Strong Solutions on a Prescribed Stochastic Basis

What Makes a Solution Strong?

An SDE specifies how a process changes through a drift term and a stochastic term driven by Brownian motion. To call a solution strong is to require that it be constructed on a probability space already carrying the prescribed Brownian motion and initial value. The process cannot be defined by choosing a different probability space or a different driving noise. Its paths must be continuous, it must be adapted to the given filtration, and it must satisfy the SDE’s integral equation.

Fix a finite horizon $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Let $\mathbf{B}_t$ be an $m$-dimensional Brownian motion relative to this filtration, and let $x_0\in\mathbb{R}^d$ be a deterministic initial value. Consider coefficients $\mu:[0,T]\times\mathbb{R}^d\to\mathbb{R}^d$ and $\sigma:[0,T]\times\mathbb{R}^d\to\mathbb{R}^{d\times m}$. The SDE in differential notation is $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,d\mathbf{B}_t$, with $X_0=x_0$.

Definition: A strong solution on the prescribed stochastic basis over $[0,T]$ is an $\mathbb{R}^d$-valued continuous adapted process $X=(X_t)_{0\le t\le T}$ such that the drift and stochastic integrals below are defined and, almost surely, for every $t\in[0,T]$, $$ X_t=x_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,d\mathbf{B}_s. $$ For the ordinary integral, require $\int_0^T|\mu(s,X_s)|\,ds<\infty$ almost surely. For the Ito integral, require each row of $s\mapsto\sigma(s,X_s)$ to be progressively measurable and square-integrable, meaning $\mathbb{E}\int_0^T|\sigma(s,X_s)|^2\,ds<\infty$. Here progressive measurability means that, for every $t$, the restriction to $[0,t]\times\Omega$ is measurable with respect to $\mathcal{B}([0,t])\otimes\mathcal{F}_t$; the integrand is thus in the Ito-integrable class $\mathcal{L}^2$.

The vector stochastic integral is taken componentwise: its $j$-th coordinate is the Ito integral of row $j$ of $\sigma(s,X_s)$ against $\mathbf{B}_s$. The filtration, Brownian motion, and deterministic initial value are fixed before $X$ is sought. Adaptedness requires $X_t$ to be $\mathcal{F}_t$-measurable at each time $t$; continuity refers to the paths of the process. The integral equation is the precise meaning of the differential notation. It is an equality of processes outside a single null set, using the continuous versions of the integral processes.

This definition is about the form of a solution, not whether one exists or is unique. In particular, it does not assert pathwise uniqueness or uniqueness in law. A weak solution allows the probability space and driving Brownian motion to be part of what is constructed; that is a different formulation. Existence theorems often use global Lipschitz and linear growth conditions on the coefficients, but those hypotheses are not part of the definition itself.

Deterministic Coefficients: A Direct Construction

Proposition: Suppose $a:[0,T]\to\mathbb{R}^d$ is measurable with $\int_0^T|a(s)|\,ds<\infty$, and $G:[0,T]\to\mathbb{R}^{d\times m}$ is deterministic and measurable with $\int_0^T|G(s)|^2\,ds<\infty$. On a filtered probability space carrying the $m$-dimensional Brownian motion $\mathbf{B}$, the process $$ X_t=x_0+\int_0^t a(s)\,ds+\int_0^t G(s)\,d\mathbf{B}_s $$ is a strong solution of the SDE with coefficients $\mu(t,x)=a(t)$ and $\sigma(t,x)=G(t)$.

Proof. Define $X$ by the displayed formula. The deterministic integral is finite by the assumed integrability of $a$. The deterministic process $G$ is progressively measurable, and its squared norm has finite integral, so each row defines an Ito integral in $\mathcal{L}^2$. The Ito integral has a continuous adapted version; adding the deterministic continuous integral and the constant $x_0$ gives a continuous adapted process $X$. Substituting the coefficients $\mu(t,X_t)=a(t)$ and $\sigma(t,X_t)=G(t)$ into the defining integral equation gives exactly the formula used to define $X_t$. Thus all requirements in the definition hold. $\square$

Worked Example: A Constant Drift

On any filtered probability space carrying a Brownian motion, consider the scalar SDE $dX_t=2\,dt$ with $X_0=1$. The diffusion coefficient is zero. Integrating the drift gives $$ X_t=1+\int_0^t2\,ds+\int_0^t0\,dB_s=1+2t. $$

The candidate is deterministic, hence adapted, and its paths are continuous. The drift integral is finite on every finite interval, and the zero stochastic integrand is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^T0^2\,ds=0$. Substitution checks the equation: $1+\int_0^t2\,ds+\int_0^t0\,dB_s=1+2t=X_t$. At $t=0$, the formula gives $X_0=1$, as prescribed.

Check: The solution is independent of the Brownian path because the diffusion coefficient is zero. The Brownian motion is still part of the prescribed basis, but it contributes no stochastic term to this equation.

Worked Example: A Constant Diffusion

Consider $dX_t=2\,dB_t$, $X_0=3$, on the given basis with Brownian motion $B_t$. Integrating both terms yields the candidate $X_t=3+2B_t$. It is continuous and adapted because Brownian motion has continuous paths and is adapted to its filtration.

The drift integrand is zero. The diffusion integrand is the constant $2$, which is progressively measurable and satisfies $\mathbb{E}\int_0^T2^2\,ds=4T<\infty$. Now substitute the candidate into the right-hand side of the integral equation: $$ 3+\int_0^t0\,ds+\int_0^t2\,dB_s=3+2B_t=X_t. $$

Check: At $t=0$, $B_0=0$, so the formula gives $X_0=3$. For each $t$, $\mathbb{E}[X_t]=3$, since $\mathbb{E}[B_t]=0$; this agrees with the zero drift, while the stochastic term accounts for the random variation.

