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.


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 $Z = \sum_i X_i$?


1 Answer 1


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 $Z=\sum_i Xi$

  • $\begingroup$ 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? $\endgroup$
    – ryu576
    Jul 13, 2021 at 3:29
  • $\begingroup$ No, $Z$ is just the sum, as in the question $\endgroup$ Jul 13, 2021 at 3:44
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    $\begingroup$ 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? $\endgroup$
    – sztal
    Jul 14, 2021 at 20:43
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 15, 2021 at 1:58
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    $\begingroup$ Because "independent" is much stronger than "zero correlation". Means plus independence do uniquely specify the distribution. Means plus zero correlation would not. $\endgroup$ Jul 16, 2021 at 23:51

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