# If $(B_t)$ is a Brownian motion and adapted to the filtration $(\mathcal F_t)$ does $B_t-B_s$ is independent of $\mathcal F_s$?

Let $$(B_t)$$ a Brownian motion adapted to the filtration $$(\mathcal F_t)$$.

Q1) If $$t> s$$, does $$B_t-B_s$$ is necessarily independent of $$\mathcal F_s$$ or $$(\mathcal F_t)$$ must be the natural filtration of $$(B_t)$$ to have $$B_t-B_s$$ independent of $$\mathcal F_s$$ ?

Q2) Does $$(B_t)$$ is necessarily a martingale w.r.t. $$(\mathcal F_t)$$ or $$(\mathcal F_t)$$ must be the natural filtration for $$(B_t)$$ being a martingale w.r.t. to $$(\mathcal F_t)$$ ?

• The answer is NO to both. Sep 11, 2019 at 9:02

The filtration $$(\mathcal{F}_t)_{t \geq 0}$$ does not need to be necessarily the natural filtration, but you can't pick just any filtration.

For instance, consider the filtration $$\mathcal{F}_t := \sigma(B_r; r \geq 0).$$ Note that $$\mathcal{F}_t$$ actually does not depend on $$t$$. It is trivial that $$(B_t)_{t \geq 0}$$ is adapted to this filtration. Moreover, you can easily show that $$B_t-B_s$$ is not independent from $$\mathcal{F}_s$$ and that $$(B_t)_{t \geq 0}$$ is not a martingale with respect to $$\mathcal{F}_t$$.

Typically, the two properties, which you mentioned, hold true if $$\mathcal{F}_t$$ is the canonical filtration enlarged by some independent events. For instance, if, say, $$U$$ is a random variable which is independent from $$(B_t)_{t \geq 0}$$, then

$$\mathcal{F}_t := \sigma(U, B_r; r \leq t)$$

is strictly larger than the canonical filtration, and both properties (independence and martingale property) hold true.

• Thank you for your answer. Just a small question then. I have a theorem that says : $(B_t)$ is a Brownian motion $\iff$ $(B_t)$ and $(B_t^2-t)$ are continuous martingales. But they didn't precise any filtration. In this context, shall I guess that it's martingale w.r.t. the natural filtration ? Sep 11, 2019 at 9:29
• @user659895 Yes, that's a convention which many authors use ("unless otherwise stated $\mathcal{F}_t$ is the canonical filtration" ... something like this).
– saz
Sep 11, 2019 at 9:47
• Let $$(\Omega ,\mathcal F,\mathbb P)$$ a probability space and $$(B_t)$$ a Brownian motion. Take $$\mathcal F_t=\mathcal F$$ for all $$t$$. Then $$(B_t)$$ won't be a martingale.
• Notice that if $$B_t-B_s$$ is independent of $$\mathcal F_s$$ for all $$t>s$$, then $$(B_t)$$ will be a martingale.

• If $$(\mathcal F_t)$$ is a filtration adapted to $$(B_t)$$ and such that $$B_t-B_s$$ is independent of $$\mathcal F_s$$ for all $$t>s$$, then $$(\mathcal F_t)$$ is called an admissible filtration.