[Math] How to make dependent Brownian motions independent

brownian motionstochastic-calculus

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?

Best Answer

$W$ 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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enter image description here

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Suppose $X$ is Gaussian with mean 0. Then $E(X^4) = 3E(X^2)^2$.

Lemma. Suppose that $Y$ is a random variable independent of $X$, and $X$ is $\mathcal N(0,1)$. Then $XY$ is a Gaussian if and only if $Y^2$ is constant almost surely.

Proof: Note that $E(XY) = 0$, $E((XY)^2) = E(Y^2)$, and $E((XY)^4) = 3 E(Y^4)$. So if $XY$ is Gaussian, then $E(Y^4) = E(Y^2)^2$. This implies $\text{Var}(Y^2) = 0$, and hence $Y^2$ is constant almost surely. The converse is obvious. $\square$

Now consider $Y_t - Y_1$ for $t > 1$. It can be written as a $\mathcal N(0,1)$ Gaussian multiplied by $$ V = \sqrt{(\sqrt t\cos W_{\log t}-1)^2 + (\sqrt t\sin W_{\log(t)})^2} = \sqrt{1+t-2\sqrt t\cos W_{\log t})} .$$ Thus $Y_t - Y_1$ is Gaussian if and only if $1+t-2\sqrt t\cos W_{\log t}$ is constant almost surely. And this is not the case.

Best Answer

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