Properties preserved under passage to augmented filtration

Question

Dear all,

generally speaking, my question is about which properties of a stochastic process are preserved when I skip from the original to the augmented filtration.

Recall that if $(\mathcal{F}_t)_{t\geq0}$is a filtration on a probability space $(\Omega,\mathcal{A},P)$, the augmented filtration $(\mathcal{F}^0_t)_{t\geq0}$ is the smallest filtration such that

• $\mathcal{F}_t\subseteq \mathcal{F}^0_t$ for all $t\geq0$

• $N\subseteq A\in\mathcal{A}$ and $P(A)=0$ imply $N\in\mathcal{F}^0_0$ (complete)

• $\mathcal{F}_t^0=\mathcal{F}_{t+}^0$ for all $t\geq0$ (right-continuous)

It is well known that in many situations it is desirable to work with a complete and right-continuous filtration (for example, this ensures that martingales have a right-continuous modification, or that certain random times are stopping times).

Now here is my question in detail: Suppose we consider a stochastic process $(X(t))_{t\geq0}$and its natural filtration $(\mathcal{F}_t)_{t\geq0}$. And suppose we know that $X$ is a (strong) Markov process and a martingale. Do these properties still hold with respect to the augmented filtration? Or what about martingale problems: Existence results for martingale problems usually refer to the natural filtration of the potential solution process (that is the problem is "find a process which solves the martingale problem with respect to its own natural filtration"). So if I have established existence with respect to the natural filtration does the solution process solve the martingale problem with respect to the augmented filtration, as well?

In fact, my question reduces to the following: If working with the augmented filtration has many advantages why still bother with natural filtrations??

I am aware that these are a lot of questions packed into one post. Sorry, for not being more concise. I would greatly appreciate any comment which helps me to clarify these issues.

Best regards

lpdbw

Answer 1

Hi lpdbw,

I think this is a very interesting questions, here at least a partial answer; It depends heavily on the little word "strong" in parentheseis.

Assuming that $(X_t)$ is strong Markov, the answer to your questions seems to be "yes": The completed filtration of an $\mathbb{R}^d$-valued strong Markov process is already right-continuous (cf. Theorem 2.7.7. of Karatzas/Shreve: Brownian Motion and Stochastic Calculus - note that they use a different terminology for augmented/completed). And just augmenting the filtration does not affect the martingale property (these are just sets of mass zero).

If you look however on Markov processes which does not necessarily has the strong Markov property, the answer seems to be negative as seen in the following counterexample: Let $(W_t)$ be a Brownian motion in its natural filtration and $\xi$ a random variable independent of the Brownian filtration, $\mathbb{P}[\xi =1] = \mathbb{P}[\xi =2] =\frac{1}{2}$, and define the process $(X_t)$ as $$ X_t = \int_0^t \xi \; \;dW_s. $$

This process is a Markov process and a martingale in it's natural filtration $\mathcal{F}_{t}$, but just passing to the right continuous filtration $\mathcal{F}_{t+} = \bigcap_{s>t}\mathcal{F}_s$ destroys the Markov property. Note that for $t>0$ we have $\mathcal{F}_t =\mathcal{F}_{t+} = \sigma(W_s; s\leq t) \vee \sigma(\xi)$, however at $t=0$ they are fundamentally different: $\mathcal{F}_0$ is trivial whereas $\mathcal{F}_{0+} = \sigma(\xi)$. Adding additional null sets changes again nothing.

Answer 2

Thanks a lot for your comments so far. Glad to hear other people are interested in the problem, as well!

I think it is an important message that the strong Markov property is so closely related to the problem.

http://books.google.de/books?id=p_Bn8QAuDf4C&printsec=frontcover has some results which do not require the strong Markov property:

If I got it right (the relevant page is not displayed) "completed filtration" here is understood in the sense of [Karatzas/Shreve] (in contrast to what [KS] refers to as "augmented"). There seems to be a bit of a confusion of terminology in the literature ...

