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Suppose $(\Omega,\mathcal F,\mathbb F,\mathbb P)$ is a probability space supporting a Brownian motion $\{W_t\}_{t=0}^T$, where the filtration is the augmented brownian filtration $\{F_t^W\}$ and $\mathcal F=F_T^W$. Let $\{S_t\}_{0 \leq t \leq T}$ be a jointly measurable process which is a martingale in its own natural filtration. I want to understand if there is condition in terms of the joint law of $(W,S)$ (on the canonical path space $\mathcal C([0,T])\times \mathbb D([0,T])$) which is sufficient for $\{S_t\}_{0 \leq t \leq T}$ to be $\mathbb F$-adapted. My first guess is that if $W$ is a martingale w.r.t. $\{F_t^W \vee F_t^S \}_{0\leq t\leq T}$, then $\{S_t\}_{0 \leq t \leq T}$ is $\mathbb F$-adapted. Unfortunately I cannot prove it.

Jinglin-Kiki Xu
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1 Answers1

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It is enough to have $W$ to be a martingale under the enlarged filtration $\mathcal F_t^{W,S}$ to guarantee that $S$ is adapted. This is because, by martingale representation theorem, any brownian martingale is a stochastic integral w.r.t. $W$. If $S$ is not adapted, then $\mathcal F_t^{W,S}$ would be a strict enlargement, which will certainly destroy the martingale property of some martingale. This implies what $W$ cannot remain a martingale under the enlarged filtration.

Ryan
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  • By "will certainly destroy the martingale property of some martingale" do you mean that (when $\mathcal F_t^{W,S}$ is strictly bigger than $\mathcal F_t^W$) there is a martingale of ${\mathcal F_t^W}$ that is not a martingale of ${\mathcal F_t^{W,S}}$? – John Dawkins Apr 08 '17 at 18:40
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    yes, you are right. – Ryan Apr 08 '17 at 19:48