Application of Lévy–Khinchine formula

Question

How can we express the characteristic functions of Wiener and Poisson processes by using the Lévy–Khinchine formula?

I don't know how to find the characteristic functions of particular Levy processes such as Wiener process by the formula. I think because I don't know how to work with the Levy measure.

P.S. Is there any Lévy–Khinchine type formula for other processes such as Markov process?

Answer 1

Depending on what you know about the Lévy triplet it is much easier to calculate the characteristic functions using the very definition of Wiener Process and Poisson process, respectively. For example, for a Wiener process $(W_t)_{t \geq 0}$, we know that $W_t \sim N(0,t)$ and since the exponential moments of the Normal distribution can be calculated explicitly, it follows easily that

$$\mathbb{E}e^{\imath \, \xi W_t} = \exp \left(- t \frac{|\xi|^2}{2} \right). \tag{1}$$

A similar argumentation applies for the Poisson process.

In contrast, if we want to use the (one-dimensional) Lévy-Khintchine formula

$$\mathbb{E}e^{\imath \, \xi X_t} = \exp \left[-t \left( \imath \, b \xi + \frac{\sigma^2}{2} |\xi|^2 + \int_{y \neq 0} (e^{\imath \, y \xi}-1-\imath \, y \xi 1_{|y| < 1}) \, \nu(dy) \right) \right],$$

then we have to interpret the Lévy triplet $(b,\sigma^2,\nu)$. This is a bit hand-waving (although it can be made rigorous), but for these easy examples it works well enough. The term $b$ corresponds to the drift, $\sigma^2$ describes the diffusion (i.e. Brownian motion-part) and $\nu$ the jumps.

1. Brownian motion: Since a Brownian motion has no drift and no jumps, it is an educated guess that $b=\nu=0$. Moreover, $\sigma^2$ equals the variation of the process and for a (Standard) Brownian motion, we therefore get $\sigma^2=1$. This agrees with $(1)$.
2. Poisson process: It is not difficult to see from the definition that a Poisson process has no drift-part and no Brownian motion part. Hence, $b=\sigma^2=0$. Consequently, $\nu$ contains the information about the intensity as well the heights of the jumps. One can show that $\nu(B)=0$ for some Borel set $B$, implies that the corresponding process has (a.s.) no jumps with height $B$. Since a Poisson process has only jumps of height $1$, we may guess that $\nu = \lambda \delta_{\{1\}}$ where $\delta_{\{x\}}$ denotes the Dirac measure and $\lambda$ some constant. Finally, $\lambda$ determines the intensity of the jumps. Consequently, the characteristic function equals $$\exp(-t \lambda (e^{\imath \, \xi}-1)).$$

The Lévy-Khintchine formula is strongly related to the Lévy-Itô decomposition which states that

$$X_t = bt + \sigma B_t + \int_0^t \!\!\! \int_{|z| \leq 1} z \, (N(dz,ds)-\nu(dz) \, ds) + \int_0^t \!\!\! \int_{|z| \geq 1} z \, N(dz,ds)$$

where $N$ denotes the jump measure of the process. Roughly speaking, this shows that any Lévy process can be written as "drift+Brownian motion+small jumps+large jumps".

In fact, there are generalizations of the Lévy-Khintchine formula for a larger class of Markov processes:

• Additive Processes: The Lévy triplet depends on the time $t$, i.e. $(b(t),\sigma^2(t),\nu(dt,dy))$.
• Lévy-Type Processes: Here the Lévy-triplet depends on the space variable $x$, i.e. $(b(x),\sigma^2(x),\nu(dx,dy))$. Basically, this means that the process behaves locally like a Lévy process, but changes its Lévy triplet when it moves on.