Stochastic Bases and Brownian Motion Relative to a Filtration

Why an SDE Needs a Stochastic Basis

An SDE is not specified by its drift and diffusion coefficients alone. It is also a statement about what is known at each time and which random fluctuations count as new noise. The filtration records that information. In particular, a process can have Brownian distributions and continuous paths while failing to be Brownian relative to a filtration that reveals information about its future.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Here $\mathcal{F}$ is a sigma-algebra of events, $\mathbb{P}$ is a probability measure, and the filtration is an increasing family of sub-sigma-algebras: $\mathcal{F}_s\subseteq\mathcal{F}_t\subseteq\mathcal{F}$ whenever $s\le t$. The information in $\mathcal{F}_t$ is the information available by time $t$.

Definition: A stochastic basis is a probability space equipped with a filtration, written $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. We use the usual convention that the filtration is complete and right-continuous: it contains all $\mathbb{P}$-null events and satisfies $\mathcal{F}_t=\bigcap_{u>t}\mathcal{F}_u$. A process is adapted to the filtration if $B_t$ is $\mathcal{F}_t$-measurable for every $t$.

The filtration matters because an increment that is independent of the past in one filtration may no longer be independent of the information in a larger one. The relevant Brownian property is therefore relative to a specified filtration, not just a statement about the distributions of sample paths.

Brownian Motion Relative to a Filtration

Definition: A process $B=(B_t)_{t\ge0}$ is a Brownian motion relative to $(\mathcal{F}_t)$ if: (i) $B_0=0$ almost surely and $B$ has continuous paths almost surely; (ii) $B$ is adapted to $(\mathcal{F}_t)$; and (iii) for every $0\le s<t$, the increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$.

Condition (iii) says more than that an increment is normally distributed. It says that knowing the entire information available at time $s$ gives no information about that increment. This is the nonanticipation property of the noise. The adaptedness condition is separate: it ensures that the value $B_t$ itself is available by time $t$.

Taking $s=0$, every increment $B_t-B_0=B_t$ is independent of $\mathcal{F}_0$. Consequently, data measurable at time zero cannot reveal any nontrivial information about the Brownian path. In particular, if a random initial value $\xi$ is $\mathcal{F}_0$-measurable, then it is independent of each $B_t$; the independent-increment property also gives independence of $\xi$ from every finite collection of Brownian increments after time zero.

Worked Example: Adding Independent Initial Information

Let $B$ be Brownian motion with its natural filtration, and let $\xi$ be independent of the entire Brownian process, with $\mathbb{P}(\xi=1)=\mathbb{P}(\xi=-1)=\tfrac12$. Enlarge the information at each time to $\mathcal{F}_t=\sigma(\xi,B_s:0\le s\le t)$, with the usual completion and right-continuous modification. Check that $B$ remains Brownian relative to this filtration.

The process is adapted because $B_t$ is among the variables generating $\mathcal{F}_t$, and its paths remain continuous. For $s<t$, $B_t-B_s$ is independent of the Brownian history up to $s$, and it is independent of $\xi$ by the assumed independence of $\xi$ and the Brownian process. Thus it is independent of the joint information $\mathcal{F}_s$, and its distribution is $N(0,t-s)$. These are precisely the required properties.

For example, take $X_t=\xi+\tfrac12t+2B_t$. This process satisfies the integral identity with initial value $\xi$, drift $\tfrac12$, and diffusion coefficient $2$. At $t=2$, $X_2=\xi+1+2B_2$. Since $\mathbb{E}[\xi]=0$, $\operatorname{Var}(\xi)=1$, $\mathbb{E}[B_2]=0$, and $\operatorname{Var}(B_2)=2$, independence gives $\mathbb{E}[X_2]=1$ and $\operatorname{Var}(X_2)=1+4\cdot2=9$.

Check: Conditional on $\mathcal{F}_0$, $\xi$ is known and $B_2$ is independent of $\mathcal{F}_0$. Hence $\mathbb{E}[X_2\mid\mathcal{F}_0]=\xi+1$ and $\operatorname{Var}(X_2\mid\mathcal{F}_0)=4\cdot2=8$. The law of total variance gives $8+\operatorname{Var}(\xi)=8+1=9$, agreeing with the unconditional calculation.

