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The Honest Truth About Causal Trees: Accuracy Limits for Heterogeneous Treatment Effect Estimation

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The Honest Truth About Causal Trees: Accuracy Limits for Heterogeneous Treatment Effect Estimation Supplemental Appendix

abstractThis supplemental appendix presents more general theoretical results encompassing those discussed in the main paper, and their proofs.

Keywords: recursive partitioning, decision trees, causal inference, heterogeneous treatment effects

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Overview

This supplement presents proofs for the results in the main paper, and several additional theoretical results. We start with a homoskedastic constant regression model in Section (ref), showing that the standard CART decision tree estimator of the (constant) conditional mean suffers from slow uniform convergence rates. In Section (ref), we then study the more challenging heterogeneous causal effect estimators discussed in the main paper: inverse probability weighting (IPW) estimator, the difference in mean (DIM) estimator, and the sum-of-square-minimization (SSE) estimator are considered in Sections (ref), (ref) and (ref), respectively. Section (ref) links the results in this supplemental appendix to those presented in the main paper.

Notations

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Sets. $\mathbb{R}$ is the set of real numbers and $\mathbb{N}$ the positive integers. For $n\in\mathbb{N}$ we write $[n]=\{1,\dots ,n\}$.

Vectors and matrices. Boldface lower‑case letters (e.g. $\mathbf{x}$) denote column vectors, and boldface upper‑case letters (e.g.\ $\mathbf{A}$) denote matrices. For a vector $\mathbf{x}$, its $i$‑th component is $x_i$; for a matrix $\mathbf{A}$, its $(i,j)$‑th entry is $A_{ij}$. Denote by $\mathbf{e}_j$ the $j$-th unit vector.

Norms. For $\mathbf{x}\in\mathbb{R}^d$, define $\|\mathbf{x}\| = (\sum_{i=1}^{d} x_i^{2})^{1/2}$, and $\|\mathbf{x}\|_\infty = \max_{i\le d}|x_i|$. For a matrix $A \in \mathbb{R}^{m \times n}$, the operator norm is $\lVert A \rVert = \sup_{\lVert \mathbf{x} \rVert = 1} \lVert A\mathbf{x} \rVert$, and the max norm is $\lVert A \rVert_{\max} = \max_{1 \leq i \leq m, 1 \leq j \leq n} |A_{ij}|$. For a bounded measurable function $g$, $\|g\|_\infty = \sup_{x}|g(x)|$. For a random variable $X$ with distribution $P_X$, denote the population $L_2$ norm by $\lVert X \rVert = (\int \lVert x \rVert^2 d P_X(x))^{1/2}$; and given a random sample $\mathcal{D} = \{X_1, \cdots, X_n\}$, denote the empirical $L_2$ norm by $\lVert X \rVert_{\mathcal{D}} = (n^{-1}\sum_{i = 1}^n \lVert X_i \rVert^2)^{1/2}$.

Asymptotics. For reals sequences $a_n \ll b_n$ (or $a_n = o(b_n)$) if $\limsup_{n\to\infty} \frac{|a_n|}{|b_n|} = 0$; $|a_n| \lesssim |b_n|$ (or $a_n = O(b_n)$) if there exists some constant $C$ and $N > 0$ such that $n > N$ implies $|a_n| \leq C |b_n|$. For sequences of random variables $a_n = o_{\mathbb{P}}(b_n)$ if $\operatorname{plim}_{n \rightarrow \infty}\frac{|a_n|}{|b_n|} = 0, |a_n| \lesssim_{\mathbb{P}} |b_n|$ if $\limsup_{M \rightarrow \infty} \limsup_{n \rightarrow \infty} \mathbb{P}[|\frac{a_n}{b_n}| \geq M] = 0$.

Other. $\mathds{1}(\cdot)$ denotes the indicator function. For two random variables $X$ and $Y$, $X \protect\mathpalette{\protect\independenT}{\perp} Y$ means $X$ and $Y$ are independent. For $x \in \mathbb{R}$, $\lfloor x \rfloor$ and $\lceil x \rceil$ denote the floor and ceiling of $x$ respectively. $\mathsf{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ denotes the Gaussian distribution with mean $\boldsymbol{\mu}$ and covariance matrix $\boldsymbol{\Sigma}$. $\mathsf{Beta}(\alpha, \beta)$ denotes the Beta distribution with parameter $(\alpha, \beta)$. A stochastic process $\{B(t), 0 \leq t \leq 1\}$ is a Brownian bridge, if $B$ is a continuous Gaussian process with $\mathbb{E}[B(t)] = 0$, and $\mathbb{E}[B(t)B(s)] = \min\{t,s\} - ts$.

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Proof of Main Paper Results

itemize• Proof of Theorem 1: The conclusions follow from Corollary (ref), Corollary (ref), Theorem (ref), Theorem (ref), Corollary (ref), and Corollary (ref). • Proof of Theorem 2: The conclusions follow from Corollary (ref), Corollary (ref), Theorem (ref), Theorem (ref), Corollary (ref), and Corollary (ref). • Proof of Theorem 3: The conclusions follow from Corollary (ref), Corollary (ref), Theorem (ref), Theorem (ref), Corollary (ref), and Corollary (ref). • Proof of Lemma 4: See Section (ref).

Constant Regression Model

This section is self-contained, and substantially improves on the results reported in Cattaneo-Klusowski-Tian_2022_arXiv. The results presented herein are of independent interest in regression estimation settings, and they also offer a gentle introduction to the more technically involved results discussed in Section (ref).

Consider the canonical regression model where the observed data $\{(y_i,\mathbf{x}_i^T) : i = 1, 2, \dots n\}$ is a random sample satisfying

equation[equation omitted — 211 chars of source]

with $\mathbf{x}_i = (x_{i1}, x_{i2}, \dots, x_{ip})^T$ a vector of $p$ covariates taking values on some support set $\mathcal{X}$.

assumption[Location Regression Model] $\mathcal{D} = \{(y_i,\mathbf{x}_i^T) : 1 \leq i \leq n\}$ is a random sample such that the following conditions hold for all $i = 1,2, \cdots, n$, satisfying Equation (ref) and the following: \begin{enumerate} • $y_i = \mu(\mathbf{x}_i)+ \varepsilon_i$, with $\mathbb{E}[\varepsilon_i | \mathbf{x}_i]=0$ and $\mathbf{x}_i \protect\mathpalette{\protect\independenT}{\perp} \varepsilon_i$. • $ \mu(\mathbf{x}) = c$ for all $\mathbf{x} \in \mathcal{X}\subseteq\mathbb{R}^p$, where $c$ is some constant. • $x_{i,1}, \ldots, x_{i,p}$ are independent and continuously distributed. • There exists $\alpha > 0$ such that $\mathbb{E}[\exp(\lambda\varepsilon_i)] < \infty$ for all $|\lambda| < 1/\alpha $ and $\sigma^2 = \mathbb{E}[\varepsilon_i^2] > 0$. \end{enumerate}

In what follows, we denote by $P_X$ the marginal distribution of $\mathbf{x}_i$.

Now we illustrate the CART estimation strategy. Given any tree $\mathsf{T}$, the CART estimator is given as follows:

definition[CART Estimate] Suppose $\mathsf{T}$ is the tree used, and $\mathcal{D}_{\mu}=\{(y_i, \mathbf{x}^\top_i):i=1,2,\dots,n_{\mu}\}$, with $n_{\mu}\leq n$, is the dataset used. Let $\mathsf{t}$ be the unique terminal node in $\mathsf{T}$ containing $\mathbf{x} \in \mathcal{X}$. The CART estimator is \begin{align*} \hat\mu(\mathbf{x}; \mathsf{T},\mathcal{D}_{\mu}) = \frac{1}{n(\mathsf{t})} \sum_{i:\mathbf{x}_i \in \mathsf{t}} y_i, \end{align*} where $n(\mathsf{t}) = \sum_{i=1}^{n_{\mu}} \mathds{1}(\mathbf{x}_i \in \mathsf{t})$ is the “local” sample sizes. In case $n(\mathsf{t}) = 0$, take $\hat\mu(\mathbf{x}; \mathsf{T},\mathcal{D}_{\mu}) = 0$.
definition[Tree Construction] Given a dataset $\mathcal{D}_{\mathsf{T}} = \{(y_i,\mathbf{x}^\top_i):i=1,2,\dots,n_{\mathsf{T}}\}$, with $n_{\mathsf{T}}\leq n$, a parent node $\mathsf{t}$ in the tree (i.e., a region in $\mathcal{X}$) is divided into two child nodes, $\mathsf{t}_{\mathtt{L}}$ and $\mathsf{t}_{\mathtt{R}}$, by minimizing the sum-of-squares error (SSE), \begin{align} \min_{1\leq j\leq p} \min_{\beta_{\mathtt{L}},\beta_{\mathtt{R}},\varsigma \in \mathbb{R}} \sum_{\mathbf{x}_i \in \mathsf{t}} \big(y_{i} - \beta_{\mathtt{L}} \mathds{1}(x_{ij} \leq \varsigma) - \beta_{\mathtt{R}} \mathds{1}(x_{ij} > \varsigma) \big)^2, \end{align} where $(\beta_{\mathtt{L}}, \beta_{\mathtt{R}}, \varsigma, j)$ denote the two child nodes outputs, split point, and split direction, respectively. With at least one split, the final CART tree is denoted by $\mathsf{T}(\mathcal{D}_{\mathsf{T}})$.
definition[Sample Splitting] Recall Definition (ref) and Definition (ref), and that $\mathcal{D} = \{(y_i, \mathbf{x}^\top_i):i=1,2,\dots,n\}$ is the available random sample. \begin{itemize} • No Sample Splitting (NSS): The dataset $\mathcal{D}$ is used for both the tree construction and the treatment effect estimation, that is, $\mathcal{D}_{\mathsf{T}} = \mathcal{D}$ and $\mathcal{D}_{\mu} = \mathcal{D}$. The CART tree estimator is \begin{align*} \hat\mu^{\mathtt{NSS}}(\mathbf{x}) &= \hat\mu(\mathbf{x}; \mathsf{T}(\mathcal{D}),\mathcal{D}). \end{align*} • Honesty (HON): The dataset $\mathcal{D}$ is divided in two independent datasets $\mathcal{D}_{\mathsf{T}}$ and $\mathcal{D}_{\mu}$ with sample sizes $n_{\mathsf{T}}$ and $n_{\mu}$, respectively, and satisfying $n \lesssim n_{\mathsf{T}}, n_{\mu} \lesssim n$. The CART tree estimator is \begin{align*} \hat{\mu}^{\mathtt{HON}}(\mathbf{x}) &= \hat{\mu}(\mathbf{x}; \mathsf{T}(\mathcal{D}_{\mathsf{T}}),\mathcal{D}_{\mu}). \end{align*} \end{itemize}
definition[$\mathbf{X}$-Adaptive Estimation] Recall Definition (ref) and Definition (ref), and that $\mathcal{D} = \{(y_i, \mathbf{x}^\top_i):i=1,2,\dots,n\}$ is the available random sample. \begin{enumerate} • The dataset $\mathcal{D}$ is divided into $K+1$ datasets $(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K},\mathcal{D}_{\mu})$, with sample sizes $(n_{\mathsf{T}_1},\ldots,n_{\mathsf{T}_K},n_\mu)$, respectively, and satisfying $n_{\mathsf{T}_1} = \cdots = n_{\mathsf{T}_K} = n_{\mu}$ (possibly after dropping $n \mod K$ data points at random). For each of the datasets $\mathcal{D}_{\mathsf{T}_j} = \{(y_i, \mathbf{x}_i^{\top}):i=1,\ldots,n_{\mathsf{T}_j}\}$, $j=1,\ldots,K$, replace $\{y_i:i=1,\ldots,n_{\mathsf{T}_j}\}$ with independent copies $\{\tilde{y}_i:i=1,\ldots,n_{\mathsf{T}_j}\}$, while keeping the same $\{\mathbf{x}_i :i=1,\ldots,n_{\mathsf{T}_j} \}$. • The maximal decision tree of depth $K$, $\mathsf{T}_K(\mathcal{D}_{\mathsf{T}_1}, \cdots, \mathcal{D}_{\mathsf{T}_K})$, is obtained by iterating $K$ times the $l \in \{\mathtt{DIM}, \mathtt{IPW}, \mathtt{SSE}\}$ splitting procedures in Definition (ref), each time splitting all terminal nodes until (i) the node contains a single data point $(y_i, \mathbf{x}_i^{\top})$, or (ii) the input values $\mathbf{x}_i$ and/or all $y_i$ within the node are the same. • The $\mathbf{X}$-adaptive estimator is \begin{align*} \hat\mu^{\mathtt{X}}(\mathbf{x}; K) &= \hat\mu(\mathbf{x}; \mathsf{T}_K(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K}),\mathcal{D}_\mu). \end{align*} \end{enumerate}

No Sample Splitting

We start from the no sample splitting ($\mathtt{NSS}$) case, and characterize the location of the first split.

Decision Stumps.

For each variable $ j = 1, 2, \dots, p $, let $\pi_j$ be the permutation such that $ x_{\pi_{j}(i),j} $ is non-decreasing in the index $ i = 1, 2, \dots, n$. Then, minimizing Equation (ref) can be equivalently recasted as maximizing the so-called impurity gain:

equation[equation omitted — 630 chars of source]

where $\bar{y}_{\mathrm{t}} = n(\mathrm{t})^{-1} \sum_{\mathbf{x}_i \in \mathrm{t}} y_i \mathds{1}(\mathbf{x}_i \in \mathrm{t})$. We can show this is also equivalent to maximizing the conditional variance given the split:

align[align omitted — 142 chars of source]

We start by considering the case when the tree is depth one ($K=1$), i.e., a decision stump. Then optimization objectives are equivalent to choosing a splitting coordinate $\hat \jmath$, and a splitting index $\hat \imath$ such that

align*[align* omitted — 223 chars of source]

The tree output can then be written as

equation[equation omitted — 247 chars of source]

where $x_{\hat\jmath}$ denotes the value of the $\hat\jmath$-th component of $\mathbf{x}$.

The following theorem formally (and very precisely) characterizes the regions of the support $\mathcal{X}$ where the first CART split index $ \hat\imath $, at the root node, has non-vanishing probability of realizing. As a consequence, the theorem also characterizes the effective sample size of the resulting cells (recall the data is ordered so that $ \hat \mu = x_{\hat\imath \hat\jmath} $ and hence $ \hat\imath = \#\{\mathbf{x}_i : x_{i\hat \jmath} \leq \hat \mu \} $).

theorem[Imbalanced Splits] Suppose Assumption (ref) holds, and let $(\hat\imath, \hat\jmath)$ be the CART split index and split direction at the root node. For each $ a,b \in (0, 1) $ with $ a < b $, and $\ell \in [p]$, we have \begin{equation} \liminf_{n\to\infty} \mathbb{P}\big( n^{a} \leq \hat\imath \leq n^{b}, \hat \jmath = \ell \big) = \liminf_{n\to\infty} \mathbb{P}\big( n-n^{b} \leq \hat\imath \leq n-n^{a}, \hat \jmath = \ell \big) \geq \frac{b-a}{2pe}, \end{equation} which implies \begin{align*} \liminf_{n\to\infty} \mathbb{P}\big( n^{a} \leq \hat\imath \leq n^{b} \big) = \liminf_{n\to\infty} \mathbb{P}\big( n-n^{b} \leq \hat\imath \leq n-n^{a} \big) \geq \frac{b-a}{2 e}. \end{align*}
theorem[Convergence Rates for Decision Stumps] Suppose Assumption (ref) holds. Suppose the CART tree has depth $K = 1$. Then for any $a,b \in (0, 1)$ with $ a < b $, we have \begin{equation} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat{\mu}^{\mathtt{NSS}}(\mathbf{x}) - \mu| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq \frac{b}{e}, \end{equation} and suppose w.l.o.g. that $\mathbf{x}_i \thicksim \mathsf{Uniform}([0,1]^p)$, then \begin{equation} \liminf_{n\to\infty} \inf_{\mathbf{x}\in \mathcal{X}_n} \mathbb{P}\Big(|\hat{\mu}^{\mathtt{NSS}}(\mathbf{x}) - \mu| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Big) \geq \frac{b-a}{2e}, \end{equation} where $ \mathcal{X}_n = \{\mathbf{x} \in [0,1]^p: x_j = o(1) n^{a-1} \text{ or } 1 - x_j = o(1) n^{a-1} \text{ for some } j \in [p]\}$.

Deep Trees.

We will show that the imbalanced split issue is inherited from the decision stumps to trees of arbitrary depth.

theorem[Convergence Rates for Deep Trees] Suppose Assumption (ref) holds. Then for any $b \in (0, 1) $, we have $$\liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat{\mu}^{\mathtt{NSS}}(\mathbf{x}) - \mu| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq b/e.$$

Therefore, decision trees grown with CART methodology cannot converge faster than any polynomial-in-$n$, when uniformity over the full support of the data $\mathcal{X}$, and over possible data generating processes, is of interest.

However, for the $L_2$-risk we still have the following positive result. This is because the small cells that leads to issues in uniform consistency will have a small measure by $\mathbb{P}_X$.

theorem[$L_2$ Consistency -- NSS] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat{\mu}^{\mathtt{NSS}}(\mathbf{x}) - \mu)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^4 \log(np)}{n}, \end{align*} where $C$ is a positive constant that only depends on $\sigma^2$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg(\int_{\mathcal{X}} (\hat{\mu}^{\mathtt{NSS}}(\mathbf{x}) - \mu)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^4 \log(np)}{n} \bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on the distribution of $\varepsilon_i$.

