Brownian Filtrations, Adaptedness, and Information

Brownian Motion and Its Natural Filtration

A Brownian path records what has happened up to each time, but a stochastic process is not just a collection of random variables: it is also important to specify what information is available at each time. A filtration formalizes that information. For Brownian motion, the natural filtration is generated by the observed path, and its key feature is that increments after a fixed time are independent of the information generated by the path up to that time.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $(B_t)_{t\ge0}$ be a standard Brownian motion with state space $\mathbb{R}$. Thus $B_0=0$ almost surely, paths are continuous almost surely, and increments over disjoint time intervals are independent; for $0\le s<t$, $B_t-B_s$ has the normal distribution with mean $0$ and variance $t-s$. A filtration $(\mathcal{F}_t)_{t\ge0}$ is an increasing family of sigma fields: $\mathcal{F}_s\subseteq\mathcal{F}_t$ whenever $s\le t$.

Definition: The raw natural filtration of Brownian motion is $\mathcal{F}_t^0=\sigma(B_r:0\le r\le t)$. It contains exactly the events determined by the Brownian path observed through time $t$. Its completion adds all subsets of probability-zero events in the ambient probability space; we denote this completed filtration by $\mathcal{F}_t^B$. Independence statements below hold for either the raw filtration or its completion.

A process $(X_t)_{t\ge0}$ is adapted to $(\mathcal{F}_t)$ if $X_t$ is $\mathcal{F}_t$-measurable for every $t$. In particular, Brownian motion is adapted to its natural filtration: $B_t$ is one of the variables used to generate $\mathcal{F}_t^0$. Adaptedness says that the value at time $t$ can be determined from information available by time $t$; it does not assert continuity, independence of increments, or any other property by itself.

Worked Example: Checking Adaptedness

Let $X_t=B_t+B_{t/2}$ for $t\ge0$. For each fixed $t$, both $B_t$ and $B_{t/2}$ are measurable with respect to $\mathcal{F}_t^0$, because $0\le t/2\le t$. Their sum is therefore $\mathcal{F}_t^0$-measurable, so $(X_t)$ is adapted.

By contrast, the process $Y_t=B_{t+1}$ is not adapted to the Brownian natural filtration in general. For example, $Y_0=B_1$ is not $\mathcal{F}_0^0=\sigma(B_0)$-measurable: $B_0=0$ almost surely, while $B_1$ has the non-degenerate normal distribution with mean $0$ and variance $1$. Thus knowing the path through time $0$ does not determine $Y_0$.

Check: The first process uses only path values at times no later than $t$; the second asks for a path value at time $t+1$, which is in the future.

Past Information and Future Increments

Theorem: Fix $s\ge0$. The future-increment process $(B_{s+u}-B_s)_{u\ge0}$ is independent of $\mathcal{F}_s^B$. In particular, for every $t\ge s$, $B_t-B_s$ is independent of $\mathcal{F}_s^B$ and has distribution $N(0,t-s)$. When $t=s$, this increment is the constant $0$.

Proof. First use the raw filtration. Take any finite collection of past times $0\le r_1,\ldots,r_m\le s$ and future offsets $u_1,\ldots,u_k\ge0$. The vector of past values $(B_{r_1},\ldots,B_{r_m})$ can be written using increments on a finite partition of $[0,s]$. The vector of future values $(B_{s+u_1}-B_s,\ldots,B_{s+u_k}-B_s)$ can be written using increments on a finite partition of $[s,s+\max_j u_j]$. These two collections of intervals have disjoint interiors, so the vectors are independent by the independent-increments property of Brownian motion. This proves independence for events determined by finitely many past values and finitely many future increments. Such cylinder events generate $\mathcal{F}_s^0$ and the sigma field generated by the future-increment process. The pi-lambda theorem extends the independence from the generating cylinder events to those two sigma fields. Finally, completion changes events only by probability-zero sets; replacing an event by another event equal to it almost surely does not change the independence identity. The result therefore holds for $\mathcal{F}_s^B$ as well. $\square$

The theorem concerns increments after a fixed time $s$. It does not say that the future value $B_t$ itself is independent of the past: $B_t=B_s+(B_t-B_s)$ includes the already observed value $B_s$. The distinction is exactly what makes conditional expectations useful. Once the past is given, the remaining uncertainty in $B_t$ comes from an independent normal increment.

