Why the Initial Value Is Part of the SDE
An SDE is not specified by its drift and diffusion coefficients alone when the initial value is random. The initial value has a probability law, must be known at time zero, and must not reveal future Brownian increments. These requirements matter in both strong and weak formulations, even though the two formulations differ in who supplies the probability space and initial random variable.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). A Brownian motion \(B_t\) relative to this filtration has continuous paths, starts at zero, and, for \(0\le s<t\), the increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). In particular, future increments are independent of all information available at time \(s\), not merely of the value \(B_s\).
For precision, consider the \(n\)-dimensional integral equation driven by a \(d\)-dimensional Brownian motion: \(X_t=X_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,d\mathbf{B}_s\). Here \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. The drift integral is a Lebesgue integral and is required to be finite almost surely on the time interval considered. The stochastic integral is the Ito integral: its integrand must be progressively measurable, and in the square-integrable formulation it belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T\|\sigma(s,X_s)\|^2\,ds<\infty\). The process \(X\) is continuous and adapted, and the equation is required to hold almost surely for all times (or, equivalently for continuous versions, simultaneously on the interval).
Strong and Weak Formulations
In either formulation, being \(\mathcal{F}_0\)-measurable is not by itself enough: Brownian motion must also be Brownian relative to the filtration that contains \(X_0\). This ensures independence between the initial data and the future noise. Conversely, merely saying that \(X_0\) and the Brownian path are independent does not make \(B\) Brownian relative to an arbitrary larger filtration. The filtration must not contain additional information that predicts future increments.
Proof. Fix times \(0=t_0<t_1<\cdots<t_k\). The first increment \(B_{t_1}-B_{t_0}\) is independent of \(\mathcal{F}_0\). For \(j\ge2\), the increment \(B_{t_j}-B_{t_{j-1}}\) is independent of \(\mathcal{F}_{t_{j-1}}\), which contains \(\mathcal{F}_0\) and all the preceding increments. Repeatedly conditioning on the preceding increments therefore shows that the vector of these increments is independent of \(\mathcal{F}_0\). Since \(X_0\) is \(\mathcal{F}_0\)-measurable, it is independent of every such finite vector. Brownian paths are continuous, so their values at nonnegative rational times determine the path; independence for these finite-dimensional distributions gives independence of \(X_0\) and the Brownian path. \(\square\)
This proposition explains a key constraint, not a construction theorem: on a prescribed strong basis the initial variable must already be available at time zero, while in a weak formulation one can choose a space that supports an initial variable independent of the driving Brownian motion.
Worked Examples
Worked Example: Solving with a Nontrivial Initial Law
Let \(X_0\) equal \(2\) with probability \(\tfrac13\) and \(-1\) with probability \(\tfrac23\), independently of \(B\). On \(0\le t\le1\), solve \(dX_t=3\,dt+2\,dB_t\), and calculate the mean and variance of \(X_1\).
Integrating the constant drift and diffusion gives \(X_t=X_0+3t+2B_t\). The stochastic integral is well-defined in \(\mathcal{L}^2\): its integrand is the constant \(2\), and \(\mathbb{E}\int_0^1 2^2\,dt=4\). For the initial variable,
Since \(\mathbb{E}[B_1]=0\), \(\mathbb{E}[X_1]=0+3+2\cdot0=3\). Independence of \(X_0\) and \(B_1\) gives zero covariance, so \(\operatorname{Var}(X_1)=\operatorname{Var}(X_0)+4\operatorname{Var}(B_1)=2+4\cdot1=6\).
Check: Directly, \(\mathbb{E}[X_1^2]=\mathbb{E}[(X_0+3+2B_1)^2]=2+9+4=15\): the cross terms vanish because \(\mathbb{E}[X_0]=\mathbb{E}[B_1]=0\) and independence gives \(\mathbb{E}[X_0B_1]=0\). Thus \(\mathbb{E}[X_1^2]-(\mathbb{E}[X_1])^2=15-9=6\), agreeing with the variance calculation.
Worked Example: An Anticipative Initial Value
Let \(B\) be Brownian motion on \([0,1]\), and try to use \(X_0=\mathbf{1}_{\{B_1>0\}}\) as an initial value. Show that the filtration containing \(X_0\) at time zero cannot make \(B\) Brownian relative to that filtration.
Take \(\mathcal{F}_0\) to contain \(\sigma(X_0)\). If \(B\) were Brownian relative to this filtration, then \(B_1-B_0=B_1\) would be independent of \(\mathcal{F}_0\), and hence independent of \(X_0\). Since \(B_1\sim N(0,1)\), symmetry gives \(\mathbb{P}(B_1>0)=\tfrac12\). Independence would require \(\mathbb{P}(B_1>0\mid\mathcal{F}_0)=\tfrac12\) almost surely. But the definition of \(X_0\) gives \(\mathbb{P}(B_1>0\mid\mathcal{F}_0)=\mathbf{1}_{\{B_1>0\}}=X_0\), which is either \(0\) or \(1\), each with probability \(\tfrac12\). It therefore cannot equal \(\tfrac12\) almost surely.