Sample Splitting

For sample splitting strategy, we also present a lower bound on uniform consistency and an upper bound on $L_2$ consistency.

theoremSuppose Assumption (ref) holds. Then for any $b \in (0, 1) $, we have \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat{\mu}^{\mathtt{HON}}(\mathbf{x}) - \mu| \geq \frac{C\mathbb{E}[|y_i - \mu|]}{ n^{b/2}}\Bigg) \geq C \frac{\mathbb{E}[|y_i - \mu|^2]}{\mathbb{V}[y_i]} b, \end{equation*} where $C$ is some constant only depending on $\liminf_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\mu}}$ and $\limsup_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\mu}}$.
theorem[$L_2$ Consistency -- HON] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat{\mu}^{\mathtt{HON}}(\mathbf{x}) - \mu)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^5 }{n}, \end{align*} provided $\rho^{-1} \leq \frac{n_{\mathsf{T}}}{n_{\mu}} \leq \rho$ for some $\rho \in (0,1)$, and $C$ is a positive constant that only depends on $\sigma^2$ and $\rho$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg(\int_{\mathcal{X}} (\hat{\mu}^{\mathtt{HON}}(\mathbf{x}) - \mu)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^5}{n} \bigg) = 0, \end{align*} where $C^{\prime}$ is some constant only depending on $\rho$ and the distribution of $\varepsilon_i$.

Compared to Theorem (ref), the lower bound on the LHS of Theorem (ref) that we characterize has one less $\sqrt{(2 + o(1)) \log \log (n)}$. Compared to Theorem (ref), the upper bound on the RHS of Theorem (ref) has $\log(np)$ replaced by $\log(n)$. These changes are due to the sample splitting strategy.

X-adaptive Tree

For X-adaptive trees, we leverage the decision stump result from Theorem (ref) using an iterative argument to infer inconsistency of trees of depth $K_n \gtrsim \log \log (n)$.

theorem[Pointwise Inconsistency] Suppose Assumption (ref) holds. If $\liminf_{n \to \infty} \frac{K_n}{\log\log(n)} > 0$, then there exists a positive constant $C$ not depending on $n$ such that $$ \liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x} \in \mathcal{X}}|\hat{\mu}^{\mathtt{X}}(\mathbf{x}; K_n)-\mu| > C \Bigg) > 0. $$

Since we keep the $\mathbf{x}_i$'s and refresh the $(d_i, y_i)$'s, the tree estimator has a simple form condition on $\mathbf{x}_i$'s. Hence a direct variance calculation gives us the following $L_2$-consistency result.

theorem[L2 Consistency -- X] Suppose Assumption (ref) holds. Then \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat{\mu}^{\mathtt{X}}(\mathbf{x}; K) - \mu)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq \frac{2^{K+1} (K+1) \sigma^2}{n + 1}. \end{align*}

Using the same argument as Theorem (ref), we can show

align*[align* omitted — 170 chars of source]

where $C$ is a positive constant that only depends on $\sigma^2$. The direct variance calculation allows us to remove extra $\operatorname{poly-log}$ terms.

Heterogeneous Causal Effect Estimation

In this section, we consider the heterogeneous causal effect estimation problem from the main paper. The assumptions on the data generating process and the definitions of causal trees are the same as in the main paper. For completeness, we include them here:

assumption[Data Generating Process] $\mathcal{D} = \{(y_i,d_i,\mathbf{x}^\top_i): 1 \leq i \leq n\}$ is a random sample, where $y_i = d_i y_i(1) + (1 - d_i) y_i(0)$, $\mathbf{x}_i = (x_{i,1}, \ldots, x_{i,p})^{\top}$, and the following conditions hold for all $d = 0,1$ and $i = 1,2, \ldots, n$. \begin{enumerate} • $(y_i(0),y_i(1),\mathbf{x}_i) \protect\mathpalette{\protect\independenT}{\perp} d_i$, and $\xi = \mathbb{P}[d_i = 1] \in (0,1)$. • $y_i(d) = \mu_d(\mathbf{x}_i) + \varepsilon_i(d)$, with $\mathbb{E}[\varepsilon_i(d)|\mathbf{x}_i]=0$ and $\mathbf{x}_i \protect\mathpalette{\protect\independenT}{\perp} \varepsilon_i(d)$. • $\mu_d(\mathbf{x}) = c_d$ for all $\mathbf{x} \in \mathcal{X}$, where $c_d$ is some constant, and $\mathcal{X}$ is the support of $\mathbf{x}_i$. • $x_{i,1}, \ldots, x_{i,p}$ are independent and continuously distributed. • There exists $\alpha > 0$ such that $\mathbb{E}[\exp(\lambda\varepsilon_i(d))] < \infty$ for all $|\lambda| < 1/\alpha$ and $\mathbb{E}[\varepsilon_i^2(d)] > 0$. \end{enumerate}

And the causal trees are constructed based on the following rules:

definition[CATE Estimators] Suppose $\mathsf{T}$ is the tree used, and $\mathcal{D}_\tau=\{(y_i,d_i,\mathbf{x}^\top_i):i=1,2,\dots,n_{\tau}\}$, with $n_{\tau}\leq n$, is the dataset used. Let $\mathsf{t}$ be the unique terminal node in $\mathsf{T}$ containing $\mathbf{x} \in \mathcal{X}$. \begin{itemize} • The Difference-in-Means (DIM) estimator is \begin{align*} \hat\tau_\mathtt{DIM}(\mathbf{x}; \mathsf{T},\mathcal{D}_\tau) = \frac{1}{n_{1}(\mathsf{t})} \sum_{i:\mathbf{x}_i \in \mathsf{t}} d_i y_i - \frac{1}{n_{0}(\mathsf{t})} \sum_{i:\mathbf{x}_i \in \mathsf{t}} (1-d_i) y_i, \end{align*} where $n_{d}(\mathsf{t}) = \sum_{i=1}^{n_{\tau}} \mathds{1}(\mathbf{x}_i \in \mathsf{t}, d_i = d)$, for $d=0,1$, are the “local” sample sizes. In case $n_0(\mathsf{t}) = 0$ or $n_1(\mathsf{t}) = 0$, take $\hat\tau_\mathtt{DIM}(\mathbf{x}; \mathsf{T},\mathcal{D}_\tau) = 0$. • The Inverse Probability Weighting (IPW) estimator is \begin{align*} \hat\tau_\mathtt{IPW}(\mathbf{x}; \mathsf{T},\mathcal{D}_\tau) = \frac{1}{n(\mathsf{t})} \sum_{i:\mathbf{x}_i \in \mathsf{t}} \frac{d_i-\xi}{\xi(1-\xi)} y_i, \end{align*} where $n(\mathsf{t}) = n_{0}(\mathsf{t}) + n_{1}(\mathsf{t}) = \sum_{i=1}^{n_{\tau}} \mathds{1}(\mathbf{x}_i \in \mathsf{t})$ is the “local” sample size. In case $n(\mathsf{t}) = 0$, take $\hat\tau_\mathtt{IPW}(\mathbf{x}; \mathsf{T},\mathcal{D}_\tau) = 0$. \end{itemize}
definition[Tree Construction] Suppose $\mathcal{D}_{\mathsf{T}} = \{(y_i,d_i,\mathbf{x}^\top_i):i=1,2,\dots,n_{\mathsf{T}}\}$, with $n_{\mathsf{T}}\leq n$, is the dataset used to construct the tree $\mathsf{T}$. \begin{itemize} • Variance Maximization: A parent node $\mathsf{t}$ (i.e., a terminal node partitioning $\mathcal{X}$) in a previous tree $\mathsf{T}^{\prime}$ is divided into two child nodes, $\mathsf{t}_{\mathtt{L}}$ and $\mathsf{t}_{\mathtt{R}}$, forming the new tree $\mathsf{T}$, by maximizing \begin{align} \frac{n(\mathsf{t}_{\mathtt{L}})n(\mathsf{t}_{\mathtt{R}})}{n(\mathsf{t})} \Big(\hat\tau_l(\mathsf{t}_{\mathtt{L}}; \mathsf{T},\mathcal{D}_{\mathsf{T}}) - \hat\tau_l(\mathsf{t}_{\mathtt{R}}; \mathsf{T},\mathcal{D}_{\mathsf{T}})\Big)^2, \qquad l \in \{\mathtt{DIM},\mathtt{IPW}\}. \end{align} With at least one split, the two final causal trees are denoted by $\mathsf{T}^{\mathtt{DIM}}(\mathcal{D}_{\mathsf{T}})$ and $\mathsf{T}^{\mathtt{IPW}}(\mathcal{D}_{\mathsf{T}})$, respectively, for $l \in \{\mathtt{DIM},\mathtt{IPW}\}$. • SSE Minimization: A parent node $\mathsf{t}$ (i.e., a terminal node partitioning $\mathcal{X}$) in the previous tree $\mathsf{T}^{\prime}$ is divided into two child nodes, $\mathsf{t}_{\mathtt{L}}$ and $\mathsf{t}_{\mathtt{R}}$, forming the next tree $\mathsf{T}$, by solving \begin{align} \min_{a_{\mathtt{L}}, b_{\mathtt{L}}, a_{\mathtt{R}}, b_{\mathtt{R}} \in \mathbb{R}} \sum_{\mathbf{x}_i \in \mathsf{t}_{\mathtt{L}}} (y_i - a_{\mathtt{L}} - b_{\mathtt{L}} d_i)^2 + \sum_{\mathbf{x}_i \in \mathsf{t}_{\mathtt{R}}} (y_i - a_{\mathtt{R}} - b_{\mathtt{R}} d_i)^2, \end{align} where only the data $\mathcal{D}_{\mathsf{T}}$ is used. With at least one split, the final causal tree is denoted by $\mathsf{T}^{\mathtt{SSE}}(\mathcal{D}_{\mathsf{T}})$. \end{itemize}
definition[Sample Splitting and Estimators] Recall Definition (ref) and Definition (ref), and that $\mathcal{D} = \{(y_i, \mathbf{x}^\top_i, d_i):i=1,2,\dots,n\}$ is the available random sample. \begin{itemize} • No Sample Splitting (NSS): The dataset $\mathcal{D}$ is used for both the tree construction and the treatment effect estimation, that is, $\mathcal{D}_{\mathsf{T}} = \mathcal{D}$ and $\mathcal{D}_{\tau} = \mathcal{D}$. The causal tree estimators are \begin{align*} \hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{DIM}}(\mathcal{D}),\mathcal{D}),\\ \hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) &= \hat\tau_{\mathtt{IPW}}(\mathbf{x}; \mathsf{T}^{\mathtt{IPW}}(\mathcal{D}),\mathcal{D}), \quad and\\ \hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{SSE}}(\mathcal{D}),\mathcal{D}), \end{align*} • Honesty (HON): The dataset $\mathcal{D}$ is divided in two independent datasets $\mathcal{D}_{\mathsf{T}}$ and $\mathcal{D}_{\tau}$ with sample sizes $n_{\mathsf{T}}$ and $n_\tau$, respectively, and satisfying $n \lesssim n_{\mathsf{T}}, n_{\tau} \lesssim n$. The causal tree estimators are \begin{align*} \hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}(\mathbf{x}) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{DIM}}(\mathcal{D}_{\mathsf{T}}),\mathcal{D}_\tau),\\ \hat\tau^{\mathtt{HON}}_{\mathtt{IPW}}(\mathbf{x}) &= \hat\tau_{\mathtt{IPW}}(\mathbf{x}; \mathsf{T}^{\mathtt{IPW}}(\mathcal{D}_{\mathsf{T}}),\mathcal{D}_\tau), \quad and\\ \hat\tau^{\mathtt{HON}}_{\mathtt{SSE}}(\mathbf{x}) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{SSE}}(\mathcal{D}_{\mathsf{T}}),\mathcal{D}_\tau). \end{align*} \end{itemize}

While the estimators $\hat\tau^{\mathtt{NSS}}_{l}(\mathbf{x})$ and $\hat\tau^{\mathtt{HON}}_{l}(\mathbf{x})$, $l\in\{\mathtt{DIM},\mathtt{IPW},\mathtt{SSE}\}$ depend on the depth of the tree construction used, our the notation does not make this dependence explicit because our results only require (at least) one single split.

X-Adaptive Trees.

definition[$\mathbf{X}$-Adaptive Estimation] Recall Definition (ref) and Definition (ref), and that $\mathcal{D} = \{(y_i, \mathbf{x}^\top_i, d_i):i=1,2,\dots,n\}$ is the available random sample. \begin{enumerate} • The dataset $\mathcal{D}$ is divided into $K+1$ datasets $(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K},\mathcal{D}_{\tau})$, with sample sizes $(n_{\mathsf{T}_1},\ldots,n_{\mathsf{T}_K},n_\tau)$, respectively, and satisfying $n_{\mathsf{T}_1} = \cdots = n_{\mathsf{T}_K} = n_{\tau}$ (possibly after dropping $n \mod K$ data points at random). For each of the datasets $\mathcal{D}_{j} = \{(y_i, d_i, \mathbf{x}_i^{\top}):i=1,\ldots,n_{\mathsf{T}_j}\}$, $j=1,\ldots,K$, replace $\{(y_i, d_i):i=1,\ldots,n_{\mathsf{T}_j}\}$ with independent copies $\{(\tilde{y}_i, \tilde{d}_i):i=1,\ldots,n_{\mathsf{T}_j}\}$, while keeping the same $\{\mathbf{x}_i :i=1,\ldots,n_{\mathsf{T}_j} \}$. • The maximal decision tree of depth $K$, $\mathsf{T}^{l}_K(\mathcal{D}_{\mathsf{T}_1}, \cdots, \mathcal{D}_{\mathsf{T}_K})$, is obtained by iterating $K$ times the $l \in \{\mathtt{DIM}, \mathtt{IPW}, \mathtt{SSE}\}$ splitting procedures in Definition (ref), each time splitting all terminal nodes until (i) the node contains a single data point $(y_i, d_i, \mathbf{x}_i^{\top})$, or (ii) the input values $\mathbf{x}_i$ and/or all $(d_i, y_i)$ within the node are the same. • The $\mathbf{X}$-adaptive estimators are \begin{align*} \hat\tau^{\mathtt{X}}_{\mathtt{DIM}}(\mathbf{x}; K) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{DIM}}_{K}(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K}),\mathcal{D}_\tau),\\ \hat\tau^{\mathtt{X}}_{\mathtt{IPW}}(\mathbf{x}; K) &= \hat\tau_{\mathtt{IPW}}(\mathbf{x}; \mathsf{T}^{\mathtt{IPW}}_{K}(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K}),\mathcal{D}_\tau), \quad and\\ \hat\tau^{\mathtt{X}}_{\mathtt{SSE}}(\mathbf{x}; K) &= \hat\tau_{\mathtt{DIM}}(\mathbf{x}; \mathsf{T}^{\mathtt{SSE}}_{K}(\mathcal{D}_{\mathsf{T}_1},\ldots,\mathcal{D}_{\mathsf{T}_K}),\mathcal{D}_\tau). \end{align*} \end{enumerate}

IPW Estimator

The transformed outcomes $y_i \frac{d_i - \xi}{\xi (1 - \xi)}, 1 \leq i \leq n,$ are i.i.d, with

align*[align* omitted — 150 chars of source]

and

align*[align* omitted — 268 chars of source]

Assumption (ref) implies $\mathbb{E}[\exp(\lambda \tilde{\varepsilon}_i)] < \infty$ for all $|\lambda| \leq 1 / \beta$ with $\beta$ only depending on $\xi$ and $\alpha$, and $\mathbb{E}[\tilde{\varepsilon}_i^2] >0$. Hence the following results are immediate corollaries from the results in Section (ref).

No Sample Splitting

coro[Imbalanced Split] Suppose Assumption (ref) holds. Then for each $a,b \in (0,1)$ with $a < b$, for every $\ell \in [p]$, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\big( n^{a} \leq \hat\imath \leq n^{b}, \hat \jmath = \ell \big) = \liminf_{n\to\infty} \mathbb{P}\big( n-n^{b} \leq \hat\imath \leq n-n^{a}, \hat \jmath = \ell \big) \geq \frac{b-a}{2pe}. \end{equation*}
coro[Stump] Suppose Assumption (ref) holds, and the tree has depth $K = 1$. Then for any $ a,b \in (0, 1) $ with $ a < b $, we have \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq \frac{b}{e}, \end{equation*} where $\sigma^2 = \mathbb{V} \Big[\frac{d_i y_i(1)}{\xi} + \frac{(1 - d_i) y_i(0)}{1 - \xi}\Big]$. Moreover, if $\mathbf{x}_i$ has a density that is continuous and positive on $[0,1]^p$, then \begin{equation*} \liminf_{n\to\infty} \inf_{\mathbf{x}\in \mathcal{X}_n} \mathbb{P}\Big(|\hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Big) \geq \frac{b-a}{2e}, \end{equation*} where $ \mathcal{X}_n = \{\mathbf{x} \in [0,1]^p: x_j = o(1) n^{a-1} \text{ or } 1 - x_j = o(1) n^{a-1} \text{ for some } j \in [p]\}$.
coro[Rates] Suppose Assumption (ref) holds. Then for any $b \in (0, 1) $ and arbitrary depth tree, we have $$\liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq \frac{b}{e}.$$
coro[$L_2$ Consistency -- NSS] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^4 \log(np)}{n}, \end{align*} where $C$ is a positive constant that only depends on the distribution of $\tilde{\varepsilon}_i = y_i \frac{d_i - \xi}{\xi(1 - \xi)} - \tau$. Moreover, \begin{align*} \limsup_{n \to \infty}\mathbb{P} \bigg(\int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{IPW}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^4 \log(np)}{n}\bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on the distribution of $\tilde{\varepsilon}_i$.

