Seven Introductory Lectures
on Actuarial Mathematics
Mogens Steffensen
31. marts 2005
Indhold
1 Compound Interest
1.1 Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Particular Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Life
2.1
2.2
2.3
Insurance Risk: Mortality
The lifetime stochastic variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Expected values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Non-life Insurance Risk: Frequency
3.1 Compound risk models . . . . . . .
3.2 Frequency risk . . . . . . . . . . .
3.3 Severity risk . . . . . . . . . . . . .
3.4 Exercises . . . . . . . . . . . . . .
13
13
14
16
17
4 Valuation principles
4.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5 Life
5.1
5.2
5.3
5.4
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23
24
26
27
Distribution
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32
34
34
and
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Valuation
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Severity
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9
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11
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Insurance: Products
Single payments . . .
Series of payments . .
Premium calculation .
Exercises . . . . . . .
and
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3
3
5
7
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6 Non-life Insurance: The Aggregate Claim
6.1 The Panjer Recursion . . . . . . . . . . .
6.2 Normal Power Approximation . . . . . . .
6.3 Ruin Theory . . . . . . . . . . . . . . . .
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6.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
7 Finance and Reinsurance: Products and Valuation
7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
40
40
2
1
1.1
Compound Interest
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 ) , . . . ,
Yt
(1 + is ) = b (1 + i1 ) (1 + i2 ) · · · (1 + it ) , . . . ,
U (t) = b
s=1
Yn
(1 + it ) = b (1 + i1 ) (1 + i2 ) · · · (1 + in ) .
U (n) = b
t=1
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,
1
b, . . . ,
1 + in
Yn
1
1
1
1
V (t) = b
=
···
b, . . . ,
s=t+1 1 + is
1 + in 1 + in−1
1 + it+1
Yn
1
1
1
1
=
···
b.
V (0) = b
t=1 1 + it
1 + in 1 + in−1
1 + i1
V (n − 1) =
If in particular it = i,
U (1) = b (1 + i) , . . . ,
U (t)
= b (1 + i)t , . . . ,
n
U (n) = b (1 + i) ,
1
b, . . . ,
V (n − 1) =
1+i
n−t
1
V (t) =
b, . . . ,
1+i
n
1
b.
V (0) =
1+i
If the interest is not accumulated with the same frequency as it is announced, one must distinguish 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
m
i(m)
.
1+
m
Equating
1+i=
i(m)
1+
m
3
m
,
we get that a nominal interest i(m) corresponds to the effective interest per period
m
i(m)
i= 1+
− 1,
m
and the effective interest per period i corresponds to the nominal interest
1/m
i(m) = m (1 + i)
−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
lim i(m) = lim m (1 + i)1/m − 1
m→∞
m→∞
=
(1 + i)
lim
m→∞
1/m
0
− (1 + i)
1/m
(1 + i)0+h − (1 + i)0
h→0
h
d
t
(1 + i)
=
dt
t=0
d t log(1+i)
=
e
dt
t=0
=
lim
=
log (1 + i) et log(1+i)
t=0
= log (1 + i) .
We now introduce the following three quantities
δ
= log (1 + i) ,
1
v =
1+i
i
d = 1−v =
1+i
• The accumulation factor from time s (< t) to time t measures at time t the value of 1 unit
at time s,
t−s
(1 + i)
= 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 time
s,
s−t
1
= v s−t = e−δ(s−t) .
1+i
• The yearly discount factor measures at time t the value of 1 unit at time t + 1,
1
= v = e−δ .
1+i
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
Xn
Xn
Xn
t−s
P V (t) =
b (s) (1 + i)
=
b (s) v −(t−s) =
b (s) eδ(t−s) .
s=0
s=0
s=0
In particular,
Xn
P V (0) =
−t
t=0
Xn
P V (n) =
t=0
b (t) (1 + i)
=
n−t
b (t) (1 + i)
Xn
=
t=0
Xn
b (t) v t =
t=0
Xn
b (t) e−δt ,
Xn
=
b (t) eδ(n−t) .
t=0
b (t) v −(n−t)
t=0
2
1
,b m
,...,b
If we replace the payment b by a discrete payment stream b = b (0) , b m
we get the present value
Xmn
Xmn
Xmn
t−s/m
P V (t) =
b (s) (1 + i)
=
b (s) v −(t−s/m) =
b (s) eδ(t−s/m) .
s=0
s=0
nm
m
s=0
In particular,
P V (0) =
P V (n) =
Xmn
t=0
Xmn
t=0
b (t) (1 + i)
−t/m
=
Xmn
b (t) (1 + i)(nm−t)/m
Xmn
b (t) v t/m =
b (t) e−δt/m ,
t=0
t=0
Xmn
Xmn
=
b (t) v −(nm−t)/m =
b (t) eδ(nm−t)/m .
t=0
t=0
If we replace the payment b by a continuous payment stream b = {b (t)}0≤t≤n such that over
the time interval [t, t + dt) the amount b (t) dt is deposited, we get the present value
Z n
Z n
Z n
t−s
b (s) eδ(t−s) ds.
b (s) v −(t−s) ds =
b (s) (1 + i)
ds =
P V (t) =
0
0
0
In particular,
n
−t
b (t) (1 + i)
0
0
1.2
Z
n
t
Z
n
b (t) e−δt dt,
Z n
Z n
Z n
n−t
b (t) eδ(n−t) ds.
b (t) v −(n−t) ds =
b (t) (1 + i)
dt =
=
P V (0) =
P V (n)
Z
dt =
b (t) v dt =
0
0
0
0
Particular Present Values
1. Annuity in arrear (Efterbetalt annuitet): b = (b (0) , . . . , b (n)) : b (0) = 0, b (1) = . . . =
b (n) = 1.
an|
sn|
= P V (0)
Xn−1
Xn
1 − vn
1 − vn
vt = v
vt = v
=
=
,
t=0
t=1
1−v
i
= P V (n)
n
Xn
Xn
(1 + i) − 1
1 − vn
=
=
.
v −(n−t) = v −n
v t = v −n
t=1
t=1
i
i
5
2. Annuity-due (Forudbetalt annuitet): b (0) = . . . = b (n − 1) = 1, b (n) = 0.
= P V (0)
Xn−1
1 − vn
1 − vn
vt =
(a) =
=
t=0
1−v
d
Xn
Xn
t−1
−1
(b) =
v
=v
v t = (1 + i) an| ,
än|
t=1
t=1
¨sn|
= P V (n)
n
Xn−1
Xn−1
1 − vn
(1 + i) − 1
v t = v −n
v −(n−t) = v −n
=
(a) =
t=0
t=0
1−v
d
Xn
Xn
−(n−(t−1))
−1
−(n−t)
(b) =
v
=v
v
= (1 + i) sn| .
t=1
t=1
3. Fractionated annuity in arrear (Efterbetalt fraktioneret annuitet): b = b
1
1
b (0) = 0, b m
= . . . = b nm
=m
.
m
(m)
an|
1
m
,b
2
m
,...,b
nm
m
= P V (0)
1
1 Xmn 1 t
1/m
−1← i .
v m = amn| (1 + i)
=
t=0
m
m
4. Continuous annuity (Kontinuert annuitet): b = {b (t)}0≤t≤n : b (t) = 1.
an|
sn|
= P V (0)
n
Z n
1 − vn
1 −δt
−δt
=
,
=
e dt = − e
δ
δ
0
0
Z n
Z n
n
(1 + i) − 1
e−δt dt =
eδ(n−t) ds = eδn
=
.
