We know that if $X,Y$ are independent, then $E(Y\mid X) = E(Y)$. But what about the converse?
2026-03-26 12:35:20.1774528520
If $E(Y\mid X) = E(Y)$, do we have $X,Y$ independent?
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No. Here is the joint distribution of a counterexample: \begin{array}{cc|ccc} &&&X&\\ &&-1&0&1\\\hline &1& 1/6& 0&1/6 &\\ Y&0& 0& 1/3& 0&\\ &-1 & 1/6&0 &1/6 \end{array} Note $E(Y|X=x)=0$ for all $x\in \{-1,0,1\}$.