Sample Splitting

coro[Honest Causal Output] Suppose Assumption (ref) holds. Then for any $b \in (0, 1) $, we have \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{HON}}_{\mathtt{IPW}}(\mathbf{x}) - \tau| \geq \frac{C\mathbb{E}[|\tilde{\varepsilon}_i|]}{8 n^{b/2}}\Bigg) \geq C \frac{\mathbb{E}[|\tilde{\varepsilon}_i|^2]}{\mathbb{V}[\tilde{\varepsilon}_i]} b, \end{equation*} where $C$ is some constant only depending on the distribution of $\tilde{\varepsilon}_i = y_i \frac{d_i - \xi}{\xi(1 - \xi)} - \tau$, $\liminf_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$ and $\limsup_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$.
coro[$L_2$ Consistency -- HON] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{IPW}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^5}{n}, \end{align*} provided $\rho^{-1} \leq \frac{n_{\mathsf{T}}}{n_{\tau}} \leq \rho$ for some $\rho \in (0,1)$, and $C$ is some constant only depending on the distribution of $\tilde{\varepsilon}_i = y_i \frac{d_i - \xi}{\xi(1 - \xi)} - \tau$ and $\rho$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg( \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{IPW}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^5}{n} \bigg) = 0, \end{align*} where $C^{\prime}$ is some constant only depending on the distribution of $\tilde{\varepsilon}_i$ and $\rho$.

X-adaptive Tree

coro[Honest CART+] Suppose Assumption (ref) holds. Suppose $\liminf_{n \to \infty} \frac{K_n}{\log \log (n)} > 0$. Then, there exists a positive constant $C$ not depending on $n$ such that \begin{align*} \liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x} \in \mathcal{X}}|\hat\tau^{\mathtt{X}}_{\mathtt{IPW}}(\mathbf{x}; K_n)-\tau| > C \Bigg) > 0. \end{align*}
coro[L2 Consistency -- X] Suppose Assumption (ref) holds. Then \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{X}}_{\mathtt{IPW}}(\mathbf{x}; K) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^{K} K \sigma^2}{n}, \end{align*} where $C$ is some constant only depending on the distribution of $\tilde{\varepsilon}_i = y_i \frac{d_i - \xi}{\xi(1 - \xi)} - \tau$.

DIM Estimator

The $\mathtt{DIM}$ estimator can not be directly written as a regression tree with transformed outcome. However, we show that it can be approximated by an $\mathtt{IPW}$-tree. More specifically, we view the split criterion with different splitting index and coordinate as an empirical process, and show that the split criterion for $\mathtt{DIM}$ and $\mathtt{IPW}$ approximate each other.

No Sample Splitting

Approximation Results on Decision Stumps.

Denote by $\pi_{\ell}$ permutation of index $[n]$ such that $x_{\pi_{\ell}(1),\ell} \leq x_{\pi_{\ell}(2),\ell} \leq \cdots \leq x_{\pi_{\ell}(n),\ell}$, $1 \leq \ell \leq p$. Consider the split criterion for the regression and ipw trees when splitting at the root note when $\#\{\mathbf{x}_{\pi_{\ell}(i)} \in t_L\} = k$: For $1 \leq \ell \leq p$, $1 \leq k \leq n$, consider

align*[align* omitted — 303 chars of source]

where

align*[align* omitted — 981 chars of source]

Notice that if we replace $\varepsilon_{\pi_{\ell}(i)}$ by $y_{\pi_{\ell}(i)}$, we would get $\hat \tau_L^\mathtt{IPW}$ (or $\hat \tau_R^\mathtt{IPW}$) instead of $\bar \tau_L^\mathtt{IPW}$ (or $\bar \tau_R^\mathtt{IPW}$). But putting $\varepsilon_{\pi_\ell(i)}$ here allows us to approximate the $\mathscr{I}^\mathtt{DIM}(\cdot, \ell)$ processes.

The optimization objective based on Definition (ref) for the regression based estimator with variance maximization is equivalent to choosing a splitting coordinate $\hat \jmath_{\text{DIM}}$, and a splitting index $\hat \imath_{\text{DIM}}$ such that

align*[align* omitted — 361 chars of source]

that maximizes

align*[align* omitted — 118 chars of source]

that is,

align*[align* omitted — 139 chars of source]

A technical aspect is to control for fluctuations of objects of the form $\frac{\sum_{i = 1}^k d_{\pi_{\ell}(i)} y_{\pi_{\ell}(i)}}{\sum_{i = 1}^k d_{\pi_{\ell}(i)}}$, for which we will use a truncation argument that requires $\sum_{i = 1}^k d_{\pi_{\ell}(i)} \geq r_n$ with $r_n \to \infty$. This gives the following lemma:

lemma[Approximation Error] Suppose Assumption (ref) holds. Let $(r_n)_{n \in \mathbb{N}}$ be a sequence of real numbers such that $r_n \rightarrow \infty$. Then \begin{align*} \max_{1 \leq \ell \leq p}\max_{r_n \leq k < n - r_n} \Big|\mathscr{I}^{\mathtt{DIM}}(k,\ell) - \bar\mathscr{I}^{\mathtt{IPW}}(k,\ell)\Big| = O_{\mathbb{P}} \bigg(\frac{\log \log (n)}{\sqrt{r_n}}\bigg). \end{align*}

We also control for the truncation error:

lemma[Truncation Error] Suppose Assumption (ref) holds. Let $\rho_n$ be a sequence taking values in $(0,1)$ such that $\limsup_{n \rightarrow \infty} \rho_n \log \log (n) = 0$, and take $s_n = \exp((\log n)^{\rho_n})$. Then \begin{align*} \max_{1 \leq \ell \leq p}\max_{1 \leq k \leq s_n, n - s_n \leq k \leq n} \Big|\mathscr{I}^{\mathtt{DIM}}(k,\ell) - \bar\mathscr{I}^{\mathtt{IPW}}(k,\ell)\Big| = O_\mathbb{P}\bigg(\rho_n \log \log (n) + \frac{s_n}{n - s_n} \log \log(n)\bigg). \end{align*}

Rates for Decision Stumps.

The previous two lemmas imply that we can study $\operatorname*{arg\,max}$ of $\mathscr{I}^\mathtt{DIM}$ in terms of $\operatorname*{arg\,max}$ of $\bar \mathscr{I}^\mathtt{IPW}$. The latter is the split criterion based on CART with transformed outcome $\frac{d_i}{\xi} \varepsilon_i(1) - \frac{1 - d_i}{1 - \xi} \varepsilon_i(0)$, and results from Section (ref) can be applied.

theorem[Imbalanced Split] Suppose Assumption (ref) holds. Then for each $a,b \in (0,1)$ with $a < b$, for every $\ell \in [p]$, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\big( n^{a} \leq \hat\imath_\mathtt{DIM} \leq n^{b}, \hat \jmath_\mathtt{DIM} = \ell \big) = \liminf_{n\to\infty} \mathbb{P}\big( n-n^{b} \leq \hat\imath_\mathtt{DIM} \leq n-n^{a}, \hat \jmath_\mathtt{DIM} = \ell \big) \geq \frac{b-a}{2pe}. \end{equation*}

The issue of imbalanced cells gives rise to the slow uniform convergence rate.

theorem[Rates for Stump] Suppose Assumption (ref) holds, and the tree has depth $K = 1$. Then for any $ a,b \in (0, 1) $ with $ a < b $, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq \frac{b}{e}, \end{equation*} where $\sigma^2 = \mathbb{V}[\tilde{\varepsilon_i}]$, with $\tilde{\varepsilon}_i = \frac{d_i}{\xi} \varepsilon_i(1) - \frac{1 - d_i}{1 - \xi} \varepsilon_i(0)$. Suppose w.l.o.g. that $\mathbf{x}_i \thicksim \mathsf{Uniform}([0,1]^p)$, then \begin{equation*} \liminf_{n\to\infty} \inf_{\mathbf{x}\in \mathcal{X}_n} \mathbb{P}\Big(|\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Big) \geq \frac{b-a}{2e}, \end{equation*} where $\mathcal{X}_n = \{\mathbf{x} \in [0,1]^p: x_j = o(1) n^{a-1} \text{ or } 1 - x_j = o(1) n^{a-1} \text{ for some } j \in [p]\}$.

Deeper Trees.

We generalize the above results on decision stumps to trees of arbitrary depths.

theorem[Deeper Trees] Suppose Assumption (ref) holds. Then for any $b \in (0, 1) $, \begin{align*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq b/e. \end{align*}

In comparison to the uniform convergence rate, for $L_2$ convergence rate we can give an upper bound as follows.

theorem[$L_2$ Consistency -- NSS] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^4 \log(np)}{n}, \end{align*} where $C$ is a positive constant that only depends on the distribution of $(d_i, \varepsilon_i(0), \varepsilon_i(1))$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg( \int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^4 \log(np)}{n} \bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on the distribution of $(d_i, \varepsilon_i(0), \varepsilon_i(1))$.

Sample Splitting

With the sample splitting strategy, we also give a lower bound on uniform convergence rate and an upper bound on $L_2$ convergence rate. The difference in rates from the rates in the previous section is due to the different sample splitting strategies.

theorem[Honest Causal Output] Suppose Assumption (ref) holds. Then for any $b \in (0, 1) $, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}(\mathbf{x}) - \tau| \geq C n^{-b/2} \Bigg) \geq C \xi(1 - \xi) b. \end{equation*} where $C$ is some positive constant only depending on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$, $\liminf_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$ and $\limsup_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$.
theorem[$L_2$ Consistency -- HON] Suppose Assumption (ref) holds. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^5}{n}, \end{align*} provided $\rho^{-1} \leq \frac{n_{\mathsf{T}}}{n_{\tau}} \leq \rho$ for some $\rho \in (0,1)$, and $C$ is a positive constant that only depends on $\rho$ and the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg( \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^5}{n} \bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on $\rho$ and the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$.

X-adaptive Tree

We leverage Theorem (ref) with an iterative argument to get

theorem[CART+] Suppose Assumption (ref) holds. Suppose $\liminf_{n \to \infty} \frac{K_n}{\log \log(K_n)} > 0$. Then \begin{align*} \liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x} \in \mathcal{X}}|\hat{\tau}^{\mathcal{X}}_{\mathtt{DIM}}(\mathbf{x}; K_n)-\tau| > C \Bigg) > 0, \end{align*} where $C$ is some positive constant not depending on $n$.

A direct variance calculation gives

theorem[L2 Consistency] Suppose Assumption (ref) holds. Then \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat{\tau}^{\mathcal{X}}_{\mathtt{DIM}}(\mathbf{x}; K) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{K \; 2^{K}}{n}, \end{align*} where $C$ is some positive constant that only depends on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$.

Using the same argument as Theorem (ref), we can show

align*[align* omitted — 188 chars of source]

where $C$ is a positive constant that only depends on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$. The direct variance calculation allows us to remove extra $\operatorname{poly-log}$ terms.

SSE Estimator

While the CATE estimators given the tree of the $\mathtt{SSE}$ strategy coincides with the $\mathtt{DIM}$ strategy, the tree construction methods differ. Similar to $\mathtt{DIM}$, for $\mathtt{SSE}$ we also characterize the distribution of split index via a Gaussian approximation. Here we show the split criterion with $\mathtt{SSE}$ strategy can be approximated by the split criterion from two transformed outcome regressions, one for treatment and one for control. A careful high dimensional Gaussian approximation with respect to the geometry of simple convex sets then enables us to characterize the limiting distribution of splitting indices.

No Sample Splitting

Decision Stump.

For each variable $ j = 1, 2, \dots, p $, the data $ \{ x_{ij}: \mathbf{x}_i \in \mathrm{t} \} $ is relabeled so that $ x_{ij} $ is increasing in the index $ i = 1, 2, \dots, n(\mathrm{t}) $, where $ n(\mathrm{t}) = \#\{ \mathbf{x}_i \in \mathrm{t} \} $. The fit-based objective is to minimize

align[align omitted — 199 chars of source]

with respect to the index $ i $ and variable $ j $. Again, the maximizers are denoted by $ (\hat\imath_\text{SSE}, \hat\jmath_\text{SSE}) $, and the optimal split point $ \hat \tau $ that maximizes (ref) can be expressed as $ x_{\hat\imath_\text{SSE}, \hat\jmath_\text{SSE}}$.

To break down the criterion (ref), denote

align*[align* omitted — 539 chars of source]

Also to denote the counts compactly, $n_0 = \sum_{i = 1}^n (1 - d_i)$, $n_{L,0}(k) = \sum_{i = 1}^k (1 - d_{\pi_{\ell}(i)})$, $n_{R,0}(k) = \sum_{i = k + 1}^n (1 - d_{\pi_{\ell}(i)})$, and $n_1 = \sum_{i = 1}^n d_i$, $n_{L,1}(k) = \sum_{i = 1}^k d_{\pi_{\ell}(i)}$, $n_{R,1}(k) = \sum_{i = k + 1}^n d_{\pi_{\ell}(i)}$. Then we can show that maximizing Equation (ref) is equivalent to maximizing

align*[align* omitted — 214 chars of source]

We want to show the above empirical process can be approximated by

align*[align* omitted — 217 chars of source]

with

align*[align* omitted — 442 chars of source]

The latter can be approximated by the summation of two independent time-transformed O-U process (which is again a time-transformed O-U process), for fixed coordinate $\ell \in [p]$. More precisely, we present the approximation lemmas:

lemma[Approximation Error] Suppose Assumption (ref) holds. Let $(r_n)_{n \in \mathbb{N}}$ be a sequence of real numbers such that $r_n \rightarrow \infty$. Then \begin{align*} \max_{1 \leq \ell \leq p}\max_{r_n \leq k < n - r_n} \Big|\mathscr{I}^{SSE}(k,\ell) - \mathscr{I}^{prox}(k,\ell)\Big| = O_{\mathbb{P}} \bigg(\frac{\log \log (n)^{3/2}}{\sqrt{r_n}}\bigg). \end{align*}
lemma[Truncation Error] Suppose Assumption (ref) holds. Let $\rho_n$ be a sequence taking values in $(0,1)$ such that $\limsup_{n \rightarrow \infty} \rho_n \log \log (n) = \infty$, and take $s_n = \exp((\log n)^{\rho_n})$. Then \begin{align*} \max_{1 \leq \ell \leq p}\max_{1 \leq k \leq s_n, n - s_n \leq k \leq n} \Big|\mathscr{I}^{SSE}(k,\ell) - \mathscr{I}^{prox}(k,\ell)\Big| = O_\mathbb{P}\bigg(\rho_n \log \log (n) + \frac{s_n}{n - s_n} \log \log(n)\bigg). \end{align*}
theoremSuppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then for each $a,b \in (0,1)$ with $a < b$, for every $\ell \in [p]$, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\big( n^{a} \leq \hat\imath_SSE \leq n^{b}, \hat \jmath_SSE = \ell \big) = \liminf_{n\to\infty} \mathbb{P}\big( n-n^{b} \leq \hat\imath_SSE \leq n-n^{a}, \hat \jmath_SSE = \ell \big) \geq \frac{b-a}{2pe}. \end{equation*}
remarkWe add the condition that $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$ so that a two-dimensional Darling-Erdos theorem horvath1993maximum can be applied. We conjecture that without $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$, the conclusion still holds with a Darling-Erdos theorem for i.n.i.d O-U process, but this is out of the scope of this paper.

Notice that although the splitting criteria is different from the regression tree, once cells are given the estimator given by the fit-based tree is exactly the same as the regression tree (see Section (ref)). Hence the following results can be proved based on Theorem (ref) and the same logic as Theorem (ref) to Theorem (ref).

coro[Rates for Stump] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. For any $ a,b \in (0, 1) $ with $ a < b $, we have \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq \frac{b}{e}, \end{equation*} and suppose w.l.o.g. that $\mathbf{x}_i \thicksim \mathsf{Uniform}([0,1]^p)$, then \begin{equation*} \liminf_{n\to\infty} \inf_{\mathbf{x}\in \mathcal{X}_n} \mathbb{P}\Big(|\hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Big) \geq \frac{b-a}{2e}, \end{equation*} where $\mathcal{X}_n = \{\mathbf{x} \in [0,1]^p: x_j = o(1) n^{a-1} \text{ or } 1 - x_j = o(1) n^{a-1} \text{ for some } j \in [p]\}$, and $\sigma^2 = \mathbb{V}[\frac{d_i y_i(1)}{\xi} + \frac{(1 - d_i) y_i(0)}{1 - \xi}]$.

Deeper Trees.

coro[Deeper Trees] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then for any $b \in (0, 1) $, for any sequence $K_n$ taking values in $\mathbb{N}$, \begin{align*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) - \tau| \geq \sigma n^{-b/2}\sqrt{(2+o(1))\log\log(n)}\Bigg) \geq b/e. \end{align*}
coro[$L_2$ Consistency -- NSS] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^4 \log(np)}{n}, \end{align*} where $C$ is a positive constant that only depends on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg( \int_{\mathcal{X}} (\hat\tau^{\mathtt{NSS}}_{\mathtt{SSE}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^4 \log(np)}{n}\bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$.

Sample Splitting

coro[Honest Causal Output] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then for any $b \in (0, 1) $, for any sequence $K_n$ taking values in $\mathbb{N}$, \begin{equation*} \liminf_{n\to\infty} \mathbb{P}\Bigg(\sup_{\mathbf{x}\in\mathcal{X}}|\hat\tau^{\mathtt{HON}}_{\mathtt{SSE}}(\mathbf{x}) - \tau| \geq C n^{-b/2} \Bigg) \geq C \xi(1 - \xi) b. \end{equation*} where $C$ is some constant only depending on the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$, and $\liminf_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$ and $\limsup_{n \to \infty} \frac{n_{\mathsf{T}}}{n_{\tau}}$.
coro[$L_2$ Consistency -- HON] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then for the depth $K$ (possibly non-maximal) causal tree, \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{SSE}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{2^K \log(n)^5}{n}, \end{align*} provided $\rho^{-1} \leq \frac{n_{\mathsf{T}}}{n_{\tau}} \leq \rho$ for some $\rho \in (0,1)$, and $C$ is a positive constant that only depends on $\rho$ and the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$. Moreover, \begin{align*} \limsup_{n \to \infty} \mathbb{P} \bigg( \int_{\mathcal{X}} (\hat\tau^{\mathtt{HON}}_{\mathtt{SSE}}(\mathbf{x}) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x}) \geq C^{\prime} \frac{2^K \log(n)^5}{n}\bigg) = 0, \end{align*} where $C^{\prime}$ is a positive constant that only depends on $\rho$ and the distribution of $(\varepsilon_i(0), \varepsilon_i(1), d_i)$.

