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annotate toolboxes/FullBNT-1.0.7/KPMstats/mc_stat_distrib.m @ 0:e9a9cd732c1e tip

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first hg version after svn
author wolffd
date Tue, 10 Feb 2015 15:05:51 +0000
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wolffd@0 1 function pi = mc_stat_distrib(P)
wolffd@0 2 % MC_STAT_DISTRIB Compute stationary distribution of a Markov chain
wolffd@0 3 % function pi = mc_stat_distrib(P)
wolffd@0 4 %
wolffd@0 5 % Each row of P should sum to one; pi is a column vector
wolffd@0 6
wolffd@0 7 % Kevin Murphy, 16 Feb 2003
wolffd@0 8
wolffd@0 9 % The stationary distribution pi satisfies pi P = pi
wolffd@0 10 % subject to sum_i pi(i) = 1, 0 <= pi(i) <= 1
wolffd@0 11 % Hence
wolffd@0 12 % (P' 0n (pi = (pi
wolffd@0 13 % 1n 0) 1) 1)
wolffd@0 14 % or P2 pi2 = pi2.
wolffd@0 15 % Naively we can solve this using (P2 - I(n+1)) pi2 = 0(n+1)
wolffd@0 16 % or P3 pi2 = 0(n+1), i.e., pi2 = P3 \ zeros(n+1,1)
wolffd@0 17 % but this is singular (because of the sum-to-one constraint).
wolffd@0 18 % Hence we replace the last row of P' with 1s instead of appending ones to create P2,
wolffd@0 19 % and similarly for pi.
wolffd@0 20
wolffd@0 21 n = length(P);
wolffd@0 22 P4 = P'-eye(n);
wolffd@0 23 P4(end,:) = 1;
wolffd@0 24 pi = P4 \ [zeros(n-1,1);1];
wolffd@0 25
wolffd@0 26