Strong Solutions Use Only Available Information
In a stochastic differential equation, the Brownian motion supplies new randomness over time. A solution at time \(t\) may depend on the noise observed up to \(t\), but not on a future increment that has not yet occurred. This restriction is expressed by adaptedness to the driving filtration. It is not an optional regularity condition: the Ito integral in the equation is constructed from integrands that are adapted to that filtration.
Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). In particular, \(B_t\) is \(\mathcal{F}_t\)-measurable, and \(B_t-B_s\) is independent of \(\mathcal{F}_s\) for \(s<t\). We take the filtration to be complete and right-continuous. A process is adapted if \(X_t\) is \(\mathcal{F}_t\)-measurable for each \(t\). A process is progressively measurable if, for each \(t\), its restriction to \([0,t]\times\Omega\) is measurable with respect to \(\mathcal{B}([0,t])\otimes\mathcal{F}_t\).
Adaptedness is relative to the filtration, not necessarily to the natural filtration generated by \(B\) alone. If the prescribed filtration contains additional information, an adapted process may use information already available at time zero as well as the Brownian history. The essential restriction is that \(X_t\) cannot depend on information outside \(\mathcal{F}_t\).
Why the Ito Integral Respects the Filtration
The Ito integral is first defined for adapted step integrands and then extended by limits. For a step process \(H_s=\sum_{j=0}^{n-1}H_j\mathbf{1}_{(t_j,t_{j+1}]}(s)\), where \(H_j\) is \(\mathcal{F}_{t_j}\)-measurable and square-integrable, its integral up to \(t\) is a sum of terms \(H_j(B_{t\wedge t_{j+1}}-B_{t\wedge t_j})\), with the intervals truncated at \(t\). Each term uses only information available by time \(t\). The Ito integral for a general progressively measurable \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), is the \(L^2(\mathbb{P})\) limit of such integrals.
Proof. For an adapted step integrand, the integral up to \(t\) is \(\mathcal{F}_t\)-measurable. Its increment from \(s\) to \(t\) has conditional expectation zero given \(\mathcal{F}_s\): each term is a coefficient measurable before its Brownian increment, and that increment has conditional mean zero given the earlier filtration. The Ito isometry gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_u^2\,du<\infty\). For general \(H\), take adapted step integrands converging to \(H\) in the norm \((\mathbb{E}\int_0^T H_u^2\,du)^{1/2}\). The Ito isometry makes their integrals converge in \(L^2\); since \(L^2(\mathcal{F}_t)\) is closed in \(L^2(\mathbb{P})\), the integral at each fixed \(t\) is \(\mathcal{F}_t\)-measurable. Passing to the limit in the conditional-expectation identity is valid because conditional expectation is a contraction in \(L^2\). The standard continuous version of the Ito integral is therefore adapted and satisfies the martingale identity. Its finite second moments verify integrability, so this is a true martingale, not merely a local martingale. \(\square\)
This proposition explains the role of the filtration in the SDE definition. The integral accumulated by time \(t\) is adapted because the integrand and Brownian increments used to build it are nonanticipating. It does not say that an arbitrary process satisfying an algebraic-looking integral expression is a valid solution: the integrand must meet the measurability and integrability requirements.
Worked Example: The Brownian Motion Itself Is a Strong Solution
On the stated basis, consider \(dX_t=dB_t\) with \(X_0=0\). The integral equation gives \(X_t=0+\int_0^t1\,dB_s=B_t\). Brownian motion is continuous and adapted, and the integrand \(H_s=1\) is progressively measurable with \(\mathbb{E}\int_0^T1^2\,ds=T<\infty\). Thus \(X_t=B_t\) is a strong solution.
Check: substituting \(X_t=B_t\) into the integral equation returns \(B_t=\int_0^t1\,dB_s\), as required. Its dependence is causal: at time \(t\), its value is the noise observed at that time, not a future increment.
Present and Past Noise: A Nonlinear Example
Worked Example: A Strong Solution Driven by the Current State
Consider \(dX_t=X_t\,dB_t\), \(X_0=1\), and propose \(X_t=\exp(B_t-t/2)\). To check the equation, apply Ito's formula to \(f(t,x)=\exp(x-t/2)\). Its derivatives are \(f_t=-\tfrac12 f\), \(f_x=f\), and \(f_{xx}=f\). Since \(dB_t\,dB_t=dt\), Ito's formula gives
Also \(X_0=\exp(0)=1\). The process is continuous and adapted because \(X_t\) is a function of \(B_t\) and the deterministic time \(t\). The integrand is in \(\mathcal{L}^2\) on every finite horizon: \(X_s^2=\exp(2B_s-s)\), and the normal moment-generating formula gives \(\mathbb{E}[X_s^2]=\exp(s)\), so \(\mathbb{E}\int_0^T X_s^2\,ds=\int_0^T e^s\,ds=e^T-1<\infty\).
