Let $X$ be a right-continuous process with values in $(E,\mathcal{E})$, defined on $(\Omega, \mathcal{F}_t,P)$. Suppose that $X$ has stationary, independent increments.

I now want to show the following with knowledge that $X$ is in fact a Markov process:

Let $\tau$ be a finite $(\mathcal{F}_t)_t$-stopping time. Then the process $X(\tau) = (X_{\tau + t} - X_\tau)_{t \geq 0}$ is independent of $\mathcal{F}_\tau$ and it is a Markov process, adapted to the filtration $(\mathcal{F}_{\tau+t})_t$. The distribution $P_\tau$ of $X(\tau)$ is the same as the distribution of $X- X_0$ under $P_0$.

My attempt of a proof:

First how could I prove independence? However when moving on we can put $Y_t = X_{\tau+t} - X_\tau$, $t \geq 0$. For $t_1<\ldots < t_n$ and functions $f_1,\ldots, f_n \in b \mathcal{E}$ (bounded functions) we have \begin{align*} \mathbb{E}_\nu \left( \prod_k^n f_k (Y_{t_k}) \mid \mathcal{F}_\tau \right) &= \mathbb{E}_\nu \left( \prod_k^n f_k (X_{\tau+t_k} - X_\tau) \mid \mathcal{F}_\tau \right) \\ &= \mathbb{E}_{X_\tau} \left( \prod_k^n f_k (X_{t_k} - X_0)\right) \end{align*} $P_\nu$-a.s., by the strong Markov property. As consequence the proof is complete once we have shown that for an arbitrary $x\in E$ $$ \mathbb{E}_x \left( \prod_{k=1}^n f_k(X_{t_k} - X_0)\right) = P_{t_1}f_1 \cdots P_{t_n-t_{n-1}} f_n(0),$$ which is the characterisation of a Markov process. Induction is needed but I can't seem to figure out the details. Any help is greatly appreciated thanks.

  • $\begingroup$ The idea is to approximate the stopping time $\tau$ by (suitable) discrete stopping times $\tau_n$ such that $\tau_n \downarrow \tau$. A proof of the claimed statement is e.g. contained in Schilling/Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 6 (the proof there is for the case of Brownian motion, but it works exactly the same way for any process with stationary+independent increments.) $\endgroup$
    – saz
    May 18, 2015 at 19:33
  • $\begingroup$ I think you refer to Theorem 6.5? Doesn't it work differently as we work with a general bounded function $f$ here? $\endgroup$
    – user155670
    May 18, 2015 at 20:01
  • $\begingroup$ Yes, I had Theorem 6.5 in mind. What do you mean by "we work with a general bounded fct $f$"? You have to prove the independence and the equality in distribution and that's exactly what's done in the proof. $\endgroup$
    – saz
    May 19, 2015 at 5:16
  • $\begingroup$ I just don't understand how I should convert the brackets and exponent with the brownian motion into the general case? $\endgroup$
    – user155670
    May 19, 2015 at 8:05
  • $\begingroup$ Okay. I'll write an answer later ... $\endgroup$
    – saz
    May 19, 2015 at 8:26

1 Answer 1


Set $\tau_j := (\lfloor 2^j \rfloor+1)/2^j$. Then $\tau_j$ is a discrete stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Fix $F \in \mathcal{F}_{\tau+}$. Then, by the right-continuity and the dominated convergence theorem,

$$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \mathbb{E}(e^{\imath \, \xi (X_{\tau_j+t}-X_{\tau_j+s}} 1_F).$$

Since $\tau_j$ is a discrete stopping time with values in $k \cdot 2^{-j}$, $k \in \mathbb{N}$, we can rewrite this as

$$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E}(e^{\imath \, \xi (X_{t+k2^{-j}}-X_{s+k 2^{-j}}} 1_{\{\tau_j = k 2^{-j}\}} 1_F).$$

Now, as $X_{t+k2^{-j}} -X_{s+k2^{-j}}$ is independent of $\mathcal{F}_{k2^{-j}}$ and $\{\tau_j = k 2^{-j}\} \cap F \in \mathcal{F}_{k 2^{-j}}$, we get by the stationarity of the increments

$$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E}e^{\imath \, \xi X_{t-s}} \mathbb{P}(F \cap \{\tau_j = k 2^{-j}\}) = \mathbb{E}e^{\imath \, \xi X_{t-s}} \mathbb{P}(F).$$

A similar argumentation works for fininitely many increments, i.e.

$$\mathbb{E}(e^{\imath \, \sum_{k=1}^n \xi_k(Y_{t_k}-Y_{t_{k-1}})} 1_F) = \prod_{k=1}^n \mathbb{E}e^{\imath \, \xi X_{t_k-t_{k-1}}} \mathbb{P}(F). \tag{1}$$

If we choose $F=\Omega$, then we get

$$\mathbb{E}(e^{\imath \, \sum_{k=1}^n \xi_k(Y_{t_k}-Y_{t_{k-1}})}) = \prod_{k=1}^n \mathbb{E}e^{\imath \, \xi X_{t_k-t_{k-1}}} \tag{2}.$$

Plugging this again into $(1)$, we find

$$\mathbb{E}(e^{\imath \, \sum_{k=1}^n \xi_k(Y_{t_k}-Y_{t_{k-1}})} 1_F) = \mathbb{E}(e^{\imath \, \sum_{k=1}^n \xi_k(Y_{t_k}-Y_{t_{k-1}})}) \mathbb{P}(F). \tag{3}$$

This shows that

  1. By $(3)$, $Y_{t_n}-Y_{t_{n-1}},\ldots,Y_{t_1}-Y_{t_0}$ is independent from $\mathcal{F}_{\tau+}$. Since $\tau_0<\ldots <t_n$ are arbitrary, this means that $\sigma(Y_t; t \geq 0)$ and $\mathcal{F}_{\tau+}$ are independent.
  2. It follows from $(2)$ that $$\mathbb{E}e^{\imath \, \sum_{k=1}^n (Y_{t_k}-Y_{t_{k-1}})} = \prod_{k=1}^n \mathbb{E}e^{\imath \, \xi (Y_{t_k}-Y_{t_{k-1}})},$$ i.e. $(Y_t)_{t \geq 0}$ has independent increments. This implies that $(Y_t)_{t \geq 0}$ is a Markov process.
  3. $(2)$ also shows that $(Y_t)_{t \geq 0}$ and $(X_t)_{t \geq 0}$ are equal in distribution.

Remark: This proof is adapted from the proof of Theorem 6.5 in

René Schilling, Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.

  • $\begingroup$ Thanks alot saz! It has finally become clear to me now. $\endgroup$
    – user155670
    May 19, 2015 at 16:35

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