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I am reading Karatzas and Shreve's Brownian Motion and Stochastic Calculus. Let $M_t$ be a continuous local martingale. On page 157, it wrote that "because $\langle M\rangle_t = t$, we have $M \in \mathcal{M}_2^c$", where $\mathcal{M}_2^c$ means the collection of continuous square integrable martingale. Can you tell me why it is true?

It is true that a local martingale of class DL is a martingale. However, I do not think that the condition there is concerned with class DL or uniformly integrability, because as you know, even a continuous, local martingales with uniformly integrability fail to be martingale.

Sincerely.

saz
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Mark
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  • It's been a long time since I studied this, but my guess is that square integrability gives you more than uniform integrability could. See Lemma 3 here: http://almostsure.wordpress.com/2010/03/29/quadratic-variations-and-the-ito-isometry/ –  Jun 09 '12 at 18:34
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    the fact that martingale is continuous means that you can localize by bounding it, so you have a nice, square integrable martingale up to $T$, with QV $min(T,t)$ then for the localized guy $X_{min(T,t)}^2 - min(T,t)$ is a bounded martingale, $\mathbb E X_{min(T,t)}^2 = \mathbb E min(T,t)$, let $T \rightarrow \infty$ se Fatou to get that $X_t$ is square integrable martingale. Spoiler: it turns out to be brownian motion. – mike Jun 09 '12 at 19:27
  • @mike Thank you for your answer. What I confused now is that you write "with QV $\min(T,t)$", but what in the book is "$_t$ = t". Your explanation is very helpful, but the thing is not clear for me. – Mark Jun 10 '12 at 11:56
  • @Kovalev Thank you for your advice. I read through the page, of course the Lemma 3. Since the definition of "integrability" is not in the link, I can't grab the Lemma so well. If the defintion of "integrability" is as usual, then "t" fail to be integrable on the non-compact intervel like $[0, \infty)$. Then the logic in the book is not correct. How do you think about it? – Mark Jun 10 '12 at 12:06
  • I am localizing to get off the ground. Since I know I am working with a bounded martingale, it is $\mathbb L^2$ bounded , u.i., etc, however it is a stopped version of the original and its QV is $\langle X \rangle_{T \wedge t}$. This is sort of obvious, when you stop the process you stop its QV as well, but a rigorous proof is also easy. But the main idea is to use continuity to bootstrap from a bounded martingale to the one with QV t. – mike Jun 15 '12 at 17:39

2 Answers2

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This is much easier than your comments suggest:

$$\mathbb{E}(M_t^2)=\mathbb{E}(\langle M \rangle_t)=t<\infty$$

John Fernley
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    It should be $$\operatorname E\left[\left(M_t-M_0\right)^2\right]=\operatorname E\left[\langle M\rangle_t\right];;;\text{for all }t\ge0;.$$ You're implicitly assuming $M_0=0$ almost surely. – 0xbadf00d May 23 '17 at 15:35
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According to Levy's characterisation of Brownian motion, any continuous local martingale $M$, with $M_0=0$ and $\langle M\rangle_t=t$, for all $t\geq 0$, is a standard Brownian motion. In particular, it is thus a square-jntegrable martingale.

HardyHulley
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