Let $f : \{0, 1\}^{n} \rightarrow \{-1, 1\}$ be a Boolean function. Let the Fourier coefficients of this function be given by

$$ \hat f(z) = \frac{1}{2^{n}} \sum_{x \in \{0, 1\}^{n}} f(x)(-1)^{x \cdot z}$$

for each $z \in \{0, 1\}^{n}$, where $x \cdot z$ is the bitwise inner product between $x$ and $z$. Let me choose a function uniformly at random from the set of all Boolean functions $$\{f : \{0, 1\}^{n} \rightarrow \{-1, 1\}\}. $$

What is the distribution that each Fourier coefficient $\hat f(z)$ is distributed as?


Each Fourier coefficient on its own is the average of $2^n$ independent uniformly random $\pm1$ variables. Its distribution is roughly normal with mean $0$ and variance $2^{-n}$.

  • $\begingroup$ Why is the mean $1$ and variance $2^{-n}$? $\endgroup$
    – BlackHat18
    Oct 6 '20 at 19:35
  • 1
    $\begingroup$ Oops, mean 0 of course. $\endgroup$ Oct 6 '20 at 20:10

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