Changing the Filtration

A smaller filtration contains less information, so independence of an increment from the original past can be inherited by the smaller past. But the process must still be adapted to the smaller filtration. Without that condition, it cannot be Brownian relative to it.

Proposition: Suppose $B$ is Brownian relative to $(\mathcal{F}_t)$, and $(\mathcal{G}_t)$ is a filtration such that $\mathcal{G}_t\subseteq\mathcal{F}_t$ for every $t$. If $B$ is also adapted to $(\mathcal{G}_t)$, then $B$ is Brownian relative to $(\mathcal{G}_t)$.

Proof. The initial value and path continuity do not depend on the filtration, so $B_0=0$ almost surely and $B$ has continuous paths almost surely. Adaptedness to $(\mathcal{G}_t)$ is an explicit hypothesis. For $s<t$, $B_t-B_s$ has distribution $N(0,t-s)$, since this distribution is part of the Brownian property relative to $(\mathcal{F}_t)$. It is independent of $\mathcal{F}_s$; because $\mathcal{G}_s\subseteq\mathcal{F}_s$, it is also independent of $\mathcal{G}_s$. Thus all parts of the definition hold for $(\mathcal{G}_t)$. $\square$

The adaptedness hypothesis cannot be dropped. For example, let $\mathcal{G}_t=\mathcal{F}_0$ for every $t$, with $(\mathcal{F}_t)$ the natural Brownian filtration. Then $\mathcal{G}_t\subseteq\mathcal{F}_t$, but for $t>0$, $B_t$ is not $\mathcal{F}_0$-measurable. So $B$ is not adapted to $(\mathcal{G}_t)$ and is not Brownian relative to it.

Enlarging a filtration can be more problematic: new information may disclose a future increment. For a concrete illustration, let $T>0$ and, for $0\le t<T$, add $B_T$ to the information available at time $t$. Then $B_T-B_t$ is $\mathcal{F}_t$-measurable. It is nondegenerate, with distribution $N(0,T-t)$, so it cannot be independent of $\mathcal{F}_t$: a nondegenerate random variable measurable with respect to a sigma-algebra is not independent of that sigma-algebra. Thus $B$ does not have the Brownian increment property for this enlarged information structure.

Worked Example: Future Information Breaks the Brownian Property

Let the information at time $t=\tfrac12$ include $B_1$. Compute the conditional expectation of the increment $B_1-B_{1/2}$, and compare it with the value required for a Brownian motion relative to this enlarged filtration.

Because $B_1$ is known at time $\tfrac12$, the increment is also known then. Therefore $\mathbb{E}[B_1-B_{1/2}\mid\mathcal{F}_{1/2}]=B_1-B_{1/2}$. If it were Brownian relative to this filtration, the increment would be independent of $\mathcal{F}_{1/2}$, have mean zero, and so its conditional expectation would be $0$. But $B_1-B_{1/2}$ has distribution $N(0,\tfrac12)$, so it is not zero almost surely. The required conditional expectation identity fails.

Check: The unconditional mean is still $0$, and the unconditional variance is still $\tfrac12$. Those facts do not restore the Brownian property: the failure is specifically dependence on information available at time $\tfrac12$.

How the Filtration and Initial Data Enter an SDE

On a fixed stochastic basis carrying a Brownian motion $B_t$ relative to $(\mathcal{F}_t)$, an SDE specifies a process using that noise and information available along the filtration. In the scalar notation used here, the differential notation $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$ denotes the integral identity below; it is not an ordinary differential equation.

Definition: For a fixed time horizon $T>0$, a candidate process $X$ with initial value $\xi$ satisfies the integral form of the SDE on $[0,T]$ if $X$ is adapted and has continuous paths, $\xi$ is $\mathcal{F}_0$-measurable, the drift integral is finite almost surely, and the stochastic integral is defined, with $$ X_t=\xi+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s,\qquad 0\le t\le T. $$ For the Ito integral here, the integrand $H_s=\sigma(s,X_s)$ must be progressively measurable and belong to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. A sufficient drift condition for the displayed Lebesgue integral is $\int_0^T|\mu(s,X_s)|\,ds<\infty$ almost surely. The identity is required to hold almost surely for all $t$ on the chosen time interval.

