DEFINITION OF LAPLACE-STIELTJES TRANSFORM FOR THE ERGODIC DISTRIBUTION OF THE SEMI-MARKOV RANDOM PROCESS T. I. NASIROVA, B.G.SHAMILOVA, U.Y. KERIMOVA Abstract. In this paper we will investigate the semi-markov random processes with positive tendency, negative jumps and delaying boundary at zero in this article. The Laplace transform on time, Laplace-Stieltjes transform on phase of the conditional and unconditional distributions and Laplace-Stieltjes transform of the ergodic distribution are defined. The characteristics of the ergodic distribution will be calculated on the basis of the final results. Mathematics Subject Classification 6A1, 6J25, 6G5 1. Introduction There are number of works devoted to definition of the distribution of the semi-markov processes and its main boundary functionals. Some authors are used the asymptotic, factorization and etc. methods (2,5,6,12) But other authors narrowing the class of distributions of walking are found the evident form for Laplace transforms for distributions and its main characteristics. In 7 The Laplace transformation for the distribution of the time of the system sojourn within a given band and its first and second moments are found.in 8 a model of inventory control is considered. It is described by a semi-markov random walk with a negative drift at an angle of < α < 9, with positive random jumps, a delay, an absorbing screen at zero, and a reflecting screen for a > at an angle α. The Laplace transformation is found for the distribution of the first moment storehouse exhaustion, and the first and the second moments are explicitly obtained. In 9 The Laplace-Stieltjes transform with respect to phase, the Laplace transform with respect to time, the conditional distribution, the unconditional distribution, and the Laplace- Stieltjes transform of the ergodic distribution of the process of semi-markov random walk with negative drift, nonnegative jumps, delays, and boundary screen at zero are obtained. In 1 The first passage of the zero level of the semi-markov process with positive tendency and negative jumps will be included as a random variable. The Laplace transform for the distribution of this random variable is defined. In 11 for the step process of semi-markov random walk with delaying boundary in a > the evident form of Laplace transform by time was found. The presented work explicitly defines the Laplace transform on time, Laplace-Stieltjes transform on phase of the conditional and unconditional distributions and Laplace-Stieltjes transform of the ergodic distribution for the semi-markov random processes with positive tendency, negative jumps and delaying boundary at zero. Key words and phrases. Laplace-Stieltjes transform, semi-markov random process, ergodic distribution, process with positive tendency and negative jumps. SYLWAN., 158(6). ISI Indexed 1 5
2 T. I. NASIROVA, B.G.SHAMILOVA, U.Y. KERIMOVA 2. Problem Let s assume that in probability space {Ω, F, P ( )} is given the sequence of independent, equally distributed and independent themselves positive random variables ξ k and ζ k, k = 1,. Using these random variables we will derive the following semi-markov random process: k 1 X 1 (t) = z t ζ i, if k 1 ξ i t < k ξ i, k = 1,, X 1 (t) is called semi-markov random processes with positive tendency and negative jumps. General form of process semi-markov random walk with delaying boundary is given by A.A. Borovkov1. If process X 1 (t) is some process without boundary, then process X 1 (t) with delaying boundary at zero is defined following: X(t) = X 1 (t) inf (, X s t 1 (s)) or X(t) = X 1(t) min (, inf X 1(s)). s t Idea of construction