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. ?
Asked
Active
Viewed 1,845 times
6
saz
- 123,507
chandu1729
- 3,871
-
I deleted my answer. A remark in Revuz-Yor says that a $L^1$ bounded martingale may not have integrable supremum. If you have that $M_t \log^+ M_t$ is bounded in $L^1$ then the expectation of the supremum is bounded and you can apply my argument. Maybe something more subtle is needed to prove the general case – Bunder Oct 17 '13 at 07:49
2 Answers
6
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.
Nate Eldredge
- 101,664
-
-
Nate Eldredge: Indeed very nice proof. Thank's I've learned something today. Best regards – TheBridge Oct 21 '13 at 08:54
-
Hi, may I ask why is $M_t^2 - \langle{M}\rangle_t$ a martingale? It doesn't seem to be true even if we consider discrete-time martingales. Let $X$ be a random variable such that $\mathbb{E}[|X|] < \infty$ but $\mathbb{E}[X^2] = \infty$. Then let $M_t = X$ for all $t$. Then clearly $M_t$ is a martingale which is $L^1$-bounded, but $M_t^2 - \langle{M}\rangle_t$ is not $L^1$ bounded, so it is not a martingale (of course in this case it's trivial that the quadratic variation is bounded, but I'm saying that your argument appears to break down). – Clement Yung Feb 14 '21 at 14:41
-
@ClementYung: I'm assuming here that $M_t$ is an $L^2$ martingale. This is normally assumed whenever you talk about quadratic variation, because you need $L^2$ for it to even make sense. So to be clear, we are assuming that each $E[M_t^2]$ is finite (but they could be unbounded in $t$), and that $\sup_t E[|M_t|] < \infty$ ($L^1$ bounded). And for $L^2$ martingales it is true that $M_t^2 - \langle M \rangle_t$ is a martingale; this is often taken as the definition of $\langle M \rangle_t$. – Nate Eldredge Feb 14 '21 at 15:55
1
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.
mike
- 1,319