Algorithmic problems in discrete time risk model Текст научной статьи по специальности «Математика»

ALGORITHMIC PROBLEMS IN DISCRETE TIME RISK MODEL

G.Tsitsiashvili

640041, Russia, Vladivostok, Radio str. 7, IAM, FEB RAS, e-mail: [email protected].

ABSTRACT

In this paper we consider some algorithmic problems which appear in a calculation of a ruin probability in discrete time risk models with an interest force which creates stationary and reversible Markov chain. These problems are connected as with a generation of the Markov chain by its stationary distribution so with a calculation of the ruin probability.

Keywords: ruin probability, transportation problem, asymptotic formula, enumeration problem

1. INTRODUCTION

In this paper we consider some algorithmic problems which occur in a calculation of a ruin probability in discrete time risk model with an interest force which creates stationary and reversible Markov chain. Such model of the interest force is suggested by A.A. Novikov. Algorithmic problems are connected as with accuracy calculation of the ruin probability so when we deal with its asymptotic analysis. First numerical experiments show that without a solution of these algorithmic problems it is impossible to construct programs of numerical calculation of the ruin probability.

Markov chain generation reduces to a definition of permissible solutions of appropriate transportation problems. The ruin probability calculation is connected as with a definition of special sums of exponents so with a calculation of ruin probability asymptotic. These procedures also need to solve some auxiliary algorithmic problems: of convenient designations and symbolic calculations, some enumeration problem and so on. A specific of these problems is that primitive variants of their solution have very high complexity. Moreover some calculation procedures can not be realized without a solution of these problems.

2. 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) :

S0 = X , Sn = BnSn-1 + An , n = l2^ (1)

Here annual 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 [1] Xn =-An is called insurance risk and Yn = B-1 is called financial risk. In this model with initial capital x ruin time is defined by formula

t(x) = inf {n = 1,2,...: Sn < 0S0 = x}

and finite time ruin probability y (x, n) - by formula

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

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

S0 = x ,Sn = xllBj +ZA, n Bj , n = 1,2..., (2)

j=i i=i j=i+i

where n "=n+1 = 1 by convention. According to the notation above, we can rewrite the discounted value of the surplus Sn in (2) as

&0 = x, S n = Sn n Yj = x -ix, UY, = x - Wn . j=i ,=i j=i

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

y(x,n) = P(Un > x), Un = max{o,max Wk}, Uo = 0. (3)

I i<k<n )

Suppose that the sequence {Yn,n > 0} is stationary and reversible Markov chain with state set [rq~\q e Q}, Q = {i,...,m} consisting of different positive numbers and transition matrix [tc^I , . It means that the following formulas are true

P (Yn = rql )= Pq , 0 < Pq , i Pq = ^ Pq'^q'q = Pq ^qq' , q, 4 6 Q , n > 0

qeQ

and consequently [2,Theorem 2.4]

(d)

(Yi,..., Yn ) = (Yn ,...,Yi), n > i. (4)

Assume that the random sequence {an, n > 0} consists of independent and identically distributed random variables (i.i.d.r.v/s) with uniform distribution on interval [0,i]. Suppose that random sequences {Yn,n > 0}, {an,n > 0} are independent. Introduce distribution functions (d.f/s) Fq, q e Q

and designate F_i (ra), 0 < co < i, inverse function to distribution function F(t), -<x> < t < <x>. Denote Zn = FY-1 (ran), n > 0, then from the formula (4) we obtain the formula

(d)

((Xi,Yi),..., (Xn ,Yn )) = ((Xn ,Yn ),..., (Xi, Yi)), n > i. (5)

In such a way it is possible to introduce dependence between financial and insurance risks provided financial risks create stationary and reversible Markov chain. Define another random sequence

= 0, Vn = Yn max (0, Xn + ) , n = i,2,.... (6)

Using recurrent formula (6) we introduce Markov chain(Yn,Vn), n = i,2,... and designate

yn,q (x) = P (Yn = r~\Vn > x), q e Q , x > 0, n > 0.

