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Can someone explain to me how to take 2 correlated Brownian motions and make them independent? I can't seem to grasp this process.

Just assuming $dB_1(t)dB_2(t) = \rho dt $

From what was explained to me, but I dont understand entirely:
You set the Brownian motions equal to the following functions of new uncorrelated brownian motions:
$B_1(t) = W_1(t)$
$B_2(t) = \rho W_1(t) + \sqrt{1-\rho ^2} W_2(t)$
Then if this is written in matrix form:
$dB(t) = \begin{bmatrix}1 & 0\\ \rho & \sqrt{1-\rho ^2}\\ \end{bmatrix} dW(t)$
So now that right hand side could be used as an indepedent Brownian motion?

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  • $\begingroup$ No, $(W_1,W_2)$ are independent, that is, $(B_1,(B_2-\rho B_1)/\sqrt{1-\rho^2})$ are independent Brownian motions. $\endgroup$
    – Did
    May 23, 2013 at 5:31
  • $\begingroup$ so are you saying that dB(t) multiplied by the inverse of the correlation matrix on left hand side is independent? $\endgroup$
    – riotburn
    May 23, 2013 at 5:52

1 Answer 1

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$W$ is a pair of independent Brownian motions, but $B$ is not. The trick is just to assume that $B_1$ is a certain Brownian motion (e.g. $W_1$) and that $B_2$ is a linear combination of $B_1$ and a Brownian motion that is independent from $B_1$, e.g. $W_2$. Coefficients of this linear combination are computed based on the correlation condition $\mathrm dB_1 \mathrm dB_2 = \rho\mathrm dt.$

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