δ
0
0
Note that
(m)
lim an|
m→∞
=
=
1
1/m
amn| (1 + i)
−1←i
m
1 − vn
lim
1/m
m→∞
m (1 + i)
−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.
k an|
= P V (0)
Xk+n
Xn
(a) =
vt = vk
v t = v k an|
t=k+1
t=1
Xk+n
Xk
(b) =
vt −
v t = ak+n| − ak| .
t=1
t=1
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.
k än|
= P V (0)
6
:
(a) =
(b) =
Xn−1
v t = v k än|
vt = vk
t=0
t=k
Xk−1
Xk+n−1
v t = äk+n| − äk+n| .
vt −
Xk+n−1
t=0
t=0
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.
= P V (0)
Z k+n
Z
(a) =
e−δt dt = e−δk
k an|
k
(b) =
1.3
Z
k+n
e−δt dt −
0
Z
n
0
e−δt dt = e−δk an|
k
0
e−δt dt = ak+n| − ak| .
Exercises
Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Exercises 2.4.52.4.8.
Exercise 1 Consider first the accumulation factor B (t) = v −t and the discount factor B −1 (t) =
v t . Show the following difference equations and side conditions
B (t)
= B (t − 1) (1 + i) ,
B (0) = 1,
1
B −1 (t − 1) ,
1+i
B −1 (0) = 1.
B −1 (t)
=
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
Xt
U (t) = B (t)
B −1 (j) (c (j) − b (j)) ,
j=0
Xn
V (t) = B (t)
B −1 (j) (b (j) − c (j)) .
j=t+1
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 condition
U (n) = 0.
Show that this condition implies the relations
V (0) = c (0) − b (0) ,
U (t) = V (t) ,
7
(1)
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
B (t) = δB (t) ,
dt
B (0) = 1,
d −1
B (t) = −δB −1 (t) ,
dt
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
V (t)
δt
= e
t
Z
Z0 n
t
e−δs (c (s) − b (s)) ds,
e−δs (b (s) − c (s)) ds.
Show the ordinary differential equations with side conditions
d
U (t)
dt
d
V (t)
dt
= δU (t) + c (t) − b (t) , U (0) = 0,
= 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
2.1
Life Insurance Risk: Mortality
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.
t qx
=
=
t px
P ({T ≤ x + t} ∩ {T > x})
P (T > x)
F (x + t) − F (x)
P (x < T ≤ x + t)
=
P (T > x)
1 − F (x)
= P ( T ≤ x + t| T > x) =
F (x) − F (x + t)
,
F (x)
= P ( T > x + t| T > x) =
=
P ({T > x + t} ∩ {T > x})
P (T > x)
P (T > x + t)
F (x + t)
=
.
P (T > x)
F (x)
Note that
t q0
= F (t) =⇒ t qx =
t p0
= F (t) =⇒ t px =
F (x + t) − F (x)
=
1 − F (x)
F (x + t)
=
F (x)
x+t p0
x p0
x+t q0
− x q0
,
1 − x q0
.
1
1
P ( T < t + ∆t| T > t)
t+∆t qt = lim
∆t→0 ∆t
∆t→0 ∆t
1
F (t + ∆t) − F (t)
1 F (t + ∆t) − F (t)
=
lim
= lim
∆t→0 ∆t
∆t
F (t)
F (t) ∆t→0
1 d
1 d
F (t) = −
=
F (t)
F (t) dt
F (t) dt
d
= − log F (t) .
dt
µ (t) =
lim
This is a differential equation for log F (t) with the side condition log F (0) = 0, such that the
Rt
solution becomes log F (t) = − 0 µ (s) ds or equivalently
F (t) = e−
9
Rt
0
µ(s)ds
.
Note that
R x+t
R x+t
Rt
e− 0 µ(s)ds
= − R x µ(s)ds = e− x µ(s)ds = e− 0 µ(x+s)ds .
t px =
e 0
x p0
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
x+t p0
l0
= 100.000
lx
= l0 F (x) = l0 x p0
dx
= lx − lx+1
F (x + t)
lx+t
=
lx
F (x)
t px
=
t qx
= 1 − t px =
F (x) − F (x + t)
lx − lx+t
=
lx
F (x)
Example: Gompertz-Makeham G82M,
µ = α + βeγt
= 5 · 10−4 + 7.5858 · 10−5 et 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,
Z ω
d
F (t) dt
E [G (T )] =
G (t)
dt
0
Z ω
d
F (t) dt
= −
G (t)
dt
0
Z ω
d
G (t) dt
= −G (ω) F (ω) + G (0) F (0) +
F (t)
dt
0
Z ω
d
= G (0) +
F (t)
G (t) dt
dt
0
Taking G (t) = tk we get
E Tk = k
and, in particular
e0 = E [T ] =
Z
Z
ω
tk−1 F (t) dt,
0
ω
F (t) dt =
0
Z
ω
e−
Rt
0
µ(s)ds
dt
0
One can also show that the expected remaining life time of an x years old individual is
(a) =
ex
Z ω−x
0
(b) =
Z
ω−x
0
Z
1 ω−x
Tx
lx+t
dt =
lx+t dt ≡
l
l
lx
x
x
0
0
Z ω−x
Rt
R x+t
e− 0 µ(x+s)ds dt
e− x µ(s)ds dt =
t px dt =
Z
ω−x
0
10
Definitions
Lx
Tx
=
Z
1
lx+t dt = l0
0
Z
1
F (x + t) dt
0
= the expected group life time a year from x + t,
Z ω−x
Z ω−x
Xω−x
=
lx+t dt =
F (x + t) dt
Lx+t = l0
t=0
0
0
= 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 3.6.23.6.6.
Exercise 3
• The exponential mortality law is characterized by a constant force of mortality,
µ (t) = λ. Show that
F (t) = e−λt ,
d
F (t) = λe−λt .
f (t) =
dt
Introduce the conditional survival function
t px
=
F (x + t)
.
F (x)
Show that
t px
= t p0 ,
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−β tβ−1 , α, β > 0.
11
Show that
β
F (t) = e−( α ) .
t
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 t px .
• The Gompertz-Makeham mortality law is characterized by the mortality intensity
µ = α + βeγt , α, β > 0.
Show that
F (t) = e(−αt−β(e
γt
−1)/γ )
.
Show that if γ > 0, then µ is an increasing function of t. Calculate t px .
12
3
3.1
Non-life Insurance Risk: Frequency and Severity
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 assumptions 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=
XN
i=1
Yi = Y 1 + Y 2 + . . . + Y N .
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
X∞
E [f (X)] =
E [ f (X)| N = n] p (n)
n=0
X∞
=
E [f (Y1 + Y2 + . . . + Yn )] p (n) .
n=0
In order to go on from here, we need to say more about p, F, and f , i.e. the claim number distribution, 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
X∞
X∞
E [X] =
E [ X| N = n] p (n) =
E [Y1 + Y2 + . . . + Yn ] p (n)
n=0
n=0
X∞
X∞ Xn
nE [Y1 ] p (n) = E [Y1 ] E [N ] .
E [Yi ] p (n) =
=
n=0
n=0
i=1
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
X
Xn
=
i=1
i
13
X
X
this gives that
E X2
=
=
=
=
=
=
=
=
i,j
i6=j
=
=
X X
i
j
X X
i
j:j6=i
,
h
i
X∞
2
E (Y1 + Y2 + . . . + Yn ) p (n)
E X 2 N = n p (n) =
n=0
n=0
i
hX
X∞ X
X∞
Yi Yj p (n) =
E [Yi Yj ] p (n)
E
i,j
n=0
i,j
n=0
X∞ X
X∞ X X
E [Yi Yj ] p (n) +
E [Yi Yj ] p (n)
n=0
i6=j
n=0
i
j:j=i
X∞ X
X∞ X E [Yi ] E [Yj ] p (n) +
E Yi2 p (n)
n=0
i6=j
n=0
i
X∞
X∞
nE Y12 p (n)
n (n − 1) E 2 [Y1 ] p (n) +
n=0
n=0
X∞
X∞
np (n)
E 2 [Y1 ]
n2 − n p (n) + E Y12
n=0
n=0
E 2 [Y1 ] E N 2 − E [N ] + E Y12 E [N ]
E 2 [Y1 ] E N 2 + E [N ] V ar [Y1 ] .