Theorem 1. The formula

yn (x) = i yn,q (x), n = 0,i,..., x > 0, (7)

qe

is true.

Proof . The result (7) is trivial for the case when n = 0 . Now we aim at (7) for each n = 1,2,.... Let n > 1 be fixed. In view of the equality (5) we replace Xi and YJ in Un respectively by Xn+1-i and Yn+1_ j in deriving the following relations:

f k i ] (d) r k i

Un = max \ 0, max 2 Xi n Y \ = max \ 0, max 2 Xn+1_, n Yn+i_ , I 1< k <ni=1 j=1 J [ 1<k <n/=1 j=1

= maxf 0, max 2 X* n Y* \ = maxf 0, max 2 X.. n Y.. I. (8)

\<k<nf=n+1—k ' j*=i* j J [ 1<k* <n,*=k

If we write the right-hand side of (8) as Vn, which satisfies the recurrence equation

Vn* = Yn max (0, Xn + ), n = 1,...,

which is just the same as (6). So we immediately conclude that Vn* = Vn for each n = 1,.... Finally, it follows from (8) that (7) holds for each n = 1,.... This ends the proof of Theorem 1.

3. RECURRENT ALGORITHMES OF RUIN PROBABILITY CALCULATIONS

Introduce m -dimensional vectors 1q =(¿1, ,...,Sm, ) where S,j is Kronecker symbol and R = ( 1,...,rm) , K = (ki,...,km), rt > 0, k, e {0,1,...}, i = 1,...,m ,

and denote

RK = n rk , |K| =2 kq .

qeQ qeQ

Redefine the function exp(_t) so that for t < 0 we have exp (_t) = 1 and for t > 0 this function is defined as usual. Introduce the function

- (t )={:•; > 00.

Suppose that

Fq (t) = 2 aq,i exp (_X,t), n > 1, t > 0, i=1

with

l

_<>< a i <<, i = 1,..., l, 2 ®q,i = 1, q e Q .

i=1

Theorem 2. Suppose that

RK X, ,1 < i, j < l, 1 < |K| . (9)

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

B1,',q = pqaq,exp(—X'.), B,!q = 0, q, q' É g, q * q'. (10)

and recurrent formulas for q e g :

B1'

n+1.

,',q ^ pq'nq ,q

q'eQ

I BK,qa ^ ^ n,j,q '

1<| k| <n j =1RK X j — A.

RK X, exp (—X' ) + B„Vq,' exp (—X. )

Bu i, q = q * q', (ii)

i bk-iq a X

Bj = -I Pqnq.ql ( >0 ) I jJ J exp (-RK-1q Xt ), i < |K| < n + 1, (12)

q'eQ J=1R q Xi -XJ

so that

i

I

1<|K|<s i=1

where

Vs,q(t)= I T.B£i,q exp (-RK X,t) + B0qE (t), s > 0, (13)

B0,q = Pq - i I.B?M . (i4)

i<|K\<s i=i

Proof . If positive random variables ¿¡,n are independent and

P (%> t) = exp (-/), P (n> t) = exp (-At), A,/> 0, A^/, then it is easy to obtain that

P(Z + n> t ) = /exp (-At )-Aexp ). (i5)

/- A

Calculating for t > 0

P (Yi = rq"\Yi (Zi - i)> t)) PqP (F- (ra )-i > Rlqt ) = PqP |( F- (ra, )> Rlqt + i) =

Pq ¿«q,i exp(--A■ ((qt +1)) = Pq iaq,i exp(-A,)exp(-A,Rlqt)

we obtain that

W,q (t) = Pq£aqJ exp(-X,)exp(-AlRlqt) + B°qE(t) = IB1,qexp(-A,Rlqt) + BlqE(t) .

i

I

i=1

So the formula (13) is true for s = 1 with the initial conditions (10) and the equality (14). Suppose that the formula (13) takes place for s = n and using the formula (15) calculate

