Weak Solutions and Their Probability Spaces

What a Weak Solution Specifies

An SDE is not specified by its drift and diffusion coefficients alone. A solution also lives on a probability space, with a filtration describing the information available over time and a Brownian motion relative to that filtration. The distinction between a weak and a strong solution is primarily about what may be chosen and what is prescribed: a weak solution may use a probability space constructed for the problem, whereas a strong solution must be built using a prescribed Brownian motion and initial data.

Fix a finite time horizon $T>0$, measurable coefficients $\mu(t,x)$ and $\sigma(t,x)$, and an initial probability law $\nu$ on $\mathbb{R}$. All probabilities and expectations below are under $\mathbb{P}$; no change to a risk-neutral measure $\mathbb{Q}$ is involved. For every stochastic integral in this tutorial, the integrand is progressively measurable and belongs to $\mathcal{L}^2$: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The drift integral is an ordinary time integral, required to be finite almost surely.

Definition: A weak solution of $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$ with initial law $\nu$ consists of a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, a Brownian motion $B_t$ relative to that filtration, and a continuous adapted process $X_t$, such that $X_0$ has law $\nu$, $\int_0^T|\mu(t,X_t)|\,dt<\infty$ almost surely, $\sigma(t,X_t)$ is progressively measurable and in $\mathcal{L}^2$, and, almost surely for every $t\in[0,T]$, $$ X_t=X_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s. $$

The Brownian property is relative to the stated filtration: $B_t-B_s$ must be independent of $\mathcal{F}_s$ and have distribution $N(0,t-s)$ whenever $s<t$. Adaptedness of $X$ alone does not guarantee this property. In particular, a filtration revealing future Brownian values would generally make $B$ fail to be Brownian relative to that filtration.

Definition: A strong solution, relative to a prescribed Brownian motion $B_t$ and initial variable $\xi$, is a solution on the specified probability space that is adapted to the usual augmentation of the filtration generated by $\xi$ and $B$. The initial variable is measurable at time zero and is independent of the future Brownian increments, so that $B$ is Brownian relative to this filtration. A weak solution allows the space and Brownian motion to be chosen; a strong solution does not.

Strong existence and pathwise uniqueness are separate questions. Pathwise uniqueness asks whether two solutions driven by the same Brownian motion and the same initial variable agree almost surely. Uniqueness in law instead asks whether the distributions of solutions agree. Neither phrase is another name for strong existence.

Two Facts About the Probability Space

Proposition: Suppose $\xi$ is independent of a Brownian motion $B_t$. The process $B_t$ is Brownian relative to the usual augmentation of $\mathcal{G}_t=\sigma(\xi,B_u:0\le u\le t)$.

Proof. First consider the unaugmented filtration $\mathcal{G}_t$. For $s<t$, the increment $B_t-B_s$ is independent of the Brownian history $\sigma(B_u:0\le u\le s)$ by the independent-increments property. It is also independent of $\xi$. Since $\xi$ is independent of the entire Brownian motion, the pair consisting of $\xi$ and the Brownian history is independent of $B_t-B_s$. Thus the increment is independent of $\mathcal{G}_s$, and its distribution is $N(0,t-s)$. Brownian paths are continuous. The standard usual-augmentation theorem for Brownian filtrations says that completing this filtration and making it right-continuous preserves the Brownian property. Applying that theorem gives the stated result. $\square$

This proposition is useful when constructing a weak solution with an independent random initial value. It does not say that every filtration containing the Brownian history is suitable: adding information about future increments can invalidate the required independence.

Theorem: If $(X,B)$ is a weak solution with constant drift $a$ and constant diffusion coefficient $c$, then $$ X_t=X_0+at+cB_t. $$ If $X_0$ is square-integrable, then $\mathbb{E}[X_t]=\mathbb{E}[X_0]+at$ and $\operatorname{Var}(X_t)=\operatorname{Var}(X_0)+c^2t$.

Proof. The weak solution equation in this case is $X_t=X_0+\int_0^t a\,ds+\int_0^t c\,dB_s$. The time integral is $at$, and the Itô integral of the constant integrand $c$ is $cB_t$, giving the displayed identity. The integrand is in $\mathcal{L}^2$ because $\mathbb{E}\int_0^T c^2\,ds=c^2T<\infty$. Since $B$ is Brownian relative to the filtration, $B_t$ is independent of $\mathcal{F}_0$, and hence of $X_0$, with mean zero and variance $t$. Taking expectations gives $\mathbb{E}[X_t]=\mathbb{E}[X_0]+at$. Independence gives $\operatorname{Var}(X_t)=\operatorname{Var}(X_0)+c^2\operatorname{Var}(B_t)=\operatorname{Var}(X_0)+c^2t$. $\square$

Worked Examples

Worked Example: A Strong Solution on a Prescribed Space

Let $B_t$ be Brownian motion on a given filtered probability space, and solve $dX_t=2\,dt+3\,dB_t$, $X_0=1$, on $[0,T]$.

Define $X_t=1+2t+3B_t$. The drift integral is $\int_0^t2\,ds=2t$. The stochastic integrand is the constant $3$, which is progressively measurable and satisfies $\mathbb{E}\int_0^T3^2\,ds=9T<\infty$. Its Itô integral is $\int_0^t3\,dB_s=3B_t$. Therefore the integral equation holds: $$ 1+\int_0^t2\,ds+\int_0^t3\,dB_s=1+2t+3B_t=X_t. $$ Because $X_t$ is adapted to the given Brownian filtration, this is a strong solution on the prescribed space.

Check: Brownian motion has mean $0$ and variance $t$, so the formula gives $\mathbb{E}[X_t]=1+2t$ and $\operatorname{Var}(X_t)=9t$. These agree with the constant-coefficient theorem using $X_0=1$, $a=2$, and $c=3$.

