this work generalize those of [Singh et al., 2019] and depart from those of
[Darolles et al., 2011, Hall and Horowitz, 2005].

The source conditions in this paper most closely resemble those in [Singh et
al., 2019], who study the NPIV problem without exogenous covariates. In
particular, the integral equation studied in that work is

$\gamma_{0}(z)=\int h_{0}(w)\mathrm{d}\mathbb{P}(w|z)=\langle
h_{0},E_{0}^{*}\phi(z)\rangle_{\mathcal{H}}.$

As such, the source conditions in [Singh et al., 2019] are a special case of
the source conditions in this work.

[Darolles et al., 2011] also study the NPIV problem without exogenous
covariates, but assume a different source condition. Whereas I assume two
source conditions for $E_{0}$ and $h_{0}$, [Darolles et al., 2011] assume one
source condition for $E_{0}$ and $h_{0}$. Consider the NPIV problem without
exogenous covariates. If $E_{0}$ is compact, then its singular value
decomposition is $\\{e_{j}^{w},\eta_{j},e_{j}^{z}\\}_{j=1}^{\infty}$ where
$e_{j}^{w}\in\mathbb{L}_{2}(\mathbb{P}(w))$ and
$e_{j}^{z}\in\mathbb{L}_{2}(\mathbb{P}(z))$. The authors assume

$h_{0}\in\mathcal{H}^{\beta}:=\left\\{f=\sum_{j=1}^{\infty}f_{j}e_{j}^{z}:\;\sum_{j=1}^{\infty}\frac{f_{j}^{2}}{\eta_{j}^{2\beta}}<\infty,\;f_{j}=\langle
f,e_{j}^{w}\rangle_{\mathbb{L}_{2}(\mathbb{P}(w))}\right\\},\quad\beta\in(0,2].$

This source condition involves both the smoothness of $E_{0}$, encoded by the
rate at which the singular values $\\{\eta_{j}\\}$ decay, and the smoothness
of $h_{0}$, encoded by the rate at which the Fourier coefficients
$\\{f_{j}\\}$ decay.

While [Hall and Horowitz, 2005] do not explicitly place a source condition,
the assumptions in that work are closely related to the source condition in
[Darolles et al., 2011]. Specifically, [Hall and Horowitz, 2005] directly
place assumptions on the rates at which $\\{\eta_{j}\\}$ and $\\{f_{j}\\}$ and
decay: $\eta_{j}\sim j^{-s}$ and $f_{j}\sim j^{-r}$. Note that if these
conditions hold and $\beta<s^{-1}(r-1/2)$ then $h_{0}\in\mathcal{H}^{\beta}$.
Moreover, $s$ essentially pins down the degree of smoothness of the joint
density of $(W,Z)$.

###  D.3 Alternatives to source conditions

Two recent works prove uniform consistency of estimators in ill posed
settings. Whereas this work proposes machine learning estimators with finite
sample guarantees for the negative control problem, previous work proposes
series estimators with asymptotic guarantees for related problems: NPIV [Chen
and Christensen, 2018] and panel proxy control [Deaner, 2018]. I now compare
the source conditions with the alternatives assumptions in these related
works.

[Chen and Christensen, 2018] assume that $(W,Z)$ have compact rectangular
supports and bounded densities. Denoting by $J$ the dimension of the series
and $\tau_{J}$ the series measure of ill posedness, the authors place
restrictions on how $J$ evolves relative to $n$, $\tau_{J}$, and the norms of
inverse covariances applied to the series. The main assumption is a collection
of high level conditions about ill posedness, the projection of $E_{0}$ onto
the series, and the projection of $h_{0}$ onto the series [Chen and
Christensen, 2018, Assumption 4]. The high level conditions are satisfied if
the series are taken to be the singular functions $(e_{j}^{w})$ and
$(e_{j}^{z})$ of the conditional expectation operator $E_{0}$. In such case,
the stated conditions amount to joint assumptions on $\\{f_{j}\\}$ and
$\\{\eta_{j}\\}$ as before, which we have seen to be a type of source
condition.

The conditions of [Deaner, 2018] are again in terms of series analysis.
[Deaner, 2018] assumes (i) the series approximates Hölder functions well; (ii)
joint densities are bounded away from zero and one; (iii) several ratios of
joint over marginal densities are Hölder smooth, as is the reduced form
$\gamma_{0}$; (iv) the series dimension $J$ grows at a certain rate relative
to $n$, the approximation quality, and the norms of inverse covariances
applied to the series. The smoothness of joint over marginal density ratios is
akin to assumptions on $\\{\eta_{j}\\}$. The restrictions on norms of inverse
covariances applied to the series is akin to assumptions on $\\{f_{j}\\}$. In
this sense, the assumptions of [Chen and Christensen, 2018, Deaner, 2018]
resemble the assumptions of [Hall and Horowitz, 2005].

##  Appendix E Algorithm derivation

###  E.1 Overview

I present an overview of the end-to-end procedure, first in principle then in
practice. For simplicity, I focus on the dose response curve
$\theta_{0}^{ATE}(d)=\int h_{0}(d,x,w)\mathrm{d}\mathbb{P}(x,w)$, where
$h_{0}$ is the confounding bridge and $\mathbb{P}(x,w)$ is the counterfactual
distribution with which we wish to reweight $h_{0}$. To further simplify the
discussion, I abstract from sample splitting. By Theorem 2,

$\theta_{0}^{ATE}(d)=\langle
h_{0},\phi(d)\otimes\mu\rangle_{\mathcal{H}},\quad\mu:=\int[\phi(x)\otimes\phi(w)]\mathrm{d}\mathbb{P}(x,w)$

where $\mu$ is the mean embedding of the counterfactual distribution
$\mathbb{P}(x,w)$. This representation suggests an estimator of the form

$\hat{\theta}^{ATE}(d)=\langle\hat{h},\phi(d)\otimes\hat{\mu}\rangle_{\mathcal{H}},\quad\hat{\mu}:=\int[\phi(x)\otimes\phi(w)]\mathrm{d}\hat{\mathbb{P}}(x,w)$

combining appropriate estimators $\hat{h}$ and $\hat{\mu}$. Estimation of
$\hat{\mu}$ is simply an average:
$\hat{\mu}=n^{-1}\sum_{i=1}^{n}\phi(x_{i})\otimes\phi(w_{i})$. Estimation of
$\hat{h}$ is more involved.

Towards an estimator $\hat{h}$, I recast $h_{0}$ as a nonparametric
instrumental variable (NPIV) regression. The confounding bridge is defined as
the solution to the integral equation

$\gamma_{0}(d,x,z)=\int
h_{0}(d,x,w)\mathrm{d}\mathbb{P}(w|d,x,z),\quad\gamma_{0}(d,x,z):=\mathbb{E}[Y|D=d,X=x,Z=z].$

For expositional purposes, define the residual $\epsilon:=Y-\gamma_{0}(D,X,Z)$
so that

$Y=\int
h_{0}(D,X,w)\mathrm{d}\mathbb{P}(w|D,X,Z)+\epsilon,\quad\mathbb{E}[\epsilon|D,X,Z]=0.$

Furthermore,

$Y=h_{0}(D,X,W)+\epsilon,\quad\mathbb{E}[\epsilon|D,X,Z]=0.$

This representation is precisely the NPIV problem where $W$ is endogenous and
$Z$ is the instrument.

In this light, I propose a two stage estimator for $\hat{h}$ that generalizes
two stage least squares (2SLS) and appeals to the RKHS construction. In 2SLS,
an analyst would estimate $\hat{h}^{2SLS}$ by (i) projecting $W$ onto
$(D,X,Z)$, obtaining an estimator $\hat{W}(D,X,Z)$; then (ii) projecting $Y$
onto $(D,X,\hat{W}(D,X,Z))$. I propose something similar. Define the mean
embedding

$\mu_{w}(d,x,z):=\int\phi(w)\mathrm{d}\mathbb{P}(w|d,x,z)$

which encodes the distribution $\mathbb{P}(w|d,x,z)$ in the integral equation
we wish to solve. I propose estimating $\hat{h}$ by (i) projecting $\phi(W)$
onto $\phi(D)\otimes\phi(X)\otimes\phi(W)$, obtaining an estimator
$\hat{\mu}_{w}(D,X,Z)$; then (ii) projecting $Y$ onto
$\phi(D)\otimes\phi(X)\otimes\hat{\mu}_{w}(D,X,Z)$.

######  Algorithm 5 (End-to-end: Principles).

Given $n$ observations of outcome $Y$, treatment $D$, covariates $X$, negative
control outcome $W$, and negative control treatment $Z$,

  1. 1.

Specify the kernels
$k_{\mathcal{D}},k_{\mathcal{X}},k_{\mathcal{W}},k_{\mathcal{Z}}$.

  2. 2.

Specify the regularization hyperparameters $(\lambda,\xi)$.

  3. 3.

Estimate the confounding bridge $\hat{h}$ in two stages, using
$(\lambda,\xi)$.

    1. (a)

Estimate the distribution in the integral equation $\hat{\mathbb{P}}(w|d,x,z)$
via its mean embedding $\hat{\mu}_{w}(d,x,z)$ with regularization $\lambda$.

    2. (b)

Regress $Y$ onto $\phi(D)\otimes\phi(X)\otimes\hat{\mu}_{w}(D,X,Z)$ with
regularization $\xi$.

  4. 4.

Estimate the counterfactual distribution $\hat{\mathbb{P}}(x,w)$ via its mean
embedding $\hat{\mu}$.

  5. 5.

Estimate the dose response $\hat{\theta}^{ATE}(d)$ by combining $\hat{h}$ and
$\hat{\mu}$ according to
$\hat{\theta}^{ATE}(d)=\langle\hat{h},\phi(d)\otimes\hat{\mu}\rangle_{\mathcal{H}}.$

With these principles in mind, I now fill in the details. Objects in an RKHS
are infinite dimensional, and I have reasoned about them abstractly in
Algorithm 5. To actually compute the estimator, I must express the procedure
exclusively as inner products of RKHS objects, i.e. as scalar evaluations of
kernels. I provide such details in Algorithm 3 in the main text.

###  E.2 Representation

###### Proof of Theorem 2.

In Assumption 4, I impose that the scalar kernels are bounded. This assumption
has several implications. First, the feature maps are Bochner integrable
[Steinwart and Christmann, 2008, Definition A.5.20]. Bochner integrability
permits the exchange of expectation and inner product. Second, the mean
embeddings exist. Third, the product kernel is also bounded and hence the
tensor product RKHS inherits these favorable properties. Since
$h_{0}\in\mathcal{H}$,

$\displaystyle\gamma_{0}(d,x,z)$ $\displaystyle=\int
h_{0}(d,x,w)\mathrm{d}\mathbb{P}(w|d,x,z)$ $\displaystyle=\int\langle
h_{0},\phi(d)\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}\mathrm{d}\mathbb{P}(w|d,x,z)$
$\displaystyle=\left\langle
h_{0},\phi(d)\otimes\phi(x)\otimes\int\phi(w)\mathrm{d}\mathbb{P}(w|d,x,z)\right\rangle_{\mathcal{H}}$
$\displaystyle=\langle
h_{0},\phi(d)\otimes\phi(x)\otimes\mu_{w}(d,x,z)\rangle_{\mathcal{H}}.$

Next, I generalize [Singh et al., 2020, Theorem 3.2], replacing the prediction
function with the confounding bridge $h_{0}$. By Theorem 1 and linearity of
expectation,

