Dividend problems in the dual risk model · Dividend problems in the dual risk model Lourdes B. Afonso, Rui M.R. Cardoso1 & Alfredo D. Egídio dos Reis2 CMA and FCT, New University

Dividend problems in the dual risk model

Lourdes B. Afonso, Rui M.R. Cardoso1 & Alfredo D. Egídio dos Reis2

CMA and FCT, New University of Lisbon &CEMAPRE and ISEG, Technical University of Lisbon

ASTIN 2011 Madrid

1This work was partially supported by CMA/FCT/UNL, under Financiamento Base2009 ISFL-1-297 from FCT/MCTES/PT

2The authors gratefully acknowledge �nancial support from FCT-Fundação para aCiência e a Tecnologia (Project reference PTDC/EGE-ECO/108481/2008)LRA (FCT, UNL & ISEG, UTL) Dividends Dual Risk Model ASTIN 2011 Madrid 1 / 26

Dual Risk Model

Key reference: Avanzi, Gerber & Shiu (2007).

U(t) = u � ct + S(t), t � 0

Accumulated reserve up to time t : U(t)u : initial reserve, surplus;Individual random gains: Xi _ P(x), E[Xi ] = p1;

Accumulated gains up to t : S(t) = ∑N (t)i=0 Xi , X0 � 0. fXig, i.i.d.

fXig independent of N(t);fS(t), t � 0g :compound Poisson process, parameters λ and p.d.f.p(x);

c : constant rate of expenses, µ = E [S(1)]� c = λp1 � c > 0,λp1 > c .

b � u: (Upper) dividend barrier (absorbing);Ruin barrier (absorbing).

LRA (FCT, UNL & ISEG, UTL) Dividends Dual Risk Model ASTIN 2011 Madrid 2 / 26

Dual Risk Model

U(t) = u � ct + S(t), t � 0


Classical �vs´ Dual


Problem: Dividend calculation

We have 2 absorbing barriers: Dividend barrier b and Ruin level �0�.A dividend can be paid if the process is not ruined.

b is �xed. Once reached a dividend is paid and process re-starts, fromb.

If the process is ruined, the investment is insolvent.

Let δ (> 0) be the force of interest.

Problem: Calculate the discounted future dividends, e.g., Expected.

Dividends can be multiple.

There are results on expected, moments of discounted dividends:Avanzi et al. (2007), Cheung & Drekic (2008), Ng (2009, 2010)

We go further, with a new approach: connection the classical and thedual


Solutions for the dividend calculation

Probability of ruin, with no dividend barrier: known.

Setting a dividend barrier, we evaluate:

Expected discounted future dividend payments:

known, but new approach;higher moments also available;

The probability of getting a dividend payment;Distribution for the amount of a dividend payment;Distribution for the number of dividends (in�nite time).


De�nitions

Let, process free of the barrier, and continue if it is ruined:Time to ruin, from initial surplus x :

τx = inf ft > 0 : U(t) = 0jU(0) = xg ; (τx = ∞ if U(t) � 0 8t � 0)Probability of ultimate ruin

ψ(x) = Pr fτx < ∞jU(0) = xg = e�Ru

where R > 0: λ�R ∞0 e

�Rxp(x)dx � 1�= �cR.

Time to reach an upper level b � x � 0, from x :

Tx = inf ft > 0 : U(t) > bjU(0) = xgProbability of ruin before reaching b: ξ(u, b) = Pr (Tx > τx ) .Probability of reaching b before ruin: χ(u, b) = Pr (Tx < τx )Single dividend amount: Du = fU(Tu)� b ^ Tu < τug. D.f.:

G (u, b; x) = Pr(Tu < τu and U(Tu) � b+ x)ju, b)M: number of dividends to be distributed.


Total discounted dividends

Total discounted dividends: D(u, b, δ)

V (u; b) = E[D(u, b, δ)] = E

"∞

∑i=1e�δ(∑i

j=1 T(j))D(i )

#, 0 � u � b .


