Since $(Z_j)_{j \in \mathcal{P}_a}$ are i.i.d. centered random variables, one can check that

$$ (99) \quad \mathbb{E} \left[ \left( Z_i (1 - |\mathcal{P}_a|^{-1}) - \frac{1}{|\mathcal{P}_a|} \sum_{j \in \mathcal{P}_a: j \neq i} Z_j \right)^2 \right] = (1 - |\mathcal{P}_a|^{-1}) \mathrm{Var}(Z_i), $$

which, in turn, implies that

$$ \begin{aligned} \mathbb{E} \left[ \| \tilde{L}^{N,ab} \|_2^2 \right] &= |\mathcal{P}_a|(1 - |\mathcal{P}_a|^{-1}) \mathrm{Var}(Z_i) \\ &= |\mathcal{P}_a|(1 - |\mathcal{P}_a|^{-1}) \mathrm{Var}(L_i^{N,ab}) = r_a^N r_b^N p(1-p) - N^{-1} r_b^N p(1-p). \end{aligned} $$

Combining the above identity with Jensen's inequality (and the assumption we made on the sequence of fractions $r_a^N$), we deduce that Equation (98) is satisfied.

Next, notice that $(|\mathcal{P}_a| - 1)^{-1} Y_{a,bb_1}^N$ can be seen as the empirical covariance of the random variables $L_i^{N,ab}$ and $L_i^{N,ab_1}$, $i \in \mathcal{P}_a$. In this perspective, $\mathbb{E} \left[ \left| Y_{a,bb_1}^N - \mathbb{E} [Y_{a,bb_1}^N] \right|^2 \right]$ corresponds to $(|\mathcal{P}_a| - 1)^2$ times the variance of the empirical covariance of the random variables $L_i^{N,ab}$ and $L_i^{N,ab_1}$, $i \in \mathcal{P}_a$. It is well-known (e.g., see (Casella and Berger, 2024, page 363, exercise 7.45b) for the case of the empirical variance) that the latter is equal to

$$ |\mathcal{P}_a|^{-1} \left( \mu_{2,2} + \frac{\mu_{2,0}\mu_{0,2} - (|\mathcal{P}_a| - 2)\mu_{1,1}^2}{|\mathcal{P}_a| - 1} \right), $$

where $\mu_{r,s} = \mathbb{E} \left[ \left( L_i^{N,ab} - \mathbb{E} [L_i^{N,ab}] \right)^r \left( L_i^{N,ab_1} - \mathbb{E} [L_i^{N,ab_1}] \right)^s \right]$. We have $\mu_{2,2} = \mathbb{E} \left[ (L_i^{N,ab} - bpr_b^N)^2 (L_i^{N,ab_1} - b_1pr_b^N)^2 \right] \le (\mathbb{E}[(L_i^{N,ab} - bpr_b^N)^4] \mathbb{E}[(L_i^{N,ab_1} - b_1pr_b^N)^4])^{1/2}$ by Cauchy-Schwarz inequality, $\mu_{2,0} = \mathbb{E}[(L_i^{N,ab} - bpr_b^N)^2]$, $\mu_{0,2} = \mathbb{E}[(L_i^{N,ab_1} - b_1pr_b^N)^2]$ and $\mu_{1,1} = \delta_{bb_1} \mathbb{E}[(L_i^{N,ab} - bpr_b^N)^2]$ by independence. By applying Lemma F.7, we know that, for some universal constant $K$, $\mu_{2,2} \le KN^{-2}$ and $\mu_{2,0} + \mu_{0,2} + \mu_{1,1} \le KN^{-1}$. Finally,

$$ \mathbb{E} \left[ \left| Y_{a,bb_1}^N - \mathbb{E} [Y_{a,bb_1}^N] \right|^2 \right] \le K \frac{(|\mathcal{P}_a| - 1)^2}{|\mathcal{P}_a|} \left( N^{-2} + \frac{N^{-2} - (|\mathcal{P}_a| - 2)N^{-2}}{|\mathcal{P}_a| - 1} \right) \le KN^{-1}, $$

which implies the result. $\square$

<figure><img src="image_17.png" /></figure>

**F.5. Proof of Lemma F.4.** It suffices to show that the inequalities hold for all $N$ sufficiently large. Recall that $r_a^N = |\mathcal{P}_a|/N$, for $a \in \{-, +\}$. Throughout the proof, we assume that $N$ is large enough ($N \ge N_0$) ensuring that $r_+^N \wedge r_-^N \ge r_{\min} > 0$ for some $r_{\min}$ sufficiently small depending only on the choice of $r_+$ and $r_-$. In what follows, we shall denote $K$ a constant which may depend on $\lambda$ and which may change from one line to another. The proof is divided in the 12 steps below.

