1
$\begingroup$

I have recently got myself confused about CIA and wondered if somebody could perhaps help me disentangle my thoughts. Consider the regression

$Y_i = \alpha_0 + \Delta D_i + \beta X_i + \eta_i$

Where $X_i$ are controls/covariates and $D_i$ is the treatment dummy. The CIA says that $\eta_i \perp D_i |X_i$. If this holds the causal effect of the treatment can be estimated with $E[Y_{1i} - Y_{0i}|X_i] = \Delta$.

Is the CIA the same thing as saying $E[\eta_iD_i] = 0$? (considering $\eta_i$ should have mean zero because a constant is included in the regression?) Which then is the same thing as $Cov(\eta_i, D_i) = 0$? How is this related to $E[\eta_i|X_i] = 0$ Gauss-Markov assumption? Does this have anything to do with the exclusion restriction for IV which looks very similar?

$\endgroup$
  • $\begingroup$ What does CIA stand for? $\endgroup$ – Brennan Apr 27 at 21:29
1
$\begingroup$

The CIA assumption is equivalent to $E\{ \nu_i D_i \vert X_i \} = 0$. However, it is not the case that the unconditional expectation is equal to zero. The whole reason we need the CIA is that we don't have random assignment, but potentially we have random assignment conditional on the covariates we can observe.

I think it may help to think of $\nu_i$ as an error term that is composed of other variables $Z_i$ with effect $\gamma Z_i$. In this case, $\nu_i \perp D_i \vert X_i$ means that $Z_i$ does not predict treatment after conditioning on the observed covariates $X_i$.

To answer your IV question, this is indeed similar to the exclusion restriction in that you are making assumptions about the unobserved variables $Z_i$ in your data generating process and your treatment/instrumental variable. But I wouldn't try to connect the ideas too much or else you may come to wrong conclusions.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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