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Let $(\Omega,\mathcal{F},\mathbb{P})$ be a complete probability space. We consider a Wiener process $W$ with respect to his standard filtration $(\mathcal{F}_t^W)_{t \geq 0}$ and a process $X$ with his standard filtration $(\mathcal{F}_t^X)_{t \geq 0}$. We assume that $W$ and $X$ are independent. Let $(\mathcal{F}_t)_{t \geq 0}$ the filtration defined by $$\mathcal{F}_t = \sigma(\mathcal{F}_t^W \cup \mathcal{F}_t^X),$$ for all $t \geq 0$. I want to prove that $W$ is also a Wiener process with respect to this new filtration $(\mathcal{F}_t)_{t \geq 0}$. Obviously, $W$ is $(\mathcal{F}_t)_{t \geq 0}$-adapted, $\mathbb{P}$-almost surely continuous, $W_0 = 0$ $\mathbb{P}$-almost surely and stationary. So, the last thing to prove is: for $0 \leq s < t$, $W_t - W_s$ is independent of $\mathcal{F}_s = \sigma(\mathcal{F}_s^W \cup \mathcal{F}_s^X)$, i.e. $\sigma(W_t-W_s)$ is independent of $\sigma(\mathcal{F}_s^W \cup \mathcal{F}_s^X)$.

I know we have $\sigma(W_t-W_s)$ is independent of $\mathcal{F}_s^W$, $\sigma(W_t-W_s)$ is independent of $\mathcal{F}_s^X$, $\mathcal{F}_s^W$ is independent of $\mathcal{F}_s^X$, by definition. So, how can I conclude that $\sigma(W_t-W_s)$ is independent of $\sigma(\mathcal{F}_s^W \cup \mathcal{F}_s^X)$?

Thank you very much for your help!

1 Answers1

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The proposition at the end is wrong: to assume that $\mathcal G$, $\mathcal H$ and $\mathcal K$ are pairwise independent sigma-algebras does not guarantee that $\mathcal G$ is independent of $\sigma(\mathcal H\cup \mathcal K)$. One needs a stronger hypothesis, namely that $\mathcal G$ and $\mathcal H$ are independent and that $\mathcal K$ is independent of $\sigma(\mathcal G\cup \mathcal H)$. Then indeed, $\mathcal G$ is independent of $\sigma(\mathcal H\cup \mathcal K)$.

Use this with $\mathcal G=\sigma(W_t-W_s)$, $\mathcal H=\mathcal F^W_s$ and $\mathcal K=\mathcal F^X_s$.

Did
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  • Thank you! So, here the $\sigma$-algebras are pairwise independent, but how can I prove (if it is possible) that $\mathcal{F}_s^X$ is independent of $\sigma(\sigma(W_t-W_s) \cup \mathcal{F}_s^W )$? Do you know a good reference about independence between $\sigma$-algebras? –  Apr 19 '13 at 10:07
  • These are respectively sub-sigma-algebras of $\mathcal F^X$ and $\mathcal F^W$, which are independent. // Reference: David Williams, Probability with martingales. – Did Apr 19 '13 at 10:53
  • Great, that's exactly what I was thinking. But my problem is now: how can I prove that $\sigma(\sigma(W_t-W_s) \cup \mathcal{F}_s^W)$ is a sub-$\sigma$-algebra of $\mathcal{F}^W$? To be sure: $\mathcal{F}^W$ is the smallest $\sigma$-algebra for which $W_t$ is measurable, for all $t \geq 0$, right? Then we have well $\mathcal{F}_s^W \subseteq \mathcal{F}^W$ because $\mathcal{F}_s^W$ is the smallest $\sigma$-algebra for which $W_u$ is measurable, for all $0 \leq u \leq s$. So the question is: why $\sigma(W_t-W_s) \subseteq \mathcal{F}^W$ ? Thank you. –  Apr 19 '13 at 11:49
  • Well, finally, I think I got it and it is quite obvious, no? Because $W_t$ and $W_s$ are both $\mathcal{F}^W$-measurable, it is the same for $W_t - W_s$, then $\sigma(W_t-W_s) \subseteq \mathcal{F}^W$, isn't it? –  Apr 19 '13 at 13:26
  • Yes it is. $ $ $ $ – Did Apr 19 '13 at 17:50