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Let ($\xi_t)_{t \geq 0}$ an infinititely divisible cadlag process on $[0,\infty)$ and denote by $p$ its jump measure. Define a measure $p_t$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ by $$p_t(B) := p((0,t] \times B) \qquad (B \in \mathcal{B}(\mathbb{R}))$$ Let $$\eta_t := \int x \cdot (1-1_{(-a,a)}(x)) \, p_t(dx)$$ where $a>0$. Then $\eta_t$ is infinitely divisble, cadlag, with $b=\sigma=0$ and Lévy measure $\Lambda(\Gamma \backslash (-a,a))$ where $\Lambda$ denotes the Lévy measure of $\xi_t$.

The last assertion refers to Khinchin's formula $$\mathbb{E}e^{\imath \, \lambda \cdot X_t} = \exp \left[ t \cdot \left( \int (e^{\imath \, \lambda \cdot x}-1- \imath \, \lambda \cdot \sin x) \, \Lambda(dx) + \imath \, b \cdot \lambda - \sigma^2 \cdot \frac{\lambda^2}{2} \right) \right] \tag{1} $$

My question is the following: The author proves $$\mathbb{E}e^{\imath \, \lambda \cdot \eta_t} = \exp \left(t \cdot \int_{\mathbb{R} \backslash (-a,a)} (e^{\imath \, \lambda \cdot x}-1) \, \Lambda(dx) \right)$$ But as far as I can see, this does not imply that $(1)$ holds with $b=\sigma=0$ and Lévy measure $\Lambda(\Gamma \backslash (-a,a))$. Instead, I would conclude $$\mathbb{E}e^{\imath \, \lambda \cdot \eta_t} = \exp \bigg[ t \cdot \bigg( \int_{\mathbb{R} \backslash (-a,a)} (e^{\imath \, \lambda \cdot x}-1-\imath \, \lambda \cdot \sin x) \, \Lambda(dx) + \imath \, \lambda \cdot \underbrace{\int_{\mathbb{R} \backslash (-a,a)} \sin x \, \Lambda(dx)}_{=:b} \bigg) \bigg]$$ i.e. $(1)$ holds with Lévy measure $\Lambda(\Gamma \backslash (-a,a))$, $\sigma=0$ and $b \not= 0$. Am I right or are there properties of the Lévy measure $\Lambda$ which imply $b=0$?

(Literature: "Introduction to the Theory of Random Processes" - N.V. Krylov (chapter 5, section 3))

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I also agree with you. Indeed, $b$ may fail to vanish even for the Poisson process of parameter $1$, whose Lévy measure being equal to $\Lambda(dx) = \delta_1(dx)$. – Sangchul Lee Mar 8 '13 at 17:55
@sos440 Thanks. The Poisson process is indeed a nice example to show that $b$ is not necessarily equal to 0. – saz Mar 8 '13 at 18:10
I guess that the author tried to figure out that the drift coefficient of the resulting process $\eta_t$ vanishes. This makes sense since $\eta_t$ has bounded variation. – Sangchul Lee Mar 8 '13 at 19:59
@sos440 Why does the drift coefficient vanish? I thought, $b$ (as in $(1)$) is called drift coefficient...? – saz Mar 9 '13 at 9:37
@sos440 Thank you for the clarification. So, in this case, the drift coefficient is equal to 0, since $\eta$ is a pure jump process. Am I right? – saz Mar 9 '13 at 10:18

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