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Poisson process $N(t)$ with density $\lambda$, could generate a compensated Poisson Process $$M(t) = N(t) - \lambda t,$$ $M(t)$ is a martingale with mean of $0$.

Now, how could I calculate the volatility of this compensated Poisson process $M(t)$?

ps. volatility = standard deviation, or, let mean $$\mu := E\{f(t)\}$$, then volatility

$$\sigma := E\{(f(t)-\mu)^2\} $$.

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How to you define volatility? –  Davide Giraudo May 27 '12 at 10:09
    
volatility = standard deviation –  athos May 27 '12 at 14:08
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1 Answer

up vote 1 down vote accepted

You know that the mean of the process is $0$, that's the point of being compensated. Hence, variation is just a second moment: $$ V(t) = \mathsf E[(N_t- \lambda t)^2] =\mathsf E[N_t^2] - 2\lambda t \mathsf E[N_t]+\lambda^2t^2 = (\lambda t + \lambda^2t^2)-2\lambda^2t^2+\lambda^2t^2 = \lambda t $$ so that $\sigma(t) = \sqrt{\lambda t}$.

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thanks, now i get it! –  athos May 28 '12 at 15:46
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