From Martingale Problems to Weak SDEs

From the Generator to the Noise

A diffusion martingale problem describes a process through the martingales produced by applying test functions to it. A weak SDE describes the same process by specifying its drift and its Brownian noise. The bridge between these descriptions is the continuous local martingale part of the process: the generator determines its quadratic covariation, and an invertible diffusion matrix lets us transform that martingale part into Brownian motion.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). Let \(X\) be a continuous adapted process with values in \(\mathbb{R}^d\), and let \(b:[0,\infty)\times\mathbb{R}^d\to\mathbb{R}^d\) and \(\sigma:[0,\infty)\times\mathbb{R}^d\to\mathbb{R}^{d\times d}\) be Borel functions. Set \(a=\sigma\sigma^\top\). Assume the coefficients are locally bounded, the integrals below are finite almost surely on finite time intervals, and, for each \(T<\infty\),

Here \(\|\cdot\|_F\) is the Frobenius norm. The generator acting on a twice continuously differentiable function \(f:\mathbb{R}^d\to\mathbb{R}\) is

The Brownian integral here has progressively measurable integrand \(\sigma(s,X_s)\) in the class \(\mathcal{L}^2\): the stated expectation bound ensures \(\mathbb{E}\int_0^T\|\sigma(s,X_s)\|_F^2ds<\infty\). A weak solution need not be built from a preassigned Brownian motion; that is the distinction from a strong solution. Neither the existence of a weak solution nor this construction asserts pathwise uniqueness or uniqueness in law.

Extracting the Local Martingale and Its Bracket

Proof. Let \(\tau_R=\inf\{t\ge0:|X_t|\ge R\}\), with the infimum of the empty set taken as infinity. Choose smooth compactly supported functions that agree with \(x_i\), and then with \(x_ix_j\), on the ball of radius \(R\). Applying the martingale problem to the first set and stopping at \(\tau_R\) shows that \(X^i_{t\wedge\tau_R}-X^i_0-\int_0^{t\wedge\tau_R}b_i(s,X_s)\,ds\) is a martingale. As \(R\) increases, \(\tau_R\) tends to infinity on every finite time interval because \(X\) has continuous paths; hence \(M\) is a continuous local martingale.

For the bracket, apply the martingale problem to the test function equal to \(x_ix_j\) on the ball, and stop at \(\tau_R\). On that ball its generator is \(b_i x_j+b_j x_i+a_{ij}\). On the other hand, the product formula for the stopped coordinates gives the same drift terms and the additional term \([M^i,M^j]_{t\wedge\tau_R}\). Comparing the finite-variation terms yields \([M^i,M^j]_{t\wedge\tau_R}=\int_0^{t\wedge\tau_R}a_{ij}(s,X_s)\,ds\). Letting \(R\) increase proves the bracket identity. Finally, the assumed finite expectation of \(\int_0^T\|\sigma(s,X_s)\|_F^2ds\) implies \(\mathbb{E}\operatorname{tr}[M]_T<\infty\). A continuous local martingale with finite expected quadratic variation on \([0,T]\) is square-integrable there: its stopped versions satisfy the \(L^2\) isometry, and their bounded second moments give uniform integrability as the stopping levels increase. Thus \(M\) is a true martingale on \([0,T]\), so its increments have conditional expectation zero. \(\square\)

This identifies the covariance of the noise, but it does not yet produce a Brownian motion. That step uses invertibility.

Constructing Brownian Motion When \(\sigma\) Is Invertible

Proof. The inverse is progressively measurable along \(X\). This integral is an integral against the continuous local martingale \(M\); its integrability is checked using the bracket, rather than by requiring \(\mathbb{E}\int_0^T\|\sigma^{-1}(s,X_s)\|_F^2ds<\infty\), which need not hold. Indeed, writing \(H_s=\sigma(s,X_s)^{-1}\), the bracket identity gives

Thus the integral is well-defined, and its bracket is \([B]_t=\int_0^t H_s a(s,X_s)H_s^\top ds=tI_d\), where \(I_d\) is the identity matrix. It starts at zero. Lévy’s characterization says that a continuous local martingale starting at zero with this bracket is Brownian motion relative to the given filtration; this standard characterization is used here, not reproved.

Associativity of stochastic integration gives \(\int_0^t\sigma(s,X_s)\,dB_s=\int_0^t\sigma(s,X_s)H_s\,dM_s=M_t\), since \(\sigma H=I_d\) almost everywhere along the paths. The SDE follows from the definition of \(M\). Its Brownian integral is in \(\mathcal{L}^2\) by the expectation assumption. The Ito isometry therefore gives a square-integrable martingale integral, whose conditional expectation of each future increment given \(\mathcal{F}_s\) is zero. This proves the required representation and martingale property. \(\square\)

The inverse-integrand integral and the final Brownian integral have different integrators. In particular, the first one is justified by its bracket calculation; it is not necessary, and generally would be wrong, to assume the inverse matrix is square-integrable with respect to ordinary time.

What Changes When the Diffusion Is Singular?

For a singular diffusion matrix, a Brownian representation may still be possible, but it requires a separate argument; one cannot silently replace the inverse by a pseudoinverse and claim the same conclusion. The key issue is that the bracket determines only the covariance carried by the active directions. The theorem above uses invertibility precisely to recover all \(d\) Brownian coordinates.

Common Pitfall: Losing a Cross-Covariation

When expanding the bracket of a linear combination, the cross-covariation term appears with the coefficient dictated by bilinearity. For \(B^1=M^1-\tfrac12M^2\), the correct expansion is \([B^1]=[M^1]-[M^1,M^2]+\tfrac14[M^2]\). A tempting but incorrect substitution is \(2t-2(2t)+\tfrac14(4t)\), which evaluates to \(2t-4t+t=-t\), not \(t\). The correction is to use \([M^1,M^2]_t=2t\), not \(2(2t)\), in that formula: \(2t-2t+t=t\). A negative value for \([B^1]\) is also an immediate warning, since quadratic variation cannot be negative.

The general procedure is therefore: read the drift from the first-order part of the generator; use coordinate and product tests to identify the local martingale and its bracket; then, if the diffusion matrix is invertible, integrate its inverse against that martingale part. The result is a Brownian motion on the same filtered probability space and hence a weak SDE representation. This direction complements the calculation from an SDE to its martingale problem: one moves from the generator’s action on test functions back to a noise-driven process.