of the process semi-markov random walk is following : Let X 1 () = z. Process X(t) is equally to process X 1 (t) until, the process X 1 (t) is positive. Let X 1 (t) ; then X(t) is equally to zero until, the process X 1 (t) will not have positive jump. In moment of jump of the process X 1 (t), process X(t) will be have jump, such is equally to jump of the process X 1 (t). The obtained process is called a process of a semi-markov random walk with positive tendency, negative jumps and delaying boundary at zero. The aim of the present study is to find an evident form of the Laplace-Stieltjes transform of the ergodic distribution for X(t). 3. Definition of Laplace transform on time for the distribution of the process X(t) In accordance with formula of total probability for x we have P {X(t) < x X() = z} = P {X(t) < x; ξ 1 > t X() = z} P {X(t) < x; ξ 1 < t X() = z} = P {z t < x; ξ 1 > t} t s= y= P {ξ 1 ds; X(s) dy X() = z.} P {X(t s) < x X() = y} We denote R(t, x z) = P {X(t) < x X() = z}, x > R(θ, x z) = t= e θt R(t, x z)dt θ > R(θ, α z) = e α x d x R(t, x z), α >. In this case equation (1) will be as follows : R(t, x z) = P {z t x < } P {ξ 1 > t} SYLWAN., 158(6). ISI Indexed 51 (1), (2)
DEFINITION OF LAPLACE-STIELTJES TRANSFORM 3 t P {ξ 1 ds} d y P {max(, z s ζ 1 ) < y} R(t s, x y). y= s= Both sides of this equation we applied Laplace transform by t y= e θ t R(t, x z)dt = t= e θ t ε(x z t)p {ξ 1 > t} dt e θ t d y P {max(, z t ζ 1 ) < y} dp {ξ 1 < t}, where, θ >. Further R(θ, x z) = ε(x z) e θ t P {ξ 1 > t} dt e θ t d y P { < y} P {z t ζ 1 < y} d P {ξ 1 < t}. y= t= After some simplifications we will get: y= = ε(x z) R(θ, x z) = ε(x z) = ε(x z) y= If take into account t= y= e θ t P {ξ 1 > t} dt e θ t d y ε(y) 1 P {ζ 1 < z t y} dp {ξ 1 < t} = e θ t P {ξ 1 > t} dt R(θ, x ) t= e θ t dp {ξ 1 < t} e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t} = e θ t P {ξ 1 > t} dt R(θ, x ) t=max(,y z) e θ t dp {ξ 1 < t} e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t}. that is, why we get {, if y < z max {, y z} = y z, if y > z R(θ, x z) = ε(x z) z y= y=z R(θ, x ) t= t=y z e θ t P {ξ 1 > t} dt SYLWAN., 158(6). ISI Indexed 52 e θ t dp {ξ 1 < t} e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t} e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t}. (3) Both sides of this equation we applied Laplace transform by x see (2)
4 T. I. NASIROVA, B.G.SHAMILOVA, U.Y. KERIMOVA R (θ,α z) = x= e α x d x ε(x z) t= e θ t P {ξ 1 > t} dt R (θ,α ) e θ t dp {ξ 1 < t} z Take into consideration y= R (θ,α y) t= e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t} y=z R (θ,α y) t=y z e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t} x= e α x d x ε(x z) t= e θ t P {ξ 1 > t} dt = x= e α x ε(x z) d x t= e θ t P {ξ 1 > t} dt x= e α x t= e θ t P {ξ 1 > t} dt d x ε(x z) = x=z e α x d x t= e θ t P {ξ 1 > t} dt = = e α x e θ(x z) P {ξ z 1 > x z} dx. At last we received the following integral equation for R (θ,α z)when ξ k and ζ k, k = 1, equally distributed and independent themselves positive random variables R (θ,α z) = e θ z z e (αθ) x P {ξ 1 > x z} dx R (θ, α ) e θ t P {ζ 1 > z t} dp {ξ 1 < t} z y= R (θ, α y) t= e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t} (4) y=z R (θ, α y) t=y z e θ t d y P {ζ 1 < z t y} dp {ξ 1 < t}. We will solve this integral equation in special case. Let s assume that ξ 1 random variable has the Erlangian distribution of n order, while ζ 1 random variable has the single order Erlangian distribution: where { } P {ξ 1 (ω) < t} = 1 1 t (t)2 (t)n 1 e t ε(t), >, 2! (n 1)! P {ζ 1 (ω) < t} = 1 e λt ε(t), λ >, ε(t) = In this case equation (4) will be as follows: {, t <, 1, t >. SYLWAN., 158(6). ISI Indexed 53