X∞
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 X 2 − E 2 [X]
= E 2 [Y1 ] E N 2 + E [N ] V ar [Y1 ] − E 2 [Y1 ] E 2 [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
b
p (n) = a +
p (n − 1) .
n
• The Poisson distribution is characterized for λ > 0 by
p (n) = e−λ
λn
, n = 0, 1, . . . .
n!
That the Poisson distribution belongs to the (a, b)-class is seen by
p (n) = e−λ
λ
λ
λn
λn−1
= e−λ
= p (n − 1) ,
n!
n
(n − 1)!
n
14
such that
a = 0,
b = λ, λ > 0.
• The negative binomial distribution is characterized for 0 < q < 1 and α > 0 by
!
α+n−1
α
p (n) =
q n (1 − q) , n = 0, 1, . . . .
n
That the negative binomial distribution belongs to the (a, b)-class is seen by
!
α+n−1
α
p (n) =
q n (1 − q)
n
Γ (α + n) n
α
q (1 − q)
n!Γ (α)
q (α + n − 1) Γ (α + n − 1) n−1
q
(1 − q)α
=
n
(n − 1)!Γ (α)
q (α − 1) Γ (α + n − 1) n−1
α
=
q+
q
(1 − q) ,
n
(n − 1)!Γ (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
!
m
m−n
p (n) =
pn (1 − p)
, n = 0, 1, . . . , m.
n
That the binomial distribution belongs to the (a, b)-class is seen by
!
m
m−n
p (n) =
pn (1 − p)
n
=
=
p (m − n + 1)
m!
m−(n−1)
pn−1 (1 − p)
n (1 − p) (n − 1)! (m − n + 1)!
p
(m + 1) p
−
p (n − 1) ,
+
1−p
n (1 − p)
such that
p
,0 < p < 1
1−p
p
b = (m + 1)
= − (m + 1) a.
1−p
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 characterizing 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) =
δ γ γ−1 −δy
y
e , y ≥ 0.
Γ (γ)
For the Gamma distribution we exploit that
Z ∞ γ
δ
y γ−1 e−δy dy = 1,
Γ
(γ)
0
and we know this since this is how the Gamma function is actually defined. We can then
calculate
Z ∞
δ γ γ−1 −δy
y k e−ay
y
e dy
E Y k e−aY
=
Γ (γ)
0
Z ∞
δγ
=
y k+γ−1 e−(δ+a)y dy
Γ (γ) 0
Γ (γ + k)
δγ
.
=
Γ (γ) (δ + a)γ+k
• The Pareto (λ, α) distribution is characterized for α > 0 by the distribution function
α
λ
F (y) = 1 −
, y > λ,
y
which gives the density
f (y) =
α
λ
α+1
λ
, y > λ.
y
16
For the Pareto we exploit that
Z
∞
λ
α
λ
α+1
Z
λ
α
dy = λα+1
y
λ
distribution we can for k < α calculate
Z
E Yk =
Z
=
∞
y −α−1 dy =
λ
∞
yk
λ
∞
λ
∞
α
λ
∞
α α+1
1
=1
λ
− y −α
λ
α
λ
α+1
λ
dy
y
α λα+1
dy
λ y α−k+1
α − k λα−k+1
α
λk
dy
λ y α−k+1
λ α−k
Z ∞
α
α − k λα−k+1
λk
dy
α−k
λ y α−k+1
λ
α
λk .
α−k
Z
=
=
=
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 Pareto (λ, α) distributed, then the conditional distribution of Y given Y > y is again Pareto
distributed with parameters (y, α). Thus,
m (y) =
αy
y
−y =
, y > λ.
α−1
α−1
• The lognormal α, σ 2 distribution is characterized by the density
f (y) =
1 1 − 12 ( log σy−α )2
√
e
, y > 0.
σ 2π y
The moments of the lognormal distribution can be calculated to
σ2 2
E Y k = eαk+ 2 k .
3.4
Exercises
Exercise 4 Assume that N is Poisson distributed with parameter λ. Show that
E [N ] = λ,
E N 2 = λ2 + λ,
V ar [N ] = λ,
and use this to show that
E [X] = λE [Y ] ,
E X 2 = λE Y 2 + λ2 E 2 [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 N 2 = q 2 E N 2 + q 2 (2α + 1) E [N ] + q 2 α2 + α + E [N ]
and calculate that
E [N ] =
E N2 =
=
=
V ar [N ] =
=
qα
,
1−q
q2
1 + q 2 (2α + 1)
E [N ] +
α2 + α
2
2
1−q
1−q
2
qα
1 + q (2α + 1)
+
q
(α
+
1)
1 − q2
1−q
1 + q 2 α + qα + q
qα
,
1 − q2
1−q
!
1 − q2 q
1 + q 2 (2α + 1)
+ q (α + 1) −
2 α
1−q
(1 − q)
qα
1 − q2
qα
(1 − q)
2.
Exercise 6 We introduce the mean, variance, coefficient of variation, and skewness,
µ = E [Y ] ,
σ2
= E 2 [Y − E [Y ]]
= E Y 2 − E 2 [Y ] ,
σ
,
κ =
µ
E 3 [Y − E [Y ]]
ν =
3
3σ
E Y
3
1
− − 3.
=
3
σ
κ κ
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
=
σ2
=
κ =
αλ
,
λ−1
αλ2
(α − 2) (α − 1)2
1
p
.
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
Xn
Xi > nπ (X) = 1,
n→∞
i=1
Xn
π (X) > E [X] : lim P
Xi > nπ (X) = 0.
π (X) < E [X] : lim P
n→∞
4.
i=1
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 ≥ 0
E [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
1.
p
V ar [X].
π (X + Y ) ≤ π (X) + π (Y )
p
V ar [X + Y ] =
≤
=
=
p
V arX + V arY + 2Cov [X, Y ]
q
p
p
V arX + V arY + 2 V ar [X] V ar [Y ]
r
√
2
√
V arX + V arY
√
√
V arX + V arY
where we use that
V ar [X + Y ]
= E (X + Y )2 − E 2 (X + Y )
= EX 2 + EY 2 + 2E [XY ] − (EX + EY )
2
= EX 2 − E 2 X + EY 2 − E 2 Y + 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 )]
q
q
2
2
≤
E (X − EX) E (Y − EY )
= 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
2
= pm2 − (pm) = m2 p (1 − p)
p
g (p) = π (X) = m p + a p (1 − p)
V arX
!
1 a (1 − 2p)
g (p) = m 1 + p
2 p (1 − p)
1 a (1 − 2)
lim g 0 (p) = m 1 +
= −∞
p→1
2
0
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
nπ
X
n
lim nπ
n→∞
1
1
X+ X
2
2
≤ 2π
1
X
2
a
V ar [X]
n
1
= π (X) − a 1 −
V ar [X]
n
< π (X)
= E [X] +
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)
21
−1
.
Exponential principle
π (X) =
Esscher principle
1
log E eaX
a
E XeaX
π (X) =
E [eaX ]
eaX
= E X
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) =
δ γ γ−1 −δx
x
e
.
