Quadratic Variation of Brownian Motion

definition: A subdivision of $[0,t]$ is a strictly increasing finite sequence $t_0,\ldots,t_n \in [0,t]$ such that $t_0 = 0, t_n = t$ .

definition: The quadratic variation of a real function $x$ over a subdivision $s_0,\ldots,s_n$ of $[0,t]$ is $v(x,s) = \sum_{i=1.n} |\Delta x(s_i)|^2$ where $\Delta r_i = r_i - r_{i-1}$ .

definition: The quadratic variation of a real function $x$ over $[0,t]$ is $v(x) = \sup \{v(x,s) : s \text{ is a subdivision of } [0,t] \}$

Let $W$ be a Brownian motion, and let $s = s_0,\ldots,s_n$ be a subdivision of $[0,t]$ . The quadratic variation of $W$ over the subdivision $s$ is $T_n(s) = \sum_{i=1..n} |\Delta W(s_i)|^2$ where $\Delta W(s_i) = W(s_i) - W(s_{i-1})$ . The expectation

$\mathbb{E} T(s) = \sum_{i=1..n} \mathbb{E} (\Delta W(s_i))^2 = \sum_{i=1..n} \Delta s_i = t$

Since the increments are independent,

$\text{var}(T(s)) = \sum_{i=1..n} \text{var} ((\Delta W(s_i))^2) = \sum_{i=1..n} 2(\Delta s_i)^2 \le 2 t ||\Delta s||_{\infty}$

Choosing a sequence of subdivisions $\left< \pi_n \right>$ such that $\sum_n ||\Delta \pi_n||_\infty < \infty$ ,

$\sum_n \text{var} (T(\pi_n)) = \sum_n \mathbb{E} (T(\pi_n) - \mathbb{E} T(\pi_n))^2 < \infty$ . Since $\sum_{n \le m} (T(\pi_n) - \mathbb{E} T(\pi_n))^2$ is a non-decreasing sequence of partial sums convergent to $\sum_n (T(\pi_n) - \mathbb{E} T(\pi_n))^2$ , the MCT states $\sum_n \text{var} (T(\pi_n)) = \lim_m \mathbb{E} \sum_{n \le m} (T(\pi_n) - \mathbb{E} T(\pi_n))^2 = \mathbb{E} \sum_n (T(\pi_n) - \mathbb{E} T(\pi_n))^2 < \infty$ and this implies that the terms of the sum converge to zero almost surely: $\lim_n (T(\pi_n) - \mathbb{E} T(\pi_n))^2 = 0$ . Hence $\lim_n T(\pi_n) = \mathbb{E} T(\pi_n) = t$ almost surely.

lemma: When $Z$ is a normal(0, $\sigma$ ) random variable, $\mathbb{E} Z^k = \sigma^2, 3 \sigma^4$ for $k = 2, 4$ and $\text{var} Z^2 = 2 \sigma^4$ .

The moment generating function of $Z$ is $\phi(t) = e^{\frac{1}{2} \sigma^2 t^2}$ . Let $u = \frac{1}{2} \sigma^2 t^2$ . Then $\phi(t) = e^u$ . Since $u' = \sigma^2 t, u'' = \sigma^2$ ,