CC BY
9
ACCURACY SOLUTION OF A.A. NOVIKOV PROBLEM
Tsitsiashvili G.Sh., Osipova M.A.
Institute for Applied Mathematics, Far Eastern Branch of RAS 690041, Vladivostok, Radio str. 7, e-mail: [email protected], [email protected]
Introduction
In this paper we consider the Laplas model described by the following autoregressive random sequence
X, = RXk-X , Xo = 0, t = inf (k : Xk > X),
(1)
p(n >t) =, t>o, p(n <t) = expW, t<0,.
Our problem is to calculate a distribution of a reaching moment t . This problem was put before the authors of this paper by A.A. Novikov. The problem origins in the risk theory and in the reliability theory. M. Jacobson found approximate solution of this problem by martingale technique. V.V. Mazalov suggested to solve the problem for R < 1 approximately by some recurrent procedure which includes a compressing operator. In this paper we apply some recurrent integral equalities to find accuracy solution represented by mixtures of exponentials. This solution is illustrated by numerical calculations. Our solution may be used for an arbitrary R and when as R so X depend on k .
1. Main results
Denote X, = 0, X,-s = XRk-s, s = 1,...,k, and designate fork > 1
Tk (x) = P (Xk > x, t> k), x > 0, Sk (x) = P (Xk < x,t> k) , x < 0, P (t = k) = Tk (X) .
Denote
B (k, S, j) = exp{-IX^2-s { j -1)), B2 (k,s, j) = exp^X^2-* ( j +1)),
B3 (k, s, j) = exp {-1X^-s { j +1)), B4 (k, s, j) = exp {xxi+[-s ( j -1)), A (k, s, j) = B (k, s, j) - B4-1 (k, s, j), A2 (k, s, j) = B2 (k, s, j) - B3-1 (k, s, j), A3 (k, s, j) = B2-1 (k, s, j) - B3 (k, s, j), A4 (k, s, j) = B1-1 (k, s, j) - B4 (k, s, j).
Theorem 1. The following formulas are true for k > 0, and for x > 0
k- s f ix ) k- s f ix)
Tk (x)= 2 akk-sj exp rj j+ s bkk-sj exp [rjj+ckk-s (2)
with Xk-s+1 < x < Xk-sfor some s e{1,...,k} and for x < 0
Sk (x)= k-? dk] exp {X) (3)
and
P (t = k) = ak ooexp (-XX). (4)
Here
and
aioo = 2' = 0' Cl° = 0 :
-^10 = 2' ci -i = 0
ak+1 k+1-s j = akk-s j-1
R2j
1 - R21 R2 j
7 _ 7
bk+1 k+1-s j = ~bk k-s j-1 1 R 2 j ■
1 ( k -1 d
4k+1 k+1-s 0
kj
s -1 k-t
E E
2 V j=0 1 + RJ+1 t=1 j
, 0 < j < k - s +1, 1 < s < k, 0 < j < k - s +1, 1 < s < k,
A ( k, t, j ) akk-t j A2 (k't' j) bkk-t j
1 - RJ+ 1 + RJ+1
k-s
l-E j=0
B1 (k's' j) "kk-s j , B2 (k's' j) bkk-
s J
b
k+1 k+1-s 0
1 ( k k-s
E E
Vt=s+1j=0
2
1 - RJ+ 1 + RJ+1
A3 (k't' j) akk-s j + A4 (k't' j) bkk-s j
1 < s < k + 1,
1 + Rj+1
1 - Rj+1
k-s
E
j=0
B3 (k's' j)akk-s j B4 (k's' j)bkk-
's J
1 < s < k' bk+100 = 0'
lk+10
1 + Rj+1 1 - RJ+1
Ck+1 k+1-s = Ck k-s - ak 0 0 exp (-AX)' 1 < s < k' ck+10 = ck+1 -1 = 0 '
1 (kE Jj
2 V jE0 1 - RJ+1 ' s=1 jt0 1 + RJ+1
k k-s akk s j k k-s bkk s , ^
E E A3 (k, s, J) + E E T^A (k' s, J)
s=1 J=0 1 - RJ+1
R2(J+1) 4+1 j+1 = -dk j 1 _ r2( j+1)
0 < j < k -1.
(5)
(6) (7)
(8)
(9) (10) (11)
(12)
2. Theorem 1 proof
It is obvious that
and
T1 (x ) = exp (~ÄX), x > 0, $ ( x ) = exp (AX), x< 0,
P (T = 1) = exp (-AX) .
