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Let's consider two random variables $X$ and $Y$ following a Bernoulli distribution such that: $$ P(X=1) = p\\ P(Y=1) = q $$ The correlation coefficient $\rho$ is given and my goal is to compute $P(X \cap Y)$: $$ \rho = \frac{E[XY]-E[X] E[Y]}{\sqrt{Var(X)*Var(Y)}} = \frac{P(X\cap Y)-pq}{\sqrt{p\cdot (1-p) \cdot q\cdot (1-q)}} \\ P(X \cap Y) = \rho \cdot \sqrt{p \cdot (1-p) \cdot q\cdot (1-q)} + pq\\ $$ When $X$ and $Y$ and perfectly anti-correlated, $\rho = -1$ and I expect to have $P(X \cap Y)=0$. But a quick numerical application taking $p=q=\frac{1}{1000}$ gives: $$ P(X \cap Y)=-0.000999 $$ What am I missing?

bfgt
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    You have proven that two such distributions cannot be perfectly anticorrelated. At https://stats.stackexchange.com/a/160669/919 I proved generally, for any two finite-variance random variables $X$ and $Y,$ that perfect anticorrelation implies $X$ and $Y$ must be linearly related (with a negative coefficient): but this is impossible unless $p=1-q.$ – whuber Feb 24 '22 at 15:59
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    Indeed that makes sense now. – bfgt Feb 24 '22 at 16:14
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    The general principle that whuber is illustrating is known as the Frechet-Hoeffding bounds. The marginal distributions of two random variables impose some (weak) constraints on the joint distribution, that prevents examples like the one you presented. – ecnmetrician Feb 24 '22 at 16:16

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