I'm working on this problem similar to one from Bass' book on stochastic processes and I cannot for the life of me make any progress.

Let $X$ be a stochastic process and suppose that each sample path of $X$ is cadlag. Put $\mathcal{F}^X_t = \sigma(X_s:s \leq t)$, $\Delta X_t = X_{t} - X_{t^-}$, $\mathcal{F}_{\infty} = \sigma(\mathcal{F}^X_t:t\geq 0 )$, and $A_c = \{\omega \in \Omega: \text{ for some }t > 0\text{, } \Delta X_t(\omega) > c\}$.

Show that $A_c \in \mathcal{F}_{\infty}$.

So far I know that for $t>0$, $\Delta X_t(\omega) > c$ if and only if there exists $K \in \mathbb{N}$ such that for each $n \in \mathbb{N}$, there exists $m \in \mathbb{N}$ such that for each $t_1 \in (0,t) \cap \mathbb{Q}, t_2 \in (t,\infty)\cap \mathbb{Q}$, if $|t_1 - t_2| < 1/m$, then $X_{t_2}(\omega) - X_{t_1}(\omega)>c + 1/K -1/n$.

If I put $B_t = \{\omega \in \Omega: \Delta X_t(\omega) > c\}$, $R_{t_1, t_2, m} = \{\omega \in \Omega: |t_1 - t_2| < 1/m \}$, $S_{K, t_1,t_2,n} = \{\omega \in \Omega: X_{t_2}(\omega) - X_{t_1}(\omega) > c +1/K - 1/n\}$, then $R_{t_1, t_2, m}$, $S_{K, t_1,t_2,n} \in \mathcal{F}_\infty$. So $B_t = \bigcup\limits_{K \in \mathbb{N}}\bigcap\limits_{n \in \mathbb{N}} \bigcup\limits_{m \in \mathbb{N}} \bigcap\limits_{t_1 \in (0,t)\cap \mathbb{Q}} \bigcap\limits_{t_2 \in (t, \infty)\cap \mathbb{Q}} R_{t_1, t_2, m}^c \cup S_{K, t_1, t_2, n}$ and hence $B_t \in \mathcal{F}_{\infty}$.

Now I'm stuck trying to answer the question since $A_c = \bigcup\limits_{t>0}B_t$ and I cannot figure out how to get a countable union.

I thought a little about how cadlag functions have at most countably many discontinuities, but I couldn't figure a way to work that in.

I'm sort of expecting this entire process to be an unhelpful direction to begin at, but it was the best that I could come up with. The question reminds me of exercise 1.7 in Karatzas and Shreve. However in that problem I was able to get a uniform characterization of continuity to be able to get a countable union/intersection. With this one, I'm not sure if that's possible, so feel free to offer any other approach to solving the problem. Thanks everyone.

  • 1
    $\begingroup$ I think there is something off with your characterization of $\Delta X_t(\omega)>c$. For instance the function $f(x) := 1_{[1,\infty)}(x)$ satisfies for $t:=1$ and $c=1$ your second condition, but clearly $\Delta f(1)=1$ is not strictly larger than $c=1$. I think you are actually characterizing $\Delta X_t(\omega) \color{red}{\geq} c$. $\endgroup$ – saz May 15 '18 at 18:59
  • 1
    $\begingroup$ Regarding your original problem: Take a look at the lemma in this answer. $\endgroup$ – saz May 15 '18 at 19:08
  • $\begingroup$ saz, you are correct. I will fix that. I'll need to look at that other answer you referenced. Thank you. $\endgroup$ – Ceeerson May 15 '18 at 19:30
  • $\begingroup$ saz, I just read the lemma you referenced and you just answered my question. I was stuck on this problem for a while and am still getting used to these new techniques. So thank you very much, this was very helpful! :) $\endgroup$ – Ceeerson May 15 '18 at 19:41
  • 1
    $\begingroup$ That $B_t$ is $\mathcal{F}_t$-measurable is not difficult: For, notice that both $X(t,\cdot)$ and $X(t-,\cdot)$ are both $\mathcal{F}_t$-measurable. $\endgroup$ – Danny Pak-Keung Chan May 15 '18 at 20:06

Define $A^{n,k}_{p,q}$ for $p,q\in\mathbb{Q}$ with $p<q$, $n\in\mathbb{N}$, $0\leq k<2^n$ as the set of all $\omega\in\Omega$ such that:

  1. $X_{p+(q-p)i/2^n}(\omega)<X_{p+(q-p)j/2^n}(\omega)-c$ for all $i,j$ with $0\leq i\leq k < j\leq 2^n$,

  2. $|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)|<c/4$ for all $i,j$ with $0\leq i\leq j\leq k$,

  3. $|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)|<c/4$ for all $i,j$ with $k<i\leq j\leq 2^n$.

This represents a subdivision of the interval $(p,q)$ where the $X$-value at the first $k+1$ points are more than $c$ smaller than the $X$-values at the remaining points. Define $A_{p,q}^n$ by the (disjoint) union $$A_{p,q}^n:=\bigcup_{k=0}^{2^n-1} A^{n,k}_{p,q}$$ and $A_{p,q}$ by $$A_{p,q} := \bigcup_{m=1}^\infty \bigcap_{n\geq m} A_{p,q}^n.$$ Here $A^n_{p,q}$ allows $k$ to occur at any of the subdivision points except the last, and $A_{p,q}$ requires $\omega$ to eventually exist in $A_{p,q}^n$ for all sufficiently large $n$. Then $$A_c=\bigcup_{\substack{p,q\in\mathbb{Q}\\p<q}} A_{p,q}.$$ Hopefully this is clear.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.