The effect of dependence on life insurance. Prace Naukowe Uniwersytetu Ekonomicznego we Wrocławiu = Research Papers of Wrocław University of Economics, 2011, Nr 230, s. 60-76

Zagadnienia statystyki

aktuarialnej

Wydawnictwo Uniwersytetu Ekonomicznego we Wrocławiu

Wrocław 2011

pod redakcją

Joanny Dębickiej

Redakcja wydawnicza: Joanna Świrska-Korłub Redakcja techniczna: Barbara Łopusiewicz Korektor: Barbara Cibis Łamanie: Adam Dębski Projekt okładki: Beata Dębska Publikacja jest dostępna na stronie www.ibuk.pl Streszczenia opublikowanych artykułów są dostępne w The Central and Eastern European Online Library www.ceeol.com, a także w adnotowanej bibliografii zagadnień ekonomicznych BazEkon http://kangur.uek.krakow.pl/bazy_ae/bazekon/nowy/index.php Informacje o naborze artykułów i zasadach recenzowania znajdują się na stronie internetowej Wydawnictwa www.wydawnictwo.ue.wroc.pl Kopiowanie i powielanie w jakiejkolwiek formie wymaga pisemnej zgody Wydawnictwa © Copyright by Uniwersytet Ekonomiczny we Wrocławiu Wrocław 2011 ISSN 1899-3192 ISBN 978-83-7695- 240-6 Wersja pierwotna: publikacja drukowana Druk: Drukarnia TOTEM

Wstęp ... 7

Joanna Dębicka: Indeksacja przepływów pieniężnych w ubezpieczeniach

wielostanowych ... 9

Stanisław Heilpern: Wyznaczanie wielkości renty w zależnych grupowych

ubezpieczeniach na życie ... 30

Aleksandra Iwanicka:

Wpływ zewnętrznych czynników ryzyka na prawdo-podobieństwo ruiny w agregacji dwóch klas ubezpieczeń ... 49

Anna Nikodem-Słowikowska: The effect of dependence on life insurance . 60 Katarzyna Ostasiewicz: Modele progowe i ich zastosowanie w socjologii

i ekonomii ... 77

Stanisława Ostasiewicz, Katarzyna Ostasiewicz: Modelowanie trwania

życia w populacjach niejednorodnych ... 99

Katarzyna Sawicz: Uwagi o finansowaniu systemu ochrony zdrowia

w Polsce ... 123

Janusz L. Wywiał, Agnieszka Źrubek:

O dokładności analitycznego wy-znaczania mocy pewnego testu na normalność rozkładu prawdopodo-bieństwa ... 131

Summaries

Joanna Dębicka, Indexing cash flows in multistate insurance contracts ... 29

Stanisław Heilpern, Calculation of pensions in the multiple life insurances 48

Aleksandra Iwanicka, Influence of some outside risk factors on a ruin

probability in the aggregated two-classes risk model ... 59

Anna Nikodem-Słowikowska, Wpływ zależności na ubezpieczenia na

życie ... 76

Katarzyna Ostasiewicz, Threshold models and their application to sociology

and economics ... 98

Stanisława Ostasiewicz, Katarzyna Ostasiewicz, Approximation of survival

function for heterogeneity population ... 122

Katarzyna Sawicz, Some comments on the financing of health care system

in Poland ... 130

Janusz L. Wywiał, Agnieszka Źrubek, On estimation of the power of a

Zagadnienia statystyki aktuarialnej ISSN 1899-3192

Anna Nikodem-Słowikowska

Wroclaw University of Economics

THE EFFECT OF DEPENDENCE

ON LIFE INSURANCE

* Summary: In the classical life insurance it is assumed that the risks are independent. This assumption is not appropriate in many practical situations. For example the lifelengths of two insured persons (such as a husband and a wife) are dependent because they share a common way of life and more or less are exposed to the same risks. In the case of the group life in-surance for persons that work for the same company, the mortality is dependent on a certain event (i.e. explosion, breakdown). In this paper models for modelling the dependence will be presented. The influence of the dependence on the aggregate claims distribution and the net premium in the life insurance will be demonstrated on numerical examples. Keywords: individual risk model, dependent risks, aggregate claims distribution, recursion formula, net premium.

