Quadratic variation of $X(s)=W_{s+\epsilon}-W_{s}$ Let $W_s$ be a standard Wiener process. The quantity $W_{s+\epsilon}-W_{s}$ is another standard Wiener process when regarded as a function of $\epsilon$. Therefore, the quadratic variation of $Y(\epsilon)=W_{s+\epsilon}-W_s$ is just $\epsilon$.
I'm interested in the behavior of $W_{s+\epsilon}-W_{s}$ as a function of $s$. In particular, I want to find the quadratic variation of $X(s)=W_{s+\epsilon}-W_{s}$. I know that it cannot simply be $s$ because the distribution of $X(s)$ is independent of time. However, I'm a bit stuck getting further than this.
 A: The proof is rather similar to the case of the Brownian motion. Let's fix some notation first:
$$S^{\Pi}(f,s) := \sum_{j=1}^n |f(s_j)-f(s_{j-1})|^2$$
where $\Pi = \{0=s_0<\ldots<s_n=s\}$ is a partition of $[0,s]$ and $f: \mathbb{R} \to \mathbb{R}$ an arbitrary function.
Let $\Pi$ a partition of $[0,s]$ such that $\text{mesh} \, \Pi := \max |s_j-s_{j-1}| < \varepsilon$. We have
$$\begin{align*} \mathbb{E}(S^{\Pi}(X,s)) &= \sum_{j=1}^n \mathbb{E} \big[ \big| (W_{s_j+\varepsilon}-W_{s_{j-1}+\varepsilon})-(W_{s_j}-W_{s_{j-1}}) \big|^2 \big] \\ &= \sum_{j=1}^n \mathbb{E}[|W_{s_j+\varepsilon}-W_{s_{j-1}+\varepsilon}|^2]+ \mathbb{E}[|W_{s_j}-W_{s_{j-1}}|^2] \\ &= \sum_{j=1}^n (s_j-s_{j-1})+(s_j-s_{j-1}) = 2s \end{align*}$$
where we used $W_t-W_r \sim N(0,t-r)$, $t>r$, and the independence of $W_{s_j+\varepsilon}-W_{s_{j-1}+\varepsilon}$ and $W_{s_j}-W_{s_{j-1}}$ (they are independent because the mesh size of $\Pi$ is smaller than $\varepsilon$). This shows that $2s$ is a good candidate for the quadratic variation of $X$.
Now we want to show that $\mathbb{E}[(S^{\Pi}(X,s)-2s)^2] \to 0$ as $\text{mesh} \, \Pi \to 0$. To do so, we have to calculate expressions of the form
$$C_{j,k} := \mathbb{E} \big[ \big( (\Delta(s_{j-1}+\varepsilon,s_j+\varepsilon)-\Delta(s_{j-1},s_j))^2-2(s_{j-1}-s_j) \big) \cdot \big( (\Delta(s_{k-1}+\varepsilon,s_k+\varepsilon)-\Delta(s_{k-1},s_k))^2-2(s_k-s_{k-1}) \big) \big]$$
where $\Delta(r,t) := W_t-W_r$. A straight-forward calculation shows that $C_{j,k} =0$ for $j \neq k$. Thus, we obtain
$$\begin{align*} \mathbb{E}[(S^{\Pi}(X,s)-2s)^2] &= \mathbb{E} \left[ \left( \sum_{j=1}^n \big( \Delta(s_{j-1}+\varepsilon,s_j+\varepsilon)-\Delta(s_{j-1},s_j) \big)^2 - 2 (s_j-s_{j-1}) \right)^2 \right] \\ &= \sum_{j=1}^n \mathbb{E} \bigg[ \bigg( (\underbrace{( \Delta(s_{j-1}+\varepsilon,s_j+\varepsilon)-\Delta(s_{j-1},s_j)}_{\sim N(0,2 (s_j-s_{j-1}))})^2 -2(s_j-s_{j-1}) \bigg)^2 \bigg] \end{align*}$$
The sum converges to $0$ as $\text{mesh} \Pi \to 0$. This follows as in the case of the Browian motion by applying scaling property, i.e. $N(0,2(s_j-s_{j-1})) \sim \sqrt{2(s_j-s_{j-1})} N(0,1)$.
(There are a lot of calculations, so don't hesitate to ask if you don't get along with it.)