Expected discounted dividends

V (u; b) = E[D(u, b, δ)]From Avanzi et al.(2007), solutions from solving,

0 = cV 0(u; b) + (λ+ δ)V (u; b)� λZ b�u

0V (u + y , b)p(y)dy (1)

� λZ ∞

b�uV (u � b+ y , b)p(y)dy � λV (b, b) [1� P(b� u)]

Also, Laplace transforms from there;Instead, we propose solving

V (u; b) = E[D(u, b, δ)] = E

"∞

∑i=1e�δ(∑i

j=1 T(j))D(i )

#n�T(i ),D(i )

�o∞

i=2, sequence of i.i.d. random pairs.�

T(i ),D(i )�d= (Tb ,Db)

Independent of�T(1),D(1)

�. Let

�T(1),D(1)

�d= (Tu ,Du).


Expected discounted dividends

Independence gives

V (u; b) = E

"∞

∑i=1e�δ(∑i

j=1 T(j))D(i )

#=

∞

∑i=1

Ehe�δ(∑i

j=1 T(j))D(i )i

= E�e�δT(1)D(1)

�+E

�e�δT(1)

�E�e�δT(2)D(2)

�+E

�e�δT(1)

�E�e�δT(2)

�E�e�δT(3)D(3)

�+ ...

Now, (Tb ,Db)d=�T(i ),D(i )

�, i = 2, 3, ...; (Tu ,Du)

d=�T(1),D(1)

�,

V (u; b) = E�e�δTuDu

�+E

�e�δTu

�E�e�δTbDb

��1+E

�e�δTb

�+E

�e�δTb

�2+ ...

�= E

�e�δTuDu

�+E

�e�δTu

� E�e�δTbDb

�1�E (e�δTb )

.


2nd moment of discounted dividends

Similarly,

V2(u; b) = E[D(u, b, δ)2] =∞

∑i=1

E[Z 2i ] + 2∞

∑i=1

∞

∑j=i+1

E[ZiZj ]

Zi = e�δ(∑ij=1 T(j))D(i )

∞

∑i=1

E[Z 2i ] = E�e�2δTuD2u

�+

E�e�2δTu

�E�e�2δTbD2b

�1�E (e�2δTb )

∞

∑i=2

∞

∑j=i+1

E[ZiZj ] =E�e�2δTu

�E�e�2δTbDb

�E�e�δTbDb

�[1�E (e�δTb )] [1�E (e�2δTb )]


Discounted Dividend Expectations

We need the expectations E�e�δTuDu

�, E

�e�δTbDb

�, E

�e�δTu

�and E

�e�δTu

�...

Adapt from the classical severity, Dickson & Waters (2004),

φn(u�, b, δ) = E[e�δTuDnu ], n = 0, 1, 2, ..., u� = b� u

ddu� φn(u

�, b, δ) = λ+δc φn(u

�, b, δ)� λc

R u�0 φn(u

� � y , b, δ)p(y)dy

�λc

R ∞u� (y � u�)np(y)dy .

Laplace transform of φn(u�, b, δ) w.r.t. u� ( � 0, extended )

φn(s, b, δ) =cφn(0, b, δ)� λpn

1�pn(s)s

cs � (λ+ δ) + λp(s). (2)

with boundary condition (φn(b, b, δ) = 0) we �nd φn(0, b, δ).LRA (FCT, UNL & ISEG, UTL) Dividends Dual Risk Model ASTIN 2011 Madrid 12 / 26

Probability of a single dividend payment

De�nition

χ(u, b) = Pr [Tu < τu ] and ξ(u, b) = Pr [Tu > τu ] , u � b,

Solutions through Integro-di¤erential equations or Laplace transforms:

Standard approach, (t0 : u � ct0 = 0), 0 < u < b.

ξ(u, b) = e�λt0 +Z t0

0λe�λt

Z b�(u�ct)

0p(x)ξ(u � ct + x , b) dxdt,

di¤erentiating and rearranging (Exact solutions in some cases)

λξ(u, b) + cddu

ξ(u, b) = λZ b�u

0p(x)ξ(u + x , b) dx (3)

ddu

log ξ(u, b) =λ

c

�Z b�u

0p(x)ξ(x , b� u) dx � 1

�Boundary condition: ξ(0, b) = 1.


Probability of a single dividend payment

Laplace transform:Change of variable: z = b� u and E(z , b) = ξ(b� z , b) = ξ(u, b)

λE(z , b)� c ∂

∂zE(z , b)� λ

Z z

0p(z � y)E(y , b) dy = 0.