Γ (γ)
π (X) = E [X] + aV ar [X]
Γ (γ + 2)
Γ2 (γ + 1)
1
Γ (γ + 1) + a
−a
=
Γ (γ) δ
δ
Γ (γ) δ
1
π (X) =
log E eaX
a
δ
γ
log
=
a
δ−a
aX E Xe
π (X) =
E [eaX ]
Γ (γ + 1)
=
Γ (γ) (δ − a)
Exercise 8 4.4
E Yk =
V ar [X] = pλ2 α
α
λk , k < α
α−k
α
1
−p
2
α−2
(α − 1)
!
π (X) = (1 + a) E [X]
α
λ
= (1 + a) p
α−1
π (X) = E [X] + aV ar [X]
α
= p
λ + apλ2 α
α−k
22
1
α
−p
α−2
(α − 1)2
!
π (X) = E [X] + a
p
V ar [X]
v
u
u
α
λ + atpλ2 α
= p
α−1
α
1
−p
α−2
(α − 1)2
!
π (X) = min {m : P (X > m) ≤ a}
= min {m : pP (Y > m) ≤ a}
α
a
λ
≤
= min m :
m
p
p 1/a
= min m :
λ≤m
a
p 1/a
λ
=
a
Exercise 9 4.5
π (X) = min {(1 + a) E [kX] + (1 + b) E [(1 − k) X]}
k
= min {(1 + a) kE [X] + (1 + b) (E [X] − kE [X])}
k
= (1 + b) E [X] + E [X] (a − b) min {k}
k
(k = 1) = (1 + a) E [X]
o
n
p
p
π (X) = min E [kX] + a V ar [kX] + E [(1 − k)] + b V ar [(1 − k) X]
k
p
= E [X] + V ar [X] b + (a − b) min {k}
k
p
(k = 1) = E [X] + a V ar [X]
π (X) = min {E [kX] + aV ar [kX] + E [(1 − k)] + bV ar [(1 − k) X]}
k
o
n
2
= E [X] + V ar [X] min ak 2 + b (1 − k)
k
o
n
= E [X] + V ar [X] min ak 2 + b (1 − k)2
k
b
ab
b
k=
= E [X] + V ar [X]
b
+
a
+
2
2
a+b
a
(a + b)
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 = v n b = e−δn b
The expected present value at time 0 of the stochastic variable B at time n,
E e−δt B (t) = e−δt E [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(s<Tx <t)
Recall that
Rt
R x+t
µ(s)ds
µ(x+u)du
− e−
= 1 − t px = 1 − e− 0 µ(x+s)ds = 1 − e−
R
d
− 0t µ(x+s)ds
fx (t) =
µ (x + t)
t qx = e
dt
t qx
x
But then
E 1(Tx >t)
E 1(s<Tx <t)
=
Z
∞
1(u>t) fx (u) du =
0
Z
∞
fx (u) du
t
Rt
= P (Tx > t) = t px = e− 0 µ(x+s)ds
= e−µt
Z ∞
Z t
=
1(s<u<t) fx (u) du =
fx (u) du
0
s
= P (s < Tx < t) = s px − t px = e−
Rt
Rs
= e− 0 µ(x+u)du 1 − e− s µ(x+u)du
R x+t
Rs
= e− 0 µ(x+u)du 1 − e− x+s µ(u)du
=
Rs
0
Rt
0
µ(x+u)du
p
qx+s
s x t−s = e−µs 1 − e−(t−s)µ
For discrete distributions we have that
X
E 1(X=x) =
1(y=x) P (X = y) = P (X = x)
y
hX
i
X
X
X xX
E
bx 1(X=x)
=
ax E 1(X=x) =
a
1(y=x) P (X = y) =
ax P (X = x)
x
x
x
y
x
i.e. a similar formula when the payment can take other values than 0 and b.
• Pure Endowment insurance
B (n) = 1(Tx >n)
−δn
E 1(Tx >n) = e−δn n px
n Ex = e
Rn
Rn
= e−δn e− 0 µ(x+s)ds = e− 0 δ+µ(x+s)ds
Rn
= e− 0 δ+µds = e−(δ+µ)n .p149
5.2
Series of payments
The present value at time 0 of the series of payments over 0, . . . , n according to lecture 1,
Xn
Xn
Xn
b (t) e−δt
b (t) v t =
b (t) (1 + i)−t =
P V (0) =
t=0
t=0
t=0
The expected present value at time 0 of the series of stochastic payments over 0, . . . , n
h Xn
i Xn
E [P V (0)] = E
B (t) e−δt =
e−δt E [B (t)]
t=0
t=0
24
• Term insurance
B (t) = 1(t−1<Tx ≤t) ,t ≤ n
Xn
e−δt E 1(t−1<Tx ≤t)
A1xn =
t=1
Xn
=
e−δt t−1 px qx+t−1
t=1
Xn
e−δt e−µ(t−1) 1 − e−µ
( =
t=1
Xn
= 1 − e−µ eµ
e−(δ+µ)t
t=1
Xn
= 1 − e−µ eµ e−(δ+µ)
e−(δ+µ)(t−1)
t=1
Xn−1
= 1 − e−µ e−δ
e−(δ+µ)t
t=0
Xn−1
= 1 − e−µ e−δ
e−(δ+µ)t
t=0
=
1−e
)
1 − e−(δ+µ)n
e−δ
1 − e−(δ+µ)
−µ
for n = ∞
Ax
=
X∞
t=1
e−δt
= 1−e
t−1 px
−µ
e
qx+t−1
1
−δ
1 − e−(δ+µ)
p145
• Endowment insurance
B (t) = 1(t−1<Tx ≤t) , t ≤ n,
+B (n) = 1(Tx >n)
Axn
=
=
(
Xn
t=1
Xn
t=1
e−δt E 1(t−1<Tx ≤t) + e−δn E 1(Tx >n)
e−δt
t−1 px
qx+t−1 + e−δn n px
1 − e−(δ+µ)n
1 − e−µ e−δ
+ e−(δ+µ)n
1 − e−(δ+µ)
−(δ+µ)n
1 − e−(δ+µ)n
−(δ+µ) 1 − e
−
e
+ e−(δ+µ)n
= e−δ
1 − e−(δ+µ)
1 − e−(δ+µ)
−(δ+µ)n
e−(δ+µ)n 1 − e−(δ+µ)
1 − e−(δ+µ)n
−(δ+µ) 1 − e
= e−δ
−
e
+
1 − e−(δ+µ)
1 − e−(δ+µ)
1 − e−(δ+µ)
−(δ+µ)
−(δ+µ)n
−(δ+µ)n
e
−e
1−e
−
= e−δ
1 − e−(δ+µ)
1 − e−(δ+µ)
)p149.
=
• Temporary life annuity(-due)
B (t)
= 1(Tx >t)
25
äxn
Xn−1
=
t=0
Xn−1
=
t=0
=
In particular for n = ∞:
e−δt E 1(Tx >t)
e−δt t px
Xn−1
t=0
X∞
äx =
5.3
e
−δt −µt
t=0
e
e
−δt
Xn−1
=
t=0
t px
=
e
−(δ+µ)t
1 − e−(δ+µ)n
=
1 − e−(δ+µ)
1
1 − e−(δ+µ)
p149
p144.
Premium calculation
Put expected present value of premiums equal to the expected present value of benefits (the
equivalence principle).
