independence two stochastic processes 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 and V are independent brownian motions if and only if $X_t$ and $Y_t$ are indenpendent 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 geomtric 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?)
 A: $$
E[XY] = E[X]E[Y]
$$
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.
A: 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.
