Every decision tree has an influential variable
title is self-explanatory
Read Paper →This summary focuses exclusively on the technical aspects of the paper concerning boolean functions \(f:\{-1,1\}^{n}\rightarrow\{-1,1\}\), including definitions, notations, the main theorem, and its proof as they apply to this specific case.
Definitions and Notations
- Input Domain: The domain is \(\{-1,1\}^{n}_{(p)}\), which represents \(n\) binary input variables. The measure is taken with respect to an arbitrary product measure \(p\) on \(\{-1,1\}^{n}\). When written simply \(\{-1,1\}^{n}\), the uniform measure case where \(p=1/2\) is implied.
- Function Output: The output is binary, \(f:\{-1,1\}^{n}\rightarrow\{-1,1\}\).
- Variance (\(Var[f]\)): For a function \(f:\{-1,1\}_{(p)}^{n}\rightarrow\{-1,1\}\), the variance is defined as \(Var[f] = E[f^2] - E[f]^2 = 4 \Pr[f=1]\Pr[f=-1]\). This measures the “balance” of the function.
- Influence (\(Inf_{i}(f)\)): The influence of the \(i\)-th coordinate on \(f\) is defined as \(Inf_{i}(f) = 2 \Pr_{x,x^{(i)}}[f(x) \neq f(x^{(i)})]\). Here, \(x\) is drawn from \(\{-1,1\}_{(p)}^{n}\) and \(x^{(i)}\) is formed by rerandomizing the \(i\)-th coordinate of \(x\) using the probability measure \(p_i\). In the uniform measure case (\(p=1/2\)), this definition aligns with \(Inf_{i}[f] = \Pr[f(x) \ne f(x \oplus i)]\).
- DDT: A decision tree.
- Probability of Querying a Variable (\(\delta_{i}(T)\)): For a DDT \(T\) computing a function \(f:\{-1,1\}_{(p)}^{n}\rightarrow\{-1,1\}\), \(\delta_{i}(T) = \Pr_{x\in\{-1,1\}_{(p)}^{n}}[T \text{ queries } x_{i}]\).
- Expected Cost (\(\Delta(T)\)): For a DDT \(T\), the expected cost is \(\Delta(T) = \sum_{i=1}^{n} \delta_{i}(T) = E[\text{#coords T queries on } x]\).
- Minimum Expected Cost (\(\Delta(f)\)): \(\Delta(f)\) denotes the minimum of \(\Delta(T)\) over all DDTs \(T\) computing \(f:\{-1,1\}_{(p)}^{n}\rightarrow\{-1,1\}\).
Main Theorems and Proofs
Theorem 1.1: Let \(f:\{-1,1\}_{(p)}^{n}\rightarrow\{-1,1\}\) and let \(T\) be a DDT computing \(f\). Then
\[Var[f]\le\sum_{i=1}^{n}\delta_{i}(T)Inf_{i}(f).\]Proof of Theorem 1.1: Let \(x\) and \(y\) be random inputs chosen independently from \(\Omega = (\{-1,1\}^n, \mu_{(p)})\). For a subset \(J\subseteq[n]\), let \(x_{J}y\) denote the hybrid input that agrees with \(x\) on coordinates in \(J\) and with \(y\) on coordinates in \([n]\setminus J\).
Let \(i_1, \dots, i_s\) be the sequence of variables queried by \(T\) on input \(x\). \(s\) is a random variable representing the number of queried variables. For \(t \ge 0\), let \(J[t] = \{i_r : s \ge r > t\}\). It is the dimensions of the last \(s-t\) node in the root to \(x\) path.
last Define \(u[t] = x_{J[t]}y\). \(u[t]\) means the datapoint agreeing with \(x\) on the last \(s-t\) dimensions’ of the nodes in the root to \(x\) path and other dimensions are set to match that of \(y\).
We are slowly transforming \(x\) to \(y\) by slowly changing its dimension one by one to match that of \(y\). We are doing it in a way that if in a step the label changed, that is counted in influence of some node. This is why we start from the bottom and go up. This way when a dimension changes if the labels also change, that is counted in influence of that node.
