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?
Best Answer
It is true that $\langle M\rangle^n_1 \to \langle M\rangle_1$ in $L^1$ when $M$ is a continuous square-integrable martingale. In fact, the result holds even if $M$ is cadlag, as long as $\langle M\rangle$ is continuous.
Indeed, $M^2$ is then a submartingale of class D and so, since
$$E[(M_{t_{i+1}} - M_{t_{i}})^2 |\mathcal{F}_{t_i}]=E[M_{t_{i+1}}^2 - M_{t_{i}}^2 |\mathcal{F}_{t_i}],$$
the result follows from the analogous results for the predictable component of the Doob decomposition of a submartingale of class D, which was proved in
and which constitutes Theorem 31.2, chapter 6 of
When $M$ is cadlag (and $\langle M\rangle$ is not continuous), $\langle M\rangle^n_1$ converges to $\langle M\rangle_1$, but only in the $\sigma\left(L^{1}, L^{\infty}\right)$-topology, see for example
In this case, it follows from Mazur's lemma that one can obtain (strong) convergence in $L^1$ by replacing $(\langle M\rangle^n_1)_n$ with a forward convex combinations thereof. In fact, taking forward convex combinations, one can obtain convergence in $L^1$ simultaneously at all times $t\in [0,1]$, as I showed in