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Quadratic Variation of a Poisson Process?

  1. Jan 30, 2010 #1
    Hey guys,

    This is my first post on PhysicsForums; my friend said that this was the best place to ask questions about math.

    Anyways, I have to find the Quadratic Variation of a Poisson Process.

    My professor doesn't have a class textbook (just some notes that he's found online), and although I can find a general formula for quadratic variation, I can't seem to plug it in, and get a non-zero answer. My professor said that this question should be pretty easy, but I'm totally lost.

    If you guys could offer any help, that would be great. If you guys have any more questions, let me know!

    Thanks!
     
  2. jcsd
  3. Jan 30, 2010 #2
    So here is what I have so far:

    The quadratic variation is the sum of all [N_t_(i+1)-N_t_i]^2, with the max |N_t_(i+1)-N_t_i| --> 0.

    Since Poisson distributions are independent, and we want to find N(t), partition the interval [0,t] to n subintervals. Let h = max |N_t_(i+1)-N_t_i|. So, since I'm taking the limit as h becomes arbitrarily tiny, I can rewrite this sum as n*(E(N_h))^2 right? But then, I get n*(lam*(1\n))^2, which goes to zero. This can't be right, so I must have made at least one error here right?
     
  4. Jan 31, 2010 #3
    First, are you trying to find the expected value of the quadratic variation? Second, are you sure it's not max ti+1-ti --> 0. In other words, the largest subinterval of time goes to zero? If it is N(ti+1) - N(ti) --> 0, then yeah, it looks like the answer would have to be zero.

    I am not familiar with the term quadratic variation, but I would guess that it is supposed to result in a Riemann sum, or integral.
     
  5. Jan 31, 2010 #4

    gel

    User Avatar

    A Poisson process has quadratic variation equal to itself.

    This is because it is a pure jump process with jump sizes equal to 1. If Nt is the Poisson process then the only contribution to the quadratic variation [N] comes from the jumps,

    [tex]
    [N]_t=\sum_{s\le t} \Delta N_s^2 = \sum_{s\le t,\Delta N_s\not=0}1 = \sum_{s\le t}\Delta N_s=N_t.
    [/tex]
     
  6. Jan 31, 2010 #5

    gel

    User Avatar

    Just to add - the distribution of the process doesn't matter. Once you know that it is piecewise constant then you can conclude that the quadratic variation is just the sum of the squares of the jumps, which is easily calculated as I did above.
     
  7. Jan 31, 2010 #6
    That makes a lot of sense. Thank you!
     
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