# Dependence of summands of $\sum_{r,c}W_rW_c$

I have the following problem and I will be too happy to help me find a solution. Consider a random variable $$K_{ij}= \sum_{r=1}^N\sum_{c=1}^N W_{ir}W_{jc}=\sum_{r=1}^N\sum_{c=1}^N V^{ij}_{rc}$$ where $i \neq j$ and $W \in {+1,-1}$ are i.i.d. random variables $\operatorname{Bernoulli}(0.5)$, my question is that I can claim $K_{ij}$ is the summation of i.i.d. random variables, I mean we can say that new random variables $V$ s are independent? Thanks a lot in advance

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Clearly not: $K_{ij}=K_{ji}$ implies strong dependence. If you add the restriction $j\gt i$, note that the expectation of the product of all $V^{12}_{rc}, 1\le r,c\le 2$ is $1$, which does not equal the product of their expectations, showing lack of independence. If this is not obvious, write everything out in detail for $N=2$. – whuber Jun 2 '11 at 13:22
@whuber, thanks for your answer, I did not get what do you mean, in fact, I am wondering whether $K_{ij}$ itself can be considered as the summation of independent random variables or not? – Farzad Jun 2 '11 at 14:06
No, it is not. You can work this out from first principles by writing the probabilities in the case $N=2$ (because there are only four variables involved for $2^4=16$ equally probably outcomes). – whuber Jun 2 '11 at 14:14
@whuber, thanks a lot for your answer, I have two more question, (1) in this case can you tell me how can evaluate the moment generating function of $K_{ij}$?, (2) and If I use $W \sim \mathcal{N}(0,1)$ instead of $\operatorname{Bernoulli}(0.5)$, can I say $K_{ij}$ itself as the summation of independent random variables or not? – Farzad Jun 2 '11 at 14:26
Related to math.stackexchange.com/questions/43032/… – JavaMan Jun 3 '11 at 15:12