Laws of Weak Solutions and Uniqueness in Law

Comparing Weak Solutions Through Their Path Laws

A weak solution of an SDE is built on a probability space that is allowed to vary. That flexibility makes it important to specify what it means for two such solutions to be the same. Equality of the probability spaces is not required; instead, one can compare the distributions of the solution paths. This comparison is the basis of uniqueness in law.

Fix a finite time horizon $T>0$. Throughout, a weak solution is defined on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to that filtration. For an SDE with deterministic initial value $x_0$, drift $\mu$, and diffusion coefficient $\sigma$, the solution is a continuous adapted process $X_t$ satisfying its integral equation almost surely. The stochastic integrand $s\mapsto \sigma(s,X_s)$ is required to be progressively measurable and in $\mathcal{L}^2$ on each finite horizon: $\mathbb{E}\int_0^T \sigma(s,X_s)^2\,ds<\infty$. The drift integral is assumed to be defined, for example through $\int_0^T|\mu(s,X_s)|\,ds<\infty$ almost surely.

Definition: Let $C([0,T],\mathbb{R})$ be the space of continuous real-valued paths on $[0,T]$, equipped with its Borel sigma-field for the uniform topology. The law of $X$ on path space is the probability measure $\mathcal{L}_{\mathbb{P}}(X)$ defined by $$ \mathcal{L}_{\mathbb{P}}(X)(A)=\mathbb{P}\big((X_t)_{0\leq t\leq T}\in A\big), \qquad A\subseteq C([0,T],\mathbb{R})\ \text{Borel}. $$ An SDE has uniqueness in law on $[0,T]$, for the specified initial condition, if every pair of weak solutions has the same path-space law. Equality of laws means equality on every Borel set $A$, not merely equality of distributions at individual times.

Several notions of uniqueness must be kept separate. Pathwise uniqueness says that if two solutions are defined on the same filtered probability space, driven by the same Brownian motion, and have the same initial value, then their paths agree almost surely at all times. Uniqueness in law compares distributions even when the solutions live on different spaces. A strong solution is adapted to the filtration generated by the driving Brownian motion (with the usual completion); a weak solution only requires a space carrying a Brownian motion and an adapted solution. These properties are distinct definitions: one cannot infer equality of solution paths from equality of their laws, or equality of laws merely from the fact that two equations use Brownian motions.

Brownian Motion: The Law Is Fixed by Its Definition

Worked Example: The Path Law of Brownian Motion

Consider the SDE $dX_t=dB_t$, $X_0=x_0$, on $[0,T]$. Its integral form is $X_t=x_0+\int_0^t 1\,dB_s$. The integrand $H_s=1$ is adapted, progressively measurable, and satisfies $\mathbb{E}\int_0^T H_s^2\,ds=T<\infty$. By the definition of the Ito integral, $\int_0^t1\,dB_s=B_t-B_0=B_t$, so every weak solution has $X_t=x_0+B_t$.

Since $B$ is standard Brownian motion, its path law is the standard Wiener measure on $C([0,T],\mathbb{R})$. Thus the law of $X$ is the distribution of the translated path $t\mapsto x_0+B_t$. For example, for times $0\leq t_1<\cdots<t_n\leq T$, the increments $X_{t_j}-X_{t_{j-1}}$ are independent normal random variables with means $0$ and variances $t_j-t_{j-1}$, and $X_0=x_0$. These finite-dimensional distributions are the same in every weak realization because the driving process in each realization is Brownian motion.

Check: At any fixed $t$, $X_t$ has distribution $N(x_0,t)$. This single-time calculation is consistent with the path law, but the path law specifies more: it also records the joint distributions at all finite collections of times and the continuity of paths.

Geometric Brownian Motion: An Explicit Law

For constants $\mu,\sigma\in\mathbb{R}$ and $s_0\in\mathbb{R}$, consider $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$, $S_0=s_0$. Here $\mu S_t$ is the drift coefficient and $\sigma S_t$ is the diffusion coefficient. The following calculation both verifies the explicit solution and shows that any weak solution must have the same path law.

