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I'm trying to prove the following:

Given $𝐸[𝑋|π‘Œ = 𝛽] = 𝐸[𝑋]$ for any value of $\beta$, prove that $\operatorname{Cov}(𝑋,π‘Œ) = 0$;

So I was thinking to start with the definition of $\operatorname{Cov}(𝑋,π‘Œ)$ and use the tower rule that is $𝐸[𝑋] = 𝐸[𝐸[𝑋|π‘Œ]$, such that:

$$\operatorname{Cov}(𝑋, π‘Œ) = 𝐸[(𝑋 βˆ’ 𝐸[𝑋]) β‹… (π‘Œ βˆ’ 𝐸[π‘Œ])] = 𝐸[𝐸[(𝑋 βˆ’ 𝐸[𝑋]) β‹… (π‘Œ βˆ’ 𝐸[π‘Œ])|π‘Œ]].$$

However, I am not sure this leads me anywhere and how to continue from here.

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    $\begingroup$ You're on the right track. If you multiply out the expression for the covariance you get $cov(X,Y) = E[XY] - E[X] \cdot E[Y]$. If you can show the first part equals the second, you are done, and you can do that by using the tower rule on $E[XY].$ $\endgroup$
    – CloseToC
    Nov 15 '18 at 15:01
  • $\begingroup$ So I can write 𝐸[π‘‹π‘Œ] as 𝐸[𝐸[π‘‹π‘Œ|π‘Œ]] . How can I simplify this further to show that it equals 𝐸[𝑋]⋅𝐸[π‘Œ]? $\endgroup$
    – D.Cohen
    Nov 16 '18 at 8:55
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    $\begingroup$ A key property of conditional expectations is the following: $E[f(Y)\cdot X|Y] = f(Y)E[X|Y]$ for any function of Y. Conditional on Y, the value of some function of Y isn't a random variable but a constant, and can be taken out of the expectation $\endgroup$
    – CloseToC
    Nov 16 '18 at 9:12
  • $\begingroup$ πΆπ‘œπ‘£(𝑋, π‘Œ) = 𝐸[π‘‹π‘Œ] - 𝐸[𝑋]⋅𝐸[π‘Œ] = 𝐸[𝐸[π‘‹π‘Œ|π‘Œ]] - 𝐸[𝑋]⋅𝐸[π‘Œ] = 𝐸[π‘ŒπΈ[𝑋|π‘Œ]] - 𝐸[𝑋]⋅𝐸[π‘Œ] = 𝐸[π‘ŒπΈ[𝑋]] - 𝐸[𝑋]⋅𝐸[π‘Œ] = 𝐸[π‘Œ]⋅𝐸[𝑋] - 𝐸[𝑋]⋅𝐸[π‘Œ] = 0; Thank you very much for the very helpful guidance. $\endgroup$
    – D.Cohen
    Nov 16 '18 at 9:43

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