• Term insurance
πäxn
A1
= A1xn ⇒ π = xn
äxn


−(δ+µ)n
−µ
(1 − e ) e−δ 1−e
1−e−(δ+µ)
=
= 1 − e−µ e−δ  p145
1−e−(δ+µ)n
1−e−(δ+µ)
note that this is independent of the time horizon, i.e. the same formula for a whole life term
insurance.
V (0−) = A1xn − πäxn = 0
V (t)
= A1x+t n−t − πäx+t n−t
V (n) = 0
∆V (t)
= V (t) − V (t − 1) = ...
• Pure endowment insurance
πäxn
=
n Ex
=
V (0−) =
V (t)
=
n Ex
⇒π=
e
n Ex
äxn
!
−(δ+µ)n
1−e−(δ+µ)n
1−e−(δ+µ)
− πäxn = 0
n−t Ex+t
− πäx+t n−t
V (n) = 0
• Endowment insurance
πäxn
= Axn ⇒ π =
Axn
äxn
−
e−δ 1−e
1−e−(δ+µ)
−(δ+µ)n
( =
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
−(δ+µ)
e
− e−(δ+µ)n
= e−δ −
1 − e−(δ+µ)n
V (0−) = Axn − πäxn = 0
V (t)
= Ax+t n−t − πäx+t n−t
V (n) = 0
∆V (t)
5.4
= V (t) − V (t − 1) = ...
Exercises
Atkinson, M.E. and Dickson, D.C.M. (2000): An Introduction to Actuarial Studies, Exercises 5.6.15.6.6.
Exercise 10 We consider a so-called deferred temporary life annuity(-due) (opsat ophørende forudbetalt livrente) where
B (t) = 1(Tx >t,k≤t<k+n) = 1(Tx >t) 1(k≤t<k+n)
with present value
P V (0) =
Xk+n
t=0
B (t) e−δt .
a) Explain this product in words and argue why one may be interested in such a product in
practice.
b) Calculate E [P V (0)] in general in terms of t px and show that
E [P V (0)] =
k Ex
äx+k
n| .
c) Calculate E [P V (0)] in the special cases where µ is constant, where n = ∞, and the combination where µ is constant and n = ∞.
d) Suggest an actuarial notation for the deferred temporary life annuity and for the deferred
life annuity (n = ∞), inspired by the notation you already know.
e) Assume that the annuity benefit is paid for by a level premium π in k periods (løbende præmie)
with expected present value äxk| . Calculate the premium according to the equivalence principle.
f ) Caluclate the premium in the special cases where µ is constant, where n = ∞, and the
combination where µ is constant and n = ∞.
Solution: a) The product pays 1 unit at the beginning of every year from year k to year k + n
as long as the insured is alive. The product is relevant for pension saving, where the the annuity
typically runs from the age of retirement until a high age or infinity (n = ∞).
b)
Xk+n
e−δt E [B (t)]
e−δt E 1(Tx >t) 1(k≤t<k+n)
=
t=0
Xk+n
e−δt 1(k≤t<k+n) E 1(Tx >t)
=
E [P V (0)] =
t=0
Xk+n
t=0
27
=
=
=
=
Xk+n−1
e−δt E 1(Tx >t)
t=k
Xk+n−1
e−δt k px t−k px+k (her får de nok problemer)
t=k
Xk+n−1
e−δt t px
t=k
k px
Xk+n−1
t=k
Xn−1
=
k px
=
e−δk
=
k Ex
e−δt
t−k px+k
e−δ(t+k) t px+k
Xn−1
e−δt t px+k
k px
t=0
t=0
äx+k
n|
c)
Xk+n−1
t=k
e−δt t px
=
=
Xk+n−1
t=k
Xk+n−1
t=k
= e−(δ+µ)k
= e−(δ+µ)k
= e−(δ+µ)k
X∞
t=k
X∞
t=k
e−δt t px
= ... =
e−δt t px
= e−(δ+µ)k
e−δt e−µt
e−(δ+µ)t
Xk+n−1
t=k
Xn−1
e−(δ+µ)(t−k)
e−(δ+µ)t
t=0
−(δ+µ)n
1−e
, (∗)
1 − e−(δ+µ)
k Ex
äx+k ,
1 − e−(δ+µ)∞
e−(δ+µ)k
=
.
1 − e−(δ+µ)
1 − e−(δ+µ)
(∗) kunne selvfølgelig også ses ved at sætte ind i k Ex äx+k n| med konstant µ.
d) Combine the deferred temporary annuity and the temporary life annuity into
k äxn|
e)
πäxk| =
k äxn|
⇒π=
k äxn|
äxk|
f)
π
=
π
=
π
=
k äxn|
äxk|
k äx
äxk|
k äx
äxk|
e−(δ+µ)k 1−e
1−e−(δ+µ)
−(δ+µ)n
=
1−e−(δ+µ)k
1−e−(δ+µ)
= e−(δ+µ)k
1 − e−(δ+µ)n
,
1 − e−(δ+µ)k
,
=
e−(δ+µ)k
.
1 − e−(δ+µ)k
Exercise 11 Assume a linear mortality intensity, i.e. µ (t) = µt. Show that
e−
Rt
s
µ(x+u)du
= e−µ(x(t−s)+ 2 (t
1
28
2
−s2 ))
.
Show that
n Ex
A1xn|
Axn|
äxn|
Solution:
e−
1
2
= e−(µx+δ)n−µ 2 n ,
Xn
2
1
1
e−δt e−µ(x(t−1)+ 2 (t−1) ) 1 − e−µ(x+t− 2 ) ,
=
t=1
Xn
2
1
1
−δn −µ(xn+ 21 n2 )
= e
e
+
e−δt e−µ(x(t−1)+ 2 (t−1) ) 1 − e−µ(x+t− 2 ) ,
t=1
Xn−1
1 2
=
e−(µx+δ)t−µ 2 t
t=0
Rt
s
µ(x+u)du
= e−
R x+t
x+s
x+t
= e
= e
= e−µ x+s udu
2
2
1
1
= e−µ( 2 (x+t) − 2 (x+s) )
−µ[ 21 u2 ]
x+s
”
“
2
−µ 21 (x2 +t2 +2xt)− 12 (x2 +s2 +2xs)
= e−µ(x(t−s)+ 2 (t
1
n Ex
A1xn|
R x+t
µ(u)du
2
= e−µ( 2 t
1 2
+xt− 21 s2 −xs)
−s2 ))
n
1 2
1 2
= e−δn e− 0 µ(x+s)ds = e−δn e−µ(xn+ 2 n ) = e−(µx+δ)n−µ 2 n
Xn
=
e−δt t−1 px qx+t−1
t=1
Xn
e−δt t−1 px (1 − px+t−1 )
=
t=1
R1
R t−1
Xn
e−δt e− 0 µ(x+u)du 1 − e− 0 µ(x+t−1+u)du
=
t=1
Rt
R t−1
Xn
=
e−δt e− 0 µ(x+u)du 1 − e− t−1 µ(x+u)du
t=1
Xn
2
2
2
1
1
e−δt e−µ(x(t−1)+ 2 (t−1) ) 1 − e−µ(x(t−(t−1))+ 2 (t −(t−1) ))
=
t=1
Xn
2
1
1
=
e−δt e−µ(x(t−1)+ 2 (t−1) ) 1 − e−µ(x+t− 2 )
R
t=1
Axn|
äxn|
=
1
n Ex + Axn|
Xn
2
1
1 2
1
= e−δn e−µ(xn+ 2 n ) +
e−δt e−µ(x(t−1)+ 2 (t−1) ) 1 − e−µ(x+t− 2 )
t=1
Xn−1
−δt −µ(xt+ 21 t2 )
=
e
e
t=0
Xn−1
1 2
=
e−(µx+δ)t−µ 2 t .
t=0
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 T x 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−rn E [B (n)] = e−rn E 1(Tx ≤n) 1(Ty >n)
= e−rn E 1(Tx ≤n) E 1(Ty >n)
= e−rn n qx n py
c)
πäxn| = e−rT
6
n qx n p y
⇒π=
e−rT
n qx n p y
.
