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Exponential of (Markov Chain) Transition matrix

  1. Oct 6, 2009 #1
    Hi,

    I have a (markov chain) transition matrix X which I understand. In particular each row of this matrix sums to 1.

    I have used this transition matrix to construct it's generator, Y. I.e. Y is the continuously compounded transition matrix,

    X = exp(Y)
    X*X = exp(2Y), etc

    both X and Y are matrices.

    I am told that the sums of Y must sum to 0, but I can not see why this should be the case. Is it obvious?

    Many Thanks.
     
  2. jcsd
  3. Oct 7, 2009 #2
    Possible hint maybe? [itex]e^0=1[/itex]
     
  4. Oct 8, 2009 #3
    Thanks.

    But I have trouble visualizing this for a matrix. Is there aproof somewhere, or a text that you can recommend which gives examples and talks about the applications of taking exponentials or logs of matrices?

    Many thanks,
     
  5. Oct 8, 2009 #4
    It is a http://en.wikipedia.org/wiki/Matrix_exponential" [Broken] right ? If you write down the power series for it you will get a pattern.

    [tex]
    X = I + Y + \frac{Y^2}{2!} + \ldots
    [/tex]
    Now if you sum up the rows of X it is 1. On the right hand side you already get 1 from the identity matrix. So all contributions from the remaining terms must be zero right? So I will let you think if your condition is sufficient or necessary.
     
    Last edited by a moderator: May 4, 2017
  6. Oct 8, 2009 #5
    Cool. Thanks a lot!

    Thinking about matrices a little further. How does one raise a matrix to a decimal power. I know that in the case of an integer power:

    X^n = S.M^n.S^-1

    where S is the eigen vector matrix and M is the matrix with eigen values along the diagonal. Now if n is non integer, then does this still hold? Could we also do it via:

    Y = X^n
    ln(Y) = n.ln(X)
    Y = exp(n.ln(X))?

    much appreciated.
     
  7. Oct 8, 2009 #6
    yes, yes

    also:
    if [tex] X = S M S^{-1}[/tex] then [tex] e^X = S e^M S^{-1}[/tex], [tex] \log(X) = S \log(M) S^{-1}[/tex], etc.
     
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