In the book Multidimensional diffusion processes, of Stroock and Varadhan one reads (page 23):

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This is the proof of $(i)$. Here the authors say

Define $f_t$ on $(\{\tau \leq t\}, \mathcal{F}_t [\{\tau \leq t\}])$

What is the $\sigma$-algebra $\mathcal{F}_t [\{\tau \leq t\}]$? Is it $\{A \cap \{\tau \leq t\}\mid A \in \mathcal{F}_t\}$? Why do we consider such a sigma algebra instead of $\mathcal{F}_t$?

after proving $(ii)$ the authors say (page 24)

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The proof of $(iii)$ is readily seen by the following reasoning:

$$A \in \mathcal{F}_{\sigma}: \quad A \cap \{\tau \leq t\} =A \cap \{\sigma \leq t\} \cap \{\tau \leq t\} \in \mathcal{F}_t \Rightarrow A \in\mathcal{F}_{\tau}. $$

But I can't see why this result follows from $(ii)$

Using $(ii)$ the sole thing I arrived at was $$A \in \mathcal{F}_{\sigma}: \quad A =A \cap \{\sigma \leq \tau\} \in \mathcal{F}_{\tau \wedge \sigma} = \mathcal{F}_{\sigma}, $$

which is not the desired result.

What am I missing here?


Regarding your second point I think this is clearer if you know that $\mathcal{F}_{\sigma\wedge\tau}=\mathcal{F}_{\sigma}\cap\mathcal{F}_{\tau}$ because then instead of writing :

$$A \in \mathcal{F}_{\sigma}: \quad A =A \cap \{\sigma \leq \tau\} \in \mathcal{F}_{\tau \wedge \sigma} = \mathcal{F}_{\sigma}$$

You would write :

$$A \in \mathcal{F}_{\sigma}: \quad A =A \cap \{\sigma \leq \tau\} \in \mathcal{F}_{\tau \wedge \sigma}= \mathcal{F}_{\sigma}\cap\mathcal{F}_{\tau}\subseteq \mathcal{F}_{\tau},$$

For the first point, the notation corresponds to the trace sigma algebra which is what you have written, it turns out that the argument is build to guarantee that the form of preimage of a set for $\theta\circ f_t$ is $A\cap \{\tau\leq t\}$ ensuring the $\mathcal{F}_\tau$-measurability.

Best regards

  • $\begingroup$ first point: maybe a sigma algebra of $X$ must be made of subsets of $X$ and that wouldn't be the case if we consider $\mathcal{F}_t$. Note that since $X = [\tau \leq t]$ It is always going to be the case that the pre-image is of the form $A\cap \{\tau\leq t\}$. I am still thinking about the second point. I am not sure if at that point one can know that $\mathcal{F}_{\sigma\wedge\tau}=\mathcal{F}_{\sigma}\cap\mathcal{F}_{\tau}$ It seems that we would have to use the point we are trying to prove, that is $\tau\wedge \sigma \leq \tau, \sigma$ and use $(i) + (iii)$ $\endgroup$ – Conrado Costa Sep 1 '15 at 12:38

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