$\displaystyle\theta_{0}^{ATE}(d)$ $\displaystyle=\int
h_{0}(d,x,w)\mathrm{d}\mathbb{P}(x,w)$ $\displaystyle=\int\langle
h_{0},\phi(d)\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}\mathrm{d}\mathbb{P}(x,w)$
$\displaystyle=\left\langle
h_{0},\phi(d)\otimes\int[\phi(x)\otimes\phi(w)]\mathrm{d}\mathbb{P}(x,w)\right\rangle_{\mathcal{H}}$
$\displaystyle=\langle h_{0},\phi(d)\otimes\mu\rangle_{\mathcal{H}};$
$\displaystyle\theta_{0}^{DS}(d,\tilde{\mathbb{P}})$ $\displaystyle=\int
h_{0}(d,x,w)\mathrm{d}\tilde{\mathbb{P}}(x,w)$ $\displaystyle=\int\langle
h_{0},\phi(d)\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}\mathrm{d}\tilde{\mathbb{P}}(x,w)$
$\displaystyle=\left\langle
h_{0},\phi(d)\otimes\int[\phi(x)\otimes\phi(w)]\mathrm{d}\tilde{\mathbb{P}}(x,w)\right\rangle_{\mathcal{H}}$
$\displaystyle=\langle h_{0},\phi(d)\otimes\nu\rangle_{\mathcal{H}};$
$\displaystyle\theta_{0}^{ATT}(d,d^{\prime})$ $\displaystyle=\int
h_{0}(d^{\prime},x,w)\mathrm{d}\mathbb{P}(x,w|d)$ $\displaystyle=\int\langle
h_{0},\phi(d^{\prime})\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}\mathrm{d}\mathbb{P}(x,w|d)$
$\displaystyle=\left\langle
h_{0},\phi(d^{\prime})\otimes\int[\phi(x)\otimes\phi(w)]\mathrm{d}\mathbb{P}(x,w|d)\right\rangle_{\mathcal{H}}$
$\displaystyle=\langle
h_{0},\phi(d^{\prime})\otimes\mu(d)\rangle_{\mathcal{H}};$
$\displaystyle\theta_{0}^{CATE}(d,v)$ $\displaystyle=\int
h_{0}(d,v,x,w)\mathrm{d}\mathbb{P}(x,w|v)$ $\displaystyle=\int\langle
h_{0},\phi(d)\otimes\phi(v)\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}\mathrm{d}\mathbb{P}(x,w|v)$
$\displaystyle=\left\langle
h_{0},\phi(d)\otimes\phi(v)\otimes\int[\phi(x)\otimes\phi(w)]\mathrm{d}\mathbb{P}(x,w|v)\right\rangle_{\mathcal{H}}$
$\displaystyle=\langle
h_{0},\phi(d)\otimes\phi(v)\otimes\mu(v)\rangle_{\mathcal{H}}.$

∎

###  E.3 Confounding bridge

Let $n$ be the number of observations of $(d_{i},x_{i},w_{i},z_{i})$ used to
estimate the conditional mean embedding $\mu_{w}(d,x,z)$ by kernel ridge
regression with regularization parameter $\lambda$. Let $m$ be the number of
observations of $(\dot{y}_{i},\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})$ used to
estimate the confounding bridge $h_{0}$ by kernel ridge regression with
regularization parameter $\xi$. This notation allows the analyst to use
different quantities of observations $(n,m)$ to estimate $(E_{0},h_{0})$, or
to reuse the same observations.

###### Derivation of Algorithm 1.

I proceed in steps, generalizing the derivation of [Singh et al., 2019,
Algorithm 1].

  1. 1.

Closed form for stage 1.

By [Singh et al., 2019, Algorithm 1], the closed form solution for the stage 1
conditional mean embedding is

$\hat{\mu}_{w}(d,x,z)=\sum_{i=1}^{n}\beta_{i}(d,x,z)\phi(w_{i})$

where

$\beta(d,x,z)=(K_{DD}\odot K_{XX}\odot K_{ZZ}+n\lambda I)^{-1}[K_{Dd}\odot
K_{Xx}\odot K_{Zz}]\in\mathbb{R}^{n}.$

Slightly abusing notation, one may write

$\hat{\mu}_{w}(d,x,z)=K_{\cdot W}\beta(d,x,z).$

  2. 2.

Closed form for stage 2.

Next, I argue that
$\hat{h}=\sum_{i=1}^{m}\alpha_{i}[\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})]$
for some $\alpha\in\mathbb{R}^{m}$. Write the objective as

$\hat{h}=\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\mathcal{E}_{\xi}^{m}(h),\quad\mathcal{E}_{\xi}^{m}(h)=\frac{1}{m}\sum_{i=1}^{m}\|\dot{y}_{i}-\langle
h,\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}\|^{2}_{\mathcal{Y}}+\xi\|h\|^{2}_{\mathcal{H}}.$

Due to the ridge penalty, the stated objective is coercive and strongly convex
with respect to $h$. Hence it has a unique minimizer that obtains the minimum.

Write $\hat{h}=\hat{h}_{m}+\hat{h}_{m}^{\perp}$ where $\hat{h}_{m}\in
span\\{\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\\}$
and $\hat{h}_{m}^{\perp}$ is an element of the orthogonal complement.
Therefore

$\mathcal{E}_{\xi}^{m}(\hat{h})=\mathcal{E}_{\xi}^{m}(\hat{h}_{m})+\xi\|\hat{h}_{m}^{\perp}\|^{2}_{\mathcal{H}}$

which implies
$\mathcal{E}_{\xi}^{m}(\hat{h})\geq\mathcal{E}_{\xi}^{m}(\hat{h}_{m})$. Since
$\hat{h}$ is the unique minimizer, $\hat{h}=\hat{h}_{m}$.

  3. 3.

Substitution.

I substitute the functional form of $\hat{h}$ into its objective. Note that

$\displaystyle\|\hat{h}\|^{2}_{\mathcal{H}}$
$\displaystyle=\langle\hat{h},\hat{h}\rangle_{\mathcal{H}}$
$\displaystyle=\left\langle\sum_{i=1}^{m}\alpha_{i}[\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})],\sum_{j=1}^{m}\alpha_{j}[\phi(\dot{d}_{j})\otimes\phi(\dot{x}_{j})\otimes\hat{\mu}_{w}(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})]\right\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i}\alpha_{j}k(\dot{d}_{i},\dot{d}_{j})k(\dot{x}_{i},\dot{x}_{j})\beta(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})^{\top}K_{WW}\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})$
$\displaystyle=\alpha^{\top}M\alpha$

where in the last line I define the matrix

$M=K_{\dot{D}\dot{D}}\odot K_{\dot{X}\dot{X}}\odot\\{\dot{A}^{\top}(A+n\lambda
I)^{-1}K_{WW}(A+n\lambda I)^{-1}\dot{A}\\}\in\mathbb{R}^{m\times m}$

with

$A=K_{DD}\odot K_{XX}\odot K_{ZZ}\in\mathbb{R}^{n\times
n},\quad\dot{A}=K_{D\dot{D}}\odot K_{X\dot{X}}\odot
K_{Z\dot{Z}}\in\mathbb{R}^{n\times m}.$

Further define

$B=(A+n\lambda I)^{-1}\dot{A}\in\mathbb{R}^{n\times m}$

whose $j$-th column is $\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j}).$ In terms
of $B$,

$M=K_{\dot{D}\dot{D}}\odot K_{\dot{X}\dot{X}}\odot\\{B^{\top}K_{WW}B\\}.$

Next note that

$\displaystyle\langle\hat{h},\phi(d)\otimes\phi(x)\otimes\hat{\mu}_{w}(d,x,z)\rangle_{\mathcal{H}}$
$\displaystyle=\left\langle\sum_{j=1}^{m}\alpha_{j}[\phi(\dot{d}_{j})\otimes\phi(\dot{x}_{j})\otimes\hat{\mu}_{w}(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})],\phi(d)\otimes\phi(x)\otimes\hat{\mu}_{w}(d,x,z)\right\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{j=1}^{m}\alpha_{j}\left\langle\phi(\dot{d}_{j})\otimes\phi(\dot{x}_{j})\otimes\hat{\mu}_{w}(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j}),\phi(d)\otimes\phi(x)\otimes\hat{\mu}_{w}(d,x,z)\right\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{j=1}^{m}\alpha_{j}k(\dot{d}_{j},d)k(\dot{x}_{j},x)\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})^{\top}K_{WW}\beta(d,x,z).$

Hence

$\displaystyle\langle\hat{h},\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{j=1}^{n}\alpha_{j}k(\dot{d}_{j},\dot{d}_{i})k(\dot{x}_{j},\dot{x}_{i})\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})^{\top}K_{WW}\beta(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})$
$\displaystyle=\sum_{j=1}^{m}\alpha_{j}k(\dot{d}_{i},\dot{d}_{j})k(\dot{x}_{i},\dot{x}_{j})\beta(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})^{\top}K_{WW}\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j}).$

In terms of $M$,

$\displaystyle[M^{\top}\alpha]_{i}$
$\displaystyle=\left[\left(K_{\dot{D}\dot{D}}\odot
K_{\dot{X}\dot{X}}\odot\\{B^{\top}K_{WW}B\\}\right)\alpha\right]_{i}$
$\displaystyle=\left[\sum_{j=1}^{m}\alpha_{j}\left(K_{\dot{D}\dot{d}_{j}}\odot
K_{\dot{X}\dot{x}_{j}}\odot
B^{\top}K_{WW}\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})\right)\right]_{i}$
$\displaystyle=\sum_{j=1}^{m}\alpha_{j}k(\dot{d}_{i},\dot{d}_{j})k(\dot{x}_{i},\dot{x}_{j})\beta(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})^{\top}K_{WW}\beta(\dot{d}_{j},\dot{x}_{j},\dot{z}_{j})$
$\displaystyle=\langle\hat{h},\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}.$

In summary,

$\mathcal{E}_{\xi}^{m}(\hat{h})=\frac{1}{m}\|\dot{Y}-M^{\top}\alpha\|_{2}^{2}+\xi\alpha^{\top}M\alpha.$

  4. 4.

Optimization.

Express the objective, scaled by $m$, as

$\displaystyle m\mathcal{E}_{\xi}^{m}(\hat{h})$
$\displaystyle=\|\dot{Y}-M^{\top}\alpha\|_{2}^{2}+m\xi\cdot\alpha^{\top}M\alpha$
$\displaystyle=\dot{Y}^{\top}\dot{Y}-2\alpha^{\top}M\dot{Y}+\alpha^{\top}MM^{\top}\alpha+m\xi\cdot\alpha^{\top}M\alpha$
$\displaystyle=\dot{Y}^{\top}\dot{Y}-2\alpha^{\top}M\dot{Y}+\alpha^{\top}\left(MM^{\top}+m\xi
M\right)\alpha$

Solving the first order condition with respect to $\alpha$,

$\hat{\alpha}=(MM^{\top}+m\xi M)^{-1}M\dot{Y}.$

Therefore

$\displaystyle\hat{h}(d,x,w)$
$\displaystyle=\langle\hat{h},\phi(d)\otimes\phi(x)\otimes\phi(w)\rangle_{\mathcal{H}}$
$\displaystyle=\left\langle\sum_{i=1}^{m}\hat{\alpha}_{i}[\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})],\phi(d)\otimes\phi(x)\otimes\phi(w)\right\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{i=1}^{m}\hat{\alpha}_{i}\left\langle\phi(\dot{d}_{i})\otimes\phi(\dot{x}_{i})\otimes\hat{\mu}_{w}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i}),\phi(d)\otimes\phi(x)\otimes\phi(w)\right\rangle_{\mathcal{H}}$
$\displaystyle=\sum_{i=1}^{m}\hat{\alpha}_{i}k(\dot{d}_{i},d)k(\dot{x}_{i},x)\beta(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})^{\top}K_{Ww}$
$\displaystyle=\hat{\alpha}^{\top}[K_{\dot{D}d}\odot
K_{\dot{X}x}\odot\\{B^{\top}K_{Ww}\\}].$

∎

###  E.4 Treatment effects

###### Derivation of Algorithm 2.

By Algorithm 1,

$\langle\hat{h},\phi(d)\otimes\phi(\cdot)\otimes\phi(\cdot)\rangle_{\mathcal{H}}=\hat{\alpha}^{\top}[K_{\dot{D}d}\odot
K_{\dot{X}\cdot}\odot\\{B^{\top}K_{W\cdot}\\}]$

where I leave the arguments $(x,w)$ blank in order to showcase how the
confounding bridge estimator will be combined with the kernel mean embedding
estimators. By Theorem 2, it is sufficient to obtain expressions for kernel
mean embedding estimators and to substitute them in. In particular,

  1. 1.

$\theta_{0}^{ATE}$:
$\hat{\mu}=n^{-1}\sum_{i=1}^{n}[\phi(x_{i})\otimes\phi(w_{i})]=n^{-1}\sum_{i=1}^{n}k(\cdot,x_{i})k(\cdot,w_{i});$

  2. 2.

$\theta_{0}^{DS}$:
$\hat{\nu}=\tilde{n}^{-1}\sum_{i=1}^{\tilde{n}}[\phi(\tilde{x}_{i})\otimes\phi(\tilde{w}_{i})]=\tilde{n}^{-1}\sum_{i=1}^{\tilde{n}}k(\cdot,\tilde{x}_{i})k(\cdot,\tilde{w}_{i});$

  3. 3.

$\theta_{0}^{ATT}$: $\hat{\mu}(d)=[K_{\cdot X}\odot K_{\cdot
W}](K_{DD}+n\lambda_{1}I)^{-1}K_{Dd};$

  4. 4.

$\theta_{0}^{CATE}$: $\hat{\mu}(v)=[K_{\cdot X}\odot K_{\cdot
W}](K_{VV}+n\lambda_{2}I)^{-1}K_{Vv}.$

I use the $n$ observations of $(d_{i},x_{i},w_{i},z_{i})$ to estimate the
kernel mean embeddings, and I do not use the $m$ observations of
$(\dot{y}_{i},\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})$, because the former
contain the negative control outcome while the latter do not. The conditional
mean embedding expressions follow from [Singh et al., 2019, Algorithm 1].
Matching the blank arguments yields the desired result. ∎

##  Appendix F Tuning

###  F.1 Simplified setting

The dose response and heterogeneous treatment effect estimators I propose are
composed of kernel ridge regressions. The hyperparameters of the new
estimators are therefore the same hyperparameters of kernel ridge regression.
I present practical tuning procedures for the hyperparameters of (i) ridge
regression penalties and (ii) the kernel itself. This appendix is an
elaboration of [Singh et al., 2020, Appendix F].