Check: The event \(\{B_1>0\}\) has probability \(\tfrac12>0\), so the conditional probability calculation is meaningful. The failure is not a technicality about the SDE coefficients: the proposed initial information reveals the sign of a future Brownian increment, violating the definition of Brownian motion relative to the enlarged filtration.
Worked Example: Choosing a Space in a Weak Formulation
Consider \(dX_t=0\) with initial law assigning probability \(\tfrac12\) to each of \(-1\) and \(1\). Construct a weak solution, and explain why a prescribed basis with trivial \(\mathcal{F}_0\) cannot carry the required initial value.
For a weak solution, take a product probability space carrying a sign variable \(U\) with \(\mathbb{P}(U=1)=\mathbb{P}(U=-1)=\tfrac12\) and an independent standard Brownian motion \(B\). Set \(X_0=U\), and let \(\mathcal{F}_t\) be the usual augmentation of the filtration generated by \(U\) and \((B_s)_{0\le s\le t}\). Independence ensures that future Brownian increments are independent of this past, so \(B\) is Brownian relative to \((\mathcal{F}_t)\). Set \(X_t=U\) for all \(t\). Then \(X\) is continuous and adapted, and the integral equation is \(X_t=X_0\), because both coefficient terms are zero.
At every \(t\), \(\mathbb{P}(X_t=1)=\tfrac12\), \(\mathbb{P}(X_t=-1)=\tfrac12\), \(\mathbb{E}[X_t]=\tfrac12-\tfrac12=0\), and \(\operatorname{Var}(X_t)=\mathbb{E}[X_t^2]=1\). On a prescribed basis with trivial \(\mathcal{F}_0\), every \(\mathcal{F}_0\)-measurable real random variable is almost surely constant. Such a basis cannot supply an \(X_0\) with this two-point law, even though the weak construction can choose a space that does.
Check: The constructed solution preserves its initial distribution because its equation is \(X_t=X_0\). Replacing the initial law by the point mass at \(0\) would instead give \(X_t=0\) and variance \(0\). Thus changing the initial law changes the solution law, even when both drift and diffusion vanish.
What the Formulations Do—and Do Not—Say
| Feature | Strong formulation | Weak formulation |
|---|---|---|
| Probability space and filtration | Prescribed before solving | May be chosen as part of the solution |
| Initial variable | Prescribed and \(\mathcal{F}_0\)-measurable | Constructed on the chosen space, with the required law |
| Brownian requirement | Brownian relative to the prescribed filtration | Brownian relative to the chosen filtration |
| Independence constraint | Initial value independent of the Brownian path | Same constraint, achieved on the chosen space |
The labels describe how the stochastic basis and initial data are supplied. They do not, by themselves, assert that a solution exists, that it is unique, or that solutions on different spaces can be identified path by path. In particular, a weak solution does not permit dependence between the initial value and future noise: flexibility to choose a space does not remove the Brownian-motion condition.
Common mistake. One might enlarge the time-zero information by setting \(X_0=\mathbf{1}_{\{B_1>0\}}\) and then assert that the Brownian motion remains Brownian because its unconditional law is unchanged. The unconditional probability \(\mathbb{P}(B_1>0)=\tfrac12\) is indeed unchanged, but the required conditional probability is not: it is \(\mathbb{P}(B_1>0\mid\mathcal{F}_0)=X_0\), rather than \(\tfrac12\). Brownian motion relative to a filtration requires independence from its past information, so an unchanged marginal distribution does not repair this failure.
Check Your Understanding
For each question, distinguish the law of the initial value from the information available in the filtration.
- Why does \(X_0\in\mathcal{F}_0\) and Brownian motion relative to \((\mathcal{F}_t)\) imply independence of \(X_0\) and future Brownian increments?
- If \(X_0\) is independent of the Brownian path, what additional requirement must hold for \(B\) to be Brownian relative to a chosen filtration?
- For \(dX_t=2\,dt+dB_t\), write the integral solution for a general \(\mathcal{F}_0\)-measurable \(X_0\), and identify the condition that makes the stochastic integral square-integrable on \([0,T]\).
- Why can a weak formulation accommodate a non-degenerate initial law on a space where a prescribed strong formulation with trivial \(\mathcal{F}_0\) cannot?
- For the anticipative choice \(X_0=\mathbf{1}_{\{B_1>0\}}\), compute \(\mathbb{P}(B_1>0\mid\mathcal{F}_0)\) and compare it with the value required if \(B\) were Brownian relative to \((\mathcal{F}_t)\).