

∑mn

t=0b (t) e−δt/m,

PV (n) =∑mn

t=0b (t) (1 + i)(nm−t)/m =

∑mn

t=0b (t) v−(nm−t)/m =

∑mn

t=0b (t) eδ(nm−t)/m.

If we replace the payment b by a continuous payment stream b = {b (t)}0≤t≤n such that overthe time interval [t, t + dt) the amount b (t) dt is deposited, we get the present value

PV (t) =

∫ n

0

b (s) (1 + i)t−s

ds =

∫ n

0

b (s) v−(t−s)ds =

∫ n

0

b (s) eδ(t−s)ds.

In particular,

PV (0) =

∫ n

0

b (t) (1 + i)−t

dt =

∫ n

0

b (t) vtdt =

∫ n

0

b (t) e−δtdt,

PV (n) =

∫ n

0

b (t) (1 + i)n−t

dt =

∫ n

0

b (t) v−(n−t)ds =

∫ n

0

b (t) eδ(n−t)ds.

1.2 Particular Present Values

1. Annuity in arrear (Efterbetalt annuitet): b = (b (0) , . . . , b (n)) : b (0) = 0, b (1) = . . . =

b (n) = 1.

an| = PV (0)

=∑n

t=1vt = v

∑n−1

t=0vt = v

1− vn1− v =

1− vni

,

sn| = PV (n)

=∑n

t=1v−(n−t) = v−n

∑n

t=1vt = v−n

1− vni

=(1 + i)

n − 1i

.

5

2. Annuity-due (Forudbetalt annuitet): b (0) = . . . = b (n− 1) = 1, b (n) = 0.

än| = PV (0)

(a) =∑n−1

t=0vt =

1− vn1− v =

1− vnd

(b) =∑n

t=1vt−1 = v−1

∑n

t=1vt = (1 + i) an|,

s̈n| = PV (n)

(a) =∑n−1

t=0v−(n−t) = v−n

∑n−1

t=0vt = v−n

1− vn1− v =

(1 + i)n − 1

d

(b) =∑n

t=1v−(n−(t−1)) = v−1

∑n

t=1v−(n−t) = (1 + i) sn|.

3. Fractionated annuity in arrear (Efterbetalt fraktioneret annuitet): b =(

b(

1m

)

, b(

2m

)

, . . . , b(

nmm

))

:

b (0) = 0, b(

1m

)

= . . . = b(

nmm

)

= 1m .

a(m)

n|= PV (0)

=1

m

∑mn

t=0

(

v1m

)t

=1

mamn|

(

(1 + i)1/m − 1← i

)

.

4. Continuous annuity (Kontinuert annuitet): b = {b (t)}0≤t≤n : b (t) = 1.

an| = PV (0)

=

∫ n

0

e−δtdt =

[

−1δe−δt

]n

0

=1− vn

δ,

sn| =

∫ n

0

eδ(n−t)ds = eδn∫ n

0

e−δtdt =(1 + i)

n − 1δ

.

Note that

limm→∞

a(m)

n|=

1

mamn|

(

(1 + i)1/m − 1← i

)

= limm→∞

1− vn

m(

(1 + i)1/m − 1

)

=1− vn

δ= an|.

5. Deferred annuity in arrear (Opsat efterbetalt annuitet): b = (b (0) , . . . , b (k + n)) : b (0) =

. . . = b (k) = 0, b (k + 1) = . . . = b (k + n) = 1.

kan| = PV (0)

(a) =∑k+n

t=k+1vt = vk

∑n

t=1vt = vkan|

(b) =∑k+n

t=1vt −

∑k

t=1vt = ak+n| − ak|.

6. Deferred annuity-due (Opsat forudbetalt annuitet): b = (b (0) , . . . , b (k + n)) : b (0) = . . . =

b (k − 1) = b (n) = 0, b (k) = . . . = b (k + n− 1) = 1.

kän| = PV (0)

6

(a) =∑k+n−1

t=kvt = vk

∑n−1

t=0vt = vkän|

(b) =∑k+n−1

t=0vt −

∑k−1

t=0vt = äk+n| − äk+n|.

7. Deferred continuous annuity (Opsat kontinuert annuitet): b = {b (t)}0≤t≤k+n : b (t) = 0, 0 ≤t ≤ k, b (t) = 1, k ≤ t ≤ k + n.

kan| = PV (0)

(a) =

∫ k+n

k

e−δtdt = e−δk∫ n

0

e−δtdt = e−δkan|

(b) =

∫ k+n

0

e−δtdt−∫ k

0

e−δtdt = ak+n| − ak|.

1.3 Exercises

Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Exercises 2.4.5-

2.4.8.

Exercise 1 Consider first the accumulation factor B (t) = v−t and the discount factor B−1 (t) =

vt. Show the following difference equations and side conditions

B (t) = B (t− 1) (1 + i) ,B (0) = 1,

B−1 (t) =1

1 + iB−1 (t− 1) ,

B−1 (0) = 1.

Consider a general savings contract over n years specifying that at each time t = 0, . . . , n the

amount c (t) is deposited and the amount b (t) is withdrawn. Introduce the quantities

U (t) = B (t)∑t

j=0B−1 (j) (c (j)− b (j)) ,

V (t) = B (t)∑n

j=t+1B−1 (j) (b (j)− c (j)) .

Show the difference equations (recurrence equations) with side conditions

U (t) = U (t− 1) (1 + i) + c (t)− b (t) ,U (0) = c (0)− b (0) ,V (t) = (1 + i)

−1(b (t + 1)− c (t + 1) + V (t + 1)) ,

V (n) = 0.

Give an interpretation of the quantities U and V and the difference equations. Interpret the con-

dition

U (n) = 0.

Show that this condition implies the relations

V (0) = c (0)− b (0) , (1)U (t) = V (t) ,

7

and give an interpretation.

Consider a so-called bank endowment contract where a person saves a fixed amount c annually

in 15 years (t = 0, . . . , 14) and thereafter (t = 15) withdraws b = 1 (million DKK, say). Write

down the equivalence constraint (1) in terms of i and c. Suppose the annual interest rate is fixed

and equal to i = 0.045. Calculate the annual savings amount c. (solution: 0.04604).

Exercise 2 This exercise is a continuous time version of Exercise 1. Consider the accumulation

factor B (t) = eδt and the discount factor B−1 (t) = e−δt. Show the following ordinary differential

equations and side conditions

d

dtB (t) = δB (t) ,

B (0) = 1,

d

dtB−1 (t) = −δB−1 (t) ,

B−1 (0) = 1.

Consider a general savings contract over n years specifying that at time t deposit occurs at rate

c (t) and withdrawal occurs at rate b (t). Introduce the quantities

U (t) = eδt∫ t

0

e−δs (c (s)− b (s)) ds,

V (t) = eδt∫ n

t

e−δs (b (s)− c (s)) ds.

Show the ordinary differential equations with side conditions

d

dtU (t) = δU (t) + c (t)− b (t) , U (0) = 0,

d

dtV (t) = δV (t)− (b (t)− c (t)) , V (n) = 0.

Give an interpretation of the quantities U and V and the differential equations. Interpret the

condition

U (n) = 0.

Show that this condition implies the relations

V (0) = 0, (2)

U (t) = V (t) ,

and give an interpretation.

Consider a bank annuity contract where a person saves at a fixed rate c in 15 years (0 < t < 15)

and withdraws at a fixed rate b = 1 (million DKK, say) in 5 years (15 < t < 20). Write down the

equivalence constraint (2) in terms of δ and c. Suppose the interest rate is fixed and equal to

δ = log (1 + 0.045). Calculate the saving rate c.

8

2 Life Insurance Risk: Mortality

2.1 The lifetime stochastic variable

We denote the remaining lifetime of an individual by the stochastic variable T ,

F (t) = P (T ≤ t) ,F (t) = P (T > t) = 1− F (t) .

Assume that 0 < T ≤ ω (possibly ω =∞) such that

F (0) = 1− F (0) = 1,F (ω) = 1− F (ω) = 0.

Typically one thinks of time t = 0 as being the time of birth of an individual such that time equals

ages.

tqx = P (T ≤ x + t|T > x) =P ({T ≤ x + t} ∩ {T > x})

P (T > x)

=P (x < T ≤ x + t)

P (T > x)=

F (x + t)− F (x)1− F (x)

=F (x)− F (x + t)

F (x),

tpx = P (T > x + t|T > x) =P ({T > x + t} ∩ {T > x})

P (T > x)

=P (T > x + t)

P (T > x)=

F (x + t)

F (x).