In E(z , b) extend the range of z from 0 � z � b to 0 � z � ∞,resulting function by ε(z)Compute its Laplace transform, ε(s), [p(s) is transform of p(x)]

ε(s) =cε(0)

cs � λ+ λp(s)(4)

ε(0) = ξ(b, b), [ε(b) = E(b, b) = ξ(0, b) = 1].


Single dividend amount distribution

De�nition

G (u, b; x) = Pr(Tu < τu and U(Tu) � b+ x)ju, b)

Ruin probability ψ(x) = e�Rx can be written as

ψ(u) = ξ(u, b) +Z ∞

0g(u, b; x)ψ(b+ x)dx

= ξ(u, b) + ψ(b)Z ∞

0g(u, b; x)ψ(x)dx

ξ(u, b) = e�Ru � e�Rb g(u, b;R) (5)

Integro-di¤erential equation for G (u, b; x), using usual procedure,

G (u, b; x) =Z t0

0λe�λt

�Z b�(u�ct)

0p(y)G (u � ct + y , b; x) dy

+Z b�(u�ct)+x

b�(u�ct)p(y)dy

�dt.



Rearranging and di¤erentiating.

λG (u, b; x) + c∂

∂uG (u, b; x) = λ

Z b

up(y � u)G (y , b; x) dy

+λ [P(b� u + x)� P(b� u)] .

Boundary condition G (0, b; x) = 0. Similarly,

λg(u, b; x) + c∂

∂ug(u, b; x) =

= λZ b�u

0p(y)g(u + y , b; x) dy + λp(b� u + x) .

Let G(z , b; x) = G (b� z , b; x). Then G(b, b; x) = G (0, b; x) = 0



Use of Laplace transforms

λG(z , b; x)� c ∂

∂zG(z , b; x)� λ

Z z

0p(z � y)G(y , b; x) dy

�λ(P(z + x)� P(z)) = 0Let ρ(z , x) be the function resulting from extending the range of z .Taking Laplace transforms,

λρ(s, x)� c [s ρ(s, x)� ρ(0, x)]� λρ(s, x)p(s)

+λ

�p(s)s� p(s, x)

�= 0 ,

ρ(s, x) =cρ(0, x) + λ [p(s)/s � p(s, x)]

cs � λ+ λp(s)(6)

where

p(s, x) =Z ∞

0e�szP(z + x) dz = esx

Z ∞

xe�syP(y) dy .


Using the Classical Model

Consider the process continuing even if ruin occurs. The process can crossthe upper dividend level before or after ruin.

De�nition(proper) Distribution of the amount by which the process �rst upcrosses b,

H(u, b; x) = Pr [U(Tu) � b+ x ]

H(u, b; x) = H(u, b; x jTu < τu)χ(u, b) +H(u, b; x jTu > τu)ξ(u, b)

= G (u, b; x) + ξ(u, b)H(0, b; x).

Equation above simply means that H(u, b; x) equals the probability ofa dividend claim less or equal than x plus the probability of a similaramount but that cannot be a dividend


Using the Classical Model

Compute now H(u, b; x), through expressions of the severity of ruinfrom the classical model

G �(b� u; x) = G (u, b; x) + ξ(u, b)G �(b; x)

G (u, b; x) = G �(b� u; x)� ξ(u, b)G �(b; x)

For u = b,

G (b, b; x) = G �(0; x)� ξ(b, b)G �(b; x)

g(b, b; x) = g �(0; x)� ξ(b, b)g �(b; x)

Calculate Laplace transforms at R and (5)

ξ(b, b) =

�1� g �(0;R)

�e�Rb

1� g �(b;R)e�Rb

with g �(0; x) = p�11 [1� P(x)] andg �(0;R) = 1

Rp1

�1�

R ∞0 e

�Rxp(x)dx�= c/λp1.


Number of dividend payments

De�nitionLet M : number of dividends to be claimed, or the number of times theprocess upcrosses the upper level b.M follows a zero-modi�ed geometric distribution

Pr[M = 0] = ξ(u, b)

Pr[M = k ] = χ(u, b)χ(b, b)k�1ξ(b, b), k = 1, 2, ...

Hence, the total amount of dividend payments (not discounted) follows azero modi�ed compound distribution.


