The Lévy–Khintchine formula says that any Lévy process, $X=(X(t), t \geq 0)$, has a specific form for its characteristic function. More precisely, for all $t \geq 0$, $u \in \mathbb R^d$: $$ \mathbb{E}\left(e^{i(u, X(t))}\right)=e^{t \eta(u)} $$ where: \begin{equation}\label{I}\tag{I} \eta(u)=i\left(b, u\right)-\frac{1}{2}(u, a u)+\int_{\mathbb{R}^d-\{0\}}\left[e^{i(u, y)}-1- i (u,y)\chi_{0<|y|<1}(y)\right] v(d y), \end{equation} Consider the case where $v \neq 0$. If $v$ is a finite measure we can rewrite $(\ref{I})$ as: $$ \eta(u)=i\left(b^{\prime}, u\right)-\frac{1}{2}(u, a u)+\int_{\mathbb{R}^d-\{0\}}\left(e^{i(u, y)}-1\right) v(d y), $$ where $b^{\prime}=b-\int_{0<|y|<1} y v(d y)$.
Suppose $v=\lambda \delta_h$, whith $\lambda>0$ and $\delta_h$ is a Dirac mass concentrated at $h \in \mathbb{R}^d-\{0\}$. In this case we can set $X(t)=b^{\prime} t+\sqrt{a} B(t)+N(t)$, where $B=(B(t), t \geq 0)$ is a standard Brownian motion and $N=(N(t), t \geq 0)$ is an independent process for which: $$ \mathbb{E}\left(e^{i(u, N(t))}\right)=\exp \left[\lambda t\left(e^{i(u, h)}-1\right)\right] $$ Note that $N$ is a Poisson process of intensity $\lambda$ taking values in the set $\{n h, n \in \mathbb{N}\}$, so that $P(N(t)=n h)=e^{-\lambda t}\left[(\lambda t)^n / n !\right]$ and $N(t)$ counts discrete events that occur at the random times $\left(T_n, n \in \mathbb{N}\right)$.
Interpretation:
$X$ follows the path of a Brownian motion with drift from time zero until the random time $T_1$. At time $T_1$ the path has a jump discontinuity of size $|h|$. Between $T_1$ and $T_2$ we again see Brownian motion with drift, and there is another jump discontinuity of size $|h|$ at time $T_2$. We can continue to build the path in this manner indefinitely.
Now, consider $v=\sum_{i=1}^m \lambda_i \delta_{h_i}$, where $m \in \mathbb{N}, \lambda_i>0$ and $h_i \in \mathbb{R}^d-$ $\{0\}$, for $1 \leq i \leq m$. We can then write $$ X(t)=b^{\prime} t+\sqrt{a} B(t)+N_1(t)+\cdots+N_m(t), $$ where $N_1, \ldots, N_m$ are independent Poisson processes (which are also independent of $B$ ); each $N_i$ has intensity $\lambda_i$ and takes values in the set $\left\{n h_i, n \in \mathbb{N}\right\}$ where $1 \leq i \leq m$. In this case, the trajectory of $X$ follows a Brownian motion with drift, interrupted by sudden jumps occurring at unpredictable moments. However, in contrast to the previous case, the magnitude of each jump can take on any of the $m$ values $\left|h_1\right|, \ldots,\left|h_m\right|$.
The book I'm reading, titled (Lévy Processes and Stochastic Calculus), states the following:
- In the general case where $v$ is finite, we can see that we have passed to the limit in which jump sizes take values in the full continuum of possibilities, corresponding to a continuum of Poisson processes. So a Lévy process of this type is a Brownian motion with drift interspersed with jumps of arbitrary size.
- Even when $v$ fails to be finite, if we have $\int_{0<|x|<1}|x| v(d x)<\infty$ a simple exercise in using the mean value theorem shows that we can still make this interpretation.
Questions
I have two questions:
How to formalize the first item of the quote above? More specifically, when the author says that the jump size is arbitrary, I imagine that I can choose any $h \in \mathbb R$ as the jump. For this, I would have a continuum of jumps and a continuum of Poisson Processes indexed by $h\in \mathbb R$. On the other hand, this reminds me a lot of a Compound Poisson Process, where the jumps are given by a random variable $Y$: $Z(t)=\sum_{j=1}^{N(t)} Y_j$. I don't know if this makes sense, but anyway, I can't formalize it.
My second question is about the second item above. I don't have much idea how to show (using the mean value theorem) that I can have the same interpretation as the previous case if $\int_{0<|x|<1}|x| v(d x)<\infty$, Even when $\nu$ fails to be finite.
