Proof: E[X|Y]=0 implies COV[X,Y]=0
I was thinking maybe the law of total covariance or tower rule but couldn't come up with the proof
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Hint: consider the expectation of $E[X\mid Y]Y.$ – whuber Jan 25 '21 at 17:36
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It's a direct application of tower rule. – StubbornAtom Jan 25 '21 at 20:45
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Start by writing out the covariance definition out – Three Diag Jan 25 '21 at 20:49
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If you would like details, see the equation immediately preceding the "Conclusions" section in my post at https://stats.stackexchange.com/questions/71260/what-is-the-intuition-behind-conditional-gaussian-distributions/71303#71303 and let $\rho=0.$ – whuber Jan 26 '21 at 14:02