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If $P_1, P_2$ are two random variables taking support in $[0,1]$ with correlation $\rho$,

$$ \operatorname{Corr}(P_1, P_2) = \rho $$

and $X_1 \sim \operatorname{Bern}(P_1)$, $X_2 \sim \operatorname{Bern}(P_2)$,

how can I find $\operatorname{Corr}(X_1, X_2)$?

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I presume the $X_1 \sim \operatorname{Bern}(P_1)$ is intended to mean that the conditional distribution of $X_1$ given $P_1 = p_1$ is Bernoulli with parameter $p_1$, i.e. $\mathbb P(X_1 = 1 \mid P_1 = p_1) = p_1$ and $\mathbb P(X_1 = 0 \mid P_1 = p_1) = 1-p_1$, and similarly $\mathbb P(X_2 = 1 \mid P_2 = p_2) = p_2$ and $\mathbb P(X_2=0 \mid P_2 = p_2) = 1-p_2$. But that is not enough to tell us much about the dependence of $X_1$ and $X_2$ on each other. You need more assumptions. For example, perhaps $X_1$ is conditionally independent of $P_2$ and $X_2$ given $P_1$, and $X_2$ is conditionally independent of $P_1$ and $X_1$ given $P_2$. Then

$$ \eqalign{ \mathbb E [ X_1 X_2 ] &= \mathbb E \left[ \mathbb E [X_1 X_2 \mid P_1, P_2] \right]\cr &=\mathbb E \left[\mathbb E[X_1 \mid P_1, P_2]\; \mathbb E[X_2 \mid P_1, P_2]\right]\cr &= \mathbb E\left[\mathbb E[X_1 \mid P_1]\; \mathbb E[X_2 \mid P_2]\right]\cr &= \mathbb E[P_1 P_2]}$$ and then $X_1$ and $X_2$ have the same covariance as $P_1$ and $P_2$.

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  • $\begingroup$ $\ldots\,$and $$ \begin{align} \operatorname{var}(X_1) & = \operatorname{var}(\operatorname E(X_1\mid P_1)) + \operatorname E(\operatorname{var}(X_1\mid P_1)) = \operatorname{var}(P_1) + \operatorname E(P_1(1-P_1)) \\ \\ & = \operatorname{var}(P_1) + \operatorname E(P_1) - \operatorname E(P_1^2) = \operatorname E(P_1) - (\operatorname E(P_1))^2 = \operatorname E(P_1)\big( 1- \operatorname E(P_1)\big). \end{align} $$ Thus $X_1,X_2$ do not generally have the same correlation as $P_1, P_2. \qquad$ $\endgroup$ – Michael Hardy Jan 26 '18 at 23:16
  • $\begingroup$ Yes, covariance is not the same as correlation. $\endgroup$ – Robert Israel Jan 26 '18 at 23:29

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