Wn+1, q (t ) = P (Yn+1 = r-1, Yn+1 (Vn + Zn+1 -1) > t ) =

= P (( = r-\Vn + F- K+1 )> q +1)= I pq, nqqp (Yn+1 = r-\Vn + Fq-1 (an+1 )> q +1

q'eQ

As for x > 0

P (Yn = rq-1,Vn + Fq- (Cn+1 )> x ) = = I I I BKXaXJ (RK Xi exp (-| X jX ) - X j exp (-RK X,x)) + B»n4 ^ exp (-X,x),

1<|k|<ni=1 j=1RK x, -XJ i=1

so for t > 0

P(Yn = rq-1,Vn +Fq-1 (®n+1)> rqt + 1) = I I ^ ^XT^jjn (t) +

1<|K|<ni=1J=1RK Xi -XJ

+Blq i a^. exp(—X'.x)exp(—R1 X)

with

AKj,q,n (t) = RKA exp (_Aj (rqt +1)) _ A exp (_RK A, (q +1)) = = RKA exp)exp(-A^t) _ A, exp(_RkA, )exp(_RK A,q) = = RKA exp(_Aj )exp(_AR^t) _ A} exp(_RkA, )exp(_RK+1q Att) = Consequently we obtain for t > 0

i i BK a

Wn+iq (t)= 2 [pq'^q-q 2 "f T [RK A exp (_Aj) exp (_AjRlqt)

q eQ 1<|K <n i=1 j=1R a, _Aj x '

-Xj exp(—RKX' )exp(—RK+1qXtt)]+Bn0q i a^ exp(—X. )exp(—RlqXtt)

= i Pq' *q',q i ii

q' eg

l l aq,' RK Xj exp (—X' )exp (—XRV)-

1<|k <n '=1 j=1R Xj — X'

i i JfOr Xj exp (—RKX' )exp (—RK+1qX.t) +

'=1 j=1R X' —Xj v '

+ i Pqnq,q + Bn0qiaqJexp(—X.)exp(—RlqX.t) = i iB^.,qexp(—RKX/) =

— i Pq' nq' ,q i q ' eQ 1<| K\<nt=1 j =1

1<K <ni=1

l

= 2 2 B+uq exp (_R1qA,t)+ 2 iBK+uq exp (_RKA,t).

q' eQi=1 v ' 2<| K| < n+1 i=1

So the formula (13) is true for 5 = n +1. Here for i = 1,...,l, 1 < |K| < n +1 we have the recurrent formulas (12) and for i = 1,...,l, |K| = 1 the recurrent formulas (11) and #0+1 qthe equality (14).

4. ASYMPTOTIC FORMULAS

Using the complete probability formula we obtain

P(Vn > t ) = Wn (t)= 2 Wn,q.....qn (t)Pqinq1,q2 " ...nqn_1,qn (16)

q1.....qneQ

with

Wn,qi.....qn (t )= P (Vn > t / Y1 = rq_1,...,Yn = r_ ) .

(C) Suppose that Fq (t)e s, q e Q, where s is the class of subexponential distributions. Assume that for any q1, q2 e Q, q1 ^ q2 and for any positive a one of the following equalities is true

Fqx (t) = O (Fq2 (at)) or Fq2 (at) = O (Fq1 (t)) , t > 0. (17)

Then using [3, Lemma 3.2] it is possible to obtain that

Fq1 (t)*Fq2 (at)e S , Fqi (t)*Fq2 (at) ~ Fq, (t) + Fq2 (at)| , t (18)

Here F * G is a conjuncture of distributions F,G . Further we consider equivalences only for

t

Theorem 3. If the condition (C) is true then

n — f n-k+i A

yn (t) ~ Z Z q F qi [ t r"i rq_ 1 Pqinqi,q2 " ... " ^ ,q„_k+1 . (19)

k=i qi.....qneQ V i=i /

Proof . It is obvious that

yi,qi (t) = P (YiXi > t / Yi = rq-i ) = P (Xi > rqi t) = Fq, (^ t) . Using the condition (C) and the formula (i8) we obtain for n > i that

yn,qi,...,qn (t) ~ P (Vn-1 > rqJ / Yi = rq-i Yn = r- )+ P (Xn > rqnt/ Yi = rq-i ,..., Yn = q1 ) =

= yn-i,qi,...,qn-i (rqnt) + Fqn M .

