Accuracy formulas of ruin probability calculations for discrete time risk model with dependence of financial and insurance risks Текст научной статьи по специальности «Математика»

ACCURACY FORMULAS OF RUIN PROBABILITY CALCULATIONS FOR DISCRETE TIME RISK MODEL WITH DEPENDENCE OF FINANCIAL

AND INSURANCE RISKS

G.Sh. Tsitsiashvili

Institute of Applied Mathematics, Far Eastern Branch of RAS, Vladivostok, Russia

e-mail: guram@iam.dvo.ru

ABSTRACT

In this article discrete time risk model with heavy tailed losses distribution and dependence between financial and insurance risks is considered. It is shown that known asymptotic formulas work with good accuracy for sufficiently large arguments. Direct methods based on calculation of ruin probability by solution of appropriate integral equations demand large volumes of calculations and so work for sufficiently small arguments. Fast and accuracy algorithms, based on approximation of loss distribution by mixture of exponential ones, to calculate ruin probability in this interval are developed. This approximation of considered model is based on continuity theorems and analog of Bernstein theorem in L metrics.

INTRODUCTION

Discrete time risk model with dependence between financial and insurance risks is considered. In modern period of strong economical crisis such dependence may be recognized easily in different large anthropogenic catastrophes. So a problem to analyze asymptotically this dependence influence on the ruin probability is actual now.

This problem is discussed in risk theory and in queueing theory (Asmussen & Bladt 1996, Asmussen 2000, Feldmann & Whitt 1998, Dufresne 2005, Albrecher, Teugels & Tichy 2001). For heavy tailed distributions of losses it is shown that known asymptotic formulas (Embrechts & Veraverbeke 1982, Embrechts, Kluppelberg & Mikosch 1997, Tang 2004) work with good accuracy for sufficiently large arguments (Asmussen 2000, Kalashnikov 1997, Kalashnikov 1999). Direct methods based on calculation of ruin probability by solution of appropriate integral equations demand large volumes of calculations (Skvarnik 2004) and so work for sufficiently small arguments. As a result an interval of mean arguments appears. This interval is interesting practically and in it asymptotic formulas still do not work and direct methods already do not work. So it is interesting to process sufficiently fast and accuracy algorithms to calculate ruin probability in this interval.

To solve this problem an analogy with the queueing theory with an approximation of heavy tailed distribution by a mixture of exponential ones is used. It is known that waiting times in an one server queueing system which creates the Lindley chain coincide by a distribution with maximums of sequential sums and are continuous for fluctuations of distributions (Borovkov 1971, Zolotarev 1976). Analogously in discrete time risk model with dependent financial and insurance risks a finite interval ruin probability coincides with tail of distribution of some Markov chain.

In this paper finite interval ruin probability is represented by sum of exponents with unknown coefficients. To find these coefficients some recurrent procedure is suggested. It allows to consider risk model with exponential, Pareto, Weibull and some other loss distributions. We consider special model of insurance and financial risks dependence based on suggestion that a financial risk has a finite number of meanings and for each meaning an insurance risk has its own distribution.

Asymptotic formulas for the ruin probability in a case of independent financial and insurance risks have been obtained in (Tang & Tsitsiashvili 2003 (Stochastic Processes Applied)). More complicated cases with special restrictions on insurance risks dependence are considered for example in (Tang & Tsitsiashvili 2003 (Extremes), Tang & Wang 2010). First asymptotic formulas for dependent financial and insurance risks are obtained in (Tsitsiashvili 2010).

1 PRELIMINARIES

Consider recurrent discrete time risk model (with annual step) with initial capital x, x > 0, and nonnegative losses Zn, n = 1,2..., P(Zn < t) = F(t):

^0 = x, Sn = Bn Sn-1 + An, n = 1,2,... (1)

Here income An, n = 1,2,..., to end of n-th year is defined as difference between unit premium sum and loss An = 1 -Zn. Assume that Bn > 1 is inflation factor from n-1 to n year, n = 1,2,... In (Norberg 1999) Xn = -An is called insurance risk and Yn = B~l is called financial risk. Suppose that {(An,Bn), n > 1} is sequence of independent and identically distributed random vectors (i.i.d.r.v/s). In this model with initial capital x ruin time is defined by formula

t(x) = inf {n = 1,2,...: Sn <0| S0 = x} and finite time ruin probability by formula

y/n (x) = P(t(x) < n).

