# Distribution of sum of possibly non-independent Bernoulli random variables with known variance-covariance matrix

I wonder if there are any results concerning the distribution of sums of possibly non-IID Bernoulli random variables when covariances in all pairs of r.v.'s are known.

To make this more concrete consider the following problem. Let $$X_i$$ for $$i = 1, \ldots, k$$ be a sequence of non-IID Bernoulli random variables with known parameters $$p_i$$ and $$X$$ the sum over all $$X_i$$'s:

$$X = \sum_{i=1}^k X_i \quad\text{where}\quad X_i \sim \mathcal{B}(p_i)$$

Moreover, the full set of covariances in all pairs $$X_i, X_j$$ is known.

What is the distribution of $$X$$?

I know that in the case when all $$X_i$$ are independent the solution is Poisson-Binomial distribution but I am interested particularily in the case when at least in some pairs $$X_i, X_j$$ covariances are non-zero.

[EDIT]

What if I could derive bivariate distributions for all pairs $$X_i$$ and $$X_j$$? Would this allow for deriving also the exact (or at least approximate) distribution of the sum $$Z = \sum_i X_i$$?

Any distribution on the integers $$0,1,\dots,k$$ can be written as a sum of $$k$$ non-IID Bernoullis.

Let $$Z$$ be a variable with the target distribution. Now,

• $$p(X_1)= p(Z>0)$$
• if $$X_1=1$$, $$p(X_2)= p(Z>1|Z\geq 1)$$ else $$p(X_2)=0$$
• if $$X_2=1$$, $$p(X_3)=p(Z>2|Z\geq 2)$$ else $$p(X_3)=0$$
• if $$X_3=1$$, $$p(X_4)=p(Z>3|Z\geq 3)$$ else $$p(X_4)=0$$
• and so on

(Edit:) so $$Z=\sum_i Xi$$

• This is a neat result, +1; but I don't understand your example. Are we getting that $P(Z=z) = P(X_1=1 \& X_2=1 \dots X_z=1)$ somehow? Jul 13, 2021 at 3:29
• No, $Z$ is just the sum, as in the question Jul 13, 2021 at 3:44
• Maybe I do not understand something, but I do not see how this answers my question. How does it really use the information on covariances between particular $X_i$'s and $X_j$'s? Jul 14, 2021 at 20:43
• This specifies how to get any distribution as a non-IID sum. You can work out all the covariances it implies. Means and covariances aren't enough to specify the joint distribution of Bernoulli variables, though. Jul 15, 2021 at 1:58
• Because "independent" is much stronger than "zero correlation". Means plus independence do uniquely specify the distribution. Means plus zero correlation would not. Jul 16, 2021 at 23:51