X-adaptive Tree

coroSuppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Suppose $\liminf_{n \to \infty} \frac{K_n}{\log \log(n)} > 0$. Then \begin{align*} \liminf_{n\to\infty}\mathbb{P}\Bigg(\sup_{\mathbf{x} \in \mathcal{X}}|\hat\tau^{\mathtt{X}}_{\mathtt{SSE}}(\mathbf{x}; K_n)-\tau| > C \Bigg) > 0. \end{align*}
coro[L2 Consistency] Suppose Assumption (ref) holds with $\mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$. Then \begin{align*} \mathbb{E} \bigg[ \int_{\mathcal{X}} (\hat\tau^{\mathtt{X}}_{\mathtt{SSE}}(\mathbf{x}; K) - \tau)^2 dF_{\mathbf{X}}(\mathbf{x})\bigg] \leq C \frac{K \; 2^{K}}{n}, \end{align*} where $C$ is some constant not depending on $n$.

Additional Results

Squared T-statistic Estimators

The fourth method proposed by Athey-Imbens_2016_PNAS is the squared T-statistic trees, where at the root node the index and coordinate to split $(\hat \imath, \hat \jmath)$ are chosen so that the squared T-statistics metric is maximized, that is,

align*[align* omitted — 190 chars of source]

where $\hat{\tau}_L(k,\ell)$ and $\hat{\tau}_R(k,\ell)$ are the causal tree estimators for the left and right nodes respectively based on split coordinate $\ell$ and index $k$, and $S(k,\ell)^2$ is the conditional sample variance given the split, that is,

align*[align* omitted — 328 chars of source]

Putting together, we see the squared T-statistics metric is a monotone transformation of the split criterion of previously studied estimators,

align*[align* omitted — 365 chars of source]

Hence the split is always the same as the split by the split criterion studied in Section (ref) and Section (ref).

Unbiasedness under Symmetric Error

lemma[Unbiasedness] Suppose Assumption (ref) holds, and $\varepsilon_i(0)$, $\varepsilon_i(1)$ are symmetrically distributed around zero. Then \begin{align*} \mathbb{E}[\hat\tau^{q}_{l}(\mathbf{x}; K)] = \tau, \qquad l \in \{\mathtt{DIM},\mathtt{IPW},\mathtt{SSE}\}, \quad q \in \{\mathtt{NSS},\mathtt{X}\}, \quad K \geq 1, \end{align*} and suppose $\mathsf{t}$ is the node containing $\mathbf{x}$, then \begin{align*} & \mathbb{E}[\hat\tau_l^{\mathtt{HON}}(\mathbf{x};K)] = \tau - \tau \mathbb{P}(n(\mathsf{t}) = 0), \qquad l \in \{\mathtt{IPW}\}, \\ & \mathbb{E}[\hat\tau_l^{\mathtt{HON}}(\mathbf{x};K)] = \tau - \tau \mathbb{P}(n_0(\mathsf{t}) = 0 or n_1(\mathsf{t}) = 0), \qquad l \in \{\mathtt{DIM}, \mathtt{SSE}\}. \end{align*}

Proofs

Proof of Lemma 4

Taking $T = c \log(n)$ in horvath1993maximum, we have

align*[align* omitted — 182 chars of source]

Now we expand the term $\frac{z + b_d(c\log (n))}{a(c \log(n))}$. For notational simplicity, denote

align*[align* omitted — 98 chars of source]

First, we present some elementary expansions, \[

aligned\sqrt{2(A+L)} &= \sqrt{2L}\,\sqrt{1+\frac{A}{L}} = \sqrt{2L}\Bigl(1+\frac{A}{2L}-\frac{A^{2}}{8L^{2}} +\operatorname{O}\!\bigl(L^{-3}\bigr)\Bigr), \\ \frac{1}{\sqrt{2(A+L)}} &= \frac{1}{\sqrt{2L}} \Bigl(1-\frac{A}{2L}+\frac{3A^{2}}{8L^{2}} +\operatorname{O}\!\bigl(L^{-3}\bigr)\Bigr), \\ \log(L+A) &= \log L + \frac{A}{L} - \frac{A^{2}}{2L^{2}} + \operatorname{O}\!\bigl(L^{-3}\bigr).

\] Now we expand the terms for the numerator $b_d(c \log(n))$, \[

alignedN_{1} &= z + 2A + 2L + \frac{d}{2}\log \bigl(\log(c\log n)\bigr) - \log \Gamma(d/2),\\ N_{2} &= z + 2A + 2L + \frac{d}{2}\log L - \log \Gamma(d/2),\\ N_{3} &= z + A + 2L + \frac{d}{2}\log L - \log \Gamma(d/2).

\] Then

align*[align* omitted — 480 chars of source]

Since $N_1 = 2L + O(\log \log \log(n))$, we have

align*[align* omitted — 204 chars of source]

Since $a(\log(n)) = \Theta(L^{1/2})$, we have

align*[align* omitted — 436 chars of source]

where the last line follows from convergence in distribution of $a(c \log(n)) \sup_{0 \leq t \leq c \log (n)} N(t) - b_d(c\log (n))$ to a continuous distribution and Slutsky's Theorem.

Proof of Theorem (ref)

First, we introduce some notations. Recall for $\ell \in [p]$, $\pi_\ell$ denotes the permutation such that $(x_{\pi_\ell(i)}: 1 \leq i \leq n)$ is non-decreasing. Define sample mean at the left and right leave at index $k \in [n]$ based on coordinate $\ell \in [p]$ by

align*[align* omitted — 206 chars of source]

We can check that minimizing the sum of squares criterion Equation (ref) is equivalent to maximizing the split criterion

align*[align* omitted — 116 chars of source]

where

align*[align* omitted — 148 chars of source]

Moreover, under the constant conditional mean assumption, Assumption (ref) (1), we have that $\hat\mu_{L}(k,\ell) - \hat \mu_{R}(k, \ell) = \frac{1}{k} \sum_{i = 1}^k \varepsilon_{\pi_{\ell}(i)} - \frac{1}{n - k} \sum_{i = k+1}^n \varepsilon_{\pi_{\ell}(i)}$. Hence we can w.l.o.g. replace $y_i$ by $\varepsilon_i$ in the definition of $\hat \mu_L$ and $\hat \mu_R$, that is,

align*[align* omitted — 226 chars of source]

The rest of the proof is organized as follows. In Section (ref), we prove the results under $p = 1$, showing a strong approximation of the split criterion $(\mathscr{I}(k,1): k \in [n])$ by the square of a time-transformed Ornstein-Uhlenbeck (O-U) process, and studying the argmax of the split criterion through the argmax of the O-U process. In Section (ref), we generalize to allow for $p \geq 1$. We show that the split criterion over different coordinates, that is, $(\mathscr{I}(k,\ell): k \in [n])$ for different $\ell$'s, are asymptotically independent. This reduces our problem to one-dimensional calculations, and the same technique of approximation by O-U process from Section (ref) can be used.

Univariate Case

This the case when $p = 1$. For notational simplicity, define partial sums by

align*[align* omitted — 80 chars of source]

By csorgo1997limit, we can define a sequence of Brownian bridges $ \{ B_n(t) : 0 \leq t \leq 1 \} $ on a suitable probability space such that

equation[equation omitted — 362 chars of source]

where $ \epsilon_n = o_{\mathbb{P}}\big((\log\log(n))^{-1/2}\big) $. We note that while csorgo1997limit bounds the approximation error of the maximum over the full range $ 1 \leq k < n $ as in (ref), its proof, which relies on invariance principles for partial sums of i.i.d. random variables, can be generalized to bound the approximation error over $ 1 \leq k < n^{a} $, $ n^{b} < k < n $. Thus,

equation[equation omitted — 292 chars of source]

We note that the standardized Brownian bridge $ \big\{ B_n(t)/\sqrt{t(1-t)} : 0 < t < 1 \big\} $ is distributionally equivalent to a time-transformed Ornstein-Uhlenbeck (O-U) process $ \big\{ U(\log(t/(1-t))) : 0 < t < 1\big\} $, where $ \big\{U(t): t \in \mathbb{R}\big\}$ is an O-U process with mean $ \mathbb{E}[U(t)] =0 $ and covariance $\mathbb{E}[U(s)U(t)] = e^{-|s-t|/2}$ csorgo1981strong, and thus

align[align omitted — 681 chars of source]

where the last equality follows from stationarity of the process $ |U(t)| $, the square of which is a Cox-Ingersoll-Ross (CIR) process jaeschke2003survey. Continuing from (ref), for any sequence $u_n$, we have

equation[equation omitted — 785 chars of source]

Now, since $ U(t) $ is a continuous, mean-zero Gaussian process, it induces a centered Gaussian measure on the space of all continuous functions on $\big[0,\,2\log(n-1)\big] $ equipped with the supremum norm (a separable Banach space). Thus, by the Gaussian correlation inequality latala2017royen, we have that

align[align omitted — 722 chars of source]

where the last equality follows from stationarity.

remarkThe next step of our proof relies on a precise characterization of weak convergence for the suprema of a standardized empirical process, as studied in eicker1979asymptotic. However, eicker1979asymptotic is incorrectly stated: the $2\log(c)$ term appearing in the limiting probability should be $\log(c)$. This correction has important implications in our proof.

By the Darling-Erd\H{o}s Limit Theorem for the O-U process csorgo1981strong and eicker1979asymptotic, for all $ c > 0 $ and $ z \in \mathbb{R} $, we have

align[align omitted — 289 chars of source]

For a detailed proof of a generalized result on multidimensional O-U process, see Section (ref).

Let $ z^* $ maximize $ z \mapsto \exp\big(-2e^{-(z-\log(2-(b-a)))}\big) - \exp\big(-2e^{-(z-\log(2))}\big) $, and set $$ u_n = \frac{2\log\log(n) + (1/2)\log\log\log(n) + z^* - (1/2)\log(\pi)}{\sqrt{2\log\log(n)}}. $$ We combine (ref), (ref), and (ref), and employ (ref) three times with $ c = 2 $ and $ c = 2-b $, and $ c = a $, together with the fact that $ \epsilon_n = o_{\mathbb{P}}\big((\log\log(n))^{-1/2}\big) $. We have that

align[align omitted — 623 chars of source]
remarkAlternatively, for any $ 0 < A < B < C $, we have \begin{equation} \mathbb{P}\bigg( \sup_{0 \leq t \leq C} |U(t)| > \sup_{0 \leq t \leq A, \; B \leq t \leq C} |U(t)| \bigg) = \frac{B-A}{C}. \end{equation} This can readily be shown using the fact that the absolute value of a zero-mean O-U process is stationary, Markov, and has continuous paths. Consequently, ignoring the stochastic error $\epsilon_n$ from approximating the split criterion (ref) by the square of a standardized Brownian bridge (not yet justified), using (ref), we can approximate the probability $\mathbb{P} (\max_{1 \leq k \leq n} \mathscr{I}(k,1) > \max_{1 \leq k < n^a, n^b < k < n} \mathscr{I}(k,1))$ by \begin{align} \nonumber & \mathbb{P}\bigg( \sup_{ 0 \leq t \leq 2\log(n-1)} |U(t)| > \sup_{ 0 \leq t < \log(\frac{n^{a-1}(n-1)}{1-n^{a-1}}), \; \log\frac{n^{b-1}(n-1)}{1-n^{b-1}}) < t \leq 2\log(n-1)} |U(t)| \bigg) \\ & \qquad = \frac{\log\frac{n^{b-1}(n-1)}{1-n^{b-1}})-\log(\frac{n^{a-1}(n-1)}{1-n^{a-1}})}{2\log(n-1)} \rightarrow \frac{b-a}{2}, \quad n \rightarrow \infty. \end{align}

Multivariate Case

Now we prove for the general case of $p \geq 1$. As a sketch of the proof, we show that the split criterion over different coordinates, that is, $(\mathscr{I}(k,\ell): k \in [n])$ for different $\ell$'s, are asymptotically independent, which will imply

align*[align* omitted — 678 chars of source]

where in equality (*) we use asymptotic independence between $\mathscr{I}(\cdot,\ell)$ for different $\ell$'s, and the last line are one-dimensional probabilities that can be handled by O-U process approximation like in Section (ref).

To show the split criteria over different coordinates are asymptotically independent, we break down into two steps: In the first step, we show the partial sum process $n$ indices and $p$ coordinates can be approximated by another partial sum process with Gaussian increments (hence a Gaussian process), with the same covariance structure. In the second step, we show the covariance between the split criteria over any two different coordinates and any indices are vanishing. Together with Gaussianity, this implies that the split criteria over different coordinates are asymptotically independent.

center[center omitted — 69 chars of source]

For $1 \leq \ell \leq p$, denote by $H_n^\ell(\frac{k}{n})$ the scaled partial sum for the $\ell$-th coordinate evaluated at time $\frac{k}{n}$, that is,

align*[align* omitted — 310 chars of source]

where $\# \pi^{\ell}:[n] \rightarrow [n]$ is the inverse mapping of $\pi^{\ell}$.

We use a truncation argument for the proof. Fix $\varepsilon \in (0,1)$. Take $r_n = \exp((\log n)^{\varepsilon})$. And consider

align*[align* omitted — 205 chars of source]

where $\# \pi^{\ell}$ denotes the inverse mapping of $\pi^{\ell}$. Notice that we add the $\sqrt{n}$ factor for standardization. Then we can check that condition on $\mathscr{B}$, the $\sigma$-algebra generateed by the $p$ permulations $\pi^1, \cdots, \pi^p$, $\mathbf{C}_i$'s are independent, and for all $1 \leq j \leq p(n - 2 r_n)$, $1 \leq \ell \leq p$, we have

align*[align* omitted — 182 chars of source]

where we assume row $j$ in $\mathbf{C}_i$ corresponds to $\sqrt{n} \sqrt{\frac{n}{k(n-k)}} (\mathds{1}(\#\pi^{\ell}(i) \leq k) - \frac{k}{n})$. To use the coupling result chernozhukov2017central, we bound a few quantities: Suppose $K_1$ and $K_2$ are the universal constants given in the cited theorem,

align[align omitted — 514 chars of source]

For notational simplicity, denote $\mathtt{P} = p (n - 2 r_n)$. Take $\bar{L}_n = L_n$, then

align*[align* omitted — 150 chars of source]

The definition of $\mathbf{C}_i$ implies $C_{ij}$ is $\sqrt{n/r_n}$-exponential. Hence

align*[align* omitted — 1,076 chars of source]

since $r_n = \exp((\log n)^{\varepsilon})$ and $\varepsilon, p$ are fixed. Now condition on $\mathscr{B}$, let $\boldsymbol{D}_i, 1 \leq i \leq n$ to be independent mean-zero Gaussian random vectors such that

align*[align* omitted — 153 chars of source]

Then for each $1 \leq j \leq \mathtt{P}$, $1 \leq i \leq n$, we have $D_{ij}$ is $r_n^{-1}$-subGaussian. Hence the same argument implies

align*[align* omitted — 50 chars of source]

chernozhukov2017central then implies

align[align omitted — 559 chars of source]

where $\mathcal{A}^\text{re}$ is the class of all rectangles $A$ of the form

align*[align* omitted — 126 chars of source]

for some $- \infty \leq a_j \leq b_j \leq \infty$, $j = 1,2, \cdots, \mathtt{P}$. In particular, suppose $u_i, 1 \leq i \leq n$ are i.i.d $N(0, \mathbb{E}[\varepsilon_i^2])$ random variables, then $\boldsymbol{D}_i$ can be taken such that

align*[align* omitted — 199 chars of source]

The above result shows if we define

align*[align* omitted — 167 chars of source]

then Equation (ref) and unconditioning on $\mathscr{B}$, we get

align*[align* omitted — 323 chars of source]
center[center omitted — 65 chars of source]

For $1 \leq \ell \leq p$, denote by $G_n^\ell(\frac{k}{n})$ the partial sum for the $\ell$-th coordinate evaluated at time $\frac{k}{n}$, that is,

align*[align* omitted — 167 chars of source]

Then $\mathbf{G}_n = ((G_n^1(1/n),G_n^1(2/n),\cdots,G_n^1(n/n))^\top, \cdots, (G_n^p(1/n),G_n^p(2/n),\cdots,G_n^p(n/n))^\top)^\top$. Then $\mathbf{G}_n$ is a $np$-dimensional Gaussian random vector, and denote by $\boldsymbol{\Sigma}_n$ its covariance matrix. We want to show that $\boldsymbol{\Sigma}_n$ is close to one with covariance between different coordinates zero.

