Fundamental Solutions Of Homogeneous Linear SDEs

The Fundamental Solution and Its Correction Term

A homogeneous linear stochastic differential equation has no forcing term: its drift and diffusion are both proportional to the current state. The proportionality makes an exponential candidate natural, but the ordinary exponential rule is not enough. Brownian motion has quadratic variation $[B]_t=t$, so Itô's formula contributes a second-derivative term. That correction changes the drift inside the exponent and is essential to obtaining a solution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to the filtration. Fix a finite horizon $T>0$. Assume that $a_t$ and $c_t$ are progressively measurable and bounded on $[0,T]$: for deterministic constants $A,C<\infty$, $|a_t|\le A$ and $|c_t|\le C$ almost everywhere. Every Itô integral below is taken with a progressively measurable integrand $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Integrals introduced after stopping are also in this class on the finite horizon.

Definition: The normalized fundamental solution for $a_t,c_t$ is a continuous adapted process $\Phi_t$ with $\Phi_0=1$ satisfying $$ d\Phi_t=a_t\Phi_t\,dt+c_t\Phi_t\,dB_t. $$ Here $a_t$ is the drift coefficient and $c_t$ is the diffusion coefficient. The equation is homogeneous because both coefficients multiply $\Phi_t$.

The formula to be derived is $$ \Phi_t=\exp\left(\int_0^t\left(a_s-\tfrac12c_s^2\right)\,ds+\int_0^t c_s\,dB_s\right). $$ The stochastic integral is well-defined because $c$ is bounded, so $\mathbb{E}\int_0^T c_s^2\,ds\le C^2T$. The $-\tfrac12c_s^2$ term is not an optional adjustment: it compensates for the Itô correction $dB_t\,dB_t=dt$. In the Itô multiplication rules, $dB_t\,dt=0$ and $dt\,dt=0$.

Derivation and Verification

Set $Z_t=\int_0^t(a_s-\tfrac12c_s^2)\,ds+\int_0^t c_s\,dB_s$, and define $\Phi_t=e^{Z_t}$. Since $a,c$ are bounded, the time integral is finite and $Z$ is a continuous adapted process. Itô's formula applies to $f(z)=e^z$. Its derivatives are $f'(z)=e^z$ and $f''(z)=e^z$, and $d[Z]_t=c_t^2\,dt$, because the finite-variation time integral has zero quadratic variation and zero covariation with the Brownian integral. Thus the correction term is $\tfrac12 f''(Z_t)c_t^2\,dt$:

\begin{aligned} d\Phi_t &=f'(Z_t)\,dZ_t+\tfrac12 f''(Z_t)\,d[Z]_t\\ &=\Phi_t\left(a_t-\tfrac12c_t^2\right)dt+\Phi_tc_t\,dB_t +\tfrac12\Phi_tc_t^2\,dt\\ &=a_t\Phi_t\,dt+c_t\Phi_t\,dB_t. \end{aligned}

Also, $Z_0=0$, so $\Phi_0=e^0=1$. This calculation establishes the formula formally; the next result verifies that its stochastic integral in the SDE belongs to the stated square-integrable class and establishes uniqueness.

Theorem: Under the boundedness and progressive-measurability assumptions above, the displayed stochastic exponential is a solution on $[0,T]$, and it is pathwise unique among continuous adapted solutions for which the SDE's drift and Itô integrals are defined locally. In particular, the fundamental solution is positive at every time on every path where its defining integrals are finite.

Proof. Let $X_t$ denote the exponential candidate. It is continuous and adapted, and the Itô calculation gives its SDE up to localization. To check square integrability, let $\tau_n=\inf\{t\ge0:|X_t|\ge n\}$, with the infimum of the empty set defined as $+\infty$. Itô's formula for $x\mapsto x^2$, whose derivatives are $2x$ and $2$, gives $$ d(X_t^2)=(2a_t+c_t^2)X_t^2\,dt+2c_tX_t^2\,dB_t. $$ Stopped at $\tau_n$, the stochastic integrand is bounded, hence square-integrable, and its Itô integral has mean zero. With $K_0=2A+C^2$, taking expectations and applying Gronwall's inequality yields $\mathbb{E}[X_{t\wedge\tau_n}^2]\le e^{K_0t}$ for $0\le t\le T$. Since each continuous path is bounded on $[0,T]$, $\tau_n$ eventually exceeds $T$ on that path. Fatou's lemma therefore gives $\mathbb{E}[X_t^2]\le e^{K_0t}$, and integration gives $\mathbb{E}\int_0^T X_s^2\,ds\le\int_0^T e^{K_0s}\,ds<\infty$. Consequently $cX\in\mathcal{L}^2$, and $aX$ is integrable in time and expectation. Removing the stopping in the localized SDE is justified by convergence of the stopped stochastic integrals in $L^2$ and of the drift integrals in $L^1$. Thus $X$ solves the SDE on $[0,T]$.