**Step 1.** Consider the event

$$ \mathcal{A}_N = \bigcap_{a \in \{-, +\}} \bigcap_{b \in \{-, +\}} \left\{ \|L^{N,ab} - (bpr_b^N)1_{\mathcal{P}_a}\|_2 \vee \|C^{N,ab} - (apr_b^N)1_{\mathcal{P}_a}\|_2 \le N^{1/4} \right\}. $$

For later use, let us observe that Points (ii) and (iii) of Lemma F.1 imply that $\mathcal{A}_N$ is included in the event

$$ \mathcal{G}_N = \bigcap_{b \in \{-, +\}} \left\{ \begin{aligned} & \|L^{N,\bullet b} - (bpr_b^N)1_N\|_2 \vee \|L^{N,\bullet b} - p(r_+^N - r_-^N)1_{\mathcal{P}_b}\|_2 \\ & \vee \|C^{N,\bullet b} - pr_b^N(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-})\|_2 \vee \|C^{N,\bullet b} - (bp)1_{\mathcal{P}_b}\|_2 \le 2N^{1/4} \end{aligned} \right\}. $$
Combining Point (iv) of Lemma F.1 with Markov's inequality, one can show that for any $\alpha \ge 1$, there exists a constant $K_\alpha > 0$ such that

$$IP(\mathcal{G}_N) \ge IP(A_N) \ge 1 - K_\alpha N^{-\alpha/4}.$$

Hence, $A_N$ is a large probability event and we will first consider the expectations appearing in Lemma F.4 under the event $A_N$ only. This is used in Step 4.

**Step 2.** Let $a, b \in \{-, +\}$. Here, we prove that

$$(i) \quad \tilde{l}^{N,ab} = (1-\lambda) \left[ 1_{\mathcal{P}_a} \odot A^N \tilde{l}^{N,\bullet b} + \sum_{e \in \{-, +\}} (r_e^N)^{-1} \overline{\tilde{l}^{N,eb}} \tilde{L}^{N,ae} + \epsilon^{N,ab} 1_{\mathcal{P}_a} \right],$$

where

$$(ii) \quad |\epsilon^{N,ab}| \le (r_a^N N)^{-1} \|C^{N,\bullet a} - pr_a(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-})\|_2 \|\tilde{l}^{N,\bullet b}\|_2.$$

Starting from $\tilde{l}^{N,\bullet b} = Q^N 1_{\mathcal{P}_b} = (I_N - (1-\lambda)A^N)^{-1} 1_{\mathcal{P}_b}$, we get $\tilde{l}^{N,\bullet b} = 1_{\mathcal{P}_b} + (1-\lambda)A^N \tilde{l}^{N,\bullet b}$. Hence, it implies that $\tilde{l}^{N,ab} = 1_{\mathcal{P}_a} \odot (1_{\mathcal{P}_b} + (1-\lambda)A^N \tilde{l}^{N,\bullet b})$ and $(r_a^N)^{-1} \overline{\tilde{l}^{N,ab}} = \delta_{ab} + (1-\lambda)(r_a^N N)^{-1} \langle A^N \tilde{l}^{N,\bullet b}, 1_{\mathcal{P}_a} \rangle$, so that

$$\tilde{l}^{N,ab} = (1-\lambda) \left[ 1_{\mathcal{P}_a} \odot A^N \tilde{l}^{N,\bullet b} - (r_a^N N)^{-1} \langle A^N \tilde{l}^{N,\bullet b}, 1_{\mathcal{P}_a} \rangle 1_{\mathcal{P}_a} \right].$$

Then, using the substitution $\tilde{l}^{N,\bullet b} = \tilde{l}^{N,\bullet b} + \sum_{e \in \{-, +\}} (r_e^N)^{-1} \overline{\tilde{l}^{N,eb}} 1_{\mathcal{P}_e}$ in the two terms above and the fact that $1_{\mathcal{P}_a} \odot A^N 1_{\mathcal{P}_e} = L^{N,ae}$, we get point (i) with

$$\epsilon^{N,ab} = -(r_a^N N)^{-1} \langle A^N \tilde{l}^{N,\bullet b}, 1_{\mathcal{P}_a} \rangle = -(r_a^N N)^{-1} \langle \tilde{l}^{N,\bullet b}, (A^N)^T 1_{\mathcal{P}_a} \rangle.$$

Yet, $(A^N)^T 1_{\mathcal{P}_a} = C^{N,\bullet a}$ by definition and $\langle \tilde{l}^{N,\bullet b}, 1_{\mathcal{P}_a} \rangle = \langle \tilde{l}^{N,\bullet b}, 1_{\mathcal{P}_a} \rangle = 0$ by construction so that

$$\epsilon^{N,ab} = -(r_a^N N)^{-1} \langle \tilde{l}^{N,\bullet b}, C^{N,\bullet a} - pr_a(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-}) \rangle,$$

and point (ii) follows from Cauchy-Schwarz inequality.

**Step 3.** Proceeding as in Step 2, one can show that

$$(i) \quad \tilde{c}^{N,ab} = (1-\lambda) \left[ 1_{\mathcal{P}_a} \odot A^N \tilde{c}^{N,\bullet b} + \sum_{e \in \{-, +\}} (r_e^N)^{-1} \overline{\tilde{c}^{N,eb}} \tilde{C}^{N,ae} + \gamma^{N,ab} 1_{\mathcal{P}_a} \right],$$

where

$$(ii) \quad |\gamma^{N,ab}| \le (r_a^N N)^{-1} \|L^{N,\bullet a} - (apr_a) 1_N\|_2 \|\tilde{c}^{N,\bullet b}\|_2.$$

