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


2 Answers 2


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

  • 1
    $\begingroup$ Clever example. +1. $\endgroup$ Mar 12 at 12:11

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}

  • $\begingroup$ Would very much appreciate how you contrived the example or the approach to construct the same. $\endgroup$ Mar 12 at 9:05
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    $\begingroup$ @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. $\endgroup$
    – fblundun
    Mar 12 at 10:31
  • $\begingroup$ This is indeed a good approach. I would urge to add this to the answer. $\endgroup$ Mar 12 at 10:32
  • $\begingroup$ Dear, does your example have to do with a linear regression model with a constant regressor? $\endgroup$ Mar 12 at 16:53

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