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.

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

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