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 $(\mathcal F_t)$ ?

  • $\begingroup$ The answer is NO to both. $\endgroup$ Commented Sep 11, 2019 at 9:02

2 Answers 2


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.

  • $\begingroup$ 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 ? $\endgroup$
    – user659895
    Commented Sep 11, 2019 at 9:29
  • $\begingroup$ @user659895 Yes, that's a convention which many authors use ("unless otherwise stated $\mathcal{F}_t$ is the canonical filtration" ... something like this). $\endgroup$
    – saz
    Commented 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.


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