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Poisson process: compute E[N(3) |N(2),N(1)]

  1. Dec 11, 2009 #1
    note: N(t) is the number of points in [0,t] and N(t1,t2] is the number of points in (t1,t2].

    Let {N(t): t0} be a Poisson process of rate 1.
    Evaluate E[N(3) |N(2),N(1)].

    If the question were E[N(3) |N(2)], then I have some idea...
    E[N(3) |N(2)]
    =E[N(2)+N(2,3] |N(2)]
    =E[N(2)|N(2)] + E{N(2,3] |N(2)}
    =N(2)+ E{N(2,3]} (independent increments)
    =N(2) + 1
    since N(2,3] ~ Poisson(1(3-2)) =Poisson(1)

    But for E[N(3) |N(2),N(1)], how can I deal with the extra N(1)?

    Thanks for any help! :)
     
  2. jcsd
  3. Dec 11, 2009 #2
    Apply the Markov property
     
  4. Dec 12, 2009 #3
    hmm...what is Markov property? How do we use it here?
    Does it mean that E[N(3) |N(2),N(1)] = E[N(3)]?

    Also, is Poisson process considered as a Markov process? Why or why not?

    Can someone please explain more?

    Any help is greatly appreciated!
     
  5. Dec 12, 2009 #4
    The wikipedia article is a good place to start.
     
  6. Dec 13, 2009 #5
    http://en.wikipedia.org/wiki/Continuous-time_Markov_process
    "Markov property states that at any times s > t > 0, the conditional probability distribution of the process at time s given the whole history of the process up to and including time t, depends only on the state of the process at time t."

    So according to this, it does not depend on ANY of its past history, and therefore E[N(3) |N(2),N(1)] = E[N(3)]???
     
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