Note that

tq0 = F (t) =⇒ tqx =F (x + t)− F (x)

1− F (x) =x+tq0 − xq0

1−x q0,

tp0 = F (t) =⇒ tpx =F (x + t)

F (x)=

x+tp0

xp0.

µ (t) = lim∆t→0

1

∆tt+∆tqt = lim

∆t→0

1

∆tP (T < t + ∆t|T > t)

= lim∆t→0

1

∆t

F (t + ∆t)− F (t)F (t)

=1

F (t)lim

∆t→0

F (t + ∆t)− F (t)∆t

=1

F (t)

d

dtF (t) = − 1

F (t)

d

dtF (t)

= − ddt

log F (t) .

This is a differential equation for log F (t) with the side condition log F (0) = 0, such that the

solution becomes log F (t) = −∫ t

0 µ (s) ds or equivalently

F (t) = e−R

t0

µ(s)ds.

9

Note that

tpx =x+tp0

xp0=

e−R

x+t0

µ(s)ds

e−R

x

0µ(s)ds

= e−R

x+tx

µ(s)ds = e−R

t

0µ(x+s)ds.

In mortality tables, one typically specifies the probabilities in terms of so-called decrement

series. One considers a population of l0, typically set to 100.000. Then lx denotes the number of

expected survivors at age x. Then

l0 = 100.000

lx = l0 F (x) = l0 xp0

dx = lx − lx+1

tpx =F (x + t)

F (x)=

lx+tlx

tqx = 1−t px =F (x)− F (x + t)

F (x)=

lx − lx+tlx

Example: Gompertz-Makeham G82M,

µ = α + βeγt

= 5 · 10−4 + 7.5858 · 10−5et log(1.09144)

= 5 · 10−4 + 7.5858 · 10−5 · 1.09144t.

2.2 Expected values

Let T be a life time stochastic variable such that F is differentiable. Then for a differentiable

function G,

E [G (T )] =

∫ ω

0

G (t)

(

d

dtF (t)

)

dt

= −∫ ω

0

G (t)

(

d

dtF (t)

)

dt

= −G (ω) F (ω) + G (0) F (0) +∫ ω

0

F (t)

(

d

dtG (t)

)

dt

= G (0) +

∫ ω

0

F (t)

(

d

dtG (t)

)

dt

Taking G (t) = tk we get

E[

T k]

= k

∫ ω

0

tk−1F (t) dt,

and, in particular

e0 = E [T ] =

∫ ω

0

F (t) dt =

∫ ω

0

e−R

t0

µ(s)dsdt

One can also show that the expected remaining life time of an x years old individual is

ex

(a) =

∫ ω−x

0tpxdt =

∫ ω−x

0

lx+tlx

dt =1

lx

∫ ω−x

0

lx+tdt ≡Txlx

(b) =

∫ ω−x

0

e−R

x+tx

µ(s)dsdt =

∫ ω−x

0

e−R

t0

µ(x+s)dsdt

10

Definitions

Lx =

∫ 1

0

lx+tdt = l0

∫ 1

0

F (x + t) dt

= the expected group life time a year from x + t,

Tx =

∫ ω−x

0

lx+tdt =∑ω−x

t=0Lx+t = l0

∫ ω−x

0

F (x + t) dt

= the expected group residual life time from x + t.

1. Demography,

2. Stationary populations,

3. Mortality tables,

4. Longevity,

5. Sex,

6. Geography,

7. Genetic vs. behavioral vs. environmental effects. Cause vs. effects.

2.3 Exercises

Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Exercises 3.6.2-

3.6.6.

Exercise 3 • The exponential mortality law is characterized by a constant force of mortality,µ (t) = λ. Show that

F (t) = e−λt,

f (t) =d

dtF (t) = λe−λt.

Introduce the conditional survival function

tpx =F (x + t)

F (x).

Show that

tpx = tp0,

i.e., in particular, this is independent of x. Conclude that

F (x + t) = F (x) F (t)

and interpret this result.

• The Weibull mortality law is characterized by the intensity

µ (t) = βα−βtβ−1, α, β > 0.

11

Show that

F (t) = e−(tα )

β

.

Show that µ is increasing for β > 1, decreasing for β < 1, and that the Weibull law equals

the exponential law for β = 1. Calculate tpx.

• The Gompertz-Makeham mortality law is characterized by the mortality intensity

µ = α + βeγt, α, β > 0.

Show that

F (t) = e(−αt−β(eγt−1)/γ).

Show that if γ > 0, then µ is an increasing function of t. Calculate tpx.

12

3 Non-life Insurance Risk: Frequency and Severity

3.1 Compound risk models

In this lecture, we shall study the accumulated claim of a non-life insurance portfolio. Claim means

the amount that the insurance company has to pay out to its police holders. Non-life insurance

means obviously types of insurance that cannot be characterized as life insurance. These include

most types of insurances on ’things’. Portfolio means that we consider an aggregate or a collective

of insurance contracts. The ideas and models can also be interpreted on an individual insurance

level, but for this purpose there are many alternatives which we shall not consider here. The model

we consider here is known as the collective risk model or simply risk model.

The accumulated claim on the insurance company contains basically two sources of uncertainty:

What is the number of claims in the portfolio? This uncertainty is modelled by probability assump-

tions on the claim number distribution. What is the size of the claims? This uncertainty is modelled

by probability assumptions on the claim size distribution.

The claim number distribution is a discrete distribution on non-negative integers. Let N be the

number of claims occurring in a particular time interval, say, a year. Then the distribution of N is

described by the probabilities

p (n) = P (N = n) , n = 0, 1, . . . .

Let Yi denote the amount of the ith claim. We assume that Y1, Y2, . . . are positive, mutually

independent, identically distributed with common distribution function F , and independent of the

claim number N . Let X be the accumulated claim amount, that is

X =∑N

i=1Yi = Y1 + Y2 + . . . + YN .

There are many reasons for studying the total claim. This shall teach us something about the

business such that we can take appropriate actions with respect to premiums, reserves, reinsurance,

etc. In general, one is interested in calculating characteristic in the form

E [f (X)] =∑∞

n=0E [f (X)|N = n] p (n)

=∑∞

n=0E [f (Y1 + Y2 + . . . + Yn)] p (n) .

In order to go on from here, we need to say more about p, F, and f , i.e. the claim number distri-

bution, the claim size distribution, and the characteristic f . We shall here proceed a step further

for two specific choices of f and get as long as we can without specifying p and F .

Firstly, we calculate the expected value of X by letting f (x) = x. This gives

E [X ] =∑∞

n=0E [X |N = n] p (n) =

∑∞

n=0E [Y1 + Y2 + . . . + Yn] p (n)

=∑∞

n=0

∑n

i=1E [Yi] p (n) =

∑∞

n=0nE [Y1] p (n) = E [Y1] E [N ] .

This result is very intuitive. The expected accumulated claim equals the expected number of claims

multiplied by the expected claim size. Secondly, we calculate the second moment of X by letting

f (x) = x2. With the notation

∑

i=

∑n

i=1

13

∑

i,j=

∑

i

∑

j∑

i6=j=

∑

i

∑

j:j 6=i,

this gives that

E[

X2]

=∑∞

n=0E[

X2∣

∣N = n]

p (n) =∑∞

n=0E[

(Y1 + Y2 + . . . + Yn)2]

p (n)

=∑∞

n=0E[

∑

i,jYiYj

]

p (n) =∑∞

n=0

∑

i,jE [YiYj ] p (n)

=∑∞

n=0

∑

i6=jE [YiYj ] p (n) +

∑∞

n=0

∑

i

∑

j:j=iE [YiYj ] p (n)

=∑∞

n=0

∑

i6=jE [Yi] E [Yj ] p (n) +

∑∞

n=0

∑

iE[

Y 2i]

p (n)

=∑∞

n=0n (n− 1)E2 [Y1] p (n) +

∑∞

n=0nE[

Y 21]

p (n)

= E2 [Y1]∑∞

n=0

(

n2 − n)

p (n) + E[

Y 21]

∑∞

n=0np (n)

= E2 [Y1](

E[

N2]

−E [N ])

+ E[

Y 21]

E [N ]

= E2 [Y1] E[

N2]

+ E [N ] V ar [Y1] .

The first and second moments makes it possible to calculate the variance of the accumulated claim.

