Suppose $f$ and $g$ are Boolean functions without a constant term, and where every variable has the same influence. How to show every variable will have the same influence in $f \circ g$?

To me it seems like influence of a variable in $f \circ g$ is the product of influence of the outer variable in $f$ with the influence of the variable in $g$, but I'm not sure

  • $\begingroup$ Have you tried to prove your suspicion? $\endgroup$ Commented Apr 25, 2022 at 17:25
  • $\begingroup$ Yes, but I'm not sure where I'm using the fact that the functions dont have a constant term in the polynomial representation $\endgroup$ Commented Apr 25, 2022 at 17:44
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    $\begingroup$ You want a random input to $f \circ g$ to translate to a random input to $f$, which requires $g$ to be balanced. $\endgroup$ Commented Apr 25, 2022 at 17:46
  • $\begingroup$ Sorry for the late reply but could you explain this in a little more detail? I just saw your answer and I'm still a little confused $\endgroup$ Commented May 10, 2022 at 13:56

1 Answer 1


Suppose that $f\colon \{\pm1\}^n \to \{\pm1\}$ and that $g\colon \{\pm1\}^m \to \{\pm1\}$ is balanced. The composed function $f \circ g\colon \{\pm1\}^{nm} \to \{\pm1\}$ is given by $$ (f \circ g)(x) = f\bigl(g(x_{1,1},\ldots,x_{1,m}),\ldots g(x_{n,1},\ldots,x_{n,m})\bigr). $$ The influence of $x_{i,j}$ is the probability that if we sample $x \in \{\pm1\}^{nm}$ and construct $x'$ by flipping $x_{i,j}$ then $(f \circ g)(x) \neq (f \circ g)(x')$. This happens if:

  1. $g(x_{i,1},\ldots,x_{i,m}) \neq g(x'_{i,1},\ldots,x'_{i,m})$.
  2. $f(y_1,\ldots,y_n) \neq f(y'_1,\ldots,y'_n)$, where $y_i = g(x_{i,1},\ldots,x_{i,m})$ and $y'_i = g(x'_{i,1},\ldots,x'_{i,m})$.

The first property happens with probability $\operatorname{Inf}_j[g]$.

Since $g$ is balanced, the vector $(y_1,\ldots,y_n)$ is uniformly random. Therefore, given that the first property happens, the second property happens with probability $\operatorname{Inf}_i[f]$. In total, $$ \operatorname{Inf}_{i,j}[f \circ g] = \operatorname{Inf}_i[f] \operatorname{Inf}_j[g]. $$

Here is a calculational proof. Recall that $$ \operatorname{Inf}_i[f] = \sum_{i \in S} \hat{f}(S)^2. $$ The Fourier expansion of $f \circ g$ is $$ f \circ g = \sum_{S \subseteq [n]} \sum_{\substack{T_i \subseteq [m] \\ \text{for all } i \in S}} \hat{f}(S) \prod_{i \in S} \hat{g}(T_i) \prod_{i \in S} \prod_{j \in T_i} x_{i,j}. $$ Since $g$ is balanced, every monomial appears exactly once: $S$ needs to be the set of $i$ indices that appear in the monomial, and for each $i$, $T_i$ needs to be the set of $j$ indices such that $x_{i,j}$ appears in the monomial. (If $g$ were unbalanced, then $S$ could be any superset of the set of $i$ indices appearing in the monomial, with $T_i = \emptyset$ for any $i$ not appearing in the monomial.) Therefore \begin{align} \operatorname{Inf}_{i,j}[f \circ g] &= \sum_{i \in S} \sum_{\substack{T_k \, \forall k \in S \\ j \in T_i}} \hat{f}(S)^2 \prod_{k \in S} \hat{g}(T_k)^2 \\ &= \sum_{i \in S} \hat{f}(S)^2 \cdot \sum_{j \in T_i} \hat{g}(T_i)^2 \cdot \prod_{\substack{k \in S \\ k \neq i}} \sum_{T_k} \hat{g}(T_k)^2 \\ &= \sum_{i \in S} \hat{f}(S)^2 \cdot \operatorname{Inf}_j[g] \\ &= \operatorname{Inf}_i[f] \cdot \operatorname{Inf}_j[g], \end{align} using $$ \sum_T \hat{g}(T)^2 = 1, $$ since $g^2 = 1$.

  • $\begingroup$ Where can I read about that kind of function composition that you're assuming/defining? I'm only familiar with the normal one described for example here. $\endgroup$ Commented May 10, 2022 at 17:37
  • $\begingroup$ It appears in the literature on query-to-communication lifting and in the literature on the KRW conjecture, for example. Here is a video lecture on lifting. $\endgroup$ Commented May 10, 2022 at 17:48
  • $\begingroup$ Hmm, I can't find the string "lift" in Wikipedia's function composition article. I only skimmed a bit of the video, but he first uses "$\circ$" after 8 minutes in $f \circ g^n$. Are you sure you're not talking about that, instead of about $f \circ g$? $\endgroup$ Commented May 10, 2022 at 18:24
  • $\begingroup$ It's the same thing. The notation $f \circ g^n$ is more pedantic. $\endgroup$ Commented May 10, 2022 at 18:25

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