**Step 4.** Here, we prove Equation (87).

From Step 2, one can sum for $a \in \{-, +\}$ to obtain (100)

$$\tilde{l}^{N,\bullet b} = (1-\lambda) \left[ A^N \tilde{l}^{N,\bullet b} + \sum_{a \in \{-, +\}} \sum_{e \in \{-, +\}} (r_e^N)^{-1} \overline{\tilde{l}^{N,eb}} \tilde{L}^{N,ae} + \sum_{a \in \{-, +\}} \epsilon^{N,ab} 1_{\mathcal{P}_a} \right].$$
Then, using the fact that $\|1_{\mathcal{P}_a}\|_2 = N^{1/2}(r_a^N)^{1/2} \le N^{1/2}$, we get

$$ (1 - \lambda)^{-1} \|\tilde{\ell}^{N, \bullet b}\|_2 \le \|A^N\|_2 \|\tilde{\ell}^{N, \bullet b}\|_2 \\ + 4 \max_{e \in \{-, +\}} \left\{ (r_e^N)^{-1} |\overline{\ell}^{N, eb}| \right\} \max_{a, e \in \{-, +\}} \left\{ \|\tilde{L}^{N, ae}\|_2 \right\} + N^{1/2} \sum_{a \in \{-, +\}} |\epsilon^{N, ab}|. $$

Recall that $\|A^N\|_2 \le 1$ and $|\overline{\ell}^{N, eb}| \le r_e^N \lambda^{-1}$ (see inequality (84)). On the event $\mathcal{A}_N \supset \mathcal{G}_N$, we have $\|C^{N, \bullet a} - pr_a^N(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-})\|_2 \le 2N^{1/4}$, and using Point (ii) of Step 2, we have

$$ \mathbf{1}_{\mathcal{A}_N} (1 - \lambda)^{-1} \|\tilde{\ell}^{N, \bullet b}\|_2 \le \|\tilde{\ell}^{N, \bullet b}\|_2 \left[ 1 + \frac{4}{r_+^N \wedge r_-^N} N^{-1/4} \right] + 4\lambda^{-1} \max_{a, e \in \{-, +\}} \left\{ \|\tilde{L}^{N, ae}\|_2 \right\}. $$

Now, since $r_+^N \wedge r_-^N \ge r_{\min}$, it follows that $1 + \frac{4}{r_+^N \wedge r_-^N} N^{-1/4} \le 1 + 4r_{\min}^{-1} N^{-1/4} < (1 - \lambda)^{-1}$ for $N \ge N_1$ for some $N_1$ sufficiently large. As a consequence, there exists a constant $K$ such that

$$ (101) \qquad \mathbf{1}_{\mathcal{A}_N} \|\tilde{\ell}^{N, \bullet b}\|_2 \le K \max_{a, e \in \{-, +\}} \left\{ \|\tilde{L}^{N, ae}\|_2 \right\}, $$

and Equations (81) and (82) of Lemma F.1 implies that

$$ \mathbb{E} \left[ \mathbf{1}_{\mathcal{A}_N} \|\tilde{\ell}^{N, \bullet b}\|_2^4 \right] \le K \mathbb{E} \left[ \max_{a, b \in \{-, +\}} \left\{ \|\tilde{L}^{N, ab}\|_2^4 \right\} \right] \le K, $$

for all $N \ge N_1$. Taking the maximum value of $\mathbb{E} \left[ \mathbf{1}_{\mathcal{A}_N} \|\tilde{\ell}^{N, \bullet b}\|_2^4 \right]$ over all $N \in [N_0, N_1]$, we get

$$ \mathbb{E} \left[ \mathbf{1}_{\mathcal{A}_N} \|\tilde{\ell}^{N, \bullet b}\|_2^4 \right] \le K, $$

for all $N \ge N_0$. Yet, inequality 84 implies that $|\tilde{\ell}_i^{N, ab}| \le 2\lambda^{-1}$ which in turn implies that $\|\tilde{\ell}^{N, \bullet b}\|_2^4 \le 4\lambda^{-2}N^2$. Hence, to get rid of the term $\mathbf{1}_{\mathcal{A}_N}$ it suffices to write

$$ \mathbb{E} \left[ \|\tilde{\ell}^{N, \bullet b}\|_2^4 \right] \le 4\lambda^{-2} N^2 \mathbb{P}((\mathcal{A}_N)^c) + \mathbb{E} \left[ \mathbf{1}_{\mathcal{A}_N} \|\tilde{\ell}^{N, \bullet b}\|_2^4 \right] \le K, $$

where we used the last inequality of Step 1 with $\alpha = 8$.

Finally, the proof of the inequality $\mathbb{E} [\|\tilde{c}^{N, \bullet b}\|_2^4] \le K$ follows the same line and is therefore omitted.

Step 5. Here, we prove Equation (85).

