Why the Present State Can Determine the Future Law
A time-homogeneous stochastic differential equation uses the same drift and diffusion coefficients at every time. If its solution is strong and pathwise unique, then after a deterministic time \(s\), the shifted process solves the same equation, starting from \(X_s\) and driven by the Brownian increments after \(s\). Those increments are independent of the information available at time \(s\). These facts together imply that, conditional on the present state, the future has the law of a fresh solution started from that state.
We work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where \(B_t\) is a Brownian motion relative to the filtration: in particular, \(B_{s+u}-B_s\), \(u\ge0\), is independent of \(\mathcal{F}_s\). Consider the time-homogeneous SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with deterministic initial state \(X_0=x\). The drift \(\mu\) and diffusion coefficient \(\sigma\) are assumed globally Lipschitz and of at most linear growth: for some finite constants \(L,C\), \(|\mu(x)-\mu(y)|+|\sigma(x)-\sigma(y)|\le L|x-y|\) and \(|\mu(x)|+|\sigma(x)|\le C(1+|x|)\) for all \(x,y\).
A strong solution is an adapted continuous process satisfying the integral equation driven by the given Brownian motion. Each stochastic integral below is an Ito integral; its integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval, meaning \(\mathbb{E}\int_0^T H_r^2\,dr<\infty\). Under the stated growth condition, the solution has finite second moments on finite intervals, so \(\sigma(X_r)\) meets this requirement. A random initial value can also be used when it is square-integrable and measurable with respect to the initial sigma-field for the equation being solved.
These are different assertions. Uniqueness in law alone does not identify a shifted solution with a particular solution driven by the shifted Brownian path. The argument here uses strong existence and pathwise uniqueness. Global Lipschitz and linear-growth conditions are convenient sufficient assumptions for these properties; the Markov argument itself needs the properties, not the particular inequalities.
The Conditional-Law Result
Proof. For a fixed \(s\), define the shifted Brownian motion \(\widetilde B_u=B_{s+u}-B_s\) and shifted process \(\widetilde X_u=X_{s+u}\). Subtracting the SDE at time \(s\) from that at time \(s+u\) gives \(\widetilde X_u=X_s+\int_0^u\mu(\widetilde X_v)\,dv+\int_0^u\sigma(\widetilde X_v)\,d\widetilde B_v\). The shifted stochastic integral is defined with the progressively measurable integrand \(\sigma(\widetilde X_v)\); on each finite interval its squared expectation integral is finite by the growth bound and the finite second moments of \(X\). Also, \(\widetilde B\) is Brownian and is independent of \(\mathcal{F}_s\).
Strong existence gives a measurable solution construction from an initial state and a driving Brownian path; under the Lipschitz assumptions it can, for example, be obtained by Picard iteration. Apply that construction to \(X_s\) and \(\widetilde B\). The resulting process and \(\widetilde X\) have the same initial value and driving Brownian motion. Pathwise uniqueness therefore makes them equal almost surely. Thus the shifted future is the solution construction evaluated at \((X_s,\widetilde B)\).
To take its conditional law, first consider a bounded measurable function \(g(x,w)\) of a state \(x\) and a future Brownian path \(w\). Independence of \(\widetilde B\) from \(\mathcal{F}_s\) gives \(\mathbb{E}[g(X_s,\widetilde B)\mid\mathcal{F}_s]=\int g(X_s,w)\,\mathbb{P}_{\widetilde B}(dw)\). For indicator functions of rectangles \(g(x,w)=\mathbf{1}_A(x)\mathbf{1}_D(w)\), this follows directly from independence; finite linear combinations and bounded measurable approximation give the identity for bounded measurable \(g\). Take \(g\) to be \(f\) applied to the solution construction at time \(t\). The integral over Brownian paths is precisely \(Q_t f(X_s)\), proving the formula. The same argument with a bounded measurable functional of the whole solution path proves the path-law statement. \(\square\)
This is the Markov property at deterministic times: once \(X_s\) is given, conditioning on the rest of \(\mathcal{F}_s\) does not change the conditional law of the future. The theorem does not assert the strong Markov property at random stopping times; that is a distinct result.
Worked Examples
Worked Example: Brownian Motion with Drift
Let \(dX_t=a\,dt+\sigma\,dB_t\), with \(X_0=x\), where \(a,\sigma\) are constants. The solution is \(X_t=x+at+\sigma B_t\), since substituting this expression into the integral equation gives \(x+\int_0^t a\,dr+\int_0^t\sigma\,dB_r=x+at+\sigma B_t\). The stochastic integrand is constant, progressively measurable, and has \(\mathbb{E}\int_0^T\sigma^2\,dr=\sigma^2T<\infty\).
At time \(s+t\), \(X_{s+t}=X_s+at+\sigma(B_{s+t}-B_s)\). The increment is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t)\). Therefore, for bounded Borel \(f\), \(\mathbb{E}[f(X_{s+t})\mid\mathcal{F}_s]=\mathbb{E}[f(X_s+at+\sigma Z)]\), where \(Z\sim N(0,t)\). In particular, the conditional mean is \(X_s+at\), and the conditional variance is \(\sigma^2t\).
Check: Directly from the solution, \(X_{s+t}-X_s=at+\sigma(B_{s+t}-B_s)\). Its mean is \(at\), and its variance is \(\sigma^2t\), agreeing with the conditional calculation. When \(t=0\), the increment is zero and the conditional state is exactly \(X_s\).
