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I need to prove that the I and Q of a complex random variable are independent, where: $$I = \sum_n(C_n * \cos(\phi_n))$$ and $$Q = \sum_n(C_n * \sin(\phi_n))$$

$C_n$ are random variables with not further specified distribution, with all $C_n$ independent.

$\phi_n$ are uniform random variables over $[-\pi, \pi]$, all independent from each other and also independent from the $C_n$.

As the sum goes to infinity, the central limit theorem is invoked to model the $I$ and $Q$ components as having the normal distribution, with zero mean. So far, so good.

The part I don't understand however, is that the $I$ and $Q$ components are always taken to also be independent from each other, making them jointly Gaussian.

Unless I am mistaken, for each $n$, $\cos(\phi_n)$ and $\sin(\phi_n)$ are dependent. So how can their sums over all $n$ (with identical values of $C_n$ as factor in both sums) be independent ?

I'd appreciate any help on this, so many thanks in advance!

P.S. the independence of $I$ and $Q$ components is then used in the theory books to model $\sqrt{I^2 + Q^2}$ as a Rayleigh random variable.

WimC
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dirac bracket
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  • If sine and cosine were dependent you would be able to write sine/cosine as a linear combination of cosines/sines, they are very independent and are in fact completely orthogonal to one another as functions, this can be proven by performing an inner product on them – Triatticus Apr 28 '17 at 04:26
  • What about $$\sin(x) = \pm\sqrt{1 - \cos^2(x)}$$? If you know $\cos(x)$ then you know $\sin(x)$ with probability 0.5 for the + and 0.5 for the -. Thus sin(x) and cos(x) are clearly dependent, no? – dirac bracket Apr 28 '17 at 05:00
  • Independence implies that knowing cos(x) does not change the probability of sin(x), which is clearly wrong here, because the probability of sin(x) for any single value of x (which has uniform density) is zero, while that changes to 0.5 for each of the two square root values above, given that you know cos(x). Furthermore, knowing cos(x) restricts sin(x) to the values above: any other value is impossible, so cos(x) and sin(x), for x uniform are clearly dependent. – dirac bracket Apr 28 '17 at 05:06
  • $\sin(x)$ and $\cos(x)$ for &x& uniform over $0,2\pi$ are uncorrelated, but not independent. I think you are confusing uncorrelatedness and orthogonality with independence. – dirac bracket Apr 28 '17 at 05:14
  • Matt, what's your point? Triatticus wrote that sin(x) and cos(x) where clearly independent because they are orthogonal. My comment tried to show that sin(x) and cos(x) are uncorrelated, but clearly dependent because knowing cos(x) changes the probability for sin(x) dramatically from 0 to 1/2 for each of the square-root values (with opposite sign). That is exactly what statistical dependence means. So what is the point are you trying to make with regard to my problem? – dirac bracket Apr 28 '17 at 05:25
  • @Triatticus You seem to confuse statistical independence (of random variables) with linear independence (in a vector space). As a consequence, your comment is quite misleading. – Did Apr 28 '17 at 05:46

1 Answers1

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Here is a fact worth to remember:

If $(R,\Theta)$ are independent with $\Theta$ uniform on $(-\pi,\pi)$ and $R$ with PDF $re^{-r^2/2}$ on $r>0$, then $$X=R\cos\Theta\qquad Y=R\sin\Theta$$ defines a couple of random variables $(X,Y)$ that is i.i.d. standard normal.

The most surprising part in this result might be that $X$ and $Y$ would be independent. It is of course quite specific to $R$ having a Rayleigh distribution.

My guess is that your $I$ and $Q$ refer to this setting, that is, that every $C_n$ in your question is distributed like (a multiple of) the random variable $R$ above. Otherwise, such $X$ and $Y$ have no reason to be independent, basically for the reason you explain. For example, if $P(R=1)=1$, the fact that $$P(\cos\Theta>.9)P(\sin\Theta>.9)\ne0=P(\cos\Theta>.9,\sin\Theta>.9)$$ forbids the independence of the random variables $R\cos\Theta$ and $R\sin\Theta$.

Did
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  • Thanks for reacting! Unfortunately, the $C_n$ in my problem formulation have a distribution that is not specified. The whole setup is actually used to prove that for large $N$, $I$ and $Q$ are Gaussian and that $\sqrt{I^2 + Q^2}$ is Rayleigh, which requires the $I$ and $Q$ to be independent. So it's not quite the same as positing that each of the $C_n$ has a Rayleigh distribution to start with. I need to understand where the independence part comes from first. – dirac bracket Apr 28 '17 at 07:30
  • I am just trying to understand if the problem I formulated above can be formally proven right or wrong, given that the distribution of $C_n$ is not further specified. Or if I just need to move on and accept (or delude myself?) that maybe $\sqrt{I^2 + Q^2}$ was historically found to be Rayleigh distributed from empirical measurement data, which then forces the $I$ and $Q$ from the theoretical model to be independent. – dirac bracket Apr 28 '17 at 07:40
  • Quote: "Otherwise, such X and Y have no reason to be independent, basically for the reason you explain." – Did Apr 28 '17 at 12:55