how to programme to calculate transition probability matrix?

Suppose I have a sequence of states like 1,3,3,1,2,1,4,2,3,1,4,2,4,4,4,3,1,2,5,1. Transition probability matrix calculated by following equation probability=(number of pairs x(t) followed by x(t+1))/(number of pairs x(t) followed by any state). transition probability matrix calculated by 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

how to programme to obtain above transition probability matrix.

Answer 1

[James Tursa]

E.g., brute force:

m = max(x);
n = numel(x);
y = zeros(m,1);
p = zeros(m,m);
for k=1:n-1
    y(x(k)) = y(x(k)) + 1;
    p(x(k),x(k+1)) = p(x(k),x(k+1)) + 1;
end
p = bsxfun(@rdivide,p,y); p(isnan(p)) = 0;

The p matrix will be ordered by natural indexing. E.g., 1, 2, 3, 4, 5 for the above example. It will not be ordered as 1, 3, 2, 4, 5 as you have it above. I.e., the probability of going from state i to state j is p(i,j).

[Ram k]

But in that I also have to count no. of states so that I can replace m by total state number.

[James Tursa]

I don't understand your comment. Are you saying that you have arbitrary state numbers which are not necessarily 1 through m?

[Ram k]

I have a sequence in which states may not be start from 1 and also may not have subsequent numbers i.e. some numbers may be absent so sequence like this 12,14,6,15,15,15,15,6,8,8,18,18,14,14 so I want build transition probability matrix and it should be like below

                      6     8     12     14     15      18

                6     0     1/2   0       0      1/2     0

                8    0     1/2   0       0       0     1/2

                12   0      0    0       1       0      0

                14  1/2     0    0      1/2      0      0

                15  1/4     0    0       0      3/4     0

                18   0      0    0       0      1/2    1/2

I tried by above programme but matrix forms from number 1 to max number of state i.e. 18, so finally matrix becomes 18*18 order but i want matrix like above because states which are not present corresponding position becomes 0 and matrix becomes very large. In matrix redundancy occurs at places because of 0, so how to built matrix like above posted by me.

[Patrick Laux]

The brute force code above is giving 5 if you sum up all probabilities ??

Shouldn't it be: p = bsxfun(@rdivide,p,n) ?

However, this, i.e. sum(sum(p)) also gives only 0.95 (not 1). I am confused now.

Maybe: p = bsxfun(@rdivide,p,n-1) ??