# Enlargement of filtration

Let $$M_t$$ be a continuous time real valued martingale, and $$\mathcal F_t$$ its natural filtration.

Suppose that $$\mathcal F_t \setminus \mathcal F_s$$ is nonempty for all $$t > s$$.

Let $$\mathcal G$$ be a sigma algebra, and define the filtration $$\mathcal H_t := F_t \vee \mathcal G$$.

Question: Is it true that $$M$$ is a $$\mathcal H_t$$ martingale if and only if $$\mathcal G$$ is independent of $$\mathcal F_t$$ for all $$t$$?

Remark: The if direction follows from a monotone class argument.

• I you wish the condition to be necessary (I think it it not, but for the moment, I have no counterexample), you should at least assume that ALL $\mathcal{F}$-martingales are still $\mathcal{H}$-martingales. This property is called immersion of filtrations. Otherwise, the null martingale provides trivial counterexamples. Jun 29 at 14:22
• Thanks for your comment - though counterexamples like the null martingale are ruled out by the condition that $\mathcal F_t \setminus \mathcal F_s$ be nontrivial. Jun 29 at 15:34
• Ah yes, you assume that $\mathcal{F}$ is the natural filtration of $M$. Anyway, the answer is negative, see the answer below. Jun 29 at 19:49

I think that I have a counterexample. Let $$(X,Y)$$ be a Brownian motion in $$\mathbb{R}^2$$. Then $$M = \int_0^\cdot X_s \mathrm{d}Y_s$$ is a martingale, in the natural filtration of $$(X,Y)$$, in its own filtration $$(\mathcal{F}_t)_{t \ge 0}$$ and also in $$(\mathcal{F}_t \vee \sigma(X))_{t \ge 0}$$. Yet, $$X$$ is not independent of $$M$$ since $$\langle M \rangle = \int_0^\cdot X_s^2\mathrm{d}s$$ is not deterministic.
Remark: in this example, $$\mathcal{F}$$ is not immersed in $$\mathcal{H}$$ since $$X$$ is no more a martingale in $$\mathcal{H}$$.
• @Nate River. Thank you. I still wonder whether your conclusion holds if you assume that that all $\mathcal{F}$-martingales are still $\mathcal{FH$-martingales. The counterexample I gave does not answer this alternative question. Jun 30 at 6:13