1. Introduction

The individual risk model can be used in the group of life insurance. Usually it is assumed that the risks of the insurance portfolio are independent. This assumption is not appropriate in many practical situations. For example a husband and a wife sha-res a common way of life and more or less are exposed to the same risk. The persons that work for the same company are exposed to a certain event (i.e. explosion, bre-akdown). Therefore a risk that is common in every life will be considered in the model. Let the insurance portfolio consist of n + m risks. The m married couples have policies in this portfolio. Then the aggregate claim amount denoted by S can be writ-ten as (see [3]) S Xi Xi Xi i m n i m = + + = + =

∑

∑

( _{1 2}) 1 1 , (1) * This work is supported by the Polish scientific fund in years 2010-2012 as the research project No. N N111 336138.

where X_i1 + X_i2 is the claim amount for policies of couple i, X_i is the claim amount for the policy i. The claim amount is expressed as X_i = b_iI_i, where I_il = min(J_il + J₀; 1) for i = 1, …, m, l = 1, 2, (2) I_i = J_i for i = m + 1, …, n. (3) The random variables J_i, J_il, J₀ are independent and have the Bernoulli distribu-tion: P(J_i = 1) = q_i, P(J_i = 0) = 1 − q_i = p_i for i = m + 1, …, n, (4) P(J_il = 1) = q_il, P(J_il = 0) = 1 − q_il = p_il for i = 1, …, m, l = 1, 2, (5) P(J₀ = 1) = q₀, P(J₀ = 0) = 1 − q₀ = p_{0. (6)} Random variables X_i1 + X_i2 and X_i for i = 1, …, n are mutually independent, but X_i1 and X_i2 are dependent because of the common risk factor J₀. The next model can be used in the group life insurance, where m persons are exposed to the same risk. Let S be the aggregate claim amount given by S X X Xm Xi i m n = + + + + = +

∑

1 2 1 ... , (7)

where the claim amount for the policy i is expressed by X_i = bI_i. The random variable I_i is defined by

I_i = min(J_i + J₀; 1) _{for i = 1, …, m,} (8) I_i = J_{i for i = m + 1, …, n, (9)} where

J

_i and

J

₀ are independent Bernoulli random variables with

P(J_i = 1) = q, P(J_i = 0) = 1 − q = p for i = 1, …, n, (10) P(J₀ = 1) = q₀, P(J₀ = 0) = 1 − q₀ = p₀. ₍₁₁₎ Therefore random variables X_i are independent for i = m + 1,…, n and for i = 1,…, m are dependent because of dependence between the occurrence random variables I_i. In actuarial science, we are interested in the computation of the distribution func- tion of the aggregate claim amount S. In many papers we can find methods of cal-culating the aggregate claim amount with independent risks (see e.g. [2; 4; 5; 6]). In [4] the recursive method for individual risk model is shown. Various papers have also addressed the subject of dependency between risks (see e.g. [1; 3; 7]). In [1] the common shock model is described. In this paper, a model proposed by Dhaene and Goovaerts [3] is considered. The dependence between two risks is introduced via the occurrence of a random variable

(similarly as in [1]). In the next section the recursion for the aggregate claims distri- bution is presented. In the last section the influence of the dependence on the aggre-gate claims distribution and the net premium is presented by numerical examples.

2. Calculation of the aggregate claims distribution

An exact recursion formula for computing the aggregate claims distribution in the individual model with dependent risks described in (1) is given in the following the-orem: Theorem 1. The aggregate claims distribution for model described in (1) can be calculated by P S p p p_{i i} p i m i i m n ( = )=

(

)

⋅ = = +

∏

∏

0 0 1 2 1 1 , (12) sP S s v si w s i m i i m n ( = )= ( )+ ( ) = = +

∑

∑

1 1 , (13) where the coefficients w_i(s), v_i(s) are given by w si q_pi b P S s b w s b i i i i i ( )=

[

⋅ ( = − )− ( − )

]

, (14) v si q_pi b P S s b v s b q_p b P S s b i i i i i i i i ( )= 1 ⋅

[

⋅ ( = − )− ( − )

]

+ ⋅ ⋅ ( = − 1 1 1 1 2 2 2 ii i i i i i i l il il v s b p q q q p p p b P S s b 2 2 0 1 2 0 0 1 2 1 2 ) ( ) ( − − + + + ⋅ ⋅ = −

[

]

(

)

=

∑

ll i l il v s b = =

∑

− −

∑









1 2 1 2 ) ( ) (15) for s b bn bi i n = =

∑

min{ , ..., }, ...,₁ 1 and w_i(s) = v_i(s) = 0 elsewhere. Proof. The proof is based on Theorem 3 in [4]. Let

Y X

i

=

i₁

+

X

i₂ , then the pro-bability generating function of the aggregate claim amount is given by g t_S E tS E t Yi X E t Y i m i i m n i i m ( ) =

_{ }

=



=





























= = + = ∑ +∑ ∑ 1 1 1



















⋅E t= +∑ = Xi i m n 1 =

_{ }

⋅

_{ }

= ⋅ = = + = = +

∏

E tY

∏

E t

∏

g t

∏

g t i m X i m n Y i m X i m n i i i i 1 1 1 1 ( ) ( )