This obviously tells us something about the riskiness of the business. This riskiness may play a

role an many actions.

V ar [X ] = E[

X2]

−E2 [X ]= E2 [Y1] E

[

N2]

+ E [N ] V ar [Y1]−E2 [Y1] E2 [N ]= E [N ]V ar [Y1] + E

2 [Y1]V ar [N ] .

The expected value and the variance are probably the two most important characteristic of a

stochastic variable. We can now calculate these for the accumulated claim amount provided that

we have expected value and variance of the claim number and the claim size. respectively. In the

next subsections, we study particular claim number and claim size distribution which for various

reasons play special roles.

3.2 Frequency risk

In this section, we present three claim number distributions. All three distribution belongs to a

certain so-called (a, b)-class of distributions for which

p (n) =

(

a +b

n

)

p (n− 1) .

• The Poisson distribution is characterized for λ > 0 by

p (n) = e−λλn

n!, n = 0, 1, . . . .

That the Poisson distribution belongs to the (a, b)-class is seen by

p (n) = e−λλn

n!=

λ

ne−λ

λn−1

(n− 1)! =λ

np (n− 1) ,

14

such that

a = 0,

b = λ, λ > 0.

• The negative binomial distribution is characterized for 0 < q < 1 and α > 0 by

p (n) =

(

α + n− 1n

)

qn (1− q)α , n = 0, 1, . . . .

That the negative binomial distribution belongs to the (a, b)-class is seen by

p (n) =

(

α + n− 1n

)

qn (1− q)α

=Γ (α + n)

n!Γ (α)qn (1− q)α

=q

n

(α + n− 1) Γ (α + n− 1)(n− 1)!Γ (α) q

n−1 (1− q)α

=

(

q +q (α− 1)

n

)

Γ (α + n− 1)(n− 1)!Γ (α) q

n−1 (1− q)α ,

a = q, 0 < q < 1

b = q (α− 1) , αq > 0⇔ b > −a

such that

a = q, 0 < q < 1,

b = q (α− 1) > −q = −a.

The fact that a > 0 has the interpretation that, compared with the Poisson distribution, the

probability of more claims increases with the number of claims. This is spoken of as positive

contagion.

• The binomial distribution is characterized for 0 < p < 1 and m = 1, 2, . . . by

p (n) =

(

m

n

)

pn (1− p)m−n , n = 0, 1, . . . , m.

That the binomial distribution belongs to the (a, b)-class is seen by

p (n) =

(

m

n

)

pn (1− p)m−n

=p (m− n + 1)

n (1− p)m!

(n− 1)! (m− n + 1)!pn−1 (1− p)m−(n−1)

=

(

− p1− p +

(m + 1) p

n (1− p)

)

p (n− 1) ,

such that

a = − p1− p , 0 < p < 1

b = (m + 1)p

1− p = − (m + 1) a.

15

The fact that a < 0 has the interpretation that, compared with the Poisson distribution,

the probability of more claims decreases with the number of claims. This is spoken of as

negative contagion. This opposite contagion effects in the binomial distribution compared to

the negative binomial distribution explains actually the term negative binomial

3.3 Severity risk

In this section, we present three claim size distributions. There are many different ways of chara-

cterizing and categorizing claim size distributions depending one the purpose. In particular, one

is often interested in characteristics of the riskiness. We know that the variance is one measure of

risk. However, this is for some purposes and for some distributions a completely insufficient piece

of information. As we shall see, there exists highly relevant claim size distributions that behave so

wildly that the variance not even exists, i.e. is infinite. An important characteristic which we shall

not pursue here is the so-called mean residual life time (MRL)

m (y) = E [Y − y|Y > y] .

This measures the expected claim size beyond y given that the claims size is above y. Comparing

with the previous section, this quantity actually equals ey, explaining the term MRL. The MRL is

in practice often estimated to being increasing and concave. One typically discusses the ’thickness

of the tail’ which is related to measure like m (y) and how it behaves when y goes to infinity.

• The Gamma (γ, δ) distribution is characterized for γ > 0 and δ > 0 by the density

f (y) =δγ

Γ (γ)yγ−1e−δy, y ≥ 0.

For the Gamma distribution we exploit that

∫ ∞

0

δγ

Γ (γ)yγ−1e−δydy = 1,

and we know this since this is how the Gamma function is actually defined. We can then

calculate

E[

Y ke−aY]

=

∫ ∞

0

yke−ayδγ

Γ (γ)yγ−1e−δydy

=δγ

Γ (γ)

∫ ∞

0

yk+γ−1e−(δ+a)ydy

=Γ (γ + k)

Γ (γ)

δγ

(δ + a)γ+k

.

• The Pareto (λ, α) distribution is characterized for α > 0 by the distribution function

F (y) = 1−(

λ

y

)α

, y > λ,

which gives the density

f (y) =α

λ

(

λ

y

)α+1

, y > λ.

16

For the Pareto we exploit that

∫ ∞

λ

α

λ

(

λ

y

)α+1

dy =α

λλα+1

∫ ∞

λ

y−α−1dy =α

λλα+1

[

− 1α

y−α]∞

λ

= 1

distribution we can for k < α calculate

E[

Y k]

=

∫ ∞

λ

ykα

λ

(

λ

y

)α+1

dy

=

∫ ∞

λ

α

λ

λα+1

yα−k+1dy

=

∫ ∞

λ

α

α− kλk α− k

λ

λα−k+1

yα−k+1dy

=α

α− kλk

∫ ∞

λ

α− kλ

λα−k+1

yα−k+1dy

=α

α− kλk.

Note that the kth moments only exist up to α. In more dangerous lines of business, one will

often find that the estimated value of α is close to 1.

A very appealing and interesting property of the Pareto distribution is that is Y is Pa-

reto (λ, α) distributed, then the conditional distribution of Y given Y > y is again Pareto

distributed with parameters (y, α). Thus,

m (y) =αy

α− 1 − y =y

α− 1 , y > λ.

• The lognormal(

α, σ2)

distribution is characterized by the density

f (y) =1

σ√

2π

1

ye−

12 (

log y−ασ )

2

, y > 0.

The moments of the lognormal distribution can be calculated to

E[

Y k]

= eαk+σ2

2 k2

.

3.4 Exercises

Exercise 4 Assume that N is Poisson distributed with parameter λ. Show that

E [N ] = λ,

E[

N2]

= λ2 + λ,

V ar [N ] = λ,

and use this to show that

E [X ] = λE [Y ] ,

E[

X2]

= λE[

Y 2]

+ λ2E2 [Y ] ,

V ar [X ] = λE[

Y 2]

.

17

Exercise 5 Assume that N is negative binomial distributed with parameters (α, q). Show that

E [N ] = qα + qE [N ]

E[

N2]

= q2E[

N2]

+ q2 (2α + 1) E [N ] + q2(

α2 + α)

+ E [N ]

and calculate that

E [N ] =qα

1− q ,

E[

N2]

=1 + q2 (2α + 1)

1− q2 E [N ] +q2

1− q2(

α2 + α)

=qα

1− q2(

1 + q2 (2α + 1)

1− q + q (α + 1))

=qα

1− q2(

1 + q2α + qα + q

1− q

)

,

V ar [N ] =qα

1− q2

(

1 + q2 (2α + 1)

1− q + q (α + 1)−(

1− q2)

q

(1− q)2α

)

=qα

(1− q)2.

Exercise 6 We introduce the mean, variance, coefficient of variation, and skewness,

µ = E [Y ] ,

σ2 = E2 [Y −E [Y ]]= E

[

Y 2]

−E2 [Y ] ,κ =

σ

µ,

ν =E3 [Y − E [Y ]]

σ3

=E[

Y 3]

σ3− 3

κ− 1

κ3.

Argue why µ and σ are not independent of monetary unit, while κ and ν are.

Consider the Pareto claim size distribution. Show that

µ =αλ

λ− 1 ,

σ2 =αλ2

(α− 2) (α− 1)2,

κ =1

√

α (α− 2).

Assume that N is negative binomial distributed with parameters (q, β). Calculate E [X ] and

V ar [X ].

18

4 Valuation principles

4.1 Properties

1.

π (X + Y ) ≤ π (X) + π (Y ) .

It should not be possible to get a discount from splitting up risk.

2.

P (X ≤ Y ) = 1⇒ π (X) ≤ π (Y ) .

A better cover should be more expensive cover.

3.

π (X) ≥ E [X ] .

