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Say that $W(t)$ is a Brownian motion. The quadratic variation $[W,W](t)$ is defined in terms of a partition $\Pi = \{0 = t_0 < t_1 < \cdots < t_n = t\}$ by
$$ \begin{split} [W,W](t) &= \lim_{|\Pi|\to 0} \sum_{j=0}^{n-1} \Big( W(t_{j+1}) - W(t_j) \Big)^2\\ &= \lim_{|\Pi|\to 0} Q_n \end{split} $$

Here $|\Pi|\to 0$ means that $\displaystyle\mathop{\text{max}}_{0\leq j<n} (t_{j+1} - t_j) \to 0$.

One can argue that $[W,W](t) = t$ by noting that the expected value and variance of the $j$th summand are $t_{j+1} - t_j$ and $2(t_{j+1} - t_j)^2$, respectively, so that $E[Q_n] = t$. One argues that $Var[Q_n]$ is proportional to the maximum partition length, and so vanishes in the limit. The quadratic variation thus converges in mean-square to $t$: $$ \lim_{|\Pi|\to 0} E[(Q_n - t)^2] = 0. $$

1) What additional arguments, if any, are needed to locate a subsequence of $\{Q_n\}$ so that the convergence is almost-sure?

2) How do you verify that the definition of the quadratic variation is independent of the choice of partition $\Pi$?

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It need not converge almost surely, although there will always be a subsequence converging almost surely, and you always get almost sure convergence with nested partitions of Brownian motion. Also, you've just shown that it converges to $t$ independently of the partition, so I'm not sure what the question is. –  George Lowther Aug 24 '11 at 21:11
    
The almost sure result mentioned by George is Proposition (2.12) (page 62) of Continuous Martingales and Brownian Motion (1st edition) by Daniel Revuz and Marc Yor. –  Byron Schmuland Aug 25 '11 at 0:44
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Hi, The problems arise when you are taking the Quadratic Variation using the sup over all partitions and not only refining (or nested) ones. Then you don't have a.s. CV but only CV in probability. An explicit construction of partitions for which almost sure convergence is not achieved can be done. There is one such construction in a book by Peres and Mörters (available at stat.berkeley.edu/~peres/bmbook.pdf) Exercise 1.13. Regards –  TheBridge Aug 25 '11 at 11:48
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1 Answer

1) Since your quantity converges in $L^2$ it converges in Probability, so a subsequence will converge a.s. and you need not impose any further conditions.

2) The limit does not depend on the partition, because if you choose any other partition with mesh going to 0, you would still have convergence in Probability to a certain limit (your proof works for any such partition) and thus you still have a subsequence converging a.s. Thus, the two limits must be the same.

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