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Assume we have a vector of random variables $U=\langle u_1,u_2,u_3,....,u_n \rangle$. random variables of $U$ are not necessary independent random variables.

Now I have two vectors of coefficients: 1) $a= \langle a_1,a_2,a_3,...,a_n \rangle$; 2) $b=\langle b_1,b_2,b_3,...,b_n \rangle$, $a$ and $b$ are orthogonal to each other. That is, $\sum_{i=1} a_i b_i=0$

Will $a^TU$ and $b^TU$ be independent? or under what conditions these two linear combinations of $U$ will be independent? Any reference or proof to demonstrate this?

Note: I can assume there is no perfect correlation among $u$'s

Thanks,

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  • $\begingroup$ I assume that you mean $a$ and $b$ to be orthogonal to each other. $\endgroup$ – Mark Fischler Mar 9 '15 at 17:56
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Even with $a\perp b$, $a^TU$ and $b^&TU$ need not be independant. Example:

Let $u_1$ be uniform on $(0,1)$ and let $u_2 = 2u_1$ (these are perfectly correlated). Now let $a = \left( \begin{array}{c} 1\\1\end{array}\right)$ and $b = \left( \begin{array}{c} -1\\1\end{array}\right)$. Then $a\perp b$ yet $a^TU$ is uniform on $(0,3)$ while $b^TU$ is always \frac{1}{3} of the variable obtained from $a^TU$; they are still perfectly correlated, and obviously not independant.

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  • $\begingroup$ Thanks, Mark. What if I can say there is no perfect correlation among $u'$s? or Under which conditions, I can safely conclude that $a^T U$ and $b^T U$ are independent? $\endgroup$ – Vincent Mar 9 '15 at 18:11
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A really simple counterexample: Take $a$ to be a unit vector, say $a_i=1$ for $i=k$ and zero for all other $i$. Take $b$ to be a different unit vector, with value $1$ in the $j$'th position ( with $j\ne k$ ) and zero otherwise. Then $a$ and $b$ are orthogonal, and $a^TU=u_k$ and $b^TU=u_j$.

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  • $\begingroup$ thanks, Grand_chat. This is a good counter-example. $\endgroup$ – Vincent Mar 9 '15 at 18:43

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