I have trouble finding out why this condition

$\int_{\mathbb{R}\backslash\{0\}} \min(1, x^2 ) \nu(dx) < \infty$

in the Lévy-Khintchine formula is necessary. The Lévy-Khintchine formula is defined as

$\varphi_X(u) := \mathbb{E}[{e^{iu X_1}}] = \exp \Bigl( aiu - \tfrac{1}{2}\sigma^2u^2 + \int_{\mathbb{R}\backslash\{0\}} \Bigl( e^{iu x}-1 -iu x \mathbb{I}_{|x|<1} \Bigr) \,\nu(dx) \Bigr)$.

I have been told it can be done by a Taylor expansion. I have tried very hard, but I don't know how to solve it.


1 Answer 1


The condition ensures that the integral

$$\int_{\mathbb{R} \backslash \{0\}} (e^{iux}-1-iux 1_{\{|x|<1\}}) \, \nu(dx)$$

is well-defined.

Since $|e^{iux}| \leq 1$, we have

$$|e^{iux}-1| \leq 2$$

for all $x \in \mathbb{R}$. On the other hand, applying Taylor's formula

$$f(x) = f(0)+f'(0) x + \frac{f''(\xi)}{2} x^2$$

(here $\xi \in (0,x)$ is some intermediate value) for $f(x) := e^{iux}$, we obtain

$$|e^{iux}-1-iux| \leq \frac{1}{2} x^2 u^2.$$

since $$|f''(\xi)| = |i^2 u^2 e^{iu \xi}| \leq u^2$$ for any $\xi \in \mathbb{R}$. Combining both estimates, we get

$$\begin{align*} \int_{\mathbb{R} \backslash \{0\}} |e^{iux}-1-iux 1_{\{|x|<1\}}| \, \nu(dx) &= \int_{(-1,1) \backslash \{0\}} |e^{iux}-1-iux| \, \nu(dx) + \int_{|x| \geq 1} |e^{iux}-1| \, \nu(dx) \\ &\leq \frac{1}{2} u^2 \int_{(-1,1) \backslash \{0\}} x^2 \, \nu(dx) + 2 \int_{|x| \geq 1} \nu(dx) \\ &= \frac{1}{2} u^2 \int_{(-1,1) \backslash \{0\}} \min\{x^2,1\} \, \nu(dx) + 2 \int_{|x| \geq 1} \min\{x^2,1\} \nu(dx) \\ &<\infty. \end{align*}$$

  • $\begingroup$ Is there a similar way to prove this for multi-dimensional Levy processes? $\endgroup$
    – Leguan3000
    Commented Apr 12, 2022 at 22:13

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