appendix-d

Appendix D. Proof for doubly robust ATE^aipw estimator

 

In Chapter 8 we introduced the ATE^aipw estimator and saw that it is doubly robust. If you are curious about why the ATE^aipw is doubly robust, you can find the proof here. We need to check the two following conditions:

  • If the models from the T-learner are unbiased, f0(c) = E[Y| c, T=0] and f1(c) = E[Y| c, T=1], then the ATE^aipw is also unbiased, that is E[ATE^aipw] = ATE.
  • If the propensity score is unbiased, s(c) = P(T=1|c), then the ATE^aipw is also unbiased, that is E[ATE^aipw] = ATE.

D.1 DR property with respect to the T-learner

We will first check the DR property for the T-learner. So, assume that the models from the T-learner are unbiased. First, notice that ATE^aipw can be expressed in terms of the T-learner estimator ATE^t. Consider the random variables Ti, Ci and Yi. Then

\begin{equation} \displaylines{ ATE^{\hat{}}_{aipw} = \\ \frac{1}{n} \left[\sum_i f_1(C_i) - f_0(C_i)\right] + \frac{1}{n} \left[\sum_i \frac{(Y_i - f_1(C_i)) T_i}{s(C_i)} - \frac{(Y_i - f_0(C_i))(1-T_i)}{1 - s(C_i)}\right] = } \end{equation}

\[ATE^{\hat{}}_{t} + RES^{\hat{}}_t =\]

If we can see that the residual has expectation zero, E[RES^_t] = 0, then we are done, because in that case

\[E[ATE^{\hat{}}_{t}] = E[ATE^{\hat{}}_{t}] + E[RES^{\hat{}}_t] = ATE\]

Let’s see that E[RES^_t] = 0. For simplicity, we will drop the index i and calculate the expectation for only one term of the summand. Using the total law of expectation we get

D.2 DR property with respect to the IPW