I have a question, please. If I have a cadlag which could cross this increasing sequences of intervals say [0, Sn], in the paper that I am reading, it says that this cadlag cannot cross this interval infinitely many times in finite time, it should cross in infinite time. I did not understand the statement, can you explain it please ?

Thanks ]]>

Thanks a lot for the interesting blog. Concerning Theorems 1 and 2, I was looking for references in the literature, but so far I haven’t found any. Can you help here?

Marcus

]]>In note 2: I think the bit that is concerning you is where I state that “ would converge to zero”. This is true, even for increasing sequences. Fix a sample path which has left limits on S (which is true for almost all sample paths). Consider choosing a sequence of times with for k even and for k odd. The fact that this is an increasing sequence means that converges to a limit. So, converges to 0. However, by choosing very close to , we can ensure that is bounded by 1/k and, hence, tends to zero. So, tends to zero. Maybe that was a bit of a big step to make in the proof…

]]>In note 1, I meant you need right continuity for to be adapted.

In note 2, I still think you need a decreasing sequence (even if left limits exist) since you define as the limit from the right and since you are comparing the value of X_{t} with to get the contradiction. But maybe I am missing a detail and haven’t thought it through carefully enough yet 🙂

In fact, suppose that *X* is any adapted process which has a cadlag modification with respect to the completion of the filtration (or, with respect to any enlargement of the filtration). Then we can define

As *X* does have a cadlag modification, with respect to some enlargement of the filtration, it must be equal to almost surely. So, is almost-surely cadlag. Also, as almost surely (for each fixed *t*) and is -measurable, it follows that will be an adapted and almost-surely cadlag modification of *X* so long as contains all sets in with zero probability. If we want a modification which is actually cadlag rather than just almost surely cadlag, you can set to be identically 0 in the event where it is not cadlag (which can be seen -measurable by expressing it in terms of upcrossings of *X* on ).

So, this shows that: for a cadlag modification, no assumptions on the filtration are required. For a cadlag adapted modification, requiring to contain all sets in of zero probability is enough. For an adapted and almost-surely cadlag modification, requiring to contain all zero probability sets in is sufficient.

For example, let ε_{1}, ε_{2} be a sequence of Bernoulli IID random variables, each equal to 1 and -1 with probability 1/2. Let *t*_{1}, *t*_{2} be a sequence of times strictly increasing to 1. Define,

(and set *X*_{t} to zero if this sum does not converge). This is right-continuous and has left limits everywhere except, possibly, at *t* = 1 in the case where fails to converge. By martingale convergence of uniformly square integrable martingales, this converges almost surely, so *X* is almost surely cadlag – but not necessarilly cadlag. If is a modification of *X* and adapted to the natural filtration of *X*, then for each $t < 1$ (and not just almost surely, because is finite with no nonempty zero probability sets). Then, $\tilde X_t$ fails to have a limit at *t* = 1 whenever *X* does. In fact, the natural filtration here is already right-continuous, showing that even in the right-continuous case we need to enlarge the filtration by adding zero probability sets to to get a modification which is right-continuous.

For another example, let *Y* be a Poisson process and . It can be seen that *Y* is a cadlag modification of *X* but is not adapted with respect to the natural filtration of *X*. To get an adapted modification which is almost surely cadlag, you need to enlarge the filtration by adding zero probability sets of to (for almost all times *t*).

In answer to,

Note 1: No, right-continuity is not required except where explicitly stated (e.g., Theorem 5). Hopefully my comment above helps explain this – otherwise could you state where you think right continuity is required?

Note 2: Any uncountable set of times will contain a strictly decreasing subsequence. This is not required though, as the argument will also work for increasing sequences. Letting *t* be the limit of the sequence *t _{k}*, we are using the fact that

I am curious whether completeness of the filtration is required for the existence of measurable cadlag modifications. The proof you present does not seem to require such an assumption, but books such as Karatzas and Shreve include that assumption. There is a paper by Hans Follmer on exit times of supermartingales which doesn’t require completeness, but which has additional requirements on the filtration. Hence I am curious about your view on the matter (whether completeness is required).

Thank you again for these excellent posts.

Note 1: I think you need the filtration to be right-continuous in your proof in order to get an adapted modification.

Note 2: It wasn’t fully clear why the sequence of times you construct in the second to last paragraph would have a subsequence converging from the right. Convergence from the right of t_k to some limit point t seems a key requirement in order to get

(i) \tilde{X}_(t_k) to converge to \tilde{X}_(t)

and

(ii) X_(t_k) to converge to \tilde{X}_(t)

the combination of which leads to the contradiction you mention.