2
$\begingroup$

Let $X$ be a stochastic process with every sample path RCLL(right continuous on $[0,\infty)$ and with left-hand finite limits on $(0,\infty)$). Let $A$ be an event that $X$ is continuous on $[0,t_0)$. Show $A\in\mathcal{F}_{t_0}^X$.

For $(t_n)_n$ a dense sequence contained in $(0,t_0)$, if we can show

$$A=\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega),\ \ \forall n\geq 1,\ \ t_n<t\}$$

then $A\in\mathcal{F}_{t_0}^X$ since every $\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega)\}\in\mathcal{F}_{t_0}^X$. Could someone show how to prove

$$A=\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega),\ \ \forall n\geq 1,\ \ t_n<t\}$$

$\endgroup$
  • 1
    $\begingroup$ I don't think your claim is true as written. If there's an $\omega$ such that $X(\omega)$ has a jump at some $t \notin \{t_n\}$ (which is entirely possible under the given assumptions, and could even happen with probability 1), then $\omega$ is in the set on the RHS, but not in $A$. $\endgroup$ – Nate Eldredge Jun 23 '13 at 15:14
  • $\begingroup$ However, as another approach to the measurability of $A$, you might use (after first verifying) the fact that an RCLL path is continuous on $[a,b]$ iff its restriction to $\mathbb{Q} \cap [a,b]$ (or your favorite dense subset) is uniformly continuous. $\endgroup$ – Nate Eldredge Jun 23 '13 at 15:18
  • $\begingroup$ Thanks a lot for your help! $\endgroup$ – Higgs88 Jun 23 '13 at 16:10
  • 1
    $\begingroup$ If you get it worked out, you might like to post the solution here as an answer. $\endgroup$ – Nate Eldredge Jun 23 '13 at 18:01
1
$\begingroup$

Thanks to Nate Eldredge, I could have a solution here.

Set $\{t_n\}_{n\geq 1}=\mathbb{Q}\cap [0,t_0)$, then we firstly we prove $X$ is continuous on $[0,t_0)$ if and only if its restriction on $\{t_n\}_{n\geq 1}$ is uniformly continuous:

if $X$ is continuous on $[0,t_0)$, by definition we can extend $X$ on $[0,t_0]$ by defining $X_{t_0}:=\lim_{t\rightarrow t_0-}X_t$, then $X$ is continuous on $[0,t_0]$ and so it is uniformly continuous on $[0,t_0]$;

if $X$ is uniformly continous on $\{t_n\}_{n\geq 1}$, then by Cauchy's criterion, we can define $\widetilde{X}$ on $[0,t_0]$ by

$$\widetilde{X}_t:=\lim_{s\rightarrow t, s\in\{t_n\}_{n\geq 1}}X_s$$

Evidently $\widetilde{X}$ is continuous. For $X$ is right-continuous, we have necessarily $X=\widetilde{X}$, so $X$ is continuous on $[0,t_0)$.

Secondly by an argument largely used to describe the continuity of a function, we have

\begin{eqnarray} A&=&\{\omega: X(\omega) \text{ is uniformly continuous on } \{t_n\}_{n\geq 1}\}\\ &=&\bigcap_{m\geq 1}\bigcup_{k\geq 1}\bigcap_{n\geq 1}\{\omega: |X_t(\omega)-X_{t_n}(\omega)|<\frac{1}{m},\ \ \forall t\in \{t_l\}_{l\geq 1} \text{ with } |t-t_n|<\frac{1}{k}\} \end{eqnarray}

So $A$ is measuralbe $\clubsuit$

$\endgroup$

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.