Theorem: Let $0\le s\le t$, and let $g:\mathbb{R}\to\mathbb{R}$ be Borel measurable with $\mathbb{E}[|g(B_t)|]<\infty$. For $s<t$, $$ \mathbb{E}[g(B_t)\mid\mathcal{F}_s^B] = \int_{\mathbb{R}}g\bigl(B_s+\sqrt{t-s}\,z\bigr)\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz \quad\text{almost surely}. $$ For $s=t$, the conditional expectation is $g(B_s)$ almost surely.

Proof. Suppose first that $s<t$, and write $Y=B_t-B_s$. The preceding theorem says $Y$ is independent of $\mathcal{F}_s^B$, and $Y$ has the same distribution as $\sqrt{t-s}\,Z$, where $Z$ is standard normal. For any bounded Borel function $h(x,y)$, independence gives $$ \mathbb{E}[h(B_s,Y)\mid\mathcal{F}_s^B] =\int_{\mathbb{R}}h(B_s,y)\,\nu(dy), $$ where $\nu$ is the distribution of $Y$. To justify this identity, first take $h(x,y)=\mathbf{1}_C(x)\mathbf{1}_D(y)$ for Borel sets $C,D$; the conditional expectation identity follows from independence and the fact that $\mathbf{1}_C(B_s)$ is $\mathcal{F}_s^B$-measurable. Finite linear combinations give the identity for simple functions built from such rectangles, and the monotone class theorem extends it to bounded Borel $h$. Apply it to $h(x,y)=g(x+y)$, first for bounded $g$. For integrable $g(B_t)$, apply the bounded case to truncations of the positive and negative parts and pass to the limit using conditional monotone convergence. Substituting the normal density of $Y$ gives the displayed formula. If $s=t$, then $B_t$ is already $\mathcal{F}_s^B$-measurable, so its measurable function $g(B_s)$ is its own conditional expectation. $\square$

Conditional Expectations from the Brownian History

The formula turns a conditional expectation into an ordinary integral. Equivalently, for $s<t$, the conditional distribution of $B_t$ given $\mathcal{F}_s^B$ is normal with mean $B_s$ and variance $t-s$. This is a statement about a conditional distribution given the whole observed history, not merely given one earlier value. It follows here because the increment is independent of that entire history.

Worked Example: Conditional First and Second Moments

Let $0\le s<t$. Using the conditional expectation formula with $g(x)=x$, write $B_t=B_s+(B_t-B_s)$. The independent increment has mean $0$, so

\mathbb{E}[B_t\mid\mathcal{F}_s^B] =B_s+\mathbb{E}[B_t-B_s\mid\mathcal{F}_s^B] =B_s.

For the second moment, expand the square. Conditional on $\mathcal{F}_s^B$, the increment has mean $0$ and second moment $t-s$, so

\begin{aligned} \mathbb{E}[B_t^2\mid\mathcal{F}_s^B] &=\mathbb{E}\bigl[B_s^2+2B_s(B_t-B_s)+(B_t-B_s)^2\mid\mathcal{F}_s^B\bigr]\\ &=B_s^2+2B_s\cdot 0+(t-s)\\ &=B_s^2+(t-s). \end{aligned}

For example, at $s=1$ and $t=2$, these identities give $\mathbb{E}[B_2\mid\mathcal{F}_1^B]=B_1$ and $\mathbb{E}[B_2^2\mid\mathcal{F}_1^B]=B_1^2+1$. Check: the increment $B_2-B_1$ has mean $0$ and variance $2-1=1$, giving exactly the added term $1$.