_,

₍₁₆₎

where the generating function of the random variable X_i and Y_i are respectively equal to g t_X E tX E tb I E tb J p q t i i b i i i i i i i ( ) =

_{ }

=

_{ }

=

_{ }

= + (17)

and g tY E t E t E t t E t Y X X b I b I b i i i i i i i i i ( )_{= =} ( ) _{= ⋅ =}

 



1+ 2





1 1 2 2





1mmin(Ji1+J0; )1 ⋅tbi2min(Ji2+J0; )1



= = + = ⋅ = + =

∏

∏













E tb J J p E t l b J l ilmin(il 0; )1 ilmin(il ; ) 1 2 0 0 1 1 2





_

+ ⋅



_

+



_

= =

∏

q E tb J l il il 0 1 1 1 2 min( ; ) = ⋅ + ⋅ = = =

∏

∏

















 

p E tb J q E t p E t l b l b J il il il il il 0 1 2 0 1 2 0 ll b q t il l =

∏









_



+ ⋅ ∑= = 1 2 0 1 2 = ⋅

(

+

)

+ ⋅ =

∏

∑= p p_il q t_il b q t l b il l il 0 1 2 0 1 2 . (18) If the death benefit b_i is a positive and integer number, then the random variable ta-kes value 1 1

0, min{ , ..., }, ...,

n n i i

s

b

b

b

=

=

∑

and the probability

P S s

(

=

)

can be cal-culated by the following way ( )

₍₀₎

(

)

g

Ss

_!

P S s

=

=

_s

for ₁ 1

0, min{ , ..., }, ...,

n n i i

s

b

b

b

=

=

∑

. (19) Therefore P S gS g g g S Y i m X i m n i i ( = )= ( )( )_! = ( )= ( )⋅ ( )= = = +

∏

∏

0 0₀ 0 0 0 0 1 1 =



⋅

(

+

)

+ ⋅ ⋅ +





₌



_



(

=

∏

∏

p p q q ∑= p q il il b l b i m i i b il l il i 0 1 2 0 1 0 0 1 0 2

))

= +

∏

= i m n 1 =

(

)

⋅ = = +

∏

p p p_{i i}

∏

p i m i i m n 0 1 2 1 1 . (20) In order to calculate the other probabilities the derivative of order s will be taken. The logarithm of (16) is equal to 1 1

ln ( )

_S m

ln ( )

_Yi n

ln

_Xi

( )

i i m

g t

g t

g t

= = +

=

∑

+

∑

(21) and derivative of is 1 1

( )

( )

( )

( )

i_i

( )

i_i

( )

m n Y X S S i Y i m X

g t

g t

g t

g t

₌

g t

_{= +}

g t

′

′

′

=

∑

+

∑

, 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) i _{( )} i _{( )} i i i i m n Y S X S S Y X i i m V t W t g t g t g t g t g t _{g t g t} = = + ′ ⋅ ′ ⋅ ′ =

∑

+

∑

  . .

Taking the auxiliary function V_i(s) and W_i(s) we obtain 1 1

( )

m

( )

n

( )

S i i i i m

g t

V t

W t

= = +

′

=

∑

+

∑

, where V_i(s), W_i(s) are defined by 0

( )

(

1)

x i i x

V t

∞

v x

t

=

=

∑

+

, 0

( )

(

1)

x i i x

W t

∞

w x

t

=

=

∑

+

. Therefore the derivative of order s of g_s(t) is equal to ( ) ( 1) ( 1) 1 1

( )

m

( )

n

( )

s s s S i i i i m

g t

V

−

t

W

−

t

= = +

=

∑

+

∑

, where ( 1) ( 1) 1

( )

(

1) !

s x s i i x s

V

−

t

∞

v x

x t

− − = −

=

∑

+

, ( 1) ( 1) 1

( )

(

1) !

s x s i i x s

W

−

t

∞

w x

x t

− − = −

=

∑

+

. Thus for t = 0 we obtain ( ) ( 1) ( 1) 1 1

! (

)

s

(0)

m s

(0)

n s

(0)

S i i i i m

s P S s

g

V

−

W

− = = +

=

=

=

∑

+

∑

=

1

( )( 1)!

1

( )( 1)!

m n i i i i m

v s s

w s s

= = +

=

∑

−

+

∑

−

, which implies that 1 1

(

)

m i

( )

n i

( )

i i m

sP S s

v s

w s

= = +

=

=

∑

+

∑

.