###  F.2 Ridge penalty

To begin, I quote a tuning procedure for kernel ridge regression. For
simplicity, I focus on the regression of $Y$ on $A$. It is convenient to tune
$\lambda$ by leave-one-out cross validation (LOOCV), since the validation loss
has a closed form solution.

######  Algorithm 6 (Tuning kernel ridge regression; Algorithm F.1 of [Singh
et al., 2020]).

Construct the matrices

$H_{\lambda}:=I-K_{AA}(K_{AA}+n\lambda I)^{-1}\in\mathbb{R}^{n\times
n},\quad\tilde{H}_{\lambda}:=diag(H_{\lambda})\in\mathbb{R}^{n\times n}$

where $\tilde{H}_{\lambda}$ has the same diagonal entries as $H_{\lambda}$ and
off diagonal entries of 0. Then set

$\lambda^{*}=\operatorname*{\arg\\!\min}_{\lambda\in\Lambda}\frac{1}{n}\|\tilde{H}_{\lambda}^{-1}H_{\lambda}Y\|_{2}^{2},\quad\Lambda\subset\mathbb{R}.$

The same principles govern the tuning of conditional mean embeddings, which
can be viewed as vector valued regressions. Moreover, the same principles
govern the tuning of kernel instrumental variable regression, which consists
of two kernel ridge regressions. The extension of these tuning procedures to
kernel instrumental variable regression is an innovation, and it differs from
the tuning procedure in [Singh et al., 2019, Appendix A.5.2]. To simplify the
discussion, I focus on the conditional mean embedding
$\mu(a)=\int\phi(y)\mathrm{d}\mathbb{P}(y|a)$. Recall that the closed form
solution of the conditional mean embedding estimator using all observations is

$\hat{\mu}(a)=K_{\cdot Y}(K_{AA}+n\lambda I)^{-1}K_{Aa}.$

######  Algorithm 7 (Tuning conditional mean embedding).

Construct the matrices

$\displaystyle R:=K_{AA}(K_{AA}+n\lambda I)^{-1}\in\mathbb{R}^{n\times n}$
$\displaystyle S\in\mathbb{R}^{n\times n}\text{ s.t.
}S_{ij}=1\\{i=j\\}\left\\{\frac{1}{1-R_{ii}}\right\\}^{2}$ $\displaystyle
T:=S(K_{YY}-2K_{YY}R^{\top}+RK_{YY}R^{\top})\in\mathbb{R}^{n\times n}$

where $R_{ii}$ is the $i$-th diagonal element of $R$. Then set

$\lambda^{*}=\operatorname*{\arg\\!\min}_{\lambda\in\Lambda}\frac{1}{n}tr(T),\quad\Lambda\subset\mathbb{R}.$

###### Derivation.

I prove that $n^{-1}tr(T)$ is the LOOCV loss. By definition, the LOOCV loss is

$\mathcal{E}(\lambda):=\frac{1}{n}\sum_{i=1}^{n}\|\phi(y_{i})-\hat{\mu}_{-i}(a_{i})\|_{\mathcal{H}_{\mathcal{Y}}}^{2}$

where $\hat{\mu}_{-i}$ is the conditional mean embedding estimator using all
observations except the $i$-th observation.

Informally, let $\Phi$ be the matrix of features for $\\{a_{i}\\}$, with
$i$-th row $\phi(a_{i})^{\top}$, and let $Q:=\Phi^{\top}\Phi+n\lambda$. Let
$\Psi$ be the matrix of features for $\\{y_{i}\\}$, with $i$-th row
$\phi(y_{i})^{\top}$. By the regression first order condition

$\displaystyle\hat{\mu}(a)^{\top}$
$\displaystyle=\phi(a)^{\top}Q^{-1}\Phi^{\top}\Psi$
$\displaystyle\hat{\mu}_{-i}(a)^{\top}$
$\displaystyle=\phi(a)^{\top}\\{Q-\phi(a_{i})\phi(a_{i})^{\top}\\}^{-1}\\{\Phi^{\top}\Psi-\phi(a_{i})\phi(y_{i})^{\top}\\}.$

The Sherman-Morrison formula for rank one updates gives

$\\{Q-\phi(a_{i})\phi(a_{i})^{\top}\\}^{-1}=Q^{-1}+\frac{Q^{-1}\phi(a_{i})\phi(a_{i})^{\top}Q^{-1}}{1-\phi(a_{i})^{\top}Q^{-1}\phi(a_{i})}.$

Let $\beta_{i}:=\phi(a_{i})^{\top}Q^{-1}\phi(a_{i})$. Then

$\displaystyle\hat{\mu}_{-i}(a)^{\top}$
$\displaystyle=\phi(a_{i})^{\top}\left\\{Q^{-1}+\frac{Q^{-1}\phi(a_{i})\phi(a_{i})^{\top}Q^{-1}}{1-\beta_{i}}\right\\}\\{\Phi^{\top}\Psi-\phi(a_{i})\phi(y_{i})^{\top}\\}$
$\displaystyle=\phi(a_{i})^{\top}\left\\{I+\frac{Q^{-1}\phi(a_{i})\phi(a_{i})^{\top}}{1-\beta_{i}}\right\\}\\{\hat{\mu}^{\top}-Q^{-1}\phi(a_{i})\phi(y_{i})^{\top}\\}$
$\displaystyle=\left\\{1+\frac{\beta_{i}}{1-\beta_{i}}\right\\}\phi(a_{i})^{\top}\\{\hat{\mu}^{\top}-Q^{-1}\phi(a_{i})\phi(y_{i})^{\top}\\}$
$\displaystyle=\left\\{1+\frac{\beta_{i}}{1-\beta_{i}}\right\\}\\{\hat{\mu}(a_{i})^{\top}-\beta_{i}\phi(y_{i})^{\top}\\}$
$\displaystyle=\frac{1}{1-\beta_{i}}\\{\hat{\mu}(a_{i})^{\top}-\beta_{i}\phi(y_{i})^{\top}\\}$

i.e. $\hat{\mu}_{-i}$ can be expressed in terms of $\hat{\mu}$. Note that

$\displaystyle\phi(y_{i})-\hat{\mu}_{-i}(a_{i})$
$\displaystyle=\phi(y_{i})-\frac{1}{1-\beta_{i}}\\{\hat{\mu}(a_{i})-\beta_{i}\phi(y_{i})\\}$
$\displaystyle=\phi(y_{i})+\frac{1}{1-\beta_{i}}\\{\beta_{i}\phi(y_{i})-\hat{\mu}(a_{i})\\}$
$\displaystyle=\frac{1}{1-\beta_{i}}\\{\phi(y_{i})-\hat{\mu}(a_{i})\\}.$

Substituting back into the LOOCV loss

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}\left\|\phi(y)_{i}-\hat{\mu}_{-i}(a_{i})\right\|_{\mathcal{H}_{\mathcal{Y}}}^{2}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\left\|\\{\phi(y_{i})-\hat{\mu}(a_{i})\\}\left\\{\frac{1}{1-\beta_{i}}\right\\}\right\|_{\mathcal{H}_{\mathcal{Y}}}^{2}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\left\\{\frac{1}{1-\beta_{i}}\right\\}^{2}\left\|\phi(y_{i})-\hat{\mu}(a_{i})\right\|_{\mathcal{H}_{\mathcal{Y}}}^{2}.$

By arguments in [Singh et al., 2020, Appendix F],

$\beta_{i}=[K_{AA}(K_{AA}+n\lambda I)^{-1}]_{ii}$

i.e. $\beta_{i}$ can be calculated as the $i$-th diagonal element of
$K_{AA}(K_{AA}+n\lambda I)^{-1}$. Moreover

$\displaystyle\|\phi(y_{i})-\hat{\mu}(a_{i})\|_{\mathcal{H}}^{2}$
$\displaystyle=k(y_{i},y_{i})-2K_{y_{i}Y}(K_{AA}+n\lambda
I)^{-1}K_{Aa_{i}}+K_{a_{i}A}(K_{AA}+n\lambda I)^{-1}K_{YY}(K_{AA}+n\lambda
I)^{-1}K_{Aa_{i}}$ $\displaystyle=[K_{YY}-2K_{YY}(K_{AA}+n\lambda
I)^{-1}K_{AA}+K_{AA}(K_{AA}+n\lambda I)^{-1}K_{YY}(K_{AA}+n\lambda
I)^{-1}K_{AA}]_{ii}$

i.e. $\|\phi(y_{i})-\hat{\mu}(a_{i})\|_{\mathcal{H}}^{2}$ can be calculated as
the $i$-th diagonal element of a matrix as well. Substituting these results
back into the LOOCV loss gives the final expression. ∎

###  F.3 Kernel

The Gaussian kernel satisfies the requirements of Assumption 4. Formally, the
kernel

$k(a,a^{\prime})=\exp\left\\{-\frac{1}{2}\frac{\|a-a^{\prime}\|^{2}_{\mathcal{A}}}{\sigma^{2}}\right\\}$

is continuous, bounded, and characteristic. The kernel hyperparameter $\sigma$
is called the lengthscale. A simple heuristic is to set $\sigma$ as the median
interpoint distance of $\\{a_{i}\\}^{n}_{i=1}$, where the interpoint distance
between observations $i$ and $j$ is $\|a_{i}-a_{j}\|_{\mathcal{A}}$.

I use the Gaussian kernel in experiments. When the input $a$ is a vector
rather than a scalar, I use the kernel obtained as the product of scalar
kernels for each input dimension, following [Singh et al., 2019, Singh et al.,
2020]. For example, if $\mathcal{A}\subset\mathbb{R}^{d}$ then

$k(a,a^{\prime})=\prod_{j=1}^{d}\exp\left\\{-\frac{1}{2}\frac{[a_{j}-a_{j}^{\prime}]^{2}}{\sigma_{j}^{2}}\right\\}.$

Each lengthscale $\sigma_{j}$ is set as the median interpoint distance for
that input dimension. In principle, one could instead use LOOCV to tune kernel
hyperparameters as above. The LOOCV approach to tuning lengthscales
$\\{\sigma_{j}\\}$ is impractical in high dimensions, since there is a
lengthscale $\sigma_{j}$ for each input dimension.

###  F.4 Time complexity

As in classic kernel ridge regression, the time consuming step is tuning. We
see in Algorithms 6 and 7 that to choose the ridge penalty hyperparameter
$\lambda^{*}$, one must invert the matrix

$K_{AA}+n\lambda I\in\mathbb{R}^{n\times n}$

for each value $\lambda$ in the grid $\Lambda$. Inversion of such a matrix has
complexity $O(n^{3})$; the sample size $n$ is the limiting factor. The same is
true for the two ridge penalty hyperparameters in the confounding bridge
estimated by Algorithm 1, and for the additional ridge penalty hyperparameter
that appears when estimating the heterogeneous treatment effect in Algorithm
2. Therefore tuning of the heterogeneous treatment effect in Algorithm 2 takes
roughly three times as long as tuning of a kernel ridge regression, whose
runtime scales as $O(n^{3})$.

A vast literature considers how to speed up kernel methods by replacing the
kernel matrix with a low rank approximation. Appendix E.3 of [Dikkala et al.,
2020] discusses popular techniques and their implementation in NPIV. I pose as
a question for future work how to extend the main results of this paper to
accommodate kernel matrix approximations.

##  Appendix G Confounding bridge consistency proof

In this appendix, I (i) state a probability lemma, (ii) explicitly specialize
the smoothness assumptions, (iii) provide regression lemmas, (iv) prove
technical bounds, and (v) prove uniform consistency of the confounding bridge.
This is the most technically demanding appendix.

###  G.1 Probability lemma

######  Lemma 4 (Lemma 2 of [Smale and Zhou, 2007]).

Let $\xi$ be a random variable taking values in a real separable Hilbert space
$\mathcal{K}$. Suppose there exists $\tilde{M}$ and $\sigma^{2}$ such that

$\displaystyle\|\xi\|_{\mathcal{K}}$
$\displaystyle\leq\tilde{M}<\infty\quad\text{ almost
surely},\quad\mathbb{E}\|\xi\|_{\mathcal{K}}^{2}\leq\sigma^{2}.$

Then $\forall n\in\mathbb{N},\forall\eta\in(0,1)$,

$\mathbb{P}\bigg{[}\bigg{\|}\dfrac{1}{n}\sum_{i=1}^{n}\xi_{i}-\mathbb{E}\xi\bigg{\|}_{\mathcal{K}}\leq\dfrac{2\tilde{M}\ln(2/\eta)}{n}+\sqrt{\dfrac{2\sigma^{2}\ln(2/\eta)}{n}}\bigg{]}\geq
1-\eta.$

###  G.2 Smoothness assumptions

Let the symbol $\circ$ mean composition. I use it to emphasize the composition
of operators. To lighten notation, let
$\mathcal{H}_{RF}=\mathcal{H}_{\mathcal{D}}\otimes\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{Z}}$,
which is named for the reduced form.