Calculate now for x > 0
T2 (x) = ( 0 dS1(u 1-"in(f•XR)dT1 fu 11exP<-A<x -u))
XR
j dT1iui(1 -exp (A(x - u )r
min(x, XR)
(13)
As a result obtain
T2 (x) = 2(1 -R2 ) ifexP (-Ax) - R2 exP ^ )) - 2exP (-AX) ■
+ exp(Ax)exp(-AX (1 + R)) 0 < x < XR 4 (1 + R) ' :
T2 ( x ) = '
= exp (-Ax)( 1 exp (A(R -1)X)
1 - R2
2 (1 - R )
Calculate now for x < 0
$2 (x) = 1^1 (R )(1 -exp (-A2( x - u)) )+(№ V R ) - XRdTl\
XR < x < X .
u 11 exp (A(x - u))
(14)
(15)
As a result obtain
S2 {x) = exp exp(Xx)
Xx V 1
1
1
1
R ){2 4(1 + R) 4(1 -R)
1 (1 - exp {-XX (1 + R))) I, x < 0.
(16)
2 ^ 2 (1 - R) 2 (1 + R)
In an accordance with (14) - (16) assume by an induction that for x > 0 the formula (2) is true and for x < 0 the formula (3) takes place. Then for x > 0
Tk+1 (x) = J1 (x) + J2 (x) + J3(x) + J4(x).
Here
Calculate now
J1(x) = I dSk
R
exp (-X(x - u ))= exp (-Xx)k-1 dk
jS) 1 + Rj+1'
j2 (x )=-"'"(jXi+1' dTt (R) =
.exp (-X ^^ exp f-X, f-L-" ^'
t=1 j=)1 - RJ+1
ex
p (-Xx)k k-t btt.
-tj
S S 1 ,=01 + Rj+1
exp I Xu
t=1 j
RJ+1 1
RJ+1
-1
min(x, Xkk++;-t) min( x, Xkk++12-t)
min(x, X^-t)
x> 0.
Then we have for X,
k+ 2-s ^ „ ^ x^k+1-s k+1 S x S Xk+1
X!
J3 ( x) = - I dTk lRl=-ak 00exp (-XX ) + min(x,X1+1) {R )
k-s f Xx ^ k-s f Xx
akk-sj exp{-Rj+i)+ IS bkk-sj expI I + c
RJ+1
k k-s
and
X
1+1
J4 (x )= I dTk lR
min (x, X1+1) { R
exp (-X( x - u))
exp (Xx)sk-sakk_sj f X f 1 ^ min(max(x,Xk++^s ),x1+1 )
:—— S S-exp I -Xu I —— +1
2 s=1 jJ=01 + Rj+1 { '
RJ+1 )) min(max(x,Xk++12-s),X'+1)
exp (Xx )SkSsbkk-s j fX f 1 - S S :——rr exp I Xu
2 s=1 ,t01 - Rj+1
RJ+1
- 1
min( max( x, X^-s), Xlt+1) min(max(x, X^2-1), X1+1)
fors e {1,...,k},
J3 (x) = J4 (x) = 0, s = k +1.
From (17) - (22) we have for x > 0, s e {1,...,k}
T i \ i X^\,exp (-Xx)k"1 dk j ,
Tk+1 (x) = -ak 0 0 exp (-XX ) + -^-- SS 1 + rj + 1 +
min (x, Xkk++12-t)
1-j exp I-X
1
RJ+1
-1
min (x, Xkk++11-t)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
and for s = k +1
exp (-Ax)*k~t bkk-t
E E
Ä jt01 + RJ+1
exp | Au
RJ+1
+1
k+2-t\ k+1 I
min( x, X\ min(x, Xfc1-')
k-s ( — Ax 1 k-s ( Ax 1
+E0 akk-sj exp f Rj+r J+E0 bkk-sj exp V Rj+1 J+ckk-s
j=0
exp (Ax jEk;' akk-tj
"S jE0l + RJ+1'
exp | -Au
+1
min (max( x, X^1-s ), X1+1)
RJ+1 J J min(max(x^2-s ),X1+1)
k k-t b,
exp(Ax)v v 9 E E
2 t=1 j =0
exp (Au ( 1 - rJJ
min( max( x, X^ ), Xl+i)
n(max( x, Xkk++i2 - s ), Xl+i)
Tk+1 (x) = exp (-Ax) E kEt akk-tj
= exp(-Ax)
-Ax )k -l d,
2 tEnE^^ni-1
E kJ + jE=0 1 + RJ+1
min( x, Xkk++i2-')
1
exp(-Ax)
k k-t b E E
'k k-t j
t=1J=0 1 + RJ+1
exp | Au
1
RJ+1
+1
min(x, Xkk++1-') min( x, Xkk++2-')
min(x, Xtl-')
(24)
Using the formulas (2) rewrite the formulas (23), (24) as follows: for 5 e {1,...,k}
