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Let $X_{1,1},X_{1,2},X_{2,1},X_{2,2}$ be identically distributed r.v.s with distribution $\sim Be(p)$ and equally par-wise correlated, with pair-wise Pearson correlation coefficient $\rho$, e.g. $Corr[X_{1,1},X_{1,2}]=\rho$, $Corr[X_{1,1},X_{2,1}]=\rho$ and so on. We define $Y_1 = X_{1,1}\cup X_{1,2}$ and $Y_2 = X_{2,1}\cup X_{2,2}$, meaning that $Y_{1}=0$ if and only if $X_{1,1}=0,X_{1,2}=0$ and $Y_1=1$ otherwise. It follows that $Y_1$ and $Y_2$ are also Bernoulli distributed with probability $q$; and $q$ can be computed according to the discussion here, resulting in $q=\rho \cdot p (1-p)+p^2+2p(1-p)(1-\rho)$.

The question is how to calculate the correlation coefficient $\rho'$ between $Y_1$ and $Y_2$ as a function of $p$ and $\rho$.

Note: I estimate it using simulations, leading to $\rho'$ being somewhat larger than $\rho$, but I have not been able to find an analytical answer to this.

jmen
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  • The usual meaning of $A\cap B$ is that it equals $1$ only if $A=1 $ and $B=1$. Your definion would correspond to $A \cup B$ instead. Care to check that? – leonbloy Apr 27 '21 at 14:30
  • Yes. Thanks for pointing it out. It should be correct now. – jmen Apr 27 '21 at 15:04

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To compute $\rho'$ amounts to compute the full joint probability of $Y_1,Y_2$, in particular $P(Y_1=0,Y_2=0)$. This event corresponds to the four original rv being zero. But you cannot compute this because you only have the first and second moments, and this alone does not determine the joint distribution of the four original variables.

leonbloy
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  • If one has the marginal distributions of the four r.v.s. and the correlation matrix of the joint distribution, why would not you be able to compute the joint distribution? Also, what is the result obtained with simulations representing, if there is no solution? Please, read this not with an aggressive tone, but a curious one. Thanks for your comment. – jmen Apr 28 '21 at 08:00
  • I linked a question that answers precisely that. The correlation and the marginal don't determine the joint distribution, there are additional degrees of freedom, so there are many (infinite) models "Also, what is the result obtained with simulations representing, if there is no solution?" You've obviously simulated some particular distribution. – leonbloy Apr 28 '21 at 11:30
  • Perhaps you have in mind (and in your simulation) a Markov chain (for the four original rvs) ? – leonbloy Apr 28 '21 at 12:30
  • No, I sample independent realisations from the joint distribution. I believe the assumption that leads to your conclusion "simulated some particular distribution" comes from the fact that I used the Nataf transform, which uses a Gaussian copula to build the joint distribution. – jmen Apr 30 '21 at 10:04