Question: Given a real-valued continuous increasing predictable process $A$, is there a real-valued continuous locally square-integrable martingale $M$, such that $\langle M\rangle = A$ ? If so, is this $M$ unique in distribution, that is, if there is an $N$ fulfill the assumption, then $M$ and $N$ has the same law in the path space ?
The angle bracket $\langle M\rangle$ denotes the predictable quadratic variation of $M$, that is, the unique continuous increasing predictable process starting at zero such that $M^2 − \langle M\rangle$ is a local martingale.
Motivation: This question arises when I was reading the book of J. Jacod & A. N. Shiryaev. They define the characteristics of semimartingales in Section II.2. Given a truncation function $h$, the characteristics of a semimartingale $X$ associated with $h$ is a unique triplet $(B,C,\nu)$. So a natural question is: can a triplet $(B,C,\nu)$ for a fixed truncation function $h$ determine a semimartingale uniquely in the sense of distribution? The begining question is a special case and part of this question.
Part of the answer: If the increasing process $A=\{A_t\}_{t\ge0}$ in the question is an absolutely continuous function of $t$ a.s., then the martingale representation theorem asserts that there is a stochastic integral with respect to a Brownian motion such that the angle bracket of this stochastic integral is $A$.
I think you can let $B$ be a Brownian motion and set $M_t = B_{A_t}$. You need to be a bit careful about the filtrations to make that work, but that can be handled if you look at the proofs about continuous martingales as time changed Brownian motion. I'm not sure whether or not this is unique in distribution, though.
Edit: This $M$ is not unique. Let $B_t$ be a Brownian motion, set $\tau := \inf\{t : B_t = -1\}$, and let $A_t := t \wedge \tau$. The processes $M_t := B_{t \wedge \tau}$ and $N_t := - B_{t \wedge \tau}$ are both continuous locally square integrable martingales with $\langle M \rangle_t = \langle N \rangle_t = t \wedge \tau = A_t$, but $\mathbb{P}(M_t < -1) = 0$ and $\mathbb{P}(N_t < -1) > 0$.