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Suppose we have three independent ($N \times 1)$ complex vectors $\mathbf{w}$, $\mathbf{e}$ and $\mathbf{f}$ be . Assume that $\mathbf{f}$ and $\mathbf{w}$ are of unit norm and isotropically distributed. In addition, we assume that $\mathbf{e}$ is complex gaussian with zero mean and variance $= \sigma^2$.
We define $x=(\mathbf{w}^* \mathbf{f})^*$ and $y=\mathbf{e}^* \mathbf{f}$, where $(\cdot)^*$ denotes the conjugate transpose.

Can we claim that $x$ and $y$ are independent (complex) random variables? if yes, how to prove it ?

Attempt 1:
We know that $E[x]=E[y]=E[y^*]=0$.
If we take $N=2$ for example, we can write :
$Cov(x,y^*)=E[xy]=E[w_1e_1^*|f_1|^2+w_2e_2^*|f_2|^2+w_1f_1^*e_2^*f_2+w_2f_2^*e_1^*f_1]=0$.
So $x$ is independent of $y^*$ and thus of $y$ .

Attempt 2:
If we perform a change of basis such that $\mathbf{f}=[1,0,\ldots,0]$, so for $N=2$ we get $x=w_1$ and $y=e_1^*$. Since $e_1$ and $w_1$ are independent, we can say that $x$ and $y$ are independent.

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  • $\begingroup$ I asked a similar question here, but it did not get any attention! I am not sure if this is the proper way to link these questions. $\endgroup$ – tam Nov 30 '15 at 17:21
  • $\begingroup$ The preferred method is to flag your original post on Mathematics for migration here. However, contrary to your assertion, it got plenty of attention, including excellent hints about the answer. You should continue pursuing it there rather than re-posting it here. $\endgroup$ – whuber Nov 30 '15 at 19:43
  • $\begingroup$ Should I understand that my approach is incorrect ? and that $x$ and $y$ are not independent ? thank you! $\endgroup$ – tam Nov 30 '15 at 20:09
  • $\begingroup$ As to your approach - does a covariance of 0 imply independence? $\endgroup$ – jbowman Nov 30 '15 at 21:47
  • $\begingroup$ @jbowman Oh.. my fault. You are right! covariance$=0$ is just a neccessary condition for independence. Do you think they are not independent? $\endgroup$ – tam Nov 30 '15 at 22:12

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