What a Substitution Check Must Establish
A proposed formula is not verified merely because it resembles a familiar solution. For an SDE, the check has three distinct parts: compute the candidate’s Itô differential, match its drift and diffusion coefficients with those in the equation, and verify the initial value. The stochastic integral must also be defined under the stated integrability conditions.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix a finite horizon \(T>0\). All stochastic integrals below are Itô integrals. Their integrands are adapted, progressively measurable processes in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). A strong solution here means a continuous adapted process on this given space that satisfies the integral form of the SDE almost surely for every \(t\le T\). A weak solution, in contrast, may use a probability space and Brownian motion chosen as part of the solution. Pathwise uniqueness means two solutions driven by the same Brownian motion and with the same initial value agree almost surely; uniqueness in law means their distributions agree. Verifying a candidate by substitution does not by itself assert either uniqueness property.
Consider the scalar affine linear equation \(dX_t=(a_tX_t+f_t)\,dt+(c_tX_t+g_t)\,dB_t\), \(X_0=x_0\), where \(a,f,c,g\) are bounded deterministic measurable functions on \([0,T]\). Its drift is \(a_tX_t+f_t\), and its diffusion coefficient is \(c_tX_t+g_t\). To be a strong solution under the stated convention, a candidate must in particular satisfy \(\mathbb{E}\int_0^T(c_tX_t+g_t)^2\,dt<\infty\); bounded coefficients and \(\mathbb{E}\int_0^T X_t^2\,dt<\infty\) suffice. The drift is then integrable in time and expectation as well.
The equality of coefficients is understood almost everywhere, not necessarily at every time: changing a coefficient on a set of times of Lebesgue measure zero does not change either integral. The notation \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) records the quadratic-variation terms in Itô calculus; it is not the ordinary chain rule. This tutorial uses Itô integrals only, not Stratonovich integrals, for which a drift correction is required when converting to Itô form. A square-integrable Itô integral is a true martingale, not merely a local martingale: its conditional expectation satisfies \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). This follows first for elementary adapted integrands and then for all \(\mathcal{L}^2\) integrands by the Itô isometry and \(L^2\) approximation.
The Product Rule Behind Factorized Candidates
Many proposed linear-SDE solutions are products of simpler processes. Suppose \(F\) and \(Y\) are continuous Itô processes, \(dF_t=p_t\,dt+q_t\,dB_t\) and \(dY_t=u_t\,dt+v_t\,dB_t\), with the displayed integrals well defined. Their product has drift \(Fp+Yu+qv\) and diffusion \(Fv+Yq\). The term \(qv\) is the contribution of quadratic covariation and must not be omitted.
Proof. Apply Itô’s formula to the function \(h(x,y)=xy\). Its first derivatives are \(h_x=y\) and \(h_y=x\); its second derivatives are \(h_{xx}=0\), \(h_{yy}=0\), and \(h_{xy}=h_{yx}=1\). The second-order term is therefore \(d[F,Y]_t=q_tv_t\,dt\), since both processes are driven by the same Brownian motion. Substituting the first-order terms gives \(d(FY)=Y\,dF+F\,dY+d[F,Y]\), which is the stated formula. \(\square\)
This formula is an efficient verification tool. If \(F\) has finite variation, then \(q=0\), so no cross term remains. If both factors have diffusion terms, the cross term generally contributes to the drift and has to be included before matching coefficients.
Proof. For sufficiency, boundedness of \(c,g\) and the assumed second-moment integrability give \(\mathbb{E}\int_0^T(c_tX_t+g_t)^2\,dt<\infty\), using \((r+s)^2\le2r^2+2s^2\). Boundedness of \(a,f\) and Cauchy–Schwarz give \(\mathbb{E}\int_0^T|a_tX_t+f_t|\,dt<\infty\). Thus both integrals in the SDE are defined. Integrating the computed differential from \(0\) to \(t\), using the matched coefficients and \(X_0=x_0\), gives the required integral equation.
