Are the following Stopping Times?

Question

I've been working through the following list of stopping time questions. I am have problems with the final two (e and f).

I appreciate any assistance offered.

$\textbf{Question:}$

Let $S,T : \Omega \rightarrow \{0,1,2,...,\infty \}$be $(\mathcal{F}_n)_{n \ge 0}$-stopping times. Prove, or provide a counter-example disproving, the following statements:

a) $S+1$ is a stopping time.

b) $S \wedge T$ is a stopping time.

c) $S \vee T$ is a stopping time.

d) $S + T$ is a stopping time.

e) $S-1$ is a stopping time.

f) $\lambda S$ is a stopping time for $\lambda > 0$.

$\textbf{Answers:}$

a) Yes. For any $n \ge 0$, \begin{align} \{S + 1 \le n\} = \{ S \le n - 1\} \in \mathcal{F}_{n-1} \subseteq \mathcal{F}_{n}. \end{align}

b) Yes. For any $n \ge 0$, \begin{align} \{S \wedge T \le n\} = \{ S \le n \} \cup \{T \le n\} \in \mathcal{F}_{n}. \end{align}

c) Yes. For any $n \ge 0$, \begin{align} \{S \vee T \le n\} = \{ S \le n \} \cap \{T \le n\} \in \mathcal{F}_{n}. \end{align}

d) Yes. For any $n \ge 0$, \begin{align} \{S + T = n\} = \cup^{n}_{k=0} \underbrace{\{ S = k \}}_{\in \mathcal{F}_k \subset \mathcal{F}_n} \cap \underbrace{\{T = n - k\}}_{\in \mathcal{F}_{n-k} \subset \mathcal{F}_n} \in \mathcal{F}_{n}. \end{align}

e) $S-1$ - I do not think $S-1$ is a stopping time, but I'm not sure now to write a counter-example?

f) $\lambda S$ - not sure.

Appreciate any help.

Many thanks,

John

Answer 1

e) No, this is (in general) not a stopping time. Just consider the following example: Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distributed random variables such that $$\mathbb{P}(X_1 = 1) = \mathbb{P}(X_1 = -1) = \frac{1}{2}$$ and denote by $\mathcal{F}_n := \sigma(X_1,\ldots,X_n)$ the canonical filtration. Then $$S := \inf\{n \in \mathbb{N}; X_n=1\}$$ is a stopping time. On the other hand, $$\{S-1 =n\} = \{S=n+1\} = \bigcap_{j=1}^n \{X_j=-1\} \cap \{X_{n+1} =1\} \notin \mathcal{F}_n.$$ (Note that $X_{n+1}$ and $\mathcal{F}_n$ are independent; therefore, $X_{n+1}$ cannot be measurable with respect to $\mathcal{F}_n$. Hence, $\{X_{n+1}=1\} \notin \mathcal{F}_n$.)

f) Well, it depends. We have to consider the cases separately:

• $\lambda=1$: Obvious.
• $\lambda>1$: Recall that we have to ensure that $\lambda S$ takes values in $\mathbb{N} \cup \{\infty\}$. This means that $\lambda \in \mathbb{Q}$ is necessary. However, if $\lambda S$ takes values in $\mathbb{N} \cup \{\infty\}$, then it is indeed a stopping time (the argumentation is similar to a)).
• $\lambda<1$: Again, $\lambda S$ has to take values in $\mathbb{N} \cup \{\infty\}$; thus, $\lambda \in \mathbb{Q} \cap (0,1)$. However, even if $\lambda S$ takes values in $\mathbb{N} \cup \{\infty\}$, then it is (in general) not a stopping time. Can you think of a counterexample? (Hint: Choose for example $\lambda = \frac{1}{2}$ and try to construct a similar counterexample as in e)).