Why is $T_n-T_{n-1}$ independent of $\mathcal{F}_{T_{n-1}}$ where $X$ is adapted cadlag Levy process and $C=\sup_{t} |\Delta X_t|$ where $\Delta X_t=X_t- X_{t^-}$

$T_1=\inf\{t:|X_t| \geq C\} $

$\vdots$

$T_{n+1}=\inf \{t>T_n: |X_t-X_{T_n}| \geq C\}$

The author says this is a consequence of the strong Markov Property but it doesnt make any sense since that says that the Levy process renews itself at stopping times but doesnt say anything about the independence of the increments of the stopping times. Can someone give me some hints on how could I go about proving it?? enter image description here

Theorem 32 "Let $X$ be a Levy Process and $T$ a stopping time . On the set $\{T < \infty\}$ , the process $Y$ defined by $Y_t=X_{T+t}-X_T$" is a Levy Process adapted to $H_t=\mathcal{F}_{T+t}$ is independent of $\mathcal{F}_T$ and $Y$ has the same distribution as $X$

  • Can you tell us what theorem 32 states? – Blaza Feb 4 '17 at 18:06
  • @Blaza Theorem 32 "Let $X$ be a Levy Process and $T$ a stopping time . On the set $\{T < \infty\}$ , the process $Y$ defined by $Y_t=X_{T+t}-X_T$" is a Levy Process adapted to $H_t=\mathcal{F}_{T+t}$ is independent of $\mathcal{F}_T$ and $Y$ has the same distribution as $X$ – user3503589 Feb 4 '17 at 20:07
up vote 2 down vote accepted

Let $T$ be a stopping time and $Y_t := X_{T+t}-X_T$ the restarted Lévy process. Theorem 32 states that the $\sigma$-algebra generated by $(Y_t)_{t \geq 0}$ is independent of $\mathcal{F}_T$, i.e.

$$\mathcal{F}_{\infty}^Y := \sigma(Y_s; s \geq 0) \quad \text{and} \quad \mathcal{F}_T \quad \text{are independent}. \tag{1}$$

If we define a stopping time $T_n$ by

$$T_n := \inf\{t>T_{n-1}; |X_t-X_{T_{n-1}}| \geq C\}$$

and set $Y_t^{(n)} := X_{t+T_{n-1}}-X_{T_{n-1}}$, then

$$T_n := \inf\{s \geq 0; |Y_s^{(n)}| \geq C\} + T_{n-1};$$

hence

$$T_n-T_{n-1} = \inf\{s > 0; |Y_s^{(n)}| \geq C\}.$$

The right-hand side is measurable with respect to $\mathcal{F}_{\infty}^{Y^{(n)}}$, and therefore it follows from $(1)$ (with $Y$ replaced by $Y^{(n)}$ and $T$ replaced by $T_{n-1}$) that $T_n-T_{n-1}$ and $\mathcal{F}_{T_{n-1}}$ are independent.

  • So $\inf \{s>0 ; |Y_s^{(n)}| \geq C\}$ is measurable since each of $|Y_s^{(n)}| \geq C$ is measurable for each $s$ and since $Y^{(n)}$ is cadlag we can write the $\inf$ as a countable union and retain measurability? – user3503589 Feb 5 '17 at 10:51
  • 1
    @user3503589 It's not that simple, but it is well-known that for a Lévy process $(Y_t)_{t \geq 0}$ and for a closed set $C$ (in fact, for any Borel set $C$), the hitting time $$T:=\inf\{s \geq 0; Y_s \in C\}$$ is a stopping time (see this answer) which implies, in particular, that $T$ is measurable with respect to $\mathcal{F}_{\infty}^Y$. – saz Feb 5 '17 at 12:13
  • yes ofocurse. It was stupid question. You already explained this to me a few weeks ago for cadlag adapted first hitting time into a closed or even Bored(Debut theorem) implies its a stopping time and therefore measurable with respect to $\mathcal{F}_{\infty}^Y$ – user3503589 Feb 5 '17 at 12:19

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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.