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Let $\mathcal{X} \subseteq \mathbb{R}^n$ be a measurable space, Fix $m,q \in \mathbb{N}^*$, (the $m$ $x_i$'s are i.i.d and follow distribution $\mathcal{D}$)

and $X = (x_1, \dots, x_m) \sim \mathcal{D}_{\mathcal{X}}^m$.

Consider a finite set of functions $\mathcal{F} = \Big \{ h_1, \dots h_q \Big\}$ that takes the values:$\{-1,1\}$

We are interested in the following set:

$$\mathcal{F}(X) = \Big\{(h_1(x_i), \dots h_q(x_i)), i \in [m]\Big\}$$

such that at least two elements of $\mathcal{F}$ disagree on $X$ (disagree on all samples $x_1, \dots, x_m$): That is :

$$\forall l \in [m], \exists i \neq j \in [q]: h_i(x_l) \neq h_j(x_l)$$

I'm interested in the two following questions :

1- what is a tight upper bound on $|\mathcal{F}(X)|$?

2- what can we say about $\mathbb{E}[|\mathcal{F}(X)|]$?

EDIT:

Previously (before the edit), I represented $\mathcal{F}(X)$ in the matrix form by where each row represents a vector belonging to $\mathcal{F}(X)$,so it can be seen as:

$$ \mathcal{F}(X) = \begin{pmatrix} h_1(x_1) & h_2(x_1)& \cdots & h_q(x_1) \\ h_1(x_2) & h_2(x_2)& \cdots & h_q(x_2) \\ \vdots & \cdots & \cdots & \vdots \\ h_1(x_m) & h_2(x_m) & \cdots & h_q(x_m) \end{pmatrix} $$

Counting elements of $\mathcal{F}(X)$ is equivalent to counting different rows in the matrix representation.

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    $\begingroup$ You could just have a 2-column matrix with as many rows as you want where each row is (1, -1) or (1, -1), so I think you are missing some constraints from the question. $\endgroup$ Commented Dec 20, 2023 at 2:08
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    $\begingroup$ Also, for "expectation" to make sense, there should be some randomness in the construction, but I did not see any. $\endgroup$ Commented Dec 20, 2023 at 6:37
  • $\begingroup$ @GordonRoyle counting rows means counting vectors, if you repeat your row n times doesn't mean your count would increase, I'm interested in the cardinal different rows (different vectors) . $\endgroup$
    – rivana
    Commented Dec 20, 2023 at 12:12
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    $\begingroup$ rivana, when you need to clarify a question you should edit the original so that everyone doesn't have to read all the comments. "number of rows" is not the same as "number of different rows". It seems the answer to Q1 is $2^m-2$ and your example shows only 6 possibilities and not $2^3=8$. $\endgroup$ Commented Dec 20, 2023 at 12:28
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    $\begingroup$ Where are the functions $f_i$ defined? How are $h_i$ related to them? What exactly is $\mathcal{F}(X)$: is it the multiset of rows of the $q\times m$ matrix $(f_i(x_j))$? What do you mean by expectation: the functions $f_1,\ldots,f_q$ are fixed and $x_i$ are i.i.d. in $\mathcal{X}$ (then probably $\mathcal{X}$ is not only measurable, but probability space)? $\endgroup$ Commented Dec 21, 2023 at 8:21

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