Four Different Questions About an SDE
Existence and uniqueness for a stochastic differential equation are not a single property. Existence can refer either to a solution built using a specified Brownian motion or to a solution on some probability space chosen for the construction. Uniqueness can compare two solutions driven by the same Brownian path or compare the probability laws of solutions. These distinctions matter: one equation can have solutions in several senses while failing others.
Consider the scalar SDE on a fixed finite horizon \([0,T]\), with \(T>0\), initial value \(\xi\), drift \(\mu\), and diffusion coefficient \(\sigma\):
All stochastic integrals here are Ito integrals. On a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), the driver \(B_t\) must be Brownian motion relative to the filtration: in particular, its increments after time \(s\) are independent of \(\mathcal{F}_s\). A solution \(X_t\) is continuous and adapted, and satisfies the displayed integral equation almost surely for all \(t\). We require the drift integral to be finite almost surely and, on each finite interval under consideration, \(\sigma(s,X_s)\) to be progressively measurable with \(\mathbb{E}\int_0^T\sigma(s,X_s)^2\,ds<\infty\). These conditions make the Ito integral well defined. More general local formulations are possible, but the examples here meet these integrability requirements.
Definitions: Which Space, Which Comparison?
The word “strong” refers to the solution being adapted to the given noise and initial data; it does not mean that the equation has a unique solution. Weak existence allows the probability space to be selected as part of the construction. When comparing these notions, the law of the initial value must be held fixed. For simplicity, the examples below use deterministic initial values.
Pathwise uniqueness is a comparison on one common noise space. Uniqueness in law compares distributions and does not require two solutions to use the same space or the same Brownian path. Equality at each fixed time almost surely is not the definition of pathwise uniqueness; the required conclusion is indistinguishability, meaning equality for every time outside one null event.
A Direct Implication: Strong Existence Gives Weak Existence
Proof. A strong solution on the prescribed basis already comes with a probability space, filtration, initial variable, and Brownian motion relative to that filtration. By the definition of strong solution, it is adapted and satisfies the SDE's integral equation; the assumed integrability conditions ensure the displayed integrals are defined. These are exactly the ingredients required for a weak solution. Retaining the same basis and process, but now regarding the basis as one possible choice rather than a prescribed one, proves weak existence. \(\square\)
This implication is conditional on the prescribed basis being a valid stochastic basis for the equation. It does not say that a weak solution can always be realized on an arbitrary preassigned basis.
Worked Example: The Zero Equation Has All Four Properties
Take \(\xi=3\), \(\mu(t,x)=0\), and \(\sigma(t,x)=0\). Verify the solution and classify the four assertions.
The integral equation becomes \(X_t=3+\int_0^t0\,ds+\int_0^t0\,dB_s\). Both integrals vanish, so \(X_t=3\) for every \(t\). This process is adapted on any valid basis, and its stochastic integrand is progressively measurable with \(\mathbb{E}\int_0^T0^2\,ds=0\). Thus strong existence holds on every such prescribed basis, and weak existence follows as well.
For pathwise uniqueness, any solution \(Y\) on any common basis must satisfy the same equation \(Y_t=3\) for all \(t\); hence \(X_t=Y_t\) for every time. Every weak solution also has the deterministic path \(t\mapsto3\), so all its path laws are the point mass at that path. Uniqueness in law holds. Check: substitution of \(X_t=3\) into the integral equation gives \(3=3+0+0\), as required.
Strong Existence Does Not Mean Pathwise Uniqueness
Worked Example: A Deterministic SDE with Several Solution Paths
Let \(\xi=0\), \(\mu(t,x)=2\sqrt{x^+}\), and \(\sigma(t,x)=0\), where \(x^+=\max(x,0)\). On \([0,1]\), compare \(X_t=t^2\) with \(Y_t=(t-\tfrac12)_+^2\).
For \(X\), the derivative is \(2t\), while \(2\sqrt{X_t^+}=2\sqrt{t^2}=2t\) because \(t\ge0\). Also \(X_0=0\). Therefore, for each \(t\in[0,1]\),
For \(Y\), when \(0\le s\le\tfrac12\), \(Y_s=0\), so its derivative and the drift \(2\sqrt{Y_s^+}\) are both zero. When \(s>\tfrac12\), \(Y_s=(s-\tfrac12)^2\), its derivative is \(2(s-\tfrac12)\), and \(2\sqrt{Y_s^+}=2(s-\tfrac12)\). Thus
Both processes are deterministic and adapted on any basis, so strong existence holds. They differ, for example, at \(t=\tfrac14\): \(X_{1/4}=\tfrac1{16}\), whereas \(Y_{1/4}=0\). They also differ at \(t=1\): \(X_1=1\), whereas \(Y_1=\tfrac14\). Consequently pathwise uniqueness fails. Since the two paths are deterministic, their laws are distinct point masses, so uniqueness in law fails too. Check: the initial values are \(X_0=0^2=0\) and \(Y_0=0\); the integral identities above verify the equation at every time, including \(t=\tfrac12\).