Proposition: Suppose a weak solution satisfies $\mathbb{E}\int_0^T S_s^2\,ds<\infty$. Then, almost surely for all $t\in[0,T]$, $$ S_t=s_0\exp\big((\mu-\tfrac12\sigma^2)t+\sigma B_t\big). $$ Consequently the SDE has uniqueness in law on $[0,T]$ for initial value $s_0$.

Proof. First verify the candidate by Ito's formula. Set $f(t,b)=s_0\exp((\mu-\tfrac12\sigma^2)t+\sigma b)$. Its derivatives are $f_t=(\mu-\tfrac12\sigma^2)f$, $f_b=\sigma f$, and $f_{bb}=\sigma^2 f$. The Ito correction is $\tfrac12 f_{bb}\,dt=\tfrac12\sigma^2f\,dt$, so $df(t,B_t)=\mu f(t,B_t)\,dt+\sigma f(t,B_t)\,dB_t$, with $f(0,B_0)=s_0$.

For uniqueness, take any weak solution $S$ and set $Z_t=\exp(-(\mu-\tfrac12\sigma^2)t-\sigma B_t)$. Apply Ito's formula to $g(t,b)=\exp(-(\mu-\tfrac12\sigma^2)t-\sigma b)$. Its derivatives are $g_t=-(\mu-\tfrac12\sigma^2)g$, $g_b=-\sigma g$, and $g_{bb}=\sigma^2g$; hence the correction term is $\tfrac12\sigma^2g\,dt$ and $dZ_t=(-\mu+\sigma^2)Z_t\,dt-\sigma Z_t\,dB_t$.

The Ito product rule gives $d(Z_tS_t)=Z_t\,dS_t+S_t\,dZ_t+d[Z,S]_t$. The covariation term is $d[Z,S]_t=(-\sigma Z_t)(\sigma S_t)\,dt=-\sigma^2Z_tS_t\,dt$. Substitution yields drift coefficient $\mu+(-\mu+\sigma^2)-\sigma^2=0$, and diffusion coefficient $\sigma-\sigma=0$. Therefore $Z_tS_t=Z_0S_0=s_0$, which gives the asserted formula since $Z_t>0$. The integrability assumption ensures the SDE's stochastic integrand is in $\mathcal{L}^2$; the explicit candidate has this property because $\mathbb{E}[S_t^2]=s_0^2e^{(2\mu+\sigma^2)t}$, whose integral over $[0,T]$ is finite. Finally, the displayed formula is the same measurable function of a standard Brownian path in every weak realization. Brownian paths have the same law, so their images under this function have the same law. $\square$

Worked Example: A Geometric Brownian Motion Law

Take $s_0=2$, $\mu=0$, and $\sigma=1$. The formula gives $S_t=2e^{B_t-t/2}$. To check the correction directly, $f(t,b)=2e^{b-t/2}$ has $f_t=-f/2$, $f_b=f$, and $f_{bb}=f$; thus $\tfrac12f_{bb}\,dt=\tfrac12f\,dt$, which cancels $f_t\,dt$. The remaining term is $f\,dB_t=S_t\,dB_t$, as required.

At time $t$, using the normal moment-generating function $\mathbb{E}[e^{uB_t}]=e^{u^2t/2}$, we obtain $\mathbb{E}[S_t]=2e^{-t/2}e^{t/2}=2$. Also $\mathbb{E}[S_t^2]=4e^{-t}\mathbb{E}[e^{2B_t}]=4e^{-t}e^{2t}=4e^t$. In particular, $\mathbb{E}\int_0^T S_t^2\,dt=4\int_0^T e^t\,dt=4(e^T-1)<\infty$, so the diffusion integrand meets the required $\mathcal{L}^2$ condition.

Check: The candidate starts at $2e^0=2$. Its mean remains $2$, while its second moment at $t=0$ is $4$, agreeing with the deterministic initial value. The Ito correction is essential: omitting it would produce $2e^{B_t}$, whose Ito drift is $\tfrac12 S_t\,dt$, not the required zero drift.

