# 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!

• Since the 1st condition also is stronger than the usual one, it would be helpful if you could provide the reference for these two conditions. – Alecos Papadopoulos May 16 '14 at 19:53
• Wooldridge, §5.1.1 – ajohnrobertson May 16 '14 at 20:58
• Um... in Wooldridge condition 1) is $\text{Cov}(z_1,u)=0$. In my copy at least ;-) – Sergio May 16 '14 at 21:34

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.

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.

• So in order for $z_2$ to be a relevant instrument, we need to make sure that $corr(z_2, x_2)\neq0$ is not due to the fact that $corr(z_2, x_1)\neq0$ and $corr(x_1, x_2)\neq0$? – ajohnrobertson May 16 '14 at 20:55
• Yes (roughly), but the condition is clearer when you have several exogenous variables. $E(x_k\mid x_1,\dots,x_{k-1},z_k)\ne E(x_k\mid x_1,\dots,x_{k-1})$ is better because correlation is a relation between two variables, but $x_k$ and $z_k$ could be correlated to a combination of some exogenous variables. When there is only one exogenous variable, imaging a violation of condition 2 is difficult. – Sergio May 16 '14 at 21:27