######  Assumption 8 (Smoothness of conditional expectation).

Assume

  1. 1.

The conditional expectation operator $E_{0}$ is well specified as a Hilbert-
Schmidt operator between RKHSs, i.e.
$E_{0}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{W}},\mathcal{H}_{RF})$, where

$E_{0}:\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{RF},\quad
h(\cdot)\mapsto\mathbb{E}[h(W)|D=\cdot,X=\cdot,Z=\cdot].$

  2. 2.

The conditional expectation operator is a particularly smooth element of
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{W}},\mathcal{H}_{RF})$. Formally,
define the covariance operator
$T_{0}:=\mathbb{E}[\phi(D,X,Z)\otimes\phi(D,X,Z)]$ for
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{W}},\mathcal{H}_{RF})$. I assume there
exists $G_{0}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{W}},\mathcal{H}_{RF})$
such that $E_{0}=(T_{0})^{\frac{c_{0}-1}{2}}\circ G_{0}$, $c_{0}\in(1,2]$, and
$\|G_{0}\|^{2}_{\mathcal{L}_{2}(\mathcal{H}_{\mathcal{W}},\mathcal{H}_{RF})}\leq\zeta_{0}$.

######  Assumption 9 (Smoothness of confounding bridge).

Assume

  1. 1.

The confounding bridge $h_{0}$ is well specified, i.e.
$h_{0}\in\mathcal{H}_{\mu}\subset\mathcal{H}$.

  2. 2.

The confounding bridge is a particularly smooth element of
$\mathcal{H}_{\mu}$. Formally, define the covariance operator
$T:=\mathbb{E}[\mu(D,X,Z)\otimes\mu(D,X,Z)]$, where
$\mu(d,x,z)=\phi(d)\otimes\phi(x)\otimes\mu_{w}(d,x,z)$, for
$\mathcal{H}_{\mu}$. I assume there exists $g\in\mathcal{H}$ such that
$h_{0}=T^{\frac{c-1}{2}}\circ g$, $c\in(1,2]$, and
$\|g\|^{2}_{\mathcal{H}}\leq\zeta$.

######  Proposition 8.

The following assumptions are equivalent

  1. 1.

Assumption 6 with $\mathcal{A}_{0}=\mathcal{W}$ and
$\mathcal{B}_{0}=\mathcal{D}\times\mathcal{X}\times\mathcal{Z}$ is equivalent
to Assumption 8

  2. 2.

Assumption 7 is equivalent to Assumption 9

###### Proof.

The result is immediate from [Caponnetto and De Vito, 2007, Remark 2]. The
expanded expressions are more convenient for analysis. ∎

###  G.3 Regression lemmas

Let $n$ be the number of observations of $(d_{i},x_{i},w_{i},z_{i})$ used to
estimate the stage 1 conditional mean embedding
$\mu(d,x,z)=\phi(d)\otimes\phi(x)\otimes\mu_{w}(d,x,z)$ by kernel ridge
regression with regularization parameter $\lambda$. Let $m$ be the number of
observations of $(\dot{y}_{i},\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})$ used to
estimate the stage 2 confounding bridge operator $h_{0}$ by kernel ridge
regression with regularization parameter $\xi$.

######  Proposition 9.

Suppose Assumptions 2 and 4 hold, and $h_{0}\in\mathcal{H}$. Then

$\mathbb{E}[\mu(D,X,Z)Y]=Th_{0}.$

###### Proof.

Appealing to the definition of $T$, the argument in the proof of Theorem 2,
and the law of iterated expectations,

$\displaystyle Th_{0}$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)\otimes\mu(D,X,Z)]h_{0}$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)\langle\mu(D,X,Z),h_{0}\rangle_{\mathcal{H}}]$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)\gamma_{0}(D,X,Z)]$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)Y].$

∎

To facilitate analysis, define the following quantities.

######  Definition 3 (Confounding bridge risk).

Define

  1. 1.

Target bridge

$h_{0}\in\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\mathcal{E}(h),\quad\mathcal{E}(h)=\mathbb{E}[\\{Y-\langle
h,\mu(D,X,Z)\rangle_{\mathcal{H}}\\}^{2}].$

  2. 2.

Regularized bridge

$h_{\xi}=\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\mathcal{E}_{\xi}(h),\quad\mathcal{E}_{\xi}(h)=\mathcal{E}(h)+\xi\|h\|^{2}_{\mathcal{H}}.$

  3. 3.

Empirical regularized bridge

$h^{m}_{\xi}=\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\mathcal{E}^{m}_{\xi}(h),\quad\mathcal{E}^{m}_{\xi}(h)=\dfrac{1}{m}\sum_{i=1}^{m}\\{\dot{y}_{i}-\langle
h,\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}\\}^{2}+\xi\|h\|^{2}_{\mathcal{H}}.$

  4. 4.

Estimated bridge

$\hat{h}^{m}_{\xi}=\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\hat{\mathcal{E}}^{m}_{\xi}(h),\quad\hat{\mathcal{E}}^{m}_{\xi}(h)=\dfrac{1}{m}\sum_{i=1}^{m}\\{\dot{y}_{i}-\langle
h,\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}\\}^{2}+\xi\|h\|^{2}_{\mathcal{H}}$

where $\mu^{n}_{\lambda}(d,x,z)$ is an estimator of the conditional mean
embedding $\mu(d,x,z)$.

######  Proposition 10.

Suppose the conditions of Proposition 9 hold. Then

$\mathbb{E}[\langle
h-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]=\mathcal{E}(h)-\mathcal{E}(h_{0}).$

###### Proof.

Write

$\displaystyle\mathcal{E}(h)$ $\displaystyle=\mathbb{E}[\\{Y-\langle
h,\mu(D,X,Z)\rangle_{\mathcal{H}}\\}^{2}]$
$\displaystyle=\mathbb{E}[\\{\left(Y-\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}\right)+\left(\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}-\langle
h,\mu(D,X,Z)\rangle_{\mathcal{H}}\right)\\}^{2}].$

Squaring the initial terms gives $\mathcal{E}(h_{0})$. Squaring the latter
terms gives $\mathbb{E}[\langle h-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]$.
To complete the argument, I show that the cross term is zero. Write

$\displaystyle-2\mathbb{E}[\left(Y-\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}\right)\cdot\left(\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}-\langle
h,\mu(D,X,Z)\rangle_{\mathcal{H}}\right)]$
$\displaystyle=-2\mathbb{E}[\left(Y-\gamma_{0}(D,X,Z)\right)\cdot\left(\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}-\langle
h,\mu(D,X,Z)\rangle_{\mathcal{H}}\right)]$ $\displaystyle=0$

where the first equality appeals to the the argument in the proof of Theorem
2, and the second appeals to the law of iterated expectations. ∎

######  Proposition 11.

Suppose the conditions of Proposition 9 hold. Then

$h_{\xi}=\operatorname*{\arg\\!\min}_{h\in\mathcal{H}}\mathbb{E}[\langle
h-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]+\xi\|h\|^{2}_{\mathcal{H}}.$

###### Proof.

The result is a corollary of Proposition 10. ∎

######  Proposition 12 (Closed form).

$\forall\xi>0$, the solution $h^{m}_{\xi}$ to $\mathcal{E}_{\xi}^{m}$ and the
solution $\hat{h}^{m}_{\xi}$ to $\hat{\mathcal{E}}_{\xi}^{m}$ both exist, are
unique, and

$\displaystyle h_{\xi}^{m}$
$\displaystyle=(\mathbf{T}+\xi)^{-1}\mathbf{g},\quad\mathbf{T}=\dfrac{1}{m}\sum_{i=1}^{m}\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i}),\quad\mathbf{g}=\dfrac{1}{m}\sum_{i=1}^{m}\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\dot{y}_{i},$
$\displaystyle\hat{h}_{\xi}^{m}$
$\displaystyle=(\hat{\mathbf{T}}+\xi)^{-1}\hat{\mathbf{g}},\quad\hat{\mathbf{T}}=\dfrac{1}{m}\sum_{i=1}^{m}\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i}),\quad\hat{\mathbf{g}}=\dfrac{1}{m}\sum_{i=1}^{m}\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\dot{y}_{i}.$

###### Proof.

The result is a simplification of [Singh et al., 2019, Theorem 3]. ∎

######  Proposition 13 (Sampling error).

Suppose the conditions of Proposition 9 hold. Then $\forall\delta\in(0,1)$,
the following holds with probability $1-\delta$:

$\|h^{m}_{\xi}-h_{\xi}\|_{\mathcal{H}}\leq\dfrac{4\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\delta)}{\sqrt{m}\xi},\quad\kappa:=\kappa_{d}\kappa_{x}\kappa_{w}.$

###### Proof.

I generalize [Smale and Zhou, 2007, Theorem 1]. I proceed in steps.

  1. 1.

Decomposition.

$h^{m}_{\xi}-h_{\xi}=\bigg{(}\mathbf{T}+\xi
I\bigg{)}^{-1}\bigg{(}\mathbf{g}-\mathbf{T}h_{\xi}-\xi
h_{\xi}\bigg{)}=\bigg{(}\mathbf{T}+\xi
I\bigg{)}^{-1}\bigg{(}\frac{1}{m}\sum_{i=1}^{m}\dot{\xi}_{i}-\mathbb{E}[\dot{\xi}]\bigg{)}$

where

$\displaystyle\dot{\xi}_{i}=\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\dot{y}_{i}-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})h_{\xi}=\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})[\dot{y}_{i}-\langle
h_{\xi},\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\rangle_{\mathcal{H}}].$

In particular, by the definitions in Proposition 12

$\frac{1}{m}\sum_{i=1}^{m}\dot{\xi}_{i}=\mathbf{g}-\mathbf{T}h_{\xi}.$

Meanwhile

$\displaystyle\mathbb{E}[\dot{\xi}]$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)Y]-Th_{\xi}$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)Y]-Th_{\xi}-\xi h_{\xi}+\xi h_{\xi}$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)Y]-(T+\xi I)h_{\xi}+\xi h_{\xi}$
$\displaystyle=\mathbb{E}[\mu(D,X,Z)Y]-\mathbb{E}[\mu(D,X,Z)Y]+\xi h_{\xi}$
$\displaystyle=\xi h_{\xi}$

as desired.

Next, observe that

$\displaystyle\bigg{(}\mathbf{T}+\xi
I\bigg{)}^{-1}\in\mathcal{L}(\mathcal{H}),\quad\bigg{(}\dfrac{1}{n}\sum_{i=1}^{n}\dot{\xi}_{i}-\mathbb{E}\dot{\xi}\bigg{)}\in\mathcal{H}$

where $\mathcal{L}(\mathcal{H})$ is the space of bounded operators from
$\mathcal{H}$ to $\mathcal{H}$. Therefore

$\displaystyle\|h^{m}_{\xi}-h_{\xi}\|_{\mathcal{H}}\leq\dfrac{1}{\xi}\Delta,\quad\Delta=\bigg{\|}\frac{1}{m}\sum_{i=1}^{m}\dot{\xi}_{i}-\mathbb{E}\dot{\xi}\bigg{\|}_{\mathcal{H}}.$

  2. 2.

Moments.

