Defination of the Probability Characteristic of the
System from Given Region for Case (2+,1–)
Rahila Sadigova
Cybernetics Institute of ANAS, Baku, Azerbaijan
[email protected]
Abstract— In this paper has been studied the process of semimarkovian random walk with jumps and two delaying screens.
It is obtained in this clause to find the Laplace transformation of
the distribution of a random variable  ( ) .
We introduce a random variable  ( ) , meaning the
duration of the time in which process X (t ) is in a region
(0, a) .
Keywords— process of semi-markovian random walk; the
probability space; Laplas transformation
III. THE FINDING OF THE LAPLACE
TRANSFORMATION OF THE DISTRIBUTION OF A
RANDOM VARIABLE  ( )
I.
INTRODUCTION
Investigation of the ergodic distribution for process semimarkovian random walk take a special place in the theory of
random processes. In 1975 V.Smit has proved the ergodic
theorem for semi-markovian processes [1]. The general
theorem about ergodic for processes with discrete intervention
is proved in [2]. In [3] the ergodic theorem for complex semimarkovian processes with delaying screen is proved.
The purpose in this paper to find the Laplace
transformation of the distribution of a random variable  ( ) .
We denote
K (t z)    t X (0)  z.
It is obvious, that

K (t z )   inf X ( s)  0; sup X ( s)  a X (0)  z
0 s t
In clause [4] to find the Laplace transformation of the
distribution for case (1 ,1 ) of a random variable  ( ) .
II.

0 s t
  1 ( )  t 
t


0 s t
(1)
a
   ( )  ds; z  
1
1
 dy  t  s X (0)  y
Then the equation (1) will be written in following form:
m 1
 , if 
k 0

0 s t
s 0 y 0
We construct the process [5]
k
0 s t
  inf X ( s )  0; sup X ( s )  a;  1  t X (0)  z 
sequence  k , k k 1, , where  k ,  k , k  1,  , are independent
identically distributed random variables and independent
themselves,  k  0 .
k 0

K (t z )   inf X ( s )  0; sup X ( s )  a;  1  t X (0)  z 
Let on the probability space , , (.)  is given the
X 1 (t ) 

On total probability form we have:
THE PROCESS CONSTRUCTION
m 1
0 s t
m
k
 t   k ,
m  1, 
K (t z )   1 ( )  t 
k 0
t
a
   ( )  dsd 
where  0  z  0,  0  0 .

We delay process X 1 (t ) with screen in the zero (see [1]):
Let's denote:
1
y
1
 y  zK (t  s y )
(2)
  0.
(3)
s 0 y 0
X 2 (t )  X 1 (t )  inf (0, X 1 (s))
0  s1
Then we delay this process with screen in a(a  0) :
X (t )  X 2 (t )  sup(0, X 2 (s)  a).
0 s 1
This process is called the process of semimarkov random
walk with double delaying screens in the “ a ”end zero.

~
K ( z )   e t K (t z )dt,
t 0
 ( )  Ee  ,   0 .
1
If to apply the Laplace transformation both sides of the
equation (1) with respect to t :

t ~
 e K (t z)dt 
t 0

2 
e  z  e y dy 
(   ) 2
y 0
L z    ( )   ( )
t
 e  1 ( )  tdt 
z
t 0

  ( )
2 
e  z  e y L y e y dy 
(   ) 2
y 0
  ( )
2 
ez e  y dy 
(   ) 2 y 0
  ( )
2 
 2 z z
z
 y


e
e
L

y
dy


(

)
e  e  y dy 

(   ) 2

yz
yz
(4)
  ( )
 2 z z
 2  z
e  e  y L y dy  ( )
e  e  y ydy 


yz
yz
Let's solve this equation in the class for the Laplace
distributions. For example, let
  ( )
 2  z
e  e  y yL y dy.

yz
a

t
 d y 1  y  z  e
y 0

t 0
1   ( )

t
z
  ( )  dsK (t  s y) 
1
s 0
a
a
~
  ( )  K ( y)d y 1  y  z .
y 0
~
Then we have the following equation for K ( z ) :
~
1   ( )
~
K ( z ) 
  ( )  K ( y)d y 1  y  z
a
a

y 0
1  1   2  1 .
 2
t
 (   ) 2 e ,

F{1  t}  
1   [1    t ]e t ,
   

1
(5)
t  0.
a
(9)

e t , t  0,
(   ) 2
L ( z )  (2   ) L ( z )   (  2 ) L ( z ) 
 2 [1   ( )]L( z )  0.
(10)
k 3 ( )  (2   )k 2 ( )   (  2 )k ( )  2 [1   ( )]  0.
(11)
2
 1
  t
     t  e , t  0.


