I have a diffusion $X=(X_t)_{t\ge0}$ and a stopping time $\tau$. From the strong Markov property I know that for any time $t\ge0$ (or a random time independent of $X$) I get that $$\mathbb{P}^{X_0}[X_{\tau + t} \in \cdot|\mathcal{F}_\tau]=\mathbb{P}^{X_\tau}[X_t \in \cdot].$$ Now, let us in addition to $\tau$ have another stopping time $\sigma$ which we define as $$\sigma:=\inf\{s > \tau;\, X_s \in A\}$$ for some subset $A$ of the state space. Now, intuitively $$\mathbb{P}^{X_0}[X_{\sigma} \in \cdot|\mathcal{F}_\tau]=\mathbb{P}^{X_\tau}[X_{\sigma'} \in \cdot],$$ where $\sigma':=\inf\{s>0; X_s \in A\}$. However, I somehow fail to show this rigorously. It is clear that $\sigma$ is "sort of like" $\tau+\sigma'$ but the problem is that unlike the case when I have $\tau+t$,i.e. a stopping time and a constant, the $\sigma'$ depends on $X$. I feel like I am missing something fairly easy here. Thanks for any hints.

EDIT: Is it simply that if I denote by $\theta_\tau$ the shift operator at $\tau$ then $\sigma\circ \theta_\tau=\sigma'$ and $X_\sigma\circ \theta_\tau=X_{\sigma'}$? I am asking because I am not sure how to actually apply the shift operator in this case.


1 Answer 1


1. (Correction) Your first formula should read $\Bbb P^{X_0}[X_{\tau+t}\in\cdot\,|\mathcal F_\tau]=\Bbb P^{X_\tau}[X_t\in\cdot]$.

2. It is true that $\sigma(\omega)=\tau(\omega)+\sigma'(\theta_{\tau(\omega)}\omega)$. Consequently $$ \Bbb P^{X_0}[X_{\sigma}\in\cdot\,|\mathcal F_\tau]=\Bbb P^{X_\tau}[X_\sigma'\in\cdot]$$

  • $\begingroup$ Thanks! ad 1: Yeah, a stupid mistake, I edited it to include the conditioning. $\endgroup$
    – Trademark
    Commented Jan 12, 2016 at 11:29
  • $\begingroup$ @John Dawkins Are you using what this post (mathoverflow.net/questions/293557/…) calls the "really strong Markov property" here? $\endgroup$
    – No-one
    Commented Aug 1, 2023 at 13:55
  • $\begingroup$ More a consequence of the SMP in the form $$ \Bbb E^{X_0}[F(\theta_\tau)\mid\mathcal F_\tau] = \Bbb E^{X_\tau}[F], $$ for $F$ bounded and measurable. (Notice $X_\sigma = X_{\sigma'}\circ\theta_\tau$.) These are both special cases of the really really SMP $$ \Bbb E^{X_0}[F(\theta_\tau,\cdot)\mid\mathcal F_\tau](\omega) =\int_\Omega F(\omega',\omega)\,\Bbb P^{X_\tau(\omega)}(d\omega'), $$ for bounded $F$ that is $\mathcal F\otimes\mathcal F_\tau$ measurable. Either one follows from an application of the functional form of the monotone class theorem $\endgroup$ Commented Aug 1, 2023 at 15:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .