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.



Filtrations generated by cadlag martingales.
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](https://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/43841#43841) of a Markov process which is not strong Markov.




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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.


