# infimum of a set of stopping times

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.

• It is not true even if all $\sigma^a$ are deterministic. – zhoraster Feb 26 '11 at 16:41

## 1 Answer

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 $t>0$.

• 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 $[0,c]$. – zhoraster Feb 26 '11 at 17:00
• Sorry I'm new here, and I can't seem to fix the Tex at the end of my post. But since for each $t>0$, $\inf_a \theta^a <t$ a.s., then this inf is $0$ almost surely. – pgassiat Feb 26 '11 at 17:07
• @Paul: Aha, I see now. Your post unfortunately lacks some signs in it, so it was not very clear. You can check "community wiki" so that everyone is able to edit it. – zhoraster Feb 26 '11 at 17:29
• 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. – kenneth Feb 26 '11 at 18:01
• I just want to make it clear for those who are new that making a post community wiki also makes it so you cannot gain any reputation for upvotes (or lose reputation for downvotes). So you should probably not go making all your posts community wiki unless you don't mind this. Getting reputation points is occasionally useful as described in the FAQ page. – Noah Stein Feb 26 '11 at 18:05