So the sum Sn money accumulated by insurance company to n-th year end satisfies recurrent formula

S0 = x, Sn = xflBj +¿4 njBj, n = 1,2,..., (2)

j=1 i=1 j=i+\

n

where ^ Bj = 1 by convention. According to the notation above we can rewrite the discounted

j=n+1

value of the surplus Sn in (2) as

S0 = x, Sn = SnnYj = x-±Xtf\Y} = x-Wnl.

j=1 i= j=1

Hence we easily understand that for each n=0,1,...

Vn(x) = P(Un > x), Un = max {0, max Wk}, U0 = 0. (3)

V 1<k<n )

Define another Markov chain as

V = 0, Vn = Yn max(0, Xn + V^), n = 1,2,... (4)

In (Tang & Tsitsiashvili 2003 (Stochastic Processes Applied), Tsitsiashvili 2010) the following statement is proved.

Theorem 1. The formula \f/n(x) = P(Vn > x), n = 1,2,... is true.

Suppose that Q = {1,...,m} and introduce m-dimensional vectors 1 = {£1q,...,Smq} where Sij is Kroneker symbol and

R = (r„...,rm), K = (k„...,km), r > 0, k e {0,1,...}, i = 1,...,m,

and denote

RK =n ,\K\=Z kq .

qeQ qeQ

Redefine the function e ' so that for t < 0 we have e 1 = 1 and for t > 0 this function is defined as

(0, t > 0,

usual. Introduce the function E(t) = j 1 ^ < 0 Suppose that i.i.d.r.vectors (Yn, Zn), n > 1, have following distributions

CYi"1 = r, Zn = zq) = pq, P(Zq > t) = Fq (t), n > 1, q e 0. Consider disturbed Markov chain Vn, n > 0, so that

V0 = 0, Vn = Yn max(0, J + n = 1,2,... (5)

have following distributions

(Yn-1 = rq , Z = Z% n ) = Pq , P(Z n > t) = Gq (t), n > 1, q E Q. Denote <pn(x) = P(Vn > x).

2 RECURRENT ALGORITHMS OF RUIN PROBABILITY CALCULATIONS

Theorem 2. Suppose that there are real numbers aqi, q e Q, i = 1,...,l, and pq > 0, q e Q, _ i

2 Pq = 1, so that Gq (t) = 2 aqi exp(-At), q e Q, n > 1, and

i=1

R*A , 1 < i, j < l, | * | > 1. (6)

q qEQ

Then there are real numbers Bni, i = 1,...,l, 1 <| K |< n, which satisfy for n > 1, i = 1,...,l initial conditions

B,q = Pq«qi exp(-/ ), q E Q, (7)

and recurrent formulas

B>+u = pq z exp(-Ai ) + PqB« exp(-/ ), q e Q, (8)

1<|K|<n j=1 R Lj - A

i BK—« LL

=—z ^ >0)Pq z pK—1q «* L j exp(-RK—1 / ), 1 <| K | <n + 1, (9)

qEQ j=1 R L - lL

so that

where

<p. (t) = z z BK exp(-RK Li t) + B0E(t), s > 0, (10)

1<|K|<s i =1

b;=1- z zbk . (11)

s

1<|K|<s i=1

Proof. If random variables n are independent and

P(£ > t) = \ exp(-/t), P(n > t) = \ exp(-At), A, /u> 0, A * /, then it is easy to obtain that

P(^+n> t) =/exEi-^l-Ai£xEi-u/l. (12)

/- A

Calculating

p (y (Z -1) > t) = 2 Pq P(Z q -1 > Rlqt) = 2 Pq P(Z q > Rlqt+1) =

qEQ qEQ

= zpq zai exP(-^-(Rlqt +1)) = zpq zai exP(-4)exp(-A!Rlqt)

qeg i=1 q^6 i=1

we obtain that

V(t) = zPq exp(-4)exp(-ARlqt) + B?E(t) = zzBi exp(-4Rlqt) + B1(t).

qe6 '=1 9^6 ' =1

So the formula (10) is true for 5=1 with the initial conditions (7) and the equality (11).

