4
$\begingroup$

Suppose that $\left(A_n\right)$ is an adapted process, and that $B\in\mathcal{B}$. Let $L = \sup\left\{n:n\leq10;A_n\in B\right\}$, $\sup\left(\emptyset\right)=0$. Convince yourself that $L$ is NOT a stopping time (unless $A$ is freaky).

Williams, "Probability with Martingales", Ch. 10 "Martingales", Example in section 10.8 "Stopping time"

I can easily come up with a counter-example (e.g. a symmetric random walk), but is it possible to prove that under general conditions it is always the case that $L$ is not a stopping time? If so, how to prove it?

$\endgroup$

1 Answer 1

4
$\begingroup$

Suppose that $(A_n)_{n \in \mathbb{N}}$ is adapted with respect to a filtration $(\mathcal{F}_n)_{n \in \mathbb{N}}$.

$L$ is a stopping time if, and only if, $\{L=k\} \in \mathcal{F}_k$. By definition, we have

$$\{L=k\} = \{A_k \in B\} \cap \bigcap_{j=k+1}^{10} \{A_j \notin B\}.$$

Since $(A_n)_n$ is an adapted process, we know that $\{A_k \in B\} \in \mathcal{F}_k$ and $\{A_j \notin B\} \in \mathcal{F}_j$, $j=k+1,\ldots,n$. In general, $\mathcal{F}_k \subset \mathcal{F}_j$ and this means that we cannot conclude $\{L=k\} \in \mathcal{F}_k$. Therefore, $L$ is not a stopping time.

Loosely speaking: The $\sigma$-algebra $\mathcal{F}_n$ contains the information about the process upon time $n$. $L$ is a stopping time if, and only if, the information up to time $n$ allows you to decide whether the stopping occurs before or after time $n$. For the given stopping time this is obviously not true since you need the whole information up to time $n=10$.

Example Let $(A_n)_{n \in \mathbb{N}_0}$ a process such that $A_0=0$ and $A_n>0$ for any $n \geq 1$. Set $$\tau := \sup\{n \in \mathbb{N}_0; A_n =0\}.$$ Then, $\{\tau=0\} = \Omega$ and $\{\tau=n\}=\emptyset$ for $n \geq 1$ (since $A_n>0$ for $n \geq 1$). Consequently, $\tau$ is a stopping time with respect to any filtration $(\mathcal{F}_n)_n$. Now consider the process $$B_n(\omega) := \begin{cases} 0 & n <\sigma(\omega) \\ 1 & \text{otherwise} \end{cases}$$ where $\sigma$ is an arbitrary ($\mathbb{N}_0$-valued) random variable. Then, $$\{\tau=k\} = \{\sigma=k+1\}.$$ It is not difficult to see that $\tau$ is not a stopping time with respect to the canonical filtration $(\mathcal{F}_n)_n$, but a stopping time with respect to the filtration $(\mathcal{F}_{n+1})_{n}$.

Similarly, one can construct examples which show that it does not only depend on the properties of the process and the filtration, but also on the set $B$.

$\endgroup$
5
  • $\begingroup$ Thank you. What general conditions on the filtration and/or $\left(A_n\right)_{n\in\mathbb{N}}$ will ensure that $\left\{A_k\in B\right\}\cap\bigcap_{j = k + 1}^{10}\left\{A_j\notin B\right\}\notin\mathcal{F}_k$? Is the requirement $\mathcal{F}_k\subsetneq\mathcal{F}_j$ ($k < j$) enough? $\endgroup$
    – Evan Aad
    Commented Feb 1, 2014 at 20:41
  • 2
    $\begingroup$ @EvanAad No, $\mathcal{F}_k \subset \mathcal{F}_j$ is not a sufficient condition. If $(A_n)_n$ is $(\mathcal{F}_n)$-adapted, then, obviously, $(A_n)_n$ is $(\mathcal{G}_n)_n$-adapted for any filtration satisyfing $\mathcal{F}_n \subseteq \mathcal{G}_n$. Say, $\mathcal{F}_j = \mathcal{F}_k$ for some $k<j$. This means that we can always enlarge our filtration by adding sets which do not contain any information about the process $(A_n)_{n}$. I think there are no general conditions; it really depends on the process $(A_n)_n$, the set $B$ and the given filtration $(\mathcal{F}_n)_n$. $\endgroup$
    – saz
    Commented Feb 1, 2014 at 21:08
  • $\begingroup$ @EvanAad I have added an example. $\endgroup$
    – saz
    Commented Feb 1, 2014 at 21:22
  • $\begingroup$ Thanks. In the definition of $B_n$, the condition of the first branch should be $n < \sigma\left(\omega\right)$ to correspond to the later statement that $\left\{\tau = k\right\} = \left\{\sigma = k + 1\right\}$. (An "off-by-one" error) $\endgroup$
    – Evan Aad
    Commented Feb 2, 2014 at 6:21
  • $\begingroup$ @EvanAad you are right, thanks. $\endgroup$
    – saz
    Commented Feb 2, 2014 at 6:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .