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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

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

  • 0
    How to you define volatility?2012-05-27
  • 0
    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 $\sigma(t) = \sqrt{\lambda t}$.