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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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    volatility = standard deviation2012-05-27

1 Answers 1

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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 $\sigma(t) = \sqrt{\lambda t}$.