Now we can apply Proposition 3.1, to conclude
$\sqrt{N}(\hat{\theta}^{*}-\theta_{0,\textnormal{ATE}})\stackrel{{\scriptstyle\mathrm{d\,\,}}}{{\rightarrow}}\mathcal{N}(0,V_{0})$
where

$\displaystyle V_{0}$
$\displaystyle=\mathbb{E}_{0}[s_{\textnormal{AIPW}}(W;\theta_{0,\textnormal{ATE}},\nu_{0,\textnormal{AIPW}},e_{0})^{2}]$
$\displaystyle=\mathbb{E}_{0}\left[\left((\tau_{0}(X)-\theta_{0,\textnormal{ATE}})+\frac{Z(Y(1)-m_{0}(1,X))}{e_{0}(X)}-\frac{(1-Z)(Y(0)-m_{0}(0,X))}{1-e_{0}(X)}\right)^{2}\right]$
$\displaystyle=\mathbb{E}_{0}[(\tau_{0}(X)-\theta_{0,\textnormal{ATE}})^{2}]+\mathbb{E}_{0}\left[\frac{Z(Y(1)-m_{0}(1,X))^{2}}{(e_{0}(X))^{2}}\right]+\mathbb{E}_{0}\left[\frac{(1-Z)(Y(0)-m_{0}(0,X))^{2}}{(1-e_{0}(X))^{2}}\right]$
$\displaystyle=\mathbb{E}_{0}[(\tau_{0}(X)-\theta_{0,\textnormal{ATE}})^{2}]+\mathbb{E}_{0}\left[\frac{1}{e_{0}(X)}\cdot\mathbb{E}[(Y(1)-m_{0}(1,X))^{2}\mid
X]\right]$
$\displaystyle+\mathbb{E}_{0}\left[\frac{1}{1-e_{0}(X)}\cdot\mathbb{E}[(Y(0)-m_{0}(0,X))^{2}\mid
X]\right]$ $\displaystyle=V_{0,\textnormal{AIPW}}$

The third equality in the preceding display follows by noting that the three
cross terms in the expansion of the square have mean zero. The first two
vanish by conditioning on $X$ and the third because $Z(1-Z)=0$.

###  C.3 Proof of Corollary 3.2: CLT for $\hat{\theta}^{*}_{\textnormal{EPL}}$

Here we prove Corollary 3.2, the CLT for the partial linear estimator
$\hat{\theta}^{*}_{\textnormal{EPL}}$ of the regression parameter
$\theta_{0,\textnormal{PL}}$ under the linear treatment effect assumption (6).
This Corollary holds under the numbered generalizations of Appendix A with the
additional condition that the mean and variance functions are stationary. That
is, we have $\mathbb{E}_{t}[Y(z)\mid X=x]=m_{0}(z,x)$ and
$\textnormal{Var}_{t}(Y(z)\mid X=x)=v_{0}(z,x)$ for all $t=1,\ldots,T$,
$z=0,1$, and $x\in\mathcal{X}$. This condition is needed to ensure Assumption
A.2 holds.

As in the proof of Corollary 3.1, we first show that Assumption A.2 holds with
$\gamma=0$, estimand $\theta_{0}=\theta_{0,\textnormal{PL}}$, score
$s(\cdot)=s_{\textnormal{EPL}}(\cdot)$, and nuisance functions
$\nu_{0}=\nu_{0,\textnormal{EPL}}(\cdot)=(m_{0}(0,\cdot),v_{0}(0,\cdot),v_{0}(1,\cdot))$
lying in the nuisance set $\mathcal{N}=\mathcal{N}_{\textnormal{EPL}}$. Then
we show that the three numbered conditions in Proposition 3.1 hold.

Fix $e(\cdot),e^{\prime}(\cdot)\in\mathcal{F}_{0}$. For each $t=0,1,\ldots,T$,

$\displaystyle\mathbb{E}_{t,e}[Y\mid X,Z=0]$
$\displaystyle=\mathbb{E}_{t,e}[Y(0)\mid X]=m_{0}(0,X),\quad\text{and}$
$\displaystyle\mathbb{E}_{t,e}[Y\mid X,Z=1]$
$\displaystyle=\mathbb{E}_{t,e}[Y(1)\mid X]=m_{0}(1,X)$

hold by the unconfoundedness and SUTVA assumptions. Hence by (6),

$\mathbb{E}_{t,e}[Y\mid X,Z]=m_{0}(0,X)+Z\psi(X)^{\top}\theta_{0}.$ (55)

Thus for any $e^{\prime}(\cdot)\in\mathcal{F}_{\gamma}$,

$\displaystyle\mathbb{E}_{t,e}[s(W;\theta_{0,\textnormal{PL}},\nu_{0},e^{\prime})]$
$\displaystyle=\mathbb{E}_{t,e}[w(X;\nu_{0},e^{\prime})(Z-e^{\prime}(X))(Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0,\textnormal{PL}})\psi(X)]$
$\displaystyle=0$

after conditioning on $(X,Z)$ and applying (55). Integrability is not a
concern because $w(X;\nu_{0},e^{\prime})\leqslant c^{-1}$ for
$\nu_{0}\in\mathcal{N}_{\textnormal{EPL}}$. Thus Assumption A.2 is satisfied.

Now we consider the numbered conditions of Proposition 3.1 in turn.

  1. 1.

Because the predictor variables $\psi(X)$ satisfy $\|\psi(X)\|\leqslant C$

$\displaystyle(\mathbb{E}_{0}[\|s_{a}(W;\nu_{0},e_{0,N})-s_{a}(W;\nu_{0},e_{0})\|^{2}])^{1/2}$
$\displaystyle=(\mathbb{E}_{0}[\|Z\Delta(X,Z)\psi(X)\psi(X)^{\top}\|^{2}])^{1/2}$
$\displaystyle\leqslant C^{2}(\mathbb{E}_{0}[\Delta(X,Z)^{2}])^{1/2}$

where

$\displaystyle\Delta(X,Z)$
$\displaystyle=w(X;\nu_{0},e_{0})(Z-e_{0}(X))-w(X;\nu_{0},e_{0,N})(Z-e_{0,N}(X))$
$\displaystyle=(w(X;\nu_{0},e_{0})-w(X;\nu_{0},e_{0,N}))(Z-e_{0}(X))+w(X;\nu_{0},e_{0,N})(e_{0,N}(X)-e_{0}(X)).$

This $\Delta$ satisfies

$\displaystyle\begin{split}(\mathbb{E}_{0}[\Delta(X,Z)^{2}])^{1/2}&\leqslant(\mathbb{E}_{0}[(w(X;\nu_{0},e_{0})-w(X;\nu_{0},e_{0,N}))^{2}(Z-e_{0}(X))^{2}])^{1/2}\\\
&\phantom{=}+(\mathbb{E}_{0}[w^{2}(X;\nu_{0},e_{0,N})(e_{0,N}(X)-e_{0}(X))^{2}])^{1/2}.\end{split}$
(56)

Now $c\leqslant v_{0}(z,x)\leqslant\mathbb{E}[Y(z)^{2}\mid X=x]\leqslant C$
for all $z=0,1$ and $x\in\mathcal{X}$, so for any propensity $e(\cdot)$ we
have

$C^{-1}\leqslant
w(X;\nu_{0},e)=(v_{0}(0,X)e(X)+v_{0}(1,X)(1-e(X)))^{-1}\leqslant c^{-1}$

and

$\displaystyle\sup_{e\in[0,1]}\Big{|}\frac{\partial}{\partial
e}\frac{1}{v_{0}(0,X)e+v_{0}(1,X)(1-e)}\Big{|}$
$\displaystyle=\sup_{e\in[0,1]}\Big{|}\frac{v_{0}(0,X)-v_{0}(1,X)}{(v_{0}(0,X)e+v_{0}(1,X)(1-e))^{2}}\Big{|}$
$\displaystyle\leqslant\frac{2C}{c^{2}}.$

Applying these latter two facts to equation (56) yields

$\displaystyle(\mathbb{E}_{0}[\Delta(X,Z)^{2}])^{1/2}$
$\displaystyle\leqslant\left(\frac{2C}{c^{2}}+c^{-1}\right)\|e_{0,N}-e_{0}\|_{2,P_{0}^{X}}=o(1)$

by (54). Similarly,

$\displaystyle\phantom{=}\,\,(\mathbb{E}_{0}[\|s(W;\theta_{0,\textnormal{PL}},\nu_{0},e_{0,N})-s(W;\theta_{0,\textnormal{PL}},\nu_{0},e_{0})\|^{2}])^{1/2}$
$\displaystyle=(\mathbb{E}_{0}[\|\Delta(X,Z)(Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0})\psi(X)\|^{2}])^{1/2}$
$\displaystyle\leqslant C(\mathbb{E}_{0}[\Delta(X,Z)^{2}v_{0}(Z,X)])^{1/2}$
$\displaystyle\leqslant C^{3/2}(\mathbb{E}_{0}[\Delta(X,Z)^{2}])^{1/2}$
$\displaystyle=o(1).$

  2. 2.

For invertibility of $\mathbb{E}_{0}[s_{a}]$, we compute

$\displaystyle\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]$
$\displaystyle=\mathbb{E}_{0}[-w(X;\nu_{0},e_{0})(Z-e_{0}(X))Z\psi(X)\psi(X)^{\top}]$
$\displaystyle=\mathbb{E}_{0}[-w(X;\nu_{0},e_{0})e_{0}(X)(1-e_{0}(X))\psi(X)\psi(X)^{\top}]$
$\displaystyle\preccurlyeq-C^{-1}\mathbb{E}_{0}[e_{0}^{2}(X)(1-e_{0}^{2}(X))\psi(X)\psi(X)^{\top}].$

With $\mathbb{E}_{0}[e_{0}(X)(1-e_{0}(X))\psi(X)\psi(X)^{\top}]$ positive
definite by assumption, $\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]$ is strictly
negative definite and hence invertible.

For boundedness of $\mathbb{E}_{0}[\|s\|^{2}]$ with $e=e_{0}$, we compute

$\displaystyle\mathbb{E}_{0}[\|s(W;\theta_{0},\nu_{0},e_{0})\|^{2}]$
$\displaystyle=\mathbb{E}_{0}[w(X;\nu_{0},e_{0})^{2}(Z-e_{0}(X))^{2}(Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0})^{2}\|\psi(X)\|^{2}]$
$\displaystyle\leqslant\frac{C^{2}}{c^{2}}\mathbb{E}_{0}[(Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0})^{2}]$
$\displaystyle=\frac{C^{2}}{c^{2}}\mathbb{E}_{0}[v_{0}(Z,X)]\leqslant\frac{C^{3}}{c^{2}}$

  3. 3.

For boundedness of $\mathbb{E}_{0}[\|s\|^{q}]$ with $e=e_{0,N}$,

$\displaystyle(\mathbb{E}_{0}$
$\displaystyle[\|s(W;\theta_{0},\nu_{0},e_{0,N})\|^{q}])^{1/q}$
$\displaystyle=\Big{(}\mathbb{E}_{0}[|w(X;\nu_{0},e_{0,N})|^{q}|Z-e_{0,N}(X)|^{q}|Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}|^{q}\|\psi(X)\|^{q}]\Big{)}^{1/q}$
$\displaystyle\leqslant\frac{C}{c}(\mathbb{E}_{0}[|Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}|^{q}])^{1/q}$
$\displaystyle\leqslant\frac{4C^{2}}{c}$

where the final inequality follows from Minkowski’s inequality and Jensen’s
inequality as below:

$\displaystyle(\mathbb{E}_{0}[|Y-\mathbb{E}_{0}(Y\mid X,Z)|^{q}])^{1/q}$
$\displaystyle\leqslant(\mathbb{E}_{0}[|Y|^{q}])^{1/q}+(\mathbb{E}_{0}[|\mathbb{E}_{0}[Y\mid
X,Z]|^{q}])^{1/q}$
$\displaystyle\leqslant(\mathbb{E}_{0}[|Y|^{q}])^{1/q}+(\mathbb{E}_{0}[\mathbb{E}_{0}[|Y|^{q}\mid
X,Z]])^{1/q}$ $\displaystyle=2(\mathbb{E}_{0}[|Y|^{q}])^{1/q}$
$\displaystyle\leqslant
2[(\mathbb{E}_{0}[|Y(0)|^{q}])^{1/q}+(\mathbb{E}_{0}[|Y(1)|^{q}])^{1/q}]$
$\displaystyle\leqslant 4C\text{ by Assumption~{}\ref{assump:pl_regularity}}$

We now compute

$\displaystyle\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]$
$\displaystyle=\mathbb{E}_{0}[-w(X;\nu_{0},e_{0})Z(Z-e_{0}(X))\psi(X)\psi(X)^{\top}]$
$\displaystyle=\mathbb{E}_{0}[-w(X;\nu_{0},e_{0})(1-e_{0}(X))e_{0}(X)\psi(X)\psi(X)^{\top}]$
$\displaystyle=-\mathbb{E}_{0}\left[\frac{e_{0}(X)(1-e_{0}(X))}{v_{0}(0,X)e_{0}(X)+v_{0}(1,X)(1-e_{0}(X))}\psi(X)\psi(X)^{\top}\right]$

and

$\displaystyle\mathbb{E}_{0}[s(W;\theta_{0},\nu_{0},e_{0})^{\otimes 2}]$
$\displaystyle=\mathbb{E}_{0}[w^{2}(X;\nu_{0},e_{0})(Z-e_{0}(X))^{2}(Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0})^{2}\psi(X)\psi(X)^{\top}]$
$\displaystyle=\mathbb{E}_{0}[w^{2}(X;\nu_{0},e_{0})(Z-e_{0}(X))^{2}v_{0}(Z,X)\psi(X)\psi(X)^{\top}]$
$\displaystyle=\mathbb{E}_{0}[w^{2}(X;\nu_{0},e_{0})e_{0}^{2}(X)v_{0}(0,X)(1-e_{0}(X))\psi(X)\psi(X)^{\top}]$
$\displaystyle\phantom{=}\
+\mathbb{E}_{0}[w^{2}(X;\nu_{0},e_{0})(1-e_{0}(X))^{2}v_{0}(1,X)e_{0}(X)\psi(X)\psi(X)^{\top}]$
$\displaystyle=\mathbb{E}_{0}[e_{0}(X)(1-e_{0}(X))w^{2}(X;\nu_{0},e_{0})(v_{0}(0,X)e_{0}(X)+v_{0}(1,X)(1-e_{0}(X)))\psi(X)\psi(X)^{\top}]$
$\displaystyle=-\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})].$

Finally, we apply Proposition 3.1 to conclude
$\sqrt{N}(\hat{\theta}^{*}-\theta_{0,\textnormal{PL}})\rightarrow\mathcal{N}(0,V_{0})$
where

$\displaystyle V_{0}$
$\displaystyle=(\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})])^{-1}(\mathbb{E}_{0}[s(W;\theta_{0},\nu_{0},e_{0})^{\otimes
2}])(\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})])^{-1}$
$\displaystyle=-(\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})])^{-1}$
$\displaystyle=V_{0,\textnormal{EPL}}.$

###  C.4 Proof of Theorem 3.1: Pooling dominates linear aggregation

Here we prove that the oracle pooled estimators
$\hat{\theta}_{\textnormal{AIPW}}^{*}$ and
$\hat{\theta}_{\textnormal{EPL}}^{*}$ dominate the best linearly aggregated
estimators $\hat{\theta}_{\textnormal{AIPW}}^{*,(\mathrm{LA})}$ and
$\hat{\theta}_{\textnormal{EPL}}^{*,(\mathrm{LA})}$ for estimating
$\theta_{0,\textnormal{ATE}}$ and $\theta_{0,\textnormal{PL}}$, respectively.
For $\theta_{0.\textnormal{PL}}$, Theorem 3.1 generalizes to nonstationary
batches as described in Appendix A. But for $\theta_{0,\textnormal{ATE}}$, the
Theorem does not necessarily hold under the nonstationarities of Appendix A.
For example, suppose that $T=2$, $P_{1}^{X}$ is uniform on $(0,1)$, and
$P_{2}^{X}$ is the distribution with density $2x$ on $(0,1)$ with respect to
Lebesgue measure. Let $v_{0}(z,x)=x$ and suppose
$e_{1}(x)=e_{2}(x)=\kappa_{1}=1/2$ for each $x\in\mathcal{X}$. Then
$V_{0,\textnormal{AIPW}}=7/3>V^{(\mathrm{LA})}_{\textnormal{AIPW}}=16/7$. We
can however generalize the ATE result to the case of a multivariate outcome
variable, i.e., $Y(1),Y(0)\in{}^{q}$, where the conditional mean and variance
functions $m_{0}(\cdot)$ and $v_{0}(\cdot)$ take values in q and
$\mathbb{S}_{+}^{q}$, respectively.

We begin by showing the result for ATE estimation. We have
$V^{(\mathrm{LA})}_{\textnormal{AIPW}}=\left(\sum_{t=1}^{T}\kappa_{t}V_{t,\textnormal{AIPW}}^{-1}\right)^{-1}$
where we compute

$V_{t,\textnormal{AIPW}}=\mathbb{E}\left[\frac{v_{0}(1,X)}{e_{t}(X)}+\frac{v_{0}(0,X)}{1-e_{t}(X)}+(\tau_{0}(X)-\theta_{0})^{\otimes
2}\right],\quad t=1,\ldots,T$

by applying Corollary 3.1 to the observations in each single batch
$t\in\\{1,\ldots,T\\}$. Letting

$\displaystyle h(e)$
$\displaystyle=\left(\mathbb{E}\left[\frac{v_{0}(1,X)}{e(X)}+\frac{v_{0}(0,X)}{1-e(X)}+(\tau_{0}(X)-\theta_{0})^{\otimes
2}\right]\right)^{-1}$
$\displaystyle=\left(\mathbb{E}\left[\frac{v_{0}(1,X)(1-e(X))+v_{0}(0,X)e(X)}{e(X)(1-e(X))}+(\tau_{0}(X)-\theta_{0})^{\otimes
2}\right]\right)^{-1}$

for each $e\in\mathcal{F}_{\gamma}$, we have

$\displaystyle\Bigl{(}V^{(\mathrm{LA})}_{\textnormal{AIPW}}\Bigr{)}^{-1}$
$\displaystyle=\sum_{t=1}^{T}\kappa_{t}\left(\mathbb{E}\left[\frac{v_{0}(1,X)}{e_{t}(X)}+\frac{v_{0}(0,X)}{1-e_{t}(X)}+(\tau_{0}(X)-\theta_{0})^{\otimes
2}\right]\right)^{-1}$
$\displaystyle=\sum_{t=1}^{T}\kappa_{t}h(e_{t}),\quad\text{versus}$
$\displaystyle V_{0,\textnormal{AIPW}}^{-1}$
$\displaystyle=\left(\mathbb{E}\left[\frac{v_{0}(1,X)}{e_{0}(X)}+\frac{v_{0}(0,X)}{1-e_{0}(X)}+(\tau_{0}(X)-\theta_{0})^{\otimes
2}\right]\right)^{-1}$ $\displaystyle=h(e_{0}).$

Thus, to prove the desired result, it suffices to show $h(\cdot)$ is a concave
(matrix-valued) function on $\mathcal{F}_{\gamma}$. To that end, fix
$e_{1},e_{2}\in\mathcal{F}_{\gamma}$ and $\lambda\in(0,1)$. It suffices to
show that $g(\lambda)=h(e_{1}+\lambda(e_{2}-e_{1}))$ is concave on $[0,1]$,
i.e., that $g(\lambda)\succcurlyeq\lambda g(e_{2})+(1-\lambda)g(e_{1})$ for
each $\lambda\in[0,1]$. Because $g(\cdot)$ is continuous on $[0,1]$, we need
only show that $g^{\prime\prime}(\lambda)\preccurlyeq 0$ for each
$\lambda\in(0,1)$. Letting $e_{\lambda}=e_{1}+\lambda(e_{2}-e_{1})$ and
$V_{\lambda}=h(e_{\lambda})^{-1}$, we obtain

$\displaystyle
g^{\prime\prime}(\lambda)=2h(e_{\lambda})[B_{\lambda}h(e_{\lambda})B_{\lambda}-C_{\lambda}]h(e_{\lambda})$
(57)

for

$\displaystyle B_{\lambda}$
$\displaystyle=\mathbb{E}\left[(e_{2}(X)-e_{1}(X))\left(\frac{v_{0}(0,X)}{(1-e_{\lambda}(X))^{2}}-\frac{v_{0}(1,X)}{(e_{\lambda}(X))^{2}}\right)\right],\quad\text{and}$
$\displaystyle C_{\lambda}$
$\displaystyle=\mathbb{E}\left[(e_{2}(X)-e_{1}(X))^{2}\left(\frac{v_{0}(1,X)}{(e_{\lambda}(X))^{3}}+\frac{v_{0}(0,X)}{(1-e_{\lambda}(X))^{3}}\right)\right].$

To get this, we reversed the order of differentiation and expectation, which
is justified by the regularity conditions in Assumption 3.1.

Since $h(e_{\lambda})$ is symmetric and appears on the left and right in (57),
it suffices to show that $B_{\lambda}h(e_{\lambda})B_{\lambda}\preccurlyeq
C_{\lambda}$. For this we note that for $\tilde{Z}$ conditionally independent
of $(Y(0),Y(1))$ given $X$ with $\tilde{Z}\mid
X\sim\textnormal{Bern}(e_{\lambda}(X))$, we have
$h(e_{\lambda})=(\textnormal{Cov}(b_{\lambda}))^{-1}$,
$B_{\lambda}=\textnormal{Cov}(c_{\lambda},b_{\lambda})=B_{\lambda}^{\prime}$,
and $C_{\lambda}=\textnormal{Cov}(c_{\lambda})$ where

$\displaystyle b_{\lambda}$
$\displaystyle=\frac{(1-\tilde{Z})(Y(0)-m_{0}(0,X))}{1-e_{\lambda}(X)}-\frac{\tilde{Z}(Y(1)-m_{0}(1,X))}{e_{\lambda}(X)}-(\tau_{0}(X)-\theta_{0})$
$\displaystyle c_{\lambda}$
$\displaystyle=(e_{2}(X)-e_{1}(X))\biggl{(}\frac{\tilde{Z}(Y(1)-m_{0}(1,X))}{(e_{\lambda}(X))^{2}}+\frac{(1-\tilde{Z})(Y(0)-m_{0}(0,X))}{(1-e_{\lambda}(X))^{2}}\biggr{)}$

are mean zero random vectors in d. Applying Lemma B.6 completes the proof for
ATE.