In order to find functions v_i(s) and w_i(s), we take the (s − 1) order derivative of

( )

( )

( )

i

_{( )}

i X S i X

g t g t

W t

=

′

_{g t}

⋅

, that is

( )

_i

( )

_i

( )

( )

i X X S

W t g t

⋅

=

g t g t

′

⋅

. Using the Leibnitz formula for (s − 1) th derivative of a product we obtain

Ds W t g t D g t g t i X s X S i i (−1)

(

_{( )⋅ ( )}

)

₌ ( −1)

(

_{′ ( )⋅ ( )}

)

_, 1 1 1 1 0 0

1

1

( )

_i

( )

_i

( )

( )

s s k s k k s k i X X S k k

s

s

D W t D

g t

D g t D

g t

k

k

− − − − − − = =

−

−









_′

⋅

=

⋅

















∑

∑

. Inserting t =0 we have 1 1 0

1

1

(0)

(0)

(0)

(0)

1

1

i i i i s b b s i X i X i

s

s

D W

D g

D

W

D g

s

b

s

− − −

−

−









⋅

+

_

_

⋅

=



₋



_{− −}









1

1

(0)

(0)

1

i i i b s b X S i

s

D g

D g

b

− −

−





_′

=

_

_

⋅

−





. Hence 1 1 ₍₀₎ 0 _{(0) (0)} 1 ₍₀₎ 1 ₍₀₎ 1 1 i i i i i b s b s b s i X X S i i i s s D W D g D g D g D W b s b − − − − _{⋅ =}  − _{⋅ −} −    ₋   _{− −}        . From (17) we have (0)

₍₀₎

₍₀₎

i i X X i

g

=

g

=

p

. The n’th derivative of the probability generating function of

X

i equals ( ) 1

( )

(

1)(

2)...(

1)

i i b n n X i i i i

g t

₌

q b b

₋

b

₋

b n

_{− +}

t

− _, where ( ) ( )

(0) 0

for

,

(0)

( !) for

.

i i i n X i b X i i i

g

n b

g

q b

n b



=

≠





=

=



Therefore we get the following recursion formula for the function

w s

i

( )

w s s_i( )( − ⋅1)! p_i = = − − − ⋅ − = − − − − q b_{i i s b b}s s b P S s b _{b s}s i i i i i ( !)



₍ ( _)!(1)! _1)! ( )! ( ) _!(( ₁₁1₋)! − − −1







b_i)!w s b si( i)( bi)! ,, w si q_pi b P S s b w s b i i i i i ( )=

[

⋅ ( = − )− ( − )

]

_. The recursion formula for the function v_i(s) is obtained like for function w_i(s). Using the Leibnitz formula we calculate the (s − 1) order derivative of

( )

_i

( )

_i

( )

( )

i Y Y S

V t g t

⋅

=

g t g t

′

⋅

,

that is Ds V t g t D g t g t i Y s Y S i i (−1)

(

_{( )⋅ ( )}

)

₌ (−1)

(

_{′( )⋅ ( )}

)

_, s k D V t D g t s k D g t k i s k Y k s k Y i i −       ⋅ =  −      ′ ⋅ − − = −

∑

1 1 1 0 1 ( ) ( ) ( ) DDs kg t S k s − − = −

∑

1 0 1 ( )_{. (22)} Form (18) we have (0) 0 1 2

(0)

(0)

i i Y Y i i

g

=

g

=

p p p

. Taking the auxiliary function

U t

_i

( )

and

Z t

_i

( )

we obtain

g tY p pil q til q t b l U t b Z t i il i il l i ( ) ( ) ( ) = ⋅

(

+

)

+ ⋅ =

∏

∑= 0 1 2 0 1 2

 











, that is 0 0

( )

( )

( )

i Y i i

g t

=

p U t

⋅

+

q Z t

⋅

. Than the n’th derivative of the probability generating function of

Y

_i is equal to ( ) ( ) ( ) 0 0

( )

( )

( )

i n n n Y i i

g t

=

p U

⋅

t

+

q Z

⋅

t

. For the function

( )

bi1 bi2 i

Z t

₌

t

+ _{we obtain} ( )n

_{( )}

₍

_1)...(

₁₎

m n i

Z

t

₌

m m

₋

m n

_{− +}

t

− _, 1 2 i i

m b

=

+

b

, where

(

)

( ) 1 2 ( ) 1 2 1 2

(0) 0

for

,

(0)

! for

.

n i i i n i i i i i

Z

n b

b

Z

b

b

n b

b



=

≠

+





₌

₊

₌

₊



For the function U ti pil q til b l il ( ) =

(

+

)

=

∏

1 2 we have U ti( ) =

(

pi1+q ti1bi1

)

(

pi2+q ti2 bi2

)

