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?

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    $\begingroup$ There is absolutely no reason for $\mathsf{E} \left[ (M_t - M_s)^2 \, \middle| \, \mathcal{F}_s \right]$ to be almost surely finite, so your statement is certainly not true without additional integrability assumptions. That being said, I believe your statement is indeed true for martingales that are bounded in the $L^2$ norm. $\endgroup$ – Alexander Shamov Apr 24 '15 at 15:45
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    $\begingroup$ Besides, for continuous martingales $[M]$ and the properly defined $\langle M \rangle$ (i.e. $M^2 - \langle M \rangle$ is a local martingale) are equal. $\endgroup$ – Alexander Shamov Apr 24 '15 at 15:52
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    $\begingroup$ On second thought, since you define $\langle M \rangle$ in such a way that it doesn't always exist, it's not even clear what your statement means... $\endgroup$ – Alexander Shamov Apr 24 '15 at 16:06
  • $\begingroup$ Alexander, thanks for the comment. Yes, I assume that $EM_1^2<\infty$ and in this case we clearly have $[M]=\langle M\rangle$. Why this implies that $\langle M^N\rangle$ converges to $\langle M\rangle$? $\endgroup$ – Oleg Apr 30 '15 at 12:50

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