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 $t-T$.
Best Answer
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.