Independence two stochastic processes

Question

being $X, Y$ two continuous processes, $\theta \in R$

• $U_t=\sin{(\theta)}X_t+\cos{(\theta)}Y_t$

• $V_t=\cos{(\theta)}X_t-\sin{(\theta)}Y_t$

I have to show that $(U_t)$ and $(V_t)$ are independent Brownian motions if and only if $(X_t)$ and $(Y_t)$ are independent Brownian motions.

My approach has been the following...if $X_t$, $Y_t$ are Brownian motions then also $U_t$ and $V_t$ are also normal, then I can compute the var/covariance matrix and nullify all the terms but the ones on the main diagonal...is that a good approach? In this case $\theta$ would be $\pi/2$ and $X$, $Y$ must be independent.

The problem, then, asks me also for a geometric property of the 2-dim Brownian motion that should appear evident from the problem itself...(? can you post a link in which I can go deeper?)

Answer 1

$$ E[XY] = E[X]E[Y] $$ is my claim for independence.

So taking U and V lets compute $E[UV]$ $$ \begin{eqnarray} U_tV_t&=& (\sin (\theta) X_t +\cos (\theta) Y_t)(\cos (\theta) X_t -\sin (\theta) Y_t),\\ &=&\sin \theta \cos \theta X_t^2 + \cos^2(\theta)Y_tX_t - \sin^2(\theta)X_tY_t - \sin \theta \cos \theta Y_t^2,\\ \implies E[U_tV_t] &=& \sin \theta \cos \theta \left(E[X_t^2]-E[Y_t^2]\right) + \left(\cos^2(\theta) - \sin^2(\theta)\right)E[X_tY_t].\tag{1} \end{eqnarray} $$ lets compute $E[U_t]E[V_t]$ $$ \begin{eqnarray} E[U_t] &=& \sin (\theta) E[X_t] + \cos (\theta) E[Y_t],\\ E[V] &=& \cos (\theta) E[X_t] - \sin (\theta) E[Y_t].\\ \implies E[U]E[V] &=& \sin \theta \cos \theta \left(E[X_t]^2 -E[Y_t]^2\right)+ \left(\cos^2\theta - \sin^2\theta\right)E[X_t]E[Y_t].\tag{2} \end{eqnarray} $$ Now for $U_t$ and $V_t$ to be independent

$$ E[U_tV_t] = E[U_t]E[V_t] $$ equating Eq.1 and 2 we find $$ \sin \theta \cos \theta \left(E[X_t^2]-E[Y_t^2]\right) + \left(\cos^2(\theta) - \sin^2(\theta)\right)E[X_tY_t] = \\ \sin \theta \cos \theta \left(E[X_t]^2 -E[Y_t]^2\right)+ \left(\cos^2\theta - \sin^2\theta\right)E[X_t]E[Y_t] $$ we already see that the first terms respectively cancel to yield $$ \left(\cos^2(\theta) - \sin^2(\theta)\right)E[X_tY_t] = \left(\cos^2\theta - \sin^2\theta\right)E[X_t]E[Y_t] $$ or $$ E[X_tY_t] = E[X_t]E[Y_t] $$

thus to ensure that $U_t$ and $V_t$ are independent we require $X_t$ and $Y_t$ to be also.

Answer 2

First let's compute the conditions for which $U_t$, $V_t$ are two brownian motions. $U_t$ and $V_t$ are two brownian motions if: $$E[U_t U_s]=s$$ $$E[V_t V_s]=s$$ Computing these two quantities: $$E[U_t U_s]=s+\sin(\theta)\cos(\theta)(E[X_tY_s]+E[X_s Y_t])$$ $$E[V_t V_s]=s-\sin(\theta)\cos(\theta)(E[X_tY_s]+E[X_s Y_t])$$ If $X_t$ and $Y_t$ are independent brownian motions, $U_t$ and $V_t$ are brownian motions. To check independence between $U_t$ and $V_t$, we need to verify that $cov(U_t,V_t)=0$ and that every linear combination of $U_t,V_t$ is a brownian motion. $$cov(U_t,V_t)=cov(X_t,Y_t)[\cos^2\theta-\sin^2\theta]$$ (If $X_t$,$Y_t$ are independent then $cov(U_t,V_t)=0$) and $$Z_t=X_t\lambda_1(\cos\theta+\sin\theta)+Y_t\lambda_2(cos\theta-\sin\theta)$$ $Z_t$ is a brownian motion process if and only if $X_t$,$Y_t$ are two independent brownian motion processes.