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

Question

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. $E[u|x]=0$.

Answer 1

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 $U \mathrel{:=} X^2 - 1$.

Answer 2

How about:

\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}