= p pi1 i2+ p q ti1 2i bi2 +p q ti2 1i bbi1 +q q ti1 2i b bi1+i2. (23) The n’th derivative of (23) has the following form 2 1 2 1 ( ) 1 2 2 2 2 2 1 1 1 1 2 2 2 1 2 1 1 1

( )

(

1)...(

1)

(

1)...(

1)

(

1)...(

1)

i i il l b n n i i i i i i b n i i i i i b n i i il il il l l l

U

t

p q b b

b

n

t

p q b b

b

n

t

q q

b

b

b

n

t

= − − − = = = ∑





=

_

−

− +

_

+





+

_

−

− +

_

+





+



−

− +









∑ ∑

∑



and is equal to Ui p q b b i i i i ( )1 _{( )}0 _! 2 1 1 =

( )

_, Ui p q b b i i i i ( 2)_{( )}0 _! 1 2 2 =

( )

, U_ib b q q b b i i i i i i ( 1 2)_{( )}0 _! 1 2 1 2 + ₌

(

(

₊

)

)

and ( )n

_{(0) 0}

i

U

=

elsewhere. Therefore the probability generating function of

Y

_i is equal to gY p p q b b i i i i i ( )1 _{( )}0 _! 0 2 1 1 =

( )

_{, (24)} gY p p q b b i i i i i ( 2)_{( )}0 _! 0 1 2 2 =

( )

_{, (25)} gY p q q b b q b b b b i i i i i i i i i ( 1 2)_{( )}0 _{! !} 0 1 2 1 2 0 1 2 + ₌

(

(

₊

)

)

₊

(

(

₊

)

)

(26) and ( )

_{(0) 0}

i n Y

g

=

elsewhere. Putting t =0 into (22) we obtain 1

₍₀₎

0

₍₀₎

i s i Y

D V

−

_⋅

D g

₌

1 1 1 2 2 2 2 2 1 1 1 1 1 1 2 2 2 1

1

1

(0)

(0)

(0)

1

1

1

1

(0)

(0)

(0)

1

1

1

(0)

1

i i i i i i i i il il l l i b s b s b Y S i i i b s b s b Y S i i i b s b Y il l

s

s

D g

D

g

D

V

b

s

b

s

s

D g

D

g

D

V

b

s

b

s

D g

D

g

b

= = − − − − − − − = ∑ ∑





−





−





=

_

_

_

⋅

−

_

_

_

+

−

− −

















−





−





+

_

_

_

⋅

−

_

_

_

+

−

− −













−









+

_

_

⋅

−







∑



2 1 1 2 1

1

(0)

(0) .

1

il l s b S i il l

s

D

V

s

b

= − − = ∑





−







₋











_

_{− −}



_





_

_





∑



Using (24) – (26) we have v s q p b P S s b v s b q p b P S s i i i i i i i i i i ( ) ( ) ( ) ( = = ⋅ ⋅ = − − − + + ⋅ ⋅ =

[

]

1 1 1 1 1 2 2 2 −− − − + + + ⋅ ⋅ = −

[

]

(

)

=

∑

b v s b p q q q p p p b P S s b i i i i i i i l il 2 2 0 1 2 0 0 1 2 1 2 ) ( ) ( _iil l i l il v s b = =

∑

− −

∑









1 2 1 2 ) ( ) . This completes the proof.

Theorem 2. The aggregate claims distribution for the model described in (7) is satisfied by P(S = 0) = p₀ ._pn_{, (27)} 1

(

)

( )

n i

( )

( ) (

) ( )

i m

sP S s

v s

w s

v s

n m w s

= +

=

=

+

∑

=

+

−

⋅

, (28)

where the coefficients

v s

( )

and

w s

( )

are given by

v s m r q p br P S s br v s br r r ( )=



⋅ ( = − )− ( − )













_

 

_















(

)













=11 1 m−

∑

+ + + ⋅ ⋅ = − − −













(

)

p q q p p bm P S s bm v s bm m m 0 0 0 ( ) ( ) , (29) w s q p b P S s b w s b ( )=

[

⋅ ( = − −) ( − )

]

. (30) for

s b b nb

=

, 2 ...,

and

w s

( )

=

v s

( ) 0

=

elsewhere.

Proof. This proof is similar to proof of Theorem 1. Let

Y X

=

₁

+

X

₂

+ +

...

