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Say $X_1, X_2, \dots X_n$ are i.i.d. Bernoulli random variables with parameter $0 < p < 1.$ Consider the following random variable: $$Y = \sum_{i,j} c_{ij}X_iX_j $$ where $c_{ij}$ is a constant. I am interested in studying the distribution of $Y$ and would like to find a moment generating function for it. Unfortunately, while finding the expected value and variance is relatively straightforward, because of the dependence produced from the multiplication, finding the MGF has alluded me. $Y$ is similar to the square of a binomial distribution (I am not sure if the square is a well studied distribution).

Edit: If it makes the calculation easier, one can work under the simplification that $$Y = \sum_{i,j} X_iX_j $$

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    If $c_{i,j}$ is a constant, then why does it have an $i j$ subscript ... it is just $c$. – 2017-02-04
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    It is a constant dependent on $i$ and $j$. So there are $n^2$ constants. Although for simplicity we could work under the assumption that all $c_{ij}$ are 1, as it is not the difficult part of the problem! – 2017-02-04
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    You Can Approximate it by Normal Distribution if it suffices. – 2017-02-04
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    I don't think the CLT can be applied without independence, though. – 2017-02-04
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    An ugly brute force method exists because the resulting distribution is discrete. The mgf is thus $\sum p_k e^{tx_i}$ where $x_i$ is an outcome of $Y$ occurring with probability $p_i$. Probabilities will be of the form $p^k(1-p)^{n-k}$ for those outcomes with $k$ $X$'s equal to 1. The values of those outcomes can be calculated by noting that whenever $X_i = 1$, there is a $c_{ii}$ contribution, and whenever $X_i = 1, X_j = 1$ ($i \neq j$), there is a $c_{ij}$ and $c_{ji}$ contribution. There will be $2^n$ (possibly distinct) values corresponding to each outcome of $Y$. – 2017-02-04
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    Based on the form of the answer, for general choices of $c_{ij}$, there's no getting around an unsimplifiable complicated answer. However, if all the $c_{ij}$'s are $1$, you can collect a lot of terms and get a decent answer summing $0$ to $n$ exponential terms with a bunch of binomial coefficients in your answer. – 2017-02-04

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