The jumping times of a càdlàg process are stopping times.

Protter first proves this theorem:

Let $X$ be an adapted càdlàg stochastic process, and let $A$ be a closed set. Then the random variable:

$T(\omega)=\inf\{t: X_t(\omega)\in A \text{ or } X_{t^-} \in A\}$,

is a stopping time.

Then he says that it follows from this theorem :

If $K>0$ is a positive number, then:

$T^1 = \inf\{t >0 : |\Delta X_t|>k\}$

$T^{n+1}=\inf\{t >T^n : |\Delta X_t|>k\}$, are stopping times.

I've tried proving this from the above theorem, but I am not able to do it.

attempt regarding $T_1$

I need to show that $\{\omega : T^1(\omega) \le t\}\in \mathcal{F}_t$. If an $\omega$ is in this set, then since a càdlàg process only have a finite number of jumps larger than $k/2$ in the interval $[0,t+1]$, we must have that at least one jump $>k$ occurring in the interval $[0,t]$. My guess in using the theorem is that A maybe should be $[-k,k]$ ? But I don't see how this works?

attempt regarding induction

I also have a problem with the induction part. Lets say that we have that $T^1,\ldots,T^n$ are stopping times. Why is then $T^{n+1}$ a stopping time? I have that $\{T^{n} \le t\}\subset\{T^{n+1} \le t\}$. But my guess is that I must write $\{T^{n} \le t\}\cup C=\{T^{n+1} \le t\}$, where $C \in \mathcal{F}_t$. But I can't really see what $C$ this must be.

Do you see how to show this? I have seen this argument used in many books, maybe it is very easy to show. Is there a book that shows the argument?

Solution 1:

The key point is the following lemma which has nothing to do with probability theory, but is just elementary "deterministic" calculus.

Lemma: Let $f: [0,\infty) \to \mathbb{R}$ be a càdlàg function and fix $M>0$ and $0 \leq T_0<T_1<\infty$. Then the following statements are equivalent:

1. There exists $t \in (T_0,T_1]$ such that $|\Delta f(t)|>M$.

2. There exists $K \in \mathbb{N}$ such that for any $k \in \mathbb{N}$ there are $u,v \in \mathbb{Q} \cap (T_0,T_1+k^{-1}]$ such that $|u-v| \leq k^{-1}$ and $|f(u)-f(v)|>M+\frac{1}{K}$.

Applying this lemma with $T_0 =0$ and $T_1 = t$ and $f(t) := X_t(\omega)$, we find

$$\{T^1\leq t\} = \bigcup_{K \geq 1} \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{u,v \in \mathbb{Q} \cap (0,t+k^{-1}) \\ |u-v| \leq k^{-1}}} \{|X_u-X_v| > M+1/K\}.$$

Since the right-hand side is a countable union/intersection, this shows $\{T^1 \leq t\} \in \mathcal{F}_{t+}$ and so $T^1$ is an $(\mathcal{F}_{t+})$-stopping time. If the filtration is right-continuous, then $T^1$ is an $(\mathcal{F}_t)_t$-stopping time.

Similarly, it holds that

$$\{T^{n+1} \leq t\} = \bigcup_{q \in \mathbb{Q} \cap [0,t)} \{T^n \leq q\} \cap \bigg(\bigcup_{K \geq 1} \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{u,v \in \mathbb{Q} \cap (q,t+k^{-1}) \\ |u-v| \leq k^{-1}}} \{|X_u-X_v|>M+1/K\} \bigg).$$

Again, this is a countable union/intersection and therefore $T^{n+1}$ is an $(\mathcal{F}_{t+})$-stopping time.