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Diffstat (limited to 'sourcecodes/bnt-master/KPMstats/mc_stat_distrib.m')
| -rw-r--r-- | sourcecodes/bnt-master/KPMstats/mc_stat_distrib.m | 26 |
1 files changed, 26 insertions, 0 deletions
diff --git a/sourcecodes/bnt-master/KPMstats/mc_stat_distrib.m b/sourcecodes/bnt-master/KPMstats/mc_stat_distrib.m new file mode 100644 index 00000000..a5806092 --- /dev/null +++ b/sourcecodes/bnt-master/KPMstats/mc_stat_distrib.m @@ -0,0 +1,26 @@ +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]; + + |