theorem states that if $\chi_1,\chi_2,\dots,\chi_k$ each have a small correlation with $f$ and, in addition, have small pairwise correlations, then a distribution exists with respect to which $f$ is completely uncorrelated with $\chi_1,\chi_2,\dots,\chi_k.$ The latter part of the hypothesis, namely the requirement (\[eqn:diag-dominance\]) of small pairwise correlations for $\chi_1,\chi_2,\dots,\chi_k,$ may seem unnecessary at first. In actuality, it is vital. Exponential lower bounds on the weights of linear perceptrons [@myhill-kautz61; @siu91small-weights] imply, by linear programming duality, the existence of functions $f,\chi_1,\chi_2,\dots,\chi_k\colon X\to\moo$ such that $\abs{\langle f,\chi_i\rangle} =
\exp\{-\Theta(k)\},$ $i=1,2,\dots,k,$ and yet $$\begin{aligned}
f(x) \equiv \sign {\left(\sum_{i=1}^k \alpha_i\chi_i(x)\right)}
\label{eqn:linear-comb}\end{aligned}$$ for some fixed reals $\alpha_1,\dots,\alpha_k.$ In this construction, the correlation of $f$ with each $\chi_i$ is small, in fact exponentially smaller than what is assumed in Theorem \[thm:distribution-by-inversion\]; nevertheless, the representation (\[eqn:linear-comb\]) rules out a distribution $\mu$ with respect to which $f$ could have zero correlation with each $\chi_i,$ for such a distribution $\mu$ would have to obey $$\begin{aligned}
0<\Exp_\mu\left[\left|\sum_{i=1}^k
\alpha_i\chi_i(x)\right|\right]=\Exp_\mu\left[f(x)\sum_{i=1}^k
\alpha_i\chi_i(x)\right]=\sum_{i=1}^k\alpha_i\Exp_\mu[f(x)\chi_i(x)]=0.\end{aligned}$$

Consider the linear system $$\begin{aligned}
M\alpha = \gamma
\label{eqn:lin-system}\end{aligned}$$ in the unknown $\alpha\in\Re^k,$ where $M=[\langle \chi_i,\chi_j\rangle]_{i,j}$ is a matrix of order $k$ and $\gamma = (\langle f,\chi_1\rangle, \dots, \langle f,\chi_k\rangle)\in\Re^k.$ Then (\[eqn:diag-dominance\]) shows that $M$ is strictly diagonally dominant and hence 