# Show that $E [u] = 0$ and $cov(u,x_j)=0$ does not imply $E[u|x]=0$

In the regression model $$y= x'\beta + u, \quad x = (1, x_2,...,x_K)$$ with $$E[u |x]=0,$$

we know that it implies: $$E [u] = 0$$ and $$cov(u,x_j)=0$$, for $$j=1,...,K$$.

I think that the reciprocal is not true. Thinking geometrically, they seem equivalent, but I can't come up with a counterexample!

Do you have a counterexample?

Thinking geometrically

Defining $$\langle X,Y\rangle= E[XY]$$ we have that if $$X$$ or $$Y$$ is such that $$E[X]=0$$ or $$E[Y]=0$$, we have: $$\langle X,Y\rangle=cov(X,Y)$$ So, if $$cov(u,x_j)=0$$, I have orthoganility: $$\langle u,x_j\rangle=0$$ for all $$j=1,...,k$$. This seems to imply that the projection of $$u$$ onto $$1,x_2,...,x_k$$ is $$0$$, i.e. $$E[u|x]=0$$.

• Mar 13 at 7:28

One way to contrive counterexamples is to let $$X$$ follow some (non-degenerate) distribution symmetric about $$0$$ and choose $$c \in \mathbb R_{>0}$$ s.t. $$U \mathrel{:=} X^b - c$$ has expectation $$0$$ for an even number $$b \neq 0$$.

Then, symmetry of the distribution of $$X$$ about $$0$$ implies $$\mathrm{Cov}(X,U) = 0$$, and we have $$\mathbb E(U) = 0$$ and $$\mathbb E(U \,|\, X) = X^b - c$$ by construction.

An easy example that comes to my mind is $$X \sim \mathcal N(0,1)$$ and $$U \mathrel{:=} X^2 - 1$$.

• Clever example. +1. Mar 12 at 12:11

\begin{align} P(U=-1, X=0)& = 1/2\\ P(U=1, X=1) &= 1/4\\ P(U=1, X=-1) &= 1/4 \end{align}
• @User1865345 Almost any two distributions with 0 covariance and E(X)=0 will do. The first 2 lines can be anything, e.g. P(U=3, X=10) = 0.1, P(U=16,X=17) = 0.2. Then you just solve the simultaneous equation Cov(U, X) = 0; E(x) = 0 to find the third line P(U=-5,X=16.4) = 0.7. This will result in a valid counterexample unless you are very unlucky. Mar 12 at 10:31