# Filtrations generated by cadlag martingales.

Let $(\Omega,P,\mathcal{F})$ be a probability space with filtration $\mathbb{F} = (\mathcal{F}_t), t \in [0,T]$, where $T$ can be finite or infinite. Let $M$ be a cadlag (local) martingale with respect to $\mathbb{F}$, and let $\mathbb{F}^M$ be the filtration generated by $M$ and then completed with respect to $P$.

Question: Is $\mathbb{F}^M$ a right-continuous filtration?

Some facts:

1. If $X$ is a strong markov process, then the completion of $\mathbb{F}^X$ is right-continuous. This is in Karatzas and Shreve.
2. A sort of converse: If $M$ is a local martingale in a right continuous and complete filtration, it has a right continuous modification.

One possible idea: A continuous local martingale can be expressed as a time-changed Brownian Motion, which is strong markov.

• It seems to me like an affirmative answer follows directly the "càd" part of càdlàg...perhaps I'm overlooking some nuance? – Steve Huntsman Mar 3 '11 at 0:35
• Youre saying the completed filtration generated by any cadlag process is right continuous? If things aren't completed, sets like {M has a local maximum at time t} are in $\mathcal{F}_{t+}$ but not in $\mathcal{F}_t$. – weakstar Mar 3 '11 at 0:43
• Ah, silly me... – Steve Huntsman Mar 3 '11 at 1:05
• For a brownian motion, those kinds of sets have probability zero. I'm not sure if that's the only kind of obstruction, but since martingales have the same paths as brownian motion, I had hoped you might be able to say something. – weakstar Mar 3 '11 at 1:16

No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.

1. $W$ is a standard Brownian motion.
2. $U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$.

Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$.

Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov.

As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.

• Thanks. Maybe i'm being dumb, but can this counterexample be extended to a nonzero starting position? – weakstar Mar 3 '11 at 1:07
• @weakstar: I'm not quite sure what you mean. Setting $M_t=UW_t+c$ extends it to a nonzero starting position. If $W$ was started from a nonzero position and you wrote $M=UW_t$, then $U$ would be $\mathcal{F}^M_0$-measurable, so the argument breaks down. – George Lowther Mar 3 '11 at 1:26
• Right, I meant your second suggestion, for brownian motion starting away from zero. Just seeing if any affirmative answer can be salvaged. – weakstar Mar 3 '11 at 1:45
• @ George Lowther : Hi, could you please recall the version of the Bluementhal 0-1 Law, you are using to justify that sets in $\mathcal{F}_{0^+}^M$ have 0-1 measure. Best Regards – The Bridge Mar 3 '11 at 8:32
• @The Bridge: I'm not using any such law, nor am I saying that $\mathcal{F}^M\_{0+}$ only has sets with probability 0 or 1 (it doesn't). I was saying that $\mathcal{F}^M\_{0}$ only has sets of measure 0 or 1 (in the first example) because it is generated by a random variable which is identically zero (and then completed). – George Lowther Mar 3 '11 at 9:02