Two Different Comparisons Between Solutions
An SDE solution has both a path and a probability distribution. These give rise to two different questions. If two candidate solutions are driven by the same Brownian motion and start at the same random value, must their paths agree almost surely? That is the question of pathwise uniqueness. If two solutions are constructed on possibly different probability spaces, must their distributions agree? That is the question of uniqueness in law. The first compares realizations under a shared noise and initial value; the second compares distributions.
To state the distinction precisely, fix a time horizon \(T>0\), coefficients \(\mu,\sigma\), and a prescribed probability distribution for the initial value. A stochastic basis is a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to that filtration. An initial value \(\xi\) is \(\mathcal{F}_0\)-measurable; the usual formulation requires Brownian increments after time zero to be independent of \(\mathcal{F}_0\).
In the definition of pathwise uniqueness, it is the initial random variable that must be shared, not merely its distribution. Uniqueness in law instead compares distributions across solution constructions, which may live on different probability spaces. Equality in law is not a statement that two paths agree pointwise, or even that the two paths have been placed on a common probability space.
Consequences of the Definitions
Proof. Let \(X\) and \(Y\) be solutions on the same basis with the same \(\xi\) and \(B\). Pathwise uniqueness gives \(X_t=Y_t\) for every \(t\in[0,T]\) on one event of probability one. On that event, \(X\) and \(Y\) are the same element of \(C([0,T])\). Thus, for every Borel set \(A\subseteq C([0,T])\), the indicators \(\mathbf{1}_{\{X\in A\}}\) and \(\mathbf{1}_{\{Y\in A\}}\) agree almost surely, so their expectations, and hence their probabilities, agree. For the second assertion, the map \((x,b,z)\mapsto z\) from the product path space to \(C([0,T])\) is measurable. Equal joint distributions assign equal probabilities to inverse images under this map, which are exactly the events used to specify the marginal law of \(X\). \(\square\)
The first conclusion is deliberately limited to solutions on a common basis. By itself it does not compare solutions that have been constructed on different spaces. The second conclusion is a projection fact: joint-law uniqueness is stronger than marginal-law uniqueness because it also fixes how the solution is distributed together with its initial value and driving noise.
Worked Example: A Deterministic Initial Value and Brownian Noise
Consider \(dX_t=dB_t\), \(X_0=2\), on \([0,1]\). Verify the solution and compare the two notions of uniqueness.
Here \(\mu(t,x)=0\), \(\sigma(t,x)=1\), and the integral equation is \(X_t=2+\int_0^t 1\,dB_s\). The integrand is progressively measurable and belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^1 1^2\,ds=1\). By the defining integral of the constant integrand, \(X_t=2+B_t\).
For any other solution \(Y\) with this same initial value and the same Brownian motion, its integral equation reads \(Y_t=2+\int_0^t1\,dB_s=2+B_t\). Thus \(Y_t=X_t\) for every \(t\), path by path, and pathwise uniqueness holds. The solution law is also fixed: \(X_t\) has the same path distribution as a Brownian path shifted upward by \(2\).
Check: at time \(t=1\), \(X_1=2+B_1\), so its mean is \(2\) and its variance is \(1\), because \(B_1\sim N(0,1)\). The same integral equation on any other admissible basis again gives the shifted Brownian path, confirming equality of the solution laws.
Worked Example: Same Law Does Not Mean the Same Realization
Let \(U\) be Bernoulli with \(\mathbb{P}(U=0)=\mathbb{P}(U=1)=\tfrac12\), and consider \(dX_t=0\), \(X_0\) with this Bernoulli distribution, on \([0,1]\). First compare two realizations on separate spaces. Then explain why a shared-space comparison needs an additional check.
On either space the integral equation is \(X_t=X_0+\int_0^t0\,ds+\int_0^t0\,dB_s=X_0\). The drift integrand \(0\) is time-integrable, and the diffusion integrand \(0\) is progressively measurable with \(\mathbb{E}\int_0^1 0^2\,ds=0\). Therefore each solution path is constant: it is the zero path with probability \(\tfrac12\) and the one path with probability \(\tfrac12\). The two constructions have exactly the same path law.
Now put \(U\) and a Bernoulli variable \(V=1-U\) on one space, and give both solutions the same Brownian motion, independent of \(U\). The paths \(X_t=U\) and \(Y_t=V\) still have the same distribution. But \(\mathbb{P}(X_t\ne Y_t)=1\), since \(U\) is always either \(0\) or \(1\) and \(V=1-U\). This does not contradict pathwise uniqueness: \(X_0=U\) and \(Y_0=V\) are different random variables, whereas pathwise uniqueness requires the same initial random variable.
Check: each marginal probability of the one path is \(\tfrac12\), and each marginal probability of the zero path is \(\tfrac12\), so the path laws agree. Nevertheless, under this coupling, the paths differ with probability \(1\). Matching the initial distributions alone does not meet the shared-initial-value hypothesis in the definition of pathwise uniqueness.
