# Proving covariance equals zero given a specific conditional expectation

I'm trying to prove the following:

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

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

$$\operatorname{Cov}(𝑋, 𝑌) = 𝐸[(𝑋 − 𝐸[𝑋]) ⋅ (𝑌 − 𝐸[𝑌])] = 𝐸[𝐸[(𝑋 − 𝐸[𝑋]) ⋅ (𝑌 − 𝐸[𝑌])|𝑌]].$$

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 𝐸[𝑋𝑌] as 𝐸[𝐸[𝑋𝑌|𝑌]] . How can I simplify this further to show that it equals 𝐸[𝑋]⋅𝐸[𝑌]? 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
• 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸[𝑋𝑌] - 𝐸[𝑋]⋅𝐸[𝑌] = 𝐸[𝐸[𝑋𝑌|𝑌]] - 𝐸[𝑋]⋅𝐸[𝑌] = 𝐸[𝑌𝐸[𝑋|𝑌]] - 𝐸[𝑋]⋅𝐸[𝑌] = 𝐸[𝑌𝐸[𝑋]] - 𝐸[𝑋]⋅𝐸[𝑌] = 𝐸[𝑌]⋅𝐸[𝑋] - 𝐸[𝑋]⋅𝐸[𝑌] = 0; Thank you very much for the very helpful guidance. Nov 16 '18 at 9:43