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In Stochastic Calculus for Finance II: Continuous-time Models by Steve Shreve,

Theorem 3.4.3. Let $W$ be a Brownian motion. Then $[W, W](T) = T$ for all $T > 0$ almost surely.

where $[W, W](T)$ is the quadratic variation of $W$ up to time T $$ [W, W](T): = \lim_{||\Pi|| \to 0} \sum_{j=0}^{n-1} [W(t_{j+1}) - W(t_j)]^2 $$ where $$\Pi=[t_0, t_1, \dots, t_n] \text{ and } 0 \leq t_0 < t_1 < \cdots < t_n = T$$ and $$||\Pi||= \max_{j=0,\dots, n-1} (t_{j+1} -t_j).$$

One fellow student also says a stronger conclusion is also true, i.e.

$[W, W](T)$ converges to $T$ in $L^2$ or in $L^p, p>1$, for all $T > 0$.

By "stronger", I mean I heard it implies the original theorem for convergence a.e..

I wonder if there are some texts or online tutorials for proving this stronger conclusion, and/or if you could kindly provide the proof here?

Thanks!

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    An $L^2$ convergence argument is given here: http://math.stackexchange.com/questions/59547/quadratic-variation-of-brownian-motion-and-almost-sure-convergence2011-11-07
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    @ByronSchmuland: Thanks! Does the link talk about convergence in $L^2$ or a.e.? All I see is latter.2011-11-07
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    Sorry I wasn't clearer. In the problem description, before he gets to points 1) and 2), the OP gives an argument that concludes "The quadratic variation thus converges in mean-square to t". This is the $L^2$ convergence.2011-11-07
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    @ByronSchmuland: Thanks! In this case, does convergence in $L^2$ imply convergence in a.e., or the other way around, or neither implies the other?2011-11-07
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    Neither implies the other.2011-11-07
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    It does *not* converge almost everywhere. Only for certain sequences of partitions do you get a.e. convergence, such as refining sequences of partitions.2011-11-08
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    And it will converge in $L^p$ for all $p < \infty$. This implies convergence in probability but not almost sure (everywhere) convergence.2011-11-08
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    Also, "$[W,W](T)$ converges to $T$ in ..." is not a meaningful sentence, because $[W,W](T)$ does not refer to the sequence itself, so it isn't something which converges. Rather, it refers to the limit (in probability), which is just a random variable. The theorem quoted is using "almost surely" to refer to equality of the random variable $[W,W](T)$ with $T$, and is not a statement on how the sequence converges.2011-11-08
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    @GeorgeLowther: Thanks! "It does not converge almost everywhere", but the quoted theorem from Shreve's book says so. Or do I misunderstand something?2011-11-08
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    @Ethan: My previous comment should answer that.2011-11-08
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    @Ethan: I have to log off now, and can check back tomorrow night. Hopefully my comment clears up some misunderstandings, and I can post a full answer if you want.2011-11-08
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    @George: Thank you so much! Yes, it does clarify my misunderstanding. If you could post a full answer, that will be great, and I can upvote it.2011-11-08

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