Markov chain: join states in Transition Matrix

Question

I need to merge two state in transition Matrix:

For example: i have the matrix below

              A       B       C      D      E      F 
          A  0.5     0.4      0      0      0.1    0

          B  0.5     0.1      0.2    0.1    0.1    0

          C  0       0.1      0.9    0      0      0

          D  0       0        0      0.7    0.3    0

          E  0       0.2      0      0.7    0      0.1

          F  0       0        0      0.5    0      0.5

And i want to join the states D and E:

              A      B        C    (D+E)    F 
          A  0.5     0.4      0      ?      0

          B  0.5     0.1      0.2    ?      0

          C  0       0.1      0.9    ?      0

       (D+E) ?       ?        ?      ?      ?

          F  0       0        0      ?      0.5

what are the formulas to obtain the row and column (D+E)?

Using the constraint: "the sum over column must be equal to 1" is simple to calculate the elements:

(A,(D+E))=0.2

(B,(D+E))=0.2

(C,(D+E))=0.1

(F,(D+E))=0.5

how can I calculate the elements of row ((D+E),i)?

Answer 1

I think you can work it out by writing P(A | D + E) = P(A, D + E) / P(D + E) and then applying de Morgan's law and noting D and E are mutually exclusive. I get P(A | D + E) = (P(A | D) P(D) + P(A | E) P(E)) / (P(D) + P(E)), likewise for any other states. The marginal probabilities just the elements of the eigenvector with eigenvalue 1. Disclaimer: you'll want to verify this.

Answer 2

If your markov chain is aperiodic and irreducible as in your case, I think you can sum up the rows corresponding to "D" and "E" with weights "pai_D" and "pai_E". "pai_S" denotes the probability of being in State "S" after a long run. (called stationary state)