Let $k$ and $N_1$ be positive integers and set $N=kN_1$. Partition $[N] := \{1,2,\ldots,N\}$ $k$ disjoint from $G_1,\ldots,G_k$ of each of size $N_1$, and let $\mathcal T(k,N_1)$ be a *transversal* of the $G_i$'s, i.e the collection of subsets of $[N]$ which contain exactly one element from each $G_i$. Note that $\mathcal T$ is isormophic to $G_1 \times \ldots \times G_k$ in an obvious way, and thus $|\mathcal T| = N_1^k$.

Let $x \in \{\pm 1\}^N$ be a random vector vector with iid Rademacher components. Fix $\theta \in [0,1)$, and define a random vector $y=(y_1,\ldots,y_N) \in \{\pm 1\}^N$ as follows:

- Let $I_\theta$ a uniformly random subset of $[N]$ of size $\theta N$, drawn independently of $x$.
- For any $n \in [N]$, set
$$
y_n = \begin{cases}-1,&\mbox{ if }n \in I_\theta,\\
x_n,&\mbox{ else.}
\end{cases}
$$
Finally, let $z = x \odot y \in \{\pm 1\}^N$ be the component-wise product of $x$ and $y$, and define a random variable $Z$ by
$$
Z := \sum_{T \in \mathcal T} z_T,
$$
where $z_T := \prod_{t \in T} z_t$. Note that $Z$ is a **random multilinear polynomial** of total degree $k$.

>My objective is to design $N_1$ and $k$ (as a function of $N$) such that $Z$ is as large as possible (and positive !) w.p $1-o(1)$ in the limit $N \to \infty$.

Now, it is clear that we can alternately write
$$
Z = \prod_{1 \le i \le k} S_i,
$$

where $S_i := \sum_{t \in G_i} z_t$. It is clear that

- The $S_i$'s are iid.
- Each $S_i$ is itself a sum of iid random variables which take values $\pm 1$, with $\mathbb P(z_t = 1) = 1-\theta/2$ and $\mathbb E\, z_t = 1-\theta/2 - \theta/2 = a := 1-\theta \in [0,1]$. Also,  $\mathbb E S_i = a N_1$ and
$$
\begin{split}
\mathbb E S_i^2 &= \sum_{t \in G_i} \sum_{t' \in G_i} \mathbb E z_{t} \mathbb E z_{t'} = N_1 + \sum_{t' \ne t} a^2 N_1 + N_1(N_1-1)(1-\theta)^2\\
& = N_1(1-a^2) + a^2 N_1^2 = N_1(1-(1-\theta)^2) + (\mathbb E S_i)^2.
\end{split}
$$

It follows that $\mathbb E Z = (a N_1)^k$, and
$$
\begin{split}
\mathrm{var}(Z) &= \prod_{i=1}^k \mathbb E S_i^2 - \prod_{i=1}^k (\mathbb E S_i)^2 = ((aN_1)^2 + N_1(1-a^2))^k - ((a N_1)^2)^k\\
& = ((aN_1)^2)^{k}\left(\left(1 + \frac{1/a^2-1}{N_1}\right)^k - 1\right) = (\mathbb E Z)^2 R(Z),
\end{split}
$$
where $R(Z) := \mathrm{var}(Z) / (\mathbb E Z)^2 = \left(1 + \dfrac{c}{N_1}\right)^k - 1$, with $c := 1/a^2 - 1 \ge 0$. Now, one computes

$$
0 \le \left(1 + \frac{c}{N_1}\right)^k - 1= \left(\left(1 + \frac{c}{N_1}\right)^{N_1}\right)^{k/N_1} - 1 \le e^{ck/N_1} - 1.
$$

Thus, if $N_1 \to \infty$ such that $k = o(N_1)$ (i.e $k/N_1 \to 0$), then $R(Z) = o(1)$, and Chebychev's inequality gives
$$
\mathbb P(|Z-\mathbb EZ| \ge (1/2) \mathbb E Z) \le 4R(Z) = o(1).
$$

We deduce that

**Proposition 1.** *If $N_1 \to \infty$ such that $k=o(N_1)$, then $Z \asymp \mathbb E Z = (aN_1)^k$ w.p $1-o(1)$.*

**Question.** Is there a concentration inequality for $Z$ which doesn't requiring that $k=o(N_1)$ ? In fact, is it possible to concentrate $Z$ in the regime $N_1 = o(k)$ ?

My hope is that it would be possible to go beyond the "$k=o(N_1)$" barrier by computing higher moments of $Z$, and then using a Chernoff-type bound, but I don't know how to go about this (the combinatorics seem to be quite involved).