DEFINITION OF LAPLACE-STIELTJES TRANSFORM 5 R (θ,α z) = (αθ)n n (αθ) n (αθ) e α z λ λθ λθ n e λz R (θ,α ) n e λz z eλ y R (θ,α y)dy (5) λn (n 1)! e λz y=z eλ y R (θ,α y) t=y z e (λθ)t t n 1 dtdy. We will get differential equation from this integral equation. For this purpose, we will multiply both sides of equation (5) by e λz and derive on z. Then we will multiply both sides of last equation by e (λθ)z and derive on z. If repeat this process (n-1) time we have following differential equation : k=o C k n λ (k) R (θ,α z) (k 1) R (θ,α z) ( 1) n k ( θ) n k ( 1) n 1 λ n R (θ,α z) = = ( 1) n 1 (α θ) n n (λ α) (α θ) e α z. (6) 4. The general solution of the differential equation (6) The general solution of this differential equation will be R (θ,α z) = C 1 (θ, α)e k1(θ)z C 2 (θ, α)e k2(θ)z C n (θ, α)e kn(θ)z R sp (θ,α z), (7) where k i (θ), i = 1, 2,...n,-are the roots of characteristic equation of (6) R sp (θ,α z)- is the special solution of the equation (5) where R sp (θ,α z) = A e α z, A = (λ α)(α θ)n n (α θ) n α k i(θ). By finding c 1 (θ, α),..., c n (θ, α) from equation (5) we will get the following system of algebraic equations: SYLWAN., 158(6). ISI Indexed 54
6 T. I. NASIROVA, B.G.SHAMILOVA, U.Y. KERIMOVA R (θ,α ) = (αθ)n n (αθ) n (αθ) n λθ R (θ,α ) λn (n 1)! y= eλ y R (θ,α y) t=y e (λθ)t t n 1 dtdy n R (θ,α ) = α (αθ)n n λ (αθ) n (αθ) λθ R (θ,α ) λ2 n (n 1)! y= eλ y R (θ,α y) t=y e (λθ)t t n 1 dtdy λn e (θ)y (n 1)! R (θ,α y) y n 1 dy... n 1 λ R (k) (θ, α ) R (k1) (θ, α ) = Cn k k= = ( 1) n k (αθ)n n (λ α) ( 1) n 1 λ n e (θ)y (αθ) z R (θ,α y)dy By exploitation of equation (7), equation (8) becomes C i (θ, α) A = (αθ)n n (αθ) n (αθ) λθ n n C i (θ, α) A (8) n λn (n 1)! y= eλ y C i (θ, α)e ki(θ)y Ae α y t=y e (λθ)t t n 1 dtdy C i (θ, α)k i (θ) αa = α (αθ)n n (αθ) n (αθ) λ λθ n n C i (θ, α) A n λ2 n (n 1)! y= eλ y C i (θ, α)e ki(θ)y Ae α y t=y e (λθ)t t n 1 dtdy n λn e (θ)y C (n 1)! i (θ, α)e ki(θ)y Ae α y y n 1 dy Cn k k= { n n } λ k n C i (θ, α) α n A k n1 C i (θ, α) α n1 A = = ( 1) n 1 α n 1 (αθ) n n (λ α) (αθ) (9) ( 1) n 1 λ n n e (θ)y C z i (θ, α)e ki(θ)y Ae α y dy. Now we proof linear dependence of this algebraic system If to consider the following substitutions: n θ k i (θ) = ( 1) n1 λ n, n λ k i (θ) = ( 1) n1 λ n, SYLWAN., 158(6). ISI Indexed 55
DEFINITION OF LAPLACE-STIELTJES TRANSFORM 7 n θ k i (θ) = ( 1) n1 λ k j (θ) θ k j (θ) n = ( 1) n1 λ n, y equation (9) becomes: n α k i (θ) = (λ α)( θ α) n λ n, t n 1 e (λθ)t dt = yn 1 λ θ { θ k i (θ) n n } C i (θ, α) = { θ k i (θ) n n } C i (θ, α) = k=2 (n 1)! y n k (n k)! (λ θ) k e (λθ)y (1) α n (αθ) n n (αθ)λ n (λ α)(θα) n = αn A α n (αθ) n n (αθ)λ n (λ α)(θα) n = αn A... { θ k i (θ) n n α } C i (θ, α) = n (αθ) n n = (αθ)λ n (λ α)(θα) n αn A Thus, (11) is a linear dependence equations system, as (11) Then we have C 2 (θ, α) = C 3 (θ, α) = = C n (θ, α) =. α n (α θ) n n C 1 (θ, α) = (α θ)( θ k 1 (θ) n n ) λ n (λ α)( θ α) n. (12) Then the general solution of integral equation (5) will be as follows: R (θ,α z) = α n (αθ) n n (αθ)(θ k 1 (θ) n n )λ n (λ α)(θα) n e k 1(θ) z (λ α)(αθ) n n (αθ) λ n (λ α)(θα) n e α z. (13) This expression is the Laplace transform on time, Laplace-Stieltjes transform on phase for conditional distribution of the process X(t) 5. Ergodic distribution of the process. We will need to find Laplace transform on time, Laplace-Stieltjes transform on phase for unconditional distribution of the processx(t). From construction process X(t) is seen that Then we will get X() = X 1 () = ξ 1 (ω). R (θ,α) = R (θ,α z) dp{x() < z}. SYLWAN., 158(6). ISI Indexed 56
