Endogeneity & IV Consider the the following structural model:
$y=\beta_1x_1+\beta_2x_2+u$
where $u$ is an iid disturbance term.
Suppose $E(u|x_1)=0$ but $E(u|x_2)\neq 0$.For $z_2$ to be a valid instrument, it must satisfy the following two conditions:
1) $E(u|z_2)=0$
2) $E(x_2|x_1, z_2) \neq E(x_2|x_1)$
I am struggling to understand the intuition of the relevance condition 2). 
Normally for a structural model with one endogenous variable, no $x_1$ just $x_2$, the relevance condition is:
2') $Cov(x_2, z_2)\neq0$
Why is the condition different when the structural model contains one exogenous variable and one endogenous variable?
Thank-you!
 A: Since Wooldridge has written more than one books, the reference points to "Econometric Analysis of Cross Section and Panel Data". I have what it appears to be the 1st edition (2002), and in here it says that the variable $z_1$ is a candidate instrument for regressor $x_k$ if the following holds:
$$[(5.3),\; p.83]\;\; Cov(z_1,u) = 0 \;\;(\Rightarrow E(z_1u)=0)$$
$$[(5.4)-(5.5), p.83-84] \;\;\text {in}\;\; x_k = \delta_0+\delta_1x_1+...+\delta_{k-1}+\theta_1z_1+r_k,\;\; \theta_1\neq0$$
The first condition (zero covariance) is indeed the standard one made, and is weaker than mean-independence, that is written in the question. What this condition does is to make the IV estimator consistent.  
The second condition says that $z_1$ is correlated with $x_k$ even in the presence of the other regressors. Intuitively, if the other regressors do not leave room for $z_1$ to "explain" (i.e. "represent the variability of") $x_k$, then inserting $z_1$ in place of $x_k$ in the $y$-regression, all variability of the now absent $x_k$ will be "taken on" by the other $x$'s, and $z_1$ will be left powerless to help us recover the value of $\beta_1$. Formally, $\beta_1$ will no longer be identifiable, i.e. we would not have a unique solution. Wooldridge states this further down pp 85-86.
When only one variable is present, we indeed only need to assume that $z_1$ is correlated with $x_1$, since no other regressors exist. In their presence, the conditions for a valid instrument have to be stronger. 
A: If condition 2 is (as I think it is):
$$E(x_2\mid x_1,z_2)\ne E(x_2\mid x_1)$$
then it requires that $\theta\ne 0$ in:
$$x_2=\delta_0+\delta_1 x_1+\theta z_2+r$$
i.e. $z_2$ is partially correlated with $x_2$ once the other exogenous variabile $x_1$ has been netted out (see Wooldridge, §5.1.1)
If there is no $x_1$, just $x_2$, this reduces to your condition 2'. You need contidion 2 because, in general, in $y=\beta_0+\beta_1x_1+\beta_2x_2+\cdots+\beta_kx_k+u$, $z$ could be correlated to $x_k$ just because correlated to some $x_j$'s and they, not $z$, are correlated to $x_k$ (this may happen even if correlation is not always transitive). As an extreme example, if $z=x_1+x_2+\cdots+x_{k-1}$, then by replacing $x_k$ with $z$ you get perfect multicollinearity.
