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I noticed that unfoundedness is a key cornerstone of the Rubin Causal Model. That is, if outcomes are $Y$, treatment is $T$, and covariates are $X$, then we have:

$$ (Y(0),Y(1))\perp T|X $$

I am wondering why this is needed. Is it because for certain estimators, in order to prove they are unbiased, we need this assumption to factor out terms in the conditional expectation? Thanks.

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Is it because for certain estimators, in order to prove they are unbiased,

Depending on your definition of bias, no. If you are using the usual statistical definition, then this is not why we invoke this assumption. The assumption is used for identification of the effect, that is, it's used to say that it's possible to estimate what you want to estimate with observational data (considering infinite samples). It says nothing about statistical properties of specific estimators under finite samples. Of course, if what you want to estimate is not identified, then all estimators will be biased for this causal quantity, by definition. But you can still have identification of the causal quantity with finite sample biases.

More specifically, unconfoundedness (also called conditional ignorability, selection on observables etc) is usually invoked when you want to identify the average treatment effect with an observational quantity such as a traditional difference in means.

The average treatment effect is usually defined as:

$$ \tau = E[Y_i(1) - Y_i(0) ] $$

To simplify the derivation, let's also define the conditional average treatment effect:

$$ \tau(X) = E[Y_i(1) - Y_i(0) |X_i] $$

You never get to observe any of thoses quantities, because you only observe $Y_i(1)$ for those who were treated and $Y_i(0)$ for those who were not treated.

However if $\{Y_i(1), Y_i(0)\} \perp T|X$, you can rewrite $\tau(X)$ as:

$$ \begin{align} \tau(X) &= E[Y_i(1) - Y_i(0) |X_i]\\ &= E[Y_i(1)|X] - E[Y_i(0) |X_i]\\ &= E[Y_i(1)|T_i =1, X_i] - E[Y_i(0) |T_i = 0, X_i]\\ &= E[Y_i|T_i =1, X_i] - E[Y_i | T_i = 0, X_i]\\ &= \tau(X)_{diff} \end{align} $$

And this last quantity you do observe. So you can identify the average treatment effect $\tau$ with $\tau(X)_{diff}$ taking the expectation over $X$:

$$ \tau = E[Y_i(1) - Y_i(0) ] = \int_X E[Y_i(1) - Y_i(0) |X]dP(X) =\int_X\tau(X)_{diff}dP(X) \\ $$

Notice that this assumption is sufficient for identifying the average treatment effect, but it's not necessary. Also, implicit in our derivation is the assumption of common support of $X$ between the treated and control units.

Notice we did not talk about any specific estimator here. You could estimate those quantities in several ways. For identification, this estimator has to be consistent, but it could still have finite sample bias. Maybe the simplest example you could think of is a penalized estimator for $E[Y_i|T_i, X_i]$, which would probably be biased in finite samples but still consistent (under other assumptions, such as correct specification etc).

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    $\begingroup$ Wow thank you! Very clear in understanding things here. I have a similar but related follow-up question. You wrote above we can identify the average treatment effect with $E[E[Y_i|T_i =1, X_i] - E[Y_i | T_i = 0, X_i]]$. Suppose that $X \sim N(0,1)$ iid and $p_i = logit^{-1}(\beta_0 + \beta_1X)$ with $T_i \sim Bern(p_i)$ and finally $Y_i \sim N(\beta_0 + \beta_1X,1)$. In cases like these where we have multiple distributions across $Y,T,X$, would the expectations be too hard to compute and hence we need to rely on other conditions? $\endgroup$
    – user321627
    Commented Sep 16, 2017 at 2:47
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    $\begingroup$ @user321627 yes, conditional ignorability is just to get you started, that is, to say that what you want to do is possible. But after that you have all the other statistical modeling problems --- you can still get asymptotic bias with incorrect specification or very bad finite sample behavior with certain estimators etc. In the specific case you put in your comment, for instance, you are assuming linearity on $X$ for the outcome model, so you could use linear regression and related friends for example. $\endgroup$
    – Carlos Cinelli
    Commented Sep 16, 2017 at 3:02
  • $\begingroup$ @user321627 there are several estimation strategies under unfoundedness --- matching, weighting, linear regression, non parametric regression and so on. All of them rely on some other modeling assumptions. $\endgroup$
    – Carlos Cinelli
    Commented Sep 16, 2017 at 3:08
  • $\begingroup$ Good answer. One could add that identification is closely related (or actually the same as?) "having a consistent estimator". This suggests that unconfoundedness is intrinsically related to the notion of consistency, but not to finite-sample unbiasedness. $\endgroup$
    – Julian Schuessler
    Commented Sep 19, 2017 at 9:30
  • $\begingroup$ Don't mean to play the devil's advocate but you say that estimators will be unbiased but you do not show this even for a single estimator. Furthermore, the implication of unbiasedness is more general, because it allows Likelihood and Bayesian inference about distributional parameters, so inference is not limited to expectations. $\endgroup$
    – tomka
    Commented Sep 19, 2017 at 10:10

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