The author of the paper I'm viewing says two things which I don't understand,
A characteristic function is infinitely divisible if and only if it is of the form:
$$\phi(t) = \exp\bigg\{ibt - \dfrac{1}{2}\sigma^2t^2 + \int_{-\infty}^{\infty} \bigg(e^{itx} - 1 - \dfrac{itx}{1 + x^2} \bigg)dv(x) \bigg\}$$
where $v$ is a measure assigning no mass to the origin... $v$ assigns finite mass to the intervals $(-\infty,-1)$ and $(1,\infty)$ but can assign infinite mass to the interval $[-1,1]$
- I don't really understand what he means by 'assigning mass' and why $v$ can assign infinite mass to that specific interval or why it assigns no mass to the origin. Forgive me if this is simple for anyone who learns Levy processes but could somebody explain this to me?
- In addition, I want to ask, why is the representation above equivalent to the following representation (think this has something to do with 1. too), for $M$ finite on compact sets:
$$\phi(t) = \exp\bigg\{ibt + \int_{-\infty}^{\infty} \bigg(e^{itx} - 1 - \dfrac{itx}{1 + x^2} \bigg)\dfrac{dM(x)}{x^2} \bigg\}$$
Regarding Question 1: "assigns no mass to the origin" means $\nu(\{0\})=0$. Similarly, "$\nu$ assigns finite mass to a certain interval $I$" means $\nu(I)<\infty$. So the assumptions on $\nu$ can be summarized as follows: $\nu$ has to satisfy $\nu(\{0\})=0$ and $\nu(\mathbb{R} \backslash [-1,1])<\infty$ but $\nu([-1,1])$ might be infinite.
Regarding Question 2: The key point is that $M$ might assign mass to the origin, i.e. $M(\{0\})$ does not need to be $0$. Since
$$\lim_{x \to 0} \frac{e^{itx}-1-itx/(1+x^2)}{x^2} = \frac{1}{2} (itx)^2 = - \frac{1}{2} t^2$$
we have
$$\int_{\mathbb{R}} \left( e^{itx}-1-\frac{itx}{1+x^2} \right) \frac{dM(x)}{x^2} = - \frac{1}{2} t^2 M(\{0\}) + \int_{\mathbb{R} \backslash \{0\}} \left( e^{itx}-1-\frac{itx}{1+x^2} \right) \frac{dM(x)}{x^2}.$$
If we set $\sigma^2 := M(\{0\})$ and $d\nu(dx) := 1_{\mathbb{R} \backslash \{0\}}(x) dM(x)/x^2$, then we get the standard representation for $\phi$:
$$\phi(t) = \exp \left(ibt - \frac{1}{2} \sigma^2 t^2 + \int_{\mathbb{R}} \left( e^{itx}-1-\frac{itx}{1+x^2} \right) d\nu(x) \right).$$
(You can easily verify that $\nu$ has all the properties I explained in the first part of my answer.)