$\displaystyle\|\dot{\xi}_{i}\|_{\mathcal{H}}$ $\displaystyle\leq\kappa
C+\kappa^{2}\|h_{\xi}\|_{\mathcal{H}};$
$\displaystyle\mathbb{E}\|\dot{\xi}_{i}\|^{2}_{\mathcal{H}}$
$\displaystyle\leq\kappa^{2}\mathbb{E}[\\{Y-\langle
h_{\xi},\mu(D,X,Z)\rangle_{\mathcal{H}}\\}^{2}]=\kappa^{2}\mathcal{E}(h_{\xi}).$

By Proposition 11 with $h=0$

$\displaystyle\mathbb{E}[\langle
h_{\xi}-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]+\xi\|h_{\xi}\|^{2}_{\mathcal{H}}$
$\displaystyle\leq\mathbb{E}[\langle
h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]$
$\displaystyle\leq\|h_{0}\|^{2}_{\mathcal{H}}\mathbb{E}\|\mu(D,X,Z)\|_{\mathcal{H}}^{2}$
$\displaystyle\leq\kappa^{2}\|h_{0}\|_{\mathcal{H}}^{2}.$

This result has two implications:

$\displaystyle\mathbb{E}[\langle
h_{\xi}-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]$
$\displaystyle\leq\kappa^{2}\|h_{0}\|_{\mathcal{H}}^{2},\quad\|h_{\xi}\|_{\mathcal{H}}\leq\dfrac{\kappa\|h_{0}\|_{\mathcal{H}}}{\sqrt{\xi}}.$

Moreover by the definition of $h_{0}$ as the minimizer of $\mathcal{E}$,

$\mathcal{E}(h_{0})\leq\mathcal{E}(0)=\mathbb{E}[Y^{2}]\leq C^{2}$

so by Proposition 10

$\mathcal{E}(h_{\xi})=\mathcal{E}(h_{0})+\mathbb{E}[\langle
h_{\xi}-h_{0},\mu(D,X,Z)\rangle_{\mathcal{H}}^{2}]\leq
C^{2}+\kappa^{2}\|h_{0}\|_{\mathcal{H}}^{2}.$

In summary,

$\displaystyle\|\dot{\xi}_{i}\|_{\mathcal{H}}$ $\displaystyle\leq\kappa
C+\kappa^{2}\dfrac{\kappa\|h_{0}\|_{\mathcal{H}}}{\sqrt{\xi}}=\kappa(C+\kappa^{2}\|h_{0}\|_{\mathcal{H}}/\sqrt{\xi})$
$\displaystyle\mathbb{E}\|\dot{\xi}_{i}\|^{2}_{\mathcal{H}}$
$\displaystyle\leq\kappa^{2}(C^{2}+\kappa^{2}\|h_{0}\|_{\mathcal{H}}^{2}).$

  3. 3.

Concentration.

I then apply Lemma 4. With probability $1-\delta$,

$\Delta\leq\kappa(C+\kappa^{2}\|h_{0}\|_{\mathcal{H}}/\sqrt{\xi})\dfrac{2\ln(2/\delta)}{m}+\sqrt{\kappa^{2}(C^{2}+\kappa^{2}\|h_{0}\|_{\mathcal{H}}^{2})\dfrac{2\ln(2/\delta)}{m}}.$

There are two cases, which I analyze with techniques similar to [Singh et al.,
2019, Theorem 5].

    1. (a)

$\dfrac{\kappa}{\sqrt{m\xi}}\leq\dfrac{1}{4\ln(2/\delta)}$.

Because $a^{2}+b^{2}\leq(a+b)^{2}$ for $a,b\geq 0$,

$\displaystyle\Delta$ $\displaystyle<\dfrac{2\kappa
C\ln(2/\delta)}{m}+\dfrac{2\kappa^{3}\|h_{0}\|_{\mathcal{H}}\ln(2/\delta)}{m\sqrt{\xi}}+\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\sqrt{\dfrac{2\ln(2/\delta)}{m}}$
$\displaystyle=\dfrac{2\kappa
C\ln(2/\delta)}{m}+\dfrac{2\kappa^{2}\|h_{0}\|_{\mathcal{H}}\ln(2/\delta)}{\sqrt{m}}\dfrac{\kappa}{\sqrt{m\xi}}+\dfrac{\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\delta)}{\sqrt{m}}\sqrt{\dfrac{2}{\ln(2/\delta)}}$
$\displaystyle\leq\dfrac{2\kappa
C\ln(2/\delta)}{\sqrt{m}}+\dfrac{2\kappa^{2}\|h_{0}\|_{\mathcal{H}}\ln(2/\delta)}{\sqrt{m}}+\dfrac{2\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\delta)}{\sqrt{m}}$
$\displaystyle=\dfrac{4\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\delta)}{\sqrt{m}}.$

Then recall

$\|h^{m}_{\xi}-h_{\xi}\|_{\mathcal{H}}\leq\dfrac{1}{\xi}\Delta.$

    2. (b)

$\dfrac{\kappa}{\sqrt{m\xi}}>\dfrac{1}{4\ln(2/\delta)}$.

Observe that by the definition of $h_{\xi}^{m}$

$\xi\|h\|^{2}_{\mathcal{H}}\leq\mathcal{E}_{\xi}^{m}(h^{m}_{\xi})\leq\mathcal{E}_{\xi}^{m}(0)=\dfrac{1}{m}\sum_{i=1}^{m}\dot{y}_{i}^{2}\leq
C^{2}.$

It follows that

$\displaystyle\|h_{\xi}^{m}\|_{\mathcal{H}}\leq\dfrac{C}{\sqrt{\xi}}$

and therefore

$\|h_{\xi}^{m}-h_{\xi}\|_{\mathcal{H}}\leq\|h_{\xi}^{m}\|_{\mathcal{H}}+\|h_{\xi}\|_{\mathcal{H}}\leq\dfrac{C}{\sqrt{\xi}}+\dfrac{\kappa\|h_{0}\|_{\mathcal{H}}}{\sqrt{\xi}}=\dfrac{C+\kappa\|h_{0}\|_{\mathcal{H}}}{\sqrt{\xi}}.$

Finally observe that

$\dfrac{1}{4\ln(2/\delta)}<\dfrac{\kappa}{\sqrt{m\xi}}\iff\dfrac{C+\kappa\|h_{0}\|_{\mathcal{H}}}{\sqrt{\xi}}<\dfrac{4\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\delta)}{\sqrt{m}\xi}.$

∎

######  Proposition 14 (Approximation error).

Suppose Assumptions 2, 4, 5, and 9 hold. Then

$\|h_{\xi}-h_{0}\|_{\mathcal{H}}\leq\xi^{\frac{c-1}{2}}\sqrt{\zeta}.$

###### Proof.

I generalize [Smale and Zhou, 2005, Theorem 4]. By Assumption 9, there exists
a function $g\in\mathcal{H}$ such that

$g=T^{\frac{1-c}{2}}h_{0}=\sum_{k}\eta_{k}^{\frac{1-c}{2}}e_{k}\langle
e_{k},h_{0}\rangle_{\mathcal{H}}$

where $\\{\eta_{k}\\}$ are the eigenvalues and $\\{e_{k}\\}$ are the
eigenfunctions of $T$. By Proposition 9, write

$h_{\xi}-h_{0}=[(T+\xi
I)^{-1}T-I]h_{0}=\sum_{k}\bigg{(}\dfrac{\eta_{k}}{\eta_{k}+\xi}-1\bigg{)}e_{k}\langle
e_{k},h_{0}\rangle_{\mathcal{H}}.$

Therefore

$\displaystyle\|h_{\xi}-h_{0}\|^{2}_{\mathcal{H}}$
$\displaystyle=\sum_{k}\bigg{(}\dfrac{\eta_{k}}{\eta_{k}+\xi}-1\bigg{)}^{2}\langle
e_{k},h_{0}\rangle^{2}_{\mathcal{H}}$
$\displaystyle=\sum_{k}\bigg{(}\dfrac{\xi}{\eta_{k}+\xi}\bigg{)}^{2}\langle
e_{k},h_{0}\rangle^{2}_{\mathcal{H}}$
$\displaystyle=\sum_{k}\bigg{(}\dfrac{\xi}{\eta_{k}+\xi}\bigg{)}^{2}\langle
e_{k},h_{0}\rangle^{2}_{\mathcal{H}}\bigg{(}\dfrac{\xi}{\xi}\cdot\dfrac{\eta_{k}}{\eta_{k}}\cdot\dfrac{\eta_{k}+\xi}{\eta_{k}+\xi}\bigg{)}^{c-1}$
$\displaystyle=\xi^{c-1}\sum_{k}\eta_{k}^{1-c}\langle
e_{k},h_{0}\rangle^{2}_{\mathcal{H}}\bigg{(}\dfrac{\xi}{\eta_{k}+\xi}\bigg{)}^{3-c}\bigg{(}\dfrac{\eta_{k}}{\eta_{k}+\xi}\bigg{)}^{c-1}$
$\displaystyle\leq\xi^{c-1}\sum_{k}\eta_{k}^{1-c}\langle
e_{k},h_{0}\rangle^{2}_{\mathcal{H}}$
$\displaystyle=\xi^{c-1}\|g\|^{2}_{\mathcal{H}}$
$\displaystyle\leq\xi^{c-1}\zeta.$

∎

###  G.4 Bounds

######  Proposition 15.

Suppose Assumption 4 holds. Assume that $\forall
d\in\mathcal{D},x\in\mathcal{X},z\in\mathcal{Z}$,
$\|\mu^{n}_{\lambda}(d,x,z)-\mu(d,x,z)\|_{\mathcal{H}}\leq
r_{\mu}(n,\delta,c_{0})$.

  1. 1.

Then
$\|\hat{\mathbf{T}}-\mathbf{T}\|_{\mathcal{L}(\mathcal{H})}\leq\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0}).$

  2. 2.

If in addition Assumption 5 holds then
$\|\hat{\mathbf{g}}-\mathbf{g}\|_{\mathcal{H}}\leq Cr_{\mu}(n,\delta,c_{0})$.

###### Proof.

For the former result, write

$\displaystyle\hat{\mathbf{T}}-\mathbf{T}$
$\displaystyle=\dfrac{1}{m}\sum_{i=1}^{m}\\{\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\\}$
$\displaystyle=\dfrac{1}{m}\sum_{i=1}^{m}\\{\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})$
$\displaystyle\quad+\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\\}.$

Hence

$\displaystyle\|\mathbf{T}-\hat{\mathbf{T}}\|_{\mathcal{L}(\mathcal{H})}$
$\displaystyle\leq\dfrac{1}{m}\sum_{i=1}^{m}\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{L}(\mathcal{H})}$
$\displaystyle\quad+\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{L}(\mathcal{H})}.$

In the first term,

$\displaystyle\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{L}(\mathcal{H})}$
$\displaystyle=\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}\cdot\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}$
$\displaystyle\leq\left(\|\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}+\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}\right)\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}$
$\displaystyle\leq\left\\{\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0}).$

In the second term

$\displaystyle\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\otimes\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{L}(\mathcal{H})}$
$\displaystyle=\|\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}\cdot\|\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\|_{\mathcal{H}}$
$\displaystyle\leq\kappa\cdot r_{\mu}(n,\delta,c_{0}).$

In summary,

$\|\mathbf{T}-\hat{\mathbf{T}}\|_{\mathcal{L}(\mathcal{H})}\leq\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0}).$

For the latter result, write

$\hat{\mathbf{g}}-\mathbf{g}=\dfrac{1}{m}\sum_{i=1}^{m}\\{\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\\}\dot{y}_{i}$

Hence

$\|\hat{\mathbf{g}}-\mathbf{g}\|_{\mathcal{H}}\leq\dfrac{1}{m}\sum_{i=1}^{m}\|\\{\mu^{n}_{\lambda}(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})-\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\\}\|_{\mathcal{H}}\cdot|\dot{y}_{i}|\leq
Cr_{\mu}(n,\delta,c_{0}).$

∎

######  Proposition 16.

Suppose Assumption 4 holds. If $\forall
d\in\mathcal{D},x\in\mathcal{X},z\in\mathcal{Z}$,
$\|\mu^{n}_{\lambda}(d,x,z)-\mu(d,x,z)\|_{\mathcal{H}}\leq
r_{\mu}(n,\delta,c_{0})$ then

$\|(\hat{\mathbf{T}}+\xi)^{-1}-(\mathbf{T}+\xi)^{-1}\|_{\mathcal{L}(\mathcal{H})}\leq\frac{\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0})}{\xi^{2}}.$

###### Proof.

Since $A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$,

$(\hat{\mathbf{T}}+\xi)^{-1}-(\mathbf{T}+\xi)^{-1}=(\mathbf{T}+\xi)^{-1}(\mathbf{T}-\hat{\mathbf{T}})(\hat{\mathbf{T}}+\xi)^{-1}.$

Therefore

$\|(\hat{\mathbf{T}}+\xi)^{-1}-(\mathbf{T}+\xi)^{-1}\|_{\mathcal{L}(\mathcal{H})}\leq\frac{1}{\xi^{2}}\|\mathbf{T}-\hat{\mathbf{T}}\|_{\mathcal{L}(\mathcal{H})}\leq\frac{\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0})}{\xi^{2}}$

where the final inequality appeals to Proposition 15. ∎

######  Proposition 17.

Suppose Assumptions 4 and 5 hold. Then

$\|\mathbf{g}\|_{\mathcal{H}}\leq\kappa C.$

###### Proof.

$\|\mathbf{g}\|_{\mathcal{H}}=\left\|\frac{1}{m}\sum_{i=1}^{m}\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\dot{y}_{i}\right\|_{\mathcal{H}}\leq\frac{1}{m}\sum_{i=1}^{m}\left\|\mu(\dot{d}_{i},\dot{x}_{i},\dot{z}_{i})\right\|_{\mathcal{H}}|\dot{y}_{i}|\\\
\leq\kappa C.$

∎

###  G.5 Main result

######  Theorem 5 (Collecting results).

Suppose Assumptions 2, 4, 5, and 9 hold. Suppose $\forall
d\in\mathcal{D},x\in\mathcal{X},z\in\mathcal{Z}$,
$\|\mu^{n}_{\lambda}(d,x,z)-\mu(d,x,z)\|_{\mathcal{H}}\leq
r_{\mu}(n,\delta,c_{0})$ with probability $1-\delta$. Then
$\forall\delta\in(0,1)$ and $\forall\eta\in(0,1)$, the following holds with
probability $1-\eta-\delta$:

$\displaystyle\|\hat{h}_{\xi}^{m}-h_{0}\|_{\mathcal{H}}$ $\displaystyle\leq
r_{h}(n,\delta,c_{0};m,\eta,c)$
$\displaystyle:=\frac{Cr_{\mu}(n,\delta,c_{0})}{\xi}+\frac{\kappa
C}{\xi^{2}}\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0})+\dfrac{4\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\eta)}{\sqrt{m}\xi}+\xi^{\frac{c-1}{2}}\sqrt{\zeta}.$

###### Proof.

I begin with a decomposition using Proposition 12.

