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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.

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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 $$ \Bbb P^{X_0}[X_{\sigma}\in\cdot\,|\mathcal F_\tau]=\Bbb P^{X_\tau}[X_\sigma'\in\cdot]$$

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  • $\begingroup$ Thanks! ad 1: Yeah, a stupid mistake, I edited it to include the conditioning. $\endgroup$ – Trademark Jan 12 '16 at 11:29

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