X

m, then the probability generating function of S is equal to g tS E t g t g t S Y X i m n i ( )=

_{ }

= ( )⋅ ( ) = +

∏

1 , (31) where g t_X E tX E tbI E tbJ p qtb i i i i ( ) =

_{ }

=

_{ }

=

_{ }

= + , g t_{Y( ) =}E tY ₌ p _⋅ p qt₊ b m_{+ ⋅}q tbm

 

0

(

)

0 . (32) Putting t = 0 in (31) lead to (27). Taking the logarithm and the derivative of g_s(t) we obtain 1

( )

( )

n

( )

S i i m

g t

V t

W t

= +

′

=

+

∑

, (33) where 0

( )

( )

( )

(

1)

x Y

_{( )}

S Y x

g t g t

V t

∞

v x

t

_{g t}

=

′

⋅

=

∑

+

=

, 0

( )

( )

( )

(

1)

i

_{( )}

i X S x i i X x

g t g t

W t

∞

w x

t

_{g t}

=

′

⋅

=

∑

+

=

. Taking the s’th derivative of (33) and inserting t =0we have 1

(

)

( )

n i

( )

i m

sP S s

v s

w s

= +

=

=

+

∑

.

The recursion formula for w_i(s) has the same form like in the previous theorem, that is w si( )= q_p

[

b P S s b w s b⋅ ( = − −) i( − )

]

. The recursion formula for v(s) is obtained by differentiating the following expression (s − 1) times:

( )

Y

( )

Y

( )

S

( )

V t g t

⋅

=

g t g t

′

⋅

. Using the Leibnitz formula we have Ds V t g t D g t g t Y s Y S (−1)

(

_{( )⋅ ( )}

)

₌ (−1)

(

_{′( )⋅ ( )}

)

_, 1 1 1 1 0 0

1

1

( )

( )

( )

( )

s s k s k k s k Y Y S k k

s

s

D V t D

g t

D g t D

g t

k

k

− − − − − − = =

−

−









_′

⋅

=

⋅

















∑

∑

. (34) From (32) we obtain (0) 0

(0)

(0)

m Y Y

g

=

g

=

p p

⋅

. (35) If we express the probability generating function of Y by means of the auxiliary function

U t

( )

and

Z t

( )

, that is

g tY p p qtb m q t p U t q Z t U t bm Z t ( ) ( ) ( ) ( ) ( ) = ₀⋅

(

+

)

+ ⋅₀ = ₀⋅ + ⋅₀

 

 





, then the n’th derivative is equal to ( ) ( ) ( ) 0 0

( )

( )

( )

n n n Y

g t

=

p U

⋅

t

+

q Z

⋅

t

. Taking the derivative of order n we have Z (n)_{(t) = bm(bm − 1)…(bm − n + 1)t} bm−n _, where Z n bm Z bm n bm n n ( ) ( ) ( ) , ( ) ! . 0 0 0 = ≠ =

( )

=







for for (36) Consider the function U(t): U t_{( ) =}

(

p qt₊ b m

)

₌ = + −

( )

+ − −

( )

+ − − −

( )

+ pm mpm1 qtb m m 1 pm 2 qtb 2 m m m pm 3 qtb 3 2 13 2 ( ) ! ( )(! ) ....+ +m m− m r− + −

( )

+ +

( ) ( )

− + = r pm r qtb r mp qtb m qtb m ( )...( ) ! ... 1 1 1

=













−

( )

=

∑

m_r pm r qtb r r m 1 _. The n’th derivative of

U t

( )

is equal to D U t D m r p qt m r p q D t n n m r b r r m m r r n br ( ) =



=











( )

























− = −

∑

1

∑

=

(( )

= r m 1 =



− − + =











(

)













− − = −

∑

m_r p q br br br n t m r p m r r br n r m m ( 1)...( 1) 1 rr r br n r m q _{(br n}( )!br_{− )!}t − =













∑

1 . Inserting t = 0 we have U n br U m r p q br r n br m r r ( ) ( ) ( ) , ( ) ( )! , ..., 0 0 0 1 = ≠ =



=











− for for mm.











Therefore g_Y( )br _{( )0} p U( )br _{( )0} q Z( )br _{( )0} 0 0 = ⋅ + ⋅ = = = − +













− p m r p q br r m p q bm q bm m r r m 0 0 0 1 1 ( )! , ..., , ( )! ( )! for forr r m=











(37) and ( )n

_{(0) 0}

Y

g

=

elsewhere. Putting t = 0 into (34) we obtain 1 0 1 1 1 1

1

(0)

(0)

1

1

1

(0)

(0)

(0)

1

1

1

1

(0)

(0)

(0)

1

1

s Y m br s br s br Y S r bm s bm s bm Y S

s

D V

D g

s

s

s

D g

D

g

D

V

br

s

br

s

s

D g

D

g

D

V

bm

s

bm

− − − − − = − − −

−





_⋅

₌



₋











−





−





=

_

_

_

⋅

−

_

_

_

+

−

− −

















−





−





+

_

_

_

⋅

−

_

_

_

−

− −













∑

1 0 1 1 1 1

1

(0)

