I am not an expert on circuits and I wonder whether the following problem was already studied (and possibly solved). Any reference or suitable method to solve this question would be welcome.
Let $C_{n}$ be the Boolean circuit satisfying the following conditions:
- the depth of $C$ is $2n$,
- the gates of odd level are binary $\operatorname{AND}$ gates,
- the gates of even level are binary $\operatorname{OR}$ gates.
The circuit $C_{n}$ computes a function $f_n$ which associates to each binary word of
length $2^{2n}$ a Boolean value.
For instance, for $n = 1$, $f_1(1010) = 0$ and $f_1(0011) = 1$, as shown in the
figure below:
For $0 \leqslant k \leqslant 2^{2n}$, let $L_k = \{ u \in \{0, 1\}^{2^{2n}} \mid |u|_1 = k \}$,
a language of size
$$
\binom{2^{2n}}{k} = \frac{2^{2n}!}{k! (2^{2n} - k)!}
$$
Here is an instance of the problem I have in mind: if 40\% of the bits of the input
$x$ are equal to $1$, what is the expected value of $f_n(x)$? Here are a few more precise instances of this problem.
Compute the average value of $f_n(x)$ on $L_k$, that is, $$ \frac{k! (2^{2n} - k)!}{2^{2n}!} \sum_{x \in L_k} f_n(x), $$
In particular, let $c \in [0,1]$ be a real number and let $c_n = \lfloor c2^{2n}\rfloor$. Can one estimate the average value of $f_n(x)$ on $L_{c_n}$, that is, $$ \frac{c_n! (2^{2n} - c_n)!}{2^{2n}!} \sum_{x \in L_{c_n}} f_n(x)\ ? $$
Does the limit of the previous expression exists when $n$ tends to $\infty$? $$ \lim_{n \to \infty} \frac{1}{\binom{2^{2n}}{c_n}} \sum_{x \in L_{c_n}} f_n(x) = \lim_{n \to \infty} \frac{c_n! (2^{2n} - c_n)!}{2^{2n}!} \sum_{x \in L_{c_n}} f_n(x) $$
