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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?)

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  • $\begingroup$ "geomtric property of the 2-dim brownian motion" Invariance by the rotations. $\endgroup$ – Did Aug 1 '14 at 23:08
  • $\begingroup$ thx! evidently the exercise wanted me to notice the coefficients of the processes are the elements of a 2x2 rotation matrix, and link this to the rotational invariance (which I wasn't aware of) of the 2-dim brownian motion $\endgroup$ – prob89 Aug 2 '14 at 21:57
  • $\begingroup$ Funny: the accepted answer only proves a part of one implication (and the OP quite clearly says so), namely, that if X and Y are independent then so are U and V. Much more work is needed to solve the question, since it asks to show that if X and Y are independent Brownian motions then so are U and V, and the reverse implication. $\endgroup$ – Did Aug 28 '14 at 7:57
  • $\begingroup$ if $U_t, V_t$ are brownian motions, wouldn't be sufficient (and necessary) to compute the covariance between them and set equal to zero? of course, once shown $U_t$ and $V_t$ are brownian motions... $\endgroup$ – prob89 Aug 28 '14 at 8:59
  • $\begingroup$ $$E[U_t U_s]=s+\sin(\theta)\cos(\theta)(E[X_tY_s]+E[X_sY_t])$$ $$E[V_tV_s]=s-\sin(\theta)\cos(\theta)(E[X_tY_s]+E[X_sY_t])$$ I would conclude $V_t$, $U_t$ are brownian motion if and only if $X_t$, $Y_t$ are two independent brownian motions... $\endgroup$ – prob89 Aug 28 '14 at 9:28
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$$ E[XY] = E[X]E[Y] $$ 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.

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  • $\begingroup$ you're right, I was assuming the X, Y were brownian motion, while the statement doesn't tell so. The only thing I didn't understand is why $E^2[X_t]-E^2[Y_t]=E[X^2_t]-E[Y^2_t]$, for brownian motion they should be different ($0$ the squared mean, and $t$ the second moment...) $\endgroup$ – prob89 Aug 1 '14 at 22:45
  • $\begingroup$ and what about the geometric property of the 2-dim brownian motion that this exercise should make evident? ps about the question in the previous comment...ok, they're both zero eheh $\endgroup$ – prob89 Aug 1 '14 at 22:52
  • $\begingroup$ I am still trying to figure out the second part. I am not to sure how, or what they are trying to gain from the question.. $\endgroup$ – Chinny84 Aug 1 '14 at 23:01
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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.

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  • $\begingroup$ You are making so many twists and turns that it is difficult to see what the whole constitutes a proof of, in the end... For example, you start assuming... what? That X and Y are two independent Brownian motions, presumably, and you want to show that U and V are two independent Brownian motions, right? This conclusion cannot be reduced to E(U_tU_s)=s and E(V_tV_s)=s alone. As a first step, you might want to state clearly which properties would establish that U (alone) is a Brownian motion (then we'll talk about V...). $\endgroup$ – Did Aug 28 '14 at 19:41

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