Say I have two Brownian motions $X^1$ and $X^2$. Say they have constant correlation $\rho$. Then of course I know the correlation between $X^1_t$ and $X^2_t$. Furthermore I know that correlation between $X^1_t $ and $X^1_s$ is $min(t,s)$. Do I know the correlation between $X^1_t - X^1_s$ where $t> s$ and $X^2_t$ ? I can not recollect ever seeing a result regarding anything like this.


Let $$X_1(t)=B_1(t)$$$$X_2(t)=\rho B_1(t)+\sqrt{1-\rho^2}B_2(t)$$ where $B_1(t)$ and $B_2(t)$ are independent. we have $$\operatorname{cov}\left( {{X}_{t}}^{1}-X_{s}^{1}\,,\,{{X}_{2}}^{t} \right)=\operatorname{cov}\left( {{B}_{1}}(t)-{{B}_{1}}(s),\,\rho {{B}_{1}}(t)+\sqrt{1-{{\rho }^{2}}}{{B}_{2}}(t) \right)={{\rho }}t-\rho \min \{t,s\}$$ then $${{\rho }^{*}}=\frac{{{\rho }}t-\rho \min \{t,s\}}{\sqrt{t+s-2\min \{t,s\}}\sqrt{t}}=\frac{\rho\sqrt{t-s}}{\sqrt{t}}$$

  • $\begingroup$ Indeed I use Cholesky decomposition $\endgroup$ – Behrouz Maleki Jun 13 '16 at 16:43
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    $\begingroup$ Are you sure $\rho$ should be squared? $\endgroup$ – user347551 Jun 13 '16 at 17:53
  • $\begingroup$ yes you are right $\endgroup$ – Behrouz Maleki Jun 13 '16 at 18:17

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