Quadratic Variation of a Poisson Process?

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SUMMARY

The quadratic variation of a Poisson process is equal to the process itself, represented mathematically as [N]_t = N_t. This is due to the nature of Poisson processes being pure jump processes, where the only contributions to the quadratic variation come from the jumps, which are of size 1. The calculation involves summing the squares of the jumps, leading to the conclusion that the quadratic variation is simply the count of jumps up to time t, confirming that the expected value of the quadratic variation is non-zero.

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RedZone2k2
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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!
 
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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?
 
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.
 
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,

<br /> [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.<br />
 
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.
 
That makes a lot of sense. Thank you!
 

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