How to obtain transition probability matrix in matrix?

问题

Suppose I have a sequence x= 1,3,3,1,2,1,4,2,3,1,4,2,4,4,4,3,1,2,5,1 and it has five states 1 3 2 4 5. I have to obtain transition probability matrix in MATLAB by this equation, probability= (Number of observation pairs x(t) & x(t+1), with x(t) in state i and x(t+1) in state j)/(Number of observation pairs x(t) & x(t+1), with x(t) in state i and x(t+1) in any one of the states 1......s). I tried by this code but it giving error

x=[1 3 3 1 2 1 4 2 3 1 4 2 4 4 4 3 1 2 5 1]
n = length(x)-1
p = zeros(5,5)
for t = 1:n
if x(t)=x(t+1);
   a(t)=count (x(t)=x(t+1)) % Here i am trying to count how many number of times pair of that states occur in sequence.
   q(t)=sum(x==x(t)) % (Here i am trying to count Number of observation pairs x(t) & x(t+1), with x(t) in state i and x(t+1) in any one of the states 1......s)
end
for i=1:5
p(i, :) = a(t)/q(t)
end

Transition probability matrix calculated manually by me as follows

                     1      3     2    4     5

                1    0     1/5   2/5  2/5    0

                3    3/4   1/4    0    0     0

                2    1/4   1/4    0    1/4   1/4

                4     0    1/5   2/5   2/5   0

                5     1     0     0     0    0

回答1:

Since it has been a while, I think it is safe to provide an answer to this now. There is no toolbox required for either approach below. Assumes basic knowledge of a transition probability matrix of a Discrete Time Markov Chain (DTMC).

Both approaches use the unique function to find the statespace. Note that the order is different, e.g. your [1 3 2 4 5] vs. my [1 2 3 4 5] but that isn't a limiting issue. I've separated getting the transition counts from the transition probabilities to illustrate some techniques.

Approach 1: Vectorized Approach
This approach uses the unique and accumarray functions.

% MATLAB 2018b
X =[1 3 3 1 2 1 4 2 3 1 4 2 4 4 4 3 1 2 5 1];
[u,~,n] = unique(X);
NumU = length(u);     % Number of Unique Observations
Counts = accumarray([n(1:end-1),n(2:end)],1,[NumU,NumU]);

P = Counts./sum(Counts,2);   % Probability transition matrix

Verification: You can verify that sum(sum(Counts)) == length(X)-1 and the rows of P sum to one (sum(P,2)).

Notice that the counts matrix uses a 1-step offset to count the transitions. The output is a NumU x NumU array of the number of transitions in terms of indices as given in the n-output from unique.

Approach 2: Single for loop
This is a direct approach that can use any ordering of the statespace (see below).

States = unique(X);

Counts = zeros(length(States));
for k = 2:length(X)
    Counts(find(X(k-1) == States),find(X(k) == States)) = ...
        Counts(find(X(k-1) == States),find(X(k) == States)) + 1;
end

P = Counts./sum(Counts,2);   % Probability transition matrix

Using your statespace ordering: If you use Approach 2 with States = [1 3 2 4 5];, the resulting probability transition matrix, P, matches the one you manually calculated.

来源：https://stackoverflow.com/questions/37054721/how-to-obtain-transition-probability-matrix-in-matrix

标签

matlab

markov-chains