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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.

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    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.2012-06-15

2 Answers 2

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This is much easier than your comments suggest:

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

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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.2017-05-23
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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.

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    The question is exactly on the proof of the Lévy's characterisation of Brownian Motion.2017-09-28