[Math] Quadratic variation of Brownian motion and almost-sure convergence

brownian motionquadratic-variationstochastic-processes

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 $\Pi$?

Best Answer

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).

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