Now we consider the partially linear model, allowing for the generalizations
in Appendix A, as discussed above. We have
$V_{\textnormal{EPL}}^{(\mathrm{LA})}=\bigl{(}\sum_{t=1}^{T}\kappa_{t}V_{t,\textnormal{EPL}}^{-1}\bigr{)}^{-1}$
where

$\displaystyle
V_{t,\textnormal{EPL}}=\biggl{(}\mathbb{E}_{t}\left[\frac{e_{t}(X)(1-e_{t}(X))}{v_{0}(0,X)e_{t}(X)+v_{0}(1,X)(1-e_{t}(X))}\psi(X)\psi(X)^{\top}\right]\biggr{)}^{-1}$

by applying Corollary 3.2 to the observations in each single batch
$t\in\\{1,\ldots,T\\}$. Then

$\displaystyle\bigl{(}V^{(\mathrm{LA})}_{\textnormal{EPL}}\bigr{)}^{-1}$
$\displaystyle=\sum_{t=1}^{T}\kappa_{t}V_{t,\textnormal{EPL}}^{-1}$
$\displaystyle=\sum_{t=1}^{T}\mathbb{E}_{t}\left[\kappa_{t}\frac{e_{t}(X)(1-e_{t}(X))}{v_{0}(0,X)e_{t}(X)+v_{0}(1,X)(1-e_{t}(X))}\psi(X)\psi(X)^{\top}\right]$
$\displaystyle=\mathbb{E}_{0}\left[\sum_{t=1}^{T}\kappa_{t}\frac{\mathrm{d}P_{t}^{X}}{\mathrm{d}P_{0}^{X}}(X)\frac{e_{t}(X)(1-e_{t}(X))}{v_{0}(0,X)e_{t}(X)+v_{0}(1,X)(1-e_{t}(X))}\psi(X)\psi(X)^{\top}\right]$
$\displaystyle\preccurlyeq\mathbb{E}_{0}\left[\frac{e_{0}(X)(1-e_{0}(X))}{v_{0}(0,X)e_{0}(X)+v_{0}(1,X)(1-e_{0}(X))}\psi(X)\psi(X)^{\top}\right]$
$\displaystyle=(V_{0,\textnormal{EPL}})^{-1}$

where the inequality follows from the fact that for each $x\in\mathcal{X}$, we
have
$\sum_{t=1}^{T}\kappa_{t}{\mathrm{d}P_{t}^{X}}/{\mathrm{d}P_{0}^{X}}(x)=1$ and
the map

$e\mapsto\frac{e(1-e)}{v(0,x)e+v(1,x)(1-e)}\psi(x)\psi(x)^{\top}$

is concave on $e\in[0,1]$.

###  C.5 Proof of Theorem 4.1: CLT for feasible $\hat{\theta}$ in a CSBAE

Here we prove the CLT for the feasible estimator $\hat{\theta}$ in a CSBAE. It
holds under the nonstationarity conditions described in Appendix A.

We begin by constructing the non-adaptive batch experiment in the statement of
the theorem. Define the counterfactual treatment indicators via

$\tilde{Z}_{ti}=\bm{1}(U_{ti}\leqslant e_{t}(X_{ti})),\quad t=1,\ldots,T,\quad
i=1,\ldots,N_{t}$

for $U_{ti}\stackrel{{\scriptstyle\mathrm{iid}}}{{\sim}}\mathbb{U}(0,1)$. Then
the observations in the counterfactual non-adaptive batch experiment are the
vectors $\tilde{W}_{ti}=(\tilde{Y}_{ti},X_{ti},\tilde{Z}_{ti})$, where

$\tilde{Y}_{ti}=Y_{ti}(\tilde{Z}_{it})=\tilde{Z}_{it}Y_{ti}(1)+(1-\tilde{Z}_{it})Y_{ti}(0).$

The corresponding oracle estimator $\hat{\theta}^{*}$ from (11) is then

$\displaystyle\hat{\theta}^{*}$
$\displaystyle=-\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})\biggr{)}^{-1}\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{b}(\tilde{W}_{ti};\nu_{0},e_{0,N})\biggr{)}.$

Because $(\nu_{0},e_{0,N})\in\mathcal{T}_{N}$, conditions 1, 2, and 3 of
Proposition 3.1 are satisfied for this counterfactual non-adaptive batch
experiment by equations (22), (23), and (25) along with condition (a) of
Assumption 4.1. Hence the oracle CLT
$\sqrt{n}(\hat{\theta}^{*}-\theta_{0})\stackrel{{\scriptstyle\mathrm{d\,\,}}}{{\rightarrow}}\mathcal{N}(0,V_{0})$
holds with $V_{0}$ as in the conclusion of Proposition 3.1.

It remains to show that

$\hat{\theta}=\hat{\theta}^{*}+o_{p}(N^{-1/2})$

where

$\displaystyle\hat{\theta}=-\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{b}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}$

as in (18). To do so, we show that the following intermediate quantity

$\tilde{\theta}=-\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{b}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}$

satisfies both $\tilde{\theta}=\hat{\theta}^{*}+o_{p}(N^{-1/2})$ and
$\hat{\theta}=\tilde{\theta}+o_{p}(N^{-1/2})$.

####  C.5.1 Showing that $\tilde{\theta}=\hat{\theta}^{*}+o_{p}(N^{-1/2})$

We first show $\tilde{\theta}=\hat{\theta}^{*}+o_{p}(N^{-1/2})$. By score
linearity (4), we can write

$N^{1/2}(\tilde{\theta}-\hat{\theta}^{*})=N^{1/2}(\tilde{\theta}-\theta_{0})-N^{1/2}(\hat{\theta}^{*}-\theta_{0})=A_{1}B_{1}+A_{2}B_{2}$

where

$\displaystyle A_{1}$
$\displaystyle=\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})\biggr{)}^{-1}-\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1},$
$\displaystyle B_{1}$
$\displaystyle=\frac{1}{\sqrt{N}}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N}),$
$\displaystyle A_{2}$
$\displaystyle=\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}\quad\text{and}$
$\displaystyle B_{2}$
$\displaystyle=\frac{1}{\sqrt{N}}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}[s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})-s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})].$

We will prove that $A_{1}=o_{p}(1)$, $B_{1}=O_{p}(1)$, $A_{2}=O_{p}(1)$, and
$B_{2}=o_{p}(1)$.

To show $A_{1}=o_{p}(1)$, for each fold $k\in\\{1,\ldots,K\\}$ let
$N_{k}=\sum_{t=1}^{T}n_{t,k}=N/K+O(1)$ be the total number of observations in
fold $k$ across all batches $t=1,\ldots,T$. Also define the quantity

$\tilde{A}_{1}^{(k)}=\frac{1}{N_{k}}\sum_{(t,i)\in\mathcal{I}_{k}}[s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})-s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})]=\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\tilde{A}_{1,t}^{(k)}$

where

$\tilde{A}_{1,t}^{(k)}=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})-s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)}).$

For each $k=1,\ldots,K$, let $\mathcal{E}_{N,k}$ be the event that
$(\hat{\nu}^{(-k)},\hat{e}^{(-k)})\in\mathcal{T}_{N}$. This
$\mathcal{E}_{N,k}$ is $\mathcal{S}^{(-k)}$ measurable and
$\textnormal{Pr}(\mathcal{E}_{N,k})\rightarrow 1$ as $N\rightarrow\infty$ by
assumption. Then for all sufficiently large $N$,

$\displaystyle\phantom{=}\,\mathbb{E}\bigl{[}\|\tilde{A}_{1,t}^{(k)}\bm{1}(\mathcal{E}_{N,k})\|\mid\mathcal{S}^{(-k)}\bigr{]}$
$\displaystyle\leqslant\bm{1}(\mathcal{E}_{N,k})\biggl{(}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in
I_{j}}\mathbb{E}\bigl{[}\|s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})\|\mid\mathcal{S}^{(-k)}\bigr{]}\biggr{)}$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}\mathbb{E}_{t}\bigl{[}\|s_{a}(W;\nu,e)-s_{a}(W;\nu_{0},e_{0,N})\|\bigr{]}$
$\displaystyle\leqslant\kappa_{t}^{-1}\sup_{(\nu,e)\in\mathcal{T}_{N}}\mathbb{E}_{0}\bigl{[}\|s_{a}(W;\nu,e)-s_{a}(W;\nu_{0},e_{0})\|\bigr{]}+\kappa_{t}^{-1}\mathbb{E}_{0}[\|s_{a}(W;\nu_{0},e_{0,N})-s_{a}(W;\nu_{0},e_{0})\|]$
$\displaystyle\leqslant 2\kappa_{t}^{-1}\delta_{N}$

by equation (22). The second inequality above uses the fact that
$\hat{\nu}^{(-k)}$, $\hat{e}^{(-k)}$ are nonrandom given $\mathcal{S}^{(-k)}$,
but the vectors $\tilde{W}_{ti}$ in fold $k$ are independent of
$\mathcal{S}^{(-k)}$ and i.i.d. from $P_{t}$; the third inequality follows
because

$\frac{\mathrm{d}P_{t}}{\mathrm{d}P_{0}}(w)\leqslant\kappa_{t}^{-1}$ (58)

for all $w$ by the definitions of $P_{t}$ and $P_{0}$. Then
$\tilde{A}_{1,t}^{(k)}=o_{p}(1)$ by Lemma B.1. This immediately shows
$\tilde{A}_{1}^{(k)}=o_{p}(1)$ for all folds $k=1,\ldots,K$ and so

$\tilde{A}_{1}=\sum_{k=1}^{K}\frac{N_{k}}{N}\tilde{A}_{1}^{(k)}=o_{p}(1),$
(59)

too. Now recall the identity

$\|A^{-1}-B^{-1}\|=\|A^{-1}(A-B)B^{-1}\|\leqslant\|A^{-1}\|\|A-B\|\|B^{-1}\|$
(60)

for any invertible square matrices $A,B$ of the same size. By (52) we know
that

$\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})=\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]+o_{p}(1)$
(61)

and hence

$\displaystyle\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})\biggr{)}^{-1}\quad\text{and}\quad\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}$

are both $O_{p}(1)$. The preceding display along with (59) and (60) show
$A_{1}=o_{p}(1)$.

Equations (59) and (61) immediately imply that $A_{2}=O_{p}(1)$, while (C.1)
implies that $B_{1}=O_{p}(1)$. It remains to show $B_{2}=o_{p}(1)$. To that
end we consider the quantity

$\displaystyle B_{2}^{(k)}$
$\displaystyle=\frac{1}{\sqrt{N_{k}}}\sum_{(t,i)\in\mathcal{I}_{k}}s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})$
$\displaystyle=\bar{B}_{2}^{(k)}+\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\tilde{B}_{2,t}^{(k)}$

where for each $t=1,\ldots,T$

$\displaystyle\bar{B}_{2}^{(k)}$
$\displaystyle=\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}[s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)}],\quad\text{and}$
$\displaystyle\tilde{B}_{2,t}^{(k)}$
$\displaystyle=\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})$
$\displaystyle\phantom{=}-\mathbb{E}[s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)}].$

To see that $\bar{B}_{2}^{(k)}=o_{p}(1)$, we write

$\displaystyle\bar{B}_{2}^{(k)}$
$\displaystyle=N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\int
s(w;\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(w;\theta_{0},\nu_{0},e_{0,N})\mathrm{d}P_{t}(w)$
$\displaystyle=N_{k}^{1/2}\int
s(w;\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(w;\theta_{0},\nu_{0},e_{0,N})\mathrm{d}P_{0,N}(w)$
$\displaystyle\phantom{=}+N_{k}^{1/2}\sum_{t=1}^{T}\left(\frac{n_{t,k}}{N_{k}}-\frac{N_{t}}{N}\right)\int
s(w;\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(w;\theta_{0},\nu_{0},e_{0,N})\mathrm{d}P_{t}(w).$

Now

$f_{N}^{(k)}(\lambda)=\int
s(w;\theta_{0},\nu_{0}+\lambda(\hat{\nu}^{(-k)}-\nu_{0}),e_{0,N}+\lambda(\hat{e}^{(-k)}-e_{0,N}))-s(w;\theta_{0},\nu_{0},e_{0,N})\mathrm{d}P_{0,N}(w)$

is twice continuously differentiable on $[0,1]$ by regularity. Hence by
Taylor’s theorem

$\|f_{N}^{(k)}(1)-f_{N}^{(k)}(0)-f_{N}^{(k)\prime}(0)\|\leqslant\frac{1}{2}\sup_{\lambda\in(0,1)}\|f_{N}^{(k)\prime\prime}(\lambda)\|.$
(62)

By (21) we know that

$\displaystyle\sup_{\lambda\in(0,1)}\|f_{N}^{(k)\prime\prime}(\lambda)\|\bm{1}(\mathcal{E}_{N,k})$
$\displaystyle\leqslant\sup_{\lambda\in(0,1)}\sup_{(\nu,e)\in\mathcal{T}_{N}}\bigg{\|}\frac{\partial^{2}}{\partial\lambda^{2}}\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]\bigg{\|}$
$\displaystyle\leqslant N^{-1/2}\delta_{N}.$

With $f_{N}^{(k)}(0)=0$, by (20) and (62) we conclude that that

$\|f_{N}^{(k)}(1)\|\bm{1}(\mathcal{E}_{N,K})\leqslant\frac{3}{2}N^{-1/2}\delta_{N}.$

Recalling that $\textnormal{Pr}(\mathcal{E}_{N,k})\rightarrow 1$, we have
$\|f_{N}^{(k)}(1)\|=o_{p}(N^{-1/2})$ and

$\bigg{\|}N_{k}^{1/2}\int
s(w;\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(w;\theta_{0},\nu_{0},e_{0,N})\,\mathrm{d}P_{0,N}(w)\bigg{\|}=N_{k}^{1/2}\|f_{N}^{(k)}(1)\|=o_{p}(1).$

Using $N_{t}/N\to\kappa_{t}$ from (3) we have

$\displaystyle\phantom{=}\,\bigg{\|}N_{k}^{1/2}\sum_{t=1}^{T}\left(\frac{n_{t,k}}{N_{k}}-\frac{N_{t}}{N}\right)\int
s(w;\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(w;\theta_{0},\nu_{0},e_{0,N})\,\mathrm{d}P_{t}(w)\bigg{\|}$
$\displaystyle\leqslant
N_{k}^{1/2}\sum_{t=1}^{T}\bigg{|}\frac{n_{t,k}}{N_{k}}-\frac{N_{t}}{N}\bigg{|}\cdot\frac{N}{N_{t}}\|f_{N}^{(k)}(1)\|$
$\displaystyle=o_{p}(1)$

as

$\frac{\mathrm{d}P_{t}}{\mathrm{d}P_{0,N}}(w)\leqslant\frac{N}{N_{t}},\quad
t=1,\ldots,T,\quad w\in\mathcal{W}.$ (63)

Thus we have shown $\bar{B}_{2}^{(k)}=o_{p}(1)$. Finally, for each batch
$t=1,\ldots,T$, the quantity $\tilde{B}_{2,t}^{(k)}$ is a sum of $n_{t,k}$
random variables that are i.i.d. and mean 0 conditional on
$\mathcal{S}^{(-k)}$. Thus by Lemma B.7, we have

$\displaystyle\mathbb{E}\bigl{[}\|\tilde{B}_{2,t}^{(k)}\|^{2}\mid\mathcal{S}^{(-k)}\bigr{]}$
$\displaystyle=\sum_{i:(t,i)\in\mathcal{I}_{k}}\frac{1}{n_{t,k}}\bar{\mathbb{E}}\bigl{[}\|r_{ti}^{(k)}\|^{2}\bigr{]}$
$\displaystyle\leqslant\sum_{i:(t,i)\in\mathcal{I}_{k}}\frac{1}{n_{t,k}}\mathbb{E}\bigl{[}\|s(W_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti};\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}^{(-k)}\bigr{]}$

where for each $i$ such that $(t,i)\in\mathcal{I}_{k}$ we’ve defined

$\displaystyle\begin{split}r_{ti}^{(k)}&=s(W_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})\\\
&\phantom{=}\,-\mathbb{E}[s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(\tilde{W}_{ti};\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)}]\end{split}$
(64)

and used the basic variance inequality

$\mathbb{E}\bigl{[}\|X-\mathbb{E}[X\mid\mathcal{S}]\|^{2}\bigr{]}\leqslant\mathbb{E}\bigl{[}\|X\|^{2}\mid\mathcal{S}\bigr{]}$
(65)

for any random vector $X$ and $\sigma$-algebra $\mathcal{S}$. Hence by (58),
Minkowski’s inequality, and then (23), we have

$\displaystyle\bigl{(}\mathbb{E}\bigl{[}\|\tilde{B}_{2,t}^{(k)}\|^{2}\mid\mathcal{S}^{(-k)}\bigr{]}\bigr{)}^{1/2}\bm{1}(\mathcal{E}_{N,k})$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}\bigl{(}\mathbb{E}_{t}\bigl{[}\|s(W;\theta_{0},\nu,e)-s(W;\theta_{0},\nu_{0},e_{0,N})\|^{2}\bigr{]}\bigr{)}^{1/2}$
$\displaystyle\leqslant\kappa_{t}^{-1/2}\sup_{(\nu,e)\in\mathcal{T}_{N}}\bigl{(}\mathbb{E}_{0,N}\bigl{[}\|s(W;\theta_{0},\nu,e)-s(W;\theta_{0},\nu_{0},e_{0})\|^{2}\bigr{]}\bigr{)}^{1/2}$
$\displaystyle\phantom{=}\,+\kappa_{t}^{-1/2}\bigl{(}\mathbb{E}_{0,N}\bigl{[}\|s(W;\theta_{0},\nu_{0},e_{0,N})-s(W;\theta_{0},\nu_{0},e_{0})\|^{2}\bigr{]}\bigr{)}^{1/2}$
$\displaystyle\leqslant 2\kappa_{t}^{-1/2}\delta_{N}.$

Thus $\tilde{B}_{2,t}^{(k)}=o_{p}(1)$ by Lemma B.1. With

$-B_{2}=\sum_{k=1}^{K}\sqrt{\frac{N_{k}}{N}}B_{2}^{(k)}=\sum_{k=1}^{K}\sqrt{\frac{N_{k}}{N}}\biggl{(}\bar{B}_{2}^{(k)}+\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\tilde{B}_{2,t}^{(k)}\biggr{)}$

we conclude that $B_{2}=o_{p}(1)$, as desired. This establishes that
$\hat{\theta}^{*}=\tilde{\theta}+o_{p}(N^{-1/2})$.

####  C.5.2 Showing that $\hat{\theta}=\hat{\theta}^{*}+o_{p}(N^{-1/2})$

To establish that $\hat{\theta}=\tilde{\theta}+o_{p}(N^{-1/2})$, similar to
above we write

$N^{1/2}(\hat{\theta}-\tilde{\theta})=N^{1/2}(\hat{\theta}-\theta_{0})-N^{1/2}(\tilde{\theta}-\theta_{0})=A_{3}B_{3}+A_{4}B_{4}$

where

$\displaystyle A_{3}$
$\displaystyle=\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}-\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1},$
$\displaystyle B_{3}$
$\displaystyle=\frac{1}{\sqrt{N}}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)}),$
$\displaystyle A_{4}$
$\displaystyle=\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1},\quad\text{and}$
$\displaystyle B_{4}$
$\displaystyle=\frac{1}{\sqrt{N}}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}[s(\tilde{W}_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti};\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})].$

We will show that $A_{3}=o_{p}(1)$, $B_{3}=O_{p}(1)$, $A_{4}=O_{p}(1)$, and
$B_{4}=o_{p}(1)$. First we define

$\tilde{A}_{3}=\sum_{k=1}^{K}\frac{N_{k}}{N}\tilde{A}_{3}^{(k)}$

where the term $\tilde{A}_{3}^{(k)}$ for each fold is decomposed into a sum
over batches $t=1,\ldots,T$:

$\displaystyle\tilde{A}_{3}^{(k)}$
$\displaystyle=\frac{1}{N_{k}}\sum_{(t,i)\in\mathcal{I}_{k}}[s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})]$
$\displaystyle=\frac{1}{N_{k}}\sum_{(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})(s_{a}(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti}(0);\hat{\nu}^{(-k)},\hat{e}^{(-k)}))$
$\displaystyle=\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\tilde{A}_{3,t}^{(k)}.$

Here

$\displaystyle\tilde{A}_{3,t}^{(k)}$
$\displaystyle:=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})(s_{a}(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti}(0);\hat{\nu}^{(-k)},\hat{e}^{(-k)}))$

satisfies

$\displaystyle\|\tilde{A}_{3,t}^{(k)}\|$
$\displaystyle\leqslant\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bm{1}(Z_{ti}\neq\tilde{Z}_{ti})\cdot\|s_{a}(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti}(0);\hat{\nu}^{(-k)},\hat{e}^{(-k)})\|$
$\displaystyle\leqslant\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\|s_{a}(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti}(0);\hat{\nu}^{(-k)},\hat{e}^{(-k)})\|^{2}}$
$\displaystyle\phantom{=}\,\cdot\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bm{1}(Z_{ti}\neq\tilde{Z}_{ti})}$

by the Cauchy-Schwarz inequality. Recalling the $\sigma$-algebra
$\mathcal{S}_{t}^{X,(k)}$ in the definition of a CSBAE (Definition 4.1), we
have

$\textnormal{Pr}(Z_{ti}\neq\tilde{Z}_{ti}\mid\mathcal{S}_{t}^{X,(k)})=|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|$

and so by (17)

$\displaystyle\mathbb{E}\biggl{[}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bm{1}(Z_{ti}\neq\tilde{Z}_{ti})\Bigm{|}\mathcal{S}_{t}^{X,(k)}\biggr{]}$
$\displaystyle=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|$
$\displaystyle\leqslant\|\hat{e}_{t}^{(k)}-e_{t}\|_{2,P_{N,t}^{X,(k)}}$
$\displaystyle=o_{p}(1).$

Hence

$\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bm{1}(Z_{ti}\neq\tilde{Z}_{ti})}=o_{p}(1)$

by Lemma B.1. Next, for $z=0,1$ and $(t,i)\in\mathcal{I}_{k}$, by Jensen’s
inequality and (58) we have

$\displaystyle\left(\mathbb{E}\bigl{[}\|s_{a}(W_{ti}(z);\hat{\nu}^{(-k)},\hat{e}^{(-k)})\|^{2}\bm{1}(\mathcal{E}_{N,k})\mid\mathcal{S}^{(-k)}\bigr{]}\right)^{q/2}$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}(\mathbb{E}_{t}\bigl{[}\|s_{a}(W(z);\nu,e)\|^{2}\bigr{]})^{q/2}$
$\displaystyle\leqslant\kappa_{t}^{-1}\sup_{(\nu,e)\in\mathcal{T}_{N}}\mathbb{E}_{0}[\|s_{a}(W(z);\nu,e)\|^{q}]$
$\displaystyle\leqslant\kappa_{t}^{-1}C^{q}$

by (25). By Markov’s inequality, for each batch $t=1,\ldots,T$ and fold
$k=1,\ldots,K$,

$\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\|s_{a}(W_{ti}(z);\hat{\nu}^{(-k)},\hat{e}^{(-k)})\|^{2}}=O_{p}(1)$

holds for $z=0,1$. Applying Minkowski’s inequality with the
$L^{2}_{P_{N,t}^{X}}$ norm shows

$\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\|s_{a}(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s_{a}(W_{ti}(0);\hat{\nu}^{(-k)},\hat{e}^{(-k)})\|^{2}}=O_{p}(1).$

We conclude $\tilde{A}_{3,t}^{(k)}=o_{p}(1)$, hence
$\tilde{A}_{3}^{(k)}=o_{p}(1)$ and $\tilde{A}_{3}=o_{p}(1)$ as well. Now we
recall

$\displaystyle\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})$
$\displaystyle=\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\nu_{0},e_{0,N})+o_{p}(1)\text{\quad(by~{}\eqref{eq:A_1_tilde})}$
$\displaystyle=\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]+o_{p}(1)\text{\quad(by~{}\eqref{eq:oracle_s_a_lln_folds})}.$

Then

$\displaystyle\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})$
$\displaystyle=\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})-\tilde{A}_{3}$
$\displaystyle=\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]+o_{p}(1)$

as well. Thus we have both

$\displaystyle\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(\tilde{W}_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}$
$\displaystyle=(\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})])^{-1}+o_{p}(1),\quad\text{and}$
$\displaystyle\biggl{(}\frac{1}{N}\sum_{k=1}^{K}\sum_{(t,i)\in\mathcal{I}_{k}}s_{a}(W_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})\biggr{)}^{-1}$
$\displaystyle=(\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})])^{-1}+o_{p}(1).$ (66)

Subtracting these two equations gives $A_{3}=o_{p}(1)$.

