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I've been trying to understand diffusions. We can show they exist by noting they solve particular SDEs, but are they unique? More precisely:

Fix a filtered probability space satisfying the usual conditions and locally bounded measurable functions $a_{i,j}$ and $b$ from $\mathbb{R}^d$ to $\mathbb{R}$, such that $(a_{i,j}(x))_{i,j}$ is symmetric non-negative definite for all $x\in\mathbb{R}^d$. Define the operator $L$ by

$Lf(x)=\frac{1}{2}\sum_{i,j}a_{i,j}(x)\frac{\partial f}{\partial x_ix_j}+\sum_ib_i(x)\frac{\partial f}{\partial x_i}$

Suppose that the continuous, adapted process $X$ is such that, for all $f\in C^2(\mathbb{R}^d)$,

$f(X_t)-f(0)-\int_0^tLf(X_s)ds$

is a local martingale, and $X_0=0$.

Then is $X$ unique in law?

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    I think _Diffusions, Markov Processes and Martingales_ by Rogers and Williams is a good book. In Chapter V they introduce the martingale problem and later they prove equivalence of solutions to SDE's and solutions to the corresponding martingale problem. You could also take a look at _Brownian Motion and Stochastic Calculus_ by Karatzas and Shreve around page 317.2012-03-29

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Here's a common counterexample to uniqueness that usually appears in the context of SDEs, e.g. Oksendal. The equivalence of SDEs and Martingale Problems is explained in this paper by Kurtz and the more technical book Ethier & Kurtz. Put $d=1$, $a\equiv 0$, $b(x)=2|x|^{\frac{1}{2}}$ and consider the (deterministic) process $X_t= \begin{cases} 0 & \text{if }\hspace{2mm} 0\leq t \leq \tau \\ (t-\tau)^2 & \text{if }\hspace{2mm} t>\tau \end{cases}$ for any $\tau>0$. Certainly $X$ is continuous and adapted and $L$ satisfies the required hypotheses. For any $f\in C^2(\mathbb{R})$ we have $f(X_t)-f(0)-\int_0^t2\sqrt{|X_s|}f'(X_s)ds=f(0)-f(0)=0$ when $t\leq\tau$. Moreover, $f(X_t)-f(0)-\int_0^t2\sqrt{|X_s|}f'(X_s)ds=f((t-\tau)^2)-f(0)-\int_{\tau}^t2(t-\tau)f'((t-\tau)^2)ds=f((t-\tau)^2)-f(0)-\Big(f((t-\tau)^2)-f(0)\Big)=0$ when $t>\tau$. Hence $f(X_t)-f(0)-\int_0^tLf(X_s)ds\equiv 0$ is a martingale. The fact that each $\tau>0$ leads to a different solution violates uniqueness. Note that requiring $a$ be positive definite or a Lipschitz $b$ coefficient would preclude this counterexample.