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I added this question to one of my others, but hasn't gotten any response. It's quite important to me so now I ask it by it self:

Is it possible to get for a general stopping time $S$ that $S\circ\theta_{n}$ is a stopping time - actually I more specifically would like an argument that gives $\{S\circ \theta _k = n-k \}\in\mathcal{F}_n$

Where $\theta_{n}$ is the shift operator: $\theta _{n}\omega(k)=\omega(k+n)$.


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up vote 0 down vote accepted

You are asking two different questions:

  1. Is $S\circ \theta_k$ a stopping time?
  2. Is $k+S\circ \theta_k$ a stopping time?

The second one is true, but the first one false.

For question 2., first show that the shift operator $\theta_k$ is ${\cal F}_{m+k}\backslash {\cal F}_m$ measurable for any $m\geq 0$. Then $$(k+S\circ\theta_k=n)=(S\circ\theta_k=n-k)=\theta_k^{-1}S^{-1}(n-k),$$ where $ S^{-1}(n-k)\in{\cal F}_{n-k}$ because $S$ is a stopping time. Putting $m=n-k$ above, we conclude that $ (k+S\circ\theta_k=n)\in{\cal F}_n$ and hence that $k+S\circ \theta_k$ a stopping time.

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Thank you very much - I appreciate much you figured out what I needed. I was trying to ask the second question both times btw. – Henrik Aug 12 '12 at 21:15

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