For necessity, subtract the target equation from the candidate’s integral equation. Write the differences of drift and diffusion as \(\delta b_t\) and \(\delta\sigma_t\). The result is \(\int_0^t\delta b_s\,ds+\int_0^t\delta\sigma_s\,dB_s=0\) for all \(t\le T\), almost surely. The first term has finite variation, so the quadratic variation of the sum is \(\int_0^t\delta\sigma_s^2\,ds\). Since the sum is identically zero, this quadratic variation is zero; hence \(\delta\sigma=0\) almost everywhere in \(dt\,d\mathbb{P}\). The stochastic integral of \(\delta\sigma\) is then zero, leaving \(\int_0^t\delta b_s\,ds=0\) for every \(t\), which implies \(\delta b=0\) almost everywhere. This proves both coefficient equalities. \(\square\)
Worked Substitution Checks
Worked Example: A Homogeneous Geometric Candidate
Verify that \(X_t=2\exp(B_t)\) solves \(dX_t=\tfrac12X_t\,dt+X_t\,dB_t\), with \(X_0=2\).
Use Itô’s formula with \(f(t,z)=2e^z\). The derivatives are \(f_t=0\), \(f_z=2e^z=f\), and \(f_{zz}=2e^z=f\). The formula, including its second-order correction, gives
At \(t=0\), \(B_0=0\), so \(X_0=2e^0=2\). For integrability, the normal moment-generating function gives \(\mathbb{E}[X_t^2]=4\mathbb{E}[e^{2B_t}]=4e^{2t}\), and therefore \(\mathbb{E}\int_0^T X_t^2\,dt=4\int_0^T e^{2t}\,dt=2(e^{2T}-1)<\infty\). Thus the diffusion integrand \(X_t\) is in \(\mathcal{L}^2\), and the drift is integrable.
Check: The solution’s mean is \(\mathbb{E}[X_t]=2e^{t/2}\). The integral equation has mean \(2+\tfrac12\int_0^t\mathbb{E}[X_s]\,ds\), because the square-integrable stochastic integral has expectation zero. Substitution gives \(2+\int_0^t e^{s/2}\,ds=2+2(e^{t/2}-1)=2e^{t/2}\), agreeing with the candidate.
Worked Example: An Affine Mean-Reverting Equation
Check the candidate \(X_t=3+e^{-2t}\left(2+\int_0^t e^{2s}\,dB_s\right)\) for \(dX_t=2(3-X_t)\,dt+dB_t\), \(X_0=5\).
Set \(F_t=e^{-2t}\) and \(Y_t=2+\int_0^t e^{2s}\,dB_s\), so \(X_t=3+F_tY_t\). The deterministic factor satisfies \(dF_t=-2F_t\,dt\); the stochastic integral has integrand \(e^{2t}\), which is deterministic and satisfies \(\int_0^T e^{4s}\,ds=(e^{4T}-1)/4<\infty\). Hence \(dY_t=e^{2t}\,dB_t\). The product rule has no cross term because \(F\) has finite variation, and yields
The initial value is \(X_0=3+1\cdot2=5\). To check integrability, the stochastic integral in \(Y_t\) has mean zero and variance \(\int_0^t e^{4s}\,ds\), by the Itô isometry. Consequently \(\mathbb{E}[X_t^2]=(3+2e^{-2t})^2+e^{-4t}(e^{4t}-1)/4\). For \(0\le t\le T\), the first term is at most \(25\) and the second at most \(1/4\), so \(\mathbb{E}\int_0^T X_t^2\,dt\le(25+1/4)T<\infty\). The diffusion integrand is \(1\), which is in \(\mathcal{L}^2\).
Check: Taking expectations in the candidate gives \(\mathbb{E}[X_t]=3+2e^{-2t}\). Its derivative is \(-4e^{-2t}\), while \(2(3-\mathbb{E}[X_t])=2(3-3-2e^{-2t})=-4e^{-2t}\), and the initial mean is \(5\). This independently checks the drift and initial value at the level of the mean.
