Applied Mathematics
p-ISSN: 2163-1409 e-ISSN: 2163-1425
2016; 6(2): 36-39
doi:10.5923/j.am.20160602.03

Roza Shakenova , Alua Shakenova
Kazach National Research Technical University Named after K. I. Satpaev, Almaty, Kazakhstan
Correspondence to: Roza Shakenova , Kazach National Research Technical University Named after K. I. Satpaev, Almaty, Kazakhstan.
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Copyright © 2016 Scientific & Academic Publishing. All Rights Reserved.
This work is licensed under the Creative Commons Attribution International License (CC BY).
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In this paper authors proposed that the human-being’s state of health can be described as the function of three main factors: economics, ecology and politics of government. We obtained three models of the state of health from the worst to the best using Markov processes. We hope that our theoretical models can be applied in practice.
Keywords: Limiting probability, Probability of state, Markov Processes
Cite this paper: Roza Shakenova , Alua Shakenova , Applied Problems of Markov Processes, Applied Mathematics, Vol. 6 No. 2, 2016, pp. 36-39. doi: 10.5923/j.am.20160602.03.
The set of different states of one physical system with discrete states, where random process occurs, is finite or countable.![]() | (1) |
to another state
directly or through other states [2], [3]. Usually the graph of states describes the states visually, where the vertices of graph correspond to the states. If the probability of each state in the future for every time
of the system
with discrete states,,
depends on the state in the present
and does not depend on its behavior in the past
then the random process is Markov process. Let’s consider conditional probability of transferring of system S on the kth step in the state
, if it is known that it was in the state
on the (k-1)th step. Denote this probability by:![]() | (2) |
is the probability that the system remains in the state
on kth step. Probabilities
are transition probabilities of Markov chain on the kth step. Transition probabilities can be written in the form of square table (matrix) of size n. This is Stochastic matrix; the sum of all probabilities of one row is equal to
because the system can be in one of mutually exclusive states.![]() | (3) |

with their sum equal to one.Markov chain is called uniform, if transition probabilities do not depend on the step’s number (3).Let’s consider only uniform Markov chains to simplify life.
can occur together with only one of following events
which form the full group of pair wise mutually exclusive events, i.e.
and
Then the probability of
can be calculated using the formula of total probability.![]() | (4) |
are called hypotheses, and the values
- probabilities of hypotheses.Make hypothesis such that the system was in the state
at initial time with (k=0). The probability of this hypothesis is known and equal to
Assuming that this hypothesis takes place, the conditional probability of System
being in the state
on the first step is equal to transition probability
Applying formula for total probability we obtain the following:![]() | (5) |
on the first step, The probability of this hypothesis is known and equal to
Given this hypothesis, the conditional probability of the system being in the state
on the second step is equal to
Using the formula for total probability we obtain:
Applying this method several times we obtain recurrent formula:![]() | (6) |
the stationary mode sets up, at which the system continues to wander over states, but the probabilities of these states do not depend on the number of step. These probabilities are denoted final (limit) probabilities of Markov chains. The equations for such probabilities can be written using mnemonic rule: Given stationary mode, total probability flux of the system remains constant: the flow into the state s is equal to the flow out of the state s.![]() | (7) |
. Add normalization condition
to these n equations.
be human-being from certain ecological, economic sphere. It can be in the following states:
The problem: To construct the equation and find the final probabilities of human-being’s state of health. Solution: Let’s consider
on the graph. Two arrows are directed into this state; consequently, there are two terms for addition on the left side (7) for
(state
). One arrow is directed out of this state, subsequently, there is only one term on the right side (7) for
(state
). Hence, using balance condition (7), we obtain the first equation:![]() | (8) |
,The fifth equation is the normalization condition:
We rewrite the system of equations in the following way:
Let’s solve the system of equations. From 2) we find:
where
From 4) we find:
where
From 3) find:
where
From 1)find:
where
Giving corresponding values of probabilities:
Calculating the following values:
,According to the equality5) we have:
Normalization condition
works. We did not need the probabilities 