Consider two different coordinates, $\ell_1, \ell_2 \in [p]$. W.l.o.g, we can assume $\ell_1 = 1$ and $\ell_2 = 2$. Let $k,j \in [n]$. Denote by $\boldsymbol{\sigma}$ the sigma-algebra generated by $\pi_1, \cdots, \pi_p$. Then

align*[align* omitted — 903 chars of source]

To calculate $\sum_{i = 1}^k \sum_{i^\prime = 1}^j \mathbb{E}[u_{\pi_1(i)} u_{\pi_2(i^{\prime})}|\boldsymbol{\sigma}]$, we can first condition on $\pi_1$, and let $\mathcal{I} = \{\pi_1(i): 1 \leq i \leq k\}$. Observe that $\sum_{i = 1}^k \sum_{i^\prime = 1}^j \mathbb{E}[u_{\pi_1(i)} u_{\pi_2(i^{\prime})}|\boldsymbol{\sigma}] = |\{i^{\prime} \in [j]: \pi_1(i^{\prime}) \in \mathcal{I}\}|$. Now consider $$f(\pi) = \frac{n}{jk}|\{i \in [j]: \pi(i) \in \mathcal{I}\}|,$$ $\pi$ is a random permutation of $[n]$. Changing the order of the first $j$ values of $\pi$ does not change the value of $f(\pi)$, and $|f(\pi) - f(\pi^{s,t})| \leq \frac{n}{jk}$ for all $\pi$, $s \in \{1,\cdots,j\}$, $t \in \{j+1,\cdots,n\}$, where the permutation $\pi^{s,t}$ is obtained from $\pi$ by transposition of its $s$th and $t$th coordinates. We will show later that w.l.o.g. we can assume $j,k \leq \lceil n/2 \rceil$. Then by Lemma 2 from el2009transductive, for any $t \geq 0$,

align*[align* omitted — 383 chars of source]

Since $\frac{n - 1/2}{n - j}(1 - \frac{1}{2 \max(j,n-j)}) \geq 1 - \frac{1}{n}$, we can marginalize over $\pi_1$ and uncondition on $\boldsymbol{\sigma}$ to get there exists a positive constant $C$ such that for $n$ large enough, for all $j,k \in [n]$,

align*[align* omitted — 170 chars of source]

which implies

align[align omitted — 189 chars of source]

The reduction to $j,k \leq \lceil n/2 \rceil$ is because

align*[align* omitted — 301 chars of source]

Now consider a $np$-dimensional mean-zero Gaussian random vector $$\mathbf{Z}_n = ((Z_n^1(1/n),Z_n^1(2/n),\cdots,Z_n^1(n/n))^\top, \cdots, (Z_n^p(1/n),Z_n^p(2/n),\cdots,Z_n^p(n/n))^\top)^\top,$$ where for each $1 \leq \ell \leq p$, $(Z_n^\ell(1/n),Z_n^\ell(2/n),\cdots,Z_n^\ell(n/n))^\top$ has the same joint distribution as the partial sum random vector $(G_n^\ell(1/n)$,$G_n^\ell(2/n),\cdots,G_n^\ell(n/n))^\top$, and for any $\ell \neq \ell^{\prime}$ and any $j,k \in [n]$,

align*[align* omitted — 81 chars of source]

Denote by $\boldsymbol{\Gamma}_n$ the covariance matrix of $\mathbf{Z}_n$. We want to show $\boldsymbol{\Gamma}_n$ is close to $\boldsymbol{\Sigma}_n$. For a tight control on the rate of convergence, consider the truncated random vector,

align*[align* omitted — 220 chars of source]

Also by an abuse of notations, denote by $T_{r_n}(\boldsymbol{\Sigma}_n)$ and $T_{r_n}(\boldsymbol{\Gamma}_n)$ the covariance matrix of $T_{r_n}(\mathbf{G}_n)$ and $T_{r_n}(\mathbf{Z}_n)$, respectively. Then Equation (ref) implies

align[align omitted — 162 chars of source]

Additionally, we can lower bound the variance of each item of $T_{r_n}(\mathbf{Z}_n)$ by the following conditioning argument: Condition on the permutations $\pi_\ell$, $1 \leq \ell \leq p$, then

align*[align* omitted — 508 chars of source]

where in the third line, we have used the fact that condition on $\pi_\ell, 1 \leq \ell \leq p$, $(u_{\pi^\ell(i)})_{i \in [n]}$'s are i.i.d. $\mathsf{N}(0,1)$. By the Gaussian-to-Gaussian Comparison result chernozhuokov2022improved,

align*[align* omitted — 280 chars of source]

where $C$ is an absolute constant, and $T(n) = \lceil n - r_n \rceil - \lfloor r_n \rfloor$. Combining with Equation (ref) and taking $\mathbf{y} = (t_1 \mathbf{1}^\top, \cdots t_p \mathbf{1}^\top)$, $\mathbf{y} = -(t_1 \mathbf{1}^\top, \cdots t_p \mathbf{1}^\top)$ separately with $\mathbf{1}$ a vector of $T(n)$ 1's, we get

align[align omitted — 332 chars of source]
center[center omitted — 91 chars of source]

As in the previous two sections, fix $\varepsilon > 0$, and take $r_n = \exp((\log n)^{\varepsilon})$. Let $\mathcal{E} = \{\exists \ell \in [p]: \operatorname*{arg\,max}_k \mathscr{I}(k,\ell) < r_n \text{ or } \operatorname*{arg\,max}_k \mathscr{I}(k,\ell) > n - r_n\}$. Then by csorgo1997limit, and a union bound argument, we have

align*[align* omitted — 206 chars of source]

Hence we can effectively restrict the candidates of $\operatorname*{arg\,max}$ to $[r_n, n - r_n]$. W.l.o.g., we consider coordinate $\ell = 1$, and

align*[align* omitted — 966 chars of source]

Now we can using the coupling result developed previously. Using our notation, we have $\mathscr{I}(k,\ell) = (H_n^\ell(k/n))^2$. Hence

align*[align* omitted — 1,192 chars of source]

where we have used Lemma (ref) and Lemma (ref). Since we choose $r_n = \exp((\log n)^{\varepsilon})$, we have $\log(n)^{7/6} r_n^{-1/6} = o(1)$. It then follows from independence and symmetry between $Z_n^\ell$'s across different $\ell$'s that

align*[align* omitted — 895 chars of source]

where the third line is by similar calculation as in Section (ref). Putting together, we have

align*[align* omitted — 139 chars of source]

and by symmetry, we have

align*[align* omitted — 147 chars of source]

Proof of Theorem (ref)

For simplicity, we denote $\hat{\mu}^{\mathtt{NSS}}(\mathbf{x})$ by $\hat{\mu}(\mathbf{x})$. We divide the proofs into two parts, one for uniform estimation and one for pointwise results near the boundary.

center[center omitted — 73 chars of source]

For notational simplicity, introduce the partial sum based on ordering for the $\ell$'s coordinate,

align*[align* omitted — 113 chars of source]

and define the optimal index for splitting based on the $\ell$'s coordinate by

align*[align* omitted — 112 chars of source]

Consider the event

align*[align* omitted — 305 chars of source]

Consider the case $\hat \imath < n^b$ on $\mathtt{Imbalance}_{\ell}$. The other case where $\hat \imath > n - n^b$ can be dealt with by symmetry. Then

align*[align* omitted — 682 chars of source]

where the last line is because $\imath_{\ell}$ is the index that maximize the split criterion based on the $\ell$'s coordinate, i.e.,

align*[align* omitted — 338 chars of source]

Fix $\epsilon > 0$. Consider the events

align*[align* omitted — 377 chars of source]

By csorgo1997limit $\limsup_{n \rightarrow \infty} \mathbb{P}(A^{\varepsilon}_{\ell}) = \limsup_{n \rightarrow \infty} \mathbb{P}(B^{\varepsilon}_{\ell}) = 1$. Hence for any $\epsilon > 0$,

align*[align* omitted — 291 chars of source]

where we have used the fact that $\mathtt{Imbalance}_{\ell}$'s are disjoint for different $\ell$'s and Theorem (ref). Equation (ref) then follows.

center[center omitted — 73 chars of source]

Consider the event

align*[align* omitted — 277 chars of source]

Since $\pi_{\ell}$ is the uniform permutation, we have

align*[align* omitted — 186 chars of source]

Together with Theorem (ref),

align*[align* omitted — 119 chars of source]

Then on the event $\mathtt{Off}_{\ell}$ and $x_{\ell,\imath_{\ell}} \geq n^{a-1}$, for any $\mathbf{x} \in [0,1]^p$ such that $x_{\ell} \leq n^{a-1}$, we have $x_{\ell} \leq x_{\ell,\imath_{\ell}}$, and

align*[align* omitted — 533 chars of source]

By similar arguments as Part 1, we can show

align*[align* omitted — 204 chars of source]

which is Equation (ref).

Proof of Theorem (ref)

Due to the recursive splitting and Theorem (ref), the optimal split index $\hat \imath$ at the $K_n$-th split ($K_n \geq 1$) also satisfies

align*[align* omitted — 170 chars of source]

Hence the same argument as Part 1 in the proof of Theorem (ref) leads to the result.

Proof of Theorem (ref)

This follows directly from klusowski2024large, choosing $g^{\ast} \equiv \mu$ and $g \equiv \mu$, and changing the sub-Gaussian rate to the sub-exponential rate by choosing $U \asymp \log(n)$ instead of $U \asymp \sqrt{\log(n)}$ in the truncation argument step. The last statement follows from the proof of klusowski2024large.

Proof of Theorem (ref)

Throughout the proof, we abbreviate the honest tree $\hat{\mu}^{\mathtt{HON}}(\mathbf{x})$ by $\check{\mu}(\mathbf{x})$. Recall $(\hat \imath, \hat \jmath)$ denotes the optimal splitting index and coordinate for the decision stump. We use $(y_i,\mathbf{x}_i^{\top})_{i=1}^M$ to denote $\mathcal{D}_{\mathtt{HON},1}$, which we used to construct the causal tree. Denote by $(\hat \imath, \hat \jmath)$ the splitting index and coordinate at the $K_n$-th step.

Use $(\tilde{y}_i,\tilde{\mathbf{x}}_i^{\top})_{i=1}^N$ to denote $\mathcal{D}_{\mathtt{HON},2}$. By Definition (ref), $n \lesssim M,N \lesssim n$. Then

align*[align* omitted — 387 chars of source]

Since $\tilde{y}_i \protect\mathpalette{\protect\independenT}{\perp} \tilde{\mathbf{x}}_i$, condition on $\hat{\imath}, \hat{\jmath}$ and $\mathbf{X} = (\mathbf{x}_1, \cdots, \mathbf{x}_n)$, $\tilde{\mathbf{X}} = (\tilde{\mathbf{x}}_1, \cdots, \tilde{\mathbf{x}}_n)$, we have

align*[align* omitted — 340 chars of source]

where

align*[align* omitted — 141 chars of source]

By Marcinkiewicz–Zygmund inequality, for some positive absolute constant $C$, we have

align*[align* omitted — 666 chars of source]

where in the second to last line, we have used Jensen's inequality, and in the last line we have used $\tilde{\imath}$ is measurable with resepect to the $\sigma$-algebra generated by $\hat \imath, \hat \jmath, \mathbf{X}, \mathbf{X}^{\prime}$. Then by Paley–Zygmund inequality, for any $\theta \in (0,1)$,

align*[align* omitted — 1,030 chars of source]

Now we want to obtain a high probability upper bound on $\tilde \imath$ given $\imath$. Let $F$ be the cummulative distribution function of $\mathbf{x}_i$. Suppose $1 \leq k \leq N/2$. Then $F(\mathbf{x}_{(k)}) \sim \mathsf{Beta}(k, M - k + 1)$. By a Bernstein bound for Beta variables skorski_2023, we have for all $\epsilon > 0$,

align*[align* omitted — 136 chars of source]

where for large enough $n$,

align*[align* omitted — 148 chars of source]

Hence with probability at least $1 - M^{-1}$,

align*[align* omitted — 101 chars of source]

Condition on $\mathbf{X}$, $\mathds{1}(\tilde{\mathbf{x}}_i \leq \mathbf{x}_{(k)})$'s are i.i.d $\mathsf{Bernoulli}(F(\mathbf{x}_{(k)}))$. Hence condition on $\mathbf{X}$ and $\hat{\imath}$, with probability at least $1 - N^{-1}$,

align*[align* omitted — 220 chars of source]

Hence condition on the event $\hat \imath \leq M^b$, we have with probability at least $1 - 2 N^{-1}$,

align*[align* omitted — 65 chars of source]

where $C$ is some constant only depending on $\liminf_{n \to \infty}|D_{\mathtt{HON}, 1}|/|\mathcal{D}_{\mathtt{HON},2}|$ and $\limsup_{n \to \infty}|D_{\mathtt{HON}, 1}|/|\mathcal{D}_{\mathtt{HON},2}|$.

Due to the iterative partitioning, the conclusion for Theorem (ref) holds not only for decision stump, but also for the splitting index at arbitrary depth $K_n$, that is, for any $b \in (0,1)$, we have

align*[align* omitted — 99 chars of source]

Hence we have

align*[align* omitted — 640 chars of source]

This proves the conclusion.

Proof of Theorem (ref)

For notational simplicity, we use $\mathsf{T}$ to denote the data-driven decision tree. We will follow the proof strategy from klusowski2024large condition on $\mathcal{D}_{\mathsf{T}}$. Denote by $\mathscr{G}_0$ the class of constant functions. Decompose $\lVert \hat{\mu}(\mathsf{T}) - \mu \rVert^2 = E_1 + E_2$, where

align*[align* omitted — 197 chars of source]

and

align*[align* omitted — 151 chars of source]

Denote the partition for $\mathsf{T}$ by $\mathcal{P}$. Since $\mathcal{P}$ is independent to $\mathcal{D}_{\mu}$, the bound (E.27) from klusowski2024large does not apply automatically. Instead, we consider $\mathscr{G}_0$ as the reference class. Given the partitions of $\mathsf{T}$, the values of leaf nodes are obtained by least-square projection using $\mathcal{D}_{\mu}$. This immediately implies

align*[align* omitted — 182 chars of source]

for any constant function $g \in \mathscr{G}_0$. Hence for all $g \in \mathscr{G}_0$,

align*[align* omitted — 302 chars of source]

For the term $E_1$, we first assume $|y_i| \leq U$. Observe that condition on $\mathcal{D}_{\mathsf{T}}$, $\hat{\mu}(\mathsf{T})$ is still a member of the class $\mathscr{G}_{n_{\mathsf{T}}}[\mathcal{P}]$, which is the collection of all piecewise constant functions (bounded by $U$) on the partition $\mathcal{P}$. Since for any $\varepsilon \in (0,1)$,

align*[align* omitted — 265 chars of source]

we can still use Gyorfi-etal_2002_book and the same argument from Equation (B.30) to (B.33) in klusowski2024large to get

align*[align* omitted — 198 chars of source]

The result then follows choosing $\alpha \asymp \frac{U^4 2^K \log(n)}{n}$ and $\beta \asymp \frac{U^2}{n}$, and truncation argument over the sub-exponential $\varepsilon_i$'s.

Proof of Theorem (ref)

In this section, we prove Theorem (ref). First, we define some notation related to the tree construction which will be used in the proofs. Let $\tilde n_k$ be the number of observations in the node containing $x = 0$ at depth $k$, $\tilde\imath_{k} $ be the CART split index of this node, and $\tilde \jmath_{k}$ be the CART split coordinate of this node, with $\tilde n_0 = n$ and $\tilde\imath_0 = \hat\imath$ (recall that $ \hat\imath $ is the split index for the decision stump (ref)). Then, the left-most cell at the $k$-th level can be expressed as $ \mathbf{t} \cap [0, x_{ \pi_{\tilde{\jmath},\tilde{\jmath}}(\tilde\imath_{k-1})}] $ and $ \tilde n_k = \tilde \imath_{k-1}$.

lemmaThere exist $ \delta \in (0, 1) $, $c > 1$, and a positive integer $M$ such that for any depth $k\geq 1$ and $m\geq M$, we have $ \mathbb{P}(\tilde n_k \leq m) \geq (1-\delta)\cdot \mathbb{P}(\tilde n_{k-1}\leq m) + \delta \cdot \mathbb{P}(\tilde n_{k-1}\leq m^c). $
proofObserve that if $m$ is a positive integer, then $\tilde\imath_{k-1} \mid \tilde n_{k-1} = m$ has the same distribution as $\tilde\imath_0 \mid \tilde n_0 = m$, because of the honest tree construction and Assumption (ref). Therefore, we can apply (ref) to obtain \begin{equation} \mathbb{P}\big( m^a \leq \tilde\imath_{k-1} \leq m^b \mid \tilde n_{k-1} = m\big) \geq \delta > 0, \end{equation} for some $ \delta > 0 $ and sufficiently large $ m $. Hence, by (ref), we have for $ m $ sufficiently large, \begin{equation} \begin{aligned} & \mathbb{P}\big(\tilde n_k \leq m \mid m < \tilde n_{k-1} \leq m^{1/b} \big) \\ & \quad \geq \min_{m < i \leq m^{1/b} } \mathbb{P}\big(i^a \leq \tilde\imath_{k-1}\leq i^b \mid \tilde n_{k-1} = i\big)\mathbb{P}\big(\tilde n_k \leq m \mid i^a \leq \tilde\imath_{k-1} \leq i^b \big) \\ & \quad \geq \delta \min_{m < i \leq m^{1/b} } \mathbb{P}\big(\tilde n_k \leq m \mid i^a \leq \tilde\imath_{k-1} \leq i^b \big) \\ & \quad \geq \delta\min_{m^a < i \leq m } \mathbb{P} \big(\tilde n_k \leq \tilde\imath_{k-1} \mid \tilde\imath_{k-1} = i \big) \\ & \quad = \delta. \end{aligned} \end{equation} Now, taking $ c = 1/b $, note that (ref) implies Lemma (ref) since, for $ m $ sufficiently large, we have \begin{align} \mathbb{P}(\tilde n_k \leq m) & = \mathbb{P}(\tilde n_k \leq m, \; \tilde n_{k-1} > m^c) + \mathbb{P}(\tilde n_k \leq m, \; \tilde n_{k-1} \leq m^c) \\ & \geq \mathbb{P}(\tilde n_k \leq m, \; \tilde n_{k-1} \leq m^c) \\ & = \mathbb{P}(\tilde n_k \leq m, \; \tilde n_{k-1} \leq m) + \mathbb{P}(\tilde n_k \leq m, \; m < \tilde n_{k-1} \leq m^c) \\ & \geq \mathbb{P}(\tilde n_{k-1}\leq m)+ \delta\cdot\mathbb{P}(m < \tilde n_{k-1}\leq m^c) \\ & = (1-\delta)\cdot \mathbb{P}(\tilde n_{k-1}\leq m)+ \delta\cdot\mathbb{P}(\tilde n_{k-1}\leq m^c). \qedhere \end{align}