No losses from large insurance portfolios

π (X) < E [X ] : limn→∞

P(

∑n

i=1Xi > nπ (X)

)

= 1,

π (X) > E [X ] : limn→∞

P(

∑n

i=1Xi > nπ (X)

)

= 0.

4.

P (X < π (X)) < 1,

essentially equivalent with

P (X ≤ m) = 1⇒ π (X) ≤ m.

A cover should be cheaper than the largest possible loss.

Assume that π (m) = m. Then we get from 2 with Y = m,

P (X ≤ m) = 1⇒ π (X) ≤ m,

which is exactly 4’. I.e. if no 4’ and π (m) = m, then no 2.

4.2 Principles

Expected value principle

π (X) = (1 + a) E [X ]

1.

π (X + Y ) = (1 + a) E [X + Y ]

= (1 + a) (E [X ] + E [Y ])

= (1 + a) E [X ] + (1 + a) E [Y ]

= π (X) + π (Y )

19

2.

P (X ≤ Y ) = 1⇒Y −X ≥ 0

E (Y −X) = EY −EX ≥ 0E [X ] ≤ E [Y ] .

3.

(1 + a) E [X ] ≥ E [X ]

4. For a constant risk m,

π (m) = (1 + a) m > m and P (m ≤ π (m)) = 1.

Standard deviation principle

π (X) = E [X ] + a√

V ar [X ].

1.

π (X + Y ) ≤ π (X) + π (Y )

√

V ar [X + Y ] =√

V arX + V arY + 2Cov [X, Y ]

≤√

V arX + V arY + 2√

V ar [X ]√

V ar [Y ]

=

√

(√V arX +

√V arY

)2

=√

V arX +√

V arY

where we use that

V ar [X + Y ] = E (X + Y )2 −E2 (X + Y )= EX2 + EY 2 + 2E [XY ]− (EX + EY )2

= EX2 −E2X + EY 2 −E2Y + 2E [XY ]− 2EXEY= V arX + V arY + 2Cov [X, Y ] ,

Cov [X, Y ] = E [(X −EX) (Y −EY )]= E [XY − Y EX −XEY + EXEY ]= E [XY ]−EXEY.

Cov [X, Y ] = E [(X −EX) (Y −EY )]

≤√

E (X −EX)2√

E (Y −EY )2

= V ar [X ]V ar [Y ] .

2. see 4.

20

3.

π (X) ≥ E [X ]

4.

P (X = m) = p = 1− P (X = 0)EX = mp

V arX = pm2 − (pm)2 = m2p (1− p)g (p) = π (X) = m

(

p + a√

p (1− p))

g′ (p) = m

(

1 +1

2

a (1− 2p)√

p (1− p)

)

limp→1

g′ (p) = m

(

1 +1

2

a (1− 2)0

)

= −∞

exist a p < 1 such that

π (X) > π (m) = m,

i.e. 4’ is not satisfied. But then also 2 is not satisfied.

Variance principle

π (X) = E [X ] + aV ar [X ]

1.

π

(

1

2X +

1

2X

)

≤ 2π(

1

2X

)

nπ

(

X

n

)

= E [X ] +a

nV ar [X ]

= π (X)− a(

1− 1n

)

V ar [X ]

< π (X)

limn→∞

nπ

(

X

n

)

= π (X)− aV ar [X ]

= E [X ]

2. see 4.

3.

π (X) ≥ E [X ] .

4.

π (X) = mp (1 + am (1− p)) > m⇔ m > (ap)−1 .

21

Exponential principle

π (X) =1

alog E

[

eaX]

Esscher principle

π (X) =E[

XeaX]

E [eaX ]

= E

[

XeaX

E [eaX ]

]

Quantile principle

π (X) = min {m : P (X > m) ≤ a} .

General on measure transformation

π (X) = E∗ [X ]

4.3 Exercises

Exercise 7 4.2

f (x) =δγ

Γ (γ)xγ−1e−δx.

π (X) = E [X ] + aV ar [X ]

=1

Γ (γ) δ

(

Γ (γ + 1) + aΓ (γ + 2)

δ− aΓ

2 (γ + 1)

Γ (γ) δ

)

π (X) =1

alog E

[

eaX]

=γ

alog

(

δ

δ − a

)

π (X) =E[

XeaX]

E [eaX ]

=Γ (γ + 1)

Γ (γ) (δ − a)

Exercise 8 4.4

E[

Y k]

=α

α− kλk, k < α

V ar [X ] = pλ2α

(

1

α− 2 − pα

(α− 1)2

)

π (X) = (1 + a) E [X ]

= (1 + a) pα

α− 1λ

π (X) = E [X ] + aV ar [X ]

= pα

α− kλ + apλ2α

(

1

α− 2 − pα

(α− 1)2

)

22

π (X) = E [X ] + a√

V ar [X ]

= pα

α− 1λ + a

√

√

√

√pλ2α

(

1

α− 2 − pα

(α− 1)2

)

π (X) = min {m : P (X > m) ≤ a}= min {m : pP (Y > m) ≤ a}

= min

{

m :

(

λ

m

)α

≤ ap

}

= min

{

m :(p

a

)1/a

λ ≤ m}

=(p

a

)1/a

λ

Exercise 9 4.5

π (X) = mink{(1 + a) E [kX ] + (1 + b) E [(1− k) X ]}

= mink{(1 + a) kE [X ] + (1 + b) (E [X ]− kE [X ])}

= (1 + b)E [X ] + E [X ] (a− b) mink{k}

(k = 1) = (1 + a) E [X ]

π (X) = mink

{

E [kX ] + a√

V ar [kX ] + E [(1− k)] + b√

V ar [(1− k) X ]}

= E [X ] +√

V ar [X ]

(

b + (a− b) mink{k})

(k = 1) = E [X ] + a√

V ar [X ]

π (X) = mink{E [kX ] + aV ar [kX ] + E [(1− k)] + bV ar [(1− k) X ]}

= E [X ] + V ar [X ]mink

{

ak2 + b (1− k)2}

= E [X ] + V ar [X ]mink

{

ak2 + b (1− k)2}

(

k =b

a + b

)

= E [X ] + V ar [X ]ab

(a + b)2

{

b + a + 2b

a

}

5 Life Insurance: Products and Valuation

Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Chapter 5.

5.1 Single payments

The present value at time 0 of the deterministic payment b at time n according to lecture 1,

(1 + i)−n

b = vnb = e−δnb

The expected present value at time 0 of the stochastic variable B at time n,

E[

e−δtB (t)]

= e−δtE [B (t)]

23

In life insurance on the life death state of health of an individual B is always for a constant b

B (t) = b1(Tx>t)

B (t) = b1(st)fx (u) du =

∫ ∞

t

fx (u) du

= P (Tx > t) = tpx = e−

R

t

0µ(x+s)ds

(

= e−µt)

E[

1(s

• Term insurance

B (t) = 1(t−1

äxn =∑n−1

t=0e−δtE

[

1(Tx>t)]

=∑n−1

t=0e−δt tpx

(

=∑n−1

t=0e−δte−µt =

∑n−1

t=0e−(δ+µ)t =

1− e−(δ+µ)n1− e−(δ+µ)

)

p149

In particular for n =∞:

äx =∑∞

t=0e−δt tpx

(

=1

1− e−(δ+µ))

p144.

5.3 Premium calculation

Put expected present value of premiums equal to the expected present value of benefits (the

equivalence principle).

• Term insurance

πäxn = A1xn ⇒ π =

A1xnäxn

=(1− e−µ) e−δ 1−e−(δ+µ)n

1−e−(δ+µ)

1−e−(δ+µ)n

1−e−(δ+µ)

=(

1− e−µ)

e−δ

p145

note that this is independent of the time horizon, i.e. the same formula for a whole life term

insurance.

V (0−) = A1xn − πäxn = 0V (t) = A1x+t n−t − πäx+t n−tV (n) = 0

∆V (t) = V (t)− V (t− 1) = ...

• Pure endowment insurance

πäxn = nEx ⇒ π = nEx

äxn(

=e−(δ+µ)n

1−e−(δ+µ)n

1−e−(δ+µ)

)

V (0−) = nEx − πäxn = 0V (t) = n−tEx+t − πäx+t n−tV (n) = 0

• Endowment insurance

πäxn = Axn ⇒ π =Axnäxn

( =e−δ 1−e

−(δ+µ)n

1−e−(δ+µ)− e−(δ+µ)−e−(δ+µ)n

1−e−(δ+µ)

1−e−(δ+µ)n

1−e−(δ+µ)

26

=e−δ

(

1− e−(δ+µ)n)

−(

e−(δ+µ) − e−(δ+µ)n)

1− e−(δ+µ)n

= e−δ − e−(δ+µ) − e−(δ+µ)n

1− e−(δ+µ)n

V (0−) = Axn − πäxn = 0V (t) = Ax+t n−t − πäx+t n−tV (n) = 0

∆V (t) = V (t)− V (t− 1) = ...