We have already used in Step 2 that $\ell^{N, ab} = 1_{\mathcal{P}_a} \odot (1_{\mathcal{P}_b} + (1 - \lambda)A^N \ell^{N, \bullet b})$ Hence, we deduce that

$$ \begin{aligned} \overline{\ell^{N, ab}} &= r_a^N \delta_{ab} + (1 - \lambda) N^{-1} \sum_{i \in \mathcal{P}_a} \sum_{j=1}^N A^N(i, j) \ell_j^{N, \bullet b} \\ &= r_a^N \delta_{ab} + (1 - \lambda) N^{-1} \sum_{j=1}^N C_j^{N, \bullet a} \ell_j^{N, \bullet b} \\ &= r_a^N \delta_{ab} + (1 - \lambda) p r_a^N \left( \overline{\ell^{N, +b}} - \overline{\ell^{N, -b}} \right) + \eta^{N, ab}, \end{aligned} $$

where $\eta^{N, ab} = (1 - \lambda) N^{-1} \langle C^{N, \bullet a} - pr_a^N(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-}), \ell^{N, \bullet b} \rangle$. In particular, we have

$$ \left\{ \begin{array}{l} \overline{\ell^{N, +b}} - \overline{\ell^{N, -b}} = \frac{(br_b^N)}{1 - (1 - \lambda)p(r_+^N - r_-^N)} + \eta^{N, +b} - \eta^{N, -b}, \\ \overline{\ell^{N, +b}} + \overline{\ell^{N, -b}} = \frac{r_b^N + 2b(1 - \lambda)p r_+^N r_-^N}{1 - (1 - \lambda)p(r_+^N - r_-^N)} + (1 - \lambda)p(\eta^{N, +b} - \eta^{N, -b}) + \eta^{N, +b} + \eta^{N, -b}, \end{array} \right. $$
and so

$$ \overline{\ell^{N,ab}} = \frac{r_a^N \delta_{ab} + b(1-\lambda)pr_+^N r_-^N}{1 - (1-\lambda)p(r_+^N - r_-^N)} + \frac{(1-\lambda)p}{2}(\eta^{N,+b} - \eta^{N,-b}) + \eta^{N,ab}. $$

Since by our assumption, $|r_+^N - r_+| + |r_-^N - r_-| \le KN^{-1}$, we have

$$ \left| \frac{r_a^N \delta_{ab} + b(1-\lambda)pr_+^N r_-^N}{1 - (1-\lambda)p(r_+^N - r_-^N)} - \frac{r_a \delta_{ab} + b(1-\lambda)pr_+ r_-}{1 - (1-\lambda)p(r_+ - r_-)} \right| \le K \left( |r_+^N - r_+| + |r_-^N - r_-| \right) \le KN^{-1}, $$

for some constant $K$ that may depend on $\lambda$, and the result will follow once we check that $\mathbb{E}[(\eta^{N,ab})^2] \le KN^{-2}$.

To this end, we write $\eta^{N,ab} = (1-\lambda)N^{-1}(\eta_{+}^{N,ab} + \eta_{-}^{N,ab})$ where, for $e \in \{-, +\}$, $\eta_e^{N,ab} = \langle C^{N,ea} - (ep)r_a^N 1_{\mathcal{P}_e}, \ell^{N,eb} \rangle$. Remind that $\ell^{N,eb} = \tilde{\ell}^{N,eb} + (r_e^N)^{-1}\overline{\ell^{N,eb}}1_{\mathcal{P}_e}$. Then, Lemma H.1 can be applied with $V^N = C^{N,ea} - (ep)r_a^N 1_{\mathcal{P}_e}$, $v_1^N = 0$, $v_2^N = \tilde{\ell}^{N,eb}$ and $v_3^N = (r_e^N)^{-1}\overline{\ell^{N,eb}}1_{\mathcal{P}_e}$: assumption (i) is satisfied thanks to Equation (82) and the fact that $\|C^{N,ea}\|_{\infty} \le 1$, assumption (ii) is satisfied thanks to Equation (87), assumption (iii) is satisfied thanks to Equations (84) and (80) because $\langle V^N, v_3^N \rangle = (r_e^N)^{-1}\overline{\ell^{N,eb}}V^N$.

Hence, $\mathbb{E}[(\eta_e^{N,ab})^2] \le K$ which in turn implies that $\mathbb{E}[(\eta^{N,ab})^2] \le KN^{-2}$.

Step 6. Here we prove Equation (86).

Starting from

$$ c^N = (Q^N)^\intercal 1_N = (I_N - (1-\lambda)(A^N)^\intercal)^{-1} 1_N, $$

we deduce that $c^{N,+\bullet} + c^{N,-\bullet} = c^N = 1_N + (1-\lambda)(A^N)^\intercal c^N$, so that

$$ \begin{aligned} \overline{c^{N,a\bullet}} &= r_a^N + (1-\lambda)N^{-1} \sum_{i \in \mathcal{P}_a} \sum_{j=1}^N A^N(j,i) c_j^N \\ &= r_a^N + (1-\lambda)N^{-1} \sum_{j=1}^N L_j^{N,\bullet a} c_j^N \\ &= r_a^N + (1-\lambda)(apr_a^N)\overline{c^N} + \xi^{N,a}, \end{aligned} $$

where $\xi^{N,a} = (1-\lambda)N^{-1}\langle L^{N,\bullet a} - (apr_a^N)1_N, c^N \rangle$. Since $\overline{c^N} = \overline{\ell^N}$, to conclude the proof of this step it suffices to use Equation (85) and to show that $\mathbb{E}[(\xi^{N,a})^2] \le KN^{-2}$.