Next, equation (66) immediately provides $A_{4}=O_{p}(1)$, while
$B_{3}=B_{1}-B_{2}=O_{p}(1)$ as shown above. Thus it only remains to show that
$B_{4}=o_{p}(1)$. We write

$B_{4}=\sum_{k=1}^{K}\sqrt{\frac{N_{k}}{N}}\Bigl{(}B_{4}^{(k),1}+B_{4}^{(k),2}+B_{4}^{(k),3}\Bigr{)}$

where

$\displaystyle B_{4}^{(k),1}$
$\displaystyle=\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\bigl{(}s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\bigr{)},$
$\displaystyle B_{4}^{(k),2}$
$\displaystyle=N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\mathbb{E}[s(W_{ti}(1);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}]$
$\displaystyle\phantom{=}\,-N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\mathbb{E}[s(W_{ti}(0);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}]$
and $\displaystyle B_{4}^{(k),3}$
$\displaystyle=\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\bigl{(}s(W_{ti}(1);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(0);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})\bigr{)}$
$\displaystyle-B_{4}^{(k),1}-B_{4}^{(k),2}.$

It suffices to prove that for each $k=1,\ldots,K$, the terms $B_{4}^{(k),1}$,
$B_{4}^{(k),2}$, and $B_{4}^{(k),3}$ are all asymptotically negligible. First,
we have

$\displaystyle\mathbb{E}$
$\displaystyle\left[(\tilde{Z}_{ti}-Z_{ti})(s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N}))\mid\mathcal{S}_{t}^{X,(k)}\right]$
$\displaystyle=(e_{t}(X_{ti})-\hat{e}_{t}^{(k)}(X_{ti}))\mathbb{E}[s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle=(e_{t}(X_{ti})-\hat{e}_{t}^{(k)}(X_{ti}))\mathbb{E}[s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\mid
X_{ti}]$ $\displaystyle=0.$

We briefly justify each of the equalities in the preceding display:

  * •

The first equality holds because given $\mathcal{S}_{t}^{X,(k)}$, the only
randomness in $(Z_{ti},\tilde{Z}_{ti})$ for any $(t,i)\in\mathcal{I}_{k}$ is
in $U_{ti}$, which is independent of $(W_{ti}(0),W_{ti}(1))$,
$\mathcal{S}^{(-k)}$, and $\mathcal{S}_{t}^{X,(k)}$; hence

$\begin{split}(Z_{ti},\tilde{Z}_{ti})&\perp\\!\\!\\!\perp(W_{ti}(0),W_{ti}(1))\mid\mathcal{S}_{t}^{X,(k)}\quad\text{
and }\\\
(Z_{ti},\tilde{Z}_{ti})&\perp\\!\\!\\!\perp(W_{ti}(0),W_{ti}(1))\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}.\end{split}$
(67)

  * •

The second equality holds because the vectors $\\{S_{ti},1\leqslant t\leqslant
T,1\leqslant i\leqslant N_{t}\\}$ are mutually independent (Assumption A.1).

  * •

The third equality follows directly from (19).

For batches $t=1,\ldots,T$, the vectors
$\\{(\tilde{Z}_{ti},Z_{ti},W_{ti}(1),W_{ti}(0)),1\leqslant i\leqslant
N_{t}\\}$ are conditionally independent given $\mathcal{S}_{t}^{X,(k)}$.
Defining

$\mathcal{I}_{4,t}^{(k)}=\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})(s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N}))$

and letting $p$ be the real solution to $p^{-1}+2q^{-1}=1$, we apply Lemma B.7
and Holder’s inequality to get

$\displaystyle\phantom{=}\,\mathbb{E}[\|\mathcal{I}_{4,t}^{(k)}\|^{2}\mid\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}[(\tilde{Z}_{ti}-Z_{ti})^{2}\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|\mathbb{E}\big{[}\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}_{t}^{X,(k)}\big{]}$

using the conditional independence in (67) once again. Then

$\displaystyle\mathbb{E}[\|\mathcal{I}_{4,t}^{(k)}\|^{2}\mid\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle\leqslant\biggl{(}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|^{p}\biggr{)}^{1/p}$
$\displaystyle\cdot\biggl{(}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\left(\mathbb{E}[\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}_{t}^{X,(k)}]\right)^{q/2}\biggr{)}^{2/q}.$

We can assume WLOG that $q<4$ so that $p>2$. Then with
$0\leqslant|\hat{e}_{t}^{(j)}(x)-e_{t}(x)|\leqslant 1$ for all $x$, we
conclude

$\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\big{|}\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})\big{|}^{p}\leqslant\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bigl{(}\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})\bigr{)}^{2}=o_{p}(1)$
(68)

by (17). Next, from Jensen’s inequality

$\displaystyle\phantom{=}\,\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\left(\mathbb{E}\bigl{[}\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}_{t}^{X,(k)}\bigr{]}\right)^{q/2}$
$\displaystyle\leqslant\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}\left[\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{q}\mid\mathcal{S}_{t}^{X,(k)}\right]$
$\displaystyle=\mathbb{E}_{t}\bigl{[}\|s(W(1);\theta_{0},\nu_{0},e_{0,N})-s(W(0);\theta_{0},\nu_{0},e_{0,N})\|^{q}\mid
X\bigr{]}.$

But by Minkowski’s inequality, (58), and (25), we see

$\displaystyle\phantom{=}\
\bigl{(}\mathbb{E}_{t}[\|s(W(1);\theta_{0},\nu_{0},e_{0,N})-s(W(0);\theta_{0},\nu_{0},e_{0,N})\|^{q}]\bigr{)}^{1/q}$
$\displaystyle\leqslant\bigl{(}\mathbb{E}_{t}[\|s(W(1);\theta_{0},\nu_{0},e_{0,N})\|^{q}])^{1/q}+(\mathbb{E}_{t}[\|s(W(0);\theta_{0},\nu_{0},e_{0,N})\|^{q}]\bigr{)}^{1/q}$
$\displaystyle\leqslant\kappa_{t}^{-1/q}[(\mathbb{E}_{0}[\|s(W(1);\theta_{0},\nu_{0},e_{0,N})\|^{q}])^{1/q}+(\mathbb{E}_{0}[\|s(W(0);\theta_{0},\nu_{0},e_{0,N})\|^{q}])^{1/q}]$
$\displaystyle\leqslant 2C\kappa_{t}^{-1/q}.$

We conclude by Markov’s inequality that

$\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\left(\mathbb{E}[\|s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})\|^{2}\mid\mathcal{S}_{t}^{X,(k)}]\right)^{q/2}=O_{p}(1).$

Along with (68) we can conclude that $\mathcal{I}_{4,t}^{(k)}=o_{p}(1)$ by
Lemma B.1. Then also

$B_{4}^{(k),1}=\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}\mathcal{I}_{4,t}^{(k)}=o_{p}(1)$

Next, for $z=0,1$ define

$B_{4}^{(k),2}(z)=N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\mathbb{E}\bigl{[}s(W_{ti}(z);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(z);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}\bigr{]}$

so that $B_{4}^{(k),2}=B_{4}^{(k),2}(1)-B_{4}^{(k),2}(0)$. Then by the
triangle inequality $\|B_{4}^{(k),2}(z)\|$ is no larger than

$N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\tilde{Z}_{ti}-Z_{ti}|\cdot\|\mathbb{E}[s(W_{ti}(z);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(z);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}]\|.$

Taking conditional expectations of both sides yields

$\displaystyle\mathbb{E}\bigl{[}\|B_{4}^{(k),2}(z)\|\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}\bigr{]}$
$\displaystyle\leqslant
N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bigl{|}\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})\bigr{|}$
$\displaystyle\qquad\qquad\qquad\times\bigl{\|}\mathbb{E}[s(W_{ti}(z);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(z);\theta_{0},\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}]\bigr{\|}$
$\displaystyle\leqslant N_{k}^{1/2}\sum_{t=1}^{T}\frac{n_{t,k}}{N_{k}}\cdot
S_{t}^{(-k)}(z)\cdot\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti}))^{2}}$

by Cauchy-Schwarz. We have by (17) that

$\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti}))^{2}}=O_{p}(N^{-1/4}),$

Then by equation (26), we get $B_{4}^{(k),2}(z)=o_{p}(1)$ for $z=0,1$, and so
$B_{4}^{(k),2}=o_{p}(1)$ as well.

Finally, we write

$B_{4}^{(k),3}=\sum_{t=1}^{T}\sqrt{\frac{n_{t,k}}{N_{k}}}B_{4,t}^{(k),3}$

for

$\displaystyle B_{4,t}^{(k),3}$
$\displaystyle=\frac{1}{\sqrt{n_{t,k}}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\tilde{Z}_{ti}-Z_{ti})\Delta_{ti}^{(-k)},$
where $\displaystyle\Delta_{ti}^{(-k)}$
$\displaystyle=(s(W_{ti}(1);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(0);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)}))$
$\displaystyle\phantom{=}\
-(s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N}))$
$\displaystyle\phantom{=}\
-\mathbb{E}[s(W_{ti}(1);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(0);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})\mid\mathcal{S}^{(-k)},X_{ti}]$
$\displaystyle\phantom{=}\
-\mathbb{E}[s(W_{ti}(1);\theta_{0},\nu_{0},e_{0})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0})\mid\mathcal{S}^{(-k)},X_{ti}].$

For each $(t,i)\in\mathcal{I}_{k}$, we have

$(W_{ti}(0),W_{ti}(1))\perp\\!\\!\\!\perp\mathcal{S}_{t}^{X,(k)}\mid\mathcal{S}^{(-k)},X_{ti}$
(69)

since given $\mathcal{S}^{(-k)}$ and $X_{ti}$, the only remaining randomness
in $(W_{ti}(0),W_{ti}(1))$ is in the potential outcomes
$(Y_{ti}(0),Y_{ti}(1))$. These are independent of both $\mathcal{S}^{(-k)}$
and $\\{X_{tj}\mid j\neq i\\}$, the covariates of the other subjects in batch
$t$. Thus
$\mathbb{E}[\Delta_{ti}^{(-k)}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]=0$,
and by (67) we have

$\mathbb{E}[B_{4,t}^{(k),3}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]=0.$

Then applying Lemma B.7 and equations (67) and (65),

$\displaystyle\mathbb{E}[\|B_{4,t}^{(k),3}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}[(\tilde{Z}_{ti}-Z_{ti})^{2}\|\Delta_{ti}^{(-k)}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|\mathbb{E}[\|\Delta_{ti}^{(-k)}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle\leqslant\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|\mathbb{E}[\|\tilde{\Delta}_{ti}^{(-k)}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]$
$\displaystyle\leqslant\biggl{(}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}|\hat{e}_{t}^{(k)}(X_{ti})-e_{t}(X_{ti})|^{p}\biggr{)}^{1/p}\Gamma_{t}^{(k)}$

by Holder’s inequality, where $p$ is as above for

$\displaystyle\Gamma_{t}^{(k)}$
$\displaystyle=\biggl{(}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\mathbb{E}[\|\tilde{\Delta}_{ti}^{(-k)}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}])^{q/2}\biggr{)}^{2/q},\quad\text{with}$
$\displaystyle\tilde{\Delta}_{ti}^{(-k)}$
$\displaystyle=(s(W_{ti}(1);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(0);\theta_{0},\hat{\nu}^{(-k)},\hat{e}^{(-k)}))$
$\displaystyle\phantom{=}\
-(s(W_{ti}(1);\theta_{0},\nu_{0},e_{0,N})-s(W_{ti}(0);\theta_{0},\nu_{0},e_{0,N})).$

By Jensen’s inequality

$(\Gamma_{t}^{(k)})^{q/2}\leqslant\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}\bigl{[}\|\tilde{\Delta}_{ti}^{(-k)}\|^{q}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}\bigr{]}.$

Taking conditional expectations, we get

$\mathbb{E}\bigl{[}(\Gamma_{t}^{(k)})^{q/2}\mid\mathcal{S}^{(-k)}\bigr{]}=\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\mathbb{E}\bigl{[}\|\tilde{\Delta}_{ti}^{(-k)}\|^{q}\mid\mathcal{S}^{(-k)}\bigr{]}.$

Then

$\displaystyle\phantom{=}\
\left(\mathbb{E}[(\Gamma_{t}^{(k)})^{q/2}\mid\mathcal{S}^{(-k)}]\right)^{1/q}\bm{1}(\mathcal{E}_{N,k})$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}\left(\mathbb{E}_{t}[\|(s(W(1);\theta_{0},\nu,e)-s(W(0);\theta_{0},\nu,e))-(s(W(1);\theta_{0},\nu_{0},e_{0,N})-s(W(0);\theta_{0},\nu_{0},e_{0,N}))\|^{q}]\right)^{1/q}$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}\sum_{z\in\\{0,1\\}}(\mathbb{E}_{t}[\|s(W(z);\theta_{0},\nu,e)\|^{q}])^{1/q}+(\mathbb{E}_{t}[\|s(W(z);\theta_{0},\nu_{0},e_{0,N})\|^{q}])^{1/q}$
$\displaystyle\leqslant 4C\kappa_{t}^{-1/q}$

by (58) and moment boundedness. We conclude that
$\mathbb{E}[(\Gamma_{t}^{(k)})^{q/2}\mid\mathcal{S}^{(-k)}]\bm{1}(\mathcal{E}_{N,k})$
is uniformly bounded. Recalling that
$\textnormal{Pr}(\mathcal{E}_{N,k})\rightarrow 1$, Markov’s inequality then
ensures $\Gamma_{t}^{(k)}=O_{p}(1)$. In view of (68), we then have
$\mathbb{E}[\|B_{4,t}^{(k),3}\|^{2}\mid\mathcal{S}^{(-k)},\mathcal{S}_{t}^{X,(k)}]=o_{p}(1)$.
This implies $B_{4,t}^{(k),3}=o_{p}(1)$ by Lemma B.1, and hence
$B_{4}^{(k),3}=o_{p}(1)$. This establishes that
$\hat{\theta}=\tilde{\theta}+o_{p}(N^{-1/2})$ and completes the proof.

###  C.6 Proof of Corollary 4.1: CLT for feasible
$\hat{\theta}_{\textnormal{AIPW}}$ in a CSBAE

Corollary 4.1, which shows the feasible estimator
$\hat{\theta}_{\textnormal{AIPW}}$ satisfies a CLT for estimating
$\hat{\theta}_{0,\textnormal{ATE}}$ in a CSBAE, holds under the numbered
generalizations at the end of Appendix A, subject to the additional
requirement that the mean functions are stationary, as in our generalization
of Corollary 3.1 to nonstationary batches. We prove this more general result.
The proof of our generalized Corollary 3.1 shows that Assumption A.2 is
satisfied under the moment bounds in Assumption 3.1 with
$\theta_{0}=\theta_{0,\textnormal{ATE}}$,
$s(\cdot)=s_{\textnormal{AIPW}}(\cdot)$,
$\nu_{0}=\nu_{0,\textnormal{AIPW}}\in\mathcal{N}=\mathcal{N}_{\textnormal{AIPW}}$,
and any $\gamma\in(0,1/2)$. It remains to show that the further conditions of
Assumption 4.1, namely (a), (b), (c) and equations (20) through (25), are
satisfied. Then Corollary 4.1 follows by Theorem 4.1.

Condition (a)

Invertibility of $\mathbb{E}_{0}[\theta_{0}(W;\nu_{0},e_{0})]$ and existence
of $\mathbb{E}_{0}[s(W;\theta_{0},\nu_{0},e_{0})^{2}]$ follow from Assumption
3.1, as shown in the proof of Corollary 3.1.

Condition (b)

Fix $\lambda\in[0,1]$ and
$(\nu,e)\in\mathcal{T}=\mathcal{N}\times\mathcal{F}_{\gamma}$, where
$\nu(\cdot)=(m(0,\cdot),m(1,\cdot))$. Consider

$\displaystyle f_{N}(\lambda)$
$\displaystyle:=s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))$
$\displaystyle=m_{\lambda}(1,X)-m_{\lambda}(0,X)+\frac{Z(Y-m_{\lambda}(1,X))}{e_{\lambda,N}(x)}-\frac{(1-Z)(Y-m_{\lambda}(0,X))}{1-e_{\lambda,N}(x)}$

where $m_{\lambda}(z,x):=m_{0}(z,x)+\lambda(m(z,x)-m_{0}(z,x))$ and
$e_{\lambda,N}(x):=e_{0,N}(x)+\lambda(e(x)-e_{0,N}(x))$. Taking two
derivatives with respect to $\lambda$ we get

$\displaystyle f_{N}^{\prime}(\lambda)$
$\displaystyle=(m(1,X)-m_{0}(1,X))-(m(0,X)-m_{0}(0,X))$
$\displaystyle\phantom{=}\
+Z\left(\frac{m_{0}(1,X)-m(1,X)}{e_{\lambda,N}(X)}-\frac{(Y-m_{\lambda}(1,X))(e(X)-e_{0,N}(X))}{e_{\lambda,N}(X)^{2}}\right)$
$\displaystyle\phantom{=}\
-(1-Z)\left(\frac{m_{0}(0,X)-m(0,X)}{1-e_{\lambda,N}(X)}-\frac{(Y-m_{\lambda}(0,X))(e_{0,N}(X)-e(X))}{(1-e_{\lambda,N}(X))^{2}}\right),\quad\text{and}$
$\displaystyle f_{N}^{\prime\prime}(\lambda)$
$\displaystyle=2Z\left(\frac{(Y-m_{\lambda}(1,X))(e(X)-e_{0,N}(X))^{2}}{e_{\lambda,N}(X)^{3}}-\frac{(m_{0}(1,X)-m(1,X))(e(X)-e_{0,N}(X))}{e_{\lambda,N}(X)^{2}}\right)$
$\displaystyle\phantom{=}\
+2(1-Z)\left(\frac{m_{0}(0,X)-m(0,X)(e_{0,N}(X)-e(X))}{(1-e_{\lambda,N}(X))^{2}}-\frac{(Y-m_{\lambda}(0,X))(e_{0,N}(X)-e(X))^{2}}{(1-e_{\lambda,N}(X))^{3}}\right).$

We now show that $f_{N}(\cdot)$, $f_{N}^{\prime}(\cdot)$, and
$f_{N}^{\prime\prime}(\cdot)$ are upper bounded by an integrable random
variable on an open interval containing $[0,1]$, so that by the Leibniz rule,
we can swap both first and second derivatives with expectations to conclude
the function
$\lambda\mapsto\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]$
has second derivative $\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]$ for
$\lambda\in[0,1]$. Note $e_{\lambda,N}\in\mathcal{F}_{\gamma}$. Then by
repeated application of the triangle inequality, we see that for sufficiently
small $\epsilon>0$, we must have

$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}|f_{N}(\lambda)|$
$\displaystyle\leqslant 2(|m(1,X)|+|m_{0}(1,X)|+|m(0,X)|+|m_{0}(0,X)|)$
$\displaystyle+2\gamma^{-1}(|Y(1)|+|Y(0)|+|m(1,X)|+|m_{0}(1,X)|+|m(0,X)|+|m_{0}(0,X)|)$
$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}|f_{N}^{\prime}(\lambda)|$
$\displaystyle\leqslant 2(|m(1,X)|+|m_{0}(1,X)|+|m(0,X)|+|m_{0}(0,X)|)$
$\displaystyle+2\gamma^{-2}(|Y(1)|+|Y(0)|+|m(1,X)|+|m_{0}(1,X)|+|m(0,X)|+|m_{0}(0,X)|)$
$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}|f_{N}^{\prime\prime}(\lambda)|$
$\displaystyle\leqslant
2\gamma^{-3}(|Y(1)|+|Y(0)|+|m(1,X)|+|m_{0}(1,X)|+|m(0,X)|+|m_{0}(0,X)|).$

The right-hand sides above are clearly integrable under $P_{0,N}$. Hence, the
mapping
$\lambda\mapsto\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]$
is indeed twice differentiable with second derivative
$\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]$ for $\lambda\in[0,1]$. This
second derivative is continuous for such $\lambda$ by continuity of
$f_{N}^{\prime\prime}(\cdot)$ and dominated convergence.

Condition (c)

Fix $e\in\mathcal{F}_{\gamma}$. By stationarity of the mean functions
$m_{0}(z,\cdot)$, $z=0,1$, for each $t=1,\ldots,T$, $i=1,\ldots,N_{t}$ we have
(letting $\nu_{0}=\nu_{0,\textnormal{AIPW}}$ for brevity)

$\displaystyle\mathbb{E}[s(W_{ti}(1);\theta_{0},\nu_{0},e)\mid X_{ti}]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}+\mathbb{E}_{t}\left[\frac{Y(1)-m_{0}(1,X)}{e(X)}\Bigm{|}X\right]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}+(e(X))^{-1}\mathbb{E}_{t}[Y(1)-m_{0}(1,X)\\!\mid\\!X]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}$

and similarly

$\displaystyle\mathbb{E}[s(W_{ti}(0);\theta_{0},\nu_{0},e)\mid X_{ti}]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}-\mathbb{E}_{t}\left[\frac{Y(0)-m_{0}(0,X)}{1-e(X)}\Bigm{|}X\right]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}-(1-e(X))^{-1}\mathbb{E}_{t}[Y(0)-m_{0}(0,X)\\!\mid\\!X]$
$\displaystyle=m_{0}(1,X)-m_{0}(0,X)-\theta_{0}.$

Subtracting shows (19).

