Let $(M_{t})_{0\le t\le 1}$ be a continuous martingale with respect to the filtration $(\mathcal{F}_{t})_{0\le t\le 1}$. Assume that $E M_1^2<\infty$.

Fix $N$ and consider now a discrete version of this martingale, i.e., the process $(M _{n/N})_{0\le n\le N}$. Then the quadratic variation of this disrete martingale is
$$
[M^N]=\sum_{k=0}^{N-1} (M_{\frac{k+1}{N}}-M_{\frac{k}{N}})^2
$$
and its **predictable** quadratic variation (i.e., a unique increasing predictable process starting at zero such that $M^2 − \langle M\rangle $ is a martingale) is given by
$$
\langle M^N\rangle=\sum_{k=0}^{N-1} E\bigl((M_{\frac{k+1}{N}}-M_{\frac{k}{N}})^2|\mathcal{F}_{\frac{k}{N}}\bigr).
$$

Clearly, as $N\to\infty$ we have $$[M^N]\to [M].$$ My question is, is it also true that $$\langle M^N\rangle\to \langle M\rangle?$$

For example, for a Brownian motion $W$ we have $[W]_t=\langle W\rangle_t=t$ (because $W_t^2-t$ is a martingale). We also have

$$ \lim\sum_{k=0}^{N-1}(W_{\frac{k+1}{N}}-W_{\frac{k}{N}})^2=t. $$ and $$ \lim\sum_{k=0}^{N-1} E\bigl((W_{\frac{k+1}{N}}-W_{\frac{k}{N}})^2|\mathcal{F}_{\frac{k}{N}}\bigr)=\lim\sum_{k=0}^{n-1} \bigl(\frac{k+1}{N}-\frac{k}{N}\bigr)=t. $$

So the question is, is it true for any continuous martingale, or just for the Brownian motion?

continuousmartingales $[M]$ and the properly defined $\langle M \rangle$ (i.e. $M^2 - \langle M \rangle$ is alocalmartingale) are equal. $\endgroup$ – Alexander Shamov Apr 24 '15 at 15:52