$\hat{h}_{\xi}^{m}-h_{0}=\left[(\hat{\mathbf{T}}+\xi)^{-1}\hat{\mathbf{g}}-(\hat{\mathbf{T}}+\xi)^{-1}\mathbf{g}\right]+\left[(\hat{\mathbf{T}}+\xi)^{-1}\mathbf{g}-(\mathbf{T}+\xi)^{-1}\mathbf{g}\right]+\left[(\mathbf{T}+\xi)^{-1}\mathbf{g}-h_{0}\right].$

Consider the first term.

$\|(\hat{\mathbf{T}}+\xi)^{-1}(\hat{\mathbf{g}}-\mathbf{g})\|_{\mathcal{H}}\leq\frac{1}{\xi}\|\hat{\mathbf{g}}-\mathbf{g}\|_{\mathcal{H}}\leq\frac{1}{\xi}Cr_{\mu}(n,\delta,c_{0})$

by Proposition 15.

Consider the second term.

$\displaystyle\left\|\left[(\hat{\mathbf{T}}+\xi)^{-1}-(\mathbf{T}+\xi)^{-1}\right]\mathbf{g}\right\|_{\mathcal{H}}$
$\displaystyle\leq\|(\hat{\mathbf{T}}+\xi)^{-1}-(\mathbf{T}+\xi)^{-1}\|_{\mathcal{L}(\mathcal{H})}\|\mathbf{g}\|_{\mathcal{H}}$
$\displaystyle\leq\frac{1}{\xi^{2}}\left\\{2\kappa+r_{\mu}(n,\delta,c_{0})\right\\}r_{\mu}(n,\delta,c_{0})\cdot\kappa
C$

with probability $1-\delta$ by Propositions 16 and 17.

Consider the third term.

$\left\|h_{\xi}^{m}-h_{0}\right\|_{\mathcal{H}}\leq\left\|h_{\xi}^{m}-h_{\xi}\right\|_{\mathcal{H}}+\left\|h_{\xi}-h_{0}\right\|_{\mathcal{H}}\leq\dfrac{4\kappa(C+\kappa\|h_{0}\|_{\mathcal{H}})\ln(2/\eta)}{\sqrt{m}\xi}+\xi^{\frac{c-1}{2}}\sqrt{\zeta}.$

with probability $1-\eta$, appealing to triangle inequality and Propositions
13 and 14. ∎

######  Theorem 6 (Conditional mean embedding).

Suppose Assumptions 4, 5, and 8 hold. Set $\lambda=n^{-\frac{1}{c_{0}+1}}$.
Then with probability $1-\delta$, $\forall
d\in\mathcal{D},x\in\mathcal{X},z\in\mathcal{Z}$

$\|\hat{\mu}(d,x,z)-\mu(d,x,z)\|_{\mathcal{H}}\leq r_{\mu}(n,\delta,c_{0})$

where $\kappa_{RF}:=\kappa_{d}\kappa_{x}\kappa_{z}$ and

$r_{\mu}(n,\delta,c_{0}):=\kappa_{d}\kappa_{x}\cdot\kappa_{RF}\cdot\dfrac{\sqrt{\zeta_{0}}(c_{0}+1)}{4^{\frac{1}{c_{0}+1}}}\bigg{(}\dfrac{4\kappa_{RF}(\kappa_{w}+\kappa_{RF}\|E_{0}\|_{\mathcal{L}_{2}})\ln(2/\delta)}{\sqrt{n\zeta_{0}}(c_{0}-1)}\bigg{)}^{\frac{c_{0}-1}{c_{0}+1}}.$

###### Proof.

Write

$\displaystyle\hat{\mu}(d,x,z)-\mu(d,x,z)$
$\displaystyle=[\phi(d)\otimes\phi(x)\otimes\hat{\mu}_{w}(d,x,z)]-[\phi(d)\otimes\phi(x)\otimes\mu_{w}(d,x,z)]$
$\displaystyle=\phi(d)\otimes\phi(x)\otimes[\hat{\mu}_{w}(d,x,z)-\mu_{w}(d,x,z)]$

so that

$\|\hat{\mu}(d,x,z)-\mu(d,x,z)\|_{\mathcal{H}}\leq\kappa_{d}\kappa_{x}\cdot\|\hat{\mu}_{w}(d,x,z)-\mu_{w}(d,x,z)\|_{\mathcal{H}_{\mathcal{W}}}.$

The bound on the final factor follows from [Singh et al., 2019, Corollary 1],
observing that

$E_{0}:\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{RF},\quad\|\phi(w)\|_{\mathcal{H}}\leq\kappa_{w},\quad\|\phi(d,x,z)\|_{\mathcal{H}_{RF}}\leq\kappa_{RF}.$

∎

###### Proof of Theorem 3.

Summarize the bound in Theorem 5 as

$r_{h}=O\left(\frac{r_{\mu}}{\xi}+\frac{r_{\mu}}{\xi^{2}}+\frac{r^{2}_{\mu}}{\xi^{2}}+\frac{1}{\sqrt{m}\xi}+\xi^{\frac{c-1}{2}}\right)=O\left(\frac{r_{\mu}}{\xi^{2}}+\frac{1}{\sqrt{m}\xi}+\xi^{\frac{c-1}{2}}\right).$

Summarize the bound in Theorem 6 as

$r_{\mu}=O\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}}\right)=O(m^{-\frac{a}{2}}),\quad
n=m^{\frac{a(c_{0}+1)}{(c_{0}-1)}}.$

Combining results,

$r_{h}=O\left(\frac{1}{m^{\frac{a}{2}}\xi^{2}}+\frac{1}{\sqrt{m}\xi}+\xi^{\frac{c-1}{2}}\right),\quad\text{such
that}\quad\xi^{2}\geq r_{\mu},\quad m\xi^{2}\geq 1.$

This choice of $(n,m)$ ratio is the same parametrization as [Singh et al.,
2019, Theorem 4].

I have choice over $\xi$ as a function of $m$ to achieve the single stage rate
of $m^{-\frac{1}{2}\frac{c-1}{c+1}}$. I choose $\xi$ to match the bias
$\xi^{\frac{c-1}{2}}$ with the variance
$\frac{1}{m^{\frac{a}{2}}\xi^{2}}+\frac{1}{\sqrt{m}\xi}$. I set bias equal to
each term in the variance.

  1. 1.

$\xi^{\frac{c-1}{2}}=\frac{1}{m^{\frac{a}{2}}\xi^{2}}$. Rearranging,
$\xi=m^{-\frac{a}{c+3}}$. The bias term becomes

$\xi^{\frac{c-1}{2}}=\left(m^{-\frac{a}{c+3}}\right)^{\frac{c-1}{2}}$

and the remaining term becomes

$\frac{1}{\sqrt{m}\xi}=\frac{m^{\frac{a}{c+3}}}{\sqrt{m}}=m^{\frac{2a-(c+3)}{2(c+3)}}.$

Note that the former dominates the latter if and only if

${-\frac{a}{c+3}}{\frac{c-1}{2}}\geq\frac{2a-(c+3)}{2(c+3)}\iff
a\leq\frac{c+3}{c+1}.$

  2. 2.

$\xi^{\frac{c-1}{2}}=\frac{1}{\sqrt{m}\xi}$. Rearranging,
$\xi=m^{-\frac{1}{c+1}}$. The bias term becomes

$\xi^{\frac{c-1}{2}}=\left(m^{-\frac{1}{c+1}}\right)^{\frac{c-1}{2}}$

and the remaining term becomes

$\frac{1}{m^{\frac{a}{2}}\xi^{2}}=m^{-\frac{a}{2}}\left(m^{-\frac{1}{c+1}}\right)^{-2}=m^{\frac{4-a(c+1)}{2(c+1)}}.$

Note that the former dominates the latter if and only if

$-\frac{1}{c+1}\frac{c-1}{2}\geq\frac{4-a(c+1)}{2(c+1)}\iff
a\geq\frac{c+3}{c+1}.$

∎

##  Appendix H Treatment effect consistency proof

In this appendix, I (i) explicitly specialize the smoothness assumptions, (ii)
provide rates for unconditional mean embeddings, (iv) provide rates for
conditional mean embeddings, and (iv) prove uniform consistency of negative
control treatment effects.

###  H.1 Smoothness assumptions

######  Assumption 10 (Smoothness for $\theta_{0}^{ATT}$).

Assume

  1. 1.

The conditional expectation operator $E_{1}$ is well specified as a Hilbert-
Schmidt operator between RKHSs, i.e.
$E_{1}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{D}})$,
where

$E_{1}:\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{\mathcal{D}},\quad
f(\cdot,\cdot)\mapsto\mathbb{E}[f(X,W)|D=\cdot].$

  2. 2.

The conditional expectation operator is a particularly smooth element of
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{D}})$.
Formally, define the covariance operator
$T_{1}:=\mathbb{E}[\phi(D)\otimes\phi(D)]$ for
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{D}})$.
I assume there exists
$G_{1}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{D}})$
such that $E_{1}=(T_{1})^{\frac{c_{1}-1}{2}}\circ G_{1}$, $c_{1}\in(1,2]$, and
$\|G_{1}\|^{2}_{\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{D}})}\leq\zeta_{1}$.

######  Assumption 11 (Smoothness for $\theta_{0}^{CATE}$).

Assume

  1. 1.

The conditional expectation operator $E_{2}$ is well specified as a Hilbert-
Schmidt operator between RKHSs, i.e.
$E_{2}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{V}})$,
where

$E_{2}:\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{\mathcal{V}},\quad
f(\cdot,\cdot)\mapsto\mathbb{E}[f(X,W)|V=\cdot].$

  2. 2.

The conditional expectation operator is a particularly smooth element of
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{V}})$.
Formally, define the covariance operator
$T_{2}:=\mathbb{E}[\phi(V)\otimes\phi(V)]$ for
$\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{V}})$.
I assume there exists
$G_{2}\in\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{V}})$
such that $E_{2}=(T_{2})^{\frac{c_{2}-1}{2}}\circ G_{2}$, $c_{2}\in(1,2]$, and
$\|G_{2}\|^{2}_{\mathcal{L}_{2}(\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}},\mathcal{H}_{\mathcal{V}})}\leq\zeta_{2}$.

######  Proposition 18.

The following assumptions are equivalent

  1. 1.

Assumption 6 with $\mathcal{A}_{1}=\mathcal{X}\times\mathcal{W}$ and
$\mathcal{B}_{1}=\mathcal{D}$ is equivalent to Assumption 10

  2. 2.

Assumption 6 with $\mathcal{A}_{2}=\mathcal{X}\times\mathcal{W}$ and
$\mathcal{B}_{2}=\mathcal{V}$ is equivalent to Assumption 11

###### Proof.

The result is immediate from [Caponnetto and De Vito, 2007, Remark 2]. The
expanded expressions are more convenient for analysis. ∎

###  H.2 Unconditional mean embedding

######  Theorem 7 (Unconditional mean embedding).

Suppose Assumptions 4 and 5 hold. Then with probability $1-\delta$,

$\|\hat{\mu}-\mu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq
r_{\mu}(n,\delta):=\frac{4\kappa_{x}\kappa_{w}\ln(2/\delta)}{\sqrt{n}}.$

Likewise, with probability $1-\delta$

$\|\hat{\nu}-\nu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq
r_{\nu}(\tilde{n},\delta):=\frac{4\kappa_{x}\kappa_{w}\ln(2/\delta)}{\sqrt{\tilde{n}}}.$

###### Proof.

The first result follows from Lemma 4 with
$\xi_{i}=\phi(x_{i})\otimes\phi(w_{i})$, since

$\displaystyle\left\|\frac{1}{n}\sum_{i=1}^{n}[\phi(x_{i})\otimes\phi(w_{i})]-\mathbb{E}[\phi(X)\otimes\phi(W)]\right\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\leq\frac{2\kappa_{x}\kappa_{w}\ln(2/\delta)}{n}+\sqrt{\dfrac{2\kappa^{2}_{x}\kappa^{2}_{w}\ln(2/\delta)}{n}}$
$\displaystyle\leq\frac{4\kappa_{x}\kappa_{w}\ln(2/\delta)}{\sqrt{n}}.$

The argument for $\nu$ is identical, using
$\xi_{i}=\phi(\tilde{x}_{i})\otimes\phi(\tilde{w}_{i})$. ∎

###  H.3 Conditional mean embedding

######  Theorem 8 (Conditional mean embedding rate).

Suppose Assumptions 4 and 5 hold. Set
$(\lambda_{1},\lambda_{2})=(n^{-\frac{1}{c_{1}+1}},n^{-\frac{1}{c_{2}+1}})$.

