Looping for Markov Chain

2 次查看(过去 30 天)
Wendell
Wendell 2012-10-25
I need to run the following:
P = [0.942 0 0 0 0.058 0 0 0; 0 0.982 0 0 0.018 0 0 0; 0 0 0.999 0 0 0.001 0 0; 0 0 0 0.993 0 0.007 0 0; 0 0 0 0 0.987 0 0.013 0; 0 0 0 0 0 0.999 0.001 0; 0 0 0 0 0 0 0.972 0.028; 0 0 0 0 0 0 0 1]; x_0 = [1000; 800; 300; 400; 700; 200; 300; 0]; n = 100; for i = 1:n x = P*x_0; x(i) = P^(i)*x_0; end disp (x(i)) plot (x(i),n)

回答(1 个)

Kye Taylor
Kye Taylor 2012-10-25
编辑:Kye Taylor 2012-10-25
You have all the pieces... its just awkward to organize (below I use cells) and display.. Try something like
n = 100;
P = cell(1,n);
P{1} = [0.942 0 0 0 0.058 0 0 0; 0 0.982 0 0 0.018 0 0 0; 0 0 0.999 0 0 0.001 0 0; 0 0 0 0.993 0 0.007 0 0; 0 0 0 0 0.987 0 0.013 0; 0 0 0 0 0 0.999 0.001 0; 0 0 0 0 0 0 0.972 0.028; 0 0 0 0 0 0 0 1];
x_0 = [1000; 800; 300; 400; 700; 200; 300; 0];
% I expect x_0 to be a probability
x_0 = x_0/sum(x_0);
% Create P, P^2, P^3,...
for i = 2:n
P{i} = P{1}*P{i-1};
end
g = @(A)A*x_0;
evolvedX = cellfun(g,P,'un',0); % applies P^k to x_0 for k = 1,2,3,...
plot([evolvedX{:}]')
xlabel('Time-step')
ylabel('Probability')
legend('x_0(1)','x_0(2)','x_0(3)','x_0(4)','x_0(5)','x_0(6)','x_0(7)','x_0(8)')
  1 个评论
Wendell
Wendell 2012-10-26
Kye I appreciate very much the help...glad to have someone willing to share their knowledge and skill...

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Markov Chain Models 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by