www.pudn.com > HMM1.zip > mc_stat_distrib.m
function pi = mc_stat_distrib(P)
> MC_STAT_DISTRIB Compute stationary distribution of a Markov chain
> function pi = mc_stat_distrib(P)
>
> Each row of P should sum to one; pi is a column vector
> Kevin Murphy, 16 Feb 2003
> The stationary distribution pi satisfies pi P = pi
> subject to sum_i pi(i) = 1, 0 <= pi(i) <= 1
> Hence
> (P' 0n (pi = (pi
> 1n 0) 1) 1)
> or P2 pi2 = pi2.
> Naively we can solve this using (P2 - I(n+1)) pi2 = 0(n+1)
> or P3 pi2 = 0(n+1), i.e., pi2 = P3 \ zeros(n+1,1)
> but this is singular (because of the sum-to-one constraint).
> Hence we replace the last row of P' with 1s instead of appending ones to create P2,
> and similarly for pi.
n = length(P);
P4 = P'-eye(n);
P4(end,:) = 1;
pi = P4 \ [zeros(n-1,1);1];