I read from the following slides on observational studies, pg. 16, Observational Studies, Keio,

that given:

$$ ATE ≡ E[Y_i(1) − Y_i(0)] $$

They pose the following question: Can we identify the $ATE$ when $D_i$ is not randomized? Here, $D_i \in \{0,1\}$ is the treatment assignment variable.

I am wondering what it means by identification in this context. I understand generally that identification in models refers to the fact that the data alludes to only a single set of parameters, which something like a mixture model does not adhere to. In the context above, if $D_i$ is not randomized, what will the "multiple" sets of parameters referring to the same data be?

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    $\begingroup$ I think it's simply means that you can't compute ATE. If the data is not randomized, then you cannot infer or compute the ATE since $X_i$ is confounded with $D_i$ and $Y_i$. $\endgroup$ Commented Mar 6, 2019 at 18:58

1 Answer 1


what will the "multiple" sets of parameters referring to the same data be?

Identification of a causal parameter in this case means it can be uniquely computable from the data. Here the causal parameter of interest is the ATE.

Thus, if the ATE is not identified, this means that, given the data, there are several values of the ATE consistent with the observed joint probability distribution.

To see this in a simple example, let us consider the case of a gaussian linear causal model $Y= \tau D + U_{y}$, with standardized observed variables. You only observe $\{Y, D\}$ and $D$ was not randomized, thus $cov(D, U_{y}) = \omega$ may be different from zero.

Our parameter of interest is the ATE:

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

But we only observe:

$$ cov(Y, X) = \tau + \omega $$

Now suppose $cov(Y, X) = 0.5$. Note that there are infinite values of the ATE compatible with the observed data, since you can always pick an $\omega$ such that $\tau + \omega = 0.5$.

  • $\begingroup$ Thanks, may I ask why $D$ being randomized causes $cov(D, U_y)$ to be zero? Thank you. $\endgroup$
    – user321627
    Commented Mar 8, 2019 at 18:06
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    $\begingroup$ @user321627 if $D$ was properly randomized, this means its values were determined by the randomization procedure, independently of everything else, including $U_y$ $\endgroup$ Commented Mar 8, 2019 at 18:26
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    $\begingroup$ @user321627 identification and bias can mean different things, depending on context. In statistics, bias usually means the difference of the expected value of an estimator from its population counterpart. For instance, you can have an unbiased estimator for the conditional expectation E(y|x) yet this still may not identify the counterfactual expectation E(Y_{x}). This answer might help stats.stackexchange.com/questions/303403/… $\endgroup$ Commented Mar 10, 2019 at 1:00
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    $\begingroup$ @user321627 estimand is a population quantity ("no hats"), for instance the expected value of $X$, $E[X]$, is an estimand--the population average. The estimator is what you do with finite samples, for instance, the sample average, $\widehat{E[X]} = \sum_{i=1}^{n}x_i/n$ is an estimator for the estimand $E[X]$. $\endgroup$ Commented Mar 24, 2019 at 4:55
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    $\begingroup$ @user321627 no, in the slides it’s also an estimand. $\endgroup$ Commented Mar 25, 2019 at 14:59

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