$L^1$ bounded martingale

Question

If $(M_t)_{0\leq t<\infty}$ is continuous martingale and it is $L^1$ bounded, does it imply that quadratic variation $\langle M\rangle_\infty$ is finite a.s. ?

Answer 1

Yes, it does. I will flesh out what mike probably had in mind.

For $b > 0$, let $T_b = \inf\{t : |M_t| \ge b\}$. Since $M_t$ is continuous, for any $t$ we have $|M_{T_b \wedge t}| \le b$. Now $M_t^2 - \langle M \rangle_t$ is a martingale, so by optional stopping we have, for any $t$, $$E[\langle M \rangle_{T_b \wedge t}] = E[M_{T_b \wedge t}^2] \le b^2.$$

Letting $t \to \infty$ and using Fatou's lemma, we have $E[\langle M \rangle_{T_b}] \le b^2 < \infty$; in particular, $\langle M \rangle_\infty < \infty$ a.s. on $\{T_b = \infty\}$. (Note of course that $\langle M \rangle_t$ is an increasing process, so $\langle M \rangle_\infty$ makes sense.) So $\langle M \rangle_\infty < \infty$ almost surely on the event $A := \bigcup_{b \in \mathbb{N}} \{T_b = \infty\}$. But $A$ is precisely the event that $M_t$ is bounded (that there exists some $b$ which is never reached; here we have used continuity of $M_t$ again).

But $M_t$ is $L^1$ bounded, so by the martingale convergence theorem, $M_t$ converges almost surely. Thus $M_t$ is almost surely bounded, and we have $P(A) = 1$, which completes the proof.

Answer 2

yes, since it converges, being L1 bounded, its sample paths will be the same as those of the process stopped when you hit $\pm K$ for a large K with high probability, and so are the same, with high probability, as those of a $\mathbb L^2$ bounded martingale.