äxn|
Non-life Insurance: The Aggregate Claim Distribution
We shall here continue the study of the distribution of the total claim and expected values of
XN
functions of the total claim, i.e. X =
Yi and E [f (X)] .We wish to calculate the probability
i=1
distribution P (X ≤ x), such that we can calculate so-called fixed time ruin probabilities. Assume
that 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 that
the 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)
MX (s)
ΦX (s)
LX (s)
= E sX ,
= E esX ,
= E eisX ,
= E e−sX .
We shall here work with the probability generating function PX (s). We have the following probability generating functions for the claim number and claim size distributions,
X∞
p (n) sn ,
PN (s) = E sN =
n=0
PY (s) = E sY .
With these functions we get
X∞
PX (s) = E sX =
E sX N = n p (n)
" Xn=0
#
n
X∞
Yi
=
E s i=1
p (n)
n=0
X∞
E sY1 sY2 · · · sYn p (n)
=
E sY1 E sY2 · · · E sYn p (n)
n=0
X∞
n
=
E sY
p (n)
=
n=0
X∞
n=0
= PN (PY (s)) .
30
(3)
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)
X∞
P (X = x, N = n)
=
n=0
X∞
=
p (n) P ( X = x| N = n)
n=0
X∞
=
p (n) P (Y1 + . . . + Yn = x)
n=0
X∞
=
p (n) f ∗n (x) ,
n=0
where
f ∗n (x)
= P (Y1 + . . . + Yn = x)
Xx
P (Y1 + . . . + Yn = x, Yn = y)
=
y=1
Xx
f (y) P ( Y1 + . . . + Yn = x| Yn = y)
=
y=1
Xx
=
f (y) P (Y1 + . . . + Yn−1 = x − y)
y=1
Xx
=
f (y) f ∗n−1 (x − y) .
(4)
y=1
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) =
11
1
=
,
44
16
1
11 11
+
= ,
42 24
4
11 11 11
3
f ∗2 (4) = f (1) f ∗1 (3) + f (2) f ∗1 (2) + f (3) f ∗1 (1) =
+
+
= ,
44 22 44
8
..
.
1
1 1
∗3
=
,
f (3) = f (1) f ∗2 (2) =
4 16
64
..
.
f ∗2 (3) = f (1) f ∗1 (2) + f (2) f ∗1 (1) =
g (0) = p (0) = e−1 ,
g (1) = p (1) f ∗1 (1) =
1 −1
e ,
4
g (2) = p (1) f ∗1 (2) + p (2) f ∗2 (2) =
31
1 −1
1 1 −1
e +
e =
2
16 2
1
1
+
2 32
e−1 ,
g (3) = p (1) f ∗1 (3) + p (2) f ∗2 (3) + p (3) f ∗3 (3)
1 1
1 −1 1 1 −1
1 1 −1
1
e−1 .
=
e +
e +
e =
+ +
4
42
64 6
4 8 384
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,
PN0 (s) =
=
=
=
=
X∞
np (n) sn−1
b
p (n − 1) sn−1
n a+
n=1
n
n=0
X∞
X∞
n=1
X∞
n=0
X∞
n=0
(na + b) p (n − 1) sn−1
((n + 1) a + b) p (n) sn
X∞
(a + b) p (n) sn + as
np (n) sn−1
n=0
= (a + b) PN (s) + asPN0 (s) ,
which leads to the differential equation
PN0 (s) =
a+b
PN (s) .
1 − as
This differential equation has the side condition
X∞
PN (1) =
p (n) = 1
n=0
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−a
1 − as
a+b
a
.
By differentiation of (3) we get
0
PX
(s) = PN0 (PY (s)) PY0 (s)
a+b
=
PN (PY (s)) PY0 (s)
1 − aPY (s)
a+b
PX (s) PY0 (s) ,
=
1 − aPY (s)
such that
0
0
PX
(s) − aPX
(s) PY (s) = (a + b) PX (s) PY0 (s) .
32
X∞
X∞
X∞
0
f (y) sy , and
xg (x) sx−1 , PY (s) =
g (x) sx , PX
(s) =
Plugging in PX (s) =
y=1
x=0
x=0
X∞
PY0 (s) =
yg (y) sy−1 , we get
y=1
=
=
=
=
=
X∞
X∞
xg (x) sx−1
f (y) sy
xg (x) sx−1 − a
x=0
y=1
x=0
X∞
X∞
yf (y) sy−1 ⇔
g (x) sx
(a + b)
y=1
x=0
X∞
X∞ X∞
xg (x) sx−1 − a
xf (y) g (x) sx+y−1
x=1
x=0
y=1
X∞ X∞
yg (x) f (y) sx+y−1 ⇔(z=x+y)
(a + b)
x=0
y=1
X∞ X∞
X∞
(z − y) f (y) g (z − y) sz−1
xg (x) sx−1 − a
z=y
y=1
x=1
X∞ X∞
(a + b)
yg (z − y) f (y) sz−1 ⇔(switch summation)
y=1
z=y
X∞
X∞ Xz
xg (x) sx−1 − a
(z − y) f (y) g (z − y) sz−1
x=1
y=1
z=1
X∞ Xz
yg (z − y) f (y) sz−1 ⇔(rename x=z)
(a + b)
y=1
z=1
X∞ Xx
X∞
(x − y) f (y) g (x − y) sx−1
xg (x) sx−1 − a
y=1
x=1
x=1
X∞ Xx
yg (x − y) f (y) sx−1
(a + b)
X∞
y=1
x=1
from which we collect terms and summations into
X∞ Xx
(ax + by) g (x − y) f (y) sx−1 = 0
xg (x) −
x=1
y=1
If this true for all s, then
xg (x) −
Xx
y=1
(ax + by) g (x − y) f (y) = 0 ⇔
y
g (x − y) f (y)
y=1
x
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 ).
g (x) =
Xx
a+b
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) =
g (2) =
g (3) =
=
=
1 −1
e ,
4
1
1 1 −1 1
1
g (1) f (1) + g (0) f (2) =
e
+ e−1 =
2
24
4
2
1
2
g (2) f (1) + g (1) f (2) + g (0) f (3)
3
3
1 −1 1 2 1 −1 1
1
1 1
+
+
e
+ e−1
e
3 32 2
4 34
2
4
1 1
2
1 −1
1
e
+
+
+
32 2 12 24 4
33
1
1
+
32 2
e−1 ,
=
=
6.2
3
1 −1
1
e
+
+
384 24 4
1 1
1
e−1 .
+ +
4 8 384
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 such
that 1 − G (u) ≤ ε, and let z1−ε denote the 1 − ε fractile of the standard normal distribution. NP
approximation says that approximately
i
h
3
E
(X
−
E
[X])
p
1 2
z
−1
.
u1−ε = E [X] + z1−ε V ar [X] +
6 1−ε
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 u 1−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,
XN (t)
X (t) =
Yi .
i=1
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
i
i
h
h
k−1
k
.
E N k = aE (N + 1) + bE (N + 1)
b) Find simple expressions for E [N ] and V ar [N ] in terms of a and b, and show that
V ar [N ]
1
=
.