An SDE with Different Weak-Solution Laws

Worked Example: Delayed Solutions of a Non-Lipschitz Drift Equation

Let $T=1$, $x_0=0$, and consider $dX_t=2\sqrt{X_t^+}\,dt+0\,dB_t$, where $x^+=\max(x,0)$. Use any filtered probability space carrying a Brownian motion $B$; the diffusion integrand is identically zero, hence progressively measurable and satisfies $\mathbb{E}\int_0^1 0^2\,dt=0$.

For each $c\in[0,1]$, define the deterministic continuous adapted process $X_t^{(c)}=(t-c)_+^2$. When $t<c$, it is zero and its derivative is zero. When $t>c$, its derivative is $2(t-c)$, and $2\sqrt{X_t^{(c)}}=2(t-c)$. At $t=c$, both the derivative and the right-hand side are zero. Thus, at every $t\in[0,1]$, \[ X_t^{(c)}=\int_0^t 2\sqrt{(X_s^{(c)})^+}\,ds+\int_0^t0\,dB_s. \] The equation is satisfied pathwise, so each process is a weak solution.

Choose $c=0$ and $c=\tfrac12$. At $t=1$, $X_1^{(0)}=1$, whereas $X_1^{(1/2)}=(1-\tfrac12)^2=\tfrac14$. Their path-space laws are point masses at different paths, and their time-one distributions are respectively point masses at $1$ and $\tfrac14$. For example, the Borel set $\{x\in C([0,1],\mathbb{R}):x(1)>\tfrac12\}$ has probability $1$ under the first law and $0$ under the second. Therefore uniqueness in law fails.

Check: Substitution confirms the endpoint values: for $c=0$, the integral is $\int_0^1 2s\,ds=1$; for $c=\tfrac12$, it is $\int_{1/2}^1 2(s-\tfrac12)\,ds=(\tfrac12)^2=\tfrac14$. Both match the corresponding $X_1^{(c)}$.

What the Comparisons Do—and Do Not—Show

The examples separate a deterministic equation from one whose solution is driven by Brownian randomness. In the Brownian example, the solution is a fixed translation of the driver. In geometric Brownian motion, the Ito calculation identifies the solution as a fixed pathwise function of the driver. In both cases, the Brownian path law determines the solution path law. In the delayed-drift example, different solutions are possible even on one space with the same Brownian motion, because the diffusion coefficient is zero and the drift does not select a unique departure time from zero.

A common mistake is to argue that the last two delayed solutions must have the same law because they can be placed on a space carrying the same Brownian motion. That is false: sharing a driver is not a law comparison. The explicit calculation at time $1$ gives $X_1^{(0)}=1$ and $X_1^{(1/2)}=\tfrac14$, so their distributions—and therefore their path laws—are different. Conversely, two processes can have the same law without being equal path by path on a particular probability space.

Another useful distinction is between equality of one-time distributions and equality of path-space laws. The latter implies equality of all one-time distributions, since evaluation $x\mapsto x(t)$ is a measurable map on continuous path space. The reverse implication does not follow from the definition: knowing the marginal distribution at each time does not by itself specify the dependence between different times, and path laws include that dependence. In the examples above, the different endpoint distributions already disprove uniqueness in law, so no more delicate comparison is needed.

Uniqueness in law is a property of the SDE together with its coefficients, initial condition, and time horizon. The delayed example illustrates why coefficient regularity matters: $x\mapsto2\sqrt{x^+}$ is not locally Lipschitz at zero. No general uniqueness theorem is needed for the example; direct substitution constructs the solutions and their distinct laws. By contrast, the geometric Brownian motion conclusion was established directly by an Ito calculation and a transformation of an arbitrary solution.

Check Your Understanding

Use the path-space definition and the calculations above to answer these questions.

For $dX_t=dB_t$ with $X_0=x_0$, what is the solution law on $C([0,T],\mathbb{R})$?
In the geometric Brownian motion calculation, identify the $\tfrac12 f_{bb}\,dt$ correction and explain why it changes the drift.
For the delayed-drift example, which Borel set distinguishes the laws for $c=0$ and $c=\tfrac12$?
Does equality of path-space laws imply that two solutions on a common probability space are equal almost surely? Explain the distinction.
Why does the fact that two weak solutions use the same Brownian motion not establish uniqueness in law?