The first variation of function $f(t)$ on interval $[0,T]$ is defined as
$$ FV(f) = \lim_{\|\pi\| \rightarrow 0} \sum_{i=0}^{n-1} |f( t_{i+1} ) - f( t_i)|.$$
How can we estimate the first variation of Brownian Motion?
I set $Q_{\pi} = \sum_{i=0}^{n-1} |f( t_{i+1} ) - f( t_i)|$. And compute that $\mathbf{E}[Q_{\pi}] = 0$ and $\mathbf{Var}[Q_{\pi}] = T$, based on these results, how can we compute $\lim_{||\pi|| \rightarrow 0} Q_{\pi}$ ? Is it $\sqrt{T}$?
Recall that the quadratic variation of Brownian motion up to time $t$ is simply given by $t$. It follows that the first variation of Brownian motion is infinite since processes of finite first variation have $0$ quadratic variation.
Indeed, if $X$ is a continuous process with finite first variation then for any partition $\mathcal{P}$, \begin{align*}\sum_\mathcal{P} |X(t_{i+1}) - X(t_i)|^2 &\leq \max_\mathcal{P} |X(t_{i+1}) - X(t_i)| \cdot \sum_\mathcal{P} |X(t_{i+1}) - X(t_i)| \\& \leq \max_{|s-t| \leq \|\mathcal{P}\|} |X(t) - X(s)| \cdot FV(X) \to 0 \end{align*} as $\|\mathcal{P}\| \to 0$ since $X$ is continuous.