How is zero conditional mean assumption "stronger" than uncorrelated assumption?

Question

I am trying to understand what the zero conditional mean assumption ($\mathbb{E}[u\vert X]=0 $) encompasses in addition to a zero-correlation assumption ($\text{Corr}(X,u)=0$). I assume it must be "stronger" (by stronger I mean covers more cases) assumption, as otherwise we could replace it with a zero-correlation assumption. So what would be the cases where the zero-correlation assumption is not sufficient?

Answer 1

One simple contrived example:

Consider 2 discrete random variables $X \in \{-1,0,1\},Y \in \{-1,1\}$ with the following joint distribution:

\begin{array}{c|ccc|c} Y,X & -1 & 0 & 1 \\ \hline 1 & 1/12 & 4/12 & 1/12 &1/2\\ -1 & 3/12 & 0 & 3/12 & 1/2 \\ \hline &1/3&1/3&1/3 \end{array}

Then by symmetry $\mathbb{E}(X) = \mathbb{E}(Y) = 0$ and so $$\text{cor}(X,Y) = \mathbb{E}(XY) = 1 \times 1/3 -1 \times 1/3 = 0$$

Yet $\mathbb{E}(Y|X=-1) = - 1/2 \neq 0$

This is the minimal example I could think of; you should get some intuition from the construction, but might seem a bit cheeky/unmotivated. Perhaps better intuition is that correlation 0 is a single scalar constraint therefore very weak; however having zero conditional expectation gives a scalar constraint for every value of $X$ and so is a lot stronger. Another perspective might be that correlation is only a linear property, whereas conditional expectation can tell you more about non-linear relationships between random variables (like in this example).

Answer 2

The result follows from the law of iterated expectations, $$ E[X^T \epsilon]=E[X^TE[\epsilon | X ]]=E[X^T0]=E[0]=0. $$ For examples where the stronger condition is not satisfied but the weaker may be, consider Proving OLS unbiasedness without conditional zero error expectation? or Non Linear Endogeneity.