Suppose that the formula (10) takes place for s=n and using the formula (12) calculate

P(Yn+1 (V + Zn+1 -1) > t) = z PqP(Vn + Z9+1 > rqt +1). As

q^6

1 1 BK a„

P(Vn + Zn+1 > X) = z zzTTTT (RKA' exp(-V) exp(-RK\x)) + B"n zaqi exp(-Ax)

1<|K|<n i=1 j=1 R Ai - Aj i=1

1 so

P(Vn + Zn+1 > rqt +1) = z zZRrrr [RK^ expH)exp(-A;.Rlqt)-A; exp(-RKAt)exp(-RK+1qt)] +

1<|K|<n i=1 j=1 R Ai - Aj

1

1

+B0Zaqi exp(-A)exp(-ARlqt).

i=1

Consequently we obtain

' ' BKa

vn+1(t) = z Pq z zzuAV [RKA exp(-Aj) exp(-AjR lqt) -A, exp(-RKAi )exp(-RK+1)]-

qe6 1<|K|<n i=1 j=1 R A - ^

+ZPBZaqi exp(-A)exp(-ARlqt) + B^Etf) =

qe6 i=1

= zPqz z z^KAy RKAj exp(-A)exp(-ARlqt)-

qe6 i=1 1<|K|<n j=1 R A - A

1 1 DK'

-zPqz z zRay Aj exp(-RKA)exp(-RK'+1 At)] +

qe6 i=1 1<|K'|<n j=1 R A - Aj

+zPqB0±aqi exp(-Ai)exp(-ARlqt) + B°n+lE(t) = z zB+u exp(-RKAt) +B„0+1 E(t) =

qe6 i=1 1<|K |<n+1 i=1

= zzB+u exp(-R A) + z zB+1, exp(-RKAt) +B0+1 E(t).

qe6 i=1 2<|K|<n+1 i=1

So the formula (10) is true for 5=n+1. Here for i = 1,..., 1, 1 <| K |< n +1 we have the recurrent formulas (9) and for i = 1,..., 1, | K | = 1 the recurrent formulas (8) and for B°+1 the equality (11). The theorem is proved.

3 CONTINUITY OF RISK MODEL IN Li METRICS

In the sequel we assume that

m m

(Yn, Zn) = zI (in = qXr;1, Fq-\rnn)) , (Yn, Zn) = z i (n = ?)(r;\ Gq->n)). (13)

q=1 q=1

Here i.i.d.r.v's a>n, n > 1, are uniformly distributed on interval [0, 1], F_1(0), 0 < 0 < 1, is inverse to d.f. F(t) function. Then using uniform metrics

p( F, G) = sup| F (x)- G( x)|

and results on stability of queueing systems (Zolotarev 1976) it is simple to obtain following statement.

Theorem 3. For fixed n > 0 inequality

p(p„,p„) < np(F, G) (14)

is true.

Say that distribution density f (t) concentrated on [0, ro), is absolutely monotone if it has derivatives of all orders and (-1)k f(k)(t) > 0 for all t > 0 and k > 1. Example of such distribution is Pareto distribution satisfying equality F(x) = (1 + bx) a, x > 0. From Bernstein theorem (Feldmann 998) it is known that for d.f. F with absolutely monotone density there is sequence of d.f/s represented as sums of exponents

's

Fs (x) = 2 Pi (1 - exp(-4-x)) , x > 0, s > 0,

i=1

where 0 <4., psi <ro Ps1 +...+PsS =1 and p(F,Fs) —0, s — ro.