For uniqueness, let $X$ and $Y$ be two solutions with the same initial value, and put $D=X-Y$. Then $D_0=0$ and $dD_t=a_tD_t\,dt+c_tD_t\,dB_t$. Stop when $|D|$ first reaches $n$. Itô's formula for $D^2$, followed by expectation of the stopped equation, gives $$ \mathbb{E}[D_{t\wedge\tau_n}^2] =\mathbb{E}\int_0^t\mathbf{1}_{\{s\le\tau_n\}}(2a_s+c_s^2)D_s^2\,ds \le K_0\int_0^t\mathbb{E}[D_{s\wedge\tau_n}^2]\,ds. $$ The stopped stochastic integral has mean zero because its integrand is bounded on the finite horizon. Gronwall's inequality, with initial value zero, implies $\mathbb{E}[D_{t\wedge\tau_n}^2]=0$. Hence $D_{t\wedge\tau_n}=0$ almost surely for every $t$; continuity and increasing $n$ show $D_t=0$ for all $t\in[0,T]$ almost surely. Finally, an exponential of a finite real number is strictly positive, proving positivity of $X$. $\square$

Worked Forms of the Fundamental Solution

Worked Example: Constant Drift and Diffusion

Take $a_t=\tfrac14$ and $c_t=\tfrac12$ for all $t\ge0$. Find $\Phi_t$, evaluate its deterministic exponent at $t=2$, and check its mean at that time.

The correction in the exponent is $\tfrac12c^2=\tfrac12\cdot\tfrac14=\tfrac18$, so $a-\tfrac12c^2=\tfrac14-\tfrac18=\tfrac18$. The formula gives $$ \Phi_t=\exp\left(\tfrac18t+\tfrac12B_t\right). $$ At $t=2$, the deterministic part of the exponent is $\tfrac18\cdot2=\tfrac14$, so $\Phi_2=\exp(\tfrac14+\tfrac12B_2)$.

Check by substitution: Put $Z_t=\tfrac18t+\tfrac12B_t$. Then $dZ_t=\tfrac18dt+\tfrac12dB_t$ and $d[Z]_t=\tfrac14dt$. For $f(z)=e^z$, $f'(z)=f''(z)=e^z$, and the correction is $\tfrac12e^{Z_t}\cdot\tfrac14dt=\tfrac18\Phi_tdt$. Therefore $d\Phi_t=(\tfrac18+\tfrac18)\Phi_tdt+\tfrac12\Phi_tdB_t=\tfrac14\Phi_tdt+\tfrac12\Phi_tdB_t$, as required.

For an independent mean check, $B_2$ has distribution $N(0,2)$, whose moment-generating function is $\mathbb{E}[e^{uB_2}]=e^{u^2}$. Hence $\mathbb{E}[\Phi_2]=e^{1/4}\mathbb{E}[e^{B_2/2}]=e^{1/4}e^{1/4}=e^{1/2}$. This agrees with $e^{\int_0^2a_s\,ds}=e^{(1/4)2}$.

Worked Example: Deterministic Time-Varying Coefficients

Let $a_t=t$ and $c_t=1+t$ on $[0,1]$. Compute $\Phi_1$ and its expectation.

For $0\le t\le1$, $$ \int_0^t a_s\,ds=\frac{t^2}{2},\qquad \frac12\int_0^t c_s^2\,ds =\frac12\int_0^t(1+2s+s^2)\,ds =\frac{t}{2}+\frac{t^2}{2}+\frac{t^3}{6}. $$ The deterministic part of the exponent is therefore $-\tfrac{t}{2}-\tfrac{t^3}{6}$, and $$ \Phi_t=\exp\left(-\frac{t}{2}-\frac{t^3}{6}+\int_0^t(1+s)\,dB_s\right). $$ At $t=1$, this is $\Phi_1=\exp(-\tfrac23+\int_0^1(1+s)\,dB_s)$.

