annotate toolboxes/FullBNT-1.0.7/KPMstats/mc_stat_distrib.m @ 0:cc4b1211e677 tip

initial commit to HG from Changeset: 646 (e263d8a21543) added further path and more save "camirversion.m"
author Daniel Wolff
date Fri, 19 Aug 2016 13:07:06 +0200
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Daniel@0 1 function pi = mc_stat_distrib(P)
Daniel@0 2 % MC_STAT_DISTRIB Compute stationary distribution of a Markov chain
Daniel@0 3 % function pi = mc_stat_distrib(P)
Daniel@0 4 %
Daniel@0 5 % Each row of P should sum to one; pi is a column vector
Daniel@0 6
Daniel@0 7 % Kevin Murphy, 16 Feb 2003
Daniel@0 8
Daniel@0 9 % The stationary distribution pi satisfies pi P = pi
Daniel@0 10 % subject to sum_i pi(i) = 1, 0 <= pi(i) <= 1
Daniel@0 11 % Hence
Daniel@0 12 % (P' 0n (pi = (pi
Daniel@0 13 % 1n 0) 1) 1)
Daniel@0 14 % or P2 pi2 = pi2.
Daniel@0 15 % Naively we can solve this using (P2 - I(n+1)) pi2 = 0(n+1)
Daniel@0 16 % or P3 pi2 = 0(n+1), i.e., pi2 = P3 \ zeros(n+1,1)
Daniel@0 17 % but this is singular (because of the sum-to-one constraint).
Daniel@0 18 % Hence we replace the last row of P' with 1s instead of appending ones to create P2,
Daniel@0 19 % and similarly for pi.
Daniel@0 20
Daniel@0 21 n = length(P);
Daniel@0 22 P4 = P'-eye(n);
Daniel@0 23 P4(end,:) = 1;
Daniel@0 24 pi = P4 \ [zeros(n-1,1);1];
Daniel@0 25
Daniel@0 26