So an induction by n and the formula (i6) give the equivalence

yn,qi,...,qn (t) ~ Z F* [t^\rqi l .

k=i V i=k J

Consequently using the formula (i6) it is easy to obtain the equivalence

yn (t) ~ Z QPqinqi,q2 " ... ' * q^^ Fqk [ tY]rq, ) ■

qi.....qneQ k=i V i=k J

So the formula (i9) is true.

Consider the following conditions.

1) There are positive numbers cq, a , q e Q, a1 <aq, i < q < m, so that Fq (t) ~ cqt_ttq.

2) There are positive numbers cq, q e Q, a , so that Fq (t) ~ cqt"a .

3) There are positive numberscq,pq , q e Q , $ <flq, i < q < m, so that Fq (t) ~ exp(-cqA).

4) There are positive numbers cq, q e Q, ^ , so that Fq (t) ~ exp(-cqtand cr/ < cqr^, q e Q, q * i.

It is easy to prove that the family Fq (t), q e Q under each of the conditions i) - 4) satisfies the condition (C).

Using Theorem 3 it is possible to obtain the following statements. If the condition i) is true

then

_ n

yn (t) ~ ciPi (rit) ai i Sn-k+i , k=i

with

Si = i , S2 = Z S2,q2 , S2,q2 = ni,q2 rq2ai ,

q2eQ

Si = i Si,q, , Si,qi = i Si—i,q, i nqj uq,rq.^ , 2 < i . (20)

q,eQ qi-ieQ j

If the condition 2) is true then

yn (t) ~ t^iT^+i k=i

with

T = i T T = cP r_a

^ ^ i,qi,q^ ^q^qi q^q/qi '

qi,qieQ

Ti = Z Ti,qi,q, , Ti,qi,qt = Z Tj■_1,ql,qí-i ^q--i,q/q," , 1 <1 . (21)

qi,q,eQ qi,qi-ieQ

The formulas (20), (21) show that to find asymptotic constants in the conditions 1), 2) it is necessary to use number of arithmetical operations proportional to n. If the condition 3) is true then

Yn(t) ~ Aexp(-ci (rit )"A ).

If the condition 4) is true then

Yn(t) ~ Piexp (-ci (rit/).

5. GENERATION OF TRANSITION MATRICES FOR STATIONARY AND REVERSIBLE MARKOV CHAINS

Consider stationary and reversible Markov chain Yn , n| > 0, with stationary distribution pq, q e Q . Then its transition matrix ||n,q||m satisfies the equalities:

m m

A,, j = j > 0 , S A, j = P, = S j , 1 < i, j < m . (22)

j=i j=i

where A,,j = p^n,j. So symmetric matrix A,Xj-^ with positive elements is a permissible solution of the transportation problem (22) with n sources and n consumers. If we have the problem (22) solution A,jfj=1 then it is possible to find the transition matrix ||n,^^j-i using the formula

n, j = A,, j / P,.

Each permissible solution of the transportation problem (2) may be found by the following sequence of algorithms.

The algorithm {pl,...,pm} generates Au,...,Al,m so that

m

0 < Al,l < A — 0 < Al,m < pm , S Al,k = pi

k-i

and put A2,i = Al,2,..., Am,i = Ai,m and redefines pi := pi - pi = 0, p2 := p2 - Al,2,..., pm := pm - Ai,m .