Theorem 3 and Bernstein theorem allow to construct approximative algorithm for a calculation of ruin probability. But linear by n upper bound in (14) is not convenient for this aim. So we begin to reformulate these results in terms of L metrics. Denote EYn = a—1 and introduce the metrics Lj(F, G) between d.f/s F, G as follows

ro

L1(F, G) = f | F(t)- G(t)| dt. (15)

—ro

Theorem 4. If 5 = maxL(F ,G ) and a >1 then

qeQ q q

£

Lx(^n) < —-. (16)

a-1

Proof. From the formulas (4), (5) we have that E | V0- V01= 0 and

E | Vn - Vn | = E21 (in = q)r— | max(0, VH x + Fq-1 (oH )-1)-max (0, V—, + Gq 1 (©n )-1) | <

qeQ

< E 21 (in = q)r— (| Vn—! — Vn—1 | + | Fq-1 (®n) — Gq-1 (®n) | ) =

qeQ

= 2 <Pq (E | Vn—! — Vn—! | + E | Fq—1 (®n) — Gq-1 (®n) |) = E | Vn—1 — Vn—J+5, h < 1.

qeQ a

Consequently an induction by h gives the formula

E|Vn — Vn |< 552a-k ^^.

k=1 a-1

As the minimum of the complex probability metrics E | Vn — Vn | by all joint distributions which conserve marginal distributions of r.v/s Vn, Vn is Lj(^n) (Zolotarev 1976) so from Theorem 1

we obtain the inequality (16). The theorem is proved.

It is easy to obtain from (Tsitsiashvili 2004, Kalashnikov & Rachev 1988) that in conditions

ro

a > 1, max f F (t)dt = C < ro (17)

geQ 0 q

there is nonincreasing function y/(t) so that y/(0) = 1, y/(t) ^ 0, t ^ ro and

lim wn (t) = V(t), t > 0.

n—>ro

Indeed from Theorem 1 and the formula (3) the sequence \f/n (t), h > 0, satisfies the inequalities y/n+l(t) > \f/n(t), h > 0, and so it has the limit y/(t). Choosing r.v. Vro so that P(Vro > t) = y/(t),

t > 0, and applying Theorem 4 proof to the sequence EVn, n > 0, it is possible to obtain the inequality

C

EVX<--<ro

a -1

and consequently y/(t) — 0, t — ro. Theorem 5. If the conditions (17) are true then

li(¥,¥„) < ( C n-i , n > 0. (18)

(a -1) a

Proof. For n=1 the formula (18) is true. Prove the formula (18) using an induction by n. Suppose that (18) takes place for some n > 0. Introduce the following joint distribution of r.v/s Vn, Vro

which conserves their marginal distributions Vn = ^(0), VrD = y~x(a>). Here r.v 0 is independent

on r.v/s Vn, Vro and has uniform distribution on the interval [0, 1] so E | Vn - Vro | = L1(pn,p). Then

for r.v/s Zn+1,Yn+1 independent on r.v/s Vn, Vro we have the equalities

(d)

Yn+1 max(0, VH + Zn+1 -1) ^ Yn+1 max(0, V„ + Z^ -1) = V„. So from minimal property of metrics L1 we obtain using mathematical induction by n that

L (wn+1 , p) < E | Vn+1 - VJ= E | Yn+1 max(0, Vn + Z^ -1) - Y^ max(0, Vx + Z+1 -1) | <

< EYn+1E | Vn - Vro |= a "1E | Vn - Vx | < a - nE | V - VJ < ^.

a

The formula (18) is true. The theorem is proved.

Denote h(s) = inf(n:L{w,Vn)<s) then from Theorem 5 we have the inequality

h(s) < inf(n: R-n+%(p,y/x) < s)

and so

. . „ lnL1(w,w1) - lns , .

h(s) < 2 +--= h1 (s). (19)

ln R

Remark 1. The formula (19) allows to establish that if L1(p,pn(s)) < s then it is enough to find \f/n, 1 < h < h1(s). From Theorem 2 we obtain that to calculate (pn(t) it is necessary O(nm+l') arithmetical operations for h ^ro. So to find <pn(s)(t) we need O(| ln s^1) arithmetical operations for s —> 0.

