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Timeline for infimum of a set of stopping times

Current License: CC BY-SA 2.5

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Feb 27, 2011 at 2:25 history edited George Lowther CC BY-SA 2.5
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Feb 26, 2011 at 19:07 comment added pgassiat @kenneth : I am saying that for each fixed $\omega$, there will be some $\sigma$ such that $\sum_k |W_{t_{k+1}} - W_{t_k}| = \int \sigma dW$. This is different than saying that you can find $\sigma$ s.t. the equality is true for all $\omega$. For instance for $n=1$, this is just saying that $|W_t - W_0|$ is either equal to $(W_t - W_0)$ or to $(-1).(W_t - W_0)$, but of course none of these equalities are true with probability $1$.
Feb 26, 2011 at 18:01 comment added kenneth Your claim may not be true. In fact, suppose $\sigma_k$ be the value of $\sigma(r)$ in interval $(t_k^n, t_{k+1}^n)$, which must be $\mathcal{F}_{t_k^n}$ measurable random variable. Then, your claim implicitly assumed $\sigma_k \cdot (W(t_{k+1}^n) - W(t_k^n)) = |W(t_{k+1}^n) - W(t_k^n)|$. But this is only possible for $\sigma_k$, which is the difference of two indicator r.v.s on the event $W(t_{k+1}^n) - W(t_k^n)>0$ and $W(t_{k+1}^n) - W(t_k^n)<0$. As a result, such a $\sigma_k$ is not $\mathcal{F}_{t_k^n}$ measurable. Please let me know, if I made some misunderstanding.
Feb 26, 2011 at 17:46 history edited pgassiat CC BY-SA 2.5
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Feb 26, 2011 at 17:07 comment added pgassiat Since for each $t>0$, $\inf_a \theta^a <t$ a.s., then this inf is $0$ almost surely.
Feb 26, 2011 at 17:00 comment added zhoraster The argument is not completely correct. However, it is possible to make it correct. E.g. saying that $\inf \theta_a > 0$ a.s. implies $\inf \theta_a>c$ with a positive probability for some $c>0$ and further speaking of the variation on the interval $[0,c]$.
Feb 26, 2011 at 16:52 history edited pgassiat CC BY-SA 2.5
added 14 characters in body; edited body; added 6 characters in body
Feb 26, 2011 at 16:45 history answered pgassiat CC BY-SA 2.5