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Let $H = \{ 0, 1\}^d$ be the $d$-th Cartesian product of $\{0, 1\}$ in $\mathbb{R}^d$. Suppose $v_1, \ldots, v_k$ are $k$ vectors in $H$ in general position. We define function $F \colon H^{k}\rightarrow \mathbb{R}$ as \begin{align} F (u_1, u_2, \ldots, u_k) = \sum_{i,j = 1}^k f( | \langle v_i, u_j \rangle |), \end{align} where $f$ is a nondecreasing univariate function. Is it possible to characterize the level sets of $F$? That is, partition $H^k$ according to the value of $F$?

Edit: If this problem is too broad. A more concrete formulation is as follows.

Let $v_1, \ldots, v_k$ and $u_1, \ldots, u_k$ be $2k$ i.i.d. random variables that are uniform on $H = \{ 0, 1\}^d$. Let $f \colon \{ 0, \ldots, d\} \rightarrow \mathbb{R}$ be a monotone increasing function. We define a random variable $U$ as \begin{align} U = \sum_{i,j = 1}^k f( | \langle v_i, u_j \rangle|). \end{align} It is possible to say something about the distribution of $U$?

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    $\begingroup$ The hypercube is $[0,1]^d$; do you really mean $\{0,1\}^d$ (the cartesian product of the two-element set)? And from the formula for $F$, it looks like you mean the cartesian product of copies of $H$ (whatever it is), not the tensor product. Straighten out the notation, and don't use blackboard bold $H$ (this is usually reserved for the quaternions). $\endgroup$ – David Handelman Aug 24 '16 at 2:45

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