Verification by Ito’s Formula

For state-dependent diffusion, the proposed process may not be a simple sum of deterministic and Brownian terms. A useful verification method is to start with an explicit candidate, apply Ito’s formula, and match the resulting drift and diffusion coefficients to the SDE. The stochastic integral must meet its integrability condition; the calculation alone does not make an undefined integral valid.

Theorem: Let $a,b,x_0\in\mathbb{R}$, with $x_0>0$, and let $B_t$ be Brownian motion relative to the prescribed filtration. The process $$ X_t=x_0\exp\!\left(\left(a-\tfrac12b^2\right)t+bB_t\right) $$ is a strong solution on every finite interval of $dX_t=aX_t\,dt+bX_t\,dB_t$, $X_0=x_0$.

Proof. Set $f(t,y)=x_0\exp((a-\tfrac12b^2)t+by)$, so that $X_t=f(t,B_t)$. The partial derivatives are $f_t=(a-\tfrac12b^2)f$, $f_y=bf$, and $f_{yy}=b^2f$. Ito’s formula, including its second-derivative correction, gives $$ dX_t=f_t(t,B_t)\,dt+f_y(t,B_t)\,dB_t+\tfrac12 f_{yy}(t,B_t)\,dt =aX_t\,dt+bX_t\,dB_t. $$ Indeed, the total coefficient of $dt$ is $(a-\tfrac12b^2)X_t+\tfrac12b^2X_t=aX_t$. At time zero, $X_0=x_0\exp(0)=x_0$.

The formula defines a continuous adapted process. It remains to check integrability on $[0,T]$. A normal moment calculation gives $\mathbb{E}[X_s^2]=x_0^2\exp((2a+b^2)s)$, which is finite and bounded as $s$ ranges over $[0,T]$. Therefore $\mathbb{E}\int_0^T b^2X_s^2\,ds<\infty$, so the diffusion integrand is in $\mathcal{L}^2$. Also, $\mathbb{E}\int_0^T|aX_s|\,ds<\infty$, since $\mathbb{E}|X_s|\le(\mathbb{E}[X_s^2])^{1/2}$ is bounded on this finite interval. Integrating the differential identity proves the required integral equation and completes the verification. $\square$

Worked Example: Geometric Brownian Motion on the Given Basis

Take $x_0=2$, $a=\tfrac12$, and $b=1$. The theorem gives $$ X_t=2\exp\!\left(\left(\tfrac12-\tfrac12\cdot1^2\right)t+B_t\right)=2e^{B_t}. $$

To check directly, let $f(t,y)=2e^y$. Its partial derivatives are $f_t=0$, $f_y=2e^y$, and $f_{yy}=2e^y$. Ito’s formula yields $$ dX_t=0\,dt+2e^{B_t}\,dB_t+\tfrac12(2e^{B_t})\,dt =\tfrac12X_t\,dt+X_t\,dB_t. $$ Also $X_0=2e^{B_0}=2$. The stochastic integrand is square-integrable on every finite interval because $\mathbb{E}[X_s^2]=4\mathbb{E}[e^{2B_s}]=4e^{2s}$, and hence $\mathbb{E}\int_0^T X_s^2\,ds=4\int_0^T e^{2s}\,ds=2(e^{2T}-1)<\infty$.

Check: The explicit solution has mean $\mathbb{E}[X_t]=2e^{t/2}$, using the normal moment formula $\mathbb{E}[e^{B_t}]=e^{t/2}$. Taking expectations in the integral equation gives $m(t)=2+\tfrac12\int_0^t m(s)\,ds$: the Ito integral has mean zero because its integrand is in $\mathcal{L}^2$. The displayed mean satisfies this check, since $2+\tfrac12\int_0^t2e^{s/2}\,ds=2+2(e^{t/2}-1)=2e^{t/2}$.

What the Definition Does—and Does Not—Require

The word “prescribed” is essential. In the examples, the Brownian motion appearing in the formula is the Brownian motion on the stated filtered probability space. If a candidate is written using some other noise, its formula does not by itself establish a strong solution on this basis. One must verify that the candidate is adapted to the given filtration and that the integral equation uses the given Brownian motion.

Three requirements should be checked separately:

Continuity: the candidate has continuous paths, as required by the definition.
Adaptedness: its value at time $t$ is measurable with respect to the information $\mathcal{F}_t$ available at that time.
Integral identity: the drift and Ito integrals are defined and their sum with $x_0$ equals the candidate for every time, almost surely.

A common mistake is to verify only the differential notation. For example, writing $X_t=x_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s$ does not establish a strong solution if the integrand is not progressively measurable or does not satisfy the required square-integrability condition. The correction is to check those conditions, along with continuity and adaptedness, before asserting that the integral equation defines a solution. Another mistake is to infer uniqueness from a successful substitution: the definition and the verification above show that a candidate is a solution, not that no other solution exists.

Check Your Understanding

For each question, keep the prescribed filtered probability space and its Brownian motion fixed.

For $dX_t=3\,dt$ with $X_0=-1$, write the integral equation and verify the candidate process.
For $dX_t=4\,dB_t$ with $X_0=2$, what condition on the integrand ensures that the Ito integral is defined on $[0,T]$?
For $dX_t=aX_t\,dt+bX_t\,dB_t$, identify the Ito correction when applying Ito’s formula to $x_0\exp((a-\tfrac12b^2)t+bB_t)$.
Why does checking the integral equation for one candidate not establish uniqueness?
In the definition of a strong solution, which probability space and Brownian motion are fixed in advance?

Strong Solutions on a Prescribed Stochastic Basis

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