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How can I simulate this code with 1000 simulations?

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If I run a chain of n = 50 with four states, then how can I adjust the following code to run for 1000 times.
% I want to run this code 1000 times?
n = 50; % length of the chain
X = zeros(n,1); % matrix to save the chain in future
mu = [1/4 1/4 1/4 1/4]; % initial position of the chain
P = [0 1/2 0 1/2; 1/2 0 1/2 0; 0 1/2 0 1/2; 1/2 0 1/2 0]; % probability transition matrix
U = rand; % generate a random number
X_0 = min(find(U <= cumsum(mu))); % get the next state
U = rand;
X(1) = min(find(U <= cumsum(P(X_0,:))));
% do this process for the entire length (n) of the chain
for i = 1:n-1
U = rand;
X(i + 1) = min(find(U <= cumsum(P(X(i),:))));
end
How can I run this chain e.g. 1000 times
The above code is just an example, in practice, I have P = cell array. It is difficult to provide the whole data here thats why I tried to come up with a small example.
I tried to do something like this that but it did not work.
n = 50; % length of the chain
simulations = 1000
X = zeros(n,simulations); % matrix to save the chain in future
mu = [1/4 1/4 1/4 1/4]; % initial position of the chain
P = [0 1/2 0 1/2; 1/2 0 1/2 0; 0 1/2 0 1/2; 1/2 0 1/2 0]; % probability transition matrix
for k = 1:simulations
U = rand(1,m); % generate a random number
X_0 = min(find(U <= cumsum(mu),2)); % get the next state
U = rand(1,m);
X(1,m) = min(find(U <= cumsum(P(X_0,:))));
% do this process for the entire length (n) of the chain
for i = 1:n-1
U = rand;
X(i + 1) = min(find(U <= cumsum(P(X(i),:))));
end
end
  4 Commenti
susman
susman il 26 Feb 2021
I have just rephrased the questions. I tried to provide the small example to understand how does the code work. If its still not clear, I would try to rephrase agaib. Really sorry for that.
Walter Roberson
Walter Roberson il 2 Mar 2021
What is m? You have
U = rand(1,m);
X(1,m) = min(find(U <= cumsum(P(X_0,:))));
Your U is going to be a row vector, but P(X_0,:) is going to be a row vector, so you have <= between a row vector and a column vector, which is going to get you an m by size(P,2) -> m x 4 array. But then you try to store that into the scalar location X(1,m)

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