It remains to show that the out-of-fold estimators
$(\hat{m}^{(-k)}(0,\cdot),\hat{m}^{(-k)}(1,\cdot),\hat{e}^{(-k)})$ lie in some
set $\mathcal{T}_{N}$ with high probability for all sufficiently large $N$,
where this set $\mathcal{T}_{N}$ satisfies equations (20) through (25). To
construct this $\mathcal{T}_{N}$, we see that by the rate conditions on the
nuisance estimators along with equation (54), there exists a sequence
$\tilde{\delta}_{N}\downarrow 0$ so that
$\|e_{0,N}-e_{0}\|_{2,P_{0}^{X}}\leqslant\tilde{\delta}_{N}$ for all $N$.
Furthermore, with probability approaching 1 as $N\rightarrow\infty$, these
four conditions hold for $z=0,1$ and all folds $k=1,\ldots,K$:

$\displaystyle\|\hat{m}^{(-k)}(z,\cdot)-m_{0}(z,\cdot)\|_{2,P_{0}^{X}}+\|\hat{e}^{(-k)}-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant\tilde{\delta}_{N},$
$\displaystyle\|\hat{m}^{(-k)}(z,\cdot)\|_{2,P_{0}^{X}}\times\|\hat{e}^{(-k)}-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant N^{-1/2}\tilde{\delta}_{N},$
$\displaystyle\|\hat{m}^{(-k)}(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}$
$\displaystyle\leqslant C,$ $\displaystyle\hat{e}^{(-k)}(\cdot)$
$\displaystyle\in\mathcal{F}_{\gamma}.$

We then define $\mathcal{T}_{N}$ be the set of functions
$(m(0,\cdot),m(1,\cdot),e(\cdot))$ in
$\mathcal{T}=\mathcal{N}_{\textnormal{AIPW}}\times\mathcal{E}$ obeying these
conditions:

$\displaystyle\|m(z,\cdot)-m_{0}(z,\cdot)\|_{2,P_{0}^{X}}+\|e-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant\tilde{\delta}_{N},$
$\displaystyle\|m(z,\cdot)\|_{2,P_{0}^{X}}\times\|e-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant N^{-1/2}\tilde{\delta}_{N},$
$\displaystyle\|m(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}$
$\displaystyle\leqslant C,$ $\displaystyle e(\cdot)$
$\displaystyle\in\mathcal{F}_{\gamma}.$

By construction, for all $k=1,\ldots,K$ we have
$\textnormal{Pr}((\hat{m}^{(-k)},\hat{e}^{(-k)})\in\mathcal{T}_{N})\rightarrow
1$ as $N\rightarrow\infty$.

For the remainder of the proof, we take $N$ large enough so that
$1/2\leqslant{\mathrm{d}P_{0,N}^{X}}/{\mathrm{d}P_{0}^{X}}\leqslant 2$. For
such $N$, $\|f\|_{2,P_{0,N}^{X}}\leqslant\sqrt{2}\|f\|_{2,P_{0}^{X}}$ holds
for all $f\in L^{2}(P_{0}^{X})$. We will show equations (20) through (23) hold
for a sequence $\delta_{N}$ that is some constant multiple of
$\tilde{\delta}_{N}$.

Equation (20)

Fix $(\nu,e)\in\mathcal{T}_{N}$. By the calculations and notation above in the
proof of condition (b) and interchanging differentiation with expectation, we
can verify using unconfoundedness that

$\frac{\partial}{\partial\lambda}\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0},\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N})]\Big{|}_{\lambda=0}=\mathbb{E}_{0,N}[f_{N}^{\prime}(0)]=0$

so the left-hand side of the Neyman orthogonality condition (20) is 0. The
full calculation is shown in the proof of Theorem 5.1 of Chernozhukov et al.,
(2018).

Equation (21)

Once again, we fix $(\nu,e)\in\mathcal{T}_{N}$ and recall the calculations and
notations in the proof of condition (b) above. We see that

$\displaystyle\Big{|}\frac{\partial^{2}}{\partial\lambda^{2}}\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0},\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N})]\Big{|}=\bigl{|}\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]\bigr{|}$

and hence

$\displaystyle\bigl{|}\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]\bigr{|}$
$\displaystyle\leqslant\frac{2}{\gamma^{3}}\Bigl{(}\bigl{|}\mathbb{E}_{0,N}[(Y(1)-m_{\lambda}(1,X))(e(X)-e_{0,N}(X))^{2}]\bigr{|}$
$\displaystyle\phantom{=}\
+\bigl{|}\mathbb{E}_{0,N}[(Y(0)-m_{\lambda}(0,X))(e_{0,N}(X)-e(X))^{2}]\bigr{|}\Bigr{)}$
$\displaystyle\phantom{=}\
+\frac{2}{\gamma^{2}}\Bigl{(}|\mathbb{E}_{0,N}[(m_{0}(1,X)-m(1,X))(e(X)-e_{0}(X))]|$
$\displaystyle\phantom{=}\
+\bigl{|}\mathbb{E}_{0,N}[(m_{0}(0,X)-m(0,X))(e(X)-e_{0}(X))]\bigr{|}\Bigr{)}.$

By Cauchy-Schwarz and the definition of $\mathcal{T}_{N}$ we have for $z=0,1$
that

$\displaystyle\phantom{=}\
\mathbb{E}_{0,N}\bigl{[}|m_{0}(z,X)-m(z,X)|\times|e(X)-e_{0,N}(X)|\bigr{]}$
$\displaystyle\leqslant\|m_{0}(z,\cdot)-m(z,\cdot)\|_{2,P_{0,N}^{X}}\times\|e-e_{0,N}\|_{2,P_{0,N}^{X}}$
$\displaystyle\leqslant 2N^{-1/2}\tilde{\delta}_{N}.$

Furthermore for $z=0,1$ we have
$\mathbb{E}_{0,N}\bigl{[}(Y(z)-m_{0}(z,X))(e(X)-e_{0,N}(X))^{2}\bigr{]}=0$ by
conditioning on $X$. Hence, for all $\lambda\in[0,1]$,

$\displaystyle\phantom{=}\
\bigl{|}\mathbb{E}_{0,N}[(Y(z)-m_{\lambda}(z,X))(e(X)-e_{0,N}(X))^{2}]\bigr{|}$
$\displaystyle=\bigl{|}\mathbb{E}_{0,N}[\lambda(m(z,X)-m_{0}(z,X))(e(X)-e_{0,N}(X))^{2}]\bigr{|}$
$\displaystyle\leqslant\lambda\mathbb{E}_{0,N}\bigl{[}|m(z,X)-m_{0}(z,X)||e(X)-e_{0,N}(X)|\bigr{]}$
$\displaystyle\leqslant 2N^{-1/2}\tilde{\delta}_{N}$

where the first inequality uses the fact that $|e(X)-e_{0,N}(X)|\leqslant 1$.
Taking suprema over $(\nu,e)\in\mathcal{T}_{N}$ and $\lambda\in[0,1]$, we get

$\sup_{(\nu,e)\in\mathcal{T}_{N}}\sup_{\lambda\in[0,1]}\Big{|}\frac{\partial^{2}}{\partial\lambda^{2}}\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0},\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N})]\Big{|}\leqslant
N^{-1/2}\left(\frac{8}{\gamma^{3}}+\frac{8}{\gamma^{2}}\right)\tilde{\delta}_{N}$

which shows equation (21).

Equation (22)

For any $(\nu,e)\in\mathcal{T}_{N}$, trivially

$\mathbb{E}_{0}\bigl{[}|s_{a}(W;\nu,e)-s_{a}(W;\nu_{0},e_{0})|^{2}\bigr{]}=\mathbb{E}_{0}\bigl{[}|-1-(-1)|^{2}\bigr{]}=0.$

Equation (23)

We fix $(\nu,e)\in\mathcal{T}_{N}$ and write

$\displaystyle s(W;\theta_{0},\nu,e)-s(W;\theta_{0},\nu_{0},e_{0})$
$\displaystyle=\left(1-\frac{Z}{e_{0}(X)}\right)(m(1,X)-m_{0}(1,X))$
$\displaystyle\phantom{=}\
+\left(1-\frac{1-Z}{1-e_{0}(X)}\right)(m_{0}(0,X)-m(0,X))$
$\displaystyle\phantom{=}\ +Z(Y-m(1,X))(e(X)^{-1}-e_{0}(X)^{-1})$
$\displaystyle\phantom{=}\ -(1-Z)(Y-m(0,X))((1-e(X))^{-1}-(1-e_{0}(X))^{-1}).$

It now suffices to show that each of the summands has asymptotically vanishing
second moment. The definition of $\mathcal{T}_{N}$ ensures that

$\displaystyle\mathbb{E}_{0}\biggl{[}\biggl{(}1-\frac{Z}{e_{0}(X)}\biggr{)}^{2}(m(1,X)-m_{0}(1,X))^{2}\biggr{]}$
$\displaystyle\leqslant(1+\gamma^{-1})^{2}\tilde{\delta}_{N}^{2}\quad\text{and}$
$\displaystyle\mathbb{E}_{0}\biggl{[}\biggl{(}1-\frac{1-Z}{1-e_{0}(X)}\biggr{)}^{2}(m_{0}(0,X)-m(0,X))^{2}\biggr{]}$
$\displaystyle\leqslant(1+\gamma^{-1})^{2}\tilde{\delta}_{N}^{2}.$

Next

$\displaystyle\phantom{=}\
\,\mathbb{E}_{0}\bigl{[}Z^{2}(Y(1)-m(1,X))^{2}(e(X)^{-1}-e_{0}(X)^{-1})^{2}\bigr{]}$
$\displaystyle\leqslant\gamma^{-4}\mathbb{E}_{0}\bigl{[}(Y(1)-m(1,X))^{2}(e(X)-e_{0}(X))^{2}\bigr{]}$
$\displaystyle=\gamma^{-4}\mathbb{E}_{0}\bigl{[}(e(X)-e_{0}(X))^{2}v_{0}(1,X)\bigr{]}$
$\displaystyle\phantom{=}\
+\gamma^{-4}\mathbb{E}_{0}\bigl{[}(e(X)-e_{0}(X))^{2}(m(1,X)-m_{0}(1,X))^{2}\bigr{]}$
$\displaystyle\leqslant
C\gamma^{-4}\|e-e_{0}\|_{2,P_{0}^{X}}^{2}+\gamma^{-4}\|m(1,\cdot)-m_{0}(1,\cdot)\|_{2,P_{0}^{X}}^{2}$
$\displaystyle\leqslant(2C+1)\gamma^{-4}\tilde{\delta}_{N}^{2}.$

By a similar computation

$\mathbb{E}_{0}[(1-Z)^{2}(Y(0)-m(0,X))^{2}((1-e(X))^{-1}-(1-e_{0}(X))^{-1})^{2}]\leqslant(2C+1)\gamma^{-4}\tilde{\delta}_{N}^{2}.$

This completes the proof of (23).

Equation (24)

For any $(\nu,e)\in\mathcal{T}_{N}$, trivially
$\mathbb{E}\bigl{[}|s_{a}(W(z);\nu,e)|^{q}\bigr{]}=1$ for $z=0,1$.

Equation (25)

Fix $(\nu,e)\in\mathcal{T}_{N}$. We write

$\displaystyle
s(W(1);\theta_{0},\nu,e)=m(1,X)-m(0,X)-\theta_{0}+\frac{Y(1)-m(1,X)}{e(X)}.$
(70)

Since $(\mathbb{E}_{0}[|Y(z)|^{q}])^{1/q}\leqslant C$ for $z=0,1$,

$\mathbb{E}_{0}\bigl{[}|m_{0}(z,X)|^{q}\bigr{]}=\mathbb{E}\bigl{[}|\mathbb{E}_{0}[Y(z)\mid
X]|^{q}\bigr{]}\leqslant\mathbb{E}_{0}\bigl{[}|Y(z)|^{q}\bigr{]}\leqslant
C^{q}$

and then by the definition of $\theta_{0}$

$\displaystyle|\theta_{0}|$
$\displaystyle=\bigl{(}|\mathbb{E}_{0}[m_{0}(1,X)-m_{0}(0,X)]|^{q}\bigr{)}^{1/q}$
$\displaystyle\leqslant\bigl{(}\mathbb{E}_{0}\bigl{[}|m_{0}(1,X)-m_{0}(0,X)|^{q}\bigr{]}\bigr{)}^{1/q}$
$\displaystyle\leqslant 2C.$

With $\|m(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}\leqslant C$ by definition of
$\mathcal{T}_{N}$, we have

$\|m(z,\cdot)\|_{q,P_{0}^{X}}\leqslant\|m(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}+\|m_{0}(z,\cdot)\|_{q,P_{0}^{X}}\leqslant
2C$

for $z=0,1$. Therefore

$\displaystyle\sup_{(\nu,e)\in\mathcal{T}_{N}}\bigl{(}\mathbb{E}_{0}\bigl{[}|s(W(1);\theta_{0},\nu,e)|^{q}\bigr{]}\bigr{)}^{1/q}$
$\displaystyle\leqslant\sup_{(\nu,e)\in\mathcal{T}_{N}}\|m(1,\cdot)\|_{q,P_{0}^{X}}+\|m(0,\cdot)\|_{q,P_{0}^{X}}+|\theta_{0}|$
$\displaystyle\phantom{=}\
+\sup_{(\nu,e)\in\mathcal{T}_{N}}\gamma^{-1}\bigl{(}(\mathbb{E}_{0}[|Y(1)|^{q}])^{1/q}+\|m(1,\cdot)\|_{q,P_{0}^{X}}\bigr{)}$
$\displaystyle\leqslant 6C+\frac{3C}{\gamma}.$

Similarly,
$\sup_{(\nu,e)\in\mathcal{T}_{N}}\bigl{(}\mathbb{E}_{0}\bigl{[}|s(W(0);\theta_{0},\nu,e)|^{q}\bigr{]}\bigr{)}^{1/q}\leqslant
6C+{3C}/{\gamma}$ as well.

Equation (26)

We begin by considering $S_{t}^{(-k)}(z)$ for $z=1$. For each $k=1,\ldots,K$
we have

$\displaystyle
s(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(1);\nu_{0},e_{0,N})$
$\displaystyle=\bigl{(}1-\hat{e}^{(-k)}(X)^{-1}\bigr{)}\bigl{(}\hat{m}^{(-k)}(1,X_{ti})-m_{0}(1,X_{ti})\bigr{)}$
$\displaystyle\phantom{=}\
+\bigl{(}Y_{ti}(1)-m_{0}(1,X_{ti})\bigr{)}\bigl{(}\hat{e}^{(-k)}(X_{ti})^{-1}-e_{0}(X_{ti})^{-1}\bigr{)}$
$\displaystyle\phantom{=}\
-\bigl{(}\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti})\bigr{)}.$

Taking the conditional expectation given $\mathcal{S}^{(-k)}$ and $X_{ti}$
yields

$\displaystyle\phantom{=}\
\mathbb{E}\bigl{[}s(W_{ti}(1);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(1);\nu_{0},e_{0,N})\mid\mathcal{S}^{(-k)},X_{ti}\bigr{]}$
$\displaystyle=\bigl{(}1-\hat{e}^{(-k)}(X)^{-1}\bigr{)}(\hat{m}^{(-k)}(1,X_{ti})-m_{0}(1,X_{ti}))-(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti})).$

Then for $z=1$ we get

$\displaystyle S_{t}^{(-k)}(1)$
$\displaystyle\leqslant\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}\bigl{(}1-\hat{e}^{(-k)}(X_{ti})^{-1}\bigr{)}^{2}\bigl{(}\hat{m}^{(-k)}(1,X_{ti})-m_{0}(1,X_{ti})\bigr{)}^{2}}$
$\displaystyle\phantom{=}\
+\sqrt{\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti}))^{2}}$
$\displaystyle=O_{p}\bigl{(}\|\hat{m}^{(-k)}(1,\cdot)-m(1,\cdot)\|_{2,P_{0}^{X}}+\|\hat{m}^{(-k)}(0,\cdot)-m(0,\cdot)\|_{2,P_{0}^{X}}\bigr{)}$
$\displaystyle=o_{p}(N^{-1/4}).$

The first equality above follows from
$\left(1-\hat{e}^{(-k)}(X_{ti})\right)^{2}\leqslant(1+\gamma^{-1})^{2}<\infty$,
followed by an application of the conditional Markov inequality using

$\displaystyle\mathbb{E}\biggl{[}\frac{1}{n_{t,k}}\sum_{i:(t,i)\in\mathcal{I}_{k}}(\hat{m}^{(-k)}(z,X_{ti})-m_{0}(z,X_{ti}))^{2}\Bigm{|}\mathcal{S}^{(-k)}\biggr{]}$
$\displaystyle=\|\hat{m}^{(-k)}(z,\cdot)-m_{0}(z,\cdot)\|_{2,P_{t}^{X}}^{2}$
$\displaystyle\leqslant\kappa_{t}^{-1}\|\hat{m}^{(-k)}(z,\cdot)-m_{0}(z,\cdot)\|_{2,P_{0}^{X}}^{2}$

for $z=0,1$. By an identical argument, $S_{t}^{(-k)}(0)=o_{p}(N^{-1/4})$,
establishing (26).

Having shown all conditions of Assumption 4.1, the conclusion of Corollary 4.1
follows by Theorem 4.1.

###  C.7 Proof of Corollary 4.2: CLT for feasible
$\hat{\theta}_{\textnormal{EPL}}$ in a CSBAE

Corollary 4.2, which shows the feasible estimator
$\hat{\theta}_{\textnormal{EPL}}$ satisfies a CLT for estimating
$\theta_{0,\textnormal{PL}}$ in a CSBAE, holds under the numbered
generalizations at the end of Appendix A, subject to the additional
requirement that the mean and variance functions are stationary across
batches, as in Appendix C.3. We prove this more general result using the same
structure as the proof of Corollary 4.1 in Appendix C.6.

The proof of Corollary 3.2 in Appendix C.3 shows that Assumption A.2 is
satisfied with $\theta_{0}=\theta_{0,\textnormal{PL}}$,
$s(\cdot)=s(\cdot)_{\textnormal{EPL}}$,
$\nu_{0}=\nu_{0,\textnormal{EPL}}\in\mathcal{N}=\mathcal{N}_{\textnormal{EPL}}$,
and $\gamma=0$. Then it suffices to show the remaining conditions of
Assumption 4.1 to complete the proof, in view of Theorem 4.1. Throughout the
remainder of this proof we let $C_{0}$ be a generic positive finite constant;
possibly depending on $c$ and $C$ in Assumption 3.2; different appearances of
$C_{0}$ may correspond to different constants.

Condition (a)

Invertibility of $\mathbb{E}_{0}[s_{a}(W;\nu_{0},e_{0})]$ and existence of
$\mathbb{E}_{0}[\|s(W;\theta_{0},\nu_{0},e_{0})\|^{2}]$ follow from Assumption
3.2, as shown in the proof of Corollary 3.2.

Condition (b)

Fix $(\nu,e)\in\mathcal{T}$. Recall that the weight function in our estimator
$\hat{\theta}_{\textnormal{EPL}}$ is
$w(X,\nu,e)=(v(0,x)e(x)+v(1,x)(1-e(x)))^{-1}$. We compute

$\displaystyle\mathbb{E}_{0,N}$
$\displaystyle[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]$
$\displaystyle=\mathbb{E}_{0,N}[w(X;\nu_{\lambda},e_{\lambda,N})(Z-e_{\lambda,N}(X))(Y-m_{\lambda}(0,X)-Z\psi(X)^{\top}\theta_{0})\psi(X)]$
$\displaystyle=\lambda\mathbb{E}_{0,N}[w(X;\nu_{\lambda},e_{\lambda,N})(Z-e_{\lambda,N}(X))(m_{0}(0,X)-m(0,X))\psi(X)]$
$\displaystyle=\mathbb{E}_{0,N}[f_{N}(\lambda)]$

where the last two equalities follow by conditioning on $(X,Z)$ then just $X$,
and we have defined

$\displaystyle\nu_{\lambda}(\cdot)$
$\displaystyle=\nu_{0}(\cdot)+\lambda(\nu(\cdot)-\nu_{0}(\cdot)),$
$\displaystyle e_{\lambda,N}(\cdot)$
$\displaystyle=e_{0,N}(\cdot)+\lambda(e(\cdot)-e_{0,N}(\cdot))\quad\text{and}$
$\displaystyle f_{N}(\lambda)$
$\displaystyle=\lambda^{2}w(X;\nu_{\lambda},e_{\lambda,N})(e_{0,N}(X)-e(X))(m_{0}(0,X)-m(0,X))\psi(X)$
$\displaystyle=\lambda^{2}w(X;\nu_{\lambda},e_{\lambda,N})g_{N}(X)$

for $g_{N}(X)=(e_{0,N}(X)-e(X))(m_{0}(0,X)-m(0,X))\psi(X)$. We compute

$\displaystyle\frac{\partial}{\partial\lambda}w(X;\nu_{\lambda},e_{\lambda,N})$
$\displaystyle=-w^{2}(X;\nu_{\lambda},e_{\lambda,N})\Delta_{N}(\lambda,X)$
$\displaystyle\frac{\partial}{\partial\lambda}w^{2}(X;\nu_{\lambda},e_{\lambda,N})$
$\displaystyle=-2w^{3}(X;\nu_{\lambda},e_{\lambda,N})\Delta_{N}(\lambda,X)$

where

$\displaystyle\Delta_{N}(\lambda,X)$
$\displaystyle=\frac{\partial}{\partial\lambda}\bigl{(}v_{\lambda}(0,X)e_{\lambda,N}(X)+v_{\lambda}(1,X)(1-e_{\lambda,N}(X))\bigr{)}$
$\displaystyle=(1-e_{\lambda}(X))(v(1,X)-v_{0}(1,X))+e_{\lambda}(X)(v(0,X)-v_{0}(0,X))\quad\text{and}$
$\displaystyle\phantom{=}\
+(e(X)-e_{0,N}(X))(v_{\lambda}(0,X)-v_{\lambda}(1,X))$

Then

$\displaystyle f_{N}^{\prime}(\lambda)$
$\displaystyle=(-\lambda^{2}\Delta_{N}(\lambda,X)w^{2}(X;\nu_{\lambda},e_{\lambda,N})+2\lambda
w(X;\nu_{\lambda},e_{\lambda,N}))g_{N}(X)$ $\displaystyle
f_{N}^{\prime\prime}(\lambda)$
$\displaystyle=(2\lambda^{2}(\Delta_{N}(\lambda,X))^{2}w^{3}(X;\nu_{\lambda},e_{\lambda,N})-\lambda^{2}\Delta_{N}^{(2)}(\lambda,X)w^{2}(X;\nu_{\lambda},e_{\lambda,N}))g_{N}(X)$
$\displaystyle+(2w(X;\nu_{\lambda},e_{\lambda,N})-4\lambda\Delta_{N}(\lambda,X)w^{2}(X;\nu_{\lambda},e_{\lambda,N}))g_{N}(X)$

where

$\displaystyle\Delta_{N}^{(2)}(\lambda,X)$
$\displaystyle=\frac{\partial}{\partial\lambda}\Delta_{N}(\lambda,X)$
$\displaystyle=2(e(X)-e_{0,N}(X))(v(0,X)-v_{0}(0,X)+v_{0}(1,X)-v(1,X))$

We conclude that for sufficiently small $\epsilon>0$,

$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}|\Delta_{N}(\lambda,X)|$
$\displaystyle\leqslant\sup_{\lambda\in(-\epsilon,1+\epsilon)}|1-e_{\lambda}(X))||v(1,X)-v_{0}(1,X)|+|e_{\lambda}(X)||v(0,X)-v_{0}(0,X)|$
$\displaystyle\phantom{=}\
+\sup_{\lambda\in(-\epsilon,1+\epsilon)}|e(X)-e_{0,N}(X)||v_{\lambda}(0,X)-v_{\lambda}(1,X)|$
$\displaystyle\leqslant C_{0}\quad\text{and}$
$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}|\Delta_{N}^{(2)}(\lambda,X)|$
$\displaystyle\leqslant C_{0}.$

Taking $\epsilon$ smaller if necessary, we can ensure

$\sup_{\lambda\in(-\epsilon,1+\epsilon)}w(X;\nu_{\lambda},e_{\lambda,N})=\sup_{\lambda\in(-\epsilon,1+\epsilon)}(v_{\lambda}(0,X)e_{\lambda,N}(X)+v_{\lambda}(1,X)(1-e_{\lambda,N}(X)))^{-1}<2c^{-1}$

and then

$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}\|f_{N}(\lambda)\|$
$\displaystyle\leqslant C_{0}\|g_{n}(X)\|\leqslant
C_{0}C|e_{0,N}(X)-e(X)||m_{0}(0,X)-m(0,X)|$
$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}\|f_{N}^{\prime}(\lambda)\|$
$\displaystyle\leqslant C_{0}\|g_{n}(X)\|\leqslant
C_{0}C|e_{0,N}(X)-e(X)||m_{0}(0,X)-m(0,X)|$
$\displaystyle\sup_{\lambda\in(-\epsilon,1+\epsilon)}\|f_{N}^{\prime\prime}(\lambda)\|$
$\displaystyle\leqslant K\|g_{n}(X)\|\leqslant
C_{0}C|e_{0,N}(X)-e(X)||m_{0}(0,X)-m(0,X)|$

By the Leibniz integral rule, the mapping
$\lambda\mapsto\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]$
then has second derivative $\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]$
on $[0,1]$. With $f_{N}^{\prime\prime}(\lambda)$ continuous on $[0,1]$, we
conclude by dominated convergence that
$\mathbb{E}_{0,N}[f^{\prime\prime}_{N}(\lambda)]$ is continuous as well.