Worked Example: A Weak Solution with a Random Initial Value

Construct a weak solution of $dX_t=0$ whose initial value is Bernoulli with parameter $\tfrac12$, and explain why it is not a strong solution on the Brownian natural space with no initial randomness.

Take the product of $\{0,1\}$, with probabilities $\tfrac12,\tfrac12$, and a space carrying a standard Brownian motion. Let $\xi$ be the first coordinate and $B_t$ the Brownian coordinate. Use the usual augmentation of $\mathcal{G}_t=\sigma(\xi,B_u:0\le u\le t)$. By the proposition, $B_t$ remains Brownian relative to this filtration. Set $X_t=\xi$ for all $t$. It is continuous and adapted, and the equation holds because both integrals vanish: $$ X_t=\xi+\int_0^t0\,ds+\int_0^t0\,dB_s=\xi. $$ The zero stochastic integrand belongs to $\mathcal{L}^2$, and $X_0$ has the required Bernoulli law. This is a weak solution.

On the usual augmentation of the filtration generated by Brownian motion alone, the time-zero sigma-field is trivial up to null sets. Any time-zero measurable random variable is therefore almost surely constant. It cannot have the Bernoulli law, which assigns probability $\tfrac12$ to each of $0$ and $1$. Thus this prescribed Brownian-only space cannot support a strong solution with the stated random initial law. If the space is enlarged by supplying $\xi$ at time zero independently of $B$, the same solution is strong relative to that enlarged initial-data space.

Check: $\mathbb{P}(X_0=0)=\tfrac12$ and $\mathbb{P}(X_0=1)=\tfrac12$, so $\mathbb{E}[X_t]=\tfrac12$. Since $X_t$ is Bernoulli, $\operatorname{Var}(X_t)=\tfrac12(1-\tfrac12)=\tfrac14$ for every $t$, consistent with the constant-coefficient theorem with $a=c=0$.

Worked Example: A Strong Stochastic-Exponential Solution

On a prescribed Brownian space, verify that $X_t=\exp(B_t-\tfrac12t)$ solves $dX_t=X_t\,dB_t$, $X_0=1$.

Apply Itô's formula to $f(t,x)=\exp(x-\tfrac12t)$ and $B_t$. The derivatives are $f_t=-\tfrac12 f$, $f_x=f$, and $f_{xx}=f$. The formula, including its second-derivative correction, is $$ df(t,B_t)=f_t(t,B_t)\,dt+f_x(t,B_t)\,dB_t+\tfrac12f_{xx}(t,B_t)\,dt =-\tfrac12X_t\,dt+X_t\,dB_t+\tfrac12X_t\,dt =X_t\,dB_t. $$ Also $X_0=f(0,0)=1$. The process $X$ is continuous and adapted. Its diffusion integrand is in $\mathcal{L}^2$, since $B_t\sim N(0,t)$ and the normal moment-generating function gives $$ \mathbb{E}\int_0^T X_s^2\,ds=\int_0^T\mathbb{E}\big[e^{2B_s-s}\big]\,ds =\int_0^T e^s\,ds=e^T-1<\infty. $$ Thus the Itô integral is defined in the required class, and the displayed differential integrates to the SDE's integral equation. This solution is strong because it is an adapted function of the prescribed Brownian path.

Check: For $s\le t$, $X_t=X_s\exp(B_t-B_s-\tfrac12(t-s))$. The increment $B_t-B_s$ is independent of $\mathcal{F}_s$, and its normal moment-generating function gives $\mathbb{E}[\exp(B_t-B_s-\tfrac12(t-s))]=1$. Hence $\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s$. Also $\mathbb{E}[X_t]=1$, so this martingale is integrable at every time. This independently checks the first moment; it does not assert that every local martingale is a true martingale.

What the Distinction Does—and Does Not—Say

Weakness is not a property of the paths being less regular. Both definitions require a continuous adapted process satisfying the same integral equation. The distinction concerns the underlying probability space and the information available to construct the process. A weak solution can be obtained by choosing a suitable space carrying both a Brownian motion and any required initial randomness. A strong solution must use the Brownian motion and initial variable already prescribed, without additional randomness beyond their generated filtration.

A useful pitfall is to assume that any filtration containing the Brownian history is acceptable. For example, one might enlarge every $\mathcal{F}_s$, for $s<T$, to include $B_T$, and then claim that $B$ is still Brownian. This is wrong: $B_T-B_s$ is then $\mathcal{F}_s$-measurable and has nonzero variance $T-s$, so it cannot be independent of $\mathcal{F}_s$. The correction is to use a filtration that reveals only the initial variable and Brownian information up to the current time, as in the independent initial enlargement above, or otherwise verify the Brownian property directly.

When checking a proposed weak solution, verify the whole tuple, not just the algebraic SDE: identify the space and filtration, establish that $B$ is Brownian relative to that filtration, check adaptedness and continuity of $X$, and check the required integrability of the integrand. A formal expression satisfying the differential notation on a path is not by itself a weak solution.

Check Your Understanding

For each question, distinguish what is determined by the equation from what depends on the choice of probability space.

List the components that must be supplied to give a weak solution with initial law $\nu$.
Why does independence of $\xi$ and $B$ matter when $\xi$ is included in the time-zero information?
For constant coefficients $a$ and $c$, write the solution formula and state the assumptions needed for the mean and variance calculation.
Why can a Brownian natural space with trivial time-zero sigma-field fail to support a strong solution with a non-degenerate random initial law?
In the stochastic-exponential example, identify the two $dt$ terms in Itô's formula and explain why they cancel.

Weak Solutions and Their Probability Spaces

What You'll Learn