convert the ATE Average Treatment Effect from a doubly robust model into Risk Ratio How would one convert the ATE Average Treatment Effect from a doubly robust model into Risk Ratio? Can this be done directly without having to convert to odds ratio first. If so, how is this done. Can the conversion be directly applied to the confidence intervals or would bootstrapping have to be done?


You can split out the doubly robust estimator into two components:
$$\hat \Delta_{DR} = \hat \Delta_{DR,1} - \hat \Delta_{DR,0}$$
where
$$\hat\Delta_{DR,1} = \frac1n\sum_{i=1}^n \frac{Y_i}{\hat \pi_i} - \frac{A_i-\hat\pi_i}{\hat\pi_i}\hat E(Y|A = 1,X_i)$$
$$\hat\Delta_{DR,0} = \frac1n\sum_{i=1}^n \frac{Y_i}{1-\hat \pi_i} - \frac{A_i-(1-\hat\pi_i)}{1-\hat\pi_i}\hat E(Y|A = 0,X_i)$$
It is straightforward to show that (provided the substantive and regularity conditions motivating the DR estimator are true) that $\hat \Delta_{DR,1}, \hat \Delta_{DR,0}$ are separately consistent estimators of $E[Y_{i1}]$ and $E[Y_{i0}]$. Then by continuous mapping theorem, we know that
$$\frac{\hat \Delta_{DR,1}}{\hat\Delta_{DR,0}}\overset p\to \frac{E[Y_{i1}]}{E[Y_{i0}]}$$
Moreover, supposing that $\hat \Delta_{DR,1}, \hat \Delta_{DR,0}$ are asymptotically normal, i.e.
$$\sqrt n((\hat \Delta_{DR,1}, \hat \Delta_{DR,0})^T - \mu^T) \overset d\to \mathcal N(0,\Sigma)$$
where $\mu = (\mu_1,\mu_0) \equiv (E[Y_{i1}], {E[Y_{i0}]})$ and $\Sigma$ is the asymptotic variance matrix, then the delta method would yield
$$\sqrt n\left(\frac{\hat \Delta_{DR,1}}{\hat\Delta_{DR,0}} - \frac{E[Y_{i1}]}{E[Y_{i0}]}\right)\overset d\to \mathcal N(0,g \Sigma g^T)$$
where
$$g = \left(\frac{1}{\mu_0}, - \left(\frac{\mu_1}{\mu_0^2}\right)^2\right)$$
Finally, as usual, you can plug $\hat \Delta_{DR,1}, \hat\Delta_{DR,0}$ respectively for $\mu_1,\mu_0$ in the expression for $g$ and still obtain asymptotically valid inference.