Next, we use Lemma (ref) to finish the proof of Theorem (ref). The main idea is to establish that the terminal nodes in a shallow tree will be small with constant probability.

proof[Proof of Theorem (ref)] For notational simplicity, we denote $\hat \mu^{\mathtt{X}}(\mathbf{x};K)$ by $\tilde{\mu}(T_K)(\mathbf{x})$. Define $n_\ell = N^{(1/c)^{\ell}}$, where $N = n/K_n$. We will show by induction that for any $k \geq 0$ and $\ell \geq 1$ such that $n_\ell \geq M$, \begin{equation} \mathbb{P}(\tilde{n}_k \leq n_{\ell}) \geq \sum_{k'=\ell}^{k} \binom{k'-1}{\ell-1} (1-\delta)^{k'-\ell} \delta^\ell.. \end{equation} The base case of $k = 0$ is trivial since $\tilde n_0 = N$. Now, assume that for some fixed $k \geq 1$ and any $\ell' \geq 1$ such that $n_{\ell'} \geq M$, we have \begin{equation} \mathbb{P}(\tilde n_{k-1}\leq n_{\ell'}) \geq \sum_{k'=\ell'}^{k-1} \binom{k'-1}{\ell'-1} (1-\delta)^{k'-\ell'} \delta^{\ell'}. \end{equation} If $\ell \geq 2$, then substituting our induction hypothesis (ref) with $\ell' = \ell$ and $\ell'= \ell-1$ into Lemma (ref), we get that \begin{align} \mathbb{P}(\tilde n_k \leq n_\ell) & \geq (1-\delta) \sum_{k'=\ell}^{k-1} \binom{k'-1}{\ell-1} (1-\delta)^{k'-\ell} \delta^\ell + \delta \sum_{k'=\ell-1}^{k-1} \binom{k'-1}{\ell-2} (1-\delta)^{k'-\ell+1}\delta^{\ell-1} \\ & =\sum_{k'=\ell}^k \binom{k'-1}{\ell-1} (1-\delta)^{k'-\ell} \delta^\ell, \end{align} where we used Pascal's identity. This completes the inductive proof of (ref). Let $X \sim \text{NB}(L, \delta)$, i.e., the number of independent trials, each occurring with probability $\delta$, until $L$ successes. Choose $$ L = \lceil \log_c\log_c(N) - \log_c\log_c(M)-1\rceil \asymp \log\log(N), \quad n_L = N^{(1/c)^L} \in [M,M^c]. $$ By (ref) and Markov's inequality applied to the tail probability of $ X $, we have that \begin{equation} \begin{aligned} \mathbb{P}(\tilde n_K \leq n_L) & \geq \sum_{k'=L}^K \binom{k'-1}{L-1} (1-\delta)^{k'-L}\delta^L \\ & = 1 - \mathbb{P}(X \geq K + 1) \\ & \geq 1- \frac{\mathbb{E}[X]}{K+1} \\ & = 1 - \frac{L}{\delta(K+1)} \\ & \geq \frac{1}{2}, \end{aligned} \end{equation} as long as $ K \geq 2L/\delta \gtrsim \log\log(N)$. By the Paley-Zygmund inequality petrov2007lower and the fact that $ \text{Var}(\tilde \mu(T_{K})(0)) = \mathbb{E}[1/\tilde n_K] \leq 1 $, we have \begin{equation} \mathbb{P}\bigg(|\tilde \mu(T_{K})(0)| > \frac{\mathbb{E}[|\tilde \mu(T_{K})(0)|]}{2} \bigg) \geq \frac{(\mathbb{E}[|\tilde \mu(T_{K})(0)|])^2}{4Var(\tilde \mu(T_{K})(0))} \geq \frac{(\mathbb{E}[|\tilde \mu(T_{K})(0)|])^2}{4} . \end{equation} By the honest construction of the tree and (ref), we have the lower bound \begin{equation} \begin{aligned} \mathbb{E}[|\tilde \mu(T_{K})(0)|] & = \sum_{k=1}^n \mathbb{E}\bigg[\bigg|\frac{1}{k}\sum_{i=1}^k \tilde y_i \bigg|\bigg]\mathbb{P}( \tilde n_K = k) \\ & \geq \min_{ k = 1, 2, \dots, \lceil n_L \rceil }\mathbb{E}\bigg[\bigg|\frac{1}{k}\sum_{i=1}^k \tilde y_i \bigg|\bigg]\mathbb{P}( \tilde n_K \leq \lceil n_L \rceil) \\ & \geq \frac{1}{2}\min_{ k = 1, 2, \dots, \lceil n_L \rceil}\mathbb{E}\bigg[\bigg|\frac{1}{k}\sum_{i=1}^k \tilde y_i \bigg|\bigg]. \end{aligned} \end{equation} Since a sum of independent random variables is almost surely constant if and only if the individual random variables are almost surely constant, it follows that the last expression in (ref) is bounded away from zero. Returning to (ref) completes the proof.

Proof of Theorem (ref)

For simplicity, denote $\hat{\mu}^{\mathtt{X}}(\mathbf{x})$ by $\tilde \mu(T_K)(\mathbf{x})$, and $N = n/(K+1)$ denotes the sample size for each folds in the $\mathtt{X}$ sample splitting scheme.

Let $\mathsf{t}_1, \mathsf{t}_2, \cdots, \mathsf{t}_{2^K}$ denote the $2^K$ leaf nodes in the decision tree, (if a node cannot be further refined, we duplicate the split indices and values at the next level). And let $N_1, N_2, \cdots, N_{2^K}$ and $m_1, m_2, \cdots, m_{2^K}$ denote the number of observations and the Lebesgue measure of the $2^K$ leaf nodes, respectively. Note that $\Vec{N} = (N_1, \cdots, N_{2^{K}})$ are independent of the $\tilde y_i $ data by the honest condition and the $x_i $ data per Assumption (ref).

Claim: Condition on $\Vec{N}$, $m_k \sim \mathsf{Beta}(N_k, N - N_k + 1)$

Thus, the IMSE can be bounded as follows:

align*[align* omitted — 486 chars of source]

Proof of Claim: We show by induction. Base Case: K = 1. For decision stumps, for some coordinate $j \in [p]$, we have $m_1 = x_{(N_1)}$, and $m_2 = x_{(N_1 + N_2)} - x_{(N_1)} = 1 - x_{(N_1)}$. By Assumption (ref), the order statistics $x_{j,(i)}$ is independent to $\Vec{N}$. Hence $m_k \sim \mathsf{Beta}(N_k, N - N_k + 1)$, $k = 1,2$.

Induction Step: $K \geq 2$. Let $\mathsf{t}_l^{\text{prev}}$ be a $(K - 1)$-th level node, we annonate all relevant depth $K-1$ information with superscript $\text{prev}$. We already know condition on $N_1^{\text{prev}}, \cdots, N_{2^{K-1}}^{\text{prev}}$, $m_l^{\text{prev}} \sim \mathsf{Beta}(N_l^{\text{prev}}, N - N_l^{\text{prev}} + 1)$. Suppose $\mathsf{t}_l^{\text{prev}}$ is divided into $\mathsf{t}_{2l}, \mathsf{t}_{2l+1}$ with Lebesgue measure and number of observations given by $m_{2l}, m_{2l+1}$ and $N_{2l}$, $N_{2l+1}$, respectively, and the split is based on coordinate $j \in [p]$. By Assumption (ref), condition on $\mathbf{x}_i \in \mathsf{t}_l^{\text{prev}}$, $\mathbf{x}_i \sim \mathsf{Uniform}(\mathsf{t}_l^{\text{prev}})$. Hence condition on $N_l^{\text{prev}}$, $N_{2l}$ and $m_l^{\text{prev}}$, we have $m_{2l}/m_{l}^{\text{prev}} \sim \mathsf{Beta}(N_l^{\text{prev}}, N - N_l^{\text{prev}} + 1)$. Hence condition on $\vec{N} = (N_1, \cdots, N_{2^K})$, we have $m_k \sim \mathsf{Beta}(N_k, N - N_k + 1)$, $1 \leq k \leq 2^K$. Induction then concludes the proof.

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Corollary (ref)

This is an immediate corollary from Theorem (ref).

Proof of Lemma (ref)

Since the number of coordinate $p$ is fixed, we can use a union bound over the approximation error for the $p$ coordinates. Hence w.lo.g. we can assume $p = 1$ and drop the second index on the coordinate $\ell$ from $\mathscr{I}^{\mathtt{DIM}}(k,\ell)$ and $\bar \mathscr{I}^{\mathtt{IPW}}(k,\ell)$ everywhere. And throughout, we assume the data is already sorted so that

align*[align* omitted — 52 chars of source]

Expand the square, we have for any $k = 1,2,\cdots,n$,

align[align omitted — 540 chars of source]

We focus on the case where $1 \leq k \leq \frac{n}{2}$, the other case where $\frac{n}{2} < k \leq n$ follow from symmetry. Consider the term $R_2(k)$. First, consider the term corresponding to $i$ from $1$ to $k$. The other term corresponding to $i$ from $k + 1$ to $n$ can be handled similarly. Breaking down $y_i(1) = \mu_1(x_i) + \varepsilon_i(1)$ and $y_i(0) = \mu_0(x_i) + \varepsilon_i(0)$, we have

align[align omitted — 1,314 chars of source]

Notice that Assumption (ref) (ii) implies that the last term is zero. Since $x_i \protect\mathpalette{\protect\independenT}{\perp} d_i$, even though the data is ordered according to $x_i$, $\{d_i/\xi - 1: 1 \leq i \leq n\}$ are i.i.d mean-zero with bounded second moment. By Theorem A.4.1 in csorgo1997limit,

align*[align* omitted — 165 chars of source]

Take $b_i = \sum_{1 \leq \ell \leq i} d_\ell$. By Equation (8) from shorack1976inequalities, for any $\lambda > 0$,

align*[align* omitted — 473 chars of source]

The assumption that $\liminf_{n \rightarrow \infty} \rho_n \log \log (n) = \infty$ implies $\liminf_{n \rightarrow \infty} r_n = \infty$. Hence

align*[align* omitted — 133 chars of source]

Hence uncondition on $(d_i)_{1 \leq i \leq n}$, and we have

align[align omitted — 178 chars of source]

Hence

align*[align* omitted — 265 chars of source]

By similar arguments, we can show the same bound holds for other terms in the first two lines of Equation (ref). Hence

align*[align* omitted — 124 chars of source]

Under the assumption that $\mu_0 \equiv c_0$ and $\mu_1 \equiv c_1$, we have

align*[align* omitted — 642 chars of source]

By Equation (ref) and Theorem A.4.1 in csorgo1997limit for the terms $k^{-1}\sum_{i = 1}^k \xi^{-1} d_i \varepsilon_i(1)$, $k^{-1} \sum_{i = 1}^k (1 - \xi)^{-1} (1 - d_i) \varepsilon_i(0)$ and the counterparts for $t_R$, we have

align*[align* omitted — 106 chars of source]

Putting together the parts for $R_1$ and $R_2$, we have

align*[align* omitted — 174 chars of source]

Proof of Lemma (ref)

Since the number of coordinates $p$ is fixed, we can use a union bound over the approximation error for the $p$ coordinates. Hence w.lo.g. we can assume $p = 1$ and drop the second index on the coordinate $\ell$ from $\mathscr{I}^{\mathtt{DIM}}(k,\ell)$ and $\bar \mathscr{I}^{\mathtt{IPW}}(k,\ell)$ everywhere. And throughout, we assume the data is already sorted so that

align*[align* omitted — 52 chars of source]

For $1 \leq k \leq s_n$ and $n - s_n \leq k \leq n$, Equations (ref) and (ref) still hold. W.l.o.g assume $1 \leq k \leq s_n$. First, we upper bound the IPW terms. Definition of $s_n$ and Equation (A.4.3) in csorgo1997limit imply

align[align omitted — 267 chars of source]

with $u_n = \sqrt{\rho_n \log \log(n)}$. Also Equation (A.4.2) in csorgo1997limit imply

align[align omitted — 297 chars of source]

where $v_n = \sqrt{\frac{s_n}{n - s_n} \log \log (n)}$. Again Equation (A.4.3) from csorgo1997limit imply that

align*[align* omitted — 141 chars of source]

Take $b_i = \sum_{1 \leq \ell \leq i} d_\ell$. By Equation (8) from shorack1976inequalities, for any $\lambda > 0$,

align*[align* omitted — 446 chars of source]

Hence uncondition on $(d_i)_{1 \leq i \leq n}$, and we have

align*[align* omitted — 133 chars of source]

It follows that

align[align omitted — 408 chars of source]

Putting together the above equation with Equation (ref) and using a similar argument for the control group,

align[align omitted — 295 chars of source]

Apply Equation (A.4.2) in csorgo1997limit for the partial sum with at least $n - s_n$ terms and using $\max_{1 \leq k \leq s_n}|\frac{1}{n - k}\sum_{i = k+1}^n (d_i - \xi)| = o_\mathbb{P}(1)$, we have

align[align omitted — 632 chars of source]

The same bound hold for $\max_{1 \leq k \leq s_n} \sqrt{k} \cdot |\frac{\sum_{i = k+1}^n (1 - d_i)\varepsilon_i(1)}{\sum_{i = k+1}^n (1 - d_i)}|$ by a similar argument. Putting together Equations (ref),(ref), (ref), (ref), we have

align*[align* omitted — 108 chars of source]

From the decomposition in Equation (ref) and the symmetry for $k \in [1,s_n]$ and $k \in [n - s_n,n]$, the conclusion follows.

Proof of Theorem (ref)

We break down the proofs into two steps.

center[center omitted — 74 chars of source]

Let $0 < a < b < 1$. Let $\rho_n$ be a sequence of real numbers taking values in $(0,1)$ to be determined, and take $s_n = \exp((\log n)^{\rho_n})$. Then for large enough $n$, we have $s_n \leq n^a \leq n^b \leq n - s_n$. Consider the event $A_n := \{\exists \ell \in [p]: \max_{k \in [n]} \mathscr{I}^\mathtt{DIM}(k,\ell) > \max_{k \notin [s_n,n - s_n]} \mathscr{I}^\mathtt{DIM}(k,\ell)\}$. By Equation (A.4.18) from csorgo1997limit,

align*[align* omitted — 155 chars of source]

Then controlling the difference between $\bar\mathscr{I}^{\mathtt{IPW}}(k,\ell)$ and $\mathscr{I}^{\mathtt{DIM}}(k,\ell)$ by Lemma (ref),

align[align omitted — 202 chars of source]

By Lemma (ref) with the choice $r_n = s_n$,

align*[align* omitted — 470 chars of source]

Equation (A.4.20) in csorgo1997limit imply that $(2 \log \log (n))^{-1/2} \max_{1 \leq k \leq n} \sqrt{\bar\mathscr{I}^{\mathtt{IPW}}(k,\ell)} = 1 +o_{\mathbb{P}}(1)$ and $(2 \log \log (n))^{-1/2}\max_{1 \leq k \leq s_n, n - s_n \leq k \leq n} \sqrt{\bar\mathscr{I}^{\mathtt{IPW}}(k,\ell)} = \rho_n(1 + o_\mathbb{P}(1))$. Hence

align[align omitted — 243 chars of source]

Choose $\log \log \log \log(n)/\log \log(n) \ll \rho_n \ll 1$, then by Equation (ref) and (ref),

align*[align* omitted — 255 chars of source]

Hence

align*[align* omitted — 213 chars of source]

which by a union bound implies

align*[align* omitted — 69 chars of source]

Observe that on the event $A_n^c$, the argmax for $\mathscr{I}^{\mathtt{DIM}}$ should be inside $[s_n, n - s_n]$. Hence

align*[align* omitted — 962 chars of source]

Now we focus on the first term. By symmetry in the $p$ coordinates,

align*[align* omitted — 1,437 chars of source]

Then using the fact that $\bar \mathscr{I}^\mathtt{IPW}(k,\ell)$ approximates $\mathscr{I}^\mathtt{DIM}(k,\ell)$ from Lemma (ref), we have

align*[align* omitted — 851 chars of source]

where $v_n = O_\mathbb{P}(\log \log (n) s_n^{-1/2})$.

center[center omitted — 84 chars of source]

Observe that the choice $s_n = \exp(\log(n)^{\rho_n})$ for $\log \log \log \log(n)/\log \log(n) \ll \rho_n \ll 1$ implies $\upsilon_n = o_{\mathbb{P}}((\log \log (n))^{-1/2})$. Let $\epsilon > 0$. Then

align*[align* omitted — 1,167 chars of source]

Choosing $z_n(u) = \frac{2 \log \log (n) + 1/2 \log \log \log (n) + u - 1/2 \log(\pi)}{\sqrt{2 \log \log (n)}}$, and from the proof of Theorem (ref), we have

align*[align* omitted — 2,444 chars of source]

Now let $\epsilon \downarrow 0$, and then all previous steps together implies

align*[align* omitted — 497 chars of source]

Proof of Theorem (ref)

The proofs follow the essentially same logic as the proof for Theorem (ref), with some tricks for the random numerator in $\frac{\sum_{1 \leq i \leq k} d_i \varepsilon_i(1)}{\sum_{1 \leq i \leq k} d_i}$.

center[center omitted — 77 chars of source]

Denote the optimal index for splitting based on the $\ell$'s coordinate by

align*[align* omitted — 148 chars of source]