Worked Example: Additive Noise and Time-Dependent Forcing
Verify that \(X_t=1+t+e^{-t}\int_0^t e^s\,dB_s\) solves \(dX_t=(-X_t+2+t)\,dt+dB_t\), \(X_0=1\).
For a direct product-rule check, write the same candidate as \(X_t=F_tY_t\), where \(F_t=e^{-t}\) and \(Y_t=1+\int_0^t e^s(2+s)\,ds+\int_0^t e^s\,dB_s\). Indeed, \(\int_0^t e^s(2+s)\,ds=e^t(1+t)-1\), because the derivative of \(e^t(1+t)\) is \(e^t(2+t)\). Thus the product simplifies to the displayed candidate. The integrand \(e^s\) is deterministic and square-integrable on \([0,T]\), and the drift integral is finite.
Now \(dF_t=-e^{-t}dt\) and \(dY_t=e^t(2+t)dt+e^t\,dB_t\). As \(F\) has finite variation, their covariation is zero. The product rule gives
At \(t=0\), the stochastic integral is zero and \(X_0=1\). The stochastic convolution has mean zero and variance \(e^{-2t}\int_0^t e^{2s}\,ds=(1-e^{-2t})/2\), by the Itô isometry. Hence \(\mathbb{E}[X_t^2]=(1+t)^2+(1-e^{-2t})/2\le(1+T)^2+1/2\) for \(0\le t\le T\). It follows that \(\mathbb{E}\int_0^T X_t^2\,dt<\infty\); the diffusion integrand \(1\) is also in \(\mathcal{L}^2\), and the drift \(-X_t+2+t\) is integrable.
Check: The mean is \(1+t\). Its derivative is \(1\), and the expected drift is \(-\mathbb{E}[X_t]+2+t=-(1+t)+2+t=1\). At \(t=0\), the mean is \(1\), as required.
Common Pitfall: Omitting the Itô Correction
A frequent error is to use an ordinary chain rule on an exponential of Brownian motion. For the equation in the first example, the incorrect candidate \( \widetilde X_t=2e^{t/2+B_t}\) may look plausible if one simply places the drift parameter in the exponent. Apply Itô’s formula to \(f(t,z)=2e^{t/2+z}\): \(f_t=\tfrac12f\), \(f_z=f\), and \(f_{zz}=f\). Therefore
This drift is \(\widetilde X_t\), not \(\tfrac12\widetilde X_t\), so the candidate fails the coefficient check despite having the correct initial value. The correction is to use \(X_t=2e^{B_t}\), whose time derivative term is zero and whose Itô correction contributes exactly \(\tfrac12X_t\,dt\). More generally, for constant coefficients in \(dX_t=aX_t\,dt+cX_t\,dB_t\), the exponential candidate uses exponent \((a-\tfrac12c^2)t+cB_t\); the term \(-\tfrac12c^2t\) offsets the Itô correction.
When checking any proposed linear-SDE solution, write its differential before simplifying, keep drift and diffusion terms separate, and then check the initial condition and integrability. A coefficient match without the last two checks is not yet a complete verification.
Check Your Understanding
For each question, interpret coefficient equalities in the almost-everywhere sense used in the theorem.
- For \(dX_t=aX_t\,dt+cX_t\,dB_t\) with constant \(a,c\), what exponent makes \(X_t=x_0\exp(\alpha t+cB_t)\) satisfy the drift coefficient?
- In the product rule for \(F_tY_t\), which term contributes to the drift when both factors have diffusion coefficients?
- Why does \(\mathbb{E}\int_0^T X_t^2\,dt<\infty\), together with bounded \(c_t,g_t\), ensure that \(c_tX_t+g_t\) is an admissible Itô integrand?
- For the affine mean-reverting example, compute \(\mathbb{E}[X_t]\) directly from its candidate formula and verify its initial value.
- What drift does the incorrect candidate \(2e^{t/2+B_t}\) produce, and why does it fail the target SDE?