Solving Linear Semimartingale Equations by Variation of Constants

The Linear Equation and Its Integrating Factor

For an ordinary differential equation, variation of constants divides out a solution of the homogeneous equation, leaving an equation for the remaining factor. The same strategy works for a linear equation driven by a continuous semimartingale, but integration by parts contributes a covariation term. That term matters even when the forcing includes a stochastic integral: omitting it generally gives the wrong solution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Let $X$ be a continuous semimartingale, decomposed as $X=X_0+M+V$, where $M$ is a continuous local martingale and $V$ is a continuous finite-variation process. Its quadratic variation is $[X]=[M]$. All stochastic integrals below are Ito semimartingale integrals. A progressively measurable integrand $H$ can be integrated against $X$ locally when, for example, $\int_0^T |H_s|\,d|V|_s$ and $\int_0^T H_s^2\,d[M]_s$ are finite almost surely after localization. This is the general semimartingale integrability condition used here; for an integral against Brownian motion, the usual square-integrable class is $\mathcal{L}^2$, consisting of progressively measurable $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Fix constants $a,b\in\mathbb{R}$, an $\mathcal{F}_0$-measurable initial value $Y_0=y_0$, and progressively measurable forcing processes $f,g$. Assume $\int_0^T|f_s|\,ds<\infty$ almost surely and that $g$ is integrable against $X$ locally. For the formula below, assume also that the displayed integrands involving the stochastic exponential are locally integrable against $X$. These conditions are satisfied, for example, when the coefficients are locally bounded and the equation is considered up to suitable stopping times. The equation is interpreted in the Ito sense:

dY_t=aY_t\,dt+bY_t\,dX_t+f_t\,dt+g_t\,dX_t.

Here $aY_t+f_t$ is the drift relative to $dt$, while $bY_t+g_t$ is the coefficient integrated against $X$; if $X$ itself has both martingale and finite-variation parts, its integral contributes to both parts of the semimartingale dynamics. The Ito and Stratonovich conventions are not interchangeable: the equation above uses Ito integration, and converting a Stratonovich equation would require its own covariation correction.

Definition: Define the homogeneous integrating factor $Z$ and its inverse $R$ by $$ Z_t=\exp\!\left(at+b(X_t-X_0)-\tfrac12 b^2[X]_t\right), \qquad R_t=Z_t^{-1}. $$ The exponential is strictly positive for every finite $t$, so $R_t$ is well-defined.

The term $-\tfrac12b^2[X]_t$ comes from the Ito correction. In differential notation, the multiplication rules are $dX_t\,dX_t=d[X]_t$, $dX_t\,dt=0$, and $dt\,dt=0$. These rules concern covariation, not ordinary products of finite increments.

Variation of Constants: Formula and Proof

Theorem: Under the assumptions above, a solution is given by $$ Y_t=Z_t\left(y_0+\int_0^t R_s f_s\,ds+\int_0^t R_s g_s\,dX_s-b\int_0^t R_s g_s\,d[X]_s\right). $$ The solution is unique up to indistinguishability among continuous semimartingale solutions for which the equation is defined.