For notational simplicity, denote

align*[align* omitted — 877 chars of source]

and consider the event

align*[align* omitted — 305 chars of source]

Since we assume $\mu_0 \equiv c_0$ and $\mu_1 \equiv c_1$ with $c_1 - c_0 = \tau$, we have on $\mathtt{Imblce}_{\ell}^\mathtt{DIM} \cap \{\hat{\iota}_{\ell}^\mathtt{DIM} \leq n/2\}$,

align[align omitted — 533 chars of source]

Take $\bar\tau^{\mathtt{DIM}} = \frac{\hat{\imath}_{\text{DIM},\ell}}{n} \bar\tau^\mathtt{DIM}_L + \frac{n - \hat{\imath}_{\text{DIM},\ell}}{n} \bar\tau^\mathtt{DIM}_R$. Then

align*[align* omitted — 490 chars of source]

By Lemma (ref) and Lemma (ref) with $r_n = s_n = \exp((\log n)^{\rho_n})$ for $\log \log \log \log (n)/\log \log(n) \ll \rho_n \ll 1$,

align*[align* omitted — 421 chars of source]

By Theorem A.4.1 in csorgo1997limit, $\max_{1 \leq k \leq n} \bar \mathscr{I}^\mathtt{IPW}(k) = 2 \log \log (n) (1 + o_\mathbb{P}(1))$. Moreover,

align*[align* omitted — 760 chars of source]

For simplicity in showing the upper bound for $\hat{\imath}_{\text{DIM},\ell} \bar{\tau}_L^{\mathtt{DIM}}(\ell)^2 \mathds{1}(\hat{\imath}_{\text{DIM},\ell} > n/2) $, we assume $\pi$ is the identity permulation. Take $b_i = \sum_{1 \leq j \leq i} d_j$. By Equation (8) from shorack1976inequalities, for any $\lambda > 0$,

align*[align* omitted — 438 chars of source]

And since $d_i$'s are i.i.d with $\mathbb{E}[d_i] = \xi > 0$, we have

align*[align* omitted — 131 chars of source]

Hence uncondition on $(d_i)_{1 \leq i \leq n}$, and we have

align*[align* omitted — 168 chars of source]

By a similar term for control, and a symmetric argument for the right node,

align*[align* omitted — 291 chars of source]

Fix $\epsilon > 0$. Consider the events

align*[align* omitted — 552 chars of source]

The above arguments show that $\liminf_{n \rightarrow \infty} \mathbb{P}(A_{\ell}^{\epsilon}) = \liminf_{n \rightarrow \infty} \mathbb{P}(B_{\ell}^{\epsilon}) = 1$. From Theorem (ref), $$\mathbb{P}(\mathtt{Imblce}_{\ell}^\mathtt{DIM}) \geq \frac{b}{pe}.$$ It then follows from a union bound argument that

equation*[equation* omitted — 219 chars of source]
center[center omitted — 73 chars of source]

Fix $\mathbf{z} \in \mathcal{X}$ such that $z_{\ell} \leq n^{a - 1}$. Since the order statistics $x_{(n^a),\ell} = n^{a - 1}(1 + o_\mathbb{P}(1))$, on the event $n^a \leq \hat{\imath}_{\text{DIM},\ell} \leq n^b$, if $z_{\ell} \leq (1 + o_\mathbb{P}(1))n^{a-1}$, then $z_{\ell} \leq x_{(n^a)} \leq x_{(\hat{\imath}_{\text{DIM},\ell}),\ell}$, and on the event $\mathtt{Imblce}_{\ell}^\mathtt{DIM}$,

align*[align* omitted — 775 chars of source]

where the second to last line is due to a similar argument as in the proof of part 1. By a symmetry argument for the event $\{n - n^b \leq \hat{\imath}_{\text{DIM},\ell} \leq n - n^a\}$, we have

equation*[equation* omitted — 224 chars of source]

where $\mathcal{X}_n = \{\mathbf{x} \in [0,1]^p: x_j = o(1) n^{a-1} \text{ or } 1 - x_j = o(1) n^{a-1} \text{ for some } j \in [p]\}$, and $\sigma^2 = \mathbb{V}[\frac{d_i y_i(1)}{\xi} + \frac{(1 - d_i) y_i(0)}{1 - \xi}]$.

Proof of Theorem (ref)

Due to the recursive splitting and Theorem (ref), the optimal split index $\hat \imath_\mathtt{DIM}$ at the $k$-th split ($k \geq 1$) also satisfies

align*[align* omitted — 196 chars of source]

Hence the same argument as Part 1 in the proof of Theorem (ref) leads to the result.

Proof of Theorem (ref)

For notational simplicity, denote $\hat\tau^{\mathtt{NSS}}_{\mathtt{DIM}}$ by $\hat{\tau}$, the data-driven partition $\mathcal{D}_{\mathsf{T}}$ by $\mathcal{P}$.

Reduction to least square prediction error.

Observe that the leaf nodes value coincide with a least square projection given $\mathcal{P}$: For $\mathsf{t} \in \mathcal{P}$, we have $\hat{\tau}(\mathsf{t}) = \hat{b}_{\mathsf{t}}$, where

align*[align* omitted — 322 chars of source]

Consider the outcome prediction model based on partition $\mathcal{P}$:

align[align omitted — 278 chars of source]

where

align*[align* omitted — 260 chars of source]

First, we show that for $L_2$-consistency of treatment effect estimation, it is enough to look at the $L_2$ loss for outcome prediction. Denote by $P_{X,d}$ the joint distribution of $(\mathbf{x}_i, d_i)$. Since we assumed $\mathbf{x}_i$ and $d_i$ are independent, we have $P_{X,d} = P_X \times P_d$, where $P_X$ and $P_d$ are the marginal distributions of $X$ and $d$. Given Assumption (ref), the target outcome prediction model is

align*[align* omitted — 177 chars of source]

Hence

align[align omitted — 1,218 chars of source]

It follows that

align*[align* omitted — 206 chars of source]

Error Bound for Least Square Prediction.

Now, we bound the least square error $\mathbb{E}[\lVert \hat{g} - g^{\ast}\rVert^2]$ following the strategy for klusowski2024large. First, assume $|y_i(t)| \leq U$, $i = 1,2, \cdots, n$, $t = 0,1$, for some $U \geq 0$. Decompose by

align*[align* omitted — 64 chars of source]

where

align*[align* omitted — 293 chars of source]

The least square representation (ref) implies that

align[align omitted — 291 chars of source]

which implies

align*[align* omitted — 42 chars of source]

We control $E_1$ using uniform law of large number arguments. Notice that $\hat{g}$ is one member of the class $\mathscr{G}_{n} = \{A(\mathbf{x}) + d \; B(\mathbf{x}): A, B \in \mathscr{H}_n\}$, where $\mathscr{H}_n$ is the class of piecewise constant functions (bounded by $U$) on partitions $\mathbb{P} \in \Pi_n$. Here

align*[align* omitted — 183 chars of source]

is the family of all achievable partitions $\mathcal{P}$ by growing a depth $K$ binary tree on $n$ points by iteratively splitting in $\mathbf{x}$-space based on any criterion. By klusowski2024large,

align*[align* omitted — 164 chars of source]

A union bound then gives

align*[align* omitted — 166 chars of source]

where $P_{X^n}$ is the empirical measure based on $X^n = (X_1, \cdots, X_n)$, $X_i \in \mathbb{R}^p$ for all $i$. Since $\hat{g} \in \mathscr{G}_n$, we can then use Gyorfi-etal_2002_book to get

align*[align* omitted — 511 chars of source]

Choosing $\alpha \propto \frac{U^4 2^K \log(np)}{n}$, and $\beta \propto \frac{U^2}{n}$, then we have

align*[align* omitted — 128 chars of source]

where $C$ is a positive universal constant.

Now we relax the condition that $|y_i(t)| \leq U$. Take $A = \{|y_i(t)| \leq U, \forall i = 1, \cdots, n, t = 0,1\}$. Then

align[align omitted — 371 chars of source]

A union bound gives

align*[align* omitted — 164 chars of source]

Using Cauchy-Schwarz inequality,

align*[align* omitted — 284 chars of source]

Choosing $U = \max\{\mu_0, \mu_1\} + 4 \log(n)$, we have

align*[align* omitted — 98 chars of source]

for some absolute constant $C$. Putting it back to Equation (ref), we get the desired conclusion.

For the high probability bound, the same analysis as Equation (ref) in almost sure sense gives

align*[align* omitted — 183 chars of source]

almost surely. Using sub-exponentianity of $\varepsilon_i(t)$,

align*[align* omitted — 181 chars of source]

with probability at least $n^{-C_2}$, where $C_1$ and $C_2$ are some positive absolute constants. Sub-exponentianity of $\varepsilon_i(t)$, $1 \leq i \leq n$, $t = 0,1$, implies that $\mathbb{P}(A^c) = n^{-C_3}$ if we choose $U = C_4 \log(n)$, where $C_3$ and $C_4$ are positive constants only depending on the distribution of $(\varepsilon_i(0), \varepsilon_i(1))$. Combining with the previous two inequalities, we get the second conclusion.

Proof of Theorem (ref)

Recall $(\hat \imath, \hat \jmath)$ denotes the optimal splitting index and coordinate for the decision stump. Denote $\hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}(\mathbf{x})$ by $\check{\tau}(\mathbf{x})$ for simplicity. We use $(y_i,\mathbf{x}_i^{\top})_{i=1}^M$ to denote $\mathcal{D}_{\mathtt{HON},1}$, which we used to construct the causal tree. Denote by $(\hat \imath, \hat \jmath)$ the splitting index and coordinate at the $K_n$-th step, based on $\mathcal{D}_{\mathtt{HON},1}$.

Use $(\tilde{y}_i,\tilde{\mathbf{x}}_i^{\top})_{i=1}^N$ to denote $\mathcal{D}_{\mathtt{HON},2}$. Then

align*[align* omitted — 707 chars of source]

Since $(\tilde{\varepsilon}_i(0), \tilde{\varepsilon}_i(1)) \protect\mathpalette{\protect\independenT}{\perp} \tilde{\mathbf{x}}_i$, condition on $\hat{\imath}, \hat{\jmath}$ and $\mathbf{X} = (\mathbf{x}_1, \cdots, \mathbf{x}_n)$, $\tilde{\mathbf{X}} = (\tilde{\mathbf{x}}_1, \cdots, \tilde{\mathbf{x}}_n)$, we have

align*[align* omitted — 794 chars of source]

where

align*[align* omitted — 141 chars of source]

Call

align*[align* omitted — 233 chars of source]

\paragraph*{High probability lower bound on $Z$.} Denote $n_1 = \sum_{i=1}^{\tilde{\iota}} d_i, n_0 = \sum_{i=1}^{\tilde{\iota}} (1-d_i), n_0 + n_1 = \tilde{\iota}.$ And consider the weights $w_i = \frac{d_i}{n_1} - \frac{1-d_i}{n_0}$, so that $Z = |\sum_{i=1}^{\tilde{\iota}} w_i \varepsilon_i|.$ By Marcinkiewicz–Zygmund inequality and a Jensen's inequality on the square root function, for some absolute constant $c_{\mathrm{MZ}}$,

align*[align* omitted — 995 chars of source]

Moreover, Assumption (ref) implies that $$ \mathbb{E}[Z^{2} \mid \bm D, \tilde{\iota}] = \frac{\sum_{i=1}^{\tilde{\iota}} d_i \mathbb{E}[\varepsilon_i(1)^{2}]}{n_1^{2}} + \frac{\sum_{i=1}^{\tilde{\iota}} (1-d_i) \mathbb{E}[\varepsilon_i(0)^{2}]}{n_0^{2}} \ge \left(\frac{1}{n_1} + \frac{1}{n_0}\right) \min\{\mathbb{V}[\varepsilon_i(1)], \mathbb{V}[\varepsilon_i(0)]\}. $$ The Paley-Zygmund inequality implies for $\theta \in (0,1)$,

align[align omitted — 483 chars of source]

Condition on $\tilde{\iota}$, $n_0 \sim \mathsf{Bernoulli}(\tilde{\iota}, \xi)$. Hence

align*[align* omitted — 349 chars of source]

We claim that whenever $s_n \leq \hat{\iota} \leq n - s_n$,

align[align omitted — 169 chars of source]

It then follows from Equation (ref) that whenever $s_n \leq \hat{\iota} \leq M - s_n$,

align*[align* omitted — 404 chars of source]

Choose $\theta = 1/2$, and take

align*[align* omitted — 197 chars of source]

Then by Theorem (ref), we have

align*[align* omitted — 265 chars of source]

We can show via the same argument as Theorem (ref) that $\liminf_{n \to \infty}\mathbb{P}(\tilde{\iota} \leq n^b/2|\hat{\iota} \leq n^b) = 1$. Hence

align*[align* omitted — 239 chars of source]

\paragraph*{Proof of Equation (ref).} Let $F$ be the cummulative distribution function of $\mathbf{x}_i$. Suppose $1 \leq k \leq n/2$. Then $F(\mathbf{x}_{(k)}) \sim \mathsf{Beta}(k, M - k + 1)$. By a Bernstein bound for Beta variables skorski_2023, we have for all $\epsilon > 0$,

align*[align* omitted — 113 chars of source]

where for large enough $n$,

align*[align* omitted — 84 chars of source]

Hence with probability at least $1 - s_n^{-1}$,

align*[align* omitted — 81 chars of source]

Condition on $\mathbf{X} = (\mathbf{x}_1, \cdots, \mathbf{x}_n)$, $\mathds{1}(\tilde{\mathbf{x}}_i \leq \mathbf{x}_{(k)})$'s are i.i.d $\mathsf{Bernoulli}(F(\mathbf{x}_{(k)}))$. Hence condition on $\mathbf{X}$ and $\hat{\iota}$, with probability at least $1 - s_n^{-1}$,

align*[align* omitted — 219 chars of source]

It follows that on $s_n \leq \hat{\iota} \leq M - s_n$, using boundedness of $(1 - \frac{1}{\tilde{\iota}})\mathds{1}(\tilde{\iota} > 0)$,

align*[align* omitted — 491 chars of source]

Proof of Theorem (ref)

For simplicity, denote $\hat\tau^{\mathtt{HON}}_{\mathtt{DIM}}$ by $\hat{\tau}$. Since given the partition $\mathcal{P}$ chosen by $\mathcal{D}_{\mathsf{T}}$, Equation (ref) is still satisfied. We can use the same argument in the proof of Theorem (ref) condition on $\mathcal{D}_{\mathsf{T}}$ to get

align*[align* omitted — 161 chars of source]

where $C$ is a positive constant that only depends on $\xi$, $\mu$ and the distribution of $\varepsilon_i(0), \varepsilon_i(1)$. In particular, the expectation is taken with respect to $\mathcal{D}_{\tau}$ with effective sample size $n_{\tau}$.

Since condition on $\mathcal{D}_{\mathsf{T}}$, the partition $\mathcal{P}$ is fixed, we can use the same argument as in Theorem (ref) to show that $\hat{g}$ lies in a class $\mathscr{H}_{n_{\mathsf{T}}}[\mathcal{P}]$ with covering number,

align*[align* omitted — 193 chars of source]

when we assume $y_i(0)$ and $y_i(1)$ are bounded by $U$. In comparison, in the proof of Theorem (ref), we show $\hat{g}$ lies in $\mathscr{H}_{n_{\mathsf{T}}}$ with covering number

align*[align* omitted — 213 chars of source]

This improvement of covering number due to honesty means we can replace a $ \log(n_{\tau} p)$-penalty in the result of Theorem (ref) by $\log(n_{\tau})$. Now uncondition over $\mathcal{D}_{\mathsf{T}}$ and using the fact that $\rho^{-1} \leq n_{\tau}/n_{\mathsf{T}} \leq \rho$, we get the conclusion.

Proof of Theorem (ref)

The conclusion follows from Theorem (ref) and the same proof for Theorem (ref).

Proof of Theorem (ref)

For simplicity, denote $\hat{\tau}^{\mathcal{X}}_{\mathtt{DIM}}(\mathbf{x}; K)$ by $\tilde \tau(T_K)$, and $N = n/(K+1)$ denotes the sample size for each folds in the $\mathtt{X}$ sample splitting scheme.

Let $\mathsf{t}_1, \mathsf{t}_2, \cdots, \mathsf{t}_{2^K}$ denote the $2^K$ leaf nodes in the decision tree, (if a node cannot be further refined, we duplicate the split indices and values at the next level). And let $N_1, N_2, \cdots, N_{2^K}$ and $m_1, m_2, \cdots, m_{2^K}$ denote the number of observations and the Lebesgue measure of the $2^K$ leaf nodes, respectively. Note that $\Vec{N} = (N_1, \cdots, N_{2^{K}})$ are independent of the $\tilde y_i $ data by the honest condition and the $x_i $ data per Assumption (ref). As in the proof of Theorem (ref), we can show condition on $\Vec{N}$, $m_k \sim \mathsf{Beta}(N_k, N - N_k + 1)$

Thus, the IMSE can be bounded as follows: Since condition on $\Vec{N}$, $m_k$'s are independent to the refreshed samples $\tilde{d_i}, \tilde{\varepsilon}_i(0)$, $\tilde{\varepsilon}_i(1)$'s, we have

align*[align* omitted — 1,232 chars of source]

Notice that condition on $\Vec{N}$, $\sum_{\mathbf{x}_i \in \mathsf{t}_k} d_i \sim \mathsf{Bin}(N_k, \xi)$ and $\sum_{\mathbf{x}_i \in \mathsf{t}_k} 1 - d_i \sim \mathsf{Bin}(N_k, 1 - \xi)$. Using the fact that for a binomial random variable $W \sim \mathsf{Bin}(n,p)$, we have

align*[align* omitted — 94 chars of source]

where $C$ is an absolute constant, and $C_p$ is some constant that only depends on $p$. It follows that

align*[align* omitted — 332 chars of source]

Proof of Lemma (ref)

Assume w.l.o.g. $k \leq n/2$, since the case of $k > n/2$ can be dealt with by symmetry. From the proof of Lemma (ref),

align*[align* omitted — 242 chars of source]

Moreover, the proof of the term $R_1$ in Lemma (ref) implies

align*[align* omitted — 206 chars of source]

Now we consider the randomness induced by $n_0, n_{L,0}, n_{R,0}$. By Theorem A.4.1 in csorgo1997limit,

align*[align* omitted — 164 chars of source]

which implies

align*[align* omitted — 161 chars of source]

Putting together, triangle inequality implies

align*[align* omitted — 217 chars of source]

Proof of Lemma (ref)

The proof of Lemma (ref) implies that

align*[align* omitted — 213 chars of source]

where $\alpha_n = \rho_n \log \log n + \frac{s_n}{n - s_n} \log \log n$, and

align*[align* omitted — 159 chars of source]

Hence it also follows that

align*[align* omitted — 218 chars of source]

When $1 \leq k \leq s_n$, we have $\frac{n_{L,0}(k) n_{R,0}(k)}{n_0} \leq n_{L,0}(k) \leq k$. The conclusion then follows.

