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Suppose unconfoundedness holds for a set of potential outcomes $(Y(1),Y(0))$ and treatment $Z$, conditional on a set of covariates, $X$ such that:

$$ (Y(1),Y(0)) \perp Z \mid X $$

Then, is it necessarily the case that conditioning on an extra set of covariates $X'$, still causes the relationship to hold? In other words, does the equation above imply:

$$ (Y(1),Y(0)) \perp Z \mid X, X' $$

It appears that if we were working with a weaker form of unconfoundedness (focusing on just $Z=1$), where we assume the relation holds only on the expectation:

$$ E(Y(1) \mid Z=1, X) = E(Y(1) \mid X) $$

then this should hold for an extended set of covariates, where:

\begin{align} E(Y(1) \mid Z=1, X, X') &= E(Y(1) \mid X,X') \\ \end{align}

My reasoning is that if $E(Y(1) \mid Z=1, X) = E(Y(1) \mid X)$, then it implies that conditioning at levels of $X$, it is sufficient for us to compare outcomes as coming from a randomized study. If we were to condition on a deeper level, at a level of $X$ and a level of $X'$, then outcomes can still be compared, since it was already established that comparison could occur at $X$, so anything deeper should be as well.

Is the above argument sound reasoning? thank you!

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    $\begingroup$ To add to other answers, conditioning on colliders indices bias. Elwert & Winship (2014) discuss this issue. Some colliders are pre-treatment, and conditioning on them does indeed induce bias. Pearl warned against the "kitchen sink" approach of including all pre-treatment variables for this very reason. $\endgroup$ – Noah May 12 '19 at 22:57
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No, neither of these two unconfoundedness assumptions implies the other. To see, why the one that conditions on more information is not "weaker", suppose that the initial set of covariates $X$ is the empty set so that we have in particular $$ Y(1) \perp Z. $$ Conditioning on more information can destroy this independence. Suppose that the data generating process is as follow:

We first draw $X' = 0, 1$ with probability $1/2$ each. If we draw $X'=0$ we then draw $(Y(1), Z)=(1,0),(0,1)$ with probability $1/2$ each. If we draw $X'=1$ we then draw $(Y(1), Z)=(0,0),(1,1)$ with probability $1/2$ each. The unconditional distribution of $(Y(1), Z)$ assigns equal probability to each of the four possible joint outcomes.

It is easy to verify that, under this data generating process, $Y(1)$ and $Z$ are (unconditionally) independent. However, conditional independence $$ Y(1) \perp Z \mid X' $$ does not hold in this example. Quite the opposite; given $X'$ there is a deterministic relationship between the two random variables.

To make the exposition easier I have ignored $Y(0)$ in the argument above, but note that the vector $(Y(1), Y(0))$ satisfies a conditional independence assumption only if (not iff) each of its components does so individually.

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  • $\begingroup$ Thank you for your great example. I just wanted to make sure I understand your example. Is the reason in your example above, that even the weaker form of unconfoundedness (weaker in that it only holds through the conditional expectations) is not implied on an extra covariate; because, $E(Y(1) \mid Z=1, X) = E(Y(1) \mid X) = \frac{1}{2}$ but (I take $X'=1$) we have $1= E(Y(1) \mid Z=1,X,X'=1) \neq E(Y(1)\mid X, X') = \frac{1}{2}$? $\endgroup$ – user321627 May 9 '19 at 6:19
  • $\begingroup$ I am wondering if there are general conditions in one may imply or not imply the other. Extrapolating from the above, is it related to the fact $X\subset X'$? In other words, if $X\subset X'$ unconfoundedness on $X$ does not imply unconfoundedness on $X,X'$? Or is it perhaps related to the fact that conditioning on $X'$ as well restricts the available outcome/treatment pairings originally available? It would be great to hear any thoughts you have on this. Thank you!! $\endgroup$ – user321627 May 9 '19 at 6:43
  • $\begingroup$ Yes, your summary of my example is correct. What you are looking for is a sufficient condition for a "tower property" for conditional independence. I'm not sure such a condition exists. For conditional expectations we have a tower property, but you will find that this tower property is not sufficient to prove a tower property for independence. $\endgroup$ – Andreas Dzemski May 9 '19 at 6:57
  • $\begingroup$ Your question is related to the common fallacy in OLS regression that adding more controls is a guaranteed way of making the OLS exogeneity assumption "more likely". Unconfoundedness is a very powerful identification assumption but the price is that is also very restrictive. Identification breaks down if we choose the wrong set of controls and this includes the case of choosing too many controls. $\endgroup$ – Andreas Dzemski May 9 '19 at 7:02
  • $\begingroup$ Thank you for the response. It really hits at a central question I am wondering in why it is not possible to "condition away" an extra set of controls/covariates so that we can get back to the covariates that were sufficient for identification. I.e., if identification occurs at $E(Y(1)\mid Z=1, X) = E(Y(1)\mid X)$, why can we not just successively iterate it away by doing $E(E(E(Y(1)\mid Z=1, X, X')\mid Z=1,X) \mid X) = E(E(Y(1)\mid Z=1,X)\mid X)$ and so $E(E(Y(1)\mid Z=1,X)\mid X) = E(E(Y(1)\mid X)\mid X) = E(Y(1)\mid X)$? Is there a specific step that breaks down? Thank you again. $\endgroup$ – user321627 May 9 '19 at 7:11
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Not sure if this is what you're getting at because I'm not sure how $X$ is related to $X'$, but if $X$ and $X'$ are both truly covariates i.e. pre-treatment, then controlling for $X'$ is controlling for more than is necessary for identification, although it may increase estimation precision. If on the hand $X'$ is post-treatment, e.g. a variable on the path between $Z$ and $Y$, then the additional conditioning on $X'$ often undoes the conditional independence that conditioning on $X$ provided.

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  • $\begingroup$ Hi, thank you for your post. I had meant for $X$ and $X'$ to be pre-treatment variables. I am wondering if there is ever a case where if $X$ was sufficient for identification, but matching also on $X'$ induces more bias. I can imagine perhaps extra matching on $X'$ could result in greater standard errors as it is less efficient, but would there be other complications? $\endgroup$ – user321627 May 9 '19 at 3:41

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