Condition (c)

For $z=0,1$,

$\mathbb{E}_{0}[Y(z)\\!\mid\\!X]=\mathbb{E}_{0}[Y\\!\mid\\!X,Z=z]=m_{0}(0,X)+z\psi(X)^{\top}\theta_{0}.$

Fix $e(\cdot)\in\mathcal{F}_{0}$. For each $t=1,\ldots,T$, $i=1,\ldots,N_{t}$
we have

$\displaystyle\mathbb{E}$ $\displaystyle[s(W_{ti}(1);\theta_{0},\nu_{0},e)\mid
X_{ti}]$
$\displaystyle=w(X_{ti};\nu_{0},e)(1-e(X))\bigl{(}\mathbb{E}_{0}[Y(1)\\!\mid\\!X=X_{ti}]-m_{0}(0,X_{ti})-\psi(X_{ti})^{\top}\theta_{0}\bigr{)}\psi(X_{ti})$
$\displaystyle=0$ and similarly $\displaystyle\mathbb{E}$
$\displaystyle[s(W_{ti}(0);\theta_{0},\nu_{0},e)\mid X_{ti}]$
$\displaystyle=-w(X_{ti};\nu_{0},e)e(X)\bigl{(}\mathbb{E}_{0}[Y(0)\\!\mid\\!X=X_{ti}]-m_{0}(0,X_{ti})\bigr{)}\psi(X_{ti})$
$\displaystyle=0.$

These are equal so their difference is 0 as required for (19).

Now we show the out-of-fold estimates
$(\hat{m}^{(-k)}(0,\cdot),\hat{v}^{(-k)}(0,\cdot),\hat{v}^{(-k)}(1,\cdot),\hat{e}^{(-k)}(\cdot))$
lie in a set $\mathcal{T}_{N}$ with high probability for all sufficiently
large $N$, where this $\mathcal{T}_{N}$ satisfies equations (20) through (25).
By the rate and regularity conditions on the nuisance estimators in Corollary
4.2, there exists a sequence $\tilde{\delta}_{N}\downarrow 0$ and constants
$0<c<C<\infty$ so that with probability approaching 1 as $N\rightarrow\infty$,
we have

$\displaystyle\|\hat{m}^{(-k)}(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{0}^{X}}+\|\hat{v}(z,\cdot)-v_{0}(z,\cdot)\|_{2,P_{0}^{X}}+\|\hat{e}^{(-k)}-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant\tilde{\delta}_{N}$
$\displaystyle\|\hat{m}^{(-k))}(0,\cdot)-m_{0}(0,\cdot\|_{2,P_{0}^{X}}\|\hat{e}^{(-k)}-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant N^{-1/2}\tilde{\delta}_{N}$
$\displaystyle\|\hat{m}^{(-k)}(0,\cdot)-m_{0}(0,\cdot)\|_{q,p_{0}^{X}}$
$\displaystyle\leqslant C$ $\displaystyle\hat{v}^{(-k)}(z,x)$
$\displaystyle\geqslant c,\quad z=0,1$

for all folds $k=1,\ldots,K$. Then let $\mathcal{T}_{N}$ be the set of
functions $(m(0,\cdot),v(0,\cdot),v(1,\cdot),e(\cdot))$ in
$\mathcal{N}_{\textnormal{EPL}}\times\mathcal{F}_{0}$ for which

$\displaystyle\|m(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{0}^{X}}+\|v(z,\cdot)-v_{0}(z,\cdot)\|_{2,P_{0}^{X}}+\|e-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant\tilde{\delta}_{N}$
$\displaystyle\|m(0,\cdot)-m_{0}(0,\cdot\|_{2,P_{0}^{X}}\|e-e_{0,N}\|_{2,P_{0}^{X}}$
$\displaystyle\leqslant N^{-1/2}\tilde{\delta}_{N}$
$\displaystyle\|m(0,\cdot)-m_{0}(0,\cdot)\|_{q,p_{0}^{X}}$
$\displaystyle\leqslant C$ $\displaystyle v(z,x)$ $\displaystyle\geqslant
c,\quad z=0,1.$

By construction,
$\textnormal{Pr}((\hat{m}^{(-k)}(0,\cdot),\hat{v}^{(-k)}(0,\cdot),\hat{v}^{(-k)}(1,\cdot),\hat{e}^{(-k)}(\cdot))\in\mathcal{T}_{N})\rightarrow
1$ as $N\rightarrow\infty$ for all $k=1,\dots,K$. Now we show that
$\mathcal{T}_{N}$ satisfies equations (20) through (25) for all $N$ large
enough to ensure that
$1/2\leqslant{\mathrm{d}P_{0,N}^{X}}/{\mathrm{d}P_{0}^{X}}\leqslant 2$. For
such $N$, $\|f\|_{2,P_{0,N}^{X}}\leqslant\sqrt{2}\|f\|_{2,P_{0}^{X}}$ holds
for all $f\in L^{2}(P_{0}^{X})$. As in the proof of Corollary 4.1, we will
show equations (20) through (23) hold for a sequence $\delta_{N}$ that is some
constant multiple of $\tilde{\delta}_{N}$.

Equation (20)

Using the notation from our proof of condition (b) above, we see
$f_{N}^{\prime}(0)=0$ with probability 1, so for any
$(\nu,e)\in\mathcal{T}_{N}$ we have

$\frac{\partial}{\partial\lambda}\mathbb{E}_{0,N}[s(W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N}))]\Big{|}_{\lambda=0}=\mathbb{E}_{0,N}[f_{N}^{\prime}(0)]=0$

which shows the left-hand side of (20) is identically zero.

Equation (21)

Once again we recall the notation and calculations from the proof of condition
(b) above. For each $\lambda\in[0,1]$ and $(\nu,e)\in\mathcal{T}_{N}$ we see

$\displaystyle\bigg{\|}\frac{\partial^{2}}{\partial\lambda^{2}}\mathbb{E}_{0,N}\bigl{[}s$
$\displaystyle\bigl{(}W;\theta_{0},\nu_{0}+\lambda(\nu-\nu_{0}),e_{0,N}+\lambda(e-e_{0,N})\bigr{)}\bigr{]}\bigg{\|}=\bigl{\|}\mathbb{E}_{0,N}[f_{N}^{\prime\prime}(\lambda)]\bigr{\|}$

which is no larger than

$\displaystyle\mathbb{E}_{0,N}\biggl{[}\,\sup_{\lambda\in[0,1]}\|f_{N}^{\prime\prime}(\lambda)\|\biggr{]}\leqslant
C_{0}\|e-e_{0,N}\|_{2,P_{0}^{X}}\|m(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{0}^{X}}\leqslant
C_{0}N^{-1/2}\tilde{\delta}_{N}$

by the definition of $\mathcal{T}_{N}$.

Equation (22)

Recall that
$s_{\textnormal{EPL},a}(W;\nu,e)=-w(X;\nu,e(X))Z(Z-e)\psi(X)\psi(X)^{\top}$.
Fix $(\nu,e)\in\mathcal{T}_{N}$. We compute

$\displaystyle s_{a}(W;\nu,e)-s_{a}(W;\nu_{0},e_{0})$
$\displaystyle=\bigl{(}w(X;\nu_{0},e_{0})(Z-e_{0}(X))-w(X;\nu,e)(Z-e(X))\bigr{)}Z\psi(X)\psi(X)^{\top}$
$\displaystyle=w(X;\nu_{0},e_{0})(e(X)-e_{0}(X))Z\psi(X)\psi(X)^{\top}$
$\displaystyle\phantom{=}\
+(w(X;\nu_{0},e_{0})-w(X;\nu,e))(Z-e(X))Z\psi(X)\psi(X)^{\top}.$

Hence, using $\|\psi(X)\psi(X)^{\top}\|\leqslant C^{2}$,

$\displaystyle\phantom{=}\
\mathbb{E}_{0}\bigl{[}\|s_{a}(W;\nu,e)-s_{a}(W;\nu_{0},e_{0})\|^{2}\bigr{]}^{1/2}$
$\displaystyle\leqslant
C^{2}\mathbb{E}_{0}\bigl{[}(w(X;\nu_{0},e_{0})Z(e(X)-e_{0}(X)))^{2}\bigr{]}^{1/2}$
$\displaystyle\phantom{=}\
+C^{2}\mathbb{E}_{0}\bigl{[}(w(X;\nu_{0},e_{0})-w(X;\nu,e))^{2}(Z-e(X))^{2}Z^{2}\bigr{]}^{1/2}.$

. To bound the right-hand side, first note that

$\Big{(}\mathbb{E}_{0}\bigl{[}(w(X;\nu_{0},e_{0})Z(e(X)-e_{0}(X)))^{2}\bigr{]}\Big{)}^{1/2}\leqslant
c^{-1}\|e-e_{0}\|_{2,P_{0}^{X}}\leqslant c^{-1}\tilde{\delta}_{N}.$

Next

$\displaystyle\phantom{=}\ \,|w(X;\nu,e)-w(X;\nu_{0},e_{0})|$
$\displaystyle\leqslant
c^{-2}\bigl{|}v(0,X)e(X)+v(1,X)(1-e(X))-v_{0}(0,X)e_{0}(X)-v_{0}(1,X)(1-e_{0}(X))\bigr{|}$
$\displaystyle\leqslant c^{-2}\bigl{(}|v(0,X)-v_{0}(0,X)|\cdot
e(X)+v_{0}(0,X)\cdot|e(X)-e_{0}(X)|\bigr{)}$ $\displaystyle\phantom{=}\
+c^{-2}\bigl{(}|v(1,X)-v_{0}(1,X)|\cdot(1-e(X))+v_{0}(1,X)\cdot|e(X)-e_{0}(X)|\bigr{)}$
$\displaystyle\leqslant
c^{-2}\bigl{(}|v(0,X)-v_{0}(0,X)|+|v(1,X)-v_{0}(1,X)|+2C|e(X)-e_{0}(X)|\bigr{)}$

and so

$\displaystyle\phantom{=}\
\,\mathbb{E}_{0}\bigl{[}(w(X;\nu_{0},e_{0})-w(X;\nu,e))^{2}(Z-e(X))^{2}Z^{2}\bigr{]}^{1/2}$
$\displaystyle\leqslant
c^{-2}\bigl{(}\|v(0,\cdot)-v_{0}(0,\cdot)\|_{2,P_{0}^{X}}+\|v(1,\cdot)-v_{0}(1,\cdot)\|_{2,P_{0}^{X}}+2C\|e-e_{0}\|_{2,P_{0}^{X}}\bigr{)}$
$\displaystyle\leqslant C_{0}\tilde{\delta}_{N},$

which shows (22).

Equation (23)

Once again, fix $(\nu,e)\in\mathcal{T}_{N}$. Then

$\displaystyle\phantom{=}\
s(W;\theta_{0},\nu,e)-s(W;\theta_{0},\nu_{0},e_{0})$
$\displaystyle=[w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))][Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}]\psi(X)$
$\displaystyle\phantom{=}\ +w(X;\nu,e)(Z-e(X))(m_{0}(0,X)-m(0,X))\psi(X)$

so that

$\displaystyle\phantom{=}\
\,\mathbb{E}_{0}\bigl{[}\|s(W;\theta_{0},\nu,e)-s(W;\theta_{0},\nu_{0},e_{0})\|^{2}\bigr{]}^{1/2}$
$\displaystyle\leqslant
C\Bigl{(}\mathbb{E}_{0}\bigl{[}\bigl{(}w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))\bigr{)}^{2}\bigl{(}Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}\bigr{)}^{2}\bigr{]}\Bigr{)}^{1/2}$
$\displaystyle\phantom{=}\
+C\bigl{(}\mathbb{E}_{0}\bigl{[}(w(X;\nu,e)(Z-e(X)))^{2}(m_{0}(0,X)-m(0,X))^{2}\bigr{]}\bigr{)}^{1/2}.$

By conditioning on $(X,Z)$ we see that

$\displaystyle\mathbb{E}_{0}$
$\displaystyle\bigl{[}|w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))|^{2}|Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}|^{2}\bigr{]}^{1/2}$
$\displaystyle=\mathbb{E}_{0}\bigl{[}v_{0}(Z,X)|w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))|^{2}\bigr{]}^{1/2}$
$\displaystyle\leqslant
C^{1/2}\mathbb{E}_{0}\bigl{[}|w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))|^{2}\bigr{]}^{1/2}.$

Now we write

$\displaystyle\bigl{|}$ $\displaystyle
w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))\bigr{|}$
$\displaystyle\leqslant|w(X;\nu,e)-w(X;\nu_{0},e_{0})||Z-e(X)|+|w(X;\nu_{0},e_{0})||e_{0}(X)-e(X)|$
$\displaystyle\leqslant
c^{-2}(|v(0,X)-v_{0}(0,X)|+|v(1,X)-v_{0}(1,X)|+2C|e(X)-e_{0}(X)|)+c^{-1}|e_{0}(X)-e(X)|$

so that

$\displaystyle\mathbb{E}_{0}$
$\displaystyle\bigl{[}|w(X;\nu,e)(Z-e(X))-w(X;\nu_{0},e_{0})(Z-e_{0}(X))|^{2}|Y-m_{0}(0,X)-Z\psi(X)^{\top}\theta_{0}|^{2}\bigr{]}^{1/2}$
$\displaystyle\leqslant
C^{1/2}\bigl{[}c^{-2}(\|v(0,\cdot)-v_{0}(0,\cdot)\|_{2,P_{0}^{X}}+\|v(1,\cdot)-v_{0}(1,\cdot)\|_{2,P_{0}^{X}}+2C\|e-e_{0}\|_{2,P_{0}^{X}})+c^{-1}\|e-e_{0}\|_{2,P_{0}^{X}}\bigr{]}$
$\displaystyle\leqslant C_{0}\tilde{\delta}_{N}$

With $(\nu,e)\in\mathcal{T}_{N}$ arbitrary and

$\mathbb{E}_{0}\bigl{[}|w(X;\nu,e)(Z-e(X))|^{2}|m_{0}(0,X)-m(0,X)|^{2}\bigr{]}^{1/2}\leqslant
c^{-1}\|m(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{0}^{X}}\leqslant
C_{0}\tilde{\delta}_{N}$

we have shown (23).

Equation (24)

Fix $(\nu,e)\in\mathcal{T}_{N}$. For $z=0,1$ we have

$\sup_{z\in\\{0,1\\}}\|s_{a}(W(z);\nu,e)\|^{q}=\sup_{z\in\\{0,1\\}}\|w(X;\nu,e)z(z-e(X))\psi(X)\psi(X)^{\top}\|^{q}\leqslant\left(\frac{C^{2}}{c}\right)^{q}$

which immediately shows that

$\mathbb{E}_{0}\bigl{[}\|s_{a}(W(z);\nu,e)\|^{q}\bigr{]}^{1/q}\leqslant
C_{0}.$

Equation (25)

For any $(\nu,e)\in\mathcal{T}_{N}$ we have

$\mathbb{E}_{0}\bigl{[}\|s(W(z);\theta_{0},\nu,e)\|^{q}\bigr{]}^{1/q}\leqslant\frac{C}{c}\mathbb{E}\bigl{[}|Y(z)-m(0,X)-z\psi(X)^{\top}\theta_{0}|^{q}\bigr{]}^{1/q}\leqslant
C_{0}$

since

$\displaystyle\mathbb{E}_{0}$
$\displaystyle\bigl{[}|Y(z)-m(0,X)-z\psi(X)^{\top}\theta_{0}|^{q}\bigr{]}^{1/q}$
$\displaystyle\leqslant\mathbb{E}_{0}\bigl{[}|Y(z)|^{q}\bigr{]}^{1/q}+\mathbb{E}_{0}\bigl{[}|m(0,X)|^{q}\bigr{]}^{1/q}+\mathbb{E}_{0}\bigl{[}|m(1,X)|^{q}\bigr{]}^{1/q}.$

We have $(\mathbb{E}_{0}[|Y(z)|^{q}])^{1/q}\leqslant C_{0}$ by Assumption 3.2,
hence for $z=0,1$

$\displaystyle(\mathbb{E}_{0}[|m(z,X)|^{q}])^{1/q}$
$\displaystyle\leqslant(\mathbb{E}_{0}[|m_{0}(z,X)|^{q}])^{1/q}+\|m(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}$
$\displaystyle\leqslant(\mathbb{E}_{0}[|Y(z)|^{q}])^{1/q}+\|m(z,\cdot)-m_{0}(z,\cdot)\|_{q,P_{0}^{X}}$
$\displaystyle\leqslant C_{0}$

which shows (25).

Equation (26)

Fix a fold $k\in\\{1,\ldots,K\\}$. For each $z=0,1$ and
$(t,i)\in\mathcal{I}_{k}$ we have

$\displaystyle s$
$\displaystyle(W_{ti}(z);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(z);\nu_{0},e_{0,N})$
$\displaystyle=\bigl{[}w(X_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})(z-\hat{e}^{(-k)}(X_{ti}))-w(X_{ti};\nu_{0},e_{0,N})(z-e_{0,N}(X_{ti}))\bigr{]}$
$\displaystyle\phantom{=}\times[Y_{ti}(z)-m_{0}(0,X_{ti})-z\psi(X_{ti})^{\top}\theta_{0}]\psi(X_{ti})$
$\displaystyle\phantom{=}+w(X_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})(z-\hat{e}^{(-k)}(X_{ti}))(m_{0}(0,X_{ti})-\hat{m}^{(-k)}(0,X_{ti}))\psi(X_{ti}).$

Taking the conditional expectation given $\mathcal{S}^{(-k)},X_{ti}$ gives

$\displaystyle\Big{\|}\mathbb{E}$
$\displaystyle\bigl{[}s(W_{ti}(z);\hat{\nu}^{(-k)},\hat{e}^{(-k)})-s(W_{ti}(z);\nu_{0},e_{0,N})\\!\mid\\!\mathcal{S}^{(-k)},X_{ti}\bigr{]}\Big{\|}^{2}$
$\displaystyle=w^{2}(X_{ti};\hat{\nu}^{(-k)},\hat{e}^{(-k)})(z-\hat{e}^{(-k)}(X_{ti}))^{2}(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti}))^{2}\|\psi(X_{ti})\|^{2}$
$\displaystyle\leqslant\frac{C^{2}}{c^{2}}(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti}))^{2}$

and so

$S_{t}^{(-k)}(z)\leqslant
C_{0}\sqrt{\frac{1}{n_{t,k}}\sum_{(t,i)\in\mathcal{I}_{k}}(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti}))^{2}}=o_{p}(N^{-1/4})$

in view of conditional Markov’s inequality, as

$\displaystyle\mathbb{E}\biggl{[}\frac{1}{n_{t,k}}\sum_{(t,i)\in\mathcal{I}_{k}}(\hat{m}^{(-k)}(0,X_{ti})-m_{0}(0,X_{ti}))^{2}\Bigm{|}\mathcal{S}^{(-k)}\biggr{]}$
$\displaystyle=\|\hat{m}^{(-k)}(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{t}^{X}}$
$\displaystyle\leqslant\kappa_{t}^{-1}\|\hat{m}^{(-k)}(0,\cdot)-m_{0}(0,\cdot)\|_{2,P_{0}^{X}}$
$\displaystyle=o_{p}(N^{-1/4})$

by assumption.

Having shown all conditions of Assumption 4.1, we can apply Theorem 4.1 to
complete the proof of Corollary 4.2.

###  C.8 Proof of Lemma 5.1

Here we show that the information functions $\Psi_{d}(\cdot)$ and
$\Psi_{a}(\cdot)$ for D-optimality and A-optimality, respectively, both
satisfy conditions (a) through (d) of Assumption 5.2. That both
$\Psi_{d}(\cdot)$ and $\Psi_{a}(\cdot)$ are continuous, concave, and non-
decreasing on $\mathbb{S}_{+}^{p}$ is well known.

If $M$ is singular then $\Psi_{d}(M)=\Psi_{a}(M)=-\infty$; however if
$M\in\mathbb{S}_{++}^{p}$ then $\Psi_{d}(M)$ and $\Psi_{a}(M)$ are finite.
Thus condition (a) holds with $\Psi_{0}=-\infty$.

Next, we recall that for $M\in\mathbb{S}_{++}^{p}$, we have
$\nabla\Psi_{d}(M)=M^{-1}$ and $\nabla\Psi_{a}(M)=M^{-2}$. Now we fix $0<k<K$
and $A,B\in\mathbb{S}_{++}^{p}$ such that $KI\succcurlyeq A\succcurlyeq kI$
and $KI\succcurlyeq B\succcurlyeq kI$. Then

$\displaystyle\|\nabla\Psi_{d}(A)-\nabla\Psi_{d}(B)\|$
$\displaystyle=\|A^{-1}-B^{-1}\|=\|A^{-1}(B-A)B^{-1}\|$
$\displaystyle\leqslant\|A^{-1}\|\|B^{-1}\|\|B-A\|\leqslant
k^{-2}\|A-B\|\quad\text{and}$
$\displaystyle\|\nabla\Psi_{a}(A)-\nabla\Psi_{a}(B)\|$
$\displaystyle=\|A^{-2}-B^{-2}\|=\|A^{-2}(B^{2}-A^{2})B^{-2}\|$
$\displaystyle\leqslant\|A^{-2}\|\|B^{-2}\|(\|B\|\|B-A\|+\|B-A\|\|A\|)$
$\displaystyle\leqslant Kk^{-4}\|A-B\|$

which shows condition (b).

We also have $K^{-1}I\preccurlyeq\nabla\Psi_{d}(A)\preccurlyeq k^{-1}I$ and
$K^{-2}I\preccurlyeq\nabla\Psi_{a}(A)\preccurlyeq k^{-2}I$. Therefore
condition (c) holds.

