= Solution
Write $m_z(X)=\mu_z(X,\beta_z)$ and $\widetilde e(X)=e(X;\alpha)$. Conditional on $X$,
$$
\widetilde\mu_1^{\mathrm{dr}}(X)
=m_1(X)+\frac{e(X)}{\widetilde e(X)}\bigl(\mu_1(X)-m_1(X)\bigr),
$$
with the analogous control expression
$$
\widetilde\mu_0^{\mathrm{dr}}(X)
=m_0(X)+\frac{1-e(X)}{1-\widetilde e(X)}\bigl(\mu_0(X)-m_0(X)\bigr).
$$
If the propensity model is correct, both ratios are one and these equal $\mu_1(X)$ and $\mu_0(X)$, so their difference is the <conditional average treatment effect>.
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