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 $t-T$.


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.

  • $\begingroup$ Hope I can revive this question a bit with a quick question. What if we are given two not independent Brownian motions, say $X$ and $Y$. Can we then still compute $\langle X,Y\rangle_t$? Is it also zero? Thanks in advance! $\endgroup$ – Charlie Shuffler Apr 14 '19 at 9:32
  • $\begingroup$ @S.Crim good question, perhaps it should be an actual Question to fit with the usual format $\endgroup$ – Bjørn Kjos-Hanssen Apr 14 '19 at 11:23
  • $\begingroup$ Alright, I posted it as a separate question. mathoverflow.net/questions/328032/… $\endgroup$ – Charlie Shuffler Apr 14 '19 at 13:02

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