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
Best Answer
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.