8 T. I. NASIROVA, B.G.SHAMILOVA, U.Y. KERIMOVA Therefore R (θ,α) = z= α n (αθ) n n (αθ) (θ k 1 (θ) n n ) λ n (λ α)(θα) n ek 1(θ) z or (λ α)(αθ) n n (αθ) λ n (λ α)(θα) n e α z d1 e z, R (θ,α) = α n (αθ) n n (αθ)(θ k 1 (θ) n n ) λ n (λ α)(θα) n (λ α)(αθ)n n. (αθ) λ n (λ α)(θα) n α k 1 (θ) This expression is Laplace transform on time, Laplace-Stieltjes transform on phase for unconditional distribution of the process X(t). Now, we will find Laplace-Stieltjes transform for ergodic distribution of the process X(t). In 3 ( see p.363) proved a general theorem on the ergodicity of the process semi-markov random walk. The process described in this article a special case of this process. Process X(t) will be ergodic, if Eξ 1 < Eζ 1, or 1 < n λ λ < n. If process X(t)ergodic,then we can use Tauber s theorem 4 We obtined Ee α X(ω) = R(α) = lim θ θ R (θ,α). R(α) = 1 (n 1)! (α ) n n λ n (α λ)(α ) n. (14) Expression (14) is Laplace-Stieltjes transform for ergodic distribution of the process X(t). Respectively, we will get the following characteristics for λ < n: R () = EX(ω) = n, λ n R () R () 2 = DX(ω) = 2n, (λ n) n λ < n, λ < n. 6. Conclusions In this article we have defined Laplace transforms on time, Laplace-Stieltjes transforms on phase for conditional and unconditional distributions and Laplace-Stieltjes transform for the ergodic distribution. SYLWAN., 158(6). ISI Indexed 57
DEFINITION OF LAPLACE-STIELTJES TRANSFORM 9 References 1 Borovkov A.A. Veroyatnostnye protsessy v teorii massovogo obsluzhivaniya (Probabilistic processes in queuing theory). Moscow: Nauka, 1972. 2 Borovkov A.A. On the asymptotic behavior of the distributions of first-passage. Mat. Zametki, 75(1):24 39, 24. 3 Gikman I.I., Skorokhod A.V. Theory of stochastic processes, Moscow, Nauka, 1973 4 Klimov G.P. Stochastic service system, Moscow, Nauka, 1966 5 Lotov V.I.. On some boundary crossing problems for gaussian random walks. The Annals of Probab, 24:2154 2171, 1996. 6 Lebowitz J. L. and Percus J. K. Asymptotic Behavior of the Radial Distribution Function, J. Math. Phys. 4, 248 (1963) 7 Nasirova, T. I.; Sadikova, R. I. Laplace transformation of the distribution of the time of system sojourns within a band.automatic Control and Computer Sciences (29) 43: 19-194, August 1, 29 8 Nasirova, T. I.; Yapar, J.; Aliev, I. M. A model of inventory control Cybernetics and Systems Analysis (1999) 35: 553-562, July 1, 1999. 9 Nasirova T.H, Ibayev E.A., Aliyeva T.A. The Laplace transform of the ergodic distribution of the process of semimmarkovian random walk with negative drift, nonnegative jumps, delays and delaying screen at zero. Theory of Stochastic Processes, 15(??),29, 1, ISSN 321-39, pp 49-6 1 Nasirova T.H, Kerimova U.Y. Definition of Laplace ransform of the first passage of zero level of the semimarkov random process with positive tendency and negative jump. Applied mathematics, 211, 2, pp 98-912 11 OmarovaK.K Laplace transformation of ergodic distribution of the step process of semi-markov random walk with delaying screen at positive point. The Third International Conference Problems of Cybernetics and Informatics 21 12 A. M. Walker On the Asymptotic Behaviour of Posterior Distributions Journal of the Royal Statistical Society. Series B (Methodological) (pp. 8-88) Vol. 31, No. 1, 1969. Baku State University, Z.Khalilov 23, AZ1148, Baku, Azerbaijan E-mail address: nasirova.tamilla@mail.ru Baku State University, Z.Khalilov 23, AZ1148, Baku, Azerbaijan E-mail address: bahar.shamilova@rambler.ru Cybernetics Institute of Azerbaijan National Academy of Sciences, F.Agayev 9, AZ1141, Baku, Azerbaijan E-mail address: ulviyye1982@gmail.com SYLWAN., 158(6). ISI Indexed 58