Yes, I should have said “nonempty paved space”. Thanks for pointing this out, will fix it

]]>Thank you for the response.

]]>Thank you very much for your last response, George, that makes perfect sense to me. I wasn’t able to reply to your comment as the ‘Reply’ button under your comment is hidden for some reason. So I post it here

]]>This is implicit in the statement that is nonnegative. If *X* was not continuous, it could jump right past the level *n* resulting in going negative.

Actually, to be strict, I should have assumed (wlog) that *X* starts from 0, just to ensure that it does not start from above *n* [Update: I fixed the proof to address this].

For theorem 4 and lemma 5, we require the martingale to be continuous. You have given an example of a cadlag martingale for which which inferences of the theorem 4 fails.

However, in the proof of lemma 5, where have you used the fact that submartingale is continuous? For the sequence of hitting times to be stopping times we just require right continuity(Debut theorem).

]]>Sorry, I missed your comment when it was posted last year. But, I have *ψ*(*δ*) = *δ²/(**β* – 1 – *δ*)², which does vanish as *δ* → 0. Or am I misunderstanding you?

That can’t be a local martingale in general. E.g., suppose M is standard Brownian motion and is the first time at which it hits 1, or the minimum of this time and 1 (to keep it bounded).

Then is a process which starts at 0 and jumps to 1 (if ), so cannot be a local martingale.

Thank you for the prompt response! I checked Protter earlier but I missed Emery’s example then.

I see now, your example of X being a martingale is rather easy to verify. The question I specifically had in mind is as follows:

If \tau is a bounded stopping time and M is a mean zero local martingale, is then M_\tau I(\tau \leq t) a (mean zero) local martingale?

(under ‘regularity conditions’; usual hypotheses, cadlag processes, etc).

Do you think such a result might hold, or do you by chance know about any counterexamples?

Thanks again and sorry for the trouble.

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