It is widely known that the characteristic function of a compound Poisson process is $$ \phi_X(u) = \exp \left(t\lambda \int_{\mathbb{R}} (e^{iux}-1) F(dx) \right). $$ But if I try to derive it via the Levy-Khintchine formula using the triplet $(0, 0, \lambda F(dx))$, I get an additional term from the truncation term, like this:

$$ \begin{align} \phi_X(u) &= \exp \left(t\lambda \int_{\mathbb{R}} (e^{iux}-1-iux_{|x| \leq 1}) F(dx) \right)\\ &= \exp \left(t\lambda \left[ \int_{\mathbb{R}} (e^{iux}-1-iux_{|x| \leq 1}) F(dx) \right] \right)\\ &= \exp \left(t\lambda \left[ -iu\int_{-1}^1 xF(dx) + \int_{\mathbb{R}} (e^{iux}-1) F(dx) \right] \right) \end{align} $$ I don't see how the truncation term disappears, unless, of course, $F(dx)$ is symmetric around 0.

So my question is: How to derive the characteristic function correctly from the triplet?

Edit: After thinking about what the truncation term actually represents, I realise that $$ -iu\int_{-1}^1 xF(dx)$$ is actually the "small" jumps added as a drift term. Does that mean that the triplet of the process corresponding to $$ \phi_X(u) = \exp \left(t\lambda \int_{\mathbb{R}} (e^{iux}-1) F(dx) \right) $$ is actually $(0, \gamma, \lambda F(dx))$ where $\gamma$ is the "drift" corresponing to a specific truncation size?

  • 1
    $\begingroup$ Yes, exactly... $\endgroup$ – saz May 5 '14 at 16:05

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