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The doubly robust estimator is a popular method for measuring the average treatment effect with observational data (assuming no unmeasured confounders):

$$ \hat{\Delta}_{DR} = n^{-1}\sum_{i=1}^n \biggl[\frac{Z_iY_i}{e(\mathbf{X_i},\hat{\beta_i})}-\frac{(Z_i-e(\mathbf{X_i},\hat{\beta_i}))}{e(\mathbf{X_i},\hat{\beta_i})}m_1(\mathbf{X_i},\hat{\alpha_1})\biggr] \\ -n^{-1}\sum_{i=1}^n \biggl[\frac{(1-Z_i)Y_i}{1-e(\mathbf{X_i},\hat{\beta_i})}-\frac{(Z_i-e(\mathbf{X_i},\hat{\beta_i}))}{1-e(\mathbf{X_i},\hat{\beta_i})}m_0(\mathbf{X_i},\hat{\alpha_0})\biggr]\\ =\hat{\mu}_{1,DR}-\hat{\mu}_{0,DR}.$$

Consider just the $\hat{\mu}_{1,DR}$ estimate for brevity (the $\hat{\mu}_{0,DR}$ calculations are similar), which can be reduced to the following expression: $$\hat{\mu}_{1,DR}=E(Y_1)+E\biggl[ \frac{(Z-e(\mathbf{X},\mathbf{\beta}))}{e(\mathbf{X},\mathbf{\beta})}(Y_1-m_1(\mathbf{X},\mathbf{\alpha_1}))\biggr].$$

If either the propensity score model $e(\mathbf{X},\bf{\beta})$ or the regression model $m_1(\mathbf{X},\mathbf{\alpha_1})$ is unbiased, the second "augmentation" term reduces to zero and we have $\hat{\mu}_{1,DR}=E(Y_1)$.

However, if both the propensity score and regression models are incorrect, the augmentation term will equal the product of the two model's expected biases, in which case the doubly robust estimator invariably ends up being worse than either the propensity score or regression model alone, is that correct?

Adding in the biased augmentation term for $\hat{\mu}_{0,DR}$ may mitigate the bias in $\hat{\mu}_{1,DR}$, although using the doubly robust estimator still presents a gamble.

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  • $\begingroup$ @jsk Perhaps a simple solution would be to expand the doubly robust estimator formula so that the augmentation terms are both added and subtracted from the expected values of $Y_1$ and $Y_0$? $\endgroup$ – RobertF Sep 12 '19 at 15:55
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    $\begingroup$ My memory was incorrect. In the simulations, the bias for the DR estimator of the odds ratio was no worse than the worse of the two estimators of the odds ratio, but the coverage probability was a bit lower. $\endgroup$ – jsk Sep 12 '19 at 19:04
  • $\begingroup$ @jsh I'll have to read the paper, do you have a link? I need to work out the math for the expected value of the augmentation term - taking the expectation of the product of two random variables, where one variable is a fraction - is not straightforward math. :-p At any rate the DRE may not be as biased as I thought. $\endgroup$ – RobertF Sep 12 '19 at 20:56
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    $\begingroup$ 2010 ncbi.nlm.nih.gov/pubmed/23049119 is the original paper that contains the theoretical details. 2013 ncbi.nlm.nih.gov/pmc/articles/PMC3664333 is a bit easier to read. Supplemental proof to 2013 paper that proves double robustness ncbi.nlm.nih.gov/pmc/articles/PMC3664333/bin/…. The simulation results though that I was referring to from a presentation do not appear to be available online. $\endgroup$ – jsk Sep 13 '19 at 0:53
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After posting this question I dug a little deeper and discovered the issue of bias in doubly robust estimators where both the outcome regression and propensity score models are incorrect has been addressed in a number of papers:

Kang, J., Schafer, J. (2007). Demystifying Double Robustness: A Comparison of Alternative Strategies for Estimating a Population Mean from Incomplete Data. Statistical Science, 22(4), 523-539. https://www.stat.cmu.edu/~ryantibs/journalclub/kang_2007.pdf

Tan, Z. (2007). Comment: Understanding OR, PS and DR. Statistical Science, 22(4), 560-568. https://arxiv.org/pdf/0804.2969.pdf

Cao, W., Tsiatis, A., Davidian, M. (2009). Improving efficiency and robustness of the doubly robust estimator for a population mean with incomplete data. Biometrika, 96(3), 723-734. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2798744/

The Cao et al. paper offers an alternative doubly robust estimator formula that performs well when both models are misspecified.

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