Let's assume that we have a discrete martingale $(M_n)_{n\in\mathbb{N}}$ with respect to a filtration $(\mathcal{F}_n)_{n\in\mathbb{N}}$. It takes only real values.
We define the discrete bracket $\langle M\rangle_n=\sum\limits_{i=1}^{n-1} \mathbb{E}(M_{k+1}^2-M_k^2|\mathcal{F}_k)$.
Of course, if $\langle M\rangle_n$ is bounded in $L^2$ then $M_n$ converges in $L^2$.
However if we only assume that $\langle M\rangle_n$ converges almost surely toward a finite random variable, is it true that $M_n$ converges almost surely?
If it is true in the discrete case, is it true in the continuous case?
Best Answer
Yes, this is true in both the discrete and continuous setting, assuming $M$ is continuous. Let $\tau_m := \inf \{t : \langle M \rangle_t > m\}$ and $\lim_{t \rightarrow \infty} \langle M \rangle_t = \langle M \rangle_\infty$. Then $\langle M \rangle^{\tau_m}$ is bounded and in particular in $L^2$ so $\lim_{t \rightarrow \infty}M_t^{\tau_m}$ exists almost surely. But since for almost every $\omega$ we have $\langle M \rangle_\infty(\omega) < \infty$, we know $\tau_m(\omega) =\infty$ for all $m$ sufficiently large, so $\lim_{t \rightarrow \infty} M_t$ exists as well.