Worked Example: A Conditional Exponential Expectation

For a real number $\theta$, the normal moment-generating function is $\mathbb{E}[e^{\theta Z}]=e^{\theta^2/2}$ for a standard normal variable $Z$. Since $B_t=B_s+(B_t-B_s)$, independence of the increment from $\mathcal{F}_s^B$ gives

\mathbb{E}[e^{\theta B_t}\mid\mathcal{F}_s^B] =e^{\theta B_s}\mathbb{E}[e^{\theta(B_t-B_s)}] =e^{\theta B_s+\theta^2(t-s)/2}.

Take $\theta=2$, $s=1$, and $t=2$. Substitution gives $\mathbb{E}[e^{2B_2}\mid\mathcal{F}_1^B]=e^{2B_1+2}$, because $\theta^2(t-s)/2=2^2(2-1)/2=2$.

Check: the future increment has variance $2-1=1$, so its exponential expectation is $e^{2^2\cdot1/2}=e^2$; multiplying by the known factor $e^{2B_1}$ gives the same answer.

Worked Example: A Conditional Threshold Probability

For $s<t$ and $a\in\mathbb{R}$, the event $B_t>B_s+a$ is equivalent to $B_t-B_s>a$. Given $\mathcal{F}_s^B$, the increment is normal with variance $t-s$, so

\mathbb{P}(B_t>B_s+a\mid\mathcal{F}_s^B) =1-\Phi\left(\frac{a}{\sqrt{t-s}}\right),

where $\Phi(x)=\mathbb{P}(Z\le x)$ for a standard normal variable $Z$. With $s=1$, $t=2$, and $a=0$, this conditional probability is $1-\Phi(0)=1-\tfrac12=\tfrac12$. Check: the increment $B_2-B_1$ is centered and has a continuous normal distribution, so its probability of being positive is $\tfrac12$.

The threshold here is measured relative to the observed value $B_s$. For a fixed threshold $c$ instead, the same reasoning gives $\mathbb{P}(B_t>c\mid\mathcal{F}_s^B)=1-\Phi((c-B_s)/\sqrt{t-s})$.

A Common Conditioning Error

A frequent mistake is to replace a conditional expectation by an unconditional one. For instance, $B_t$ has unconditional mean $0$, but that does not imply $\mathbb{E}[B_t\mid\mathcal{F}_s^B]=0$. The correct calculation is $B_t=B_s+(B_t-B_s)$: the first term is known at time $s$, and only the second has conditional mean zero.

Wrong calculation: “$B_t$ is centered normal, so $\mathbb{E}[B_t\mid\mathcal{F}_s^B]=0$.” This loses the information in $B_s$. For a concrete check, take $s=1$ and $t=2$. The correct conditional expectation is $B_1$, and $B_1$ is not almost surely zero because it has variance $1$. Averaging the correct answer over the past does give $\mathbb{E}[B_1]=0$, consistent with the unconditional mean. Thus conditional expectation retains the observed past, while unconditional expectation averages over it.

These ideas are the information-level foundation for later uses of Brownian motion as a martingale and Markov process. Adaptedness identifies what is observable at each time; independent increments describe the fresh randomness after a fixed time; conditional expectation expresses how that new randomness combines with the observed state.

Check Your Understanding

Use the natural filtration and the conditional expectation formula to answer the following.

Is $X_t=B_{t/3}$ adapted to the Brownian natural filtration? Explain the measurability check.
For $s<t$, compute $\mathbb{E}[B_t-B_s\mid\mathcal{F}_s^B]$ and $\mathbb{E}[(B_t-B_s)^2\mid\mathcal{F}_s^B]$.
Find $\mathbb{E}[B_3^2\mid\mathcal{F}_2^B]$, and identify the variance contribution from the future increment.
For $s<t$, express $\mathbb{P}(B_t>c\mid\mathcal{F}_s^B)$ for a fixed real number $c$.
Why does independence of $B_t-B_s$ from $\mathcal{F}_s^B$ not imply that $B_t$ is independent of $\mathcal{F}_s^B$?