Theorem 3.4.2 (Equivalence of Formulations).label Let $(\Omega, \bracsn{\cf_t|t \ge 0}, \bp)$ be a filtered probability space, $B$ be a $\bracs{\mathcal{F}_t}$-Brownian motion, $\sigma: [0, \infty) \times C([0, \infty); \real^{n}) \to L(\real^{n}; \real^{n})$ and $b: [0, \infty) \times C([0, \infty); \real^{n}) \to \real^{n}$ be bounded measurable functions such that $\sigma_{t}$ is invertible for all $t \ge 0$.

Let $y \in \real^{n}$ and $X: \Omega \to C([0, \infty); \real^{n})$ be a solution to the martingale problem for $(\sigma^{*}\sigma, b)$ starting at $y$, then there exists a weak solution of the SDE

\[Y_{t} = Y_{0} + \int_{0}^{t} \sigma(s, Y) dB_{s} + \int_{0}^{t} b(s, Y)ds\]

starting at $y$ whose distribution is the same as $X$.

Proof, [Theorem 20.1, RW89]. By truncation, for each $1 \le i \le n$,

\[M_{t} = X_{t} - \int_{0}^{t} b(s, X)ds\]

and

\[M_{t}M_{t}^{T} - \int_{0}^{t} a(s, X)ds\]

are local martingales. In which case, by Lévy’s characterisation of Brownian motion,

\[\td B_{t} = \int_{0}^{t} \sigma(s, X)^{-1}dM_{s}\]

is a standard Brownian motion, and $X$ and $\td B$ satisfy the SDE.$\square$