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I'm learning markov chains in order to compute estimations of transition probabilities, and I found an example of the estimator construction for continuous time markov chains:

http://www.rinfinance.com/agenda/2015/talk/AlexanderMcNeil.pdf

(pages 10 -12)

In its presentation, the author specifies that it's necessary to fix matrix diagonal and replace it by the negative of row sums without considering it's initial value:

    D <- rep(0, dim(Lambda.hat)[2])
    Lambda.hat <-rbind(Lambda.hat,D)
    diag(Lambda.hat) <- D
    rowsums <- apply(Lambda.hat,1,sum)
    diag(Lambda.hat) <- -rowsums

After that, he computes estimated transition probabilities:

    P.hat <- expm(Lambda.hat)

The output is consistent because each row adds up to one so it seems it works. Although, I don't understand why it was necessary to make these two steps and would really appreciate any help.

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There is no "initial" value of the diagonal entries in the infinitesimal generator matrix, a.k.a. transition rate matrix, I'll call it Q, of a continuous time Markov Chain. The author shows how to estimate the off-diagonal transition entries of the infinitesimal generator matrix. As is always the case with an infinitesimal generator matrix, each row must sum to zero, as described in https://en.wikipedia.org/wiki/Transition_rate_matrix . Therefore, the diagonal entries of the infinitesimal generator matrix are set equal to the negative of the sum of the entries in the corresponding row.

As described at https://en.wikipedia.org/wiki/Continuous-time_Markov_chain#Transient_behaviour , the transition probability matrix is calculated as $e^{t Q}$, where t is the length of time, in units corresponding to the intensity parameter entries in Q. The author in your link performed such calculation with annual intensity rates and t = 1, and therefore the transition probability matrix in his example corresponds to a one year period. The mathematics works out that if the row sums of Q equal zero (as they must), then the row sums of the transition probability matrix equal one, as is the case in the example in the link.

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  • $\begingroup$ when you said "The mathematics works out that if the row sums of A equal zero (as they must)", you meant the generator matrix "Q"? $\endgroup$ – José Vallejo Apr 26 '16 at 22:11
  • $\begingroup$ @José Vallejo , Yes, Q. Thanks for catching that typo. I just fixed it. $\endgroup$ – Mark L. Stone Apr 26 '16 at 22:19

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