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Let's motivate the question by a classical result: Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$ which satisfies the usual conditions
  • $X=(X_t)_{t\ge 0}$ be a $\mathbb F$-submartingale on $(\Omega,\mathcal A,\operatorname P)$

If $t\mapsto\operatorname E[X_t]$ is right-continuous, then $X$ has a right-continuous modification which can be chosen as to be RCLL.

This satement can be found in the monograph by Karatzas and Shreve, p. 16.

Question: I know what a modification is, but what's meant by continuous modification? Does it mean, that we can find a modification which is $\operatorname P$-almost surely continuous or does it mean, that we can find a modification which is continuous (i.e. every path is continuous)?

In the latter case, the RCLL-property would reduce to the existence of left-side limits. In the former-case, the RCLL-property would imply, that we can pick a special modification, which is continuous.

(Has this anything to do with the "usual conditions"-assumption?)

It's a similar problem I've got with ($\operatorname P$-almost surely) continuous processes like Brownian motion. I know, that we can find a special probability space, such that there is a Brownian motion with continuous paths. But can we always modifiy a $\operatorname P$-almost surely continuous process, such that every path is continuous?

(What do people mean, when they say that a process is continuous? Do they actually always mean almost surely continuous?)

0xbadf00d
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1 Answers1

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If one process $X$ is a.s. continuous, then it is indistinguishable from another process which is surely continuous. Thus if $X$ is a process, $Y$ is an a.s. continuous modification of $X$, then there is a surely continuous process $Z$ which is indistinguishable from $Y$. $Z$ is also a modification of $X$. Thus in particular there is really no loss in defining Brownian motion to be surely continuous. Indeed the direct construction of the Wiener measure using cylindric sets use this definition. Other constructions, such as the construction from piecewise linear approximants with iid normal slopes, don't, because they rely on a procedure which constructs a function which, if $\omega$ is in some nonempty null subset of $\Omega$, is discontinuous.

For the issue of "modification" vs. "indistinguishable", cf. Difference between Modification and Indistinguishable

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Ian
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  • But how do we know, that we can modify $X$ in a way, that it is surely continuous? Is there no possibility, that $X$ is almost surely continuous, but cannot be modified such that it is continuous? – 0xbadf00d Dec 30 '15 at 15:39
  • @0xbadf00d We already modified $X$ to make it a.s. continuous, then we made another modification of that (in general a smaller one, since we made an indistinguishable change) to get something surely continuous. You should check that the resulting change is still a modification. – Ian Dec 30 '15 at 15:41
  • So, if $X$ is a.s. continuous, there is a $\operatorname P$-null set $N\subseteq\Omega$ such that $X(\omega)$ is continuous, for all $\omega\in\Omega\setminus N$. For some $\omega\in N$, we can set the path $t\mapsto X_t(\omega)$ to anything continuous we like, e.g. $X(\omega)\equiv 0$. Doing so for all $\omega\in N$, we're left with a surely continuous process $\tilde X$ and $$\left{X_t\ne\tilde X_t\right}\subseteq N;;;\text{for all }t\ge 0;,$$ i.e. $X$ and $\tilde X$ are indistinguishable. Right? – 0xbadf00d Dec 30 '15 at 16:32
  • @0xbadf00d That's right. Now you just need to check that $\tilde{X}$ is a modification of the original process which was not even a.s. continuous. – Ian Dec 30 '15 at 16:37
  • That should be trivial, since if $X$ and $Y$ are modifications and $Y$ and $Z$ are modifications, then $X$ and $Z$ should be modifications, too, right? – 0xbadf00d Dec 30 '15 at 16:40
  • @0xbadf00d Yes, that step is indeed trivial, but still important for getting at the result you actually want. – Ian Dec 30 '15 at 16:41
  • Do we lose adaptedness at any step? Is the "usual conditions"-assumption of any importance here? – 0xbadf00d Dec 30 '15 at 16:42
  • @0xbadf00d We don't change finite dimensional distributions, so the only possibility is that the filtration we are using is somehow too small. But I can't see how that would happen, either. – Ian Dec 30 '15 at 17:07
  • So, we don't need the assumption on $\mathbb F$, i.e. it can be any filtration, right? – 0xbadf00d Dec 30 '15 at 17:16
  • @0xbadf00d Oh, you mean in your actual theorem in the OP? In that case I'm pretty sure you need the assumptions. I thought you meant to ensure that modifying twice creates a modification. – Ian Dec 30 '15 at 17:28
  • No, I don't meant the theorem. We need right-continuity of $\mathbb F$ to show it. However, I know about few statements, where we really need the usual conditions. So, I've started to hate them, since many authors assume that the filtrations satisfy these conditions, while they use them rarely. $$$$ However, let me ask a final question. I've noticed, that I've asked a similar question some time ago (link below) and in the comments below your answer, you've said "You don't modify any paths". But that's exactly what I've suggested in my second comment. – 0xbadf00d Dec 30 '15 at 17:50
  • So, I'm weird (while I notice, that replacing $\Omega$ by $\Omega\setminus N$ leads to a surely continuous process, too, but I don't want to modify the probability space). http://math.stackexchange.com/questions/1265458/why-can-we-consider-the-brownian-motion-as-being-a-mapping-into-the-space-of-con?rq=1 – 0xbadf00d Dec 30 '15 at 17:51
  • @0xbadf00d You could actually modify paths if you want. But if you look carefully at the definitions, you can also modify the underlying probability space and still satisfy all the desired properties. – Ian Dec 30 '15 at 18:13