Independent check: because \(B_t\sim N(0,t)\), \(\mathbb{E}[X_t]=\mathbb{E}[\exp(B_t-t/2)]=\exp(t/2-t/2)=1\). More fully, for \(s\leq t\), independence of \(B_t-B_s\) from \(\mathcal{F}_s\) gives \(\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s\mathbb{E}[\exp(B_t-B_s-(t-s)/2)]=X_s\). Thus \(X\) is a true martingale; the conditional expectation calculation and finite expectation verify the claim.
Adaptedness permits dependence on the entire available history, not just the value \(B_t\). For example, \(Y_t=\int_0^t B_s\,ds\) is \(\mathcal{F}_t\)-measurable: its Riemann sums use only \(B_s\) for \(s\leq t\), and their limit is measurable with respect to \(\mathcal{F}_t\). The path integral exists because Brownian paths are continuous. Such a process records accumulated past noise. This observation concerns adaptedness; it does not assert that every adapted process is a solution of a particular SDE.
Future Noise Is Not an Admissible Input
Worked Example: Detecting Anticipation in \(B_tB_T\)
Fix \(0<t<T\). The product \(Z=B_tB_T\) is not \(\mathcal{F}_t\)-measurable. Suppose instead that it were. Since \(B_t\) is \(\mathcal{F}_t\)-measurable and \(\mathbb{P}(B_t=0)=0\), the random variable equal to \(Z/B_t\) on \(\{B_t\neq0\}\) and to \(0\) on \(\{B_t=0\}\) would be \(\mathcal{F}_t\)-measurable. It equals \(B_T\) almost surely. Completeness of the filtration would then make \(B_T\) \(\mathcal{F}_t\)-measurable.
But \(B_T=B_t+(B_T-B_t)\), where the future increment \(B_T-B_t\) is independent of \(\mathcal{F}_t\) and has variance \(T-t>0\). Consequently \(\operatorname{Var}(B_T\mid\mathcal{F}_t)=T-t\), whereas an \(\mathcal{F}_t\)-measurable square-integrable variable has conditional variance zero. This contradiction proves the claim.
The endpoint cases matter. At \(t=0\), \(B_0B_T=0\), which is \(\mathcal{F}_0\)-measurable. At \(t=T\), the product is \(B_T^2\), which is \(\mathcal{F}_T\)-measurable. The nonmeasurability statement is specifically for \(0<t<T\).
The same issue prevents \(H_s=B_T\) from serving as an Ito integrand on \([0,T]\). Although it satisfies the numerical integrability test \(\mathbb{E}\int_0^T B_T^2\,ds=T\mathbb{E}[B_T^2]=T^2<\infty\), it is not adapted for \(0\leq s<T\), so it is not in the required class of progressively measurable \(\mathcal{L}^2\) integrands. The tempting expression \(\int_0^t B_T\,dB_s=B_TB_t\) is therefore not an Ito integral under this filtration. At \(t=0\), the product is zero, and at \(t=T\), the endpoint product is measurable; neither fact repairs the failure of admissibility before \(T\).
Common Pitfalls and the Meaning of Strong
Confusing square-integrability with admissibility. A finite value of \(\mathbb{E}\int_0^T H_s^2\,ds\) is not enough. The integrand must also be progressively measurable. The future-valued process \(H_s=B_T\) passes the numerical integrability check but fails the filtration condition.
Checking only a terminal value. A process may be measurable at \(T\) and fail to be adapted at every earlier time. Adaptedness is a time-by-time condition: \(X_t\) must be \(\mathcal{F}_t\)-measurable for each \(t\), not merely \(\mathcal{F}_T\)-measurable.
Interpreting nonanticipation too narrowly. Adaptedness means dependence only on information in \(\mathcal{F}_t\). It does not by itself mean that \(X_t\) is a deterministic function of the Brownian path up to \(t\), since the prescribed filtration can contain additional information. A strong solution is tied to the specified basis and its Brownian motion; a weak formulation allows the basis itself to vary.
Check Your Understanding
For each question, identify the relevant time-indexed measurability or integrability condition.
- Why does \(X_t=B_t\) satisfy the adaptedness requirement for \(dX_t=dB_t\), \(X_0=0\)?
- For \(X_t=\exp(B_t-t/2)\), compute \(\mathbb{E}\int_0^T X_s^2\,ds\) and explain why this verifies the Ito integrability condition.
- For \(0<t<T\), why would \(\mathcal{F}_t\)-measurability of \(B_tB_T\) imply \(\mathcal{F}_t\)-measurability of \(B_T\)?
- Why does \(H_s=B_T\) fail to be an admissible Ito integrand even though \(\mathbb{E}\int_0^T H_s^2\,ds\) is finite?
- What are the values of \(B_0B_T\) at \(t=0\) and \(B_TB_T\) at \(t=T\), and why do these endpoints not contradict the result for \(0<t<T\)?