# Probability distribution of the sum of products of discrete iid uniform random variables

Here is a problem I am working on, hoping to get some guidance from the experts here:

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 sum below: \begin{equation*} A=\sum_{l=1}^N \sum_{m=1}^N \sum_{n=1}^N C_lC_mC_n+S_lS_mC_n-C_lS_mS_n+S_lC_mS_n \end{equation*} Let's assume $N$=16 (or higher). I'm trying to find the probability distribution/density of $A$ which is essentially the sum of $4N^3$ terms with each term being a triple product of discrete iid uniform rvs. It can be shown that these terms ($C_lC_mC_n, S_lS_mC_n, C_lS_mS_n, S_lC_mS_n$) are also uniform distibuted with p=1/2 for $\pm$1. Furthermore, it can be shown that they are uncorrelated (e.g. $E[(C_lC_mC_n)(S_lS_mC_n)]=0$ since they are zero mean ($E[C_i]=E[S_i]=0$) and $C_n^2=1$ ).

Edit: As pointed out in comments below, the product terms are pairwise independent however they are not all jointly (mutually) independent. Therefore the question remains unsolved for now.

So here are the questions:

1. What is the probability density/distribution of $A$? How can the pdf of $A$ be determined knowing that the product terms are identically distributed, zero mean, uncorrelated and pairwise independent but not jointly (mutually) independent?

2. If an exact solution is not tractable/feasible, are there any approximations that need to be used?

3. What is the variance of $A$? Since the product terms are uncorrelated and each has variance of 1, the variance of $A$ seems to simply be the sum of the variances of all the terms (i.e. the number of terms $4N^3$).

4. While the central limit theorem (CLT) applies for iid rvs, in this case the terms are uncorrelated, identically distributed and pairwise independent but not jointly (mutually) independent. Is there a special case of the CLT that can be applied here?

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It seems $c_lc_mc_n$ and $s_ls_ms_n$ are also independent because $E[c_lc_mc_n\,s_ls_ms_n]=E[c_lc_mc_n]E[s_ls_ms_n]$ since $E[c_lc_mc_n]=E[s_ls_ms_n]=0$? – Seyhmus Güngören Sep 13 '12 at 20:01
That's true but I'm not sure if that would help with the solution since the terms in the sum ($c_lc_mc_n, s_ls_mc_n, c_ls_ms_n, s_lc_ms_n$) haven't been shown to be independent. – Hakeem Sep 13 '12 at 22:07
they all have zero mean and correlation zero right? – Seyhmus Güngören Sep 13 '12 at 22:40
ok I see what you're saying, the terms being independent really simplifies it and the CLT can be applied without any issues. Thanks! – Hakeem Sep 13 '12 at 23:02
Isn't that first summand $(\sum C_i)^3$ etc ? If so the limiting ditribtion is divide by $n^{-\frac 32}$ and get $C^3 _ S^2C$ with S,C independent normal. – mike Sep 14 '12 at 0:24