Proceeding as before, we can write $\xi^{N,a} = (1-\lambda)N^{-1}(\xi_{+}^{N,a} + \xi_{-}^{N,a})$ where, for $e \in \{-, +\}$, $\xi_e^{N,a} = \langle L^{N,ea} - (ap)r_a^N 1_{\mathcal{P}_e}, c^{N,\bullet a} \rangle$. Remind that $c^{N,eb} = \tilde{c}^{N,eb} + (r_e^N)^{-1}\overline{c^{N,eb}}1_{\mathcal{P}_e}$. Then, Lemma H.1 can be applied with $V^N = L^{N,ea} - (ap)r_a^N 1_{\mathcal{P}_e}$, $v_1^N = 0$, $v_2^N = \tilde{c}^{N,\bullet a}$ and $v_3^N = (r_e^N)^{-1}\overline{c^{N,\bullet a}}1_{\mathcal{P}_e}$: assumption (i) is satisfied thanks to Equation (82) and the fact that $\|L^{N,ea}\|_{\infty} \le 1$, assumption (ii) is satisfied thanks to Equation (87), assumption (iii) is satisfied thanks to Equations (84) and (80) because $\langle V^N, v_3^N \rangle = (r_e^N)^{-1}\overline{c^{N,\bullet a}}V^N$.

Hence, $\mathbb{E}[(\xi_e^{N,a})^2] \le K$ which in turn implies that $\mathbb{E}[(\xi^{N,a})^2] \le KN^{-2}$.

Step 7. Here, we prove that $\mathbb{E}[B^{N,\bullet b}] \le KN^{-1/2}$ where

$$ B^{N,\bullet b} = \left| \|\tilde{\ell}^{N,\bullet b}\|_2^2 - \|(1-\lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N,ab}} \tilde{L}^{N,\bullet a}\|_2^2 \right|. $$
Denoting $y^{N, \bullet b} = \tilde{\ell}^{N, \bullet b} - (1 - \lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N, ab} \tilde{L}^{N, \bullet a}}$, one can use Equation (100) to get

$$y^{N, \bullet b} = (1 - \lambda) \left[ A^N y^{N, \bullet b} + (1 - \lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N, ab}} (A^N \tilde{L}^{N, \bullet a}) + \sum_{a \in \{-, +\}} \epsilon^{N, ab} 1_{\mathcal{P}_a} \right].$$

Using the same kind of arguments as in the beginning of Step 4 (except the use of the event $\mathcal{A}_N$), we have

$$((1 - \lambda)^{-1} - 1) \|y^{N, \bullet b}\|_2 \le 4(1 - \lambda)\lambda^{-1} \max_{a, e \in \{-, +\}} \left\{ \|A^N \tilde{L}^{N, ae}\|_2 \right\} \\ + \frac{2}{r_+^N \wedge r_-^N} N^{-1/2} \|\tilde{\ell}^{N, \bullet b}\|_2 \max_{a \in \{-, +\}} \|C^{N, \bullet a} - pr_a(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-})\|_2.$$

Furthermore, using Cauchy-Schwarz inequality, Step 4 of the current Lemma and Equation (82) of Lemma F.1, we have

$$\mathbb{E} \left[ \| \tilde{\ell}^{N, \bullet b} \|_2^2 \max_{a \in \{-, +\}} \| C^{N, \bullet a} - pr_a(1_{\mathcal{P}_+} - 1_{\mathcal{P}_-}) \|_2^2 \right] \le K.$$

Combining the two equations above by convexity of the square function and then Equation (79) of Lemma F.1, we prove that $y^{N, \bullet b}$ is negligible in the sense that

$$\mathbb{E} \left[ \|y^{N, \bullet b}\|_2^2 \right] \le K \left\{ \mathbb{E} \left[ \max_{a, e \in \{-, +\}} \left\{ \|A^N \tilde{L}^{N, ae}\|_2^2 \right\} \right] + N^{-1} \right\} \le \frac{K}{N}.$$

In turn, we can prove that $B^{N, \bullet b}$ is negligible. More precisely, we factorize

$$B^{N, \bullet b} = \left| \| \tilde{\ell}^{N, \bullet b} \|_2 - \|(1 - \lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N, ab} \tilde{L}^{N, \bullet a}} \|_2 \right| \\ \times \left( \| \tilde{\ell}^{N, \bullet b} \|_2 + \|(1 - \lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N, ab} \tilde{L}^{N, \bullet a}} \|_2 \right) \\ \le \|y^{N, \bullet b}\|_2 \left( \| \tilde{\ell}^{N, \bullet b} \|_2 + K \max_{a \in \{-, +\}} \| \tilde{L}^{N, \bullet a} \|_2 \right),$$

so that, by Cauchy-Schwarz inequality, and then using Step 4 of the current Lemma and Equations (81) and (82) of Lemma F.1, we get

$$\mathbb{E} [B^{N, \bullet b}] \le K \sqrt{\mathbb{E} [\|y^{N, \bullet b}\|_2^2]} \sqrt{\mathbb{E} [\|\tilde{\ell}^{N, \bullet b}\|_2^2 + \max_{a \in \{-, +\}} \|\tilde{L}^{N, \bullet a}\|_2^2]} \le \frac{K}{\sqrt{N}}.$$

Step 8. Using the same arguments as Step 7, one can prove that

$$\mathbb{E} \left[ \left\| \|\tilde{\ell}^N\|_2^2 - \|(1-\lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N, a\bullet} \tilde{L}^{N, \bullet a}} \right\|_2^2 \right] \le \frac{K}{\sqrt{N}}.$$

