Choosing a Formula Before Calculating
Explicit linear SDE problems often begin with a choice: is the unknown multiplied by the noise, as in geometric Brownian motion, or does the equation have additive noise, as in an Ornstein–Uhlenbeck process? A deterministic forcing term can also require variation of constants. Recognizing the form first helps avoid applying a familiar formula where its hypotheses do not hold. Once a candidate is found, its mean, variance or conditional law provides a useful independent check.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), with the filtration satisfying the usual conditions. Every stochastic integral below is an Itô integral. Its integrand is adapted and progressively measurable and belongs to \(\mathcal{L}^2\) on the finite time interval in question; that is, \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). A strong solution is a process constructed on this space using the given Brownian motion and satisfying the SDE there. The explicit solutions below are strong solutions.
A Variation-of-Constants Result
The integrating factor for the deterministic part of a linear SDE is \(\Phi(t)=\exp(\int_0^t a(u)\,du)\). It is strictly positive and solves \(\Phi'(t)=a(t)\Phi(t)\). The next result makes the choice of formula precise.
Proof. Since \(a\) is continuous on a compact interval, \(\Phi\) and \(\Phi^{-1}\) are bounded there. Since \(g\) is also continuous, \(\int_0^T\Phi(s)^{-2}g(s)^2\,ds<\infty\). The integrand is deterministic, hence progressively measurable, and this finite deterministic integral verifies the \(\mathcal{L}^2\) condition. To construct a solution, define \(Y_t=x_0+\int_0^t\Phi(s)^{-1}f(s)\,ds+\int_0^t\Phi(s)^{-1}g(s)\,dB_s\) and \(X_t=\Phi(t)Y_t\). The product rule has no quadratic covariation term because \(\Phi\) is differentiable with finite variation. Thus \(dX_t=a(t)\Phi(t)Y_t\,dt+\Phi(t)(\Phi(t)^{-1}f(t)\,dt+\Phi(t)^{-1}g(t)\,dB_t)=(a(t)X_t+f(t))\,dt+g(t)\,dB_t\), with \(X_0=x_0\). For uniqueness, if \(X\) and \(\widetilde X\) are two solutions with the same initial value and Brownian motion, their difference \(D=X-\widetilde X\) satisfies \(dD_t=a(t)D_t\,dt\), \(D_0=0\). Multiplying by \(\Phi^{-1}\) gives \(d(\Phi^{-1}D)=0\), so \(D_t=0\) for every \(t\). \(\square\)
The same formula gives the conditional law. For \(0\le s\le t\le T\), split both integrals at \(s\). The part accumulated by time \(s\) is \(X_s\); the remaining stochastic integral depends only on Brownian increments after \(s\).
Proof. The solution formula implies \(X_t=m_{s,t}+\Phi(t)\int_s^t\Phi(u)^{-1}g(u)\,dB_u\). For a deterministic step-function integrand, the integral over \([s,t]\) is a sum of independent centered normal Brownian increments, independent of \(\mathcal{F}_s\). Its variance is the integral of the squared integrand. A deterministic square-integrable integrand is approximated in \(L^2\) by step functions; the Itô isometry gives convergence of their integrals in \(L^2\). Their characteristic functions therefore converge to that of the centered normal distribution with variance \(\int_s^t\Phi(u)^{-2}g(u)^2\,du\). The integral is independent of \(\mathcal{F}_s\), as the step-function integrals are and the limit is measurable with respect to the completed sigma-field generated by the increments after \(s\). Multiplication by \(\Phi(t)\) gives the stated conditional variance and mean. If the variance is zero, the Itô isometry shows the integral is zero almost surely, giving the stated degenerate case. \(\square\)
Worked Examples: Match the Structure to the Formula
Worked Example: Checking a Geometric Brownian Candidate
Let \(S_0=3\) and \(dS_t=\tfrac14S_t\,dt+\tfrac12S_t\,dB_t\). Find the solution and its mean and second moment at \(t=2\), then verify the solution directly.
For geometric Brownian motion the candidate is \(S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)\). Here \(\mu=\tfrac14\), \(\sigma=\tfrac12\), so \(\mu-\tfrac12\sigma^2=\tfrac14-\tfrac18=\tfrac18\), and \(S_t=3\exp(t/8+B_t/2)\). Since \(B_2\sim N(0,2)\), the normal exponential moment \(\mathbb{E}[e^{cB_2}]=e^{c^2(2)/2}\) gives \(\mathbb{E}[S_2]=3e^{2/8}e^{(1/2)^2(2)/2}=3e^{1/4+1/4}=3e^{1/2}\). For the second moment, \(S_2^2=9e^{1/2+B_2}\), so \(\mathbb{E}[S_2^2]=9e^{1/2}e^{2/2}=9e^{3/2}\).