Check: The stochastic integral has mean zero and variance $\int_0^1(1+s)^2\,ds=1+1+\tfrac13=\tfrac73$. It is normally distributed because its integrand is deterministic. Its exponential therefore has mean $e^{(7/3)/2}=e^{7/6}$. Thus $\mathbb{E}[\Phi_1]=e^{-2/3}e^{7/6}=e^{1/2}$, which equals $e^{\int_0^1 s\,ds}$. Substituting the general formula in Itô's formula, with $dZ_t=(t-\tfrac12(1+t)^2)dt+(1+t)dB_t$ and correction $\tfrac12\Phi_t(1+t)^2dt$, cancels the two half-square terms and leaves drift $t\Phi_tdt$, as required.

Worked Example: Piecewise-Constant Coefficients

On $[0,2]$, set $a_t=\tfrac12$ and $c_t=1$ for $0\le t<1$, and $a_t=-\tfrac12$ and $c_t=2$ for $1\le t\le2$. Find $\Phi_2$ and check its expectation.

The drift integral is $\int_0^2a_s\,ds=\tfrac12-\tfrac12=0$. The quadratic correction is $\tfrac12\int_0^2c_s^2\,ds=\tfrac12(1+4)=\tfrac52$. The stochastic integral is $$ \int_0^2c_s\,dB_s=(B_1-B_0)+2(B_2-B_1)=2B_2-B_1, $$ using $B_0=0$. Thus $\Phi_2=\exp(-\tfrac52+2B_2-B_1)$.

Check: Write $2B_2-B_1=B_1+2(B_2-B_1)$. The two terms on the right are independent centered normal variables with variances $1$ and $4$, so their sum has variance $5$. Its exponential has mean $e^{5/2}$, and therefore $\mathbb{E}[\Phi_2]=e^{-5/2}e^{5/2}=1$. This matches $e^{\int_0^2a_s\,ds}=1$. The coefficient change at $t=1$ causes no jump in $\Phi$: its defining time and stochastic integrals are continuous.

Common Mistake: Omitting the Itô Correction

A tempting but incorrect calculation is to write $\Phi_t=\exp(\int_0^t a_s\,ds+\int_0^t c_s\,dB_s)$, as if the ordinary chain rule applied. For the constant coefficients in the first example, this would give $\widetilde{\Phi}_t=\exp(\tfrac14t+\tfrac12B_t)$. With $Z_t=\tfrac14t+\tfrac12B_t$, Itô's formula gives drift $(\tfrac14+\tfrac18)\widetilde{\Phi}_t=\tfrac38\widetilde{\Phi}_t$, not the required $\tfrac14\widetilde{\Phi}_t$. The missing $-\tfrac18t$ in the exponent is exactly what restores the desired drift. The Stratonovich integral uses a different chain rule; it must not be silently substituted for the Itô integral $ \int_0^t H_s\,dB_s$.

For deterministic bounded coefficients, the stochastic integral in the exponent is centered normal with variance $\int_0^t c_s^2\,ds$. The normal moment-generating function then gives $\mathbb{E}[\Phi_t]=e^{\int_0^t a_s\,ds}$. This expectation calculation is not a claim that $\Phi$ is always a martingale: a martingale must also satisfy the conditional-expectation identity at every pair of times, and a nonzero drift generally prevents that property. For random coefficients, the expectation formula just stated does not follow from the deterministic-integrand normal calculation.

The fundamental solution records the multiplicative evolution from initial value $1$. If a solution starts at a deterministic value $x_0$, multiplying $\Phi_t$ by $x_0$ gives a solution with initial value $x_0$, since both terms in the SDE scale by that constant. The formula therefore isolates the effect of the drift, diffusion, and quadratic-variation correction without adding a forcing term.

Check Your Understanding

Use the stochastic-exponential formula and the stated bounded-coefficient assumptions in answering these questions.

For constant $a$ and $c$, write the formula for $\Phi_t$ and identify the deterministic correction in its exponent.
If $a_t=0$ and $c_t=2$ on $[0,1]$, what is the deterministic part of the exponent at time $1$?
For $a_t=t$ and $c_t=1$ on $[0,1]$, compute the deterministic part of the exponent at time $1$.
Explain why the exponential with no quadratic-variation correction fails to solve the constant-coefficient SDE in the first worked example.
Why does boundedness of $c_t$ alone ensure $c\in\mathcal{L}^2$, but not immediately establish $c\Phi\in\mathcal{L}^2$?

Fundamental Solutions Of Homogeneous Linear SDEs

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