# Proving covariance equals zero given a specific conditional expectation

I'm trying to prove the following:

Given $$πΈ[π|π = π½] = πΈ[π]$$ for any value of $$\beta$$, prove that $$\operatorname{Cov}(π,π) = 0$$;

So I was thinking to start with the definition of $$\operatorname{Cov}(π,π)$$ and use the tower rule that is $$πΈ[π] = πΈ[πΈ[π|π]$$, such that:

$$\operatorname{Cov}(π, π) = πΈ[(π β πΈ[π]) β (π β πΈ[π])] = πΈ[πΈ[(π β πΈ[π]) β (π β πΈ[π])|π]].$$

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

• 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 $E[XY].$ Nov 15 '18 at 15:01
• So I can write πΈ[ππ] as πΈ[πΈ[ππ|π]] . How can I simplify this further to show that it equals πΈ[π]βπΈ[π]? Nov 16 '18 at 8:55
• 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 Nov 16 '18 at 9:12
• πΆππ£(π, π) = πΈ[ππ] - πΈ[π]βπΈ[π] = πΈ[πΈ[ππ|π]] - πΈ[π]βπΈ[π] = πΈ[ππΈ[π|π]] - πΈ[π]βπΈ[π] = πΈ[ππΈ[π]] - πΈ[π]βπΈ[π] = πΈ[π]βπΈ[π] - πΈ[π]βπΈ[π] = 0; Thank you very much for the very helpful guidance. Nov 16 '18 at 9:43