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Following example 2 in this paper, suppose I wanted to estimate $\psi = E[E[Y|X,A=a]] $ and I had an influence function follows: $$ IF(\psi) = \frac{A}{\pi(X)}\{Y-\mu(X)\} - \psi $$ where $\pi(X)$ is the propensity score and $\mu(X) = E[Y|X,A=a]$. With estimator $$ \hat{\psi} = \mathbb{P}_n\bigg[\hat{\mu}(X) + \frac{A\{Y-\hat{\mu}(X)\}}{\hat{\pi}(X)}\bigg] $$

To estimate this I could do cross-fitting/sample-splitting to get $\hat{\mu}$ and $\hat{\pi}$: train a model on one-half the data and simply plug in the predicted values on the other half. It is unclear to me whether this estimator's variance considers the fact that there is downstream variability in fitting the models for the $\mu$ and $\pi$ on the separate partition of data. We may not have to due to cross-fitting but I am not sure. In essence, the question is can we treat $\hat{\mu}$ and $\hat{\pi}$ as fixed due cross-fitting?

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  • $\begingroup$ This is a good question and one of the miracles of doubly-robust inference. Unfortunately I don't have the technical knowledge to explain it any further. But I believe the IF arises from the nature of the estimator, not the cross-fitting. Cross-fitting allows the estimator to be consistent for a broader class of data-generating models than otherwise. $\endgroup$
    – Noah
    Sep 22 at 5:10
  • $\begingroup$ The following chapter might be helpful. Particularly, section 2.7.2 statnav.files.wordpress.com/2017/10/… $\endgroup$
    – pzivich
    Sep 22 at 12:10
  • $\begingroup$ @Noah I guess the question surrounds the variance of the IF with and without cross-fitting. If we didn't use cross-fitting, we definitely have to account for the simultaneous estimation of the propensity score, etc. But perhaps cross-fitting allows us to instead treat these as fixed, making the variance derivation much easier. $\endgroup$
    – Roy Z
    Sep 22 at 15:04
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    $\begingroup$ I don't have the bandwidth to make a rigorous explanation at the moment, but cross-fitting is exactly what allows you to treat $\hat{\mu}$, $\hat{\pi}$ as fixed (bc they're fit on the "other" fold(s)), and make statements about asymptotic variance. See Stefan Wager's notes (Lecture 3) on DR estimators. $\endgroup$
    – chang_trenton
    Oct 1 at 23:57
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    $\begingroup$ Here's a different estimator where cross-fitting is similarly used in the variance derivation. The latter also demonstrates that, for worse-than $\sqrt{n}$-consistent estimators, no cross-fitting may cause asymptotic variance to diverge. $\endgroup$
    – chang_trenton
    Oct 1 at 23:57

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