# OLS Population Orthogonality Condition Proof

In the OLS model, we assume that $$E(X'U)=0$$ (with $$u$$ being the error term), which comes from $$E(U|X=x)=0$$, providing us that $$E(U)=0$$ and $$cov(x_i, u)=0$$ $$\forall x_i$$. I understand this argument intuitively, as $$E(U|X=x)=0$$ would be violated if $$x$$ and $$u$$ were correlated, e.g. $$u$$ increases with $$x$$ would cause $$E(U|X=x)>0$$ for large $$x$$. But I also know that this is not sufficient for a formal proof, and one is not presented in Woolridge (2010).

I am also wondering if this proof goes the other way as well, i.e. does $$E(x'u)=0 \Rightarrow E(U|X=x)=0$$ given that $$E(u)=0$$.

I'm guessing these are both fairly straightforward (thus why they were omitted in the text) but an explanation or a hint in the right direction would be appreciated.

No, $$E[UX] = 0$$ and $$E[U] = 0$$ do not imply $$E[U|X = x] = 0, ~\forall x$$. Here is a simple counter-example.

Imagine $$X$$ is standard gaussian, and that $$E[U|X = x] = x^2 - 1$$. Note $$E[U|X = x] \neq0$$ for all $$x\neq 1$$.

But,

$$E[UX] = E[E[UX|X]] = E[E[U|X]X] = E[(X^2 - 1)X] = E[X^3 - X] = 0$$

Also note $$E[U] = E[E[U|X]] = E[X^2 - 1] = 1 - 1 = 0$$.

Thus, we have $$E[UX] = 0$$, $$E[U] = 0$$, with $$E[U|X = x] = x^2 -1 \neq 0$$ in general.

PS: In an OLS model, you do not need to assume $$E[U|X] = 0$$. The population linear regression is simply the best linear approximation to the conditional expectation function $$E[Y|X]$$ (in the sense of minimizing mean squared error). You invoke exogeneity for identification of structural parameters, see here.