E [N ]
1−a
Solution:
E Nk
=
=
=
=
=
X∞
nk p (n)
b
nk a +
p (n − 1)
n=0
n
X∞
ank + bnk−1 p (n − 1)
n=1
X∞
X∞
k
k−1
a
(n + 1) p (n) + b
(n + 1)
p (n)
n=0
n=0
i
i
h
h
k−1
k
.
aE (N + 1) + bE (N + 1)
n=0
X∞
EN
= aE [N + 1] + bE [1] = aE [N ] + a + b ⇒
a+b
,
E [N ] =
1−a
h
i
2
V ar [N ] = EN 2 − E 2 N = aE (N + 1) + bE [N + 1] − E 2 N
= aEN 2 + a + 2aEN + bEN + b − E 2 N
= aV arN + aE 2 N + a + 2aEN + bEN + b − E 2 N
a+b
⇒
= aV arN +
1−a
a+b
,
V ar [N ] =
(1 − a)2
1
V ar [N ]
=
.
E [N ]
1−a
Exercise 16 The special case of the negative binomial distribution where α = 1, i.e. p (n) =
q n (1 − q) is called the geometric distribution. Assume that N is geometrically distributed. Show
that the distribution function G of X, G (x) = P (X ≤ x), satisfies the recursion formula
Xx
G (x) = (1 − q) + q
f (y) G (x − y) .
y=1
35
Solution
G (x)
=
Xx
z=0
= g (0) +
g (z)
Xx
= (1 − q) +
= (1 − q) +
(v = z − y) = (1 − q) +
z=1
Xx
g (z)
Xz
y=1
z=1
Xx
y=1
Xx
= (1 − q) + q
= (1 − q) + q
y=1
Xx
Xx
z=y
Xx−y
y=1
Xx
y=1
qg (z − y) f (y)
qg (z − y) f (y)
qg (v) f (y)
Xx−y
f (y)
g (v)
v=0
v=0
f (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,
X∞
E Xk =
xk g (x)
x=0
X∞
=
xk g (x)
x=1
X∞
=
xk−1 xg (x)
x=1
Xx
X∞
(ax + by) g (x − y) f (y)
xk−1
=
y=1
x=1
X∞ Xx
axk + byxk−1 g (x − y) f (y)
=
y=1
x=1
X∞ X∞
axk + byxk−1 g (x − y) f (y)
=
x=y
y=1
X∞ X∞ k
k−1
g (z) f (y)
a (y + z) + by (y + z)
=
z=0
y=1
!
X∞ X∞ Xk
k
= a
z l y k−l g (z) f (y)
y=1
z=0
l=0
l
!
X∞ X∞
Xk−1
k−1
z l y k−1−l g (z) f (y)
+b
y
z=0
y=1
l=0
l
!
X∞ X∞ Xk−1
X∞ X∞
k
= a
z l y k−l g (z) f (y) + a
z k g (z) f (y)
y=1
z=0
l=0
y=1
z=0
l
!
X∞ X∞ Xk−1
k−1
z l y k−l g (z) f (y)
+b
l=0
z=0
y=1
l
!
X∞
X∞
X∞
Xk−1
k X∞ k−l
f (y)
z k g (z)
z l g (z) + a
y f (y)
= a
y=1
z=0
z=0
y=1
l=0
l
!
Xk−1
X∞
k − 1 X∞ k−l
+b
y f (y)
z l g (z)
l=0
y=1
z=0
l
!
Xk−1
k
= a
E Y k−l E X l + aE X k
l=0
l
!
Xk−1
k−1
+b
E Y k−l E X l
l=0
l
36
Xk−1
= aE X k +
k
l
a
l=0
!
+b
k
l
!
k−1
l
!!
E Y k−l E X l
and conclude that
E Xk =
1 Xk−1
l=0
1−a
a
+b
k−1
l
!!
E Y k−l E X l .
Exercise 18 Consider the Normal Power approximation in case of a Poisson λ claim distribution.
Show that
!
Xk−1
k
k−1
E Y k−l E X l .
E X =λ
l=0
l
The first two moments are known from a previous exercise. Verify these results and show that
Show that
E X 3 = λE Y 3 + 3λ2 E Y 2 E [Y ] + λ3 E 3 [Y ] .
h
i
E (X − EX)3 = λE Y 3 ,
and conclude the following NP approximation formula
u1−ε = λE [Y ] + z1−ε
p
λE [Y 2 ] +
E Y3
1 2
.
z1−ε − 1
6
E [Y 2 ]
Solution: Put a = 0 and b = λ in the recursion formula. Then
But then
E [X] = λE [Y ] ,
E X 2 = λ E Y 2 + E [Y ] E [X] = λE Y 2 + λ2 E 2 [Y ] ,
E X 3 = λ E Y 3 + 2E Y 2 E [X] + E [Y ] E X 2
= λE Y 3 + 3λ2 E Y 2 E [Y ] + λ3 E 3 [Y ]
h
i
3
E (X − EX)
such that
= E X 3 − 3E [X] X 2 + 3XE 2 [X] − E 3 [X]
= EX 3 − 3E [X] EX 2 + 2E 3 [X]
= λE Y 3 + 3λ2 E Y 2 E [Y ] + λ3 E 3 [Y ]
−3λ2 E [Y ] E Y 2 − 3λ3 E 3 [Y ] + 2λ3 E 3 [Y ]
= λE Y 3 ,
u1−ε = λE [Y ] + z1−ε
Exercise 19 Verify, by means of,
p
E Y3
1
2
λE [Y 2 ] +
z
−1
.
6 1−ε
E [Y 2 ]
E [N ] =
V ar [N ] =
37
a+b
,
1−a
a+b
(1 − a)
2,
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 ] + E 2 [Y ] V ar [N ] ,
also, of course, holds when N belongs to the (a, b)-class.
Solution:
!
!!
1
1
0
1 X0
a
+b
E Y 1−0 E X 0
E X
=
l=0
1−a
0
0
a+b
E [Y ]
1−a
= E [N ] E [Y ] ,
1 X1
E X 2 − E 2 [X] =
l=0
1−a
=
=
=
=
=
=
=
=
a
2
l
!
+b
2−1
l
!!
(a + b)2 2
E Y 2−l E X l −
2 E [Y ]
(1 − a)
1
(a + b)2 2
1
(a + b) E Y 2 +
(2a + b) E [Y ] E [X] −
2 E [Y ]
1−a
1−a
(1 − a)
2
1
1
a+b
(a + b) 2
(a + b) E Y 2 +
(2a + b) E [Y ]
E [Y ] −
2 E [Y ]
1−a
1−a
1−a
(1 − a)
2
a + b 2 (a + b) (a + a + b) 2
(a + b) 2
E
[Y
]
−
E [Y ]
E Y +
1−a
(1 − a)2
(1 − a)2
a + b 2 (a + b) a 2
E Y +
2 E [Y ]
1−a
(1 − a)
a+b
a+b 2
(a + b) a 2
E [Y ]
V ar [Y ] +
E [Y ] +
1−a
1−a
(1 − a)2
(a + b) (1 − a) 2
(a + b) a 2
a+b
V ar [Y ] +
E [Y ] +
2
2 E [Y ]
1−a
(1 − a)
(1 − a)
a+b
a+b
2
V ar [Y ] +
2 E [Y ]
1−a
(1 − a)
= E [N ] V ar [Y ] + E 2 [Y ] V ar [N ]
7
Finance and Reinsurance: Products and Valuation
Risk free investment - corresponds to chapter 1
S 0 (0) = 1,
S 0 (1) = (1 + r) S 0 (0) = 1 + r
This is new
S 1 (0) = s,
S 1 (1) = s (1 + Z) ,
(
u with probability pu ,
Z =
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 S 1 (1)
depends on the outcome of S 1 (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 S 1 (1) , 1
What should be trhe price for this claim: Caluclate the expected present value
π
= E [P V (0)] + risk loading
1
=
E max S 1 (1) , 1 + risk loading
1+r
1
(0.6 ∗ 1, 4 + 0.4 ∗ 1) + risk loading
=
1.1
= 1.127273 + risk loading
Consider the situation where I can invest the premium paid in S 1 . Then we can put up two
equations with two unknowns
1.4x + (π − x) 1.1 = 1.4
0.9x + (π − x) 1.1 = 1
0.5x = 0.4 ⇒ x = 0.8
1.4 − 0.8 ∗ 0.3
0.3 ∗ 0.8 + 1.1π = 1.4 ⇒ π =
= 1.054545
1.1
π
= E ∗ [P V (0)]
1
=
E ∗ max S 1 (1) , 1
1+r
1
1.4 − 1.1
1.1 − 0.9
=
∗ 1, 4 +
∗1
1.1 1.4 − 0.9
1.4 − 0.9
1
(0.6 ∗ 1, 4 + 0.4 ∗ 1)
=
1.1
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 S 1 (1) − K, 0 ⇒ determine π for given K
39
Futures:
Φ = S 1 (1) − K : determine k such that π = 0.
