How can I calculate the cross variation between a standard Brownian motion $(B\_t)\_{t\geq 0}$ and the process $(B^T\_{t})\_{t\geq T}$ given by $B^T\_t= B\_t-B\_{t-T}$? Here $T$ is just a positive number.


I tried to use the definition of cross variation. So, let $t\geq T$ and $\Pi=\{t\_0=T, t\_1, \cdots, t\_n=t\}$ be a partition of the interval $[T,t]$. Then 


$$\langle B, {B}^{T}\rangle\_t = \lim\_{\|\Pi\| \rightarrow 0} \sum\_{i=1}^n (B\_{t\_{i}}- B\_{t\_{i-1}})({B}\_{t\_{i}}^{T}- {B}^{T}\_{t\_{i-1}})$$
where $\|\Pi\|=\max\_{1\leq i\leq n}(t\_i-t\_{i-1})$.


Then
$$\langle B, {B}^{T}\rangle\_t = \lim\_{\|\Pi\| \rightarrow 0} \sum\_{i=1}^n (B\_{t\_{i}} -B\_{t\_{i-1}})^2 - \lim\_{\|\Pi\| \rightarrow 0} \sum\_{i=1}^n (B\_{t\_{i}} -B\_{t\_{i-1}}) (B\_{t\_{i}-T}- B\_{t\_{i-1}-T})\\ = (t-T) - \lim\_{\|\Pi\| \rightarrow 0} \sum\_{i=1}^n (B\_{t\_{i}} -B\_{t\_{i-1}}) (B\_{t\_{i}-T}- B\_{t\_{i-1}-T})$$
where the last equation follows since the quadratic variation of the standard Brownian motion on $[T, t]$ is just $t-T$. 



Cross variation two not independent Brownian motions
Assume $t\ge T$.
$$
[B\_t, B\_t^T]\_{[T,t]} = \int\_T^t dB\_t dB\_t^T = \int\_T^t dB\_t (dB\_t-dB\_{t-T}) = \int\_T^t (dB\_t)^2 - \int\_T^tdB\_t dB\_{t-T}
$$
$$
=\int\_T^t dt - \int\_T^t dB\_t dB\_{t-T} = (t-T) - \int\_T^t dB\_t dB\_{t-T} = t-T
$$
using
$$
\int\_T^t dB\_tdB\_{t-T}\approx \sum\_{k=0}^{n-1} (B\_{t\_{k+1}}-B\_{t\_k})(B\_{t\_{k+1}-T}-B\_{t\_k-T})\approx 0$$
since the changes in the product are independent.


