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 $\Pi$?

  • $\begingroup$ 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. $\endgroup$ Aug 24, 2011 at 21:11
  • $\begingroup$ 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. $\endgroup$
    – user940
    Aug 25, 2011 at 0:44
  • 5
    $\begingroup$ 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 $\endgroup$
    – TheBridge
    Aug 25, 2011 at 11:48

4 Answers 4


I would like to comment on your 2nd question. The independence of $\langle M,M\rangle_T$ ($[W,W](T)$) of the partition you chose.

Let's see the fundamental steps from the beginning :

  1. Let $M$ be a martingale
  2. consider a sequence $\Pi_n = \{t^n_0 ,\ldots, t^n_{p_n}\}$ of nested partitions of $[0,T]$ ($\Pi_{n} \subset \Pi_{n+1}$) such that $\big\vert \Pi_n \big\vert \underset{n \to \infty}{\longrightarrow} 0$
  3. define $$X^n_t : = \sum_{i = 1}^{p_n} M_{t^n_{i-1}} (M_{t^n_{i-1}\wedge t} - M_{t^n_{i-1}\wedge t})$$

    Note that $X^n_t$ is a martingale $(\mathbb{E} [X^n_t \vert \mathcal{F}_s] = X^n_s)$.

    Note also that $M^n_T - X^n_T = \sum_{i=1}^{p_n} (M_{t^n_i} - M_{t^n_{i-1}})^2$ (It may take a while)

  4. Prove that $\lim_{n,m \to \infty}{(X^n_T - X^m_T)^2} = 0$ (It's a bit hard)

  5. Use Doob's $L^2$ inequality and 4 (above) $$\lim_{n,m \to \infty}\mathbb{E}[{\sup_{t \leq T}(X^n_t - X^m_t)^2}] \leq 4 \mathbb{E}[{\sup_{t \leq T}(X^n_T - X^m_T)^2}] \underset{n,m \to \infty}{\longrightarrow} 0$$

  6. There is a subsequence $X^{n_k}$ sucht that

$$\mathbb{E}[{\sup_{t \leq T}(X^{n_k}_T - X^{n_{k+1}}_T)^2}] < 2^{-k} $$

  1. Consider the events $A_k = \{\omega: \sup_{t \leq T} \vert{X^{n_k}_t - X^{n_{k+1}}_t}\vert > \frac{1}{k^2}\}$

    note (by Chebyshev's inequality) that $\mathbb{P} (A_k) \leq k^4 2^{-k}$

    Note that $\sum_k k^4 2^{-k} < \infty$

  2. Now use Borel cantelli lemma to obtain that $\mathbb{P}(A_k i.o) = 0$(the event $\{A_k i.o.\}$ is the event $\{\omega : \;\{k:\omega \in A_k\}$ is an infinite set$\}$.

  3. Now there is a full measure set $\Omega^*$ ($\mathbb{P} (\Omega^*) = 1$) such that $\omega \in \Omega^* \Rightarrow \exists \;K_0(\omega) $, $k > K_0(\omega) \Rightarrow \sup_{t \leq T} \vert{X^{n_k}_t - X^{n_{k+1}}_t}\vert < \frac{1}{k^2}$ therefore $ X^{n_k}_t$ converges uniformly (to a continuous function say $Y_t(\omega)$ on the interval $[0,T]$)

  4. note that $Y_t$ is the $L^2$ limit of $X^{n}$ this implies that $Y_t$ is also a martingale

  5. Note $$M^n_{t^n_j} - X^n_{t^n_j} = \sum_{i=1}^{j} (M_{t^n_i} - M_{t^n_{i-1}})^2$$ Therefore conclude that $M^n_t - Y_t = A_t$ is a continuous increasing process (almost surely) and since $M^n_{T} - X^n_{T} = Q_n$ (in your notation) we obtain that $\mathbb{E}[(Q_n - A_T)^2] = \mathbb{E}[(X^n_T - Y_T)^2] \to 0$

  6. Now to the main point, what if you had chosen a different sequence of partitions? Then you might get a different limit increasing process say $A'_t$. But it suffices to observe that $A_t - A'_t = M_t - A'_t - (M_t - A_t)$ is a martingale of bounded variation and therefore it is indistinguishable from 0 therefore $\{A'_t = A_t$ for all $t\}$ almost surely

And you conclude that the limit is independent of the partitions you chose.

For reference I used Le Gall's book (Mouvement Brownien, Martingales Et Calcul Stochastique - is in French).


If the diameter of the $n$-th partition converges (to zero) fast enough, namely if it is of order less than $1/\log(n)$, then the quadratic variation converges almost surely. If not, then not in general (for un-nested partitions). More precisely, Wrobel (1982) shows that that there exists a sequence of partitions with diameters of order less than $1/\log(n)$ raised to the power $\alpha$, for every positive $\alpha$ less than $1$, such that the Brownian square variation diverges almost surely.

For this and related results see also Stoyanov (2013, Sec. 24.8).

Stoyanov, J.M. (2013, 3rd edition): Counterexamples in probability. New York: Dover Publ.

Wrobel, A. (1982): On the almost sure convergence of the square variation of the Brownian motion, Probability and Mathematical Statistics, 3, 97-101. PDF from http://www.math.uni.wroc.pl/~pms/publications.php?nr=3.1


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.


I dont know whether it is correct:

{Q: [0, T] is spaced equally, as usual notation:$\Delta B_i=B_{t_i}-B_{t_{i-1}}$, $t_i-t_{i-1}=T/n$. Show $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $}

$Proof$: for all $\varepsilon$,

$\mathbb{P}\left(\{\omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-t|>\varepsilon\}\right)$

$\leq {\frac{E(\sum_{i=1}^{n}\Delta B_i^2-t)^4}{\varepsilon^4}}$(by Markov inequality)

$=12T^4(1/n^2+4/n^4)=O(1/n^2)$ .

then $\sum \mathbb{P}\left(\{ \omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-T|>\varepsilon\}\right)$ is a convergent series, satisfying the definition of c.c. (complete convergence),

So $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $, that also implies $a.s.$ (or by Borel-Cantelli lemma).

Is there any mistake? welcome someone to point it out.

  • $\begingroup$ This isn't an answer to this question since the question considers a general partition and not your specific one. $\endgroup$ Feb 27, 2019 at 17:16
  • $\begingroup$ Oh,thanks. but for this equally-spaced partition, is it an almost sure convergence? just as the proof shows? $\endgroup$
    – Xisheng Yu
    Feb 28, 2019 at 1:30
  • $\begingroup$ Yes for equally spaced, and more generally if the diameter of the n-th partition is of order less than 1/log(n). If not then not. Read my earlier answer. $\endgroup$
    – AJW
    Jul 29, 2020 at 18:21

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