Finally fix $\tilde{\Psi}_{0}>-\infty$, and suppose $0\preccurlyeq
A\preccurlyeq KI$ with $\Psi_{d}(A)\geqslant\tilde{\Psi}_{0}$. Letting
$\lambda_{1}\geqslant\ldots\geqslant\lambda_{p}$ be the eigenvalues of $A$, we
have $\lambda_{j}\leqslant K$ for all $j$ and so

$(p-1)\log(K)+\lambda_{p}\geqslant\Psi_{d}(A)=\sum_{j=1}^{p}\log\lambda_{j}\geqslant\tilde{\Psi}_{0}$

so that $\lambda_{p}\geqslant\exp(\tilde{\Psi}_{0}-(p-1)\log K)>0$, showing
condition (d) for $\Psi_{d}(\cdot)$. Similarly if
$\Psi_{a}(A)\geqslant\tilde{\Psi}_{0}$ then

$-\frac{p-1}{K}-\lambda_{p}^{-1}\geqslant\Psi_{a}(A)=-\sum_{j=1}^{p}\lambda_{j}^{-1}\geqslant\tilde{\Psi}_{0}$

which implies $-\tilde{\Psi}_{0}-(p-1)/K\geqslant\lambda_{p}^{-1}>0$.
Therefore $\lambda_{p}\geqslant(-\tilde{\Psi}_{0}-{(p-1)}/{K})^{-1}>0$,
showing condition (d) for $\Psi_{a}(\cdot)$ as well.

###  C.9 Proof of Lemma 5.2: Convergence of generic concave maximization
routine

Here we prove Lemma 5.2 about the convergence rates of our generic concave
maximization routine.

###### Proof.

Let $\tilde{\delta}=\delta/2$, with $\delta<0$ as in Assumption 5.1. We will
repeatedly use the fact that since
$[\tilde{\delta},1-\tilde{\delta}]\times\mathcal{W}\subseteq{}^{r+1}$ is
compact, Assumption 5.1 implies that $f=f(\cdot,\cdot)$ and all of its partial
derivatives up to second order are uniformly bounded above in norm on that
set. Then WLOG we can make $C$ from Assumption 5.1 larger so that

$\sup_{(k,w)\in[\tilde{\delta},1-\tilde{\delta}]\times\mathcal{W}}\|h(k,w)\|\leqslant
C,\quad\forall
h=h(\cdot,\cdot)\in\\{f,f^{\prime},f^{\prime\prime},f_{w},f_{ww},f_{w}^{\prime}\\}$
(71)

where $f_{w}=f_{w}(\cdot,\cdot)$, $f_{ww}=f_{ww}(\cdot,\cdot)$, and
$f_{w}^{\prime}=f_{w}^{\prime}(\cdot,\cdot)$ are tensors with $f_{w}$ the
partial derivative of $f$ with respect to the second argument, $f_{ww}$ the
second partial derivative of $f$ with respect to the second argument, and
$f_{w}^{\prime}$ the partial derivative of $f^{\prime}$ with respect to the
second argument.

First, we show the existence of $e^{*}(\cdot)$ satisfying (27). For each
propensity $e=e(\cdot)\in\mathcal{E}$ define $\phi(e)=\Psi(M(e))$ and
$\hat{\phi}_{n}(e)=\Psi(\hat{M}_{n}(e))$ where

$M(e)=\int_{\mathcal{X}}f(e(x),\eta(x))\mathrm{d}P(x)\quad\text{and}\quad\hat{M}_{n}(e)=\frac{1}{n}\sum_{i=1}^{n}f(e(X_{i}),\hat{\eta}(X_{i})).$

For any $e_{1}=e_{1}(\cdot),e_{2}=e_{2}(\cdot)\in\mathcal{E}$ and
$\lambda\in[0,1]$ we have

$\displaystyle M(\lambda e_{1}+(1-\lambda)e_{2})$
$\displaystyle=\int_{\mathcal{X}}f\bigl{(}\lambda
e_{1}(x)+(1-\lambda)e_{2}(x),\eta(x)\bigr{)}\,\mathrm{d}P(x)$
$\displaystyle\succcurlyeq\int_{\mathcal{X}}\bigl{(}\lambda
f(e_{1}(x),\eta(x))+(1-\lambda)f(e_{2}(x),\eta(x))\bigr{)}\,\mathrm{d}P(x)$
$\displaystyle=\lambda M(e_{1})+(1-\lambda)M(e_{2})$

where the matrix inequality follows from the fact that the function $u\mapsto
f(u,\eta(x))$ is a concave matrix-valued function on $[0,1]$ since its second
derivative is globally negative semidefinite by (29). Thus $M=M(\cdot)$ is
also a concave matrix-valued function on $\mathcal{E}$. In fact, $M$ is also
Lipschitz continuous in the sense that

$\displaystyle\|M(e_{1})-M(e_{2})\|$
$\displaystyle\leqslant\int_{\mathcal{X}}\|f(e_{1}(x),\eta(x))-f(e_{2}(x),\eta(x))\|\,\mathrm{d}P(x)$
$\displaystyle\leqslant
C\int_{\mathcal{X}}|e_{1}(x)-e_{2}(x)|\,\mathrm{d}P(x)$
$\displaystyle\leqslant C\|e_{1}-e_{2}\|_{2,P},\quad\forall
e_{1},e_{2}\in\mathcal{E}$ (72)

where the second inequality uses (71) with $h=f^{\prime}$ and Taylor’s theorem
with the Lagrange form of the remainder. With the information function
$\Psi=\Psi(\cdot)$ continuous, concave, and increasing in the semidefinite
ordering by Assumption 5.2, we conclude that $\phi=\phi(\cdot)$ is continuous
and concave on $\mathcal{E}$ (e.g., by Section 3.6 of Boyd and Vandenberghe,
(2004)).

We now consider an extension $\bar{\phi}=\bar{\phi}(\cdot)$ of $\phi$ to
$L^{2}(P)$,

$\bar{\phi}(e)=\begin{cases}\phi(e),&e\in\mathcal{E}\\\ -\infty,&\text{
otherwise.}\end{cases}$

Since $\phi(\cdot)$ is continuous and concave on $\mathcal{E}$, it is
straightforward to show that the extension $\bar{\phi}$ is concave and upper
semicontinuous on $L^{2}(P)$; the latter means that
$\bar{\phi}(e_{0})=\limsup_{e\rightarrow e_{0}}\bar{\phi}(e)$ for all
$e_{0}=e_{0}(\cdot)\in L^{2}(P)$. The function $\bar{\phi}$ is also “proper”
in that it never equals $+\infty$. Since $\mathcal{F}_{*}$ is a closed,
bounded, and convex subset of the Hilbert space $L^{2}(P)$, by Proposition
1.88 and Theorem 2.11 of Barbu and Precupanu, (2012) we conclude
$\bar{\phi}(\cdot)$, and hence $\phi(\cdot)$, attains its maximum on
$\mathcal{F}_{*}$. This shows the existence of
$e^{*}=\operatorname*{arg\,max}_{e\in\mathcal{F}}\phi(e)$. We will show
uniqueness ($P$-almost surely) later.

Next, note that compactness of $F_{n}\subseteq{}^{n}$ and continuity of the
map
$(e_{1},\ldots,e_{n})\mapsto\Psi\left(n^{-1}\sum_{i=1}^{n}f(e_{i},\hat{\eta}(X_{i}))\right)$,
which follows from continuity of the information function $\Psi$ and of
$e\mapsto f(e,\hat{\eta}(x))$ on $[0,1]$ for each $x\in\mathcal{X}$, ensure
the existence of a vector $(\hat{e}_{1},\ldots,\hat{e}_{n})$ satisfying (31).
This shows the second claim of Lemma 5.2. Then condition 2 of Assumption 5.1
ensures that there exists a propensity $\hat{e}=\hat{e}(\cdot)\in\mathcal{E}$
with $\hat{e}(X_{i})=\hat{e}_{i}$ for all $i=1,\ldots,n$. So all that remains
is to show that any such propensity $\hat{e}(\cdot)$ satisfies the rate
conditions
$\|\hat{e}-e^{*}\|_{2,P_{n}}+\|\hat{e}-e^{*}\|_{2,P}=O_{p}(n^{-1/4}+\alpha_{n})$,
along with uniqueness of $e^{*}(\cdot)$ satisfying (27) ($P$-almost surely).

We have now established existence of $e^{*}(\cdot)$,
$\hat{e}_{1},\dots,\hat{e}_{n}$ and $\hat{e}(\cdot)$. It remains to show the
desired convergence rates of $\hat{e}(\cdot)$ to $e^{*}(\cdot)$ in $L^{2}$
under both $P$ and $\hat{P}_{n}$. Before proceeding further, we list a few
useful facts. By (71) with $h=f$ and recalling that the Frobenius norm of a
matrix upper bounds its spectral norm, we have

$M(e)\preccurlyeq CI\quad\text{ and }\quad\hat{M}_{n}(e)\preccurlyeq
CI,\quad\forall e\in\mathcal{E}.$ (73)

Next, we claim that

$M(e^{*})\succcurlyeq k^{*}I$ (74)

for some $k^{*}>0$. To see this, note that
$\Psi(M(e^{*}))=\phi(e^{*})\geqslant\phi(e_{0})=\Psi(M(e_{0}))>\Psi_{0}$ by
Assumption 5.2(a) and condition 3 of Assumption 5.1. Then Assumption 5.2(d)
shows that the $k_{*}>0$ we need exists.

Now, we define the function class

$\mathcal{F}_{n}=\biggl{\\{}e\in\mathcal{E}\Bigm{|}m_{L}\leqslant\frac{1}{n}\sum_{i=1}^{n}e(X_{i})\leqslant
m_{H}\biggr{\\}}.$

This class contains any propensity $\hat{e}(\cdot)$ derived as above by
solving (31) and interpolating within the base propensity class $\mathcal{E}$.

We complete the proof by the following 4 steps:

  1. 1.

Show that $\phi$ is strongly concave on $\mathcal{F}_{*}$ with respect to the
$\|\cdot\|_{2,P}$ norm and that $\hat{\phi}_{n}$ is strongly concave on
$\mathcal{F}_{n}$ with respect to the $\|\cdot\|_{2,P_{n}}$ norm. This means
there exist nonrandom positive constants $c_{0}$, $r_{0}$, and $k_{0}$ such
that

$\displaystyle
c_{0}\min\bigl{(}r_{0},\|e^{*}-e\|_{2,P}\bigr{)}\times\|e^{*}-e\|_{2,P}\leqslant\phi(e^{*})-\phi(e),\quad\forall
e\in\mathcal{F}_{*}$ (75) and $\displaystyle
c_{0}\min\bigl{(}r_{0},\|\hat{e}-e\|_{2,P_{n}}\bigr{)}\times\|\hat{e}-e\|_{2,P_{n}}\bm{1}(A_{n})\leqslant(\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e))\bm{1}(A_{n}),\quad\forall
e\in\mathcal{F}_{n}$ (76)

where $A_{n}$ is the event that $\hat{M}_{n}(\hat{e})\succcurlyeq k_{0}I$.
Equation (75) shows that $e^{*}(\cdot)$ is unique $P$-a.s., because any $e$
that maximizes (27) makes the right hand size of (75) equal $0$ which then
makes $\|e-e^{*}\|_{2,P}=0$.

  2. 2.

Conclude by the previous step that with probability approaching 1,

$\displaystyle
c_{0}\min(r_{0},\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P})\times\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P}\leqslant\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}),$
and $\displaystyle
c_{0}\min(r_{0},\|\hat{e}-e_{n}^{*}\|_{2,P_{n}})\times\|\hat{e}-e_{n}^{*}\|_{2,P_{n}}\bm{1}(A_{n})\leqslant(\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}))\bm{1}(A_{n})$

for all $\hat{e}_{\mathcal{F}}\in\mathcal{F}$, $e_{n}^{*}\in\mathcal{F}_{n}$.

  3. 3.

Show that with probability approaching 1, there exist
$\hat{e}_{\mathcal{F}}\in\mathcal{F}$ and $e_{n}^{*}\in\mathcal{F}_{n}$
converging at the rate $O_{p}(n^{-1/2})$ in sup-norm to $\hat{e}$ and $e^{*}$,
respectively, so that by empirical process arguments we can argue that

$\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*})=O_{p}(n^{-1/2})+O_{p}(\alpha_{n}).$
(77)

Show that $\textnormal{Pr}(A_{n})\rightarrow 1$ as $n\rightarrow\infty$ and
conclude by the previous step that
$\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P}+\|\hat{e}-e_{n}^{*}\|_{2,P_{n}}=O_{p}(n^{-1/4})+O_{p}(\alpha_{n}^{-1/2})$.
Then by the definitions of $\hat{e}_{\mathcal{F}}$ and $e_{n}^{*}$ we can
conclude that

$\|\hat{e}-e^{*}\|_{2,P}+\|\hat{e}-e^{*}\|_{2,P_{n}}=O_{p}(n^{-1/4})+O_{p}(\alpha_{n}^{-1/2})$

as well. In particular, $\hat{e}$ is mean square consistent for $e^{*}$ both
in-sample and out-of-sample.

  4. 4.

Apply a “peeling” argument, similar to Theorem 3.2.5 in van der Vaart and
Wellner, (1996), to show that
$\|\hat{e}-\tilde{e}\|_{2,P_{n}}=O_{p}(\alpha_{n})$, where

$\tilde{e}\in\operatorname*{arg\,max}_{e\in\mathcal{F}_{n}}\Psi\biggl{(}n^{-1}\sum_{i=1}^{n}f(e(X_{i}),\eta(X_{i}))\biggr{)}$

is the propensity score we’d estimate with knowledge of $\eta$, i.e., by
taking $\hat{\eta}=\eta$. Conclude by the previous step that
$\|\hat{e}-e^{*}\|_{2,P_{n}}=O_{p}(n^{-1/4})+O_{p}(\alpha_{n})$, and show the
same convergence rate holds for $\|\hat{e}-e^{*}\|_{2,P}$ by empirical process
arguments.

### Step 1

Strong concavity will be proven using calculus along with Assumptions 5.1 and
5.2. First, we notice that by Assumption 5.2(c) and (74), we know
$\nabla\Psi(A)|_{A=M(e^{*})}\succcurlyeq m^{*}I$ for some $m^{*}>0$. Then by
continuity of the smallest eigenvalue function $\lambda_{\min}(\cdot)$ and of
$\nabla\Psi(\cdot)$, there exists $r_{0}>0$ such that if $e\in\mathcal{E}$
satisfies $\|e-e^{*}\|_{2,P}\leqslant r_{0}$ (which implies
$\|M(e)-M(e^{*})\|\leqslant Cr_{0}$ by (72)), then
$M(e)\succcurlyeq(k^{*}/2)I$ and
$\nabla\Psi(A)|_{A=M(e)}\succcurlyeq(m^{*}/2)I$. We now extend this argument
to provide a high probability eigenvalue lower bound on $\hat{M}_{n}(e)$ for
$e$ sufficiently close to $\hat{e}$ in $L^{2}(\mathcal{X},P_{n})$:

######  Lemma C.1.

Suppose that all conditions of Lemma 5.2 hold. Fix any $k_{0}>0$ and define
$A_{n}$ to be the event that $\hat{M}_{n}(\hat{e})\succcurlyeq k_{0}I$. Then
there exist $\tilde{r}>0$ and $0<\tilde{k}<\tilde{K}$ such that whenever
$A_{n}$ holds, for all $e\in\mathcal{E}$ with
$\|e-\hat{e}\|_{2,P_{n}}\leqslant\tilde{r}$ we have
$\hat{M}_{n}(e)\succcurlyeq(k_{0}/2)I$ and
$\tilde{K}I\succcurlyeq\nabla\Psi(\hat{M}_{n}(e))\succcurlyeq\tilde{k}I$.

###### Proof of Lemma C.1.

The function $\lambda_{\min}(\cdot)$ is uniformly continuous on the compact
subset $\mathcal{M}=\\{A:0\preccurlyeq A\preccurlyeq CI\\}$ of p×p. Hence,
when the event $A_{n}$ holds, we know that there exists (nonrandom)
$\tilde{\delta}>0$ such that $A\succcurlyeq(k_{0}/2)I$ for all
$A\in\mathcal{M}$ with $\|A-\hat{M}_{n}(\hat{e})\|\leqslant\tilde{\delta}$. By
(73), $\mathcal{M}$ contains both $\\{M(e):e\in\mathcal{E}\\}$ and
$\\{\hat{M}_{n}(e):e\in\mathcal{E}\\}$. Noting that

$\displaystyle\|\hat{M}_{n}(e_{1})-\hat{M}_{n}(e_{2})\|$
$\displaystyle\leqslant\frac{1}{n}\sum_{i=1}^{n}\|f(e_{1}(X_{i}),\hat{\eta}(X_{i}))-f(e_{2}(X_{i}),\hat{\eta}(X_{i})\|$
$\displaystyle\leqslant\frac{C}{n}\sum_{i=1}^{n}|e_{1}(X_{i})-e_{2}(X_{i})|$
$\displaystyle\leqslant C\|e_{1}-e_{2}\|_{2,P_{n}}.$ (78)

we see that whenever $e\in\mathcal{E}$ with
$\|e-\hat{e}\|_{2,P_{n}}\leqslant\tilde{r}:=\tilde{\delta}/C$, we have
$\|\hat{M}_{n}(e)-\hat{M}_{n}(\hat{e})\|\leqslant\tilde{\delta}$ and hence
$\hat{M}_{n}(e)\succcurlyeq(k_{0}/2)I$. The conclusion of Lemma C.1 follows
immediately by Assumption 5.2(c). ∎

Next, we bound directional derivatives of $\phi$ and $\hat{\phi}_{n}$.

######  Lemma C.2.

For any $e_{1},e_{2}\in\mathcal{F}$ with $M(e_{1})$ and $M(e_{2})$
nonsingular, the inequality

$\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\phi(e_{1}+t(e_{2}-e_{1}))\leqslant\textnormal{tr}\left[\nabla\Psi(M(e_{1}+t(e_{2}-e_{1})))^{\top}\left(\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}M(e_{1}+t(e_{2}-e_{1}))\right)\right]$
(79)

holds for each $t\in(0,1)$. Similarly, for any $e_{1},e_{2}\in\mathcal{F}_{n}$
with $\hat{M}_{n}(e_{1})$ and $\hat{M}_{n}(e_{2})$ nonsingular, we have the
inequality

$\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\hat{\phi}_{n}(e_{1}+t(e_{2}-e_{1}))\leqslant\textnormal{tr}\biggl{[}\nabla\Psi(\hat{M}_{n}(e_{1}+t(e_{2}-e_{1})))^{\top}\biggl{(}\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\hat{M}_{n}(e_{1}+t(e_{2}-e_{1}))\biggr{)}\biggr{]}.$
(80)

###### Proof of Lemma C.2.

Fix $e_{1}=e_{1}(\cdot)$ and $e_{2}=e_{2}(\cdot)\in\mathcal{F}$ with
$M(e_{1})$ and $M(e_{2})$ invertible. Define

$\tilde{M}(t)=M(e_{(t)})$

where $e_{(t)}=e_{1}+t(e_{2}-e_{1})$. We first show that

$\displaystyle\tilde{M}^{\prime}(t)$
$\displaystyle=\int_{\mathcal{X}}(e_{2}(x)-e_{1}(x))f^{\prime}(e_{(t)}(x),\eta(x))\,\mathrm{d}P(x),\quad\forall
t\in[0,1],\quad\text{and}$ (81) $\displaystyle\tilde{M}^{\prime\prime}(t)$
$\displaystyle=\int_{\mathcal{X}}(e_{2}(x)-e_{1}(x))^{2}f^{\prime\prime}(e_{(t)}(x),\eta(x))\,\mathrm{d}P(x),\quad\forall
t\in(0,1).$ (82)

We include the endpoints $t=0,1$ in (81) so that we can apply Taylor’s theorem
with the Lagrange form of the second order remainder to complete the proof of
Lemma C.2. We could also strengthen (82) to include those endpoints, but this
will not be needed.

Consider the difference quotient

$D_{1}(t,h;X)=\frac{f(e_{(t+h)}(X),\eta(X))-f(e_{(t)}(X),\eta(X))}{h}$

By the chain rule we know that

$\lim_{h\rightarrow
0}D_{1}(t,h;X)=\frac{\mathrm{d}}{\mathrm{d}t}f(e_{(t)}(X),\eta(X))=(e_{2}(X)-e_{1}(X))f^{\prime}(e_{(t)}(X),\eta(X)).$

Furthermore the fact that $e_{1}(x),e_{2}(x)\in[0,1]$ for all
$x\in\mathcal{X}$ indicates

$\tilde{\delta}\leqslant-h\leqslant e_{(t)}(x)\wedge e_{(t+h)}(x)\leqslant
e_{(t)}(x)\vee e_{(t+h)}(x)\leqslant 1+h\leqslant 1-\tilde{\delta}$

for all $t\in[0,1]$ and $|h|\leqslant-\tilde{\delta}$. By uniform boundedness
of $f^{\prime}$ on $[\tilde{\delta},1-\tilde{\delta}]\times\mathcal{W}$ and
Taylor’s theorem, and noting $e_{(t+h)}(x)-e_{(t)}(x)=h(e_{2}(x)-e_{1}(x))$ we
conclude that

$\sup_{0<|h|\leqslant-\tilde{\delta}}\|D_{1}(t,h;X)\|\leqslant|e_{2}(X)-e_{1}(X)|\times\sup_{e\in[\tilde{\delta},1-\tilde{\delta}]}\|f^{\prime}(e,\eta(X))\|\leqslant
C$

so by dominated convergence

$\displaystyle\lim_{h\rightarrow 0}\frac{\tilde{M}(t+h)-\tilde{M}(t)}{h}$
$\displaystyle=\lim_{h\rightarrow
0}\int_{\mathcal{X}}D_{1}(t,h;x)\,\mathrm{d}P(x)$
$\displaystyle=\int_{\mathcal{X}}(e_{2}(x)-e_{1}(x))f^{\prime}(e_{(t)}(x),\eta(x))\,\mathrm{d}P(x)$

which establishes (81).

Similarly we define the second difference quotient

$D_{2}(t,h;X)=\frac{f^{\prime}(e_{(t+h)}(X),\eta(X))-f^{\prime}(e_{(t)}(X),\eta(X))}{h}.$

By the chain rule we once again have

$\lim_{h\rightarrow
0}D_{2}(t,h;X)=\frac{\mathrm{d}}{\mathrm{d}t}f^{\prime}(e_{(t)}(X),\eta(X))=(e_{2}(X)-e_{1}(X))f^{\prime\prime}(e_{(t)}(X),\eta(X)).$

By uniform boundedness of $f^{\prime\prime}$ on
$[\tilde{\delta},1-\tilde{\delta}]\times\mathcal{W}$ we get

$\sup_{0<|h|\leqslant-\tilde{\delta}}|D_{2}(t,h;X)|\leqslant|e_{2}(X)-e_{1}(X)|\times\sup_{e\in[\tilde{\delta},1-\tilde{\delta}]}\|f^{\prime\prime}(e,\eta(X))\|\leqslant
C.$

Then in view of (81) we can apply dominated convergence to conclude that

$\displaystyle\lim_{h\rightarrow
0}\frac{\tilde{M}^{\prime}(t+h)-\tilde{M}^{\prime}(t)}{h}$
$\displaystyle=\lim_{h\rightarrow
0}\int_{\mathcal{X}}D_{2}(t,h;x)(e_{2}(x)-e_{1}(x))\,\mathrm{d}P(x)$
$\displaystyle=\int_{\mathcal{X}}(e_{2}(x)-e_{1}(x))^{2}f^{\prime\prime}(e_{(t)}(x),\eta(x))\,\mathrm{d}P(x)$

establishing (82).

