Let $f$ be a continuous bounded function. $X$ is a Feller process, and $\hat X$ is its Cadlag modification. By the definition of the modification, one can write $$\mathbb E[f(X_t)] = \mathbb E[f(\hat X_t)], \ \forall t>0$$ I wonder if it is safe to generalize the above identity to a hitting time. More precisely, we define exit time for an open set $O$ by $$\tau = \inf\{t\ge 0: X_t \notin O\}, \ \hat \tau = \inf\{t\ge 0: \hat X_t \notin O\}.$$

[Q.] Is $\mathbb E[f(X\tau)] = \mathbb E[f(\hat X_{\hat \tau})]$ true? If not, is there any counter-example?

I guess it is true and very fundamental result, since many papers on stochastic processes start with problem setup on the modification process without further discussion. For instance, at the beginning of Section III.3 of the book [D Revuz and M. Yor] of strong Markov property, they write:

"We shall consider the canonical Cadlag version of a Feller process..."

Discussion 1: However, most references discuss only the existence of such a version, but not the above question.

Discussion 2: By adopting the definition Feller process given by Page 88 of book, the existence of Cadlag modification is guaranteed by section III.2 of the book [D Revuz and M Yor].

  • $\begingroup$ There are various definitions of "Feller process" floating around, so can you please carefully and completely define what it means to you? $\endgroup$ Commented Nov 13, 2015 at 3:24
  • $\begingroup$ @NateEldredge Sure, thanks for your attention. $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 3:25
  • $\begingroup$ @NateEldredge I've changed the link to google book. I am totally confused. Take a Brownian motion $W$, it is Feller and strong Markovian and continuous. Now I manually change the value at all $t\in \mathbb Q$ such that $W(t, \omega) = \infty$ for $\omega \in N_t$ of Null set $N_t$, then the new one $W'$ is modification. It is Feller, but fails to be continuous everywhere, and strong markov? $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 4:14
  • $\begingroup$ If I understand your example correctly, it is still a.s. continuous because $\bigcup_{t \in \mathbb{Q}} N_t$ is null. So nothing really has changed. $\endgroup$ Commented Nov 13, 2015 at 4:22
  • $\begingroup$ @NateEldredge Agreed $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 5:55

1 Answer 1


I think the following example says no.

Consider the state space $\{0,1\}$. Let $U$ be a uniform random variable on $[0,1]$ and let $X_t = 1$ if $t=U$ and $X_t = 0$ otherwise. Note that $X_t$ is a.s. not cadlag, but for each $t$ we have $X_t = 0$ a.s., so $X_t$ is a modification of the cadlag process $\hat{X}_t$ that just sits at 0. Taking $\mathcal{G}_t = \sigma(X_s : s \le t)$, unless I am mistaken, $X_t$ meets the Revuz/Yor definition of a Markov process (under its natural filtration). We may take the transition semigroup to simply be the identity $P_t f = f$, which is Feller.

But if we let $\tau = \inf\{t : X_t \ne 0\}$ and let $f(0)=0$, $f(1)=1$, then $E[f(X_\tau)] = 1$ while since $\hat{\tau} = +\infty$ we might consider $E[f(\hat{X}_{\hat{\tau}})] = 0$.

By a similar trick you can get a discontinuous modification of Brownian motion, with which I think you can get an example where $\hat{\tau}$ is a.s. finite.

  • $\begingroup$ Let $L$ be generator of $X$, If we consider $L u = 0$ with Dirichlet data $f$ on $\partial O$, the probabilistic counterpart is $u(x) = \mathbb E[f(X_\tau)]$. Then I wonder if $X$ and $\tau$ in the above is of Cadlag version or original one, if the answer is NO? $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 4:23
  • $\begingroup$ @kenneth: I'm sorry, I really don't understand your comment. The generator of my process $X_t$ is the 0 operator. Note that $C(\{0,1\})$ is a vector space of dimension 2. "Dirichlet boundary conditions" don't make sense here because in my state space, every set has empty boundary. $\endgroup$ Commented Nov 13, 2015 at 4:26
  • $\begingroup$ Sorry, I confused you. Later I am out, and will be back in a few hours. $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 4:33
  • $\begingroup$ In the previous comment, I was talking about general process $X$, not the specific one in your example. For instance, a Drichlet problem underlying $\alpha$-stable process with a unit ball. $\endgroup$
    – kenneth
    Commented Nov 13, 2015 at 5:53

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