Proof of Theorem (ref)

The proof is similar to the proof of Theorem (ref), except that in Theorem (ref), we approximate the split criterion by a time-transformed O-U process, while here we approximate the split criterion by the summation of two independent time transformed O-U processes. We divide the proofs into two steps.

center[center omitted — 94 chars of source]

Let $0 < a < b < 1$. Let $\rho_n$ be a sequence of real numbers taking values in $(0,1)$ to be determined, and take $s_n = \exp((\log n)^{\rho_n})$. Then for large enough $n$, we have $s_n \leq n^a \leq n^b \leq n - s_n$. Consider the event

align*[align* omitted — 161 chars of source]

Equation (A.4.18) and (A.4.20) imply that for each $\ell \in [p]$,

gather*[gather* omitted — 248 chars of source]

Hence

align*[align* omitted — 211 chars of source]

Approximations results from Lemma (ref) (taking $r_n = s_n$) and Lemma (ref), using the same argument as step 1 in the proof of Theorem (ref), with $\log \log \log \log(n)/\log \log(n) \ll \rho_n \ll 1$, then implies

align*[align* omitted — 209 chars of source]

Using a union bound, we get $\mathbb{P}(A_n) \rightarrow 0$ as $n \rightarrow \infty$. Observe that on the event $A_n^c$, the argmax for $\mathscr{I}^{\text{SSE}}$ should be inside $[s_n, n - s_n]$. Hence

align*[align* omitted — 951 chars of source]

Now we focus on the first term. By symmetry in the $p$ coordinates,

align*[align* omitted — 1,407 chars of source]

Then using the fact that $\mathscr{I}^\text{prox}(k,\ell)$ approximates $\mathscr{I}^\text{SSE}(k,\ell)$ from Lemma (ref), we have

align[align omitted — 860 chars of source]

where $v_n = O_\mathbb{P}(\log \log (n) s_n^{-1/2})$.

center[center omitted — 77 chars of source]

Recall that

align[align omitted — 231 chars of source]

and we will show that high dimensional random vector $\Xi$ from concatenating $(\sqrt{(1 - \xi) \frac{k (n - k)}{n}}(\bar \mu_{L,0}(k,\ell) - \bar \mu_{R,0}(k,\ell)): k \in [n], \ell \in [p])$ and $(\sqrt{(1 - \xi) \frac{k (n - k)}{n}}(\bar \mu_{L,1}(k,\ell) - \bar \mu_{R,1}(k,\ell)): k \in [n], \ell \in [p])$ can be approximated by a Gaussian random vector with the same covariance structure. The proof will still be based on writing $\Xi$ as $\frac{1}{\sqrt{n}}\sum_{i = 1}^n \mathbf{C}_i$ where

align*[align* omitted — 449 chars of source]

where $\# \pi^{\ell}$ denotes the inverse mapping of $\pi^{\ell}$, as in the proof of Theorem (ref).

Notice that the random vectors are $2 n p$ dimensional. For notational simplicity, in what follows, denote by $\mathbf{e}_{t,k,\ell}$ the indicator of the position corresponding to $\sqrt{(1 - \xi) \frac{k (n - k)}{n}}(\bar \mu_{L,t}(k,\ell) - \bar \mu_{R,t}(k,\ell))$, $t = 0,1$, $k \in [n]$, $\ell \in [p]$.

However, the format of Equation (ref) induces a different geometry when approximating probabilities in Equation (ref). Instead of high dimensional CLT for hyper-rectangles, we consider the class of simple convex sets chernozhukov2017central.

Let $\mathcal{J}$ be a subset of $[n] \times [p]$. Consider the class of closed convex sets $\mathcal{A}$ containing sets of the form

align[align omitted — 227 chars of source]

where $B_2(r)$ denotes the Euclidean ball centered at $\mathbf{0}$ with radius $r$ in $\mathbb{R}^2$. That is, the class $\mathcal{A}$ contains intersections of cylinders $\{\mathbf{u} \in \mathbb{R}^{2np}: \lVert {(\mathbf{e}_{j_1}^\top \mathbf{u}, \mathbf{e}_{j_2}^\top \mathbf{u})} \rVert_2 \leq s\}$. Notice that for $z \in (0, n]$, the event in Equation (ref) (inside $\sup z$) can be characterized as the high dimensional vector $\Xi$ lies in a set in $\mathcal{A}$.

For each $A \in \mathcal{A}$, we consider its approximation by simple convex sets. For each $B_2(r)$, denote by $B_2^{\text{in},n}(r)$ and $B_2^{\text{out},n}(r)$ its inscribed and circumscribed regular $n^2$-gon. Take $m = n^2 |J|$. Then for each $A \in \mathcal{A}$ of the form (ref), take

align*[align* omitted — 225 chars of source]

and

align*[align* omitted — 237 chars of source]

Then $A^m \subseteq A \subseteq A^{m,\epsilon}$. Moreover, denote by $\mathcal{V}(A^m)$ the set consisting of $m$ unit vectors that are outward normal to the facets of $A^m$. Then $A^m$ can be alternatively characterized by

align*[align* omitted — 230 chars of source]

Then we can analogously characterize $A^{m,\epsilon}$ by

align*[align* omitted — 265 chars of source]

where $\epsilon_{\mathbf{v}} \leq n^{-1}$ for large enough $n$. This shows our class $\mathcal{A}$ is a subclass of $\mathcal{A}^\text{si}(1, 3)$ (see chernozhukov2017central). Now we check its conditions (M.1'), (M.2') and (E.1'). Let $\mathbf{v} \in \mathcal{V}(A^m)$. The definition of $A^m$ implies $\mathbf{v} = v_{0,k,\ell} \mathbf{e}_{0,k,\ell} + v_{1,k,\ell} \mathbf{e}_{1,k,\ell}$ for some $(k,\ell) \in \mathcal{J}$, and $v_{0,k,\ell}^2 + v_{1,k,\ell}^2 = 1$. Let $\mathbf{v} \in \mathcal{V}(A^m)$.

align*[align* omitted — 1,041 chars of source]

which verifies (M.1'). The fact that only two entries of $\mathbf{v}$ are nonzero and $v_{0,k,\ell}^2 + v_{1,k,\ell}^2 = 1$ implies that

align*[align* omitted — 273 chars of source]

where the last inequality is from the calculation in Equation (ref), and this verifies (M.2') for the third moment. Moreover,

align*[align* omitted — 485 chars of source]

The same logic shows that $\mathbb{E}[\exp(|\mathbf{v}^\top \mathbf{C}_i|/(K\sqrt{n / r_n}))] \leq 2$, where $K$ is an absolute constant. Putting together, we verify conditions (M.2') and (E.1') with $B_n = \sqrt{n / r_n}$. Hence by chernozhukov2017central, there exists mean-zero random vectors $\boldsymbol{D}_i \sim N(\mathbf{0}, \mathbb{E}[\mathbf{C}_i \mathbf{C}_i^\top])$ such that

align[align omitted — 225 chars of source]
center[center omitted — 69 chars of source]

Observe that for any $k_1, k_2 \in [n], \ell_1, \ell_2 \in [p]$, we have $\operatorname{Cov}[\mathbf{e}_{0,k_1,\ell_1}^\top \mathbf{C}_i, \mathbf{e}_{1, k_2, \ell_2}^\top \mathbf{C}_i] = 0$. The same calculation as Multivariate Case Step 2 for the proof of Theorem (ref) implies we can replace $\boldsymbol{D}_i$ by another mean-zero Gaussian random vector $\mathbf{Z}_i$ such that

align*[align* omitted — 368 chars of source]

We want to show $\frac{1}{\sqrt{n}}\sum_{i = 1}^n \mathbf{Z}_i$ is close to $\frac{1}{\sqrt{n}}\sum_{i = 1}^n \boldsymbol{D}_i$, measured by the probability of taking value in sets from $\mathcal{A}$ defined at Equation (ref). We omit details for simplicity, but illustrate the main skeleton here. As in Step 2, Nazarov inequality implies we only need to work on the $m$-generated convex approximation with $\epsilon$ precision $A^m = A^m(A)$ for $A \in \mathcal{A}$, for a reason given in chernozhukov2017central. Moreover, $\mathbb{P}(\frac{1}{\sqrt{n}}\sum_{i = 1}^n \mathbf{Z}_i \in A^m) = \mathbb{P}((\mathbf{v}^\top (\frac{1}{\sqrt{n}}\sum_{i = 1}^n \mathbf{Z}_i ))_{\mathbf{v} \in \mathcal{V}(A^m)} \leq \mathbf{t})$ for some $\mathbf{t} \in \mathbb{R}^m$. Hence we only need to show

align*[align* omitted — 341 chars of source]

But the definition of $\mathcal{A}$ in Equation (ref) implies for any $A \in \mathcal{A}, \mathbf{v} \in \mathcal{V}(A^m)$, there exists $\mathbf{e}_k, \mathbf{e}_j$ and $v_k^2 + v_j^2 = 1$ such that $\mathbf{v} = v_k \mathbf{e}_k + v_j \mathbf{e}_j$, with

align*[align* omitted — 364 chars of source]

and hence

align*[align* omitted — 190 chars of source]

Together with Equation (ref), we know

align*[align* omitted — 445 chars of source]

The Gaussian-to-Gaussian Comparison result chernozhuokov2022improved them implies

align[align omitted — 216 chars of source]
center[center omitted — 73 chars of source]

Now we revisit Equation (ref). Consider

align*[align* omitted — 227 chars of source]

with

align*[align* omitted — 330 chars of source]

Equations (ref) and (ref) imply that

align*[align* omitted — 955 chars of source]

The same argument as csorgo1997limit shows that there exists two independent standard Brownian bridges over $[0,1]$, $B_{n,L}$ and $B_{n,R}$, for each $n$, such that

gather*[gather* omitted — 469 chars of source]

with $\sigma^2 = \mathbb{V}[\varepsilon_i(0)] = \mathbb{V}[\varepsilon_i(1)]$ and $\epsilon_n = o_{\mathbb{P}}((\log \log n)^{-1/2})$. Let $\{U_L(t): t \in \mathbb{R} \}$ and $\{U_R(t): t \in \mathbb{R}\}$ be two independent O-U processes with $\mathbb{E}[U_j(t)] = 0$ and $\mathbb{E}[U_j(s) U_j(t)] = e^{-|s-t|}$, $j = L,R$. Then

align*[align* omitted — 199 chars of source]

Take $N(t) = \lVert (U_L(t), U_R(t)) \rVert_2, t \in \mathbb{R}$. Then a time change and stationarity of O-U process implies

align*[align* omitted — 482 chars of source]

and

align*[align* omitted — 244 chars of source]

An expansion based on horvath1993maximum (Lemma TODO) gives for any $z \in \mathbb{R}$,

align*[align* omitted — 201 chars of source]

Moreover, Gaussian correlation inequality latala2017royen and stationarity of O-U process implies

align*[align* omitted — 687 chars of source]

Putting together and choosing $z^\ast$ that maximizes $z \mapsto \exp(- 2 e^{-z + \log(2 - (b - a))}) - \exp(-2 e^{-z + \log(c)})$, we can get

align*[align* omitted — 738 chars of source]

Symmetry then implies for any $0 < a < b < 1$ and $\ell \in [p]$, we have

equation*[equation* omitted — 284 chars of source]

Proof of Corollary (ref)

Notice that although the splitting criteria is different from the regression tree, once cells are given the estimator given by the fit-based tree is exactly the same as the regression tree (see Section (ref)). Hence result can be proved based on Theorem (ref) and the same logic as Theorem (ref).

Proof of Corollary (ref)

Notice that although the splitting criteria is different from the regression tree, once cells are given the estimator given by the fit-based tree is exactly the same as the regression tree (see Section (ref)). Hence result can be proved based on Theorem (ref) and the same logic as Theorem (ref).

Proof of Corollary (ref)

Since the tree is constructed by minimizing the objective Equation (ref) iteratively. The empirical risk minimization property Equation (ref) still holds. Hence the result follows from the same argument as the proof of Theorem (ref).

Proof of Corollary (ref)

Notice that although the splitting criteria is different from the regression tree, once cells are given the estimator given by the fit-based tree is exactly the same as the regression tree (see Section (ref)). Hence result can be proved based on Theorem (ref) and the same logic as Theorem (ref).

Proof of Corollary (ref)

Since the tree is constructed by minimizing the objective Equation (ref) iteratively. The empirical risk minimization property Equation (ref) still holds. Hence the result follows from the same argument as the proof of Theorem (ref).

Proof of Corollary (ref)

Notice that although the splitting criteria is different from the regression tree, once cells are given the estimator given by the fit-based tree is exactly the same as the regression tree (see Section (ref)). Hence result can be proved based on Theorem (ref) and the same logic as Theorem (ref).

Proof of Corollary (ref)

The result follows from the same argument as Theorem (ref).

Proof of Lemma (ref)

First, we consider $\mathtt{X}$ under Assumption (ref). Since $(y_i, d_i)$'s are from dataset $\mathscr{D}_{\tau}$ independent to the dataset $\mathscr{D}_{\mathsf{T}_1}$ to $\mathscr{D}_{\mathsf{T}_K}$ for tree construction, it is easy to check that

align*[align* omitted — 258 chars of source]

Next, we consider $\mathtt{HON}$ under Assumption (ref). Denote by $\mathsf{t}(\mathbf{x})$ the node that contains $\mathbf{x}$, and denote by $n(\mathsf{t})$ the local sample size in $\mathcal{D}_{\tau}$, where $n(\mathsf{t}) = \sum_{i \in \mathcal{D}_{\tau}}\mathds{1}(\mathbf{x}_i \in \mathsf{t})$. Then

align*[align* omitted — 617 chars of source]

where in the third line, we have used the fact that $\varepsilon_i(0)$ and $\varepsilon_i(1)$ in $\mathcal{D}_{\tau}$ are independent to $\mathbf{x}_i$'s in $\mathcal{D}_{\tau}$ and the whole dataset $\mathcal{D}_{\mathsf{T}}$, with $\mathbb{E}[\varepsilon_i(0)] = \mathbb{E}[\varepsilon_i(1)] = 0$. Unconditioning over $\mathcal{D}_{\mathsf{T}}$, then we get

align*[align* omitted — 134 chars of source]

The results for $\mathtt{DIM}$ and $\mathtt{SSE}$ can be obtained by similar arguments.

Finally, we consider $\mathtt{NSS}$ under Assumption (ref) and the additional symmetric error $\varepsilon_i(0)$, and $\varepsilon_i(1)$ assumption. We will use an induction assumption.

Base case: $K = 1$. Due to the assumption that $\mu_0$ and $\mu_1$ are constant, we can rewrite the splitting criteria from Definition 2 in the main paper as

align[align omitted — 597 chars of source]

and

align[align omitted — 467 chars of source]

and

align[align omitted — 838 chars of source]

Denote the vector $\boldsymbol{\varepsilon} = (\varepsilon_1(0), \varepsilon_1(1), \cdots, \varepsilon_n(0), \varepsilon_n(1))$. Notice that for all three criteria, for any $\mathbf{d} = (d_1, \cdots, d_n)$ and $\mathsf{t}_L, \mathsf{t}_R$, $\varepsilon = \mathbf{u}$ and $\varepsilon = - \mathbf{u}$ give the same value. Hence condition on $\mathbf{d}$ and the data-driven split region $\hat{\mathsf{t}}_L$ and $\hat{\mathsf{t}}_R$, $\boldsymbol{\varepsilon}$ is symmetrically distributed around zero. It then follows from the form of the three estimators that all of them are unbiased.

Induction step: $K \geq 2$. Each leaf node $\mathsf{t}$ in layer $K-1$ is further partitioned into $\mathsf{t}_L$ and $\mathsf{t}_R$ such that Equations (ref), (ref) and (ref) are maximized. The induction hypothesis is that condition on all leaf $\mathsf{t}$ in the $K-1$ th layer and $\mathbf{d}$, $\boldsymbol{\varepsilon}$ is symmetrically distributed around zero. Again for all three criteria, given $K-1$th leaf node $\mathsf{t}$, for any $\mathbf{d} = (d_1, \cdots, d_n)$ and $\mathsf{t}_L, \mathsf{t}_R$, $\varepsilon = \mathbf{u}$ and $\varepsilon = - \mathbf{u}$ give the same value. Hence the resulting $K$th level $\hat{\mathsf{t}}_L$ and $\hat{\mathsf{t}}_R$ are such that condition on $\mathbf{d}$ and the data-driven split region $\hat{\mathsf{t}}_L$ and $\hat{\mathsf{t}}_R$, $\boldsymbol{\varepsilon}$ is symmetrically distributed around zero, making the estimators unbiased.