We start with the observation that \(Var[f] = E[f(x) \neq f(y)] = E[f(u[0]) \neq f(u[s])]\). This follows because \(y = u[s]\) and \(f(x) = f(u[0])\) as \(T\) computes \(f\).
\[E[f(u[0])\neq f(u[s])] \le E\left[\sum_{t=1}^{s} f(u[t-1])\neq f(u[t])\right]\]The above inequality only means if \(f(x)\neq f(y)\) the label should have changed somewhere in the process. Notice that \(u[t-1]\) and \(u[t]\) differ only on one dimension assume \(i(t)\). Splitting the expeted value based on \(i(t)\), we get:
\[Var[f] \leq \sum_{t=1}^{n} \sum_{i=1}^{n} E[(f(u[t-1])\neq f(u[t]))\cdot 1_{\{i_t=i\}}]\]We define \(X_t\) as the sequence of variables that \(T\) queries on \(x\) till time \(t\). \(X_t = (x_{i_1}, x_{i_2}, \cdots, x_{i_{\min(t, s)}})\)
\[Var[f] \leq \sum_{t=1}^{n} \sum_{i=1}^{n} E[E[(f(u[t-1])\neq f(u[t]))\cdot 1_{\{i_t=i\}}|X_{t-1}]]\]Obviously, given the random variable \(X_{t-1}\) we can determine the first \(t\) node in the path thus determining \(i_t\). Proof magic happened here: we can show that conditional on \(X_{t-1}\), the pair \((u[t-1], u[t])\) has the distribution \(\Omega^{(i_t)}\) if \(i_t = i \in [n]\).
This is because \(u[t-1]\) and \(u[t]\) differ only along a single dimension, which is \(i(t)\). Additionally, \(y\) is unrelated to \(X_{t-1}\), so its distribution is independent. In \(u[t] = x_{J[t]} y\), the dimensions not contained in \(J[t]\) are set to match those of \(y\); therefore, they retain their original distribution \(\Omega\). The dimensions in \(J[t]\) are set to those of \(x\), specifically the \(t, t+1, \ldots, s\)-th dimensions along the root-to-\(x\) path. These dimensions are distinct from the first \(t-1\) dimensions of \(x\) and thus also preserve their original distribution. As a result, \(u[t]\) is distributed according to \(\Omega^{(i_t)}\) whenever \(i_t = i \in [n]\). This means:
\[E[(f(u[t-1])\neq f(u[t]))\cdot 1_{\{i_t=i\}}|X_{t-1}]=E_{x, y \sim \Omega^{(i_t)}}[(f(x)\neq f(y))\cdot 1_{\{i_t(x)=i\}}]=Inf_i(f) 1_{i_t(x)=i}.\]Using the above
\[Var[f] \leq \sum_{t=1}^{n} \sum_{i=1}^{n} E[Inf_i(f)\cdot 1_{\{i_t=i\}}] = \sum_{i=1}^{n} Inf_i(f)\sum_{t=1}^{n} E[1_{\{i_t=i\}}] = \sum_{i=1}^{n} Inf_i(f) \delta_i(T)\]Corollary 1.2: For every \(f:\{-1,1\}_{(p)}^{n}\rightarrow\{-1,1\}\), we have
\[\Delta(f)\ge\frac{Var(f)}{Inf_{max}(f)}\]where \(Inf_{max}(f) = \max\{Inf_i(f) : i \in [n]\}\).
Proof of Corollary 1.2: Let \(T\) be a DDT computing \(f\). From Theorem 1.1:
\[Var[f]\le\sum_{i=1}^{n}\delta_{i}(T)Inf_{i}(f)\] \[Var[f]\le Inf_{max}(f)\sum_{i=1}^{n}\delta_{i}(T) = Inf_{max}(f)\cdot\Delta(T)\]Since this holds for any DDT \(T\) computing \(f\), it holds for the one minimizing \(\Delta(T)\), which is \(\Delta(f)\).