5.4 Exercises

Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Exercises 5.6.1-

5.6.6.

Exercise 10 We consider a so-called deferred temporary life annuity(-due) (opsat ophørende for-

udbetalt livrente) where

B (t) = 1(Tx>t,k≤tt)1(k≤tt)1(k≤t

=∑k+n−1

t=ke−δtE

[

1(Tx>t)]

=∑k+n−1

t=ke−δt tpx

=∑k+n−1

t=ke−δt kpx t−kpx+k (her f̊ar de nok problemer)

= kpx∑k+n−1

t=ke−δt t−kpx+k

= kpx∑n−1

t=0e−δ(t+k) tpx+k

= e−δk kpx∑n−1

t=0e−δt tpx+k

= kEx äx+k n|

c)

∑k+n−1

t=ke−δt tpx =

∑k+n−1

t=ke−δte−µt

=∑k+n−1

t=ke−(δ+µ)t

= e−(δ+µ)k∑k+n−1

t=ke−(δ+µ)(t−k)

= e−(δ+µ)k∑n−1

t=0e−(δ+µ)t

= e−(δ+µ)k1− e−(δ+µ)n1− e−(δ+µ) , (∗)

∑∞

t=ke−δt tpx = ... = kEx äx+k,

∑∞

t=ke−δt tpx = e

−(δ+µ)k 1− e−(δ+µ)∞1− e−(δ+µ) =

e−(δ+µ)k

1− e−(δ+µ) .

(∗) kunne selvfølgelig ogs̊a ses ved at sætte ind i kEx äx+k n| med konstant µ.d) Combine the deferred temporary annuity and the temporary life annuity into

käxn|

e)

πäxk| = käxn| ⇒ π =käxn|

äxk|

f )

π =käxn|

äxk|=

e−(δ+µ)k 1−e−(δ+µ)n

1−e−(δ+µ)

1−e−(δ+µ)k

1−e−(δ+µ)

= e−(δ+µ)k1− e−(δ+µ)n1− e−(δ+µ)k ,

π =käxäxk|

,

π =käxäxk|

=e−(δ+µ)k

1− e−(δ+µ)k .

Exercise 11 Assume a linear mortality intensity, i.e. µ (t) = µt. Show that

e−R

t

sµ(x+u)du = e−µ(x(t−s)+

12 (t

2−s2)).

28

Show that

nEx = e−(µx+δ)n−µ 12 n

2

,

A1xn| =∑n

t=1e−δte−µ(x(t−1)+

12 (t−1)

2)(

1− e−µ(x+t− 12 ))

,

Axn| = e−δne−µ(xn+

12 n

2) +∑n

t=1e−δte−µ(x(t−1)+

12 (t−1)

2)(

1− e−µ(x+t− 12 ))

,

äxn| =∑n−1

t=0e−(µx+δ)t−µ

12 t

2

Solution:

e−R

ts

µ(x+u)du = e−R

x+tx+s

µ(u)du = e−µR

x+tx+s

udu

= e−µ[ 12 u

2]x+tx+s = e−µ(

12 (x+t)

2− 12 (x+s)2)

= e−µ

“

12 (x

2+t2+2xt)− 12 (x2+s2+2xs)

2”

= e−µ(12 t

2+xt− 12 s2−xs)

= e−µ(x(t−s)+12 (t

2−s2))

nEx = e−δne−

R

n0

µ(x+s)ds = e−δne−µ(xn+12 n

2) = e−(µx+δ)n−µ12 n

2

A1xn| =∑n

t=1e−δt t−1px qx+t−1

=∑n

t=1e−δt t−1px (1− px+t−1)

=∑n

t=1e−δte−

R

t−10

µ(x+u)du(

1− e−R

10

µ(x+t−1+u)du)

=∑n

t=1e−δte−

R

t−10

µ(x+u)du(

1− e−R

tt−1

µ(x+u)du)

=∑n

t=1e−δte−µ(x(t−1)+

12 (t−1)

2)(

1− e−µ(x(t−(t−1))+ 12 (t2−(t−1)2)))

=∑n

t=1e−δte−µ(x(t−1)+

12 (t−1)

2)(

1− e−µ(x+t− 12 ))

Axn| = nEx + A1xn|

= e−δne−µ(xn+12 n

2) +∑n

t=1e−δte−µ(x(t−1)+

12 (t−1)

2)(

1− e−µ(x+t− 12 ))

äxn| =∑n−1

t=0e−δt e−µ(xt+

12 t

2)

=∑n−1

t=0e−(µx+δ)t−µ

12 t

2

.

Exercise 12 Consider an insurance contract on a couple consisting of two spouses (ægtefæller)

with initial ages x and y. We assume that the insurance company uses a unisex mortality table,

i.e. the same age dependent mortality rate µ is used for males and females. We also assume that

the lifetimes of the two spouses are independent. Consider an insurance where y receives a unit at

time n if at that time, x is dead and y is alive.

a) Formalize the claim B (n) in terms of the two stochastic variables Tx and Ty.

b) What is the expected present value at time 0 of this claim?

c) Assume that the premium is paid annually as long as x is alive. Determine the equivalence

premium.

Solution: a)

B (n) = 1(Tx≤n,Ty>n) = 1(Tx≤n)1(Ty>n)

29

b)

e−rnE [B (n)] = e−rnE[

1(Tx≤n)1(Ty>n)]

= e−rnE[

1(Tx≤n)]

E[

1(Ty>n)]

= e−rn nqx npy

c)

πäxn| = e−rT

nqx npy ⇒ π =e−rT nqx npy

äxn|.

6 Non-life Insurance: The Aggregate Claim Distribution

We shall here continue the study of the distribution of the total claim and expected values of

functions of the total claim, i.e. X =∑N

i=1Yi and E [f (X)] .We wish to calculate the probability

distribution P (X ≤ x), such that we can calculate so-called fixed time ruin probabilities. Assumethat the insurance company starts out with an initial capital c and receives the premiums π. Then

the so-called ruin probability P (X > c + π) = 1 − P (X ≤ c + π) specifies the probability thatthe initial capital plus the premium is inadequate to cover the total losses. There are also other

motivations for calculating the total claim distribution.

In the study of stochastic variables, one often studies certain functions namely the probability

generating function, the moment generating function, the characteristic function and the Laplace

transform

PX (s) = E[

sX]

,

MX (s) = E[

esX]

,

ΦX (s) = E[

eisX]

,

LX (s) = E[

e−sX]

.

We shall here work with the probability generating function PX (s). We have the following proba-

bility generating functions for the claim number and claim size distributions,

PN (s) = E[

sN]

=∑∞

n=0p (n) sn,

PY (s) = E[

sY]

.

With these functions we get

PX (s) = E[

sX]

(

=∑∞

n=0E[

sX∣

∣N = n]

p (n))

=∑∞

n=0E

[

s

∑n

i=1Yi

]

p (n)

=∑∞

n=0E[

sY1sY2 · · · sYn]

p (n)

=∑∞

n=0E[

sY1]

E[

sY2]

· · ·E[

sYn]

p (n)

=∑∞

n=0

(

E[

sY])n

p (n)

= PN (PY (s)) . (3)

30

Now, assume that the severities are positive integer-valued variables with probabilities

f (y) = P (Y1 = y) , y = 1, 2, . . . .

The distribution of X is then given by

g (x) = P (X = x)

=∑∞

n=0P (X = x, N = n)

=∑∞

n=0p (n) P (X = x|N = n)

=∑∞

n=0p (n) P (Y1 + . . . + Yn = x)

=∑∞

n=0p (n) f∗n (x) ,

where

f∗n (x) = P (Y1 + . . . + Yn = x)

=∑x

y=1P (Y1 + . . . + Yn = x, Yn = y)

=∑x

y=1f (y) P (Y1 + . . . + Yn = x| Yn = y)

=∑x

y=1f (y) P (Y1 + . . . + Yn−1 = x− y)

=∑x

y=1f (y) f∗n−1 (x− y) . (4)

Unfortunately this calculation is slowly and inefficient (One can realize that the number of

terms involved in calculation of g (x) is of order O(

x3)

). In the following two subsections we shall

study two alternatives. One is a precise recursion formula which works for certain claim number

distributions, the other is a moment-based approximation formula.

