proof of the distribution of the first holding time of a Markov jump process

Discussion in 'CT4' started by forza_bologna, Aug 20, 2012.

  1. forza_bologna

    forza_bologna Keen member

    In chapter 4 at pages 29-30 it is demonstrated that the first holding time of a time-homogenous Markov jump process with transition rates miu ij is exponentially distributed with parameter lambda i.
    It is defined the event Bn = {Xkt/2^n = X0, where k=1,2,...,2^n).
    It is mentioned that B1 includes B2, B2 includes B1, Bn-1 includes Bn and so on. (And due to this, the intersection of Bm where m takes values from 1 to infinity is Bn).
    I do not undersand this: Shouldn't it be B1 is included in B2, B2 is included in B3 and so on?

    Can somebody explain me this inclusion relationship?
    What does n represent?
     
  2. phani987

    phani987 Member

    n represents the total time you waited which means that, suppose at time 1, it is B1 and time t2 it is B2 and so on. What we should understand is that, B1 is bigger than B2 and so on. If strill you are not clear, I would elobrate on it.
     
  3. Mark Mitchell

    Mark Mitchell Member

    n does not represent the total time waited. The time period considered is from 0 to t. n reflects the number of divisions of the time interval from 0 to t.

    If n = 1, there are 2^1 = 2 divisions of the time interval (ie it is split in half)
    If n = 2, there are 2^2 = 4 divisions of the time interval (ie it is split into quarters).
    As n increases the time interval from 0 to t is split down into more and more parts.

    B1 is the set of events X(0) = X(t/2) = X(t)
    B2 is the set of events X(0) = X(t/4) = X(2t/4) = X(3t/4) = X(t)

    If an event is in B2, then it is automatically in B1. If an event is in B1, it is not automatically in B2 (as the extra requirements might not hold). So B2 is wholly within (is a subset of) B1.
     

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