Pathwise Uniqueness on a Common Noise Space

What Is Being Compared?

An SDE specifies a drift, a diffusion coefficient, an initial condition, and a driving Brownian motion. Pathwise uniqueness asks a direct comparison question: if two solutions use the same initial random variable and the same Brownian motion on the same filtered probability space, can their sample paths differ? The requirement that the noise be shared is essential. If solutions are built on different probability spaces or use different Brownian motions, their paths are not being compared in the sense of pathwise uniqueness.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). Fix a finite horizon \(T>0\). Let \(\xi\) be an \(\mathcal{F}_0\)-measurable initial value, and consider the scalar equation with drift \(\mu\) and diffusion coefficient \(\sigma\). The integral form is

Here the Ito integrand \(s\mapsto \sigma(s,X_s)\) is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T\sigma(s,X_s)^2\,ds<\infty\). A solution is understood to be continuous and adapted, and to satisfy this integral equation almost surely. The same conditions apply to any second solution \(Y\); in particular, its Ito integrand must also be progressively measurable and in \(\mathcal{L}^2\).

“The same initial condition” means the same random variable \(\xi\), not merely two random variables with the same distribution. Likewise, “the same noise” means the same process \(B\), adapted to the common filtration. These requirements make the comparison meaningful on one stochastic basis. Pathwise uniqueness is a property of the equation under this comparison; by itself it does not assert that a solution exists.

A Direct Comparison for Additive Noise

Proof. Let \(X\) and \(Y\) be solutions on the same stochastic basis with the same \(B\) and \(x_0\). Their integral equations are \(X_t=x_0+\mu t+\int_0^t\sigma\,dB_s\) and \(Y_t=x_0+\mu t+\int_0^t\sigma\,dB_s\). The integrand is the constant process \(\sigma\), which is progressively measurable and satisfies \(\mathbb{E}\int_0^T\sigma^2\,ds=\sigma^2T<\infty\). Subtracting the two equations gives \(X_t-Y_t=0\) for all \(t\) on the event where both integral equations hold. The two stochastic integrals are the same Ito integral because both use the same integrand and Brownian motion. Thus the solutions agree almost surely throughout \([0,T]\), as required. \(\square\)

Linear Multiplicative Noise: An Integrating-Factor Proof

For a nonconstant diffusion coefficient, subtracting two integral equations does not usually make the difference vanish immediately. The linear equation \(dX_t=aX_t\,dt+\sigma X_t\,dB_t\) is a useful case in which an integrating factor gives a complete comparison. Let \(a,\sigma\in\mathbb{R}\) and let \(\xi\) be a square-integrable \(\mathcal{F}_0\)-measurable initial value. The Brownian increments are independent of \(\mathcal{F}_0\), so the usual normal moment calculation gives the needed integrability for the explicit solution below.

Proof. First define \(g(t,b)=\exp((a-\tfrac12\sigma^2)t+\sigma b)\). Its derivatives are \(g_t=(a-\tfrac12\sigma^2)g\), \(g_b=\sigma g\), and \(g_{bb}=\sigma^2g\). Ito's formula, including its second-derivative correction, gives

Thus \(X_t=\xi g(t,B_t)\) satisfies the equation. It is continuous and adapted. Because \(B_t\) is independent of \(\mathcal{F}_0\), its normal exponential moment gives \(\mathbb{E}[X_t^2]=\mathbb{E}[\xi^2]\exp((2a+\sigma^2)t)\). Consequently \(\mathbb{E}\int_0^T\sigma^2X_t^2\,dt<\infty\): the integrand is continuous in \(t\) after taking expectation, and the displayed exponential is bounded on the finite interval. The diffusion integrand is progressively measurable, so the Ito integral is defined in the required class.

Now let \(X\) and \(Y\) be any two solutions with this same \(\xi\) and \(B\). Set \(R_t=\exp(-\sigma B_t+(\tfrac12\sigma^2-a)t)\). For \(r(t,b)=\exp(-\sigma b+(\tfrac12\sigma^2-a)t)\), the derivatives are \(r_t=(\tfrac12\sigma^2-a)r\), \(r_b=-\sigma r\), and \(r_{bb}=\sigma^2r\). Ito's formula gives the correction term explicitly:

Apply the Ito product rule to \(R_tX_t\). The quadratic covariation term is \((-\sigma R_t)(\sigma X_t)\,dt=-\sigma^2R_tX_t\,dt\). The drift coefficient is therefore \(a+(\sigma^2-a)-\sigma^2=0\), and the diffusion coefficient is \(\sigma-\sigma=0\). Hence \(d(R_tX_t)=0\), so \(R_tX_t=R_0X_0=\xi\). The product calculation is valid locally even if its integrands are not globally square-integrable: stop when \(|R|\) or \(|X|\) first reaches a finite bound, apply Ito's product rule to the stopped processes, whose stopped integrands are bounded on \([0,T]\) and therefore in \(\mathcal{L}^2\), and then let the bounds increase. Continuity ensures these stopping times eventually exceed any fixed time on each continuous sample path.

The same calculation applied to \(R_tY_t\) gives \(R_tY_t=\xi\). Since \(R_t\) is strictly positive for every \(t\), division yields \(X_t=Y_t\) throughout the horizon almost surely. This proves pathwise uniqueness. \(\square\)

A Pair of Solutions That Do Not Agree

Pathwise uniqueness can fail even when the driving Brownian motion is present on the stochastic basis. The following equation has zero diffusion, so the same Brownian motion is available but does not affect either solution.

What the Comparison Does—and Does Not—Say

In each uniqueness proof, the common noise is used in a concrete way. For constant coefficients, the stochastic integrals cancel because they are identical. For the linear multiplicative equation, the integrating factor is built from that shared Brownian motion and cancels the stochastic terms in the product. In the counterexample, both solutions use the same Brownian motion, but the diffusion coefficient is zero and the drift equation itself allows more than one solution.

Do not replace the common-noise requirement with a statement that two solutions have the same distribution. Pathwise uniqueness compares processes on one stochastic basis using the same initial random variable and Brownian motion. A comparison of distributions, including comparisons across different probability spaces, is a different question. Also, uniqueness is not existence: the definition only says what must happen if two solutions are available.

Common mistake: Suppose one writes “the initial distributions match, so the solutions have the same initial condition,” and then subtracts their equations. That subtraction is not justified if the initial random variables differ, even when they have the same law. The valid comparison starts with \(X_0=Y_0=\xi\) as the same random variable and uses the same \(B\). For example, two constant solutions \(X_t=U\) and \(Y_t=V\) of \(dZ_t=0\) can have identical initial distributions while differing on a sample point if \(U\) and \(V\) are distinct random variables. This does not contradict pathwise uniqueness, whose hypothesis requires the same initial random variable.

When checking a claimed pathwise uniqueness result, identify all four items before calculating: the stochastic basis, the common filtration, the exact initial random variable, and the Brownian motion shared by the solutions. Then write both integral equations and see whether the difference can be controlled or eliminated. Conversely, to disprove the property, it is enough to exhibit two admissible solutions on one such common basis and verify that they fail to agree.