Illustration

Example

p(x) = αe�αx , x > 0, (α > 0)

E�e�δTuDnu

�=

n!αn

λ

ce�r2u � e�r1u

(r1 + α)e�r2b � (r2 + α)e�r1b

V (u, b, δ) =λ

α

e�r2u � e�r1u(δ� cr2)e�r2b � (δ� cr1)e�r1b

,

where r1 < 0 e r2 > 0 are solutions of the equation (7)

s2 +�

α� λ+ δ

c

�s � αδ

c= 0. (7)


Illustration (cont�d)

χ(u, b) =λ� λe�Ru

λ� αce�Rb; ξ(u, b) =

λe�Ru � αce�Rb

λ� αce�Rb.

G (u, b; x) = (1� e�αx )λ� λe�Ru

λ� αce�Rb.

G (u, b; x)χ(u, b)

= 1� e�αx .

E�e�δTuDnu jTu < τu

�= E (Dnu jTu < τu)E

�e�δTu jTu < τu

�






λ� αce�Rb.

G (u, b; x) = (1� e�αx )λ� λe�Ru

λ� αce�Rb.

G (u, b; x)χ(u, b)

= 1� e�αx .




�






λ� αce�Rb.

G (u, b; x) = (1� e�αx )λ� λe�Ru

λ� αce�Rb.

G (u, b; x)χ(u, b)

= 1� e�αx .




�



Let λ = α = 1, c = 0.75 and δ = 0.02

u b2 3 6 10 30 40

1 2.0111 2.4873 3.3693 3.1373 0.91968 0.489643 5.6241 7.6184 7.0938 2.0795 1.10715 10.271 9.5638 2.8036 1.492610 14.261 4.1805 2.225715 5.7702 3.072020 7.9130 4.2129

Table: Values for V (u, b, 0.02) for u =1, 3, 5, 10, 15, 20; b =2, 3, 6, 10, 30, 40



Let λ = α = 1, c = 0.75 and δ = 0.02

u b2 3 6 10 30 40

1 0.45098 0.37549 0.27291 0.20558 5.7974� 10�2 3.0865� 10�23 0.84904 0.61709 0.46484 0.13109 6.9791� 10�25 0.83195 0.62669 0.17673 9.4091� 10�210 0.26352 0.1403015 0.36373 0.1936520 0.49881 0.26557

Table: Values for E�e�δTuDu

�for u =1, 3, 5, 10, 15, 20; b =2, 3, 6, 10, 30, 40



Let λ = α = 1, c = 0.75 and δ = 0.02

u b2 3 6 10 30 40 ∞

1 0.46097 0.39148 0.31549 0.29126 0.28348 0.28347 0.283473 0.87299 0.70353 0.64950 0.63214 0.63212 0.632125 0.90276 0.83342 0.81115 0.81113 0.8111210 0.99084 0.96436 0.96433 0.9643315 0.99330 0.99326 0.9932620 0.99876 0.99873 0.99873

Table: Values for χ(u, b) for u =1, 3, 5, 10, 15, 20; b =2, 3, 6, 10, 30, 40


References

Avanzi (2009). Strategies for dividend distribution: A review, NorthAmerican Actuarial Journal 13(2), 217-251.Avanzi, Gerber & Shiu (2007). Optimal dividends in the dual model,Insurance: Mathematics and Economics, 41(1), 111-123.Cheung & Drekic (2008). Dividend moments in the dual risk model:exact and approximate approaches, ASTIN Bulletin 38 (2), 399-422.Dickson & Waters (2004). Some optimal dividends problems, ASTINBulletin 34 (1), 49-74.Gerber (1979). An Introduction to Mathematical Risk Theory, S.S.Huebner Foundation for Insurance Education, University ofPennsylvania, Philadelphia, Pa. 19104, USA.Ng (2009), On a dual model with a dividend threshold, Insurance:Mathematics and Economics 44, 315-324.Ng (2010). On the upcrossing and downcrossing probabilities of adual risk model with phase-type gains, ASTIN Bulletin 40 (1).281-306.


Documents

Dividend problems in the dual risk model · Dividend problems in the dual risk model Lourdes B. Afonso, Rui M.R. Cardoso1 & Alfredo D. Egídio dos Reis2 CMA and FCT, New University