Example 13 Let N be Poisson distributed with λ = 1. Let Y be distributed such that f (1) =

1/4, f (2) = 1/2, f (3) = 1/4. Note that f ∗1 (x) = f (x). Then

f∗2 (2) = f (1) f∗1 (1) =1

4

1

4=

1

16,

f∗2 (3) = f (1) f∗1 (2) + f (2) f∗1 (1) =1

4

1

2+

1

2

1

4=

1

4,

f∗2 (4) = f (1) f∗1 (3) + f (2) f∗1 (2) + f (3) f∗1 (1) =1

4

1

4+

1

2

1

2+

1

4

1

4=

3

8,

...

f∗3 (3) = f (1) f∗2 (2) =1

4

1

16=

1

64,

...

g (0) = p (0) = e−1,

g (1) = p (1) f∗1 (1) =1

4e−1,

g (2) = p (1) f∗1 (2) + p (2) f∗2 (2) =1

2e−1 +

1

16

1

2e−1 =

(

1

2+

1

32

)

e−1,

31

g (3) = p (1) f∗1 (3) + p (2) f∗2 (3) + p (3) f∗3 (3)

=1

4e−1 +

1

4

1

2e−1 +

1

64

1

6e−1 =

(

1

4+

1

8+

1

384

)

e−1.

6.1 The Panjer Recursion

Let N belong to the (a, b)-class. We shall first derive a differential equation for the probability

generating function for the claim number process,

P ′N (s) =∑∞

n=0np (n) sn−1

=∑∞

n=1n

(

a +b

n

)

p (n− 1) sn−1

=∑∞

n=1(na + b) p (n− 1) sn−1

=∑∞

n=0((n + 1)a + b) p (n) sn

=∑∞

n=0(a + b) p (n) sn + as

∑∞

n=0np (n) sn−1

= (a + b)PN (s) + asP′N (s) ,

which leads to the differential equation

P ′N (s) =a + b

1− asPN (s) .

This differential equation has the side condition

PN (1) =∑∞

n=0p (n) = 1

If a = 0 (the Poisson distribution), the solution to this equation is

PN (s) = eb(s−1)

For a 6= 0 (the binomial and negative binomial distributions), the solution is

PN (s) =

(

1− a1− as

)a+b

a

.

By differentiation of (3) we get

P ′X (s) = P′N (PY (s)) P

′Y (s)

=a + b

1− aPY (s)PN (PY (s)) P

′Y (s)

=a + b

1− aPY (s)PX (s) P

′Y (s) ,

such that

P ′X (s)− aP ′X (s) PY (s) = (a + b) PX (s) P ′Y (s) .

32

Plugging in PX (s) =∑∞

x=0g (x) sx, P ′X (s) =

∑∞

x=0xg (x) sx−1, PY (s) =

∑∞

y=1f (y) sy, and

P ′Y (s) =∑∞

y=1yg (y) sy−1, we get

∑∞

x=0xg (x) sx−1 − a

∑∞

y=1f (y) sy

∑∞

x=0xg (x) sx−1

= (a + b)∑∞

x=0g (x) sx

∑∞

y=1yf (y) sy−1 ⇔

∑∞

x=1xg (x) sx−1 − a

∑∞

y=1

∑∞

x=0xf (y) g (x) sx+y−1

= (a + b)∑∞

y=1

∑∞

x=0yg (x) f (y) sx+y−1 ⇔(z=x+y)

∑∞

x=1xg (x) sx−1 − a

∑∞

y=1

∑∞

z=y(z − y) f (y) g (z − y) sz−1

= (a + b)∑∞

y=1

∑∞

z=yyg (z − y) f (y) sz−1 ⇔(switch summation)

∑∞

x=1xg (x) sx−1 − a

∑∞

z=1

∑z

y=1(z − y) f (y) g (z − y) sz−1

= (a + b)∑∞

z=1

∑z

y=1yg (z − y) f (y) sz−1 ⇔(rename x=z)

∑∞

x=1xg (x) sx−1 − a

∑∞

x=1

∑x

y=1(x− y) f (y) g (x− y) sx−1

= (a + b)∑∞

x=1

∑x

y=1yg (x− y) f (y) sx−1

from which we collect terms and summations into

∑∞

x=1

(

xg (x)−∑x

y=1(ax + by) g (x− y) f (y)

)

sx−1 = 0

If this true for all s, then

xg (x)−∑x

y=1(ax + by) g (x− y) f (y) = 0⇔

g (x) =∑x

y=1

(

a + by

x

)

g (x− y) f (y)

This recursive formula is considerably more efficient and faster than (4), (One can realize that the

number of terms involved in calculation of g (x) is of order O(

x2)

).

Example 14 Consider again the example where N is Poisson distributed with λ = 1, i.e. a =

0, b = 1, and Y is distributed such that f (1) = 1/4, f (2) = 1/2, f (3) = 1/4. Then

g (0) = p (0) = e−1,

g (1) = g (0) f (1) =1

4e−1,

g (2) =1

2g (1) f (1) + g (0) f (2) =

1

2

1

4e−1

1

4+ e−1

1

2=

(

1

32+

1

2

)

e−1,

g (3) =1

3g (2) f (1) +

2

3g (1) f (2) + g (0) f (3)

=1

3

(

1

32+

1

2

)

e−11

4+

2

3

1

4e−1

1

2+ e−1

1

4

=

((

1

32+

1

2

)

1

12+

2

24+

1

4

)

e−1

33

=

(

1

384+

3

24+

1

4

)

e−1

=

(

1

4+

1

8+

1

384

)

e−1.

6.2 Normal Power Approximation

In certain situation one may be willing to give up part of the accuracy in order to speed up the

calculation. We shall consider a moment-based approximation method called the Normal Power

approximation. Let u1−ε denote the 1 − ε fractile of G, i.e. u1−ε is the smallest number u suchthat 1−G (u) ≤ ε, and let z1−ε denote the 1− ε fractile of the standard normal distribution. NPapproximation says that approximately

u1−ε = E [X ] + z1−ε√

V ar [X ] +1

6

(

z21−ε − 1)

E[

(X −E [X ])3]

V ar [X ].

The total claim distribution and the normal power approximation may be used in the quantile

premium principle for valuation of a claim X . Recall the premium principle

π (X) = min {m : P (X > m) ≤ a} .

If we approximate the aggregate claim distribution by the normal power we get exactly that

π (X) = u1−a.

This is typically not thought of as a valuation principle but rather as a solvency criterion principle.

A solvency rule should protect the policy holders such that the insurance company is able to pay the

losses with a high probability. The solvency requirement is a minimum requirement on capital. One

natural solvency rule is to require from the insurance company that it has sufficiently capital such

that the probability that the aggregate losses exceed the capital is smaller than a given tolerance

level ε. An approximated solvency rule is that the company should at least hold the capital u1−ε.

6.3 Ruin Theory

Studying the probability distribution of X corresponds to studying the so-called fixed time ruin

probability. Other types of ruin probabilities are also of interest and belongs to very large discipline

of non-life insurance called risk theory or ruin theory.

Above we have been working with N as a claim number stochastic variable. This corresponds to

measuring at a fixed point in time the total number of claims considered as a function of time and

measured by a so-called counting process N (t) counting the number of claims over [0, t]. If at any

fixed point in time the claim number is Poisson distributed, the process N (t) is called a Poisson

process. Based on a counting process one can introduce the process of total claims or briefly, the

risk process,

X (t) =∑N(t)

i=1Yi.

If N is a Poisson process then X is called a compound Poisson process. Assume now that the

insurance company over [0, t] receives the premiums πt. Then for a fixed time horizon t∗, the

probability

P (X (t∗) > c + πt∗)

34

corresponds to the probability which we shall study here. The insurance company is, however, also

interested in probabilities in the forms

P (X (t) > c + πt, t < t∗) ,

P (X (t) > x + πt, ∀t) ,

i.e. the so-called finite time ruin probability that the insurance company will not be ruined until

time t∗ and the infinite time ruin probability that the insurance company will never be ruined.