Lemma 3.2.1 in the cited monograph states:

If $X$ is a Markov process w.r.t. filtration $\mathcal{F}:=(\mathcal{F}_t)_{t\geq0}$, then so it is w.r.t. to the completed filtration $\overline{\mathcal{F}}:=(\overline{\mathcal{F}}_t)_{t\geq0}$ (and the transition kernels are the same).

-> http://books.google.de/books?id=p_Bn8QAuDf4C&pg=PA291

Note that he refers to $\overline{\mathcal{F}}_+:=(\overline{\mathcal{F}}_{t+})_{t\geq0}$ as "augmented completed filtration". Moreover, $S$ is assumed to be a separable locally compact space and $S_\Delta$ is its one-point-compactification. Then the following holds:

Suppose $X$ is a right-continuous Feller process w.r.t. to a filtration $(\mathcal{F}_t)_{t\geq0}$ taking values in $S_\Delta$. Then $X$ is also a Feller process w.r.t. to the augmented completed filtration $(\overline{\mathcal{F}}_{t+})_{t\geq0}$.

-> http://books.google.de/books?id=p_Bn8QAuDf4C&pg=PA298

Of course, local compactness (if it is really needed here) is quite a restrictive assumtpion in certain situations.

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Here are some ideas concerning martingale problems (MGP):

Suppose $X$ has Polish state space $E$ and solves a certain well-posed MGP. Denote its law on the cadlag path space $D_E[0,\infty)$ by $P_x$ if $X(0)=x$. According to [Ethier/Kurtz, Theorem 4.4.2], well-posedness + measurability of $x\mapsto P_x(B)$ for any Borel subset $B\subseteq D_E[0,\infty)$ imply the strong Markov property for $X$. Now, the result from [KS] ensures that the augmented filtration (in the terminology of [KS]) is right-continuous and does not harm the strong Markov property. Hence, it would only remain to show that, given the strong Markov property, the martingale property still holds under the larger filtration ... Right now I am not sure whether this is true. After all, the counter example from above is not strong Markov, but that's not a proof of course ;-)

Answer 3

Comment on last paragraph of my previous post (cannot edit post as unregistered user):

Since the augmented filtration is automatically right-continuous here it would suffice to check whether the martingale property is preserved when subsets of null sets are added ... and here I agree with Stephan Sturm that this should be true since these sets again have zero measure.

This would mean that for well-posed MGP which satisfies the measurability condition (which in many situations is relatively easy to check) working with the augmented filtration (in the sense of my original post) does not cause any problems. Do you agree?

Answer 4

Hi,

for the sake of completeness, here is another partial answer to my question:

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If $X$ is a right-continuous martingale on the filtered probability space $(\Omega,\mathcal{A},(\mathcal{F}_t)_{t\geq 0},P)$, then $X$ is a right-continuous martingale on the augmented space $(\Omega,\mathcal{A},(\mathcal{F}^0_t)_{t\geq0},P)$.

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Here, again, "augmented" is understood in the sense of the original post (see above). The proof is quite simple. It may be checked here: Rogers/Williams Vol. 1, Lemma II.67.10 -->

http://books.google.de/books?id=eJp330pIvQ4C&pg=PA173#v=onepage&q&f=false

Getting back to the martingale problem issue, if one has a cadlag solution to a certain martingale problem, then it is still a solution if one replaces the original filtration by its usual augmentation. Of course, often the primary reason for assuming the filtration to be augmented is just to ensure that martingales have right-continuous modifications. However, there are also other motives for assuming augmented filtrations, for example, to ensure that hitting times are stopping times.

Regards

lpdbw

Answer 5

Hi,

I just came across another interesting discussion of the "usual conditions" and how they are related with certain properties of stochastic processes. I add this rerefence here for the sake of completeness, maybe it is useful for someone who comes across this thread.

Here is the link:

http://books.google.de/books?id=mJkFuqwr5xgC&pg=PA34#v=onepage&q&f=false