4
$\begingroup$

I have a question regarding exclusion restrictions in instrumental variable design. If I have an instrumental variable, which is also somewhat related to the outcome, would that (and how) cause problems? Does it bias the local average treatment effect in a way? Also, how is this LATE effect estimated? I am utterly confused. A simple numerical example would really help me understand.

$\endgroup$

1 Answer 1

4
$\begingroup$

If the instruments is in fact not predetermined, the IV estimator is inconsistent. If the instruments are also weak, this inconsistency can be more serious than that of the OLS estimator.

Consider the simple model $$y=\delta_0+\delta_1z+\epsilon,$$ where $x$ is an instrument for $z$. Assume we observe an i.i.d. sample $$(y_i,z_i,x_i),\qquad i=1,\ldots,n.$$ Then, the estimation error of the IV estimator satisfies, using the LLN, \begin{eqnarray*} \widehat{\delta}_{{IV}}-\delta&=&(X'Z)^{-1}X'\varepsilon\\ &=&\frac{1}{\frac{1}{n}\sum_{i=1}^nx_iz_i-\frac{1}{n}\sum_{i=1}^nz_i\frac{1}{n}\sum_{i=1}^nx_i}\left(% \begin{array}{cc} \frac{1}{n}\sum_{i=1}^nx_iz_i & -\frac{1}{n}\sum_{i=1}^nx_i \\ -\frac{1}{n}\sum_{i=1}^nx_i & 1 \\ \end{array}% \right)\\ &&\hspace{.5cm}\times\left(% \begin{array}{c} \frac{1}{n}\sum_{i=1}^n\epsilon_i \\ \frac{1}{n}\sum_{i=1}^nx_i\epsilon_i \\ \end{array}% \right)\\ &\to_p&\frac{1}{E(x_iz_i)-E(z_i)E(x_i)}\left(% \begin{array}{cc} E(x_iz_i) & -E(x_i) \\ -E(x_i) & 1 \\ \end{array}% \right) \times\left(% \begin{array}{c} E(\epsilon_i) \\ E(x_i\epsilon_i) \\ \end{array}% \right)\\ &=&\frac{1}{Cov(x_i,z_i)}\left(% \begin{array}{cc} E(x_iz_i) & -E(x_i) \\ -E(x_i) & 1 \\ \end{array}% \right)\times\left(% \begin{array}{c} E(\epsilon_i) \\ E(x_i\epsilon_i) \\ \end{array}% \right)\\ \end{eqnarray*} Accordingly, $$ \widehat{\delta}_{1,{IV}}-\delta_1\to_p\frac{-E(x_i)E(\epsilon_i)+E(x_i\epsilon_i)}{Cov(x_i,z_i)}= \frac{Cov(x_i,\epsilon_i)}{Cov(x_i,z_i)} $$ We therefore have that, unsurprisingly, $\widehat{\delta}_1\to_p\delta_1$ if and only if $Cov(x_i,\epsilon_i)=0$. Now, write the previous display as $$ \widehat{\delta}_{1,{IV}}-\delta_1\to_p\frac{\sigma_\epsilon}{\sigma_z}\frac{Corr(x_i,\epsilon_i)}{Corr(x_i,z_i)} $$ If the correlation between $x$ and $z$ is small, even seemingly negligible endogeneity of the instruments (in the sense of $Corr(x_i,\epsilon_i)\approx 0$) can lead to arbitrarily strong inconsistency of the IV estimator.

It may then be preferable to estimate the model with OLS, despite violation of the predeterminedness assumption . Analogously to the above derivation, one shows that $$ \widehat{\delta}_{1,{OLS}}-\delta_1\to_p\frac{\sigma_\epsilon}{\sigma_z}Corr(z_i,\epsilon_i). $$ We thus have $\widehat{\delta}_{1,{OLS}}-\delta_1<\widehat{\delta}_{1,{IV}}-\delta_1$ for $$Corr(z_i,\epsilon_i)<Corr(x_i,\epsilon_i)/Corr(x_i,z_i),$$ which is quite possible with weak and endogenous instruments. (In practice, this comparison can unfortunately not be made, as $\epsilon_i$ is unobservable.)

This example has nothing to say about LATE, it assumes a single identical coefficient for all $i$.

$\endgroup$
1
  • $\begingroup$ Thanks Chris, I was hoping that someone illustrates the failure of the restriction in terms of LATE numerically, so that I understand better $\endgroup$ Dec 19, 2015 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.