Hi Taposh,

This subject has various subtleties that I cannot do justice in a quick comment. However, in order to construct continuous or cadlag modifications, I think that you do need to include the zero probability sets in each F_t, in order to modify the process on zero probability sets at which it fails to have a left-limit. When you take right-limits to obtain X_{t+} as an F_{t+}-measurable random variable, then that should work. However, I would expect that the resulting process t -> X_{t+} would only be cadlag/continuous outside of a zero probability set.

I am not sure if you received a message from me (see below). Based on a review of your responses to other readers, I think my problem can be resolved by simply taking lim sup in place of the usual limit. With the right continuity of the filtration, X_{t+} will be adapted to F_{t+}=F_t because of the lim sup used in the definition. Let me know what you think. The only problem now with this approach is that X_{t+} defined using lim sup can take extended real values infinity or minus infinity (of course, with probability zero).

My older post (not sure if you got it):

Hi George,

I also have two questions on the completeness of F_t (or inclusion of P-null sets of F_\infty) for cadlag modifications.

1. Books like Dellacheri-Meyer (part B-page 67), Karatzas-Shreve (page 16), and Revuz-Yor (page 64) start with a super martingale (X_t, F_t) and produce another process X_{t+} by using regularization of X_t outside a set of measure zero. Without assuming the right continuity of F_t or assuming completeness in any way, they argue that (X_{t+}, F_{t+}) is a supermartingale. In my view, X_{t+} is not adapted to F_{t+} because we need the P-null sets. So, I never understand why such a result is true. In books like Rogers-Williams (page 171) and also your blog, you guys are more careful.

2. In books like Lipster and Shiryaev (page 61), there is a result of cadlag modification of a martingale that can be written as X_t = E[X|F_t]. Here F_t is right continuous but not necessarily complete (no P-null sets). Again, it is claimed that this process has a cadlag modification.

Any thoughts on these? Did I miss something here?

]]>Hi. I am not exactly what the statement is that you are refering to, but it is true that a cadlag process cannot cross an interval [a,b] for a < b infinitely often in finite time. Otherwise, you would be able to find a bounded increasing sequence of times with for even n and for odd n, which would contradict the property existence of left limits, as would not converge.

]]>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 1: we have almost surely. Combining this with completeness of the filtration, the fact that X is adapted implies that is adapted. So, right-continuity is not required.

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…

]]>Thanks for this detailed answer!

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 🙂

No, you don’t need completeness of the filtration for the existence of a cadlag modification, but it is necessary to assume this (or a similar condition) if you want the modification to be adapted. I did assume completeness throughout this post. Rather than completeness, you only need the weaker condition that contains all sets in with zero probability. Right-continuity of the filtration is not required except in special cases, such as Theorem 5 above where it is needed to guarantee that all martingales are right-continuous in probability.

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