Progressive measurability means that, for every $t$, the map $(s,\omega)\mapsto H_s(\omega)$ restricted to $[0,t]\times\Omega$ is measurable with respect to the product of the Borel sets on $[0,t]$ and $\mathcal{F}_t$. This condition ensures the integrand uses no information later than the time of integration. The class $\mathcal{L}^2$ is the class of such progressively measurable processes with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Ito integral is defined for integrands in this class.

The initial value must be specified as part of the formulation. A deterministic initial value is automatically $\mathcal{F}_0$-measurable. A random initial value must be known at time zero, and it may not encode future Brownian information if $B$ is to remain Brownian relative to the filtration. Merely writing a formal differential equation does not establish that an adapted process satisfying its integral identity exists.

Worked Example: A Random Initial Value in an Integral SDE

On the stochastic basis of the first worked example, let $\xi$ take the values $1$ and $-1$ with equal probability, independently of $B$, and define $X_t=\xi+\tfrac12t+2B_t$ for $0\le t\le2$. Verify the integral identity and the required measurability and integrability conditions.

Here $X_0=\xi$, and the coefficient processes are $\mu(s,X_s)=\tfrac12$ and $\sigma(s,X_s)=2$. Both are adapted and progressively measurable because they are constant processes. The drift integral is $\int_0^t\tfrac12\,ds=\tfrac12t$. The constant integrand $2$ belongs to $\mathcal{L}^2$ since $\mathbb{E}\int_0^2 2^2\,ds=4\cdot2=8<\infty$, and its Ito integral is $\int_0^t2\,dB_s=2B_t$. Thus $X_t=\xi+\int_0^t\tfrac12\,ds+\int_0^t2\,dB_s$, as required. Also, $\xi$ is $\mathcal{F}_0$-measurable by construction.

Check: At $t=0$, the identity gives $X_0=\xi$. At $t=2$, it gives $X_2=\xi+1+2B_2$, exactly the expression used in the earlier moment calculation.

Common Pitfalls and Why the Setup Matters

Confusing distribution with independence from the past. The statement $B_t-B_s\sim N(0,t-s)$ alone is not enough to make $B$ Brownian relative to a filtration. In the future-information example, the increment retains that normal distribution, but it is already known at time $s$, violating independence from $\mathcal{F}_s$.

Dropping adaptedness when shrinking a filtration. If $\mathcal{G}_t\subseteq\mathcal{F}_t$, independence from $\mathcal{F}_s$ implies independence from $\mathcal{G}_s$, but it does not imply that $B_t$ is $\mathcal{G}_t$-measurable. The smaller-filtration proposition therefore requires both inclusion and adaptedness.

Placing future data at time zero. Suppose one attempted to set $\mathcal{F}_0=\sigma(B_1)$ while retaining $B_0=0$. Then $B_1-B_0=B_1$ is $\mathcal{F}_0$-measurable and has distribution $N(0,1)$. It cannot be independent of $\mathcal{F}_0$, so $B$ is not Brownian relative to this filtration. A random initial value can be $\mathcal{F}_0$-measurable, but the filtration must still satisfy the Brownian increment condition.

Writing an integral without checking its admissibility. The symbol $\int_0^t\sigma(s,X_s)\,dB_s$ denotes an Ito integral only when the integrand meets the relevant measurability and integrability requirements. Likewise, the drift term needs a meaningful Lebesgue integral. These requirements are part of the SDE formulation, not optional details that can be inferred from the differential notation.

Check Your Understanding

For each question, distinguish properties of the process from properties that depend on the filtration.

State the three defining requirements for $B$ to be Brownian relative to $(\mathcal{F}_t)$.
In the smaller-filtration proposition, why is $\mathcal{G}_t\subseteq\mathcal{F}_t$ not sufficient without adaptedness to $(\mathcal{G}_t)$?
If $\xi$ is $\mathcal{F}_0$-measurable and $B$ is Brownian relative to $(\mathcal{F}_t)$, what independence property does the Brownian definition give for $B_t$ and $\xi$?
For the constant diffusion coefficient $c$ on $[0,T]$, what condition does $c$ satisfy for membership in $\mathcal{L}^2$?
Why does revealing $B_T$ at a time $s<T$ violate the Brownian increment condition at time $s$?