Is every local martingale right continuous?

Question

Is every local martingale right càdlàg (i.e. right continuous with left limits)?

At the university, in the definition of martingale we assume martingales to be right càdlàg processes.

We call an $X$ process local martingale if there exists a $\left(\tau_{n}\right)_{n},\tau_{n}\nearrow\infty$ sequence of stopping times where $X^{\tau_{n}}$ stopped process at $\tau_{n}$ is a martingale for all $n$.

Does this mean every local martingale is right càdlàg as well? Implicitly, the question is if we can only localize a right-càdlàg $X$ processes with $\tau_{n}$ stopping times in order to get a right càdlàg process (, which is a martingale in this case)?

Answer 1

Assume that your objects are defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$, and that for every $n$, $X^{\tau_n}$ is almost surely càdlàg (don't need to say "right càdlàg" the rigth continuity is included in the "càd").

This means that you can find a measurable set $E$ such that $\mathbb{P}(E)=1$ and for all $\omega\in E$,

• $\tau_n(\omega)\xrightarrow[n\to\infty]{} \infty$,
• the "deterministic" function $X(\omega)^{\tau_n(\omega)}$ is càdlàg.

So, take $\omega\in E$, $t\in \mathbb{R}_+$, and $n$ such that $\tau_n(\omega)> t$ (this is possible because of what precedes).

Thus, on a neighborhood of $t$, $X^{\tau_n}(\omega)=X(\omega)$. Càdlàgness is a local property, $X^{\tau_n(\omega)}(\omega)$ is càdlàg at $t$, hence so do $X(\omega)$. This, show that $X$ is almost surely càdlàg .

But, in full generality, a martingale does not need to be right continuous (but even so, it can be regularized).