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Let $(Y^a: a\in \Lambda)$ be a set of random processes given by $$Y^a(s) = \int_0^s \sigma^a(r) dW(r)$$ where $W$ is Brownian motion w.r.t. filtered probability space $(\Omega, \mathcal{F}, P, \mathcal{F}_t)$, $\sigma^a(\cdot)$ is uniformly bounded predictable process, s.t. $|\sigma^a(r)|<1$ for all $r$ and $a$. Let $\theta^a = \inf(s>0, Y^a(s) \ge 1)$ be a stopping time.

[Q.] Is the following true, $$\inf_{a\in \Lambda} (\theta^a) >0, \quad a.s.-P$$

When $\Lambda$ is finite set, it is clearly true. But, I am not sure otherwise. Thanks.

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  • $\begingroup$ It is not true even if all $\sigma^a$ are deterministic. $\endgroup$
    – zhoraster
    Commented Feb 26, 2011 at 16:41

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This is not true. Fix some $t>0$, and define $t^n_k = \frac{k}{n}.t$ for $0 \leq k \leq n$. Define $\Lambda$ as the set of deterministic, $\{-1,1\}$-valued processes and $\Lambda_n$ as the subset of $\Lambda$ such that $\sigma \in \Lambda_n$ if it is constant on each $(t^n_k,t^n_{k+1})$.

Since brownian motion has unbounded variation, $\lim_{n\rightarrow\infty} \sum_k|W_{t^n_{k+1}}-W_{t^n_k}|=+\infty$ a.s., in particular it is greater than $1$ for some $n$. Then notice that (for each $\omega$) this sum is equal to the brownian integral corresponding to some $\sigma \in \Lambda_n$. Hence $\inf_{a\in\Lambda}\theta^a < t$ a.s. for all $t>0$.

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  • $\begingroup$ 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 $[0,c]$. $\endgroup$
    – zhoraster
    Commented Feb 26, 2011 at 17:00
  • $\begingroup$ Since for each $t>0$, $\inf_a \theta^a <t$ a.s., then this inf is $0$ almost surely. $\endgroup$
    – pgassiat
    Commented Feb 26, 2011 at 17:07
  • $\begingroup$ 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. $\endgroup$
    – kenneth
    Commented Feb 26, 2011 at 18:01
  • $\begingroup$ @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 $1$. $\endgroup$
    – pgassiat
    Commented Feb 26, 2011 at 19:07

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