# B Shamilova - The laplace transformation of ergodic distribution of the process with two delaying screens - страница 1

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ВІСНИК ЛЬВІВ. УН-ТУ VISNYK LVIV UNIV.

Серія прикл. матем. інформ. Ser. Appl. Math. Inform.

2009. Bun. 15. C. 281-290_2009. Is. 15. P. 281-290

УДК 519.21

THE LAPLACE TRANSFORMATION OF ERGODIC DISTRIBUTION OF THE PROCESS WITH TWO DELAYING SCREENS

B. Shamilova

Baku State University e-mail: bahar-shamilova@rambler.ru

For a given two-dimensional sequence of independent and identically distributed random variables constructed semi-Markov random walk process with inhibitory screens " b " and " a " (a > b) and is the Laplace-Stiltesa ergodic distribution process, and also found its expectation and variance.

Key words: process semi-markovian random walk with delaying screen, ergodic distribution, Laplace transformation, conditional distribution.

1. PROBLEM STATEMENT

Let on a probability space (Q,F,P) be given a sequences {£ .} , {n.} , i = 1, °°, of

independent identically distributed random variables and independent between themselves

£., n., i = 1, oo. ii

We'll assume that £ > 0 and let the numbers a, b and z be given a > b > 0 and 0 < z < a - b .

Let's construct a random process

kk k+1/0 \

Xi(t) = b + z+£n, if <t< JJ£i, k > 0, X = 0 ,

i=1 i=1 i=1 V  1 )

that will be called a stepwise process of semi-markovian random walk.

The process constructed in such a way is said to be a delaying barrier process semi-markovian random walk

kk+1/0 \

X(t) = Ck,   and   Y,£t <t< , k > 0,  X = 0

i=1 i=1 V  1 J

£ k = main{a,max{b, Zk-1 +%}}^ ^ > 1, Z0 = b + z.

Notice that many important problems from the theory of store control are described and solved by means of the process semi-markovian random walk ([4], ([5]).

Investigation of ergodic distribution for semi-markovian random walk processes occupies a special place in the theory of random processes ([1], [2], [3], [6]).

The aim of the paper is finding the Laplace-Stieltjes transformation in phase and in time of the distribution of the process X (t) in the case when a random variable n1 has a complex Laplace distribution. Besides, in the paper, when n1 has a Laplace distribution the

generating function of distribution of the number of steps at which it achieves the barrier " a " is obtained and evident formulas of expectation and variance are found for it.

R(в,x/b + z) = 1 <Рвв'e(x-b -z) + p(-)R(в,x/b)P{n <-z} +

a

+ р(в) J R(в,x/y)dP{?]1 < y-b-z} +

y=b

(3)

+ p(-)R (в, x/a)P{n > a -b - z}. Applying to the both sides of (3) the Laplace-Stieltjes transformation with respect to x, we get an integral equation for R (в, a/ b + z).

R (в,a/b + z) l_ppf^e-a{b+z)+p(e)P{ril <-z}R(-,a/b)+

+ р(в) J R(в,a/y)dyP{7]x < y-b -z}­(4)

y=b

+ p(-)P{n1 > a - b - z}R (в,a/a). Corollary 1 Let tj1 = П1+ - П-, where n1+ - has an Erlang distribution of first order with parameter Я, n1- -has an Erlang distribution of second order with parameter и , i.e.

Л(Л + 2ц

P fa < x}

Uix\eux, x < 0, и> 0,

1--L_ e^x,

x > 0,Я> 0.

(Я + и/

Then we immediately find from (4)

K >       вР + /и{ Я + и

м[1]р(в) (Я+и)[2]

-Я(а -b-z)

R (в,a/a )+-

Я/и[3]

(я+и/

-p(e)e-u/b+z) J R (в,a/y)euydy-

y=b

+ Яи\[Ь + z) p(-)e-u/b++^ J euyR(-,a/y)dy+-риp(-)epb++ р + и y=b + и)

(5)

y=b

f e^yR(-,a/y)dy-Яи2р(в e^(b+z) f ye"yR(-,a/y)dy.

y=b+z Я+и y=b

From integral equation (5) we get an inhomogeneous differential equation with constant coefficients

R"(e,a/b + z )-(Я-2и)R"(в,a/b + z)-ц(2P-ц)R'(в,a/ b + z)

-Яи[4]\}-Р(в)] R (в,a/b + z) 1 -Р(в)

в

(P + a)(a-ц)[5] e

-a(b+z)

with characteristic equation

к3 (в) - (Я- 2и[6] (в) - и (2Я -М(в) - Яц[7] [1 -<(-)) = 0 and general solution

(6)

R (-,a/b + z) = £ с (e,a)ek^b+z) +1 (в)(Я+a))a-и/ e-a(b+z), {1)

==i в na+к (0)]

From (6) we find:

к:(0)==2^р-(0)+12Я^1)Р(0)1

К (0 ) =-2Яц2   ,        Р(0),

Я2 + 4Яи + (Я- 2и))и2 + 4Яи

к3 (0 )=-2Яц2    ■ Р(0),

и2 + 4Яи-(Я-2и))и2 + 4Яи

Now we have to find ct (в,a),   i = 1,2,3 . From integral equation (5) we find the following boundary conditions:

R (-a/b ) = X-pPele-ab + Я(Я + 2иР (-) R (e,at b e-b)R (e,at a) +

+рииЩrePb "[e-^R(-a/y)dy,

+ и) y=b

R'Ma/b) = -al1-PpPe±e-ab -Яи2р(в}R(-a/b^ЇЩє-*-b)R(-,a/a) +

+ЯЯ^єяь Je-R(-a/y)dy,

R:(e,a/b) = a21-PPPe±e-ab -Яир(в2)R(-,a/b^^ІЖe-*-b)R(-,a/a) +

в        (Я+и) (Я+и) + рpрвje-,R(-ay)dy,

+ и) y=b

On the other hand, from the solution of differential equation we have:

R (-a/b ) = І с, (e,a)ek(e)b +1 -р(в)(Я3+a)(a-и)2 e-ab,

i=1        в na+к, (0)]

i=1

R;(-,a/b) = ±n,(-,a)k,(в)ek(-)b -1 -p(e)a(^+a)(a-и)2 e-ab, i=1 -       n[a+к, (в)]

=1

Rf(e,a/b) = ±n,(e,a)k2(e)ek(-)b +1 (-) a (Я + a)(a-и)2 e-ab. i=1 -       n[a+к, (в)]

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