- #1
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I recently encountered a problem, which I, after several other dead ends, managed to formulate in terms of the matrix exponential of an infinite matrix. As I'm not terribly familiar with the mathematics involved, I wasn't quite able to do the exponentiation. Numerically, however, I did find some interesting recurrences, but couldn't get the general case written out with pen and paper, and thus couldn't quite access the asymptotic behaviour as the size of the matrix goes to infinity.
Researching online, I encountered queueing theory, and lo and behold, I found that my problem is exactly that of calculating the phase-type distribution: http://en.wikipedia.org/wiki/Phase-type_distribution.
So, my matrix has a block Toeplitz form, to be more specific, the diagonal is filled by a repeating 2x2 block and off to the right, I've got a 2x1 matrix. The rest of the elements of the transition rate matrix, as it is apparently called, are zero.
So basically I'm looking for an easy way do the calculation, as it turned out to be rather time-consuming trying to understand the established lingo in the literature. I couldn't even figure out if a general formula actually exists or not, although given my numerical trials, I'd suspect one does. I'd content if somebody told me that it doesn't, though.
Researching online, I encountered queueing theory, and lo and behold, I found that my problem is exactly that of calculating the phase-type distribution: http://en.wikipedia.org/wiki/Phase-type_distribution.
So, my matrix has a block Toeplitz form, to be more specific, the diagonal is filled by a repeating 2x2 block and off to the right, I've got a 2x1 matrix. The rest of the elements of the transition rate matrix, as it is apparently called, are zero.
So basically I'm looking for an easy way do the calculation, as it turned out to be rather time-consuming trying to understand the established lingo in the literature. I couldn't even figure out if a general formula actually exists or not, although given my numerical trials, I'd suspect one does. I'd content if somebody told me that it doesn't, though.