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Given the arrays $C=[C_1,C_2,...,C_N]$ and $S=[S_1,S_2,...,S_N]$ of lengths $N$ with elements that are discrete iid uniform distributed with equal probability (p=1/2) of being $\pm$ 1

Consider the random variables (for a given $l, n, m$):





It can be shown that these random variables ($W, X, Y, Z$) are zero mean, uniform distributed with equal probability (p=1/2) of being $\pm$ 1. Furthermore, it can be shown that they are uncorrelated (e.g. $E[WX]=E[(C_lC_mC_n)(S_lS_mC_n)]=0$ since they are zero mean ($E[C_i]=E[S_i]=0$) and $C_i^2=S_i^2=1$ ).

Now how can one go about showing that the random variables ($W, X, Y, Z$) are pairwise independent? (The expectation $E[WX]=E[(C_lC_mC_n)(S_lS_mC_n)]=0=E[W]E[X]=0$ but that doesn't necessarily mean they are independent). So I'm wondering what is a good way to show independence, any tricks one can use?

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For Bernoulli variables, non-correlation implies independence. – leonbloy Oct 1 '12 at 21:01
More is true: any three random variables amongst W, X, Y and Z are independent (while (W,X,Y,Z) is not since, for example, WXYZ=+1 with probability 1). – Did Oct 1 '12 at 21:26
Thanks did for sharing your further observations. – Hakeem Oct 2 '12 at 5:32
up vote 1 down vote accepted

Consider two Bernoulli variables (with values in $\{0,1\}$) $X,Y$. We have

$$Cov(X,Y) = E(X,Y) - E(X) E(Y) = P(X=1,Y=1) - P(X=1) P(Y=1)$$

Then, $Cov(X,Y) = 0$ implies $P(X=1,Y=1) = P(X=1) P(Y=1)$ which further implies $P(X,Y) = P(X)P(Y)$. That is, for Bernoulli variables non-correlation implies independence.

Now, your $X,Y,Z,W$ are Bernoulli variables (except for a scaling-shifting that doesn't alter covariance/independence) pairwise uncorrelated, hence they are pairwise independent.

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Thanks leonbloy, appreciate it! Nice to see such a simple proof using the {0,1} Bernoulli variables. – Hakeem Oct 2 '12 at 5:30

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