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

  1. #1
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    Poisson process: compute E[N(3) |N(2),N(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!

    [note: also under discussion in Talk Stats Forum]
    Last edited by kingwinner; December 13th 2009 at 04:03 AM.
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  2. #2
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    I think this might have something to do with "Markov property"?
    So I guess E[N(3) |N(2),N(1)] = E[N(3)]??

    But to use Markov property, it has to be a Markov process. Is Poisson process considered to be a Markov process? Why or why not?

    Thanks for any input.
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  3. #3
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    Quote Originally Posted by kingwinner View Post
    Is Poisson process considered to be a Markov process?
    It is not "considered" to be a Markov process. It is a Markov process! That results (with some care) from the memoryless property of exponential distributions.

    But you don't need this here. When you write E[N(2)+N(2,3)|N(2)]=N(2)+E[N(2,3)], that is also true conditionally to (N(1),N(2)) for the exact same reasons.
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    Quote Originally Posted by Laurent View Post
    It is not "considered" to be a Markov process. It is a Markov process! That results (with some care) from the memoryless property of exponential distributions.

    But you don't need this here. When you write E[N(2)+N(2,3)|N(2)]=N(2)+E[N(2,3)], that is also true conditionally to (N(1),N(2)) for the exact same reasons.
    E[N(3) |N(2),N(1)]
    =E[N(2)+N(2,3] |N(2),N(1)]
    =E[N(2)|N(2),N(1)] + E{N(2,3] |N(2),N(1)}

    1) N(2,3] is independent of N(2) and N(2,3] is independent of N(1), but N(2) is NOT independent of N(1), and so N(2,3],N(2),N(1) are NOT mutually independent.
    So can we still claim that E{N(2,3] |N(2),N(1)}=E{N(2,3]} ? Why or why not?

    2) What is the fastest way to calculate E[N(2)|N(2),N(1)] ?

    Thanks for your help!
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  5. #5
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    1) I was told that there is difference between "pairwise" indepdence and "mutual" independence. In the situation above, I think we need "mutual" independence of N(2,3],N(2),and N(1) in order to claim that E{N(2,3] |N(2),N(1)}=E{N(2,3]}, is that right?

    2) I am still working on computing E[N(2)|N(2),N(1)], but I looked in my statistics textbook and I can't even find the definition. The textbook gives definition of the following: let X and Y both be discrete random variables, then E(Y|X=x)=∑ y P(Y=y|X=x), but it never defined what conditioning on TWO discrete random variables mean??

    Thanks for any help!
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  6. #6
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    Quote Originally Posted by Laurent View Post
    When you write E[N(2)+N(2,3)|N(2)]=N(2)+E[N(2,3)], that is also true conditionally to (N(1),N(2)) for the exact same reasons.
    When I wrote that, I litteraly meant that E[N(2)+N(2,3)|N(2),N(1)]=N(2)+E[N(2,3)] holds.

    The term N(2) should be obvious: it is a function of the variable we are conditioning by, therefore the conditional expectation acts trivially. (Conditioning by N(1),N(2) is like conditioning by the \mathbb{N}^2-valued random variable (N(1),N(2)), or by the sigma-algebra of events depending on N(1) and N(2))

    The second term is because N(2,3) is independent of (N(1),N(2)) (it is even independent of (N(t))_{0\leq t\leq 2}). There is no mutual independence between N(2,3),N(1),N(2): it would mean that N(1),N(2) are independent...

    As you see, these are the same arguments as for 1 random variable.
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  7. #7
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    Quote Originally Posted by Laurent View Post
    When I wrote that, I litteraly meant that E[N(2)+N(2,3)|N(2),N(1)]=N(2)+E[N(2,3)] holds.

    The term N(2) should be obvious: it is a function of the variable we are conditioning by, therefore the conditional expectation acts trivially. (Conditioning by N(1),N(2) is like conditioning by the \mathbb{N}^2-valued random variable (N(1),N(2)), or by the sigma-algebra of events depending on N(1) and N(2))

    The second term is because N(2,3) is independent of (N(1),N(2)) (it is even independent of (N(t))_{0\leq t\leq 2}). There is no mutual independence between N(2,3),N(1),N(2): it would mean that N(1),N(2) are independent...

    As you see, these are the same arguments as for 1 random variable.
    I am still struggling to understand why E[N(2)|N(2),N(1)]=N(2).
    I know that E[f(Y)|Y]=f(Y), so we have that E[N(2)|N(2)]=N(2). This is OK!
    But for E[N(2)|N(2),N(1)], it is conditional on N(2) AND also conditional on N(1). And N(2) is not independent of N(1), so we can't remove the condition N(1) on the right side.
    So how can we compute it?

    Thank you for your time and patience for explaining it!!
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  8. #8
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    Quote Originally Posted by kingwinner View Post
    I am still struggling to understand why E[N(2)|N(2),N(1)]=N(2).
    If you know N(1) and N(2), then in particular you know N(2)... I don't know what to add.

    The expectation is the best approximation given the knowledge of some r.v.; here you can do better than barely approximate since you know the variable itself...

    This is an instance of E[f(Y)|Y]=f(Y), where Y=(N(1),N(2)) and f(u,v)=v...
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