Let $(X_t)_{t\in\mathbb{R}}$ be a Levy Process, i.e. $X_0 = 0$ a.s., $X$ has independent and stationary increments, and almost all paths $t\mapsto X_t(\omega)$ are right continuous with left hand limits. Let $\mathcal{F}$ be the natural filtration of $X$, i.e. $\mathcal{F}_t = \sigma(X_s)_{s\leq t}$, and analogous $\mathcal{G}_t = \sigma (X_s)_{0\leq s \leq t}$ for $t\geq0$.

It is a well known fact (can't think of a concrete reference right now, but the proof should be for example in Kallenberg's "Foundations of Modern Probability Theory") that the one-sided process $(X_t)_{t\geq 0}$ is strong $\mathcal{G}$-Markov, i.e. for any $\mathcal{G}$-stopping time $\tau$ with a.s. $\tau<\infty$, any $t>0$ and any Borel set $B$ the equation \begin{equation*} \mathbb{P}\left[ X_{\tau+t}\in B \;\vert\; \mathcal{G}_\tau \right] = \mathbb{P}\left[ X_{\tau+t}\in B \;\vert\; X_\tau \right] \qquad\qquad\text{a.s.} \end{equation*} is fulfilled, i.e. for any $G\in\mathcal{G}_\tau$ \begin{equation*} \mathbb{E}\left[ 1G \cdot \mathbb{P} \left[ X_{\tau+t}\in B \; \vert \; X_\tau \right] \right] = \mathbb{P} G\cap \{ X_{\tau+t}\in B \} \end{equation*} As far as i can remember, this proof used the independence of $\mathcal{G}_t$ and $(X_{t+s}-X_t)_{s\geq 0}$, which follows from $X_0 = 0$ a.s. and the independent increments of $X$. One can also proof the independence of $\mathcal{G}_\tau$ and $(X_{\tau+s}-X_\tau)_{s\geq0}$.

In the double-sided case of $(X_t)_{t\in\mathbb{R}}$ and $\mathcal{F}$, this independence is not given, i.e. for $t<0$ the $\sigma$-Algebra $\mathcal{F}_t$ is not independent of $(X_{t+s}-X_t)_{s\geq0}$, because $X_{t+s}-X_t = -X_t$ a.s. for $s=-t$. Nevertheless, one can proof that $X$ is weakly $\mathcal{F}$-Markov, i.e. the above a.s. equation is fulfilled for all constant $\tau\equiv t\in\mathbb{R}$. This follows from the elementary properties of the conditional expected value.

My question is now: Does the strong $\mathcal{F}$-Markov property also hold for the double-sided case? I would guess yes (heuristically you should still only need to know $X_\tau$ in order to determine the distribution of $(X_{\tau+s}-X_\tau)_{s\geq0}$, even if the latter process needs not be distributed as $(X_t)_{t\geq 0}$). Like in the proof of the one-sided case, I would be happy for an explanation in the case $1<\vert \tau\Omega \vert < \infty$.


EDIT: The more I think about it, the more I believe this to be untrue; heuristically only knowing $X_\tau$ doesn't tell you if $\tau >0$ or $\tau<0$, which I believe is important for the distribution of "$X_{\tau+\vert \tau\vert}-X_\tau$".

Q: Is my thinking sound? Can someone produce a rigorous proof?


EDIT: As requested, the Definition I use in my question (if this is not the standard definition, please inform me).

By a double-sided (or two-sided; don't know the correct terminology) Levy Process I mean a Stochastic Process $X_t$ on the whole real line, i.e. $t\in\mathbb{R}$ (I can only find the case $t\in\mathbb{R}_+$ in the literature) with the following properties:

(a). $X_0 = 0$ almost surely;

(b). for almost every $\omega\in\Omega$ the path $X\omega : t\mapsto X_t\omega$ is an element of $D$, the space of all real functions $f:t\in\mathbb{R} \mapsto ft\in\mathbb{R}$, which are right-continuous, i.e. $ft = \lim_{\varepsilon\downarrow 0} f(t+\varepsilon)$, and have left limits, i.e. $f(t-) := \lim_{\varepsilon\downarrow 0} f(t-\varepsilon)$ exists and is real.

(c). $X$ has independent increments, i.e. for all $n\in\mathbb{N}$ and $t_0,t_1,\dots,t_n\in\mathbb{R}$ with $t_0<\cdots<t_n$ the increments $X_{t_1}-X_{t_0},\dots,X_{t_n}-X_{t_{n-1}}$ are independent.

(d). $X$ has stationary increments, i.e. for a fixed $c>0$ the increments $X_{t+c}-X_t$ and $X_{s+c}-X_s$, $s,t\in\mathbb{R}$, have the same distribution.

This definition is actually a direct copy of the one-sided case $t\in\mathbb{R}_+$.

The problem I have is the following: In the one-sided case, one can easily show the independence of $(X_t)_{t\in[0,T]}$ and $(X_{T+t}-X_t)_{t\in\mathbb{R}_+}$ for $T\in\mathbb{R}_+$, from which the strong Markov property follows (basic proof, which can be found in literature about Levy Processes; there is even a discussion about this proof on SE). As mentioned above, this independence doesn't follow in the double-sided case, which makes the proof of the strong MP non-transferable.

  • Is the definition of a stopping time the same as in the one-sided case (i.e. $\{\tau \leq t\} \in \mathcal{F}_t$)? – saz Dec 6 '15 at 11:29
  • You never really state the definition of a double sided Levy process (or maybe I'm just not seeing it?). – Olorun Dec 8 '15 at 4:27
  • 1
    updated my question – user138862 Dec 9 '15 at 16:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.