As a result the transportation problem (2) with n sources and n consumers is transformed into the transportation problem

m

A,,j = Aji > 0 , S A,,j = pj, 2 < i, j < m . (23)

j=2

with n-i sources and n-i consumers. So the algorithm {pi,...,pm},{p2,...,pm},...,{pm-i,...,pm} generates arbitrary solution of the transportation problem (22).

The algorithm {pi,...,pm} consists of m steps.

Step i. Define Ai,i from the inequalities0 < Au <pi, pi - Ai,i <p2 +... + pm and put

pi:= pi - Ai,i.

Step 2. Define Ai 2 from the inequalities0 < Ai 2 <pi, 0 < Ai 2 <p2, pi - Ai 2 <p3 +... + pm and

put pi:= pi - Ai,2 .

Step m-i. Define Ai,m-i from the inequalities 0 < Ai,m-i <pi, 0 < Ai,m-i <pm-i, pi - Ai,m-i <pm

and put pi := pi - Ai,m-1.

Step m. Define Ai m-i = pi

The algorithm {pl,..., pm},{p2,..., pm},...,{pm-1,..., pm} is a modification of the algorithm for constructing the routing matrix for an open network [4, p.177].

6. ENUMERATION PROBLEM

Assume that vector K consists of components 0,1,... and its dimension is dim dimK. Introduce the sets of vectors Kj = {K :dim K = j,\ K \= i} and designate | K j | number of vectors in the set K j,i > 0, j > 1. Our purpose is to enumerate all vectors K of the set

{K: dim K = m,1 <\ K \< n} = U Km .

i=1

It is easy to construct algorithm to define the set Km from the set Km-1as follows: Km = {K +1 : KeKm-1, q = 1,...,m}. But a complexity of this algorithm is proportional to mn and it

may generate coinciding vectors. So it is worthy to construct more efficient algorithm for example with power by n complexity.

It is obvious that K j+1 is a union of nonintersecting sets

K j+1 = U {(K, kj+1): K ej J (24)

kj+1 =0

and consequently

\ Kj+1 \= = \ K- \= = \ Kj \ where j > 1,i > 0. (25)

t=0 t=0

Here K0 consists of single j - dimensional vector with zero components, \ K0 \= 1 and K¡j < K/ <...< Kj. From the formula (25) we have by induction that

\ K/+1\= E \Kj \< (i + 1)K{ < (i +1)j+1. (26)

t=0

As K1i consists of single one dimensional vector i so to find the set Knm using the formula (24) we construct the sequence of the sets

T^ 1 T^ 2 xr m .

j

K1 2 t^ m . 1, K1 ,..., K1 ;

1 ^2 K

n • 5

m+1

k1 k 2 K m •

n ? n' •• n • 5

Consequently complexity of this algorithm is not larger than (n +1)

7. SOLUTION OF SMALL DENOMINATORS PROBLEM

Consider how to find s1,...,sm,v1,...,vl so that for any fixed s> 0 the inequalities

\ S1 - r1 \<s,...,\sm - rm \<s, \ V1 -X1 \<s,...,\ vi -Xi \<s (27)

are true and for any K we have an analogy of the condition (9):

SKvi 1 <i, j<l ,1 <, (28)

where S = (s1,...,sm). For this aim we take integer b so that 2b < s and choose fractions

sj = ^r, i = 1,...,m, v. = -b, j = 1,...,i, with odd numerators vt so that the formula (27) is true. Then the

2b J 2b

formula (28) takes place.

REFERENCES

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2. Serfozo R. Introduction to Stochastic Networks. Springer Verlag. New York. i999.

3. Tang Q.; Tsitsiashvili G. 2003. Precise estimates for the ruin probability in finite horizon in a discrete-time model with heavy-tailed insurance and financial risks. Stochast.Process. Appl. Vol. i08, no. 2, 299-325.

4. Tsitsiashvili G.Sh., Osipova M.A. New Product Theorems for Queuing Networks// Problems of Information Transmission. 2005. Vol. 4i. No. 2. P. i7i-i8i.