6.4 Exercises

Exercise 15 a) For N belong to the (a, b)-class, show that

E[

Nk]

= aE[

(N + 1)k]

+ bE[

(N + 1)k−1]

.

b) Find simple expressions for E [N ] and V ar [N ] in terms of a and b, and show that

V ar [N ]

E [N ]=

1

1− a .

Solution:

E[

Nk]

=∑∞

n=0nkp (n)

=∑∞

n=0nk(

a +b

n

)

p (n− 1)

=∑∞

n=1

(

ank + bnk−1)

p (n− 1)

= a∑∞

n=0(n + 1)

kp (n) + b

∑∞

n=0(n + 1)

k−1p (n)

= aE[

(N + 1)k]

+ bE[

(N + 1)k−1]

.

EN = aE [N + 1] + bE [1] = aE [N ] + a + b⇒

E [N ] =a + b

1− a ,

V ar [N ] = EN2 −E2N = aE[

(N + 1)2]

+ bE [N + 1]−E2N

= aEN2 + a + 2aEN + bEN + b−E2N= aV arN + aE2N + a + 2aEN + bEN + b−E2N= aV arN +

a + b

1− a ⇒

V ar [N ] =a + b

(1− a)2,

V ar [N ]

E [N ]=

1

1− a .

Exercise 16 The special case of the negative binomial distribution where α = 1, i.e. p (n) =

qn (1− q) is called the geometric distribution. Assume that N is geometrically distributed. Showthat the distribution function G of X, G (x) = P (X ≤ x), satisfies the recursion formula

G (x) = (1− q) + q∑x

y=1f (y)G (x− y) .

35

Solution

G (x) =∑x

z=0g (z)

= g (0) +∑x

z=1g (z)

= (1− q) +∑x

z=1

∑z

y=1qg (z − y) f (y)

= (1− q) +∑x

y=1

∑x

z=yqg (z − y) f (y)

(v = z − y) = (1− q) +∑x

y=1

∑x−y

v=0qg (v) f (y)

= (1− q) + q∑x

y=1f (y)

∑x−y

v=0g (v)

= (1− q) + q∑x

y=1f (y) G (x− y) .

Exercise 17 Make sure that you follow all the steps in the following derivation of the Panjer

recursion for the moments of the total claim distribution,

E[

Xk]

=∑∞

x=0xkg (x)

=∑∞

x=1xkg (x)

=∑∞

x=1xk−1xg (x)

=∑∞

x=1xk−1

∑x

y=1(ax + by) g (x− y) f (y)

=∑∞

x=1

∑x

y=1

(

axk + byxk−1)

g (x− y) f (y)

=∑∞

y=1

∑∞

x=y

(

axk + byxk−1)

g (x− y) f (y)

=∑∞

y=1

∑∞

z=0

(

a (y + z)k

+ by (y + z)k−1)

g (z) f (y)

= a∑∞

y=1

∑∞

z=0

∑k

l=0

(

k

l

)

zlyk−lg (z) f (y)

+b∑∞

y=1

∑∞

z=0y∑k−1

l=0

(

k − 1l

)

zlyk−1−lg (z) f (y)

= a∑∞

y=1

∑∞

z=0

∑k−1

l=0

(

k

l

)

zlyk−lg (z) f (y) + a∑∞

y=1

∑∞

z=0zkg (z) f (y)

+b∑∞

y=1

∑∞

z=0

∑k−1

l=0

(

k − 1l

)

zlyk−lg (z) f (y)

= a∑k−1

l=0

(

k

l

)

∑∞

y=1yk−lf (y)

∑∞

z=0zlg (z) + a

∑∞

z=0zkg (z)

∑∞

y=1f (y)

+b∑k−1

l=0

(

k − 1l

)

∑∞

y=1yk−lf (y)

∑∞

z=0zlg (z)

= a∑k−1

l=0

(

k

l

)

E[

Y k−l]

E[

X l]

+ aE[

Xk]

+b∑k−1

l=0

(

k − 1l

)

E[

Y k−l]

E[

X l]

36

= aE[

Xk]

+∑k−1

l=0

(

a

(

k

l

)

+ b

(

k − 1l

))

E[

Y k−l]

E[

X l]

and conclude that

E[

Xk]

=1

1− a∑k−1

l=0

(

a

(

k

l

)

+ b

(

k − 1l

))

E[

Y k−l]

E[

X l]

.

Exercise 18 Consider the Normal Power approximation in case of a Poisson λ claim distribution.

Show that

E[

Xk]

= λ∑k−1

l=0

(

k − 1l

)

E[

Y k−l]

E[

X l]

.

The first two moments are known from a previous exercise. Verify these results and show that

E[

X3]

= λE[

Y 3]

+ 3λ2E[

Y 2]

E [Y ] + λ3E3 [Y ] .

Show that

E[

(X −EX)3]

= λE[

Y 3]

,

and conclude the following NP approximation formula

u1−ε = λE [Y ] + z1−ε√

λE [Y 2] +1

6

(

z21−ε − 1) E

[

Y 3]

E [Y 2].

Solution: Put a = 0 and b = λ in the recursion formula. Then

E [X ] = λE [Y ] ,

E[

X2]

= λ(

E[

Y 2]

+ E [Y ] E [X ])

= λE[

Y 2]

+ λ2E2 [Y ] ,

E[

X3]

= λ(

E[

Y 3]

+ 2E[

Y 2]

E [X ] + E [Y ] E[

X2])

= λE[

Y 3]

+ 3λ2E[

Y 2]

E [Y ] + λ3E3 [Y ]

But then

E[

(X −EX)3]

= E[

X3 − 3E [X ]X2 + 3XE2 [X ]−E3 [X ]]

= EX3 − 3E [X ]EX2 + 2E3 [X ]= λE

[

Y 3]

+ 3λ2E[

Y 2]

E [Y ] + λ3E3 [Y ]

−3λ2E [Y ]E[

Y 2]

− 3λ3E3 [Y ] + 2λ3E3 [Y ]= λE

[

Y 3]

,

such that

u1−ε = λE [Y ] + z1−ε√

λE [Y 2] +1

6

(

z21−ε − 1) E

[

Y 3]

E [Y 2].

Exercise 19 Verify, by means of,

E [N ] =a + b

1− a ,

V ar [N ] =a + b

(1− a)2,

37

and the recursion formula for moments, that the simple general relations

E [X ] = E [Y ] E [N ] ,

V ar [X ] = E [N ]V ar [Y ] + E2 [Y ] V ar [N ] ,

also, of course, holds when N belongs to the (a, b)-class.

Solution:

E[

X1]

=1

1− a∑0

l=0

(

a

(

1

0

)

+ b

(

0

0

))

E[

Y 1−0]

E[

X0]

=a + b

1− aE [Y ]

= E [N ]E [Y ] ,

E[

X2]

−E2 [X ] = 11− a

∑1

l=0

(

a

(

2

l

)

+ b

(

2− 1l

))

E[

Y 2−l]

E[

X l]

− (a + b)2

(1− a)2E2 [Y ]

=1

1− a (a + b) E[

Y 2]

+1

1− a (2a + b) E [Y ] E [X ]−(a + b)2

(1− a)2E2 [Y ]

=1

1− a (a + b) E[

Y 2]

+1

1− a (2a + b) E [Y ]a + b

1− aE [Y ]−(a + b)

2

(1− a)2E2 [Y ]

=a + b

1− aE[

Y 2]

+(a + b) (a + a + b)

(1− a)2E2 [Y ]− (a + b)

2

(1− a)2E2 [Y ]

=a + b

1− aE[

Y 2]

+(a + b) a

(1− a)2E2 [Y ]

=a + b

1− aV ar [Y ] +a + b

1− aE2 [Y ] +

(a + b)a

(1− a)2E2 [Y ]

=a + b

1− aV ar [Y ] +(a + b) (1− a)

(1− a)2E2 [Y ] +

(a + b)a

(1− a)2E2 [Y ]

=a + b

1− aV ar [Y ] +a + b

(1− a)2E2 [Y ]

= E [N ]V ar [Y ] + E2 [Y ] V ar [N ]

7 Finance and Reinsurance: Products and Valuation

Risk free investment - corresponds to chapter 1

S0 (0) = 1,

S0 (1) = (1 + r) S0 (0) = 1 + r

This is new

S1 (0) = s,

S1 (1) = s (1 + Z) ,

Z =

{

u with probability pu,

d with probability pd.

38

but it is not new to have such a two point distribution. Recall e.g.