  1. 1.

If in addition Assumption 10 holds then with probability $1-\delta$, $\forall
d\in\mathcal{D}$

$\|\hat{\mu}(d)-\mu(d)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq
r^{ATT}_{\mu}(n,\delta,c_{1})$

where

$r^{ATT}_{\mu}(n,\delta,c_{1}):=\kappa_{d}\cdot\dfrac{\sqrt{\zeta_{1}}(c_{1}+1)}{4^{\frac{1}{c_{1}+1}}}\bigg{(}\dfrac{4\kappa_{d}(\kappa_{x}\kappa_{w}+\kappa_{d}\|E_{1}\|_{\mathcal{L}_{2}})\ln(2/\delta)}{\sqrt{n\zeta_{1}}(c_{1}-1)}\bigg{)}^{\frac{c_{1}-1}{c_{1}+1}}.$

  2. 2.

If in addition Assumption 11 holds then with probability $1-\delta$, $\forall
v\in\mathcal{V}$

$\|\hat{\mu}(v)-\mu(v)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq
r^{CATE}_{\mu}(n,\delta,c_{2})$

where

$r^{CATE}_{\mu}(n,\delta,c_{2}):=\kappa_{v}\cdot\dfrac{\sqrt{\zeta_{2}}(c_{2}+1)}{4^{\frac{1}{c_{2}+1}}}\bigg{(}\dfrac{4\kappa_{v}(\kappa_{x}\kappa_{w}+\kappa_{v}\|E_{2}\|_{\mathcal{L}_{2}})\ln(2/\delta)}{\sqrt{n\zeta_{2}}(c_{2}-1)}\bigg{)}^{\frac{c_{2}-1}{c_{2}+1}}.$

###### Proof.

The proof immediately follows from [Singh et al., 2019, Corollary 1],
observing that

$E_{1}:\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{\mathcal{D}},\quad\|\phi(x)\otimes\phi(w)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq\kappa_{x}\kappa_{w},\quad\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\leq\kappa_{d}$

and

$E_{2}:\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}\rightarrow\mathcal{H}_{\mathcal{V}},\quad\|\phi(x)\otimes\phi(w)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}\leq\kappa_{x}\kappa_{w},\quad\|\phi(v)\|_{\mathcal{H}_{\mathcal{V}}}\leq\kappa_{v}.$

∎

###  H.4 Main result

In summary, when reusing samples

$\displaystyle\|\hat{h}-h_{0}\|_{\mathcal{H}}$
$\displaystyle=O_{p}\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}\frac{c-1}{c+3}}\right)$
$\displaystyle\|\hat{\mu}-\mu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle=O_{p}\left(n^{-\frac{1}{2}}\right)$
$\displaystyle\|\hat{\nu}-\nu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle=O_{p}\left(\tilde{n}^{-\frac{1}{2}}\right)$
$\displaystyle\|\hat{\mu}(d)-\mu(d)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle=O_{p}\left(n^{-\frac{1}{2}\frac{c_{1}-1}{c_{1}+1}}\right)$
$\displaystyle\|\hat{\mu}(v)-\mu(v)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle=O_{p}\left(n^{-\frac{1}{2}\frac{c_{2}-1}{c_{2}+1}}\right).$

###### Proof of Theorem 4.

I generalize the argument in [Singh et al., 2020, Theorem 3.3]. I write out
the finite sample bounds. Consider the decomposition

$\displaystyle\hat{\theta}^{ATE}(d)-\theta_{0}^{ATE}(d)$
$\displaystyle=\langle\hat{h},\phi(d)\otimes\hat{\mu}\rangle_{\mathcal{H}}-\langle
h_{0},\phi(d)\otimes\mu\rangle_{\mathcal{H}}$
$\displaystyle=\langle\hat{h},\phi(d)\otimes[\hat{\mu}-\mu]\rangle_{\mathcal{H}}+\langle[\hat{h}-h_{0}],\phi(d)\otimes\mu\rangle_{\mathcal{H}}$
$\displaystyle=\langle[\hat{h}-h_{0}],\phi(d)\otimes[\hat{\mu}-\mu]\rangle_{\mathcal{H}}+\langle
h_{0},\phi(d)\otimes[\hat{\mu}-\mu]\rangle_{\mathcal{H}}+\langle[\hat{h}-h_{0}],\phi(d)\otimes\mu\rangle_{\mathcal{H}}.$

Therefore with probability $1-2\delta-\eta$

$\displaystyle|\hat{\theta}^{ATE}(d)-\theta_{0}^{ATE}(d)|$
$\displaystyle\leq\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\hat{\mu}-\mu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}+\|h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\hat{\mu}-\mu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\quad+\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\mu\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\leq\kappa_{d}\cdot r_{h}(n,\delta,c_{0};m,\eta,c)\cdot
r_{\mu}(n,\delta)+\kappa_{d}\cdot\|h_{0}\|_{\mathcal{H}}\cdot
r_{\mu}(n,\delta)$ $\displaystyle\quad+\kappa_{d}\kappa_{x}\kappa_{w}\cdot
r_{h}(n,\delta,c_{0};m,\eta,c)$
$\displaystyle=O\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}\frac{c-1}{c+3}}\right).$

By the same argument as for $\theta_{0}^{ATE}$, with probability
$1-2\delta-\eta$

$\displaystyle|\hat{\theta}^{DS}(d,\tilde{\mathbb{P}})-\theta_{0}^{DS}(d,\tilde{\mathbb{P}})|$
$\displaystyle\leq\kappa_{d}\cdot r_{h}(n,\delta,c_{0};m,\eta,c)\cdot
r_{\nu}(\tilde{n},\delta)+\kappa_{d}\cdot\|h_{0}\|_{\mathcal{H}}\cdot
r_{\nu}(\tilde{n},\delta)$
$\displaystyle\quad+\kappa_{d}\kappa_{x}\kappa_{w}\cdot
r_{h}(n,\delta,c_{0};m,\eta,c)$
$\displaystyle=O\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}\frac{c-1}{c+3}}+\tilde{n}^{-\frac{1}{2}}\right).$

Next, I turn to the nonparametric treatment effects with conditional mean
embeddings. Consider the decomposition

$\displaystyle\hat{\theta}^{ATT}(d,d^{\prime})-\theta_{0}^{ATT}(d,d^{\prime})$
$\displaystyle=\langle\hat{h},\phi(d^{\prime})\otimes\hat{\mu}(d)\rangle_{\mathcal{H}}-\langle
h_{0},\phi(d^{\prime})\otimes\mu(d)\rangle_{\mathcal{H}}$
$\displaystyle=\langle\hat{h},\phi(d^{\prime})\otimes[\hat{\mu}(d)-\mu(d)]\rangle_{\mathcal{H}}+\langle[\hat{h}-h_{0}],\phi(d^{\prime})\otimes\mu(d)\rangle_{\mathcal{H}}$
$\displaystyle=\langle[\hat{h}-h_{0}],\phi(d^{\prime})\otimes[\hat{\mu}(d)-\mu(d)]\rangle_{\mathcal{H}}+\langle
h_{0},\phi(d^{\prime})\otimes[\hat{\mu}(d)-\mu(d)]\rangle_{\mathcal{H}}$
$\displaystyle\quad+\langle[\hat{h}-h_{0}],\phi(d^{\prime})\otimes\mu(d)\rangle_{\mathcal{H}}.$

Therefore with probability $1-2\delta-\eta$

$\displaystyle|\hat{\theta}^{ATT}(d,d^{\prime})-\theta_{0}^{ATT}(d,d^{\prime})|$
$\displaystyle\leq\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d^{\prime})\|_{\mathcal{H}_{\mathcal{D}}}\|\hat{\mu}(d)-\mu(d)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\quad+\|h_{0}\|_{\mathcal{H}}\|\phi(d^{\prime})\|_{\mathcal{H}_{\mathcal{D}}}\|\hat{\mu}(d)-\mu(d)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\quad+\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d^{\prime})\|_{\mathcal{H}_{\mathcal{D}}}\|\mu(d)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\leq\kappa_{d}\cdot r_{h}(n,\delta,c_{0};m,\eta,c)\cdot
r_{\mu}^{ATT}(n,\delta,c_{1})+\kappa_{d}\cdot\|h_{0}\|_{\mathcal{H}}\cdot
r_{\mu}^{ATT}(n,\delta,c_{1})$
$\displaystyle\quad+\kappa_{d}\kappa_{x}\kappa_{w}\cdot
r_{h}(n,\delta,c_{0};m,\eta,c)$
$\displaystyle=O\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}\frac{c-1}{c+3}}+n^{-\frac{1}{2}\frac{c_{1}-1}{c_{1}+1}}\right).$

Similarly, consider the decomposition

$\displaystyle\hat{\theta}^{CATE}(d,v)-\theta_{0}^{CATE}(d,v)$
$\displaystyle=\langle\hat{h},\phi(d)\otimes\phi(v)\otimes\hat{\mu}(v)\rangle_{\mathcal{H}}-\langle
h_{0},\phi(d)\otimes\phi(v)\otimes\mu(v)\rangle_{\mathcal{H}}$
$\displaystyle=\langle\hat{h},\phi(d)\otimes\phi(v)\otimes[\hat{\mu}(v)-\mu(v)]\rangle_{\mathcal{H}}+\langle[\hat{h}-h_{0}],\phi(d)\otimes\phi(v)\otimes\mu(v)\rangle_{\mathcal{H}}$
$\displaystyle=\langle[\hat{h}-h_{0}],\phi(d)\otimes\phi(v)\otimes[\hat{\mu}(v)-\mu(v)]\rangle_{\mathcal{H}}$
$\displaystyle\quad+\langle
h_{0},\phi(d)\otimes\phi(v)\otimes[\hat{\mu}(v)-\mu(v)]\rangle_{\mathcal{H}}$
$\displaystyle\quad+\langle[\hat{h}-h_{0}],\phi(d)\otimes\phi(v)\otimes\mu(v)\rangle_{\mathcal{H}}.$

Therefore with probability $1-2\delta-\eta$

$\displaystyle|\hat{\theta}^{CATE}(d,v)-\theta_{0}^{CATE}(d,v)|$
$\displaystyle\leq\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\phi(v)\|_{\mathcal{H}_{\mathcal{V}}}\|\hat{\mu}(v)-\mu(v)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\quad+\|h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\phi(v)\|_{\mathcal{H}_{\mathcal{V}}}\|\hat{\mu}(v)-\mu(v)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\quad+\|\hat{h}-h_{0}\|_{\mathcal{H}}\|\phi(d)\|_{\mathcal{H}_{\mathcal{D}}}\|\phi(v)\|_{\mathcal{H}_{\mathcal{V}}}\|\mu(v)\|_{\mathcal{H}_{\mathcal{X}}\otimes\mathcal{H}_{\mathcal{W}}}$
$\displaystyle\leq\kappa_{d}\kappa_{v}\cdot
r_{h}(n,\delta,c_{0};m,\eta,c)\cdot
r_{\mu}^{CATE}(n,\delta,c_{2})+\kappa_{d}\kappa_{v}\cdot\|h_{0}\|_{\mathcal{H}}\cdot
r_{\mu}^{CATE}(n,\delta,c_{2})$
$\displaystyle\quad+\kappa_{d}\kappa_{v}\kappa_{x}\kappa_{w}\cdot
r_{h}(n,\delta,c_{0};m,\eta,c)$
$\displaystyle=O\left(n^{-\frac{1}{2}\frac{c_{0}-1}{c_{0}+1}\frac{c-1}{c+3}}+n^{-\frac{1}{2}\frac{c_{2}-1}{c_{2}+1}}\right).$

∎

##  Appendix I Simulation details

###  I.1 General simulation design

A single observation is generated as follows. Recall that
$(Y,D)\in\mathbb{R}$, $X\in\mathbb{R}^{dim(X)}$, $Z\in\mathbb{R}^{dim(Z)}$,
and $W\in\mathbb{R}^{dim(W)}$.