This example establishes that strong existence alone implies neither pathwise uniqueness nor uniqueness in law. It also illustrates why the two uniqueness questions must be stated separately: here both fail, but they test different comparisons. The example does not show that one failure always entails the other.
Weak Existence and Uniqueness in Law Do Not Guarantee Strong Existence
Worked Example: The Tanaka Equation as a Standard Counterexample
Consider the Tanaka SDE with deterministic initial value zero, drift \(\mu(x)=0\), and diffusion coefficient \(\sigma(x)=1\) for \(x\ge0\) and \(\sigma(x)=-1\) for \(x<0\):
The coefficient satisfies \(\sigma(x)^2=1\) for every \(x\), so any solution has \(\mathbb{E}\int_0^T\sigma(X_s)^2\,ds=T<\infty\). This equation is a standard counterexample with weak existence and uniqueness in law, but without a strong solution on the Brownian filtration and without pathwise uniqueness. These properties of the Tanaka equation are a standard result; their proof is beyond this tutorial. The point here is the classification: weak existence and uniqueness in law can both hold while strong existence and pathwise uniqueness fail.
The distinction is not a matter of the stochastic integral convention: this is an Ito SDE. Nor does uniqueness in law assert that two solutions driven by one Brownian motion coincide. It says only that every weak solution has the same distribution. Check: the coefficient's square is \(1\) on both branches, including at zero; hence the stated square-integrability condition is exactly \(\mathbb{E}\int_0^T1\,ds=T\).
A Theorem That Connects the Notions
The properties are separate assertions, but a precise combination of them has strong consequences. The following is the standard Yamada–Watanabe theorem in its usual one-dimensional Brownian formulation. It is stated here as a standard result; its proof is beyond this tutorial.
This theorem does not identify the definitions. Instead, it provides a nontrivial implication from a combination of two properties. In particular, one should not reverse it: strong existence by itself does not supply pathwise uniqueness, as the deterministic example showed. Nor should one infer strong existence from weak existence alone, as the Tanaka example rules out. Technical existence and uniqueness theorems often add assumptions such as global Lipschitz continuity and linear growth of the coefficients; those are sufficient-condition frameworks, not part of the definitions above.
A reliable classification begins by fixing the initial law and the basis being considered, then asking four distinct questions:
- Does a solution exist on the prescribed basis?
- Does at least one weak solution exist on some admissible basis?
- Do any two solutions on the same basis with the same driver agree indistinguishably?
- Do all weak solutions have the same path distribution?
Common Mistakes and a Classification Checklist
Mistake: “There is a solution, therefore the solution is unique.” For the branching equation, \(X_t=t^2\) is a strong solution, but \(Y_t=(t-\tfrac12)_+^2\) is another strong solution on the same basis with the same Brownian motion. The assertion of existence has no uniqueness conclusion.
Mistake: “The laws agree, therefore solutions driven by the same noise agree.” This confuses uniqueness in law with pathwise uniqueness. The former compares distributions; the latter requires almost-sure equality of two processes on a common basis. The Tanaka equation is a standard example where uniqueness in law holds but pathwise uniqueness does not.
Mistake: “A weak solution can be transferred to any chosen Brownian basis.” Weak existence allows the basis to vary. A strong solution must be adapted to the prescribed filtration and built from the prescribed initial data and Brownian motion. The two existence questions therefore have different requirements.
| Assertion | What it asks | What it does not assert by itself |
|---|---|---|
| Strong existence | A solution on a specified basis | Pathwise uniqueness or uniqueness in law |
| Weak existence | A solution on at least one admissible basis | Existence on a prescribed basis |
| Pathwise uniqueness | Same-basis solutions with the same data coincide | Existence of any solution |
| Uniqueness in law | All weak solution path laws agree | Almost-sure equality on a common basis |
Check Your Understanding
For each question, keep the initial law and the role of the probability space explicit.
- Which property compares two solutions driven by the same Brownian motion on the same filtered probability space?
- For the equation \(dX_t=2\sqrt{X_t^+}\,dt\), \(X_0=0\), verify that \(X_t=(t-a)_+^2\) solves the equation for a fixed \(a\ge0\).
- Why does strong existence on a valid prescribed basis imply weak existence, but not conversely by definition?
- Can uniqueness in law be concluded merely from the existence of one strong solution?
- State the two hypotheses in the Yamada–Watanabe theorem used here and its two stated consequences.