Worked Example: Geometric Brownian Motion
Consider \(dS_t=\alpha S_t\,dt+\sigma S_t\,dB_t\) with \(S_0=x>0\). Its solution is \(S_t=x\exp((\alpha-\tfrac12\sigma^2)t+\sigma B_t)\). To verify it, set \(F(t,b)=x\exp((\alpha-\tfrac12\sigma^2)t+\sigma b)\). Its derivatives are \(F_t=(\alpha-\tfrac12\sigma^2)F\), \(F_b=\sigma F\), and \(F_{bb}=\sigma^2F\). Ito's formula, including its correction term, gives \(dF(t,B_t)=F_t\,dt+F_b\,dB_t+\tfrac12F_{bb}\,dt=\alpha F\,dt+\sigma F\,dB_t\). The integrand \(\sigma S_t\) is progressively measurable; \(\mathbb{E}[S_t^2]=x^2e^{(2\alpha+\sigma^2)t}\), so \(\mathbb{E}\int_0^T\sigma^2S_t^2\,dt<\infty\) for finite \(T\).
Starting instead from a present value \(S_s=z>0\), the same solution formula driven by future increments gives \(S_{s+t}=z\exp((\alpha-\tfrac12\sigma^2)t+\sigma(B_{s+t}-B_s))\). Thus the conditional law depends on the past through \(z=S_s\) only. For example, using the moment generating function of a centered normal variable of variance \(t\), \(\mathbb{E}[S_{s+t}\mid\mathcal{F}_s]=S_s e^{(\alpha-\tfrac12\sigma^2)t}e^{\sigma^2t/2}=S_s e^{\alpha t}\).
Check: The conditional second moment computed from the same formula is \(S_s^2e^{2\alpha t+\sigma^2t}\). Dividing this by the square of the conditional mean, \(S_s^2e^{2\alpha t}\), gives \(e^{\sigma^2t}\), the lognormal second-moment ratio. The drift in the SDE is \(\alpha S_t\); the term \(-\tfrac12\sigma^2\) in the exponential is the Ito correction, not a change to that drift.
Worked Example: The Ornstein–Uhlenbeck Process
Let \(dX_t=-\kappa X_t\,dt+\eta\,dB_t\), where \(\kappa>0\). To solve from time \(s\), apply Ito's formula to \(F(u,x)=e^{\kappa u}x\). Its derivatives are \(F_u=\kappa e^{\kappa u}x\), \(F_x=e^{\kappa u}\), and \(F_{xx}=0\), so the correction term \(\tfrac12F_{xx}\,du\) is zero. The product calculation yields \(d(e^{\kappa u}X_{s+u})=\eta e^{\kappa u}\,d\widetilde B_u\). Consequently, \(X_{s+t}=e^{-\kappa t}X_s+\eta\int_0^t e^{-\kappa(t-u)}\,d\widetilde B_u\). The integrand is deterministic and progressively measurable, with squared integral \(\eta^2\int_0^t e^{-2\kappa(t-u)}du<\infty\).
The stochastic integral is centered normal with variance given by the Ito isometry: \(\eta^2\int_0^t e^{-2\kappa(t-u)}du=\eta^2(1-e^{-2\kappa t})/(2\kappa)\). Thus the conditional law is normal with mean \(e^{-\kappa t}X_s\) and that variance. Check: at \(t=0\), the mean is \(X_s\) and the variance is \(0\). As \(t\) increases, the coefficient of the initial state decays, while the variance tends to \(\eta^2/(2\kappa)\).
A Common Mistake: Keeping the Initial State
A tempting but incorrect calculation is to use the unconditional solution formula from time zero and treat the whole Brownian value \(B_{s+t}\) as future randomness independent of \(\mathcal{F}_s\). For drifted Brownian motion, this would give \(\mathbb{E}[X_{s+t}\mid\mathcal{F}_s]=x+a(s+t)\), which is generally wrong. The correct decomposition is \(X_{s+t}=X_s+at+\sigma(B_{s+t}-B_s)\), and hence the correct conditional mean is \(X_s+at\). The past is summarized by the current state; it is the increment after \(s\), not \(B_{s+t}\), that is independent of \(\mathcal{F}_s\).
Pathwise uniqueness is what makes this reasoning more than a formal shift of the SDE: it ensures that the shifted solution agrees with the solution constructed from its current state and the shifted Brownian motion. Independent increments then remove the remaining dependence on the past. Both ingredients are essential to this proof.
Check Your Understanding
In each question, take the conditioning sigma-field to be \(\mathcal{F}_s\), with deterministic \(s,t\ge0\).
- For \(dX_t=a\,dt+\sigma\,dB_t\), compute the conditional mean and variance of \(X_{s+t}\).
- Explain why uniqueness in law alone does not identify the shifted solution with the solution driven by \(B_{s+u}-B_s\).
- For the Ornstein–Uhlenbeck process, what happens to the conditional mean and variance when \(t=0\)?
- In the geometric Brownian motion calculation, identify the Ito correction term and explain why it does not alter the drift coefficient in the SDE.
- State which Brownian quantity is independent of \(\mathcal{F}_s\): \(B_{s+t}\) or \(B_{s+t}-B_s\)?