# Characteristic function of Levy process with stopping time

I am trying to prove a Levy process "renews itself" at stopping time, which implies the strong Markov property of Levy process. And I have a difficulty to verify the following result:

Let $X$ be a Levy process and $T$ be a bounded stopping time. Show $$\frac{E[e^{iuX_{T+t}}]}{E[e^{iuX_{T+s}}]}=E[e^{iuX_{t-s}}],\quad t>s.$$

First I can't use $X_{T+t}-X_{T+s}$ is independent of $\mathcal{F}_{T+s}$ and has the same distribution as $X_{t-s}$ since this is the result I am trying to prove. Hence I think I should prove it from the property of Levy process.

By definition $E[e^{iuX_{T+t}}]=\int_\Omega e^{iuX_{T(\omega)+t}(\omega)}\,dP$. Also I know $E[e^{iuX_{t}}]/E[e^{iuX_{s}}]=E[e^{iuX_{t-s}}]$ for any $t>s$. But I don't know how to connect the desired result to this identity.

## 1 Answer

It is not difficult to show hat

$$T_j(\omega) := k 2^{-j} \qquad \text{for} \, \, \omega \in \{(k-1) 2^{-j} \leq T < k 2^{-j}\}, k \geq 1,$$

defines a sequence of discrete-valued stopping times such that $T_j \downarrow T$. By the right-continuity of the sample paths of $(X_t)_{t \geq 0}$, this implies by the dominated convergence theorem

$$\mathbb{E}e^{i u X_{T+t}} = \lim_{j \to \infty} \mathbb{E}e^{iu X_{T_j+t}} = \lim_{j \to \infty}\sum_{k \geq 1} \mathbb{E} \left( e^{iu X_{k 2^{-j}+t}} 1_{\{T_j = k 2^{-j}\}} \right).$$

Writing

$$X_{k 2^{-j}+t} = \big( X_{k2^{-j}+t}-X_{k2^{-j}+s} \big) +X_{k 2^{-j}+s}$$

and conditioning on $\mathcal{F}_{k2^{-j}+s}$ we get by the independence and stationarity of the increments

$$\mathbb{E}e^{iu X_{T+t}} = \mathbb{E}e^{iu X_{t-s}} \lim_{j \to \infty} \underbrace{\sum_{k \geq 1} \mathbb{E}(e^{iu X_{k 2^{-j}+s}} 1_{\{T_j = k 2^{-j}\}})}_{\mathbb{E}e^{iu X_{T_j+s}}}.$$

Using again the dominated convergence theorem, we conclude

$$\mathbb{E}e^{iu X_{T+t}} = \mathbb{E}e^{iu X_{t-s}} \mathbb{E}e^{iu X_{T+s}}.$$