Now we differentiate $\tilde{\phi}(t):=\phi(e_{(t)})$. Note $M(e_{(t)})\succ
0$ for all $t\in[0,1]$ by concavity of $M(\cdot)$, shown previously. Then
using (81) and (82), we apply the chain rule to get

$\displaystyle\tilde{\phi}^{\prime}(t)$
$\displaystyle=\textnormal{tr}\bigl{[}\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime}(t)\bigr{]},\quad
t\in[0,1].$

Similarly for all $t\in(0,1)$

$\displaystyle\tilde{\phi}^{\prime\prime}(t)$
$\displaystyle=D^{2}\Psi(\tilde{M}(t))(\tilde{M}^{\prime}(t),\tilde{M}^{\prime}(t))+\textnormal{tr}[\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime\prime}(t)]$
$\displaystyle\leqslant\textnormal{tr}\bigl{[}\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime\prime}(t)\bigr{]},$

which shows (79). Here $D^{2}\Psi(\tilde{M}(t))$ is the second derivative
mapping of $\Psi$ evaluated at $\tilde{M}(t)$, viewed as a bilinear function
from ${}^{p\times p}\times{}^{p\times p}$ to ; the inequality in the preceding
display follows from concavity of $\Psi$.

Equation (80) follows by a very similar calculation, though the argument is
simplified, since dominated convergence is no longer needed as we are dealing
with finite sums instead of integrals. Instead we immediately perform term-by-
term differentiation to conclude

$\displaystyle\tilde{M}_{n}^{\prime}(t)$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}(e_{2}(X_{i})-e_{1}(X_{i}))f^{\prime}(e_{(t)}(X_{i}),\hat{\eta}(X_{i})),\quad\forall
t\in[0,1],\quad\text{and}$ (83) $\displaystyle\tilde{M}_{n}^{\prime\prime}(t)$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}(e_{2}(X_{i})-e_{1}(X_{i}))^{2}f^{\prime\prime}(e_{(t)}(X_{i}),\hat{\eta}(X_{i})),\quad\forall
t\in(0,1)$ (84)

where $\tilde{M}_{n}(t):=\hat{M}_{n}(e_{(t)})$, and then use the chain rule as
above. ∎

We are now ready to prove (75). We apply Lemma C.2 with $e_{1}=e^{*}$ and any
$e_{2}\in\mathcal{F}$ with $\|e_{2}-e^{*}\|_{2,P}\leqslant r_{0}$. Note that
our definition of $r_{0}$ (in the paragraph before the statement of Lemma C.1)
along with (74) ensures that $M(e_{1})$ and $M(e_{2})$ are nonsingular. Also,
note that $\tilde{\phi}^{\prime}(0)\leqslant 0$ by optimality of $e^{*}$.
Lemma C.2 along with Taylor’s theorem with the Lagrange form of the remainder
then enables us to conclude

$\phi(e_{2})=\tilde{\phi}(1)=\tilde{\phi}(0)+\tilde{\phi}^{\prime}(0)+\frac{1}{2}\tilde{\phi}^{\prime\prime}(t)\leqslant\phi(e^{*})+\frac{1}{2}\textnormal{tr}\bigl{[}\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime\prime}(t)\bigr{]}$

for some $t\in(0,1)$, where $\tilde{M}(t)=M(e_{1}+t(e_{2}-e_{1}))$ as in the
proof of Lemma C.2. By (29) and (82), we know that
$\tilde{M}^{\prime\prime}(t)\preccurlyeq 0$. Recalling that the trace of the
product of two symmetric positive semidefinite matrices is nonnegative, we
have

$0\geqslant\textnormal{tr}[(\nabla\Psi(\tilde{M}(t))-(m^{*}/2)I)^{\top}\tilde{M}^{\prime\prime}(t)]=\textnormal{tr}[\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime\prime}(t)]-\frac{m^{*}}{2}\textnormal{tr}(\tilde{M}^{\prime\prime}(t))$

since $\nabla\Psi(\tilde{M}(t))\succcurlyeq(m^{*}/2)I$. Then

$\displaystyle\textnormal{tr}[\nabla\Psi(\tilde{M}(t))^{\top}\tilde{M}^{\prime\prime}(t)]$
$\displaystyle\leqslant\frac{m^{*}}{2}\textnormal{tr}(\tilde{M}^{\prime\prime}(t))$
$\displaystyle\leqslant-\frac{cm^{*}}{2}\int_{\mathcal{X}}(e^{*}(x)-e_{2}(x))^{2}\,\mathrm{d}P(x)$
$\displaystyle=-\frac{cm^{*}}{2}\|e^{*}-e_{2}\|_{2,P}^{2}$

where the second inequality follows by (29) and (82) once again. We conclude
that whenever $e_{2}\in\mathcal{F}$ with $\|e_{2}-e^{*}\|_{2,P}\leqslant
r_{0}$ we have

$\phi(e^{*})\geqslant\phi(e_{2})+\frac{cm^{*}}{4}\|e_{2}-e^{*}\|_{2,P}^{2}.$

Now take any $e_{2}\in\mathcal{F}$ with $\|e_{2}-e^{*}\|_{2,P}>r_{0}$. Define
$t=1-r_{0}/\|e_{2}-e^{*}\|_{2,P}\in(0,1)$ and consider
$\tilde{e}_{2}=te^{*}+(1-t)e_{2}$ so that
$\|\tilde{e}_{2}-e^{*}\|_{2,P}=r_{0}$. Note $\tilde{e}_{2}\in\mathcal{F}$ by
convexity of $\mathcal{F}_{*}$, so by the preceding display

$\phi(e^{*})\geqslant\phi(\tilde{e}_{2})+\frac{cm^{*}r_{0}^{2}}{4}\geqslant
t\phi(e^{*})+(1-t)\phi(e_{2})+\frac{cm^{*}r_{0}^{2}}{4}$

where the second inequality is by concavity of $\phi$. Rearranging we have

$\phi(e^{*})\geqslant\phi(e_{2})+\frac{cm^{*}r_{0}^{2}}{4(1-t)}=\phi(e_{2})+\frac{cm^{*}r_{0}}{4}\|e^{*}-e_{2}\|_{2,P}.$

Letting $c_{0}=cm^{*}/4$, we conclude that for all $e\in\mathcal{F}$ we have

$\phi(e^{*})\geqslant\phi(e)+c_{0}\min(r_{0},\|e^{*}-e\|_{2,P})\|e^{*}-e\|_{2,P}$

which shows (75).

The proof of (76) is quite similar. Take $k_{0}>0$ such that with $k^{*}$ as
in (74), whenever $0\preccurlyeq A\preccurlyeq CI$ with
$\Psi(A)\geqslant\inf_{B\succcurlyeq(k^{*}/4)I}\Psi(B)$, we have
$A\succcurlyeq k_{0}I$. Such a $k_{0}$ exists by Assumptions 5.2(a) and
5.2(d). By Lemma C.1, on the truncation event $A_{n}$ that
$\hat{M}_{n}(\hat{e})\succcurlyeq k_{0}I$, we have
$\tilde{K}I\geqslant\nabla\Psi(\hat{M}_{n}(e))\geqslant\tilde{k}I$ whenever
$\|e-\hat{e}\|_{2,P_{n}}\leqslant\tilde{r}$, for some $\tilde{r}>0$ and
$0<\tilde{k}<\tilde{K}$. Now we apply (80) with $e_{1}=\hat{e}$ and any
$e_{2}\in\mathcal{F}_{n}$ with $\|e_{2}-\hat{e}\|_{2,P_{n}}\leqslant\tilde{r}$
(note we must have $\hat{M}_{n}(e_{1})$ and $\hat{M}_{n}(e_{2})$ nonsingular).
Defining $\tilde{\phi}_{n}(t):=\hat{\phi}_{n}(e_{(t)})$ for
$e_{(t)}=e_{1}+t(e_{2}-e_{1})$, by optimality of $\hat{e}$ we must have
$\tilde{\phi}_{n}^{\prime}(0)\leqslant 0$. Taylor’s theorem and the second
part of Lemma C.2 then allow us to conclude that

$\displaystyle\hat{\phi}_{n}(e_{2})\bm{1}(A_{n})=\tilde{\phi}_{n}(1)\bm{1}(A_{n})$
$\displaystyle=\Bigl{(}\tilde{\phi}_{n}(0)+\tilde{\phi}_{n}^{\prime}(0)+\frac{1}{2}\tilde{\phi}_{n}^{\prime\prime}(t)\Bigr{)}\bm{1}(A_{n})$
$\displaystyle\leqslant\Bigl{(}\hat{\phi}_{n}(\hat{e})+\frac{1}{2}\textnormal{tr}\bigl{[}\nabla\Psi(\tilde{M}_{n}(t))^{\top}\tilde{M}_{n}^{\prime\prime}(t)\bigr{]}\Bigr{)}\bm{1}(A_{n})$
(85)

for some $t\in(0,1)$, where $\tilde{M}_{n}(t)=\hat{M}_{n}(e_{(t)})$ as in the
proof of Lemma C.2. With $\nabla\Psi(\tilde{M}_{n}(t))\succcurlyeq\tilde{k}I$
for all $t\in(0,1)$ whenever $A_{n}$ holds, we have by (84) that

$0\geqslant\textnormal{tr}\bigl{[}(\nabla\Psi(\tilde{M}_{n}(t))-\tilde{k}I)^{\top}\tilde{M}_{n}^{\prime\prime}(t)\bigr{]}\bm{1}(A_{n})=\bigl{(}\textnormal{tr}[(\nabla\Psi(\tilde{M}_{n}(t))^{\top}\tilde{M}_{n}^{\prime\prime}(t)]-\tilde{k}\textnormal{tr}[\tilde{M}_{n}^{\prime\prime}(t)]\bigr{)}\bm{1}(A_{n}).$

Hence by (29)

$\textnormal{tr}[(\nabla\Psi(\tilde{M}_{n}(t))^{\top}\tilde{M}_{n}^{\prime\prime}(t)]\bm{1}(A_{n})\leqslant\tilde{k}\textnormal{tr}[\tilde{M}_{n}^{\prime\prime}(t)]\bm{1}(A_{n})\leqslant-\tilde{k}c\|\hat{e}-e\|_{2,P_{n}}^{2}\bm{1}(A_{n}).$

Then by (C) we conclude that whenever
$\|e-\hat{e}\|_{2,P_{n}}\leqslant\tilde{r}$ we have

$(\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e))\bm{1}(A_{n})\geqslant
c\frac{\tilde{k}}{2}\|\hat{e}-e\|_{2,P_{n}}^{2}\bm{1}(A_{n}).$

Redefining $r_{0}$ to be the minimum of the $r_{0}$ appearing in the proof of
(75) and $\tilde{r}$, since $\hat{\phi}_{n}$ is always concave on
$\mathcal{F}_{n}$. we can repeat the argument at the end of the proof of (75)
to conclude that

$(\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e))\bm{1}(A_{n})\geqslant\left(c_{0}\min(r_{0},\|\hat{e}-e\|_{2,P_{n}})\|\hat{e}-e\|_{2,P_{n}}\right)\bm{1}(A_{n})$

for all $e\in\mathcal{F}_{n}$; here $c_{0}:=c\tilde{k}/2>0$.

### Step 2

Fix $\hat{e}_{\mathcal{F}}\in\mathcal{F}_{*}$ and
$e_{n}^{*}\in\mathcal{F}_{n}$. The result of Step 1 shows that for some
positive constants $c_{0}$ and $r_{0}$, we have

$\displaystyle
c_{0}\min(r_{0},\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P})\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P}$
$\displaystyle\leqslant\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}}),\quad\text{and}$
$\displaystyle
c_{0}\min(r_{0},\|\hat{e}-e_{n}^{*}\|_{2,P_{n}})\|\hat{e}-e_{n}^{*}\|_{2,P_{n}}\bm{1}(A_{n})$
$\displaystyle\leqslant(\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}))\bm{1}(A_{n}).$

With $\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})\geqslant 0$ and
$\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*})\geqslant 0$ by the
definitions of $e^{*}$ and $\hat{e}$, we can further upper bound the right-
hand sides by

$\displaystyle\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})$
$\displaystyle\leqslant\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}),\quad\text{and}$
$\displaystyle(\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}))\bm{1}(A_{n})$
$\displaystyle\leqslant\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*}).$

### Step 3

For brevity, in this section we introduce the empirical process notation

$Pe=\int_{\mathcal{X}}e(x)\,\mathrm{d}P(x)$

for all $e\in\mathcal{E}$. For instance, with $P_{n}$ the empirical measure
induced by $X_{1},\ldots,X_{n}$, we have $P_{n}e=n^{-1}\sum_{i=1}^{n}X_{i}$
for all $e\in\mathcal{E}$.

We first show that we can choose particular
$\hat{e}_{\mathcal{F}}=\hat{e}_{\mathcal{F}}(\cdot)\in\mathcal{F}$ and
$e_{n}^{*}(\cdot)\in\mathcal{F}_{n}$ that are very close in sup norm to
$\hat{e}\in\mathcal{F}_{n}$ and $e^{*}\in\mathcal{F}$, respectively, with high
probability.

######  Lemma C.3.

Under the conditions of Lemma 5.2, there exist
$\hat{e}_{\mathcal{F}}=\hat{e}_{\mathcal{F}}(\cdot)\in\mathcal{F}$ and
$e_{n}^{*}=e_{n}^{*}(\cdot)\in\mathcal{F}_{n}$ such that
$\sup_{x\in\mathcal{X}}|\hat{e}(x)-\hat{e}_{\mathcal{F}}(x)|+\sup_{x\in\mathcal{X}}|e^{*}(x)-e_{n}^{*}(x)|=O_{p}(n^{-1/2})$.

###### Proof of Lemma C.3.

With the functions in $\mathcal{E}$ uniformly bounded by 1, by Lemma B.3 we
conclude

$\mathbb{E}_{P}\left[\sup_{e\in\mathcal{E}}\Big{|}P_{n}e-Pe\Big{|}\right]\leqslant
KCn^{-1/2}$

so that $\sup_{e\in\mathcal{E}}\big{|}P_{n}e-Pe\big{|}=O_{p}(n^{-1/2})$ by
Markov’s inequality. In view of the fact that $m_{L}\leqslant
P_{n}\hat{e}\leqslant m_{H}$ and $m_{L}\leqslant Pe^{*}\leqslant m_{H}$ since
$\hat{e}\in\mathcal{F}_{n}$ and $e^{*}\in\mathcal{F}$, we have

$\displaystyle m_{L}-\sup_{e\in\mathcal{E}}\big{|}P_{n}e-Pe\big{|}$
$\displaystyle\leqslant P_{n}e^{*}\leqslant
m_{H}+\sup_{e\in\mathcal{E}}\big{|}P_{n}e-Pe\big{|},\quad\text{and}$ (86)
$\displaystyle m_{L}-\sup_{e\in\mathcal{E}}\big{|}P_{n}e-Pe\big{|}$
$\displaystyle\leqslant P\hat{e}\leqslant
m_{H}+\sup_{e\in\mathcal{E}}\big{|}P_{n}e-Pe\big{|}.$ (87)

Next, with $e_{L}=e_{L}(\cdot)$ and $e_{H}=e_{H}(\cdot)$ as in the assumptions
of Lemma 5.2, define

$e_{n}^{*}(x)=\begin{cases}e^{*}(x),&m_{L}\leqslant P_{n}e^{*}\leqslant
m_{H}\\\ e^{*}(x)+\lambda_{n}(e_{L}(x)-e^{*}(x)),&P_{n}e^{*}<m_{L}\\\
e^{*}(x)+\lambda_{n}(e_{H}(x)-e^{*}(x)),&P_{n}e^{*}>m_{H}\end{cases}$

where

$\lambda_{n}=\begin{cases}\dfrac{m_{L}-P_{n}e^{*}}{P_{n}[e_{L}-e^{*}]},&P_{n}e^{*}<m_{L}\\\\[8.61108pt]
\dfrac{P_{n}e^{*}-m_{H}}{P_{n}[e^{*}-e_{H}]},&P_{n}e^{*}>m_{H}\\\\[8.61108pt]
0&\text{otherwise.}\end{cases}$

On the event

$A_{n}=\left\\{P_{n}e_{L}\geqslant\frac{Pe_{L}+m_{L}}{2},P_{n}e_{H}\leqslant\frac{Pe_{H}+m_{H}}{2}\right\\}$

we must have $0\leqslant\lambda_{n}\leqslant 1$ since $Pe_{L}>m_{L}$ and
$Pe_{H}<m_{H}$, and so when $A_{n}$ holds we know $e_{n}^{*}\in\mathcal{E}$ by
convexity of $\mathcal{E}$. Furthermore we have
$P_{n}e_{n}^{*}=\max(m_{L},\min(m_{H},\mathbb{E}_{P_{n}}[e^{*}(X)]))$ so that
in fact $e_{n}^{*}\in\mathcal{F}_{n}$. But with
$M=2\max\left((Pe_{L}-m_{L})^{-1},(m_{H}-Pe_{H})^{-1}\right)$ we have

$\displaystyle 0\leqslant\lambda_{n}\bm{1}(A_{n})$ $\displaystyle\leqslant
M\bm{1}(A_{n})[(m_{L}-P_{n}e^{*})\bm{1}(P_{n}e^{*}<m_{L})+(P_{n}e^{*}-m_{H})\bm{1}(P_{n}e^{*}>m_{H})]$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:P_n_e_star}}}{{\leqslant}}2M\sup_{e\in\mathcal{E}}|P_{n}e-P_{e}|\bm{1}(A_{n}).$

With $\textnormal{Pr}(A_{n})\rightarrow 1$ as $n\rightarrow\infty$ by the law
of large numbers, we get $\lambda_{n}=O_{p}(n^{-1/2})$, and hence for each
$x\in\mathcal{X}$ we have

$|e_{n}^{*}(x)-e^{*}(x)|\leqslant\lambda_{n}(|e_{L}(x)-e^{*}(x)|\vee|e_{H}(x)-e^{*}(x)|)\leqslant\lambda_{n}=O_{p}(n^{-1/2}).$

Next, define

$\hat{e}_{\mathcal{F}}(x)=\begin{cases}\hat{e}(x),&m_{L}\leqslant
P\hat{e}\leqslant m_{H}\\\
\hat{e}(x)+\tilde{\lambda}_{n}(e_{L}(x)-\hat{e}(x)),&P\hat{e}<m_{L}\\\
\hat{e}(x)+\tilde{\lambda}_{n}(e_{H}(x)-\hat{e}(x)),&P\hat{e}>m_{H}\end{cases}$

with

$\tilde{\lambda}_{n}=\begin{cases}\dfrac{m_{L}-P\hat{e}}{m_{H}-\hat{e}},&P\hat{e}<m_{L}\\\\[8.61108pt]
\dfrac{P\hat{e}-m_{H}}{P\hat{e}-m_{L}},&P\hat{e}>m_{H}\\\\[8.61108pt]
0,&\text{ otherwise}\end{cases}$

so that $0\leqslant\tilde{\lambda}_{n}\leqslant 1$ and
$\hat{e}_{\mathcal{F}}\in\mathcal{F}$ always. In fact

$\displaystyle\tilde{\lambda}_{n}$
$\displaystyle\leqslant\frac{M}{2}\bigl{[}(m_{L}-P\hat{e})\bm{1}(P\hat{e}<m_{L})+(P\hat{e}-m_{H})\bm{1}(P\hat{e}>m_{H})\bigr{]}$
$\displaystyle\leqslant M\sup_{e\in\mathcal{E}}|P_{n}e-Pe|$

so that $\tilde{\lambda}_{n}=O_{p}(n^{-1/2})$ as well and the lemma follows
since for all $x\in\mathcal{X}$

$|\hat{e}(x)-\hat{e}_{\mathcal{F}}(x)|\leqslant\tilde{\lambda}_{n}(|e_{L}(x)-\hat{e}(x)|\vee|e_{H}(x)-\hat{e}(x)|)\leqslant\tilde{\lambda}_{n}.$

∎

We are now ready to prove consistency. Taking $\hat{e}_{\mathcal{F}}$ as in
Lemma C.3, we upper bound the right-hand side of the inequalities in step 2:

$\displaystyle\begin{split}\phi(e^{*})-\phi(\hat{e}_{\mathcal{F}})+\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(e_{n}^{*})&\leqslant|\phi(e^{*})-\phi(e_{n}^{*})|+|\phi(e_{n}^{*})-\hat{\phi}_{n}(e_{n}^{*})|\\\
&\phantom{=}\
+|\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})|+|\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})-\phi(\hat{e}_{\mathcal{F}})|.\end{split}$
(88)

In view of (72) and (78), Lemma C.3 shows that

$\displaystyle\|M(e^{*})-M(e_{n}^{*})\|$
$\displaystyle=O_{p}(n^{-1/2}),\quad\text{and}$ (89)
$\displaystyle\|\hat{M}_{n}(\hat{e})-\hat{M}_{n}(\hat{e}_{\mathcal{F}})\|$
$\displaystyle=O_{p}(n^{-1/2}).$ (90)

We now use (89) to show that

$|\phi(e^{*})-\phi(e_{n}^{*})|=O_{p}(n^{-1/2}).$ (91)

As shown at the start of step 1, whenever $\|e_{n}^{*}-e^{*}\|_{2,P}\leqslant
r_{0}$ we have $M(e_{n}^{*})\succcurlyeq(k^{*}/2)I$, so that

$tM(e^{*})+(1-t)M(e_{n}^{*})\succcurlyeq(k^{*}/2)I\quad\forall t\in[0,1].$

Applying Taylor’s theorem to $\Psi(\cdot)$ we have for some $K<\infty$ that

$\displaystyle|\phi(e^{*})-\phi(e_{n}^{*})|\bm{1}(\tilde{A}_{n})$
$\displaystyle=|\Psi(M(e^{*}))-\Psi(M(e_{n}^{*}))|\bm{1}(\tilde{A}_{n})$
$\displaystyle\leqslant\sup_{t\in[0,1]}\Big{|}\textnormal{tr}[\nabla\Psi(tM(e^{*})+(1-t)M(e_{n}^{*}))]^{\top}[M(e^{*})-M(e_{n}^{*})])\Big{|}\bm{1}(\tilde{A}_{n})$
$\displaystyle\leqslant\sup_{t\in[0,1]}\|\nabla\Psi(tM(e^{*})+(1-t)M(e_{n}^{*}))\|\cdot\|M(e^{*})-M(e_{n}^{*})\|\bm{1}(\tilde{A}_{n})$
$\displaystyle\leqslant
K\sqrt{p}\cdot\|M(e^{*})-M(e_{n}^{*})\|\bm{1}(\tilde{A}_{n})$ (92)

where $\tilde{A}_{n}$ is the event $\|e_{n}^{*}-e^{*}\|_{2,P}\leqslant r_{0}$
and the last inequality follows from Assumption 5.2(c) and the fact that
$\|A\|\leqslant\sqrt{p}\lambda_{\max}(A)$ for any $A\in\mathbb{S}_{+}^{p}$.
Here $\lambda_{\max}(A)$ denotes the largest eigenvalue of $A$. Hence
$|\phi(e^{*})-\phi(e_{n}^{*})|\bm{1}(A_{n})=O_{p}(n^{-1/2})$ by (89). But
$\textnormal{Pr}(\tilde{A}_{n})\rightarrow 1$ by (89), so indeed
$|\phi(e^{*})-\phi(e_{n}^{*})|=O_{p}(n^{-1/2})$.

