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Given a probability space $S=(\Omega, \mathcal{A}, P)$, a filtration $\mathcal{F}=(\mathcal{F}_t)_{t\in[0,\infty)}$ on $\mathcal{A}$, and a separable normed space $H$, whose induced topology shall be denoted by $\tau$, a $\mathcal{F}/H$-Lévy process w.r.t. $P$ is a stochastic process $X=(X_t)_{t\in [0,\infty)}$, in which $X_t$ is $\mathcal{F}_t/\sigma(\tau)$-measurable, for every $t\in[0,\infty)$, and such that

a. $X$'s paths are càdlàg,

b. $\sigma(X_t-X_s) \perp_{P} \mathcal{F}_s$, for all $s,t\in[0,\infty)$ such that $s < t$,

c. $P^{X_t-X_s}=P^{X_{t-s}}$, for all $s,t\in[0,\infty)$ such that $s\leq t$.

If we write "a $\Omega/H$-Lévy process w.r.t. $P$" the underlying filtration is to be taken to be the natural filtration induced by $X$.

Now, suppose $X=(X_t)_{t\in[0,\infty)}$ is a $\Omega/H$-Lévy process w.r.t. $P$, and denote by $\mathcal{F}=(\mathcal{F}_t)_{t\in[0,\infty)}$ the natural filtration induced by $X$. Is $X$ also a $\mathcal{A}/\sigma(\tau)$, $\mathcal{F}^+$-adapted Lévy process w.r.t. $P$? The question boils down to whether $$ \sigma(X_t-X_s) \perp_{P} \mathcal{F}^+_s $$ for all $s,t\in[0,\infty)$ such that $s < t$.

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The answer is affirmative. In fact, we will prove a more general proposition, where the filtration is not constrained to be the natural one. We will show the following.

Suppose $X=(X_t)_{t\in[0,\infty)}$ is a $\mathcal{F}/H$-Lévy process w.r.t. $P$, then $X$ also a $\mathcal{F}^+/H$-Lévy process w.r.t. $P$.

The proof derives straightforwardly from the Markov property of Lévy processes, which we state next (a proof may be found, for instance, in Jochen Wengenroth's "Wahrscheinlichkeitstheorie" (de Gruyter 2008), where this is Theorem 7.14 on p. 144).

The Markov property of Lévy processes Let $S=(\Omega, \mathcal{A}, P)$ be a probability space, let $\mathcal{F}=(\mathcal{F}_t)_{t\in[0,\infty)}$ be a filtration on $\mathcal{A}$, and let $H$ be a separable normed space, whose induced topology shall be denoted by $\tau$.

Let $X=(X_t)_{t\in[0,\infty)}$ be a $\mathcal{F}/H$-Lévy process w.r.t. $P$, and let $\zeta$ be a real-valued $\mathcal{F}$-stopping time.

Then, for every $t\in[0,\infty)$, the function $X_{\zeta+t}:\Omega\rightarrow H$, $X_{\zeta+t}(\omega):=X_{\zeta(\omega)+t}(\omega)$, is $\mathcal{F}_\infty/\sigma(\tau)$-measurable, and the indexed collection $Y=(Y_t)_{t\in[0,\infty)}$ of $\mathcal{F}_\infty/\sigma(\tau)$-measurable functions $Y_t:=X_{\zeta+t}-X_\zeta$ is a $\Omega/H$-Lévy process w.r.t. $P$. Furthermore, $P^Y=P^X$ and $\sigma(Y) \perp_P \mathcal{F}^+_\zeta$.

Let $s, t\in[0,\infty)$ be such that $s < t$. Define $\zeta:=s\mathbb{1}_\Omega$. $\zeta$ is a $\mathcal{F}$-stopping time. Hence, by the Markov property of Lévy processes, the indexed collection $Y=(Y_z)_{z\in[0,\infty)}$, $Y_z:=X_{s+z}-X_s$ is a $\Omega/H$-Lévy process w.r.t $P$, and $\sigma(Y)\perp_P \mathcal{F}^+_s$. In particular, $\sigma(Y_{t-s}-Y_0)\perp_P \mathcal{F}^+_s$. But $Y_{t-s} - Y_0 = X_t-X_s$, Q.E.D.

Incidentally, this shows that

in the statement of the Markov property of Lévy processes we may relax the stipulation that $\zeta$ be a $\mathcal{F}$-stopping time, and require only that $\zeta$ be a weak $\mathcal{F}$-stopping time.

Indeed, if $\zeta$ is a weak $\mathcal{F}$-stopping time, then it is a $\mathcal{F}^+$-stopping time (see Wengenroth's theorem 7.13.1 on p. 142), and we may apply the version of the Markov property of Lévy processes cited above to conclude that $Y$ is a $\Omega/H$-Lévy process w.r.t. $P$, and that $P^Y=P^X$ and $\sigma(Y)\perp_P\left(\mathcal{F}^+_\zeta\right)^+$. But $\mathcal{F}^+_\infty=\mathcal{F}_\infty$ (verification left to the reader) and $\left(\mathcal{F}^+_\zeta\right)^+=\mathcal{F}^+_\zeta$ (as stated in Wengenroth in the paragraph preceding theorem 7.13 on p. 142), Q.E.D.

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