Step 9. Here we prove, for all $a_1, a_2, b \in \{-, +\}$, that

$$\mathbb{E} \left[ \left| \frac{\overline{\ell^{N, a_1 b}}}{r_{a_1}^N} \frac{\overline{\ell^{N, a_2 b}}}{r_{a_2}^N} \langle \tilde{L}^{N, \bullet a_1}, \tilde{L}^{N, \bullet a_2} \rangle - \delta_{a_1 a_2} \frac{\overline{\ell^{\infty, a_1 b}}}{r_{a_1}} \frac{\overline{\ell^{\infty, a_2 b}}}{r_{a_2}} r_{a_1} p(1-p) \right| \right] \le \frac{K}{\sqrt{N}},$$
where $\overline{\ell^{\infty,a\bar{b}}} := \frac{r_a \delta_{ab} + b(1-\lambda)pr_+r_-}{1-(1-\lambda)p(r_+-r_-)}$ is the limit appearing in Equation (85).

Since the coordinates of $\ell^{N,ab}$ are bounded by inequality (84), we know that there exists a constant $K$ such that

$$ \left| \frac{\overline{\ell^{N,a_1 b}}}{r_{a_1}^N} \frac{\overline{\ell^{N,a_2 b}}}{r_{a_2}^N} - \frac{\overline{\ell^{\infty,a_1 b}}}{r_{a_1}} \frac{\overline{\ell^{\infty,a_2 b}}}{r_{a_2}} \right| \le K \left( \left| \frac{\overline{\ell^{N,a_1 b}}}{r_{a_1}^N} - \frac{\overline{\ell^{\infty,a_1 b}}}{r_{a_1}} \right| + \left| \frac{\overline{\ell^{N,a_2 b}}}{r_{a_2}^N} - \frac{\overline{\ell^{\infty,a_2 b}}}{r_{a_2}} \right| \right). $$

Hence, using Equation (85), Cauchy-Schwarz inequality twice and finally Equations (81) and (82) of Lemma F.1, we have

$$ \begin{aligned} \mathrm{IE} \left[ \left| \frac{\overline{\ell^{N,a_1 b}}}{r_{a_1}^N} \frac{\overline{\ell^{N,a_2 b}}}{r_{a_2}^N} - \frac{\overline{\ell^{\infty,a_1 b}}}{r_{a_1}} \frac{\overline{\ell^{\infty,a_2 b}}}{r_{a_2}} \right| \left| \langle \tilde{L}^{N, \bullet a_1}, \tilde{L}^{N, \bullet a_2} \rangle \right| \right] & \\ & \le \frac{K}{N} \mathrm{IE} \left[ \left| \langle \tilde{L}^{N, \bullet a_1}, \tilde{L}^{N, \bullet a_2} \rangle \right|^2 \right]^{1/2} \le \frac{K}{N} \max_{a \in \{-, +\}} \mathrm{IE} \left[ \| \tilde{L}^{N, \bullet a} \|_2^2 \right]^{1/2} \le \frac{K}{N}. \end{aligned} $$

Finally, we conclude the step by combining the above equation with Equation (78) of Lemma F.1.

Step 10. Using the same arguments as in Step 9, one can prove that, for all $a_1, a_2 \in \{-, +\}$,

$$ \mathrm{IE} \left[ \left| \frac{\overline{\ell^{N,a_1 \bullet}}}{r_{a_1}^N} \frac{\overline{\ell^{N,a_2 \bullet}}}{r_{a_2}^N} \langle \tilde{L}^{N, \bullet a_1}, \tilde{L}^{N, \bullet a_2} \rangle - \delta_{a_1 a_2} \frac{\overline{\ell^{\infty,a_1 \bullet}}}{r_{a_1}} \frac{\overline{\ell^{\infty,a_2 \bullet}}}{r_{a_2}} r_{a_1} p(1-p) \right| \right] \le \frac{K}{\sqrt{N}}, $$

$$ \text{where } \overline{\ell^{\infty,a\bullet}} := \overline{\ell^{\infty,a+}} + \overline{\ell^{\infty,a-}} = \frac{r_a}{1-(1-\lambda)p(r_+-r_-)}. $$

Step 11. Here we prove Equation (88). According to Step 7, the limit of $\|\overline{\ell^{N,\bullet b}}\|_2^2$ is related to the limit of $\|(1-\lambda) \sum_{a \in \{-, +\}} (r_a^N)^{-1} \overline{\ell^{N,ab}} \tilde{L}^{N,\bullet a} \|_2^2$ which can be expanded as

$$ (1-\lambda)^2 \sum_{a_1, a_2 \in \{-, +\}} \frac{\overline{\ell^{N,a_1 b}}}{r_{a_1}^N} \frac{\overline{\ell^{N,a_2 b}}}{r_{a_2}^N} \langle \tilde{L}^{N, \bullet a_1}, \tilde{L}^{N, \bullet a_2} \rangle. $$

Hence, combining Steps 7 and 9, we have

$$ \mathrm{IE} \left[ \left\| \overline{\ell^{N, \bullet b}} \right\|_2^2 - (1-\lambda)^2 p(1-p) \sum_{a \in \{-, +\}} \left( \frac{\overline{\ell^{\infty,ab}}}{r_a} \right)^2 r_a \right] \le \frac{K}{\sqrt{N}}. $$