Convergence of the remaining three terms in (88) depends on the following
result:

$\sup_{e\in\mathcal{E}}\|\hat{M}_{n}(e)-M(e)\|=O_{p}(n^{-1/2})+O_{p}(\alpha_{n}).$
(93)

To show this, define

$M_{n}(e)=\frac{1}{n}\sum_{i=1}^{n}f(e(X_{i}),\eta(X_{i}))$

which replaces the estimated nuisance function $\hat{\eta}$ in the definition
of $\hat{M}_{n}$ with the true $\eta$. First note that by uniform boundedness
of $\|f_{w}\|_{2}$,

$\frac{1}{n}\sum_{i=1}^{n}\sup_{e\in[0,1]}\|f(e,\hat{\eta}(X_{i}))-f(e,\eta(X_{i}))\|\leqslant
Cn^{-1}\sum_{i=1}^{n}\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}\leqslant
C\|\hat{\eta}-\eta\|_{2,P_{n}}.$

Then by (34) we have

$\displaystyle\sup_{e\in\mathcal{E}}\|\hat{M}_{n}(e)-M_{n}(e)\|$
$\displaystyle=\sup_{e\in\mathcal{E}}\bigg{\|}\frac{1}{n}\sum_{i=1}^{n}f(e(X_{i}),\hat{\eta}(X_{i}))-f(e(X_{i}),\eta(X_{i}))\bigg{\|}$
$\displaystyle\leqslant\frac{1}{n}\sum_{i=1}^{n}\sup_{e\in[0,1]}\|f(e,\hat{\eta}(X_{i}))-f(e,\eta(X_{i}))\|$
$\displaystyle=O_{p}(\alpha_{n}).$ (94)

Now for $i,j\in\\{1,\ldots,p\\}$ define the class of functions

$\mathcal{G}_{ij}=\\{x\mapsto f_{ij}(e(x),\eta(x))\mid e\in\mathcal{E}\\}$

where $f_{ij}$ denotes the $(i,j)$-th entry of the function $f$. By Lemma B.4

$\displaystyle
n^{-1/2}\int_{0}^{1}\sqrt{\log\mathcal{N}(r,\mathcal{G}_{ij},L^{2}(P_{n}))}\,\mathrm{d}r$
$\displaystyle=Cn^{-1/2}\int_{0}^{C^{-1}}\sqrt{\log\mathcal{N}(C\epsilon,\mathcal{G}_{ij},L^{2}(P_{n}))}\,\mathrm{d}\epsilon$
$\displaystyle\leqslant
Cn^{-1/2}\int_{0}^{C^{-1}}\sqrt{\log\mathcal{N}(\epsilon,\mathcal{E},L^{2}(P_{n}))}\,\mathrm{d}\epsilon.$

Let

$D(e)=\frac{1}{n}\sum_{i=1}^{n}f(e(X_{i}),\eta(X_{i}))-\int_{\mathcal{X}}f(e(x),\eta(x))\,\mathrm{d}P(x)=M_{n}(e)-M(e).$

Then Lemma B.3 and Assumption 5.1 indicate that
$\sup_{e\in\mathcal{E}}|D_{ij}(e)|=O_{p}(n^{-1/2})$ and so

$\sup_{e\in\mathcal{E}}\|D(e)\|=\sup_{e\in\mathcal{E}}\biggl{(}\,\sum_{i=1}^{p}\sum_{j=1}^{p}D_{ij}(e)^{2}\biggr{)}^{1/2}\leqslant\biggl{(}\,\sum_{i=1}^{p}\sum_{j=1}^{p}\sup_{e\in\mathcal{E}}|D_{ij}(e)|^{2}\biggr{)}^{1/2}=O_{p}(n^{-1/2}).$
(95)

The result (93) follows by the triangle inequality in view of (94) and (95).

We are finally ready to bound the remaining terms on the right-hand side of
(88), and show that $\textnormal{Pr}(A_{n})\rightarrow 1$ as
$n\rightarrow\infty$ where $A_{n}$ is the event
$\hat{M}_{n}(\hat{e})\succcurlyeq k_{0}I$ for $k_{0}>0$ defined in Step 1.
Choose $\delta>0$ so that for any $A_{1}$, $A_{2}$ in
$\mathcal{M}=\\{A\in\mathbb{S}_{+}^{p}:0\preccurlyeq A\preccurlyeq CI\\}$,
with $\|A_{1}-A_{2}\|\leqslant\delta$, we have
$|\lambda_{\min}(A_{1})-\lambda_{\min}(A_{2})|\leqslant\min(k_{0}/2,k^{*}/4)$,
where $k^{*}$ satisfies (74). Such $\delta$ exists by uniform continuity of
$\lambda_{\min}(\cdot)$ on the compact subset $\mathcal{M}$ of p×p (cf. the
proof of Lemma C.1). Also define the event $B_{n}$ that all of the following
are true:

$\displaystyle\|e_{n}^{*}-e^{*}\|_{2,P}$ $\displaystyle\leqslant r_{0},$ (96)
$\displaystyle\sup_{e\in\mathcal{E}}\|\hat{M}_{n}(e)-M(e)\|$
$\displaystyle\leqslant\delta\quad\text{and}$ (97)
$\displaystyle\|\hat{M}_{n}(\hat{e})-\hat{M}_{n}(\hat{e}_{\mathcal{F}})\|$
$\displaystyle\leqslant\delta.$ (98)

We claim that $B_{n}$ implies the following conditions:

$M(e_{n}^{*})\succcurlyeq\frac{k^{*}}{2}I,\quad\hat{M}_{n}(e_{n}^{*})\succcurlyeq\frac{k^{*}}{4}I,\quad\hat{M}_{n}(\hat{e})\succcurlyeq
k_{0}I\quad\text{and}\quad\hat{M}_{n}(\hat{e}_{\mathcal{F}})\succcurlyeq\frac{k_{0}}{2}I.$

We prove these statements briefly. Assume $B_{n}$ holds. First, note that
$M(e_{n}^{*})\succcurlyeq(k^{*}/2)I$ holds by definition of $r_{0}$ and (96).
Next, the definition of $\delta$ immediately ensures by (97) that
$\hat{M}_{n}(e_{n}^{*})\succcurlyeq(k^{*}/4)I$. But then

$\Psi(\hat{M}_{n}(\hat{e}))\geqslant\Psi(\hat{M}_{n}(e_{n}^{*}))\geqslant\inf_{B\succcurlyeq(k^{*}/4)I}\Psi(B)$

so that $\hat{M}_{n}(\hat{e})\succcurlyeq k_{0}I$ by the definition of
$k_{0}$, and in particular we have shown $B_{n}\subseteq A_{n}$. Finally (98)
shows $\hat{M}_{n}(\hat{e}_{\mathcal{F}})\succcurlyeq(k_{0}/2)I$.

Now take $\tilde{K}<\infty$ to be as derived from Assumption 5.2(c) with
$k=\min(k^{*}/4,k_{0}/2)$ and $K=C$. Then repeated applications of arguments
analogous to (92) show that

$\displaystyle|\phi(e_{n}^{*})-\hat{\phi}_{n}(e_{n}^{*})|\bm{1}(B_{n})$
$\displaystyle=|\Psi(M(e_{n}^{*}))-\Psi(\hat{M}_{n}(e_{n}^{*}))|\bm{1}(B_{n})$
$\displaystyle\leqslant\sup_{0\preccurlyeq
A\preccurlyeq\tilde{K}I}|\textnormal{tr}(A^{\top}[M(e_{n}^{*})-\hat{M}_{n}(e_{n}^{*})])|$
$\displaystyle\leqslant\tilde{K}\sqrt{p}\cdot\|M(e_{n}^{*})-\hat{M}_{n}(e_{n}^{*})\|$
$\displaystyle=O_{p}(n^{-1/2})+O_{p}(\alpha_{n})$

by (93) and similarly

$\displaystyle|\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})|\bm{1}(B_{n})\leqslant\tilde{K}\sqrt{p}\cdot\|\hat{M}_{n}(\hat{e})-\hat{M}_{n}(\hat{e}_{\mathcal{F}})\|=O_{p}(n^{-1/2})$

by (90). Also

$\displaystyle|\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})-\phi(\hat{e}_{\mathcal{F}})|\bm{1}(B_{n})\leqslant\tilde{K}\sqrt{p}\cdot\|\hat{M}_{n}(\hat{e}_{\mathcal{F}})-M(\hat{e}_{\mathcal{F}})\|=O_{p}(n^{-1/2})+O_{p}(\alpha_{n})$

by (93).

However, by (89), (72), (90), and (93) we know that
$\textnormal{Pr}(B_{n})\rightarrow 1$ as $n\rightarrow\infty$. Since
$B_{n}\subseteq A_{n}$ we also have $\textnormal{Pr}(A_{n})\rightarrow 1$. We
conclude from the preceding displays that

$|\phi(e_{n}^{*})-\hat{\phi}_{n}(e_{n}^{*})|+|\hat{\phi}_{n}(\hat{e})-\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})|+|\hat{\phi}_{n}(\hat{e}_{\mathcal{F}})-\phi(\hat{e}_{\mathcal{F}})|=O_{p}(n^{-1/2})+O_{p}(\alpha_{n}).$

Then by step 2, (88), and (91), we conclude that

$\displaystyle\|e^{*}-\hat{e}_{\mathcal{F}}\|_{2,P}$
$\displaystyle=O_{p}(n^{-1/4})+O_{p}(\alpha_{n}^{-1/2})\quad\text{and}$
$\displaystyle\|\hat{e}-e_{n}^{*}\|_{2,P_{n}}$
$\displaystyle=O_{p}(n^{-1/4})+O_{p}(\alpha_{n}^{-1/2}).$

But by Lemma C.3 we know that
$\|\hat{e}-\hat{e}_{\mathcal{F}}\|_{2,P}+\|e^{*}-e_{n}^{*}\|_{2,P_{n}}=O_{p}(n^{-1/2})$.
So by the triangle inequality we conclude
$\|e^{*}-\hat{e}\|_{2,P}+\|e^{*}-\hat{e}\|_{2,P_{n}}=O_{p}(n^{-1/4})+O_{p}(\alpha_{n}^{-1/2})$
as well.

### Step 4

The final step in the argument to derive our best convergence rates is a
variation of a standard “peeling” argument used in deriving convergence rates
of $M$-estimators. The main argument requires deriving a bound on the “locally
centered empirical process” as in our next result.

######  Lemma C.4.

For each $e\in\mathcal{E}$, let $\phi_{n}(e)=\Psi(M_{n}(e))$ and take
$\tilde{e}\in\operatorname*{arg\,max}_{e\in\mathcal{F}_{n}}\phi_{n}(e)$. Then
there exists $\beta>0$ and a universal constant $C_{0}<\infty$ such that for
all $u\leqslant\beta$

$\sup_{e\in\mathcal{E}:\|e-\tilde{e}\|_{2,P_{n}}\leqslant
u}\bigl{[}(\hat{\phi}_{n}(e)-\phi_{n}(e))-(\hat{\phi}_{n}(\tilde{e})-\phi_{n}(\tilde{e}))\bigr{]}\bm{1}(B_{n})\leqslant
C_{0}\bigl{(}u\|\hat{\eta}-\eta\|_{2,P_{n}}+\|\hat{\eta}-\eta\|_{2,P_{n}}^{2}\bigr{)}$
(99)

for some sequence of events $B_{n}$ with $\textnormal{Pr}(B_{n})\rightarrow 1$
as $n\rightarrow\infty$.

###### Proof of Lemma C.4.

Let $U_{n}(e)=\hat{\phi}_{n}(e)-\phi_{n}(e)$ for each $e\in\mathcal{E}$. By
Taylor’s theorem, for each $e\in\mathcal{E}$

$U_{n}(e)=\Psi(\hat{M}_{n}(e))-\Psi(M_{n}(e))=\nabla\Psi(R_{n}(e))^{\top}(\hat{M}_{n}(e)-M_{n}(e))$

for some $R_{n}(e)$ lying on the line segment between $M_{n}(e)$ and
$\hat{M}_{n}(e)$. Write

$\displaystyle U_{n}(e)-U_{n}(\tilde{e})$
$\displaystyle=(\nabla\Psi(R_{n}(e))-\nabla\Psi(R_{n}(\tilde{e})))^{\top}(\hat{M}_{n}(e)-M_{n}(e))$
$\displaystyle\phantom{=}\
+\nabla\Psi(R_{n}(\tilde{e}))^{\top}[(\hat{M}_{n}(e)-M_{n}(e))-(\hat{M}_{n}(\tilde{e})-M_{n}(\tilde{e}))].$
(100)

Because $\hat{\eta}=\eta$ trivially satisfies the conditions of Lemma 5.2, all
of our results in steps 1–3 apply if we replace $\hat{M}_{n}$ with $M_{n}$. In
particular with $k_{0}>0$ as chosen in step 1, we have
$\textnormal{Pr}(M_{n}(\tilde{e})\succcurlyeq k_{0}I)\rightarrow 1$ as
$n\rightarrow\infty$ by the argument at the end of step 3. Take $\tilde{r}$,
$\tilde{k}$, and $\tilde{K}$ derived from Lemma C.1 with this choice of
$k_{0}$. By the proof of Lemma C.1, we know that for all $0\preccurlyeq
A\preccurlyeq CI$ with $\|A-M_{n}(\tilde{e})\|\leqslant C\tilde{r}$, we must
have $A\succcurlyeq(k_{0}/2)I$ and
$\tilde{K}I\succcurlyeq\nabla\Psi(A)\succcurlyeq\tilde{k}I$. Let $B_{n}$ be
the intersection of the events $M_{n}(\tilde{e})\succcurlyeq k_{0}I$ and
$\sup_{e\in\mathcal{E}}\|\hat{M}_{n}(e)-M_{n}(e)\|\leqslant C\tilde{r}/3$. Now
$\textnormal{Pr}(B_{n})\rightarrow 1$ as $n\rightarrow\infty$ by (94). Then
for any $e\in\mathcal{E}$ with
$\|e-\tilde{e}\|_{2,P_{n}}\leqslant\tilde{r}/3$, in view of (78) we have

$\displaystyle\|\hat{M}_{n}(e)-M_{n}(\tilde{e})\|\bm{1}(B_{n})$
$\displaystyle\leqslant\|\hat{M}_{n}(e)-\hat{M}_{n}(\tilde{e})\|+\|\hat{M}_{n}(\tilde{e})-M_{n}(\tilde{e})\|\bm{1}(B_{n})$
$\displaystyle\leqslant\frac{2C\tilde{r}}{3}\quad\text{and}$
$\displaystyle\|M_{n}(e)-M_{n}(\tilde{e})\|\bm{1}(B_{n})$
$\displaystyle\leqslant\|M_{n}(e)-\hat{M}_{n}(e)\|\bm{1}(B_{n})+\|\hat{M}_{n}(e)-M_{n}(\tilde{e})\|\bm{1}(B_{n})$
$\displaystyle\leqslant C\tilde{r}.$

We conclude by the preceding display that whenever $B_{n}$ holds, for all
$e\in\mathcal{E}$ with $\|e-\tilde{e}\|_{2,P_{n}}\leqslant\tilde{r}/3$ we have
$(k_{0}/2)I\preccurlyeq\hat{M}_{n}(e)\preccurlyeq CI$ and
$(k_{0}/2)I\preccurlyeq M_{n}(e)\preccurlyeq CI$, and thus
$(k_{0}/2)I\preccurlyeq R_{n}(e)\preccurlyeq CI$ along with
$\tilde{K}I\succcurlyeq\nabla\Psi(R_{n}(e))\succcurlyeq\tilde{k}I$. Then by
Assumption 5.2(b), for all $u\leqslant\tilde{r}/3$ there exists a constant
$K_{0}<\infty$ (independent of $u$) for which

$\displaystyle\|\nabla\Psi(R_{n}(e))-\nabla\Psi(R_{n}(\tilde{e}))\|\bm{1}(B_{n})$
$\displaystyle\leqslant K_{0}\|R_{n}(e)-R_{n}(\tilde{e})\|\bm{1}(B_{n})$
$\displaystyle\leqslant
K_{0}(\|R_{n}(e)-\hat{M}_{n}(e)\|+\|\hat{M}_{n}(e)-\hat{M}_{n}(\tilde{e})\|+\|\hat{M}_{n}(\tilde{e})-R_{n}(\tilde{e})\|)$

The preceding inequality holds for all $e\in\mathcal{E}$ with
$\|e-\tilde{e}\|_{2,P_{n}}\leqslant u$. Taking a supremum over such
$e(\cdot)$, another application of (78) and the fact that
$\|R_{n}(e)-\hat{M}_{n}(e)\|\leqslant\|M_{n}(e)-\hat{M}_{n}(e)\|$ for all
$e\in\mathcal{E}$ show that

$\sup_{e\in\mathcal{E}:\|e-\tilde{e}\|_{2,P_{n}}\leqslant
u}\|\nabla\Psi(R_{n}(e))-\nabla\Psi(R_{n}(\tilde{e}))\|\bm{1}(B_{n})\leqslant
K_{0}\bigl{(}Cu+2S_{n}(\mathcal{E})\bigr{)}$

where $S_{n}(\mathcal{E})=\sup_{e\in\mathcal{E}}\|\hat{M}_{n}(e)-M_{n}(e)\|$.
Then by Cauchy-Schwarz

$\begin{split}\phantom{=}\
&\sup_{e\in\mathcal{E}:\|e-\tilde{e}\|_{2,P_{n}}\leqslant
u}\bigl{|}(\nabla\Psi(R_{n}(e))-\nabla\Psi(R_{n}(\tilde{e})))^{\top}(\hat{M}_{n}(e)-M_{n}(e))\bigr{|}\bm{1}(B_{n})\\\
&\leqslant
K_{0}\bigl{(}Cu+2S_{n}(\mathcal{E})\bigr{)}S_{n}(\mathcal{E})\end{split}$
(101)

for all $u\leqslant\tilde{r}/3$.

Next, define

$c(x;e,\tilde{e},\eta,\hat{\eta})=\bigl{[}f(e(X_{i}),\hat{\eta}(X_{i}))-f(e(X_{i}),\eta(X_{i}))\bigr{]}-\bigl{[}f(\tilde{e}(X_{i}),\hat{\eta}(X_{i}))-f(\tilde{e}(X_{i}),\eta(X_{i}))\bigr{]}$

for all $x\in\mathcal{X}$ so that

$\bigl{(}\hat{M}_{n}(e)-M_{n}(e)\bigr{)}-\bigl{(}\hat{M}_{n}(\tilde{e})-M_{n}(\tilde{e})\bigr{)}=\frac{1}{n}\sum_{i=1}^{n}c(X_{i};e,\tilde{e},\eta,\hat{\eta}).$

Fix $k,\ell\in\\{1,\ldots,p\\}$. By Taylor’s theorem with the Lagrange form of
the remainder,

$\displaystyle|c(X_{i};e,\tilde{e},\eta,\hat{\eta})|$
$\displaystyle=|[f(e(X_{i}),\hat{\eta}(X_{i}))-f(e(X_{i}),\eta(X_{i}))]-[f(\tilde{e}(X_{i}),\hat{\eta}(X_{i}))-f(\tilde{e}(X_{i}),\eta(X_{i}))]|$
$\displaystyle=|f_{w}(e(X_{i}),\eta_{1}(X_{i}))^{\top}(\hat{\eta}(X_{i})-\eta(X_{i}))-f_{w}(\tilde{e}(X_{i}),\eta_{2}(X_{i}))^{\top}(\hat{\eta}(X_{i})-\eta(X_{i}))|$
$\displaystyle\leqslant\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}\|f_{w}(e(X_{i}),\eta_{1}(X_{i}))-f_{w}(e(X_{i}),\eta(X_{i}))\|_{2}$
$\displaystyle\phantom{=}\
+\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}\|f_{w}(e(X_{i}),\eta(X_{i}))-f_{w}(\tilde{e}(X_{i}),\eta(X_{i}))\|_{2}$
$\displaystyle\phantom{=}\
+\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}\|f_{w}(\tilde{e}(X_{i}),\eta(X_{i}))-f_{w}(\tilde{e}(X_{i}),\eta_{2}(X_{i}))\|_{2}$
$\displaystyle\leqslant
2C\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}^{2}+C\|\hat{\eta}(X_{i})-\eta(X_{i})\|_{2}|\tilde{e}(X_{i})-e(X_{i})|.$

We have omitted subscripts $k\ell$ on $c$ and $f$ everywhere in the preceding
display for brevity (i.e., $c$ above denotes $c_{k\ell}$ and $f$ above denotes
$f_{k\ell}$). The functions $e_{1}(x)$ and $e_{2}(x)$ are somewhere on the
line segment between $e(x)$ and $\tilde{e}(x)$, and he functions $\eta_{1}(x)$
and $\eta_{2}(x)$ are somewhere on the line segment between $\eta(x)$ and
$\hat{\eta}(x)$, and the final inequality follows from uniform boundedness of
$f_{ww}$ and $f_{w}^{\prime}$. We conclude

$\displaystyle\sup_{e\in\mathcal{E}:\|e-\tilde{e}\|_{2,P_{n}}\leqslant
u}\|(\hat{M}_{n}(e)-M_{n}(e))-(\hat{M}_{n}(\tilde{e})-M_{n}(\tilde{e}))\|$
$\displaystyle\leqslant\frac{1}{n}\sum_{i=1}^{n}\|c(X_{i};e,\tilde{e},\eta,\hat{\eta})\|$
$\displaystyle\leqslant
Cp^{2}\bigl{(}2\|\hat{\eta}-\eta\|_{2,P_{n}}^{2}+u\|\hat{\eta}-\eta\|_{2,P_{n}}\bigr{)}$

where the last inequality follows by Cauchy-Schwarz. Recalling
$\nabla\Psi(R_{n}(\tilde{e}))\preccurlyeq\tilde{K}I$ whenever $B_{n}$ holds,
we conclude

$\displaystyle\sup_{e\in\mathcal{E}:\|e-\tilde{e}\|_{2,P_{n}}\leqslant
u}\nabla\Psi(R_{n}(\tilde{e}))^{\top}[(\hat{M}_{n}(e)-M_{n}(e))-(\hat{M}_{n}(\tilde{e})-M_{n}(\tilde{e}))]\bm{1}(B_{n})$
$\displaystyle\leqslant\tilde{K}Cp^{2}(2\|\hat{\eta}-\eta\|_{2,P_{n}}^{2}+u\|\hat{\eta}-\eta\|_{2,P_{n}})$

for all $u\leqslant r/3$. The preceding display and (LABEL:eq:peeling_sup_1)
imply Lemma C.4 in view of the decomposition (100). ∎

Continuing with the proof of Step 4, let $C_{n}=A_{n}\cap B_{n}$ and
$r=r_{0}\wedge\beta>0$, where $A_{n}$ and $r_{0}$ are as in (76) with