Suppose that the condition (6) of Theorem 2 is not true then it is possible to approximate Gq (t), q e Q, in metrics L1 so that the condition becomes true. We formulate this statement in the

following way.

4 SMALL DENOMINATORS PROBLEM

Suppose that the condition (6) of Theorem 2 is not true and so we deal with zero denominators in recurrent formulas (8), (9). Then it is possible to approximate Gq (t), q e Q, in

metrics L1 so that the condition becomes true. We formulate this statement in the following way. Theorem 6. Assume that for some 8 > 0 positive numbers A1,...,A1 satisfy the condition

| -A} | > 38, 1 < i * j < l. Suppose that rq = tq / T where tq, T are coprimes and Tis prime, q e Q. Then for any s > 0 there are positive and rational numbers A1,...,Al so that

| At - % | <s, A * RK Xj, 1 < i * j < /,| K | > 0. Proof. Fix s, 0<s<5. There are integers N, s\,...,s{ so that

1S

■< —,

4 - ^

! N

S <—. 2

NT 2'

Choose A = (s.T +1) /NT then | Ai — A | < s and s.T +1, T are coprimes, 1 < i < l, so rational number

nkq sTT +1

tq T\K\ qeQ 1

can not be integer and consequently

sT +1

s.T +1 tl'-^KT, 1 < i * j < l,| K|> 0, K = (k,..., km).

qeQ 1

The theorem is proved.

Remark 2. Fix s > 0. If rq > 0 and rq is noninteger then there is prime T and rational noninteger number rq = tq / T so that | rn — r* | < s, q e Q.

q q \ q q \ ■> 1

Introduce Markov chain V* = 0, V; = Y* max(0, In + VH—1), h = 1,2,..., (pn (x) = P(V; > x).

Theorem 7. Suppose that | rq — rq* | < s, L1(Fq, Gq) < s, q e Q. If the condition (17) is true and 1/ a < d < 1, 0 <s< d — 1/ a then

r / sD ^ ^ aC + ad — 1

L>(p„) -, h > 0, D =---<ro. (20)

1 — d a — 1

Proof. From Theorem 7 condition we have that EZn < C + d — 1/ a and from Theorem 5 proof we obtain EVro < (C + d — 1/ a) /(a — 1). Assume that P(Yn = rq, Yn* = r*) = pq, q e Q, then

e | Vn+1 - c |< EYn+1 e | Vn - v; |+e | Yn+1 — yh+1 | (v;+Zn+1) < +

a

+s(C + d — 1/a + EVn + E | Vn — V* |) < E | Vn — V* | d + s(C + d — 1/a + EVro) < E | Vn — Vj | d + sD. Using mathematical induction by n and minimal property of metrics L1 we obtain the formula (20).

5 BERSTEIN THEOREM IN L1 METRICS

Bernstein theorem allows for any q e Q to approximate d.f. Fq (t) by a mixture of exponential distributions in uniform metrics. But we need analogous approximation in L1 metrics. Suppose that d.f. F (t), F (t) = 1 — F (t) concentrated on [0, ro) has mean

M = J F (t)dt <œ (21)

and continuous positive density f (t ) so that for any T > 0

inf f (t) = — > 0. (22)

0<t <Tjyj A(t )

Lemma 1. If d.f. F satisfies the conditions (21), (22) then for any s > 0 it is possible to choose discrete d.f. Gn with finite number n of positive atoms so that L1(F, Gn)< 2s. Proof. Fix positive s and using the condition (21) find Ts so that

J F (t)dt

< s.