Proof. First apply Ito's formula to $Z_t=\exp(at+b(X_t-X_0)-\tfrac12b^2[X]_t)$. Its exponent has differential $a\,dt+b\,dX_t-\tfrac12b^2d[X]_t$, and its continuous martingale part is $b\,dM_t$, with quadratic variation $b^2d[X]_t$. The exponential Ito correction is therefore $\tfrac12b^2Z_t\,d[X]_t$, which cancels the term $-\tfrac12b^2Z_t\,d[X]_t$ from the exponent. Thus $dZ_t=aZ_t\,dt+bZ_t\,dX_t$. Applying Ito's formula to $R_t=Z_t^{-1}$, or differentiating the reciprocal with its second derivative correction, gives $$ dR_t=R_t\bigl(-a\,dt-b\,dX_t+b^2\,d[X]_t\bigr). $$ For any solution $Y$, the continuous semimartingale product rule gives $d(RY)=R\,dY+Y\,dR+d[R,Y]$. The martingale parts relative to $M$ are $-bR\,dM$ for $R$ and $(bY+g)\,dM$ for $Y$. Hence $d[R,Y]=-bR(bY+g)\,d[X]$. Substituting the equation for $dY$ and the formula for $dR$, the $aRY\,dt$ terms cancel, the $bRY\,dX$ terms cancel, and the terms $b^2RY\,d[X]$ cancel. What remains is $$ d(R_tY_t)=R_tf_t\,dt+R_tg_t\,dX_t-bR_tg_t\,d[X]_t. $$ Integrating from $0$ to $t$, and using $R_0=1$, yields the expression in parentheses in the theorem. Multiplication by $Z_t$ proves the asserted formula and existence. For uniqueness, let $Y^{(1)}$ and $Y^{(2)}$ be two solutions with the same initial value and forcing. Their difference $D=Y^{(1)}-Y^{(2)}$ satisfies the homogeneous equation and $D_0=0$. The same product calculation with $g=f=0$ gives $d(RD)=0$; therefore $R_tD_t=0$ for every $t$, up to indistinguishability. Since $R_t>0$, $D_t=0$ for every $t$, proving uniqueness. $\square$

The covariation term in the formula has a specific origin: it compensates for the covariation between the inverse homogeneous factor and the $g\,dX$ part of the forcing. If $g=0$, that correction vanishes. If $b=0$, the equation is additive in its $dX$ forcing, and again there is no correction.

Worked Solutions

Worked Example: A Homogeneous Brownian-Driven Equation

Let $X_t=B_t$, $a=\tfrac12$, $b=2$, and $Y_0=3$, with $f=g=0$. Since $[B]_t=t$, the integrating factor is $$ Z_t=\exp\!\left(\tfrac12t+2B_t-\tfrac12\cdot4t\right) =\exp(2B_t-\tfrac32t). $$ The formula gives $Y_t=3\exp(2B_t-\tfrac32t)$.

Check by Ito's formula with $h(t,x)=3e^{2x-3t/2}$. The derivatives are $h_t=-\tfrac32h$, $h_x=2h$, and $h_{xx}=4h$. Since $d[B]_t=dt$, Ito's formula gives $$ dh(t,B_t)=\left(h_t+\tfrac12h_{xx}\right)dt+h_x\,dB_t =\left(-\tfrac32h+2h\right)dt+2h\,dB_t =\tfrac12Y_t\,dt+2Y_t\,dB_t. $$ This is the required equation, and $Y_0=3e^0=3$. As an independent check, the Brownian exponential $e^{2B_t-2t}$ has expectation $1$, so $\mathbb{E}[Y_t]=3e^{t/2}$; this also follows directly from the displayed solution and the normal moment generating function.

Worked Example: Forcing Through the Semimartingale Term

Take $X_t=B_t$, $a=0$, $b=1$, $Y_0=0$, $f=0$, and $g_t=1$. The equation is $dY_t=Y_t\,dB_t+dB_t$. Here $Z_t=e^{B_t-t/2}$ and $R_t=e^{-B_t+t/2}$. The formula gives $$ Y_t=Z_t\left(\int_0^t R_s\,dB_s-\int_0^t R_s\,ds\right). $$ To evaluate the bracket, Ito's formula applied to $R_s=e^{-B_s+s/2}$ gives $dR_s=-R_s\,dB_s+R_s\,ds$: its exponent contributes $-R_s\,dB_s+\tfrac12R_s\,ds$, and the second-derivative correction contributes another $\tfrac12R_s\,ds$. Thus $R_s\,dB_s-R_s\,ds=-dR_s$. Integrating, the bracket equals $R_0-R_t=1-R_t$, and consequently $Y_t=Z_t(1-R_t)=Z_t-1$.

Check by substitution: $Y=Z-1$, and $dZ=Z\,dB$ by the homogeneous exponential equation. Therefore $dY=Z\,dB=(Y+1)\,dB=Y\,dB+dB$, as required. Also $Y_0=Z_0-1=0$. This calculation shows concretely why the covariation correction in the general formula cannot simply be dropped.

