Calculating Probability of Markov Chain State in Discrete Time

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SUMMARY

This discussion focuses on calculating the probability of a specific state in a discrete-time Markov chain, denoted as P(X_i = 3) for i ranging from 0 to 8. The user initially sought a manageable expression for this probability using the transition matrix P = [pij] and the initial probability distribution p^(0). The solution was found by transforming state 3 into an absorbing state, leading to the conclusion that P(∪_{i=0}^8 X_i = 3) simplifies to P(X_8 = 3).

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Zaare
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Let [tex]\left( {X_n } \right)_{n \ge 0}[/tex] be a Markov chain (discrete time).

I have

[tex]{\bf{P}} = \left[ {pij} \right]_{i,j} = \left[ {P\left( {X_1 = j|X_0 = i} \right)} \right]_{i,j}[/tex],

and the initial probability distribution [tex]{\bf{p}}^{\left( 0 \right)}[/tex].

I need to calculate

[tex]P\left( {\mathop \cup \limits_{i = 0}^8 X_i = 3} \right)[/tex]

I can use Matlab for the numerical calculations but I need to find an expression for this probability. The only expression I have been able to find consists of too many terms to be considered reasonable.
Any suggestions on how to find an expression, that at least in some way is repetitive so that the numerical calculations can be done by the computer, would be appreciated.
 
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Hehe, nevermind, I figuered it out. If the state 3 is turned into a absorbing state, then:

[tex]P\left( {\bigcup\limits_{i = 0}^8 {X_i = 3} } \right) = 1 - P\left( {\bigcup\limits_{i = 0}^8 {X_i \ne 3} } \right) = 1 - \left( {1 - P\left( {X_8 = 3} \right)} \right) = P\left( {X_8 = 3} \right)[/tex]
 
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