I'm reading this lecture note about Brownian motion.

Definition 3.2 (Filtration). A discrete filtration of a set $\Omega$ is a collection $\left\{\mathcal{F}_n\right\}$ of $\sigma$-algebras of subsets of $\Omega$ such that $\mathcal{F}_n \subset \mathcal{F}_{n+1}$ for all $n \in \mathbb{N}$. A continuous filtration of a set $\Omega$ is a collection $\left\{\mathcal{F}_n\right\}$ of $\sigma$-algebras of subsets of $\Omega$ such that $\mathcal{F}_s \subset \mathcal{F}_t$ for all $s<t$

Definition 3.3 (Natural Filtration). In particular, for a discrete stochastic process $X_n$, the natural filtration $\left\{\mathcal{F}_n\right\}$ is a filtration such that each $\mathcal{F}_n$ is the $\sigma$-algebra generated by $X_1, \ldots, X_n$. Namely, $\mathcal{F}_n$ contains all the information in $X_1, \ldots, X_n$

Definition 3.4 (Martingale). A sequence of random variables $M_0, M_1, \ldots$ is called a martingale with respect to the filtration $\left\{\mathcal{F}_n\right\}$ if:

  • (1) For each $n, M_n$ is an $\mathcal{F}_n$-measurable random variable with $\mathbb{E}\left[\left|M_n\right|\right]<\infty$
  • (2) If $m<n$, then $\mathbb{E}\left[M_n \mid \mathcal{F}_m\right]=M_m$. Or we can write it in another form: $$ \mathbb{E}\left[M_n-M_m \mid \mathcal{F}_m\right]=0 . $$

Moreover, a martingale $M_t$ is called continuous martingale if with probability one the function $t \rightarrow M_t$ is a continuous function.

Definition 4.1 (Brownian Motion). A stochastic process $B_t$ is a Brownian motion with respect to filtration $\left\{\mathcal{F}_t\right\}$ if it has the following three properties:

  • (1) For $s<t$, the distribution of $B_t-B_s$ is normal with mean 0 and variance $t-s$. We denote this by $B_t-B_s \sim N(0, t-s)$
  • (2) If $s<t$, the random variable $B_t-B_s$ is independent of $\mathcal{F}_s$
  • (3) With probability one, the function $f: t \mapsto B_t$ is a continuous function of t.

Then the author presents a theorem and its proof.

Theorem 4.3. A standard Brownian motion $B_t$ is a continuous martingale with respect to filtration $\left\{\mathcal{F}_t\right\}$

Proof. Let $s<t$. Using properties of conditional expectation, we have that $$ \mathbb{E}\left[B_t \mid \mathcal{F}_s\right] = \mathbb{E}\left[B_s \mid \mathcal{F}_s\right]+\mathbb{E}\left[B_t-B_s \mid \mathcal{F}_s\right]=B_s+\mathbb{E}\left[B_t-B_s\right]=B_s . $$ By the third property of Brownian motion, we have proved $B_t$ is a continous martingale.

I think for $\mathbb{E}\left[B_s \mid \mathcal{F}_s\right] = B_s$ to hold, we need $B_s$ is $\mathcal{F}_s$-measurable. However, I could not find this condition in the Def 4.1.

Could you confirm that Def 4.1 lacks the condition that $(B_t)$ is adapted to the filtration $(\mathcal{F}_t)$?


1 Answer 1


Some details are missing from the notes. A filtration $(\mathscr{F}_t)_{t\geq0}$ is called admissible if $\sigma(B_s,s\leq t)=:\mathscr{F}_t^B\subseteq \mathscr{F}_t,\,\forall t\geq 0$ and $B_t-B_s\perp \mathscr{F}_s$ for $0\leq s\leq t$. Then, given a Brownian motion $(B_t)_{t \geq 0}$, we have that $B_t$ is a $\mathscr{F}_t$-martingale if $(\mathscr{F}_t)_{t\geq0}$ is admissible. Indeed: $E[B_t-B_s|\mathscr{F}_s]=E[B_t-B_s]=0$ and $E[B_s|\mathscr{F}_s]=B_s$ because $B_s$ is $\mathscr{F}_s$-measurable by admissibility.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .