I have a question about standard assumptions that are made on the Levy measure when defining an infinitely divisible distribution on $\mathbb{R}$. The characteristic function is of the form: $$ \hat{\mu}(z) = \text{exp}[i\gamma z - \frac{\sigma^2 z^2}{2} + \int_{\mathbb{R}}(e^{izx} - 1 - izx \mathbb{1}_{|x|\leq 1})\nu(dx)], $$ where $(\sigma, \nu, \gamma)$ is a characteristic triple.
The typical assumptions that are made on $\nu$ are as follows: $$ \nu(\{0\})=0 \quad \text{and}\quad\int_{\mathbb{R}}(1\wedge x^2)\nu(dx)<\infty $$
My questions are as follows:
- Why do we need the assumption $\nu(\{0\})=0$?
- From my understanding the assumption $\int_{\mathbb{R}}(1\wedge x^2)\nu(dx)<\infty$ guarantees that the integral $\int_{\mathbb{R}}(e^{izx} - 1 - izx \mathbb{1}_{|x|\leq 1})\nu(dx)$ is finite. The only explanation I found states that function $e^{izx} - 1 - izx \mathbb{1}_{|x|\leq 1}$ behaves like $x^2$ for $x\rightarrow 0$ and like a constant as $x\rightarrow\infty$. It is not clear to me.
Levy measure is about the distribution of jumps of different sizes, and having jumps of size zero is just pointless.
If the measure isn't integrable at infinity, then there are infinitely many large jumps in any interval, so the process is just not well defined. However, there still can be infinitely many jumps of small (vanishing) sizes; integrability at zero ensures that they are summable (basically through the Kolmogorov one-series theorem).