What Each Notion Does—and Does Not—Say
Pathwise uniqueness is a statement about pairs of solutions that satisfy a specific coupling requirement: they share the driving noise and the initial random variable. It does not mean that every two solutions with the same initial distribution must agree when placed together arbitrarily. Uniqueness in law asks a different question: whether the probability of every measurable set of solution paths is the same in every admissible solution construction with the prescribed initial law.
There are two common versions of the law comparison. Equality of the marginal law of \(X\) records the distribution of the solution path alone. Equality of the joint law of \((\xi,B,X)\) records, in addition, how that path is related in distribution to its initial value and its driving Brownian motion. The stronger joint formulation can distinguish constructions that happen to give the same marginal law for \(X\). Always check which version is intended before using a uniqueness-in-law statement.
| Question | Pathwise uniqueness | Uniqueness in law |
|---|---|---|
| What is compared? | Two solution paths | Distributions of solution paths, or of the specified joint tuple |
| Must the basis be shared? | Yes | No |
| Must the Brownian motion be shared? | Yes | No; the law comparison may include its distribution jointly with the solution |
| Must the initial random variable be shared? | Yes | No; the prescribed initial law must match |
| Conclusion | Indistinguishable paths under the required coupling | Equal probabilities for measurable path events |
The notions are genuinely distinct, not alternative phrasings of a single property. A classical example is the Tanaka equation \(dX_t=\operatorname{sgn}(X_t)\,dB_t\), \(X_0=0\), with \(\operatorname{sgn}(0)=0\): it has uniqueness in law but fails pathwise uniqueness. This is a standard example whose construction and proof require additional stochastic-analysis results; they are not established here. Its role is to warn against inferring pathwise uniqueness merely from a statement about distributions.
Common Pitfalls
Wrong comparison: “These two solutions have the same distribution, so they must agree almost surely.” This conclusion does not follow. For example, the Bernoulli-valued constant paths \(X_t=U\) and \(Y_t=1-U\) above have identical path laws, but disagree at every time with probability \(1\). Equality of distributions concerns probabilities of events, not equality of coupled random elements.
Correction: To test pathwise uniqueness, first verify that the two solutions use the same stochastic basis, Brownian motion, and initial random variable. To test uniqueness in law, compare the distributions under the stated convention, without requiring the solutions to be coupled. The two tests have different hypotheses and different conclusions.
A second frequent error is to say that two solutions have “the same initial condition” when only their initial distributions agree. Pathwise uniqueness requires \(X_0=Y_0\) almost surely as random variables on the common space. In the Bernoulli example, both initial values have the same distribution, but \(U\ne1-U\) almost surely. That distinction determines whether the pair is eligible for a pathwise-uniqueness comparison.
Worked Example: Testing a Proposed Uniqueness Claim
Suppose a claim says: “For \(dX_t=0\), any two solutions whose initial values each have a Bernoulli distribution with parameter \(\tfrac12\) are indistinguishable.” Decide whether the claim is valid, and state the corrected versions.
On a common space take \(U\) Bernoulli with parameter \(\tfrac12\), set \(X_t=U\), and set \(Y_t=1-U\). Both satisfy the integral equation because \(\int_0^t0\,ds=0\) and \(\int_0^t0\,dB_s=0\). Their initial distributions are both Bernoulli with parameter \(\tfrac12\), but \(\mathbb{P}(X_t=Y_t)=\mathbb{P}(U=1-U)=0\). Therefore the proposed claim is false.
The corrected pathwise-uniqueness statement requires \(X_0=Y_0\) almost surely, as well as a shared Brownian motion and basis. Under those conditions, both integral equations reduce to \(X_t=X_0\) and \(Y_t=Y_0\), so \(X_t=Y_t\) for every \(t\). The corrected uniqueness-in-law statement is that all solutions with the prescribed Bernoulli initial law have the same path distribution: each is constant at its initial value, so the zero and one paths each have probability \(\tfrac12\).
Check: for the counterexample, \(X_t=Y_t\) would require \(U=1-U\); neither possible value \(U=0\) nor \(U=1\) satisfies that equation. For the corrected pathwise statement, substituting \(X_0=Y_0\) into the two constant-path equations gives equality at each \(t\), including \(t=0\).
These distinctions matter in existence and uniqueness theory because the construction of a solution, the comparison of two solutions under common noise, and the comparison of solution distributions are separate tasks. When reading or proving a uniqueness statement, identify the objects being compared, the coupling assumptions, and whether the law refers to \(X\) alone or to a joint tuple. That precision prevents a distributional conclusion from being mistaken for a pathwise one.
Check Your Understanding
For each question, state explicitly whether it concerns a shared realization or a comparison of distributions.
- In the definition of pathwise uniqueness, which random objects must be shared by the two solutions?
- Why does equality of the marginal path laws of \(X\) and \(Y\) not imply that \(X_t=Y_t\) almost surely?
- For \(dX_t=0\), \(X_0\) Bernoulli with parameter \(\tfrac12\), what are the probabilities of the zero path and the one path?
- Does placing two Bernoulli initial values on the same space with the same distribution suffice for a pathwise-uniqueness comparison? Explain.
- What additional information does equality of the joint law of \((\xi,B,X)\) specify beyond the marginal law of \(X\)?