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A question about Poisson process (waiting online)

  1. Jul 19, 2015 #1
    Hey guys, I encounter a question (maybe a silly one )that puzzles me. Nt is a Poisson process and λ is the jump intensity.Since the quadratic variation of Poisson process is [N,N]t=Nt, and Nt2-[N,N]t is a martingale, it follows that E[Nt2]=E[[N,N]t]=λ*t. On the other hand, the direct calculation of E[Nt2] can be found in https://proofwiki.org/wiki/Variance_of_Poisson_Distribution, which indicates E[Nt2]=(t*λ)2+t*λ. The two results are different. I really appreciate it if somebody can help me.
     
    Last edited: Jul 19, 2015
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  3. Jul 19, 2015 #2

    mfb

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    It would help to know the context of that calculation and the meaning of the individual symbols.
     
  4. Jul 19, 2015 #3
    Nt is a Poisson process and λ is the jump intensity.Since the quadratic variation of Poisson process is [N,N]t=Nt, and Nt2-[N,N]t is a martingale, it follows that E[Nt2]=E[[N,N]t]=λ*t. On the other hand, the direct calculation of E[Nt2] can be found in https://proofwiki.org/wiki/Variance_of_Poisson_Distribution
     
  5. Jul 19, 2015 #4

    mathman

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    It looks like someone has confused the second moment with the variance.
     
  6. Jul 19, 2015 #5
    Hi dude, seems like you know the answer. Could you explain?
     
  7. Jul 20, 2015 #6

    mathman

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    General formulas for a random variable X:
    second moment: [itex]E(X^2)[/itex]
    variance: [itex]E(X^2)-(E(X))^2[/itex]
     
  8. Jul 20, 2015 #7
    Yeah, I know that. But my question is, why would I get two different results of E[Nt2] using two different methods?
     
  9. Jul 21, 2015 #8

    mathman

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    My guess - a mistake in the derivation. [itex]E[N_t^2]=\lambda t[/itex] looks wrong (unless the mean=0). It is the variance.
     
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