(0)

1

1

1

(0)

(0)

(0)

1

1

1

1

(0)

(0)

(0)

1

1

s Y m br s br s br Y S r bm s bm s bm Y S

s

D V

D g

s

s

s

D g

D

g

D

V

br

s

br

s

s

D g

D

g

D

V

bm

s

bm

− − − − − = − − −

−





_⋅

₌



₋











−





−





=

_

_

_

⋅

−

_

_

_

+

−

− −

















−





−





+

_

_

_

⋅

−

_

_

_

−

− −













∑

Using (35), (36), (37) we have

v s

m

r

q

p

br P S s br v s br

r

( )

=



(

)

(

)













_





_











_

 ⋅

(

= −

−

−

)

















+

+

+

⋅













(

⋅

= −

−

−

)

= −

∑

r m m m

p q

q

p p

bm P S s bm v s bm

1 1 0 0 0

(

)

(

) .

This completes the proof.

3. Influence of the dependence on the aggregate claims

distribution and the premium

In this section, influence of the dependence on the aggregate claims distribution and the net premium is presented by numerical examples. Example 1 Consider a portfolio consisting of 80 policies purchased by 40 married couples and additionally by 39 men and 12 women. The probability of natural death for men is 0,00 0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09 0 10 20 30 40 50 independent 0,005 0,01 0,05 0,1 s P(S = s) Fig. 1. The distribution function of S Source: autor’s own study. equal to 0.2 and for women 0.15. The probability of death as a result of the common risk is equal to 0.005, 0.01, 0.05 and 0.1. In Fig. 1. the exact values of the distribution function of S for different values of q₀ can be found. Those values determine the degree of dependence between the risks. For q₀ = 0 we have the case where the risks are independent.

From Fig. 1. one can see that for small values of q₀ (such as 0.005) the curves are very close to the curve in the case of independent risks. For larger values the difference is significant, the distribution function is more flattened and modal values are larger. Example 2 Consider the portfolio consisting of 98 policies. The 20 persons are exposed to the same risk. The probability of natural death is equal to 0.2. The probability of death as a result of the common risk takes value of 0.005, 0.01, 0.05 and 0.1. The aggrega-te claims distribution of S is shown in Fig. 2. 0,00 0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09 0,10 0 10 20 30 40 50 independent 0,01 0,05 0,1 s P(S = s) Fig. 2. The distribution function of S Source: autor’s own study. From Fig. 2. one can see that the curve for small value of q₀ (such as 0.01) is very close to the curve in case of independent risk. We also observe that for larger values of q₀ the distribution function of S became the bimodal distribution. In the next examples there is considered the influence of the dependence on the net premium in the presented models. In the model the net premium for the portfolio is given by 2 1 2 1 1 1 1 1 ( ) m ( i i ) n ( )i m ( )il n ( )i i i m i l i m E S E X X E X E X E X = = + = = = + =

∑

+ +

∑

=

∑∑

+

∑

_, where E X( )_il =E b I( _{il il})=b E I_il

( )

_il =b E_il

(

min(J_il +J₀, )1

)

= =b p Eil

[

0

(

min(Jil+0 1, )

)

+q E0

(

min(Jil +1 1, )

)

]

= =b p E J_il

[

₀

( )

_il +q₀

]

=b p q_il

[

_{0 il} +q₀

]

_{for i = 1, …, m}

and E X( )_i =E b I( _{i i})=b E J_i

( )

_i =b q_{i i} for i = m + 1, …, n. For model (7) the net premium E(S) for portfolio is equal to E S E X E Xm E Xi i m n ( )= ( ) ...+ + ( )+ ( ) = +

∑

1 1 , where E X( )i =E bI( )i =bE I

( )

i =bE

(

min(Ji+J0, )1

)

= =b p E

[

0

(

min(Ji+0 1, )

)

+q E0

(

min(Ji +1 1, )

)

]

= =b p E J

[

₀

( )

_i +q₀

]

=b p q q

[

₀ + ₀

]

for i = 1, …, m, E X( )_i =E bI( )_i =bE J

( )

_i =bq for i = m + 1, …, n. Hence the net premium E(S) depends on the probability of natural death and also probability of death as a result of the common risk. Example 3 Consider the portfolio described in Example 1. The value of the net premium for dependent risks E(S) are compared with the premium for independent model E(S)*_. In Table 1. the net premium and the relative error are given for small values of q₀. The relative error is equal to (E(S)−E(S)*_{) ⁄ E(S)}*_.

