This question is related to the equilibrium probability distribution of a markov chain. I have: $ \vec {π} * \vec {P} = \vec {π}$
$ [π_0,π_1,π_2,π_3] $ [ $\begin{matrix} 1/3 & 2/3 & 0 & 0 \\ 1/6 & 1/2 & 1/3 & 0 \\ 0 & 4/9 & 4/9 & 1/9\\ 0 & 0 & 5/6 & 1/6 \end{matrix}$ ] = $ [π_0,π_1,π_2,π_3] $
Which produces:
$π_0=1/3 π_0 + π_11/6 \\\\\\\\\\\\\\\\π_1=2/3 π_0+π_11/2+π_2 4/9 \ \ \ \ \ \ \ \ \\\\\\π_2=π_11/3+π_24/9+ π_35/6 \\\\π_3 = π_2 1/9 + π_3 1/6 $
I also have the constraint that $\sum_{i=0}^3 π_i = 1$
How do I solve for each π?, I really only need to solve three of them and the fourth is the difference between the sum of the others and one.
Cheers.
Use Gaussian elimination to find all $\pi_i$ are 0.25.