How do choices of probabilities and covariance matrices constrain each other for a correlated multivariate Bernoulli random variable?

Question

I have a correlated multivariate Bernoulli random variable $\textbf{X} = (X_1, ..., X_N)$, where the $X_i$ are Bernoulli random variables with parameters $p_i$ and $N \times N$ covariance matrix $\textbf{C}$.

How do choices of $p_i$ constrain choices of $C$ and vice-versa?

In one extreme case, where the $X_i$ are all independent, all choices of $p_i$ are valid. In another extreme case, where the $X_i$ are all perfectly correlated, the $p_i$ must be identical. But I would like to better understand the intermediate cases; both intuitive and more rigorous answers would be much appreciated.

Answer 1

First there is the obvious constraint on the variances: $c_{ii} = p_i(1-p_i)$. The offdiagonals are more complicated. Let $p_{ij} = \Pr(X_i = X_j = 1)$. Then $c_{ij} = p_{ij}-p_i p_j$, and the bounds on $c_{ij}$ are most easily expressed in terms of bounds on $p_{ij}$. The upper bound is straightforward: $p_{ij} \leq \min(p_i,p_j)$. The lower bound uses the same logic, but the algebra is messier: $\Pr(X_i=1,X_j=0)=p_i - p_{ij} \leq \min(p_i,(1-p_j))$, which leads to the lower bound $p_i - \min(p_i,1-p_j) \leq p_{ij}$. Subtracting $p_i p_j$ from those bounds gives the bounds on $c_{ij}$.

But that's for only a pair of variables. We also need $\textbf{C}$ to be Gramian (i.e., positive definite or semidefinite). I know of no simultaneous bounds for the $N(N-1)/2$ covariances that are necessary and sufficient for $\textbf{C}$ to be Gramian. Moreover, it's not clear to me that a Gramian $\textbf{C}$ is sufficient to guarantee that an $N$-variate Bernoulli distribution exists that has covariance matrix $\textbf{C}$.