Table 1. The relative error for the net premium 0 q p0 E S( ) * * ( ) ( ) ( ) E S E S E S − 0.001 0.999 10.8116 0.11% 0.002 0.998 10.8233 0.22% 0.003 0.997 10.8350 0.32% 0.004 0.996 10.8468 0.43% 0.005 0.995 10.8587 0.54% 0.006 0.994 10.8706 0.65% 0.007 0.993 10.8825 0.76% 0.008 0.992 10.8945 0.88% 0.009 0.991 10.9066 0.99% Source: autor’s own study.

In Fig. 3 the relative error for the net premium is shown for q₀ from 0 to 1. 0% 50% 100% 150% 200% 250% 300% 350% 400% 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 Th e re la tiv e er ro r q0 Fig. 3. The relative error for the net premium Source: autor’s own study. Example 4 Consider the portfolio described in Example 2. The value of the net premium for dependent risks E(S) is compared with the premium E(S)* . In Table 2 the net pre-mium and the relative error are given for small q₀. In Fig. 4 the relative error for the net premium is shown for q₀ varying from 0 to 1. Table 2. The relative error for the net premium 0 q p0 E S( ) * * ( ) ( ) ( ) E S E S E S − 0.001 0.999 19.616 0.08% 0.002 0.998 19.632 0.16% 0.003 0.997 19.648 0.24% 0.004 0.996 19.664 0.33% 0.005 0.995 19.680 0.41% 0.006 0.994 19.696 0.49% 0.007 0.993 19.712 0.57% 0.008 0.992 19.728 0.65% 0.009 0.991 19.744 0.73% Source: autor’s own study.

0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 Th e re la tiv e er ro r q0 Fig. 4. The relative error for the net premium Source: autor’s own study. From Fig. 1-4 we can see that the dependence influences the probability distri-bution of S and the net premium E(S). The value of the net premium increases with q₀ (see Table 1, 2 and Fig. 3, 4). Example 3 and 4 show that for small value of q₀ the relative error does not exceed one percent. If the value of the probability of the natural death is less than the value in example 3 and 4, the relative error is larger. It is shown in the next example. Example 5 Let the aggregated claim amount have form (7). The probability of death as a result of common risk is fixed. In Fig. 5 the relative error for the net premium is shown for the different values of the probability of natural death. 0% 250% 500% 750% 1000% 1250% 1500% 1750% 2000% 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

The relative error

q = 0,2 q = 0,1 q = 0,05 q = 0,01

q0

Fig. 5. The relative error for the net premium with fixed q₀ Source: autor’s own study.

From Fig. 5 we can see that for smaller values of the probability of natural death q the relative errors are larger. For the larger values of probability of death, as a result of the common risk, the difference between the curves is larger.

Literature

Cossette H., Gaillardetz P., Marceau E., Rioux J.,

[1] On two dependent individual risk models,

“Insurance: Mathematics and Economics” 2002,30, s. 153-166. Daykin C.D., Pentikainen T., Pesonen M.,

[2] Practical Risk Theory for Actuaries, Chapman &

Hall, London 1994. Dhaene J., Goovaerts M. J.,

[3] On the dependency of risks in the individual life model, “Insurance:

Mathematics and Economics” 1997, 19(3), s. 243-253. Dhaene J.,Vandebroek M.,

[4] Recursions for the individual model, “Insurance: Mathematics and

Economics” 1995, 16, s. 31-38.

Kaas R., Goovaerts M., Dhaene J., Denuit M.,

[5] Modern Actuarial Risk

Theory, Kluwer Aca-demic Publishers, Boston 2001.

Modele aktuarialne

[6] , ed. W. Ostasiewicz, UE, Wrocław 2000. Ribas C., Marin-Solano J., Alegre A.,

[7] On the computation of the aggregate claims distribution in the individual life model with bivariate

dependencies, “Insurance: Mathematics and Econom-ics”, 2003, 32, s. 201-215.

WPŁYW ZALEżNOśCI NA UbEZPIECZENIA NA żYCIE

Streszczenie: W klasycznych ubezpieczeniach na życie zakłada się, że wielkości szkód są niezależne. Założenie to w wielu sytuacjach jest niewłaściwe. Na przykład długości życia męża i żony są zależne, gdyż dzielą oni wspólne życie oraz mniej lub bardziej narażeni są na to samo ryzyko. W grupie ubezpieczonych osób, które pracują w tej samej fabryce, śmiertelność zależy od pewnych wydarzeń (tj. wybuch, zawalenie hali). W artykule przedstawione będą modele uwzględniające zależność między szkodami. Zbadany zostanie wpływ zależności na rozkład zagregowanych szkód i na składkę netto.

Słowa kluczowe: model indywidualnego ryzyka, rozkład zagregowanej wielkości szkód, wzór rekurencyjny, składka netto.