# A question on conditional expectation leading to zero covariance and vice versa

In my probability class I was tackled with this seemingly weird question involving conditional expectation:

Let X,Y be two random variables (it is not mentioned whether or not they are discrete or continuous) and we are asked the following:

1. For all constants $\beta$ we have $E[X | Y = \beta] = E[X]$

2. $cov(X,Y) = E[XY]-E[X]E[Y] = 0$

The problem: we are asked to prove or give a counterexample to 1 leads to 2 and to 2 leads to 1

I have tried to prove these two directions but got nothing just by looking at the definitions and so I thought maybe they are false and we are to give counterexamples but I got nothing there either. Looking at the covariance formula I see Y is involved but I cannot really seem to incorporate it from conditional expectation given, so I am stuck and need help. Thanks to all.

• If 1. holds then $E(XY)=E(E(X\mid Y)Y)=$ $____$ hence $\mathrm{cov}(X,Y)=$ $__$. In the other direction, choose $X=Y^2$ and assume that the distributions of $Y$ and $-Y$ coincide then $E(XY)=E(Y)=0$ hence $\mathrm{cov}(X,Y)=0$ but $E(X\mid Y)=$ $__$ $\ne E(X)$. – Did Mar 25 '16 at 14:23
• @Did: Thanks I get the one direction but regarding the counterexample, how do you mean choose Y such that -Y and Y have the same distribution? – kroner Mar 25 '16 at 14:25
• The first condition is called "mean independence." See, e.g., math.stackexchange.com/questions/967641/… – symplectomorphic Mar 25 '16 at 14:25
• @zbigniew2015 Well, what I mean is "choose Y such that -Y and Y have the same distribution". No example which springs to mind? – Did Mar 25 '16 at 14:26
• Are you asking me whether Y and -Y are identically distributed when Y is standard normal? – Did Mar 25 '16 at 14:29