Using the condition (22) define integer n so that

A(Ts)

< s

n

(23)

(24)

and put v = F(Ts). Define t, 1 < i < n, from the equalities

F(t.) = i(1-V),1 <i <n, F(tn) = F(Ts) = 1 -v. n

Suppose that discrete d.f. Gn satisfies equalities

Gn (t ) =

0, 0 < t < tj, F(t2), tj < t < t2, F(t3), t2 < t < t3,

(25)

F (tn), tn_i < t < T„ 1, TE < t < ro.

Using the formulas (15), (23) - (25) it is easy to obtain the inequality L1(F, Gn)< 2s. The lemma is proved.

Theorem 8. If d.f. F satisfies the conditions (21), (22) then for any s > 0 it is possible to choose d.f. Rn (t) concentrated on [0, ro) with tail

F(t) = Z ai exp(-ô;t), t > 0, - œ < ai < œ, 0 < bi < œ,

(26)

so that Zj( F, )< 4s.

Proof. From Lemma 1 it is easy to obtain that Gn (t) is probability mixture of point distributions

Gn (t) = —[21(t - O + £ 1(t - tr)] + v1(t - Ts)

n ¿=2

where 1(t - u) is d.f. concentrated in real point u.

Fix s > 0 and natural k. Following (Dufresne 2005, Ko & Ng 2007) let Erl(m,A) denote the Erlang distribution with density function

Amtm-1 exp(-At) t > 0 (m -1)! , ,

which describes the sum of m independent r.v/s with exponential distribution with parameter A. Then Erl(m,A) has mean m/A and variance m/A2. To approximate a degenerate distribution at k > 0 we consider Erl(m, m / k) and let m tend to infinity. So we may choose ms so that variance of d.f. Hk = Erl(m, m /k)is smaller than s2and consequently Z-j(1(t - k), Hk (t)) <s. If Erlang d.f/s H1 (t),..., Hn (t) satisfy the inequalities

L (1(t -1!), Hx (t)) < s,..., A (1(t - tn), Hn (t)) < s

then L1 (Gn, Qn) < s where

1 — v | n

n-j

Qn (t) =-[2 Hj (t) + Z Hi (t)] + v Hn (t).

i=2

i=j

Following (Ko & Ng 2007) let are i.i.d. r.v/s with exponential d.f. and with

m

parameter X . Then random sum has d.f. Erl(m, X). Suppose that independent r.v/s £,...,£m

i=1

with exponential distributions and with parameters X1,...,Xm so that

z

1 _ 1 x ~ T

<e, X * X , 1 < i * j < m,

and there are i.i.d. r.v/s a>n, n > 1, uniformly distributed on interval [0,1] so that

ln 0), „ ln 0), 1

n=—X, £=—X, 1 <i <m

Denote S (t) = P £ < t I. Then we have the inequality

E

i=1 i =1

<z e | n

If we replace in last inequality complex probability metric E | X - 71 by its minimum L1(P(X < t),P(7 < t)) among all joint distributions which conserve marginal distributions of P( X < t), P(7 < t) then we obtain L1(Erl(m, X), S) <s. Using the formula (12) it is easy to represent d.f. S (t) in the form (26).

So it is possible to find d.f. S concentrated on [0, ro) with the tail S(t) which has the form (26) so that L1(Erl(m, X),S) <s. If d.f/s S1(t),...,Sn(t) which has the form (26) satisfy the inequalities

Lj( Hj, Sj) <s,..., Lj(Hn, Sn) <s.

Then L1(Qn, Rn) <s where

Rn (t) = — [2 Sj (t) + £ S, (t)] + v Sn (t)

n i=2

and d.f. Rn (t) satisfies the condition (26) also.

Consequently from Lemma 1 we obtain the inequalities

Lj(F,Rn) < Lj(F,Gn) + Lj(Gn,Qn) + Lj(Qn,Rn) < 4s.

The theorem is proved.

The author thanks A.A. Novikov for a large help in a formulation of a problem. This paper is supported by Far Eastern Branch of Russian Academy of Science, projects 09-1-P2-07, 09-1-OMN-07.

i=1

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