Conditional Laws in the Yamada-Watanabe Argument

Why Condition on the Driving Brownian Path?

A weak solution is allowed to live on a probability space that contains randomness beyond the Brownian motion. The key question in the Yamada–Watanabe argument is whether that extra randomness can actually affect the solution. Conditioning on the entire Brownian path isolates the possible randomness left in the solution after the driver is known. If pathwise uniqueness holds, two solutions sampled independently from that conditional law must nevertheless agree. That forces the conditional law to put all its mass on a single path.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider the scalar SDE with deterministic initial value \(x_0\), \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), on \(0\le t\le T\). For this tutorial, assume \(\mu\) and \(\sigma\) are Borel measurable and bounded. The drift integral is an ordinary time integral; the Ito integral is defined for progressively measurable integrands \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Boundedness ensures that \(\mu(t,X_t)\) and \(\sigma(t,X_t)\) meet this integrability requirement. A weak solution is a continuous adapted process \(X\), on some filtered probability space carrying \(B\), that satisfies the integral SDE almost surely for every \(t\in[0,T]\).

Regard \(B\) and \(X\) as random elements of the continuous-path spaces \(C([0,T],\mathbb{R})\). These are Polish spaces, so a regular conditional distribution of \(X\) given \(B\) exists. Denote it by \(K(w,dx)\), where \(w\) is a Brownian path and \(x\) is a candidate solution path. Thus, for Borel sets \(A\) and \(C\), \(\mathbb{P}(B\in A,X\in C)=\int_A K(w,C)\,\mathbb{P}_B(dw)\), where \(\mathbb{P}_B\) is Brownian path measure.

The Conditional-Law Coupling

There is an important filtration issue in this construction. Conditioning on the entire Brownian path must not make future Brownian increments observable at earlier times. Adaptedness of the original solution prevents this. For each \(t\), the conditional distribution of the solution up to time \(t\), given the full Brownian path, depends only on the Brownian path up to time \(t\). Indeed, future Brownian increments are independent of \(\mathcal{F}_t\), which contains \(X\) up to time \(t\). This nonanticipation property is what allows the coupled Brownian motion to remain Brownian relative to the joint past of both solution paths.

Here and below, the regular conditional distribution theorem for random elements of Polish spaces is used as a standard result. We also use the standard transfer property of stochastic integrals: if a pair of continuous adapted processes has the same joint law, the Ito integral, defined by limits of adapted simple-integrand sums, transfers with that law. The conditional-law argument itself is given explicitly.

Proof. Let \(\mathbb{P}_B\) be the law of the Brownian path and define a probability measure on triples \((w,x^1,x^2)\) by \(\mathbb{P}_*(dw,dx^1,dx^2)=\mathbb{P}_B(dw)K(w,dx^1)K(w,dx^2)\). Under \(\mathbb{P}_*\), conditional on \(w\), the paths \(x^1\) and \(x^2\) are independent and each has conditional law \(K(w,\cdot)\). In particular, each pair \((w,x^i)\) has the same joint law as \((B,X)\) under the original weak solution.

Equip the triple with the filtration generated by \(w_s,x^1_s,x^2_s\) for \(s\le t\), with the usual completion. The nonanticipation property stated above implies that, given the past of all three paths, the conditional law of \(w\)'s future increments is its ordinary Brownian law given \(w\) up to time \(t\). Thus \(w\) is Brownian in this filtration. Each \(x^i\) is adapted to it, and each pair \((x^i,w)\) retains the original weak-solution law. Therefore both \(x^1\) and \(x^2\) solve the SDE with the same Brownian motion \(w\). The integral identities transfer because the Ito integrals are limits of adapted sums and the joint laws of the respective pairs agree with the original pair. Pathwise uniqueness now gives \(x^1=x^2\), \(\mathbb{P}_*\)-almost surely.

Disintegrating this last equality with respect to \(w\) shows that for \(\mathbb{P}_B\)-almost every \(w\), two independent draws from \(K(w,\cdot)\) agree almost surely. A probability measure on a Polish space with this property is a point mass: if it were not a point mass, there would be a Borel set \(C\) with probability \(p\) strictly between \(0\) and \(1\), and two independent draws would fall on opposite sides of \(C\) with probability \(2p(1-p)>0\), contradicting their almost-sure equality. Hence \(K(w,\cdot)=\delta_{F(w)}\) for almost every \(w\), for a measurable choice of its unique support point \(F(w)\).

It follows that \(X=F(B)\) almost surely. Moreover, the conditional law of \(X\) up to time \(t\) depends only on \(B\) up to time \(t\); since that conditional law is a point mass, the restriction \(F(B)|_{[0,t]}\) is measurable with respect to the completed Brownian filtration at time \(t\). Thus \(F(B)\) is adapted and is a strong solution. Finally, any weak solution with these coefficients and initial value has the same argument: its conditional law given its Brownian path is a point mass, and the resulting solution functional must agree almost surely with the first one when driven by the same Brownian motion, by pathwise uniqueness. In particular, the solution law is unique. \(\square\)

Three Calculations Behind the Argument

What the Argument Does—and Does Not—Assume

The construction does not assume that a solution is already a function of the Brownian motion. It begins with a conditional distribution that may assign positive probability to many solution paths. The two-copy coupling tests whether this remaining randomness is compatible with pathwise uniqueness. If it is not, the conditional distribution collapses to one path for almost every driver.

Common mistake. It is not valid to say that two solutions with the same Brownian motion are conditionally independent and therefore must be different. Conditional independence alone gives no such conclusion: two draws from a point mass are independent and equal. The relevant order is to construct conditionally independent copies, apply pathwise uniqueness to show they agree almost surely, and then infer that the conditional law is a point mass.

Nor should one identify pathwise uniqueness with uniqueness in law. Pathwise uniqueness compares solutions on the same space with the same Brownian motion. Uniqueness in law compares distributions of solutions, potentially constructed on different spaces. In this argument, the common-driver coupling first uses pathwise uniqueness; the point-mass conditional law then yields a strong solution and, in the stated setting, uniqueness in law.