Tk+1(x) =
p (-Ax) (kEr dkj E1 E
2 V JE0l + RJ+1 ¿iE
A (k, t, j) akk-t j . A2 ( k, *, j ) bk k -
tj
1 - RJ+1
1 + RJ+1
k-s
+ E
j=0
B1 (k, s' j) "k k-s j B2 ( k, S' j ) bkk-si
1 - RJ+1 1 + Rj+1
1 k - s
- 2 E
2 J =0
exp
Ax
' Rj+1 I Uk k-s j 1 - RJ+1
exp
Ax
Rj+1 I bkk - s j 1 + RJ+1
k-s ( Ax 1 k-s ( Ax
akk-sj exp f- j J + IE bkk-sj exp | —^ 1 + Ckk-s - ak 00exp (-AX
exp (Ax) ( E kEs
2 Vt=s+ij=0
J- V RJ+1
A3 ( k, ^, j ) ak k - s j A4 ( k, ^,j ) bk k - s j
1 + RJ+1
k-s
-E
J=0
B3 (k, s' j)akk-sj B4 (k, ^ j)bkk-sj
1 + Rj+r 1 - Rj+1
1 k-s
- 2 E
2 J=0
exp
1 - RJ+1
Ax 1
-Rj+i Jakk-sj
1 + RJ+1
exp
Ax
R j + 1 I kk - s j 1 - RJ+1
(25)
- k +E- s (_ Ax 1 k+r-s, (Ax 1
- E ak+1 k+1-sj exp i _T)7 I+ E bk+1 k+1-sj exp i tj I + ck+1 k+1-s ,
j=0 V RJ J j=0 V RJ.
for s = k + 1
Tk+1(x) =
= exp (-Ax)( k-r dkj s-1 k-t
S E
Vj
=01 + RJ+1 t=i j
A (k, t, j) ak k-t j A(k, t,j) bk k-t j
1 - RJ+1 1 + Rj+1
(26)
k-s
E
j=0
Bi (k, s' j ) "k k-s j B2 (k, S'j)bkk-
s J
1 - RJ+1
1 + RJ+1
= ak+i0 0exp (-Ax ).
Calculate now for x < 0
2
Sk+1 (x ) = jdSk {R |f1 -exp (-X2(x - u)) U0 dSk IR )-? dTk IR ))exp (X(x - u »■
(27)
k-1 j f Xx
= jS dkJexplj
1-
1
1
A
2 (1 + RJ+1) 2 (1 - RJ+1)
1 k-1 dk J exp (Xx) exp (Xx) k k~s ak k_,
2 S
2 J=0
-s j
1 - RJ+1
S1 £1 + RJ+1'
s =1 J
exp I -Xu
k k—s bu
exp (Xx l^bj exp | Xu
s =1 J
1 ,t01 - RJ+1
RJ+1
-1
1
RJ+1
\r>
x
x
k + 2-s k+1
X
k+1-s k+1
k +2-s k +1
V x
k +1-s k +1
k-1 j f Xx
= S dkjexp| j=0
RJ+1
1-
1___1
2 (1 + RJ+1) 2 (1 - RJ+1)
A3 (k,s,j)akk-s j + A4 (k,s,j)bkk-s j
_ 1 + RJ+1 1 - RJ+1
From the equalities (13) - (16), (25) - (27) we have the formulas (4) - (12). Theorem 1 is proved.
exp (Xx) l s dkj + * kSs 2 {J=01 - RJ+1 s=1 j=0
1 | ^^+1 j exp ff ^
Remark 1. Obtained formulas remain true in a case of variable boundary and interest rate: Xk = Rk_1Xk_1 _1, t = inf (k: Xk > X (k))
then we rewrite
Xt = 0, Xk = X (k) , Xj = min (XjiRk-1, X (k)) , R°° = 1, Rj = R^R^, and replace RJ+1 by RjJ and RJ by R]k+l in previous formulas, 1 < j < k -1, k > 1, without assumption
Rk-1 < 1.
Remark 2. The proof of Theorem 1 contains sufficiently complicated and long symbol transformations. The transformations create manifold mistakes. To avoid these mistakes we examined the transformations by numerical calculations.
Remark 3. Suppose that X=1, R=0.5, X=0.4491 then in an accordance with Theorem 1 we obtain
Table 1.
k N
10 0.03052
20 0.00512968
30 0.000861841
40 0.000144798
50 0.0000243276
60 4.08729* 10-6
70 6.86708* 10-7
80 1.15374* 10-7
90 1.93841* 10-8
100 3.25672* 10-9
Table 1.
The authors thank A.A. Novikov for useful discussions.