The insurance version:
S 0 (0) = 1,
S 0 (1) = (1 + r) S 0 (0) = 1 + r
This is new
S 1 (0) = π r ,
S 1 (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 = 0
0 ∗ x + (π − 0.1x) 1.1 = 1
x = −1
−1 (1 − 0.1 ∗ 1.1) + π1.1 = 0 ⇒ π =
π
=
=
=
1
E ∗ 1(Tx >1)
1+r
1
0.1 ∗ 1.1 − 0
1 − 0.1 ∗ 1.1
∗0+
∗1
1.1
1−0
1−0
1 − 0.1 ∗ 1.1
= 0.8090909
1.1
and not
π=
7.1
Exercises
7.2
Exercises
1 − 0.1 ∗ 1.1
= 0.8090909
1.1
1
0.9
px =
= 0.8181818
1.1
1.1
Exercise 20 Assume that N is Poisson distributed with parameter λ. Show that
E [N ] = λ,
E N 2 = λ2 + λ,
V ar [N ] = λ,
and use this to show that
E [X] = λE [Y ] ,
E X 2 = λE Y 2 + λ2 E 2 [Y ] ,
V ar [X] = λE Y 2 .
40
Solution 21
E [N ] =
=
=
E N2
=
=
=
=
V ar [N ] =
X∞
λn
λn
=
ne−λ
n=0
n=1
n!
n!
n−1
X∞
X∞
λn
λ
=λ
e−λ
λ
e−λ
n=0
n=1
(n − 1)!
n!
λ,
X∞
λn
n2 e−λ
n=0
n!
X∞
X∞
λn
λn
(n − 1) e−λ
e−λ
+
n=1
n=1
(n − 1)!
(n − 1)!
n
n
X∞
X
∞
λ
λ
λ
ne−λ
+λ
e−λ
n=0
n=0
n!
n!
λ2 + λ,
E N 2 − E 2 [N ] = λ,
X∞
ne−λ
E [X] = E [Y ] E [N ] = λE [Y ] ,
2
E X 2 = E [N ] V ar [Y ] + E 2 [Y ] E N 2 = λE [Y ] + E 2 [Y ] λ2 ,
V ar [X] = E X 2 − E 2 [X] = E 2 [Y ] λ2 + E Y 2 λ − λ2 E 2 [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 N 2 = q 2 E N 2 + q 2 (2α + 1) E [N ] + q 2 α2 + α + E [N ]
and calculate that
qα
,
1−q
q2
1 + q 2 (2α + 1)
E
[N
]
+
α2 + α
2
2
1−q
1−q
1 + q 2 (2α + 1)
qα
+ q (α + 1)
1 − q2
1−q
qα
1 + q 2 α + qα + q
,
1 − q2
1−q
E [N ] =
E N2 =
=
=
qα
1 − q2
qα
V ar [N ] =
=
(1 − q)
!
1 − q2 q
1 + q 2 (2α + 1)
+ q (α + 1) −
2 α
1−q
(1 − q)
2.
Solution 23
(α + n − 1)! n
q (1 − q)α
n! (α − 1)!
X∞
(α + n − 1)! n
α
=
q (1 − q)
n
n=1
n! (α − 1)!
X∞
(α + n − 1)! n
α
q (1 − q)
=
n=1 (n − 1)! (α − 1)!
X∞
(α + n)! n
= q
q (1 − q)α
n=0 n! (α − 1)!
E [N ] =
X∞
n=0
n
41
(α + n) (α + n − 1)! n
α
q (1 − q)
n=0
n! (α − 1)!
X∞
X∞ (α + n − 1)!
(α + n − 1)! n
q n (1 − q)α + q
q (1 − q)α
qα
n
n=0
n=0 n! (α − 1)!
n! (α − 1)!
qα + qE [N ] ,
X∞
(α + n − 1)! n
α
n2
q (1 − q)
n=0
n! (α − 1)!
X∞
(α + n − 1)! n
n
q (1 − q)α
n=1 (n − 1)! (α − 1)!
X∞
(α + n)! n
α
q
q (1 − q)
(n + 1)
n=0
n! (α − 1)!
X∞
X∞
(α + n)! n
(α + n)! n
α
α
q (1 − q) + q
q (1 − q)
q
n
n=0 n! (α − 1)!
n=0 n! (α − 1)!
X∞
(α + n)! n
n
q
q (1 − q)α + E [N ]
n=1 n! (α − 1)!
X∞ (α + n + 1)!
α
q2
q n (1 − q) + E [N ]
n=0 n! (α − 1)!
X∞ (α + n + 1) (α + n) (α + n − 1)!
α
q n (1 − q) + E [N ]
q2
n=0
n! (α − 1)!
X∞
(α + n − 1)! n
q (1 − q)α + E [N ]
n2 + (2α + 1) n + α2 + α
q2
n=0
n! (α − 1)!
q 2 E N 2 + q 2 (2α + 1) E [N ] + q 2 α2 + α + E [N ] .
= q
=
=
E N
2
=
=
=
=
=
=
=
=
=
X∞
Exercise 24 We introduce the mean, variance, coefficient of variation, and skewness,
µ = E [Y ] ,
σ2
= E 2 [Y − E [Y ]]
= E Y 2 − E 2 [Y ] ,
σ
κ =
,
µ
E 3 [Y − E [Y ]]
ν =
3
3σ
E Y
1
3
=
− − 3.
σ3
κ κ
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
=
σ2
=
κ =
Solution 25
E Yk
=
αλ
,
λ−1
αλ2
(α − 2) (α − 1)
1
p
.
α (α − 2)
α
λk ,
α−k
42
2,
µ = E [Y ] =
σ2
αλ
,
α−1
α
α2 λ 2
λ2 −
α−2
(α − 1)2
2
αλ2 (α − 1) − (α − 2) α
= E Y 2 − E 2 [Y ] =
=
2
(α − 2) (α − 1)
αλ2
2,
(α − 2) (α − 1)
q
αλ2
σ
(α−2)(α−1)2
κ =
=
=
αλ
µ
α−1
r r
1
α
=
2
α
α−2
1
.
= p
α (α − 2)
=
E [X] =
V ar [X] =
qβ αλ
.
1−qλ−1
qβαλ2
1−q
1
2
(α − 2) (α − 1)
43
λ
α−1
q
α
α−2
αλ
α−1
+
α
(1 − q) (λ − 1)
2
!