In order to conclude this step, it suffices to simplify

$$ \begin{aligned} \sum_{a \in \{-, +\}} \left( \frac{\overline{\ell^{\infty,ab}}}{r_a} \right)^2 r_a &= \sum_{a \in \{-, +\}} \left( \frac{\delta_{ab} + b(1-\lambda)p(1-r_a)}{1-(1-\lambda)p(r_+-r_-)} \right)^2 r_a \\ &= \frac{(1+b(1-\lambda)p(1-r_b))^2 r_b + (b(1-\lambda)pr_b)^2 (1-r_b)}{(1-(1-\lambda)p(r_+-r_-))^2} \\ &= \frac{r_b + 2b(1-\lambda)pr_b(1-r_b) + (1-\lambda)^2 p^2 r_b(1-r_b)}{(1-(1-\lambda)p(r_+-r_-))^2} \\ &= \frac{r_b + (1-\lambda)pr_+r_-(2b+(1-\lambda)p)}{(1-(1-\lambda)p(r_+-r_-))^2}. \end{aligned} $$
Step 12. Following the lines of Step 11 (and using Steps 8 and 10), one can prove Equation (89). Let us mention that the final simplification here is

$$ \sum_{a \in \{-, +\}} \left( \frac{\overline{\ell^{\infty, a\bullet}}}{r_a} \right)^2 r_a = \sum_{a \in \{-, +\}} \left( \frac{1}{1 - (1-\lambda)p(r_+ - r_-)} \right)^2 r_a \\ = (1 - (1-\lambda)p(r_+ - r_-))^{-2}. $$

**F.6. Proof of Lemma F.6.** It suffices to show that these inequalities hold for all $N$ sufficiently large. Recall that $r_a^N = |P_a|/N$, for $a \in \{-, +\}$. As in the previous proofs, in what follows we assume that $N$ is large enough ($N \ge N_0$) ensuring that $r_+^N \wedge r_-^N \ge r_{\min} > 0$ for some $r_{\min}$ sufficiently small depending only on the choice of $r_+$ and $r_-$. Also, we shall denote $K$ a constant which may depend on $\lambda$ and which may change from one line to another. The proof is divided in several steps.

We will first prove (91). To see that, we start by observing that

$$ \begin{aligned} \|\overline{\ell^N} - \overline{\ell^N} 1_N\|_2^2 - \|\tilde{\ell}^N\|_2^2 &= \sum_{a \in \{-, +\}} \sum_{i \in P_a} (\ell_i^{N, a\bullet} - \overline{\ell^N})^2 - (\ell_i^{N, a\bullet} - (r_a^N)^{-1} \overline{\ell^{N, a\bullet}})^2 \\ &= \sum_{a \in \{-, +\}} ((r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N}) \sum_{i \in P_a} (\ell_i^{N, a\bullet} - \overline{\ell^N}), \\ &= N \sum_{a \in \{-, +\}} r_a^N ((r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N})^2, \end{aligned} $$

where in the second equality we have used that $x^2 - y^2 = (x - y)(x + y)$ for all $x, y \in \mathbb{R}$ and $\sum_{i \in P_a} (\ell_i^{N, a\bullet} - (r_a^N)^{-1} \overline{\ell^{N, a\bullet}}) = 0$. Then, we combine (85), (90) and (84) together with Jensen inequality to obtain that

$$ N \mathcal{I} \mathcal{E} \left[ \left( (r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N} \right)^2 \right] \le K N \left( \mathcal{I} \mathcal{E} \left[ \left( (r_a)^{-1} \overline{\ell^{N, a\bullet}} - \frac{1}{1 - (1 - \lambda)p(r_+ - r_-)} \right)^2 \right] + \mathcal{I} \mathcal{E} \left[ |r_a^N - r_a|^2 \right] + \mathcal{I} \mathcal{E} \left[ \left( \overline{\ell^N} - \frac{1}{1 - (1 - \lambda)p(r_+ - r_-)} \right)^2 \right] \right) \le K N^{-1}, $$

so that

$$ \mathcal{I} \mathcal{E} \left[ \left\| \ell^N - \overline{\ell^N} 1_N \right\|_2^2 - \left\| \tilde{\ell}^N \right\|_2^2 \right] \le K N^{-1} \sum_{a \in \{-, +\}} r_a^N = K N^{-1}, $$

which proves (91).

We will now prove (92). To that end, first note that

$$ (102) \qquad (1 - \lambda)(Q^N L^{N, \bullet} - \overline{Q^N L^{N, \bullet}}) = \ell^{N, \bullet} - \overline{\ell^{N, \bullet}} 1_N + r_-^N 1_{P_+} - r_+^N 1_{P_-}. $$