  1. 1.

Draw unobserved noise as

$\\{\epsilon_{i}\\}_{i\in[3]}\overset{i.i.d}{\sim}\mathcal{N}(0,1),\quad\nu_{z}\sim\mathcal{U}[-1,1]^{dim(Z)},\quad\nu_{w}\sim\mathcal{U}[-1,1]^{dim(W)}.$

  2. 2.

Set unobserved confounders as $u_{z}=\epsilon_{1}+\epsilon_{3}$ and
$u_{w}=\epsilon_{2}+\epsilon_{3}$.

  3. 3.

Set the negative controls as

$Z=\nu_{z}+0.25\cdot u_{z}\cdot 1_{dim(Z)},\quad W=\nu_{w}+0.25\cdot
u_{w}\cdot 1_{dim(W)},$

where $1_{p}\in\mathbb{R}^{p}$ is the vector of ones of length $p$.

  4. 4.

Draw covariates $X\sim\mathcal{N}(0,\Sigma)$ where the covariance matrix
$\Sigma\in\mathbb{R}^{dim(X)\times dim(X)}$ is such that $\Sigma_{ii}=1$ and
$\Sigma_{ij}=\frac{1}{2}\cdot 1\\{|i-j|=1\\}$ for $i\neq j$.

  5. 5.

Then set treatment as

$D=\Lambda(3X^{\top}\beta_{x}+3Z^{\top}\beta_{z})+0.25\cdot u_{w}.$

$\beta_{x}\in\mathbb{R}^{dim(X)}$ and $\beta_{z}\in\mathbb{R}^{dim(Z)}$ are
quadratically decaying coefficients, e.g. $[\beta_{x}]_{j}=j^{-2}$. $\Lambda$
is the truncated logistic link function
$\Lambda(t)=(0.9-0.1)\frac{\exp(t)}{1+\exp(t)}+0.1$.

  6. 6.

Finally set the outcome as

$Y=\theta_{0}^{ATE}(D)+1.2(X^{\top}\beta_{x}+W^{\top}\beta_{w})+DX_{1}+0.25\cdot
u_{z},$

where $\beta_{w}\in\mathbb{R}^{dim(W)}$ is a quadratically decaying
coefficient, i.e. $[\beta_{w}]_{j}=j^{-2}$.

For the quadratic design, $\theta_{0}^{ATE}(d)=d^{2}+1.2d$ as in [Colangelo
and Lee, 2020]. For the sigmoid design,
$\theta_{0}^{ATE}(d)=\ln(|16d-8|+1)\cdot sign(d-0.5)+1.2d$ similar to [Singh
et al., 2019]. Finally, for the peaked design,
$\theta_{0}^{ATE}(d)=2\\{d^{4}/600+\exp(-4d^{2})+d/10-2\\}+1.2d$ similar to
[Singh et al., 2019].

I implement the estimator $\hat{\theta}^{ATE}(d)$ (`N.C.`) described in
Section 4, with the tuning procedure described in Appendix F. Specifically, I
use ridge penalties determined by leave-one-out cross validation, and product
Gaussian kernel with lengthscales set by the median heuristic. I implement the
continuous treatment effect estimator of [Singh et al., 2020] (`T.E.`) using
the same principles. The latter tuning procedure is simpler; whereas the new
estimator involves reweighting a confounding bridge (with two ridge penalty
hyperparameters), the previous estimator involves reweighting a regression
(with one ridge penalty hyperparameter).

###  I.2 Robustness to tuning

As explained in Appendix F, the tuning procedure for the ridge penalties
involves leave-one-out cross validation. I now confirm that the proposed
estimator’s performance is robust to improper tuning. Figures 7(a) and 7(b)
summarize results.

(a) Forcing $\lambda$

(b) Forcing $\xi$

(c) Unfounded design

Figure 7:  Robustness studies

I conduct two robustness studies, each corresponding to a ridge penalty
hyperparameter. In the first robustness study (Figure 7(a)), I force $\lambda$
to take a particular value in a grid, then tune $\xi$ by leave-one-out cross
validation: $\xi=\xi^{*}(\lambda)$. In the second robustness study (Figure
7(b)), I tune $\lambda=\lambda^{*}$ by leave-one-out cross validation, then
force $\xi$ to take a particular value in a grid. Across all levels of
improper tuning of `N.C.`, it continues to outperform the properly tuned
`T.E.` estimator.

###  I.3 No unobserved confounding

Next, I study a setting where there is no unobserved confounding. I modify the
data generating process as follows.

  1. 1.

Draw unobserved noise as

$\\{\epsilon_{i}\\}_{i\in[4]}\overset{i.i.d}{\sim}\mathcal{N}(0,1),\quad\nu_{z}\sim\mathcal{U}[-1,1]^{dim(Z)},\quad\nu_{w}\sim\mathcal{U}[-1,1]^{dim(W)}.$

  2. 2.

Set unobserved confounders as $u_{z}=\epsilon_{1}+\epsilon_{2}$ and
$u_{w}=\epsilon_{3}+\epsilon_{4}$.

  3. 3.

Set the negative controls as

$Z=\nu_{z}+0.25\cdot u_{z}\cdot 1_{dim(Z)},\quad W=\nu_{w}+0.25\cdot
u_{w}\cdot 1_{dim(W)}.$

  4. 4.

Draw covariates $X\sim\mathcal{N}(0,\Sigma)$.

  5. 5.

Then set treatment as

$D=\Lambda(3X^{\top}\beta_{x})+0.25\cdot u_{w}.$

  6. 6.

Finally set the outcome as

$Y=\theta_{0}^{ATE}(D)+1.2(X^{\top}\beta_{x})+DX_{1}+0.25\cdot u_{z}.$

Figure 7(c) visualizes results. As expected, the previously existing method,
which assumes no unobserved confounding, outperforms the proposed method. The
proposed method solves an ill posed inverse problem. One pays a cost in terms
of statistical efficiency when the ill posed inverse problem is unnecessary.
From a practical perspective, an analyst should only use negative controls
when the analyst firmly believes that unobserved confounding is present and
that the negatives controls satisfy Assumptions 1 and 2.

###  I.4 Discrete treatment

So far, I have focused on the case with continuous treatment. I now study
empirical performance when treatment is discrete. I modify the data generating
process as follows.

  1. 1.

Draw unobserved noise as

$\\{\epsilon_{i}\\}_{i\in[3]}\overset{i.i.d}{\sim}\mathcal{N}(0,1),\quad\nu_{z}\sim\mathcal{U}[-1,1]^{dim(Z)},\quad\nu_{w}\sim\mathcal{U}[-1,1]^{dim(W)}.$

  2. 2.

Set unobserved confounders as $u_{z}=\epsilon_{1}+\epsilon_{3}$ and
$u_{w}=\epsilon_{2}+\epsilon_{3}$.

  3. 3.

Set the negative controls as

$Z=\nu_{z}+0.5\cdot u_{z}\cdot 1_{dim(Z)},\quad W=\nu_{w}+0.5\cdot u_{w}\cdot
1_{dim(W)}.$

  4. 4.

Draw covariates $X\sim\mathcal{N}(0,\Sigma)$.

  5. 5.

Then set treatment as

$D~{}\sim Bernoulli(\Lambda(X^{\top}\beta_{x}+Z^{\top}\beta_{z}+u_{w})).$

  6. 6.

Finally set the outcome as

$Y=2.2+1.2(X^{\top}\beta_{x}+W^{\top}\beta_{w})+DX_{1}+0.5\cdot u_{z}.$

By construction, $\theta_{0}^{ATE}(1)-\theta_{0}^{ATE}(0)=2.2-0=2.2$. Table 1
summarizes results. The proposed method (N.C.) outperforms the previous
approach (T.E.) that ignores unobserved confounding, when the sample size is
sufficiently large. Importantly, the sample size $n$ must be at least 1000 for
the estimator to detect and correct for the unobserved confounding.
Substantial bias remains, suggesting that the debiased semiparametric
estimators subsequently proposed by [Kallus et al., 2021, Ghassami et al.,
2021] may be more appropriate when treatment is discrete.

Sample size | Mean | S.D. | M.S.E.
---|---|---|---
100 | 2.61 | 0.23 | 0.05
500 | 2.59 | 0.11 | 0.01
1000 | 2.55 | 0.06 | 0.00
5000 | 2.42 | 0.03 | 0.00
(a)

Sample size | Mean | S.D. | M.S.E.
---|---|---|---
100 | 3.07 | 0.27 | 0.07
500 | 2.62 | 0.11 | 0.01
1000 | 2.42 | 0.09 | 0.01
5000 | 1.99 | 0.05 | 0.00
(b)

Table 1: Discrete design

##  Appendix J Application details

###  J.1 Dose response of cigarette smoking

As described in the main text, I estimate the dose response curves for the
subpopulations of white, black, and Hispanic mothers who smoke. Formally, I
estimate $\theta_{0}^{CATE}(d,v)$ where $D\in\mathbb{R}$ is the number of
cigarettes smoked per day, and $V\in\mathbb{R}^{2}$ concatenates mother’s race
$V_{1}$ and mother’s smoking status $V_{2}$. Observe that race is, for our
purposes, a discrete variable with three values while smoking status is a
binary variable. Consider the subpopulation of white mothers who smoke, i.e.
$v=(\text{white},1)$. I implement the product of indicator kernels

$k_{\mathcal{V}}(v,v^{\prime})=k_{\mathcal{V}_{1}}(v_{1},v_{1}^{\prime})k_{\mathcal{V}_{2}}(v_{2},v_{2}^{\prime})=\mathbbm{1}\\{v_{1}=v_{1}^{\prime}\\}\mathbbm{1}\\{v_{2}=v_{2}^{\prime}\\}.$

Therefore the estimator of the confounding bridge $h_{0}(d,v,x,w)$ at the
value $v=(\text{white},1)$ simply zeroes out observations with
$V\neq(\text{white},1)$. Since the covariate of interest is discrete rather
than continuous, a natural choice of the estimator $\hat{\mu}(v)$ that encodes
$\mathbb{P}(x,w|v)$ simply takes the average for the corresponding
subpopulation, e.g.

$\hat{\mu}(\text{white},1)=\frac{1}{|\\{i:V_{i}=(\text{white},1)\\}|}\sum_{i:V_{i}=(\text{white},1)}\phi(x_{i})\otimes\phi(w_{i}).$

Observe that this estimator also zeroes out observations with
$V\neq(\text{white},1)$.

In summary, when $V$ is discrete, the described estimator of
$\theta_{0}^{CATE}(d,v)$ is equivalent to the following procedure: (i) subset
to the observations such that $V=v$, then (ii) implement
$\hat{\theta}^{ATE}(d)$. I implement the estimator $\hat{\theta}^{ATE}(d)$
(`N.C.`) described in Section 4, with the tuning procedure described in
Appendix F, on the appropriate subset of observations. I implement the
continuous treatment effect estimator of [Singh et al., 2020] (`T.E.`) using
the same principles.

To speed up computation, I incorporate the `kernlab` package in `R`. Due to
the large sample size for nonhispanic white women $(n=73,834)$, I split the
observations according to the year (1989, 1990, 1991) then take the average.

###  J.2 Variable classification

Table 2 summarizes the variable classification. The reason why this choice of
negative controls $(Z,W)$ is appropriate when $U$ is income, but inappropriate
when $U$ is stress, is the relevance condition detailed in Proposition 1.
Based on clinical expertise and previously published results such as [Hobel et
al., 2008], we posit that parental education and family size are relevant for
income in the sense that variation in income $U$ can be recovered from
variation in parental education $Z$ and family size $W$. The same cannot be
said for stress; it is less justified by the existing literature that
variation in stress $U$ can be recovered from variation in parental education
$Z$ and family size $W$. Figure 8 visualizes a representative DAG.

Table 2: Variable classification Symbol | Definition | Empirical application
---|---|---
$Y$ | outcome | infant birth weight (grams)
$D$ | treatment | # cigarettes smoked per day during pregnancy
$U$ | unobserved | income
| confounding | stress
$Z$ | n.c. treatment | mother’s educational attainment (years)
|  | father’s educational attainment (years)
$W$ | n.c. outcome | infant birth order
|  | infant sex
|  | Rh sensitization
$X$ | covariates | mother’s demographics: age, marriage status, foreign born
|  | father’s demographics: age, race
|  | alcohol consumption during pregnancy
|  | # prenatal visits during pregnancy
|  | trimester of first prenatal care visit
|  | hypertension: chronic, gestational
|  | other medical: diabetes, herpes, eclampsia, weight gain
|  | county
|  | year
$V$ | covariate of interest | mother’s race
|  | mother’s smoking status

$X$$W$ptbirth order,sex,Rh$Z$ptedu-cation$D$smoking$Y$birth
weight$U_{1}$income$U_{2}$stress

Figure 8: Smoking DAG with stress