关于求解马尔科夫链的平稳分布，大牛进！！！
本帖最后由 yangfan715 于 2012-3-18 12:26 编辑知道马尔科夫链的转移流量图，如何求解马尔科夫链的平稳分布？或者说，知道N*N矩阵中各元素之间的关系，如何求各元素的值？在线等。。。。。。
例如：状态（i，j），0<i,j<N，转移流程图如下：
如何求状态（i，j）的平稳分布？
{:{39}:} 一个马尔科夫链的matlab程序
function =markov(T,n,s0,V);
%function =markov(T,n,s0,V);
%chain generates a simulation from a Markov chain of dimension
%the size of T
%
%T is transition matrix
%n is number of periods to simulate
%s0 is initial state (initial probabilities)
%V is the quantity corresponding to each state
%state is a matrix recording the number of the realized state at time t
%
% Original author: Tom Sargent
% Comments added by Qiang Chen
=size(T); % r is # of rows, c is # of columns of T
if nargin == 1;% "nargin" refers to "number of arguments in". So only T is provided in this case
V=;
s0=1;
n=100;
end;
if nargin == 2;% both T and n are provided
V=;
s0=1;
end;
if nargin == 3;% T, n and S0 are provided
V=;
end;
% check if the transition matrix T is square
if r ~= c;
disp('error using markov function');
disp('transition matrix must be square');
return;% break the program and return
end;
% check if each row of T sums up to 1
for k=1:r;
if sum(T(k,:)) ~= 1;
disp('error using markov function')
disp(['row ',num2str(k),' does not sum to one']);% "num2str" converts numbers to a string.
disp(' it sums to :');
disp([ sum(T(k,:)) ]);
disp(['normalizing row ',num2str(k),'']);
T(k,:)=T(k,:)/sum(T(k,:));
end;
end;
=size(V);
if v1 ~= 1 | v2 ~=r % "|" means "or"
disp('error using markov function');
disp(['state value vector V must be 1 x ',num2str(r),''])
if v2 == 1 &v2 == r;
disp('transposing state valuation vector');
V=V';% change it to a column vector
else;
return;
end;
end
if s0 < 1 |s0 > r;
disp(['initial state ',num2str(s0),' is out of range']);
disp(['initial state defaulting to 1']);
s0=1;
end;
% The simulation of Markov chain formally starts from here
%T
%rand('uniform');
X=rand(n-1,1);% generate (n-1) random numbers drawn from uniform distribution on , each number to be used in one simulation.
s=zeros(r,1); % initiate the state vector "s" to be a rx1 zero vector
s(s0)=1; % change the "s0"th element of "s" to 1
cum=T*triu(ones(size(T)));
% "triu(ones(size(T)))" generates an upper triangular matrix with all elements equal to 1
% cum is a rxr matrix whose ith column is the cumulative sum from the 1st column to the ith column
% the ith row of cum is the cumulative distribution for the next period given the current state.
for k=1:length(X);% "length(X)" returns the size of the longest dimension of X. "k" indicates the kth simulation.
state(:,k)=s; % state is a matrix recording the number of the realized state at time k
ppi=; % this is the conditional cumulative distribution for the next period given the current state s
s=((X(k)<=ppi(2:r+1)).*(X(k)>ppi(1:r)))';
% compares each element of ppi(2:r+1) or ppi(1:r) with a scalar X(k), and
% returns 1 if the inequality holds and 0 otherwise
% this formula assigns 1 when both inequalities hold, and 0 otherwise
end;
chain=V*state;
我想问一下这个转移流量图是怎么建立的啊?{:{13}:}
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