Worked Example: A Time-Dependent Drift Forcing

Let $X_t=B_t$, $a=0$, $b=1$, and $Y_0=2$. Choose the forcing $f_t=Z_t=e^{B_t-t/2}$ and $g_t=0$. The integrating factor is $Z_t=e^{B_t-t/2}$, so $R_tf_t=1$ at every time. The variation-of-constants formula is therefore $$ Y_t=Z_t\left(2+\int_0^t1\,ds\right)=(2+t)e^{B_t-t/2}. $$ For a direct check, set $h(t,x)=(2+t)e^{x-t/2}$. Its derivatives are $h_t=e^{x-t/2}\bigl(1-\tfrac12(2+t)\bigr)$, $h_x=h$, and $h_{xx}=h$. The Ito drift is $h_t+\tfrac12h_{xx}=e^{x-t/2}$, while the diffusion term is $h\,dB_t$. Hence $dY_t=Z_t\,dt+Y_t\,dB_t=f_t\,dt+Y_t\,dB_t$, exactly the stated equation.

Check the initial value: $Y_0=(2+0)e^0=2$. The independent expectation check uses $\mathbb{E}[e^{B_t-t/2}]=1$, so $\mathbb{E}[Y_t]=2+t$. Taking expectations in the verified differential equation gives the same result because the Brownian integral has mean zero here: its integrand $Y$ is square-integrable on each bounded time interval, as follows from the finite second moment of $(2+t)e^{B_t-t/2}$.

Why the Correction Matters

A common mistake is to treat $R$ as though it were an ordinary differentiable integrating factor and to write $d(RY)=R\,dY+Y\,dR$, omitting $d[R,Y]$. For the second example, this would miss the term $-R\,d[B]$ in $d(RY)$. The incorrect formula would give $Y=Z\int_0^tR_s\,dB_s$, instead of $Y=Z(\int_0^tR_s\,dB_s-\int_0^tR_s\,ds)$. The discrepancy is not a matter of notation: since $d[B]_s=ds$, the omitted correction is an actual time integral. The check $Y=Z\int R\,dB$ by substitution does not satisfy $dY=Y\,dB+dB$.

The method is useful because it separates the multiplicative dynamics from the forcing. The homogeneous coefficient determines $Z$; the remaining terms are integrals against $dt$, $X$, and $[X]$. It also makes the relevant assumptions visible: $X$ must be a continuous semimartingale for the product and Ito calculations used here, and the displayed integrals must exist locally. A local martingale arising in an intermediate stochastic integral need not be a true martingale without additional integrability; no such martingale claim is needed for the pathwise semimartingale identity proved above.

Equation feature	Contribution to the transformed equation
Drift forcing $f_t\,dt$	$R_tf_t\,dt$
Semimartingale forcing $g_t\,dX_t$	$R_tg_t\,dX_t$
Covariation of the inverse factor with forcing	$-bR_tg_t\,d[X]_t$

Check Your Understanding

Use the formula and the product rule, paying particular attention to the covariation term.

For $dY_t=3Y_t\,dt+2Y_t\,dB_t$, $Y_0=1$, compute the explicit homogeneous solution and verify its Ito drift.
In the general formula, explain why the correction disappears when $b=0$, even if $g$ is not zero.
For $X=B$, $a=0$, $b=1$, $f=0$, and $g=1$, derive $dR_t=-R_t\,dB_t+R_t\,dt$ and use it to evaluate the transformed forcing.
State the local integrability requirements for $\int Rg\,dX$ when $X=X_0+M+V$.
Explain which product-rule term is lost in the incorrect ordinary integrating-factor calculation, and identify its value when $X=B$.

Equation feature	Contribution to the transformed equation
Drift forcing \(f_t\,dt\)	\(R_tf_t\,dt\)
Semimartingale forcing \(g_t\,dX_t\)	\(R_tg_t\,dX_t\)
Covariation of the inverse factor with forcing	\(-bR_tg_t\,d[X]_t\)

Solving Linear Semimartingale Equations by Variation of Constants

What You'll Learn