# Defining Levy processes via characteristic functions. How do we prove independence

Following the notes of Liggett- Continuous Time Markov Chains pg 95 one finds the continuation in page 96 defines a Levy process I am having trouble in exercise 3.11.

I first noted that $\Bbb{E}[\exp\{i u \xi_{t+s}\}] =\exp\{(t+s)\psi(u)\}= \Bbb{E}[\exp\{i u \xi_{t}\}]\Bbb{E}[\exp\{i u \xi_{s}\}]$

But this is not enough to prove independence.

We can construct Levy processes from another perspective (see http://page.math.tu-berlin.de/~papapan/papers/introduction.pdf ):   We obtain the characteristic function we had on Liggett.

As the Characteristic function determines the distribution and the second construction yields the same characteristic function as the first one. Is it safe to say that the incremenets of the Levy Process are stationary and independent?

How can we prove this result without having to use this second construction? I mean, is there a way to prove independence and stationarity of increments using properties of the characteristic function itself?

## 1 Answer

Denote by $(X_t)_{t \geq 0}$ the Markov process associated with $(T_t)_{t \geq 0}$ and by $\mathcal{F}_t := \sigma(X_s; s \leq t)$ the canonical filtration. For any $s \leq t$, we have by the Markov property

$$\mathbb{E}(e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s) = \mathbb{E}^y e^{i \xi (X_{t-s}-y)} \big|_{y=X_s}.$$

Using the translational invariance, that is $$T_t f(x) = \mathbb{E}^x f(X_t) = \mathbb{E}f(x+X_t), \tag{1}$$ we get

$$\mathbb{E}(e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s) = \mathbb{E}e^{i \xi (X_{t-s}-y+y)} \big|_{y=X_s} = \mathbb{E}e^{i \xi X_{t-s}}.$$

Taking expectation on both sides shows that $X_t-X_s$ equals in distribution $X_{t-s}$, i.e. $(X_t)_t$ has stationary increments. Moreover, for any $F \in \mathcal{F}_s$, we get by the tower property

$$\mathbb{E}(e^{i \xi (X_t-X_s)} \cdot 1_F) = \mathbb{P}(F) \cdot \mathbb{E}e^{i \xi (X_t-X_s)}.$$

This implies that $(X_t)_{t \geq 0}$ has independent increments.

Remarks:

• Note that, for this proof, we do not need to know how the characteristic function of $X_t$ looks like, i.e. we do not need the Lévy-Khintchine formula. The only thing we use (apart from Markov property) is the translational invariance $(1)$ of the semigroup. This shows that Lévy processes are the only Markov processes with a translational invariant semigroup.
• Since your question is also about characterizing Lévy processes via characteristic functions, let me mention the following theorem:

Let $(X_t)_{t \geq 0}$ be a stochastic process with càdlàg sample paths adapted to a filtration $(\mathcal{F}_t)_{t \geq 0}$ and suppose that $X_0 = 0$. Then $(X_t)_{t \geq 0}$ is a Lévy process if, and only if, there exists a function $\psi$ such that $$\mathbb{E} \left( e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s \right) = e^{-(t-s) \psi(\xi)}$$ for all $s \leq t$ and $\xi \in \mathbb{R}^d$.