1(Tx>1)

which also has a two point distribution.

claim, benefit, amount that you receive

Φ(

S1 (1))

depends on the outcome of S1 (1). Compare with the benefit 1(Tx>1) which also depends on (is

even equal to) the outcome of 1(Tx>1).

Example: r = 0.1, s = 1, u = 0.4, d = 0.9, pu = 60%, pd = 40%.

Consider the claim

max(

S1 (1) , 1)

What should be trhe price for this claim: Caluclate the expected present value

π = E [PV (0)] + risk loading

=1

1 + rE[

max(

S1 (1) , 1)]

+ risk loading

=1

1.1(0.6 ∗ 1, 4 + 0.4 ∗ 1) + risk loading

= 1.127273 + risk loading

Consider the situation where I can invest the premium paid in S1. Then we can put up two

equations with two unknowns

1.4x + (π − x) 1.1 = 1.40.9x + (π − x) 1.1 = 1

0.5x = 0.4⇒ x = 0.80.3 ∗ 0.8 + 1.1π = 1.4⇒ π = 1.4− 0.8 ∗ 0.3

1.1= 1.054545

π = E∗ [PV (0)]

=1

1 + rE∗[

max(

S1 (1) , 1)]

=1

1.1

(

1.1− 0.91.4− 0.9 ∗ 1, 4 +

1.4− 1.11.4− 0.9 ∗ 1

)

=1

1.1(0.6 ∗ 1, 4 + 0.4 ∗ 1)

In general it seems very restrictive two work with only two outcomes, but if we have more

outcomes we can compensate by more trading dates. The continuous (lognormally distributed)

outcome combined with continuous trading was solved in 1973 and was in 1997 awarded the Nobel

prize in economics.

Optioner:

Φ = max(

S1 (1)−K, 0)

⇒ determine π for given K

39

Futures:

Φ = S1 (1)−K : determine k such that π = 0.The insurance version:

S0 (0) = 1,

S0 (1) = (1 + r) S0 (0) = 1 + r

This is new

S1 (0) = πr,

S1 (1) = 1(Tx≤1).

Claim

Φ = 1(Tx>1).

Example r = 0.1, px = 0.9, qx = 0.1,πr = 0.1. Now put up two equations with two unknowns

1 ∗ x + (π − 0.1x) 1.1 = 00 ∗ x + (π − 0.1x) 1.1 = 1

x = −1−1 (1− 0.1 ∗ 1.1) + π1.1 = 0⇒ π = 1− 0.1 ∗ 1.1

1.1= 0.8090909

π =1

1 + rE∗[

1(Tx>1)]

=1

1.1

(

0.1 ∗ 1.1− 01− 0 ∗ 0 +

1− 0.1 ∗ 1.11− 0 ∗ 1

)

=1− 0.1 ∗ 1.1

1.1= 0.8090909

and not

π =1

1.1px =

0.9

1.1= 0.8181818

7.1 Exercises

7.2 Exercises

Exercise 20 Assume that N is Poisson distributed with parameter λ. Show that

E [N ] = λ,

E[

N2]

= λ2 + λ,

V ar [N ] = λ,

and use this to show that

E [X ] = λE [Y ] ,

E[

X2]

= λE[

Y 2]

+ λ2E2 [Y ] ,

V ar [X ] = λE[

Y 2]

.

40

Solution 21

E [N ] =∑∞

n=0ne−λ

λn

n!=∑∞

n=1ne−λ

λn

n!

= λ∑∞

n=1e−λ

λn−1

(n− 1)! = λ∑∞

n=0e−λ

λn

n!

= λ,

E[

N2]

=∑∞

n=0n2e−λ

λn

n!

=∑∞

n=1(n− 1) e−λ λ

n

(n− 1)! +∑∞

n=1e−λ

λn

(n− 1)!

= λ∑∞

n=0ne−λ

λn

n!+ λ

∑∞

n=0e−λ

λn

n!

= λ2 + λ,

V ar [N ] = E[

N2]

−E2 [N ] = λ,E [X ] = E [Y ] E [N ] = λE [Y ] ,

E[

X2]

= E [N ]V ar [Y ] + E2 [Y ] E[

N2]

= λE [Y ]2

+ E2 [Y ] λ2,

V ar [X ] = E[

X2]

−E2 [X ] = E2 [Y ] λ2 + E[

Y 2]

λ− λ2E2 [Y ] = E[

Y 2]

λ

Exercise 22 Assume that N is negative binomial distributed with parameters (α, q). Show that

E [N ] = qα + qE [N ]

E[

N2]

= q2E[

N2]

+ q2 (2α + 1) E [N ] + q2(

α2 + α)

+ E [N ]

and calculate that

E [N ] =qα

1− q ,

E[

N2]

=1 + q2 (2α + 1)

1− q2 E [N ] +q2

1− q2(

α2 + α)

=qα

1− q2(

1 + q2 (2α + 1)

1− q + q (α + 1))

=qα

1− q2(

1 + q2α + qα + q

1− q

)

,

V ar [N ] =qα

1− q2

(

1 + q2 (2α + 1)

1− q + q (α + 1)−(

1− q2)

q

(1− q)2α

)

=qα

(1− q)2.

Solution 23

E [N ] =∑∞

n=0n

(α + n− 1)!n! (α− 1)! q

n (1− q)α

=∑∞

n=1n

(α + n− 1)!n! (α− 1)! q

n (1− q)α

=∑∞

n=1

(α + n− 1)!(n− 1)! (α− 1)!q

n (1− q)α

= q∑∞

n=0

(α + n)!

n! (α− 1)!qn (1− q)α

41

= q∑∞

n=0

(α + n) (α + n− 1)!n! (α− 1)! q

n (1− q)α

= qα∑∞

n=0

(α + n− 1)!n! (α− 1)! q

n (1− q)α + q∑∞

n=0n

(α + n− 1)!n! (α− 1)! q

n (1− q)α

= qα + qE [N ] ,

E[

N2]

=∑∞

n=0n2

(α + n− 1)!n! (α− 1)! q

n (1− q)α

=∑∞

n=1n

(α + n− 1)!(n− 1)! (α− 1)!q

n (1− q)α

= q∑∞

n=0(n + 1)

(α + n)!

n! (α− 1)!qn (1− q)α

= q∑∞

n=0n

(α + n)!

n! (α− 1)!qn (1− q)α + q

∑∞

n=0

(α + n)!

n! (α− 1)!qn (1− q)α

= q∑∞

n=1n

(α + n)!

n! (α− 1)!qn (1− q)α + E [N ]

= q2∑∞

n=0

(α + n + 1)!

n! (α− 1)! qn (1− q)α + E [N ]

= q2∑∞

n=0

(α + n + 1) (α + n) (α + n− 1)!n! (α− 1)! q

n (1− q)α + E [N ]

= q2∑∞

n=0

(

n2 + (2α + 1) n + α2 + α) (α + n− 1)!

n! (α− 1)! qn (1− q)α + E [N ]

= q2E[

N2]

+ q2 (2α + 1)E [N ] + q2(

α2 + α)

+ E [N ] .

Exercise 24 We introduce the mean, variance, coefficient of variation, and skewness,

µ = E [Y ] ,

σ2 = E2 [Y −E [Y ]]= E

[

Y 2]

−E2 [Y ] ,κ =

σ

µ,

ν =E3 [Y − E [Y ]]

σ3

=E[

Y 3]

σ3− 3

κ− 1

κ3.

Argue why µ and σ are not independent of monetary unit, while κ and ν are.

Consider the Pareto claim size distribution. Show that

µ =αλ

λ− 1 ,

σ2 =αλ2

(α− 2) (α− 1)2,

κ =1

√

α (α− 2).

Solution 25

E[

Y k]

=α

α− kλk,

42

µ = E [Y ] =αλ

α− 1 ,

σ2 = E[

Y 2]

−E2 [Y ] = αα− 2λ

2 − α2λ2

(α− 1)2

=αλ2

(

(α− 1)2 − (α− 2) α)

(α− 2) (α− 1)2

=αλ2

(α− 2) (α− 1)2,

κ =σ

µ=

√

αλ2

(α−2)(α−1)2

αλα−1

=

λα−1

√

αα−2

αλα−1

=

√

1

α2

√

α

α− 2

=1

√

α (α− 2).

E [X ] =qβ

1− qαλ

λ− 1 .

V ar [X ] =qβαλ2

1− q

(

1

(α− 2) (α− 1)2+

α

(1− q) (λ− 1)2

)

43