For the verification, use Itô’s formula with \(f(t,x)=3e^{t/8+x/2}\). Its derivatives are \(f_t=\tfrac18f\), \(f_x=\tfrac12f\), and \(f_{xx}=\tfrac14f\). The Itô correction is \(\tfrac12 f_{xx}\,dt=\tfrac18f\,dt\), so \(df(t,B_t)=(\tfrac18f+\tfrac18f)\,dt+\tfrac12f\,dB_t=\tfrac14S_t\,dt+\tfrac12S_t\,dB_t\). Also \(f(0,0)=3\). As a check on the moments, the formula \(\mathbb{E}[S_t]=3e^{\mu t}\) gives \(3e^{1/2}\) at \(t=2\), agreeing with the direct normal calculation.
Worked Example: Ornstein–Uhlenbeck Mean and Variance
Solve \(dX_t=(3-X_t)\,dt+2\,dB_t\), \(X_0=0\), and find the mean and variance at \(t=1\).
Here \(a=-1\), \(f=3\), \(g=2\), and \(\Phi(t)=e^{-t}\). The variation-of-constants formula gives \(X_t=e^{-t}(3\int_0^t e^s\,ds+2\int_0^t e^s\,dB_s)=3(1-e^{-t})+2e^{-t}\int_0^t e^s\,dB_s\). The integral is centered and has variance \(\int_0^t e^{2s}\,ds=(e^{2t}-1)/2\), by the Itô isometry. Consequently, \(\mathbb{E}[X_1]=3(1-e^{-1})\) and \(\operatorname{Var}(X_1)=4e^{-2}(e^2-1)/2=2(1-e^{-2})\).
The conditional law from time \(s\) to \(t\) has mean \(3+(X_s-3)e^{-(t-s)}\) and variance \(4\int_s^t e^{-2(t-u)}\,du=2(1-e^{-2(t-s)})\). Setting \(s=0,t=1,X_0=0\) recovers the mean and variance just computed. In particular, this is a Gaussian transition law, not a lognormal one: the noise enters additively rather than multiplying \(X_t\).
Worked Example: A Vasicek Conditional Distribution
For \(dX_t=2(\tfrac12-X_t)\,dt+\tfrac12\,dB_t\), suppose \(X_1=\tfrac14\). Find the conditional law of \(X_2\) given \(\mathcal{F}_1\).
This is the Vasicek form \(dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t\) with \(\kappa=2\), \(\theta=\tfrac12\), and \(\sigma=\tfrac12\). Its conditional mean over an interval of length \(h\) is \(\theta+(X_s-\theta)e^{-\kappa h}\); its conditional variance is \(\sigma^2(1-e^{-2\kappa h})/(2\kappa)\). With \(s=1,t=2,h=1\), the mean is \(\tfrac12+(\tfrac14-\tfrac12)e^{-2}=\tfrac12-\tfrac14e^{-2}\). The variance is \((1/4)(1-e^{-4})/4=\tfrac1{16}(1-e^{-4})\). Thus \(X_2\mid\mathcal{F}_1\) is normal with those parameters.
Check: At \(h=0\), the general conditional mean reduces to \(X_s\), since \(e^0=1\), and the variance is zero, since \(1-e^0=0\). For positive \(h\), the variance is nonnegative because \(e^{-2\kappa h}\le1\) when \(\kappa>0\). These checks test both the time-zero limit and the sign of the variance.
A Common Error: Losing the Itô Correction
A frequent mistake is to use the ordinary chain rule for an exponential involving Brownian motion. For \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), writing \(S_t=S_0e^{\mu t+\sigma B_t}\) omits the Itô correction. For the parameters in the first example, that incorrect candidate at \(t=2\) would be \(\widetilde S_2=3e^{1/2+B_2/2}\). Its mean is \(3e^{1/2+(1/2)^2(2)/2}=3e^{3/4}\), not the correct \(3e^{1/2}\). This numerical mismatch exposes the error; the correct exponent is \((\mu-\tfrac12\sigma^2)t+\sigma B_t\).
The underlying multiplication rules are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). They explain why the second derivative contributes to the drift in Itô’s formula. The calculations here use Itô integrals throughout; replacing one with a Stratonovich integral would change the conversion correction and would require a separate calculation.
For a mixed problem, identify whether the unknown multiplies the noise, whether the noise is additive, and whether a deterministic forcing term is present. Then use the corresponding formula and check its initial value, drift and diffusion. Finally, compare a moment or conditional law derived from the formula with an independent calculation. These checks are particularly effective for catching sign errors in the integrating factor and missing Itô corrections.
Check Your Understanding
For each question, state any assumptions needed for the conditional law or moment calculation.
- For \(dX_t=(2X_t+1)\,dt+3\,dB_t\), \(X_0=0\), write the variation-of-constants solution and identify its deterministic and stochastic integrals.
- For \(dS_t=\tfrac12 S_t\,dt+S_t\,dB_t\), \(S_0=2\), find the exponential solution and compute \(\mathbb{E}[S_1]\).
- In the Ornstein–Uhlenbeck example, find the conditional mean and variance over an interval of length \(h\), given \(X_s=x\).
- Explain why the conditional variance in the Vasicek example is zero when the interval length is zero.
- For an exponential candidate \(f(t,B_t)\), which second derivative produces the Itô correction, and what term does it add to the drift?