How can I calculate the cross variation between a standard Brownian motion $(B_t)_{t\geq 0}$ and the process $(B_t^{\tau})_{t\geq \tau}$ defined as $B_t^{\tau}= B_t-B_{t-\tau}$? Here $\tau$ is just a positive fixed number.
I tried to use the definition of cross variation. So, let $\Pi=\{t_0=\tau, t_1, \cdots, t_n=t\}$ be a partition of the interval $[\tau,t]$. Then, for all $t\geq \tau$
$$\langle B, {B}^{\tau}\rangle_t = \lim_{\|\Pi\| \rightarrow 0} \sum_{i=1}^n (B_{t_{i}}- B_{t_{i-1}})({B}_{t_{i}}^{\tau}- {B}^{\tau}_{t_{i-1}})$$ where $\|\Pi\|=\max_{1\leq i\leq n}(t_i-t_{i-1})$.
Then $$\langle B, {B}^{\tau}\rangle_t = \lim_{\|\Pi\| \rightarrow 0} \sum_{i=1}^n (B_{t_{i}} -B_{t_{i-1}})^2 - \lim_{\|\Pi\| \rightarrow 0} \sum_{i=1}^n (B_{t_{i}} -B_{t_{i-1}}) (B_{t_{i}-\tau}- B_{t_{i-1}-\tau})\\ = (t-\tau) - \lim_{\|\Pi\| \rightarrow 0} \sum_{i=1}^n (B_{t_{i}} -B_{t_{i-1}}) (B_{t_{i}-\tau}- B_{t_{i-1}-\tau}) $$ where the last equation follows since the quadratic variation of the standard Brownian motion on $[\tau,t]$ is just $t-\tau$.