How would one convert the ATE Average Treatment Effect from a doubly robust model into Odds Ratio? Or is it better to discuss this by converting into risk ratio? If so, how is this done.
Propensity Scores – Converting ATE from a Doubly Robust Model into Odds Ratio
odds-ratiopropensity-scores
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You don't need the survey package or anything complicated. Wooldridge (2010, p. 920 onwards) "Econometric Analysis of Cross Section and Panel Data" has a simple procedure from which you can obtain the standard errors in order to construct the confidence intervals.
Under the assumption that you have correctly specified the propensity score which we denote as $p(\textbf{x}_i,\textbf{$\gamma$})$, define the score from the propensity score estimation (i.e. your first logit or probit regression) as $$\textbf{d}_i = \frac{\nabla_\gamma p(\textbf{x}_i,\textbf{$\gamma$})'[Z_i-p(\textbf{x}_i,\textbf{$\gamma$})]}{p(\textbf{x}_i,\textbf{$\gamma$}){[1-p(\textbf{x}_i,\textbf{$\gamma$})]}} $$ and let $$\text{ATE}_i = \frac{[Z_i-p(\textbf{x}_i,\textbf{$\gamma$})]Y_i}{p(\textbf{x}_i,\textbf{$\gamma$}){[1-p(\textbf{x}_i,\textbf{$\gamma$})]}}$$ as you have it in your expression above. Then take the sample analogues of these two expressions and regress $\widehat{\text{ATE}}_i$ on $\widehat{\textbf{d}}_i$. Make sure you include an intercept in this regression. Let $e_i$ be the residual from that regression, then the asymptotic variance of $\sqrt{N}(\widehat{\text{ATE}} - \text{ATE})$ is simply $\text{Var}(e_i)$. So the asymptotic standard error of your ATE is $$\frac{\left[ \frac{1}{N}\sum^N_{i=1}e_i^2 \right]^{\frac{1}{2}}}{\sqrt{N}}$$
You can then calculate the confidence interval in the usual way (see for example the comments to the answer here for a code example). You don't need to adjust the confidence interval again for the inverse propensity score weights because this step was already included in the calculation of the standard errors.
Unfortunately I am not an R guy so I can't provide you with the specific code but the outlined procedure above should be straight forward to follow. As a side note, this is also the way in which the treatrew command in Stata works. This command was written and introduced in the Stata Journal by Cerulli (2014). If you don't have access to the article you can check his slides which also outline the procedure of calculating the standard errors from inverse propensity score weighting. There he also discusses some slight conceptual differences between estimating the propensity score via logit or probit but for the sake of this answer it was not overly important so I omitted this part.
As some of the information you provided states, the two are not the same. I like better the terminology of conditional (on covariates) and unconditional (marginal) estimates. There is a very subtle language problem that clouds the issue greatly. Analysts who tend to love "population average effects" have a dangerous tendency to try to estimate such effects from a sample with no reference to any population distribution of subject characteristics. In this sense the estimates should not be called population average estimates but instead should be called sample average estimates. It is very important to note that sample average estimates have a low chance of being transportable to the population from which the sample came or in fact to any population. One reason for this is the somewhat arbitrary selection criteria for how subjects get into studies.
As an example, if one compared treatment A and treatment B in a binary logistic model adjusted for sex, one obtains a treatment effect that is specific to both males and females. If the sex variable is omitted from the model, a sample average odds ratio effect for treatment is obtained. This in effect is a comparison of some of the males on treatment A with some of the females on treatment B, due to non-collapsibility of the odds ratio. If one had a population with a different female:male frequency, this average treatment effect coming from a marginal odds ratio for treatment, will no longer apply.
So if one wants a quantity that pertains to individual subjects, full conditioning on covariates is required. And these conditional estimates are the ones that transport to populations, not the so-called "population average" estimates.
Another way to think about it: think of an ideal study for comparing treatment to no treatment. This would be a multi-period randomized crossover study. Then think about the next best study: a randomized trial on identical twins where one of the twins in each pair is randomly selected to get treatment A and the other is selected to get treatment B. Both of these ideal studies are mimicked by full conditioning, i.e., full covariate adjustment to get conditional and not marginal effects from the more usual parallel group randomized controlled trial.
Related Question
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Best Answer
If your outcome is binary, this is straightforward by noting that that the estimator is the difference between two counterfactual probabilities, and the odds ratio is just a ratio of the odds instead of of the probabilities. So, if
$$\hat{P}(Y^1=1) = n^{-1}\sum_{i=1}^n \left\{ \frac{A_i}{\hat{\pi}_i} Y_i - \frac{A_i - \hat{\pi}_i}{\hat{\pi}_i}\hat{E}(Y|A=1,X_i) \right\}$$ and $$\hat{P}(Y^0=1) = n^{-1}\sum_{i=1}^n \left\{ \frac{1-A_i}{1-\hat{\pi}_i} Y_i - \frac{A_i - \hat{\pi}_i}{1-\hat{\pi}_i}\hat{E}(Y|A=0,X_i) \right\}$$
and we know $\text{odds}(Y=1)=\frac{P(Y=1)}{1-P(Y=1)}$, then the DR estimator for the marginal causal odds ratio is simply
$$\frac{\hat{\text{odds}}(Y^1=1)}{ \hat{\text{odds}}(Y^0=1)} = \frac{\frac{\hat{P}(Y^1=1)}{1-\hat{P}(Y^1=1)}}{\frac{\hat{P}(Y^0=1)}{1-\hat{P}(Y^0=1)}}$$
To get confidence interval you can bootstrap or use the delta method or M-estimation to estimate the marginal log odds ratio from the marginal probabilities.