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 $z \in \{0, 1, \ldots, 2^{n} - 1\}$, where $x \cdot z$ is the bitwise inner product between the binary representations of $x$ and $z$. Let me choose a function uniformly at random from the set of all Boolean functions $$\{f : \{0, 1\}^{n} \rightarrow \{-1, 1\}\}. $$

Is there any "nice" name/form to the distribution of the $2^{n}$-tuple

\begin{equation} \mathbf{\hat f} =\bigg(\hat f(0)^{2}, \hat f(1)^{2}, \ldots, \hat f(2^{n} - 1)^{2}\bigg)? \end{equation}

More specifically, given $k$ samples $ \big(z_{1}, z_{2}, \ldots, z_{k})$ from the categorical distribution $\operatorname{Categotical}\big(2^{n},\mathbf{\hat f}\big)$, I am trying to find

\begin{equation} \mathrm{E}\bigg(\hat f(z)^{2} \mid \big(z_{1}, z_{2}, \ldots, z_{k} \big)\bigg), \end{equation}

for a fixed $z$.

## My attempt:

We know that \begin{equation} \sum_{z = 0}^{2^{n} - 1} \hat f(z)^{2} = 1. \end{equation} There are a few more properties that might be of interest. It can be seen that \begin{equation} \frac{2^{n}}{2}\big(\hat f(z) + 1\big) \end{equation} follows a binomial distribution with $2^{n}$ trials and success probability $\frac{1}{2}$, for each $z$. From there, one can reason that each $\hat f(z)^{2}$ is identically distributed (but not independent) and compute the mean of each $\hat f(z)^{2}$ to be $\frac{1}{2^{n}}$ and variance to be $\frac{2^{n} - 1}{2^{3n-1}}$. But I can't get much beyond this point.

Is there anything else we can say about the distribution and, finally, the expected value? Maybe it is a "discrete-variant" of a Dirichlet distribution, but I don't know how to formalize that.