1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ $E[u\mid X]$ is a function of possible values of $X$ whereas $\operatorname{Cor}(X,u)$ is a single number. When $X$ is non-constant, perforce the restriction on the former is stronger than any restriction on the latter. $\endgroup$
    – whuber
    Commented Jul 24, 2022 at 17:12

2 Answers 2

1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ I think that expected value is -1/2, not -1/12? $\endgroup$ Commented Apr 30, 2023 at 3:40
  • $\begingroup$ agreed, edited - thanks! $\endgroup$ Commented May 1, 2023 at 8:40
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.