## Lecture 15. Quadratic variation of the Brownian motion paths

If $\Delta_n [0,t]=\left\{ 0=t^n_0 \le t^n_1 \le ...\le t^n_n=t\right\}$, is a subdivision of the time interval $[0,t]$, we denote by $\mid\Delta_n [0,t] \mid=\max \{ \mid t^n_{k+1}-t^n_k \mid , k=0,...,n-1 \}$, the mesh of this subdivision.

Proposition. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion. Let $t \ge 0$. For every sequence $\Delta_n [0,t]$ of subdivisions such that $\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0$, the following convergence takes place in $L^2$ (and thus in probability),
$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2=t.$
As a consequence, almost surely, Brownian paths have an infinite variation on the time interval $[0,t]$.

Proof.

Let us denote

$V_n=\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2.$

Thanks to the stationarity and the independence of Brownian increments, we have:
$\mathbb{E} \left( (V_n-t)^2\right)=\mathbb{E} \left(V_n^2\right)-2t\mathbb{E} \left( V_n\right)+t^2$
$=\sum_{j,k=1}^n\mathbb{E} \left( \left( B_{t^n_j} -B_{t^n_{j-1}}\right)^2\left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2\right)-t^2$
$=\sum_{k=1}^n\mathbb{E} \left( \left( B_{t^n_j}-B_{t^n_{j-1}}\right)^4\right)+2\sum_{1\le j
$=\sum_{k=1}^n (t^n_k-t^n_{k-1})^2 \mathbb{E} \left( B_1^4\right)+2\sum_{1\le j
$=3\sum_{k=1}^n (t^n_j-t^n_{j-1})^2+2\sum_{1\le j
$=2\sum_{k=1}^n (t^n_k-t^n_{k-1})^2$
$\le 2t\mid\Delta_n [0,t]\mid \rightarrow_{n \rightarrow +\infty} 0.$

Let us now prove that, as a consequence of this convergence, the paths of the process $(B_t)_{t\ge 0}$ almost surely have an infinite variation on the time interval $[0,t]$. It suffices to prove that there exists a sequence of subdivisions $\Delta_n [0,t]$ such that almost surely

$\lim_{n \rightarrow +\infty} \sum_{k=1}^n \mid B_{t^n_k} -B_{t^n_{k-1}}\mid=+\infty.$

Reasoning by absurd, let us assume that the supremum on all the subdivisions of the time interval $[0,t]$ of the sums

$\lim_{n \rightarrow +\infty} \sum_{k=1}^n \mid B_{t^n_k} -B_{t^n_{k-1}}\mid$

may be bounded from above by some positive $M$. From the above result, since the convergence in probability implies the existence of an almost surely convergent subsequence, we can find a sequence of subdivisions $\Delta_n [0,t]$ whose mesh tends to $0$ and such that almost surely,

$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2=t.$

We get then
$\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2 \le M \sup_{1\le k \le n} \mid B_{t^n_k} -B_{t^n_{k-1}} \mid \rightarrow_{n \rightarrow +\infty} 0,$
which is clearly absurd $\square$

Exercise. Let $(B_t)_{t\ge 0}$ be a Brownian motion.

1. Show that for $t \ge 0$, almost surely
$\lim_{n \rightarrow +\infty}\sum_{k=1}^{2^n} \left(B_{\frac{kt}{2^n}} -B_{\frac{(k-1)t}{2^n}}\right)^2=t.$
2. Show that there exists a sequence of subdivisions $\Delta_n [0,t]$ whose mesh tends to $0$ and such that almost surely

$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2=+\infty.$

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