Then the common of (10) solution will be
(6)
3
L( z )   d i ( )e k ( )
(12)
i
i 1
From (9) we can find the initial conditions for differential
equation (10) :
~
1   ( )
K  z  
  ( )

  ( )
a
The characteristic equation of (10) will be the following form
and
  ( )
a
From (9) we can receive the differential equation:
t  0,
Hence we have



p (t )   2
 
  

a

a
2
(   ) 2
e z

~
2 
e  z  e y K  y e y dy 
2
(   )
y 0
z
 z z y ~
e  e K ( y )dy 

н 0
ф
2

~
e y K  y dy 
yz
(7)


2  z y ~
  ( )
e e yK  y dy.
   y z
a

We denote:

   t X (0)  zdt  E ( X (0)  z)
0

~
K ( z )   e t   t X (0)  zdt,
0
~
L( z )  1  K ( z ) .
a

2 
e  a e y dy 
 L( a )   ( )   ( )
2
(   )
y 0

a

2
  ( )   2 e  a e y L( y ) dy,
(   )

y 0

a
2
  2  a y

e
e dy 
(13)
 L( a )   ( )
(   ) 2
y 0

a

2
2
  ( )   e  a e y L( y )dy,

(   ) 2
y 0

 L( a )   L( a ).


From (14) we can receive the following of system of the linear
algebraic equations to d 1 ( ), d 2 ( ) and d 3 ( ) .
(8)
Than (7) we can write, the equation (8) in the following form
[(    ) 2  (   k 2 ( ))(   k 3 ( ))]e k ( ) a 

a
(   k 2 )(   k 3 )e d1 ( ) 
[(    ) 2  (   k ( ))(   k ( ))]e k ( ) a 
1
3

 (   k 1)(   k 3 )e a d 2 ( ) 

k ( ) a
2

[(    )  (   k1 ( ))(   k 2 ( ))]e
 (   k ( ))(   k ( )) e a d ( ) 
1
2
3

  ( )( 2   2  2 e  a ),

k ( ) a
2

[(    ) k1 ( )   (   k 2 ( ))(   k 3 ( ))]e

a
  (   k 2)(   k 3 )e d1 ( ) 
[(    ) 2 k ( )   (   k ( ))(   k ( ))]e k ( ) a 
2
1
3

  (   k 1)(   k 3 )e a d 2 ( ) 

k ( ) a
2

[(    ) k 3 ( )   (   k1 ( ))(   k 2 ( ))]e
  (   k )(   k )e a d ( ) 
1
2
3

 2  ( )(1  e  a ),

k ( ) a
 (   k 2 ( )) k 2 ( )e k ( ) a 
(   k1 ( )) k1 ( )e
 (   k ( )) k ( )e k ( ) a  0.
3
3

From (15) we find that
1
L(0)  
2  3   b

1
(    b)(  3 (2    b) 2  44  ) 25 (2    b)e 2 a 

 4(  2  )
2  3   b
 (2    b)a 2
 (2    b) 2  44 e 2 a 

8(  2 )
2
3

2  3  b
1
(    b)(  3 (2    b) 2  44  ) 25 (2    b)e 2 a 
4(  2 )

1
(14)
2
2  3  b
 (2    b)a 2
 (2    b) 2  44 e 2 a 
8(  2  )
 2  3 (    b) 23  4 (    b)a   3 2b a



e
  2
(  2 )(   b) 

 2  3 (    b) 23  4 (    b)a   3 2b a



e
  2
(  2  )(   b) 

3
1
 a
2
2
2 (0)
 (0)e a 
 (0)e a 
 (0) 

  2
  2
  2
f1



 3 (2   )b  a  3 (2  3 2 )b
e 
 4 ab  5 a 2 b e (  2  ) a 
  2
   2


We know that
2
D ( )  L (0)  [ L (0)] 2 .
3
Proved that
To find d i ( ), i  1,3 we must find d i (0),i  1,3 .
IV.
THE FINDING L( )
It is obvious, that
a
L( ) 

L( z )dPmin( a,  1 )  z 
  2
E ()
a0
 (0)  0
a
2
 (0)  0
  2
z 0
 L( z)d 1  Pmin( a,
a


1
)  z 
(15)
z 0
 L( 0) P  a

1
a
 L( z)d P
z
 z.

1
REFERENCES
[1]
z 0
L( )  L a (1  a)e
[2]
a
 a

2
 ze
 z
L( z )dz.
z 0
For applications we find the expectation and variance of the
distribution of the random variable  ( ) . We know that
E ( )  L(0)
[3]
[4]
[5]
B.L. Smit, “The renewall processes”, // Mathematics, 1961, vol. №3б,
p.5.
I.I. Gihman, A.V. Skorohod., “Theory of random processes”, M., Nauka,
1973, vol.2, 639 p.
T.H. Nasirova “Complex process of semimarkovian random walk with
screen”, B.:Science, 1988, 50 p., (Russian).
T.H Nasirova, R.H. Sadiqova, “The Laplas transformation of the
distribution of the length of the time of the stay in given region”, AVT,
ISSN 0132-4160, Riga, 2009, No.4, pp.30-36., (Russian).
A.A. Borovkov, “The stochastic process in the Queuing Theory”,
Moscow, Nauka, 1972, 368 p., (Russian).