1 \(p\)-Variation
The \(p\) -variation of a real-valued stochastic process \((X_{t})\) on \([0,T]\) is a process \(\left\langle X,X \right\rangle_{t}^{(p)}\) such that for all \(t>0\) and sequence of partitions \[ \Delta^{(n)}:=\{ 0 = t_{0}^{(n)}<t_{1}^{(n)}<\dots<t_{K_{n}}^{(n)}=t \}, \] of interval \([0,t]\) such that \[ \delta^{(n)}:=\max_{1\leq i\leq K_{n}}\lvert t_{i}^{(n)}-t_{i-1}^{(n)} \rvert \to 0\quad \text{as}\quad n \to \infty, \] the following holds \[ S_{n}^\Delta = \sup_{\Delta}\sum_{i=0}^{K_{n}-1} \lvert X_{t_{i}}-X_{t_{i-1}} \rvert ^p \stackrel{\mathcal{L}_{2}~\&~\mathbb{P}}\to \left\langle X,X \right\rangle_{t}^{(p)}. \]
For \(p=1\) we call this the total variation. For \(p=2\) we call this the quadratic variation and often will drop the \(p\) notation \(\left< X,X \right>_{t}\). We primarily deal with quadratic variance when we are working in the \(\mathcal{L}_{2}\) space and is key to our construction of an integral in this space.
Importantly, if a process is differentiable on \([0,T]\) then its total variation will be finite. To show that BM is not differentiable we take \(T=n\cdot\Delta t\) and find that \[ \sum_{i}\lvert B_{t_{i}}-B_{t_{i-1}} \rvert \sim n\mathbb{E}\lvert B_{\Delta t} \rvert \sim \frac{T}{\Delta t}\sqrt{ \Delta t }\to \infty\quad \text{as}\quad n \to \infty,~\Delta t \to 0. \] ## \(p\)-Variation of Brownian Motion
For BM we have \(\left< B,B \right>_{t}=t\), that is, for partition \(\Delta^{(n)}\) of an interval \([a,b]\) such that \(\delta^{(n)}\to 0\) as \(n \to \infty\) then \(S_{n}^\Delta \stackrel{\mathcal{L}_{2}~\&~\mathbb{P}}\to b-a\).
PROOF: We wish to show that \(\mathbb{E}(S_{n}^\Delta)^2\to(b-a)^2\) as \(n \to \infty\). First note that \[ \mathbb{E}S_{n}^\Delta = \mathbb{E}\left[ \sum_{i=0}^{K_{n}-1} \left\lvert B_{t_{i+1}^{(n)}}-B_{t_{i}^{(n)}} \right\rvert^2 \right]=\sum_{i=0}^{K_{n}-1}(t_{i+1}^{(n)}-t_{i}^{(n)})=b-a. \] We proceed to compute \[ \begin{align}\mathbb{E}\left[(S_{n}^{\Delta})^2\right] & =\mathbb{E}\left[ \left( \sum_{i=0}^{K_{n}-1}\lvert B_{t_{i+1}^{(n)}}-B_{t_{i}^{(n)}} \rvert ^2 \right)^2 \right] \\ & =\sum_{i=0}^{K_{n}-1}\mathbb{E}\left\lvert B_{t_{i+1}^{(n)}}-B_{t_{i}}^{(n)} \right\rvert ^4+2\sum_{i<j}\mathbb{E}\left[ \left\lvert B_{t_{i+1}^{(n)}}-B_{t_{i}}^{(n)} \right\rvert^2\cdot \left\lvert B_{t_{j+1}^{(n)}}-B_{t_{j}}^{(n)} \right\rvert \right] \\ & = \sum_{i}3\left(t_{i+1}^{(n)}-t_{i}^{(n)}\right)^2+2\sum_{i<j}\left(t_{i+1}^{(n)}-t_{i}^{(n)}\right)\cdot \left(t_{j+1}^{(n)}-t_{j}^{(n)}\right) \\ & = 2\sum_{i}\left( t_{i+1}^{(n)}-t_{i}^{(n)} \right) ^2+\left( \sum_{i}\left( t_{i+1}^{(n)}-t_{i}^{(n)} \right) \right) ^2 \\ & \leq 2\delta^{(n)}(b-a)+(b-a)^2 \\ & \to(b-a)^2\quad\text{as}\quad n \to \infty,\end{align} \] where we have used \(\mathbb{E}X^4=3[(\sigma^2_{X})^2]\) and the stationary independent increments of Brownian motion. We have showed that \[ \forall\varepsilon>0,~\exists N:\forall n\geq N,~\mathbb{E}[(S_{n}^{\Delta})^2]\leq (b-a)^2+\varepsilon. \] Additionally, we have shown that \(\mathbb{E}[(S_{n}^\Delta)^2]-(b-a)^2=\text{Var}(S_{n}^\Delta)\geq 0\) i.e. \(\mathbb{E}[(S_{n}^\Delta)^2]\geq (b-a)^2\) hence we obtain equality. Hence we have convergence in \(\mathcal{L}_{2}\) which gives convergence in probability. \(\square\)