Update: An idea 
---
We can further write $z_t = 2 b_t - 1$, where $b_t$ is Bernoulli with parameter $p=1-\theta/2 \in [1/2,1]$. Thus, $S_i = \sum_{t \in G_i} (2b_t - 1) = 2 B_i - N_1$, where $B_i := \sum_{t \in G_i} b_t \sim \mathrm{Bin}(N_1,p)$. By well-known concentration results, 

$$
\begin{split}
\mathbb P(B_i \ge (1+t)N_1 p) &\le e^{-\frac{t^2p^2N_1^2}{2+t}},\text{ for all }t \gt 0,\\
\mathbb P(B_i \le (1-t)N_1 p) &\ge e^{-\frac{t^2p^2N_1^2}{2}},\text{ for all }0 \lt t \lt 1.
\end{split}
$$

We deduce that,
$$
\begin{split}
\mathbb P(S_i \ge (2p(1+t) - 1)N_1) &\le e^{-\frac{t^2p^2N_1^2}{2+t}},\text{ for all }t \gt 0,\\
\mathbb P(S_i \le (2p(1-t)-1)N_1) &\ge e^{-\frac{t^2p^2N_1^2}{2}},\text{ for all }0 \lt t \lt 1.
\end{split}
$$

Taking $t = q / \phi(N_1)$ with $q := \sqrt 2 / p$, we obtain for any $i$, it holds w.p $1-e^{-N_1^2/\phi(N_1)^2}$ that $S_i \ge ((2p-1)-q/\phi(N_1)) N_1 = (a-q/\phi(N_1))N_1$. A union bound then gives: w.p $1-\delta(N_1) = 1 - ke^{-N_1^2/\phi(N_1)^2}$ it holds that
$$
\frac{Z}{(aN_1)^k} \ge \left(1-\frac{q/a}{\phi(N_1)}\right)^k = \left(\left(1-\frac{q/a}{\phi(N_1)}\right)^{\phi(N_1)}\right)^{k/\phi(N_1)} =: R(N_1)
$$

Now, we want $k$ to be as large as possible, and for the RHS of the above to be as large as possible too. We can achieve this by designing the function $\phi:\mathbb R_+ \to \mathbb R_+$ such that in the limit $N_1 \to \infty$,

- $\delta(N_1) = e^{-N_1^2/\phi(N_1)^2 + \log k} \to 0$, and
- $\phi(N_1) \to \infty$ as fast as possible
- $k = \phi(N_1)$.

To satisfy the above constraints it suffices to take

$$
\phi(N_1) = N_1/\sqrt{C\log N_1},
$$

This gives $Z \gtrsim (aN_1)^k b^{k\sqrt{\log N_1}/N_1} = (aN_1e^{-(q/a)\sqrt{\log N_1}/N_1})^k$, with $b := e^{-(q/a)\sqrt C} \in (0,1)$. Take $C \gt 1$ and $k = N_1^{C-1}$. This gives
$$
N_1^2 / \phi(N_1)^2 + \log k = C\log N_1 - (C-1)\log N_1 = \log N_1,
$$
and so $\delta(N_1) = 1/N_1$, and
$$
Z \ge (a N_1)^ke^{-q/a}\text{ w.p } 1 - O(1/N_1) = 1-o(1).
$$

Now, take $N_1 = N^{\alpha}$, $k = N^{1-\alpha}$ (for some $\alpha \in (0,1/2)$), to get $C=(1-\alpha)/\alpha=1/\alpha-1 \gt 1$. This gives
$$
Z \gtrsim (a N^\alpha \cdot b^{\sqrt{\alpha \log N}/N^\alpha})^{N^{1-\alpha}} \asymp (a N^\alpha)^{N^{1-\alpha}} \gg e^{N^{1-\alpha}}.
$$
 We have thus established the following result.

>**Proposition 2.** *For any $\alpha \in (0,1/2)$, and set $N_1 = N_1(N)$ and $k=k(N)$ as above. In the limit $N \to \infty$, it hold w.p $1-1/N^\alpha =1-o(1)$ that*
$$
Z \gg e^{N^{1-\alpha}}.
$$

Note that **Proposition 2** is a net improvement on **Proposition 1**: the latter only gave a lower bound of order $e^{\sqrt N}$, while the former gives a lower-bound of order $e^{N^{1-\alpha}}$, which is infinitely larger since $\alpha \in (0,1/2)$. However, this bound is still not good enough: ideally, I'd like to have a lower-bound of the form $Z \gtrsim e^{N/\log(N)^c}$ w.p $1-o(1)$.