= Covariance induced by a shared latent variable
{title2=$\operatorname{Cov}(A,B)=\operatorname{Var}(g(T))$}
Let $A,B$ be <indicator random variables> with <conditional independence> given a <latent variable> $T$, and suppose $\mathbb E[A\mid T]=\mathbb E[B\mid T]=g(T)$. Then
$$
\operatorname{Cov}(A,B)=\operatorname{Var}(g(T))\geq0.
$$
For a short proof, <conditional independence> gives $\mathbb E[AB\mid T]=g(T)^2$. The <law of total expectation> gives $\mathbb E[AB]=\mathbb E[g(T)^2]$ and $\mathbb E[A]=\mathbb E[B]=\mathbb E[g(T)]$. Subtract their product to obtain the claimed <variance>. Since $0\leq g(T)\leq1$, all the required <expected values> exist.
In the two-environment example of <conditional independence>, $g(T)$ is $0.1$ or $0.9$ with equal <probabilities>. Hence $\mathbb E[g(T)]=0.5$ and $\operatorname{Var}(g(T))=((0.1-0.5)^2+(0.9-0.5)^2)/2=0.16$. The two trials have zero <conditional covariance> in either environment, yet their unconditional <covariance> is positive because their success <probabilities> rise and fall together.
More generally, <square-integrable> <conditionally independent> <random variables> satisfy
$$
\operatorname{Cov}(A,B)
=\operatorname{Cov}\bigl(\mathbb E[A\mid T],\mathbb E[B\mid T]\bigr)
$$
by the <law of total covariance>. Equal conditional means reduce this to the displayed <variance>; unequal conditional means need not give a nonnegative <covariance>. The equal-mean result explains correlated component evolutionary states in a <coeval binary population> without correlation between initial masses: variation in a shared environment can supply the entire <covariance>.
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