Then, note that

$$ \begin{aligned} \langle \overline{\ell^N} - \overline{\ell^N} 1_N, r_-^N 1_{P_+} - r_+^N 1_{P_-} \rangle &= r_-^N \langle \ell^{N, +\bullet} - \overline{\ell^N} 1_{P_+}, 1_{P_+} \rangle - r_+^N \langle \ell^{N, -\bullet} - \overline{\ell^N} 1_{P_-}, 1_{P_-} \rangle \\ &= N r_-^N \left[ \overline{\ell^{N, +\bullet}} - \overline{\ell^N} r_+^N \right] - N r_+^N \left[ \overline{\ell^{N, -\bullet}} - \overline{\ell^N} r_-^N \right]. \end{aligned} $$
Next, by using that $\ell^N - \overline{\ell^N} 1_N = \tilde{\ell}^N + \sum_{a \in \{-, +\}} 1_{\mathcal{P}_a}((r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N})$ and that $\ell^{N, \bullet} - \overline{\ell^{N, \bullet}} 1_N = \tilde{\ell}^{N, \bullet} + \sum_{a \in \{-, +\}} 1_{\mathcal{P}_a}((r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}})$, one can check that

$$ \langle \ell^N - \overline{\ell^N} 1_N, \ell^{N, \bullet} - \overline{\ell^{N, \bullet}} 1_N \rangle = \langle \tilde{\ell}^N, \tilde{\ell}^{N, \bullet} \rangle + N \sum_{a \in \{-, +\}} r_a^N \overline{[(r_a^N)^{-1} \ell^{N, a\bullet} - \overline{\ell^N}]} \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}}]} $$

Now, observe that using the fact that

$$ \frac{\delta_{a-} - (1-\lambda)pr_+r_-(r_a)^{-1}}{1 - (1-\lambda)p(r_+ - r_-)} + ar_{-a} = \frac{r_- - 2(1-\lambda)pr_-r_+}{1 - (1-\lambda)p(r_+ - r_-)}, $$

we can write

$$ [(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N}] [(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}}] = -ar_{-a} [(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N}] + \xi_a^N, $$

where

$$ \xi_a^N = [(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N}] \left[ \left( (r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \frac{(\delta_{a-} - (1-\lambda)pr_+r_-r_a^{-1})}{1 - (1-\lambda)p(r_+ - r_-)} \right) + \left( \frac{(r_- - 2(1-\lambda)pr_+r_-)}{1 - (1-\lambda)p(r_+ - r_-)} \right) \right]. $$

Combining (85) and (90) with Jensen inequality, one can show that $\mathbb{E}[|\xi_a^N|] \le KN^{-2}$, so that

$$ N \sum_{a \in \{-, +\}} r_a^N \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^N}]} \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}}]} = -Nr_- \overline{[\overline{\ell^{N, +\bullet}} - \overline{\ell^N} r_+^N]} + Nr_+ \overline{[\overline{\ell^{N, -\bullet}} - \overline{\ell^N} r_-^N]} + \xi^N $$

where $\xi^N = N \sum_{a \in \{-, +\}} r_a^N \xi_a^N$ satisfies $\mathbb{E}[|\xi^N|] \le KN^{-1}$. Therefore, putting together all previous identities, we deduce that

$$ \langle \overline{\ell^N - \ell^N} 1_N, (1-\lambda)(Q^N L^{N, \bullet} - \overline{Q^N L^{N, \bullet}}) \rangle - \langle \tilde{\ell}^N, \tilde{\ell}^{N, \bullet} \rangle = N(r_-^N - r_-) \overline{[\overline{\ell^{N, +\bullet}} - \overline{\ell^N} r_+^N]} + N(r_+^N - r_+) \overline{[\overline{\ell^{N, -\bullet}} - \overline{\ell^N} r_-^N]} + \xi^N, $$

and the result follows from (85), (90) and the assumption that $|r_a^N - r_a| \le KN^{-1}$.

Hence, it remains to prove only (93). Starting from (102), one can check that (recall the definition of $\tilde{\ell}^{N, ab}$ given in (83))

$$ (1 - \lambda)(Q^N L^{N, \bullet} - \overline{Q^N L^{N, \bullet}}) = \tilde{\ell}^{N, \bullet} + \sum_{a \in \{-, +\}} 1_{\mathcal{P}_a} \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}} + ar_{-a}^N]} $$

which together with the fact that $\langle 1_{\mathcal{P}_+}, 1_{\mathcal{P}_-} \rangle = \langle 1_{\mathcal{P}_a}, \tilde{\ell}^{N, \bullet} \rangle = 0$ implies that

$$ (1 - \lambda)^2 \|(Q^N L^{N, \bullet} - \overline{Q^N L^{N, \bullet}})\|_2^2 - \tilde{\ell}^{N, \bullet} = \sum_{a \in \{-, +\}} |\mathcal{P}_a| \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}} + ar_{-a}^N]} \|^2 $$

Proceeding similarly as in the proof of (92), one can show that

$$ \sum_{a \in \{-, +\}} |\mathcal{P}_a| \mathbb{E} \left[ \overline{[(r_a^N)^{-1} \overline{\ell^{N, a\bullet}} - \overline{\ell^{N, \bullet}} + ar_{-a}^N]} \right]^2 \le KN^{-1}, $$

concluding the proof of the lemma.
**F.7. Proof of Proposition 4.2.** We are now able to prove the three inequalities.

PROOF OF INEQUALITY (35). First, remind that $\overline{m^N} = \mu \overline{\ell^N} - (1 - \lambda) \overline{Q^N L^{N, \bullet-}}$. Thanks to Remark F.5, it suffices to prove that

$$ \mathrm{IE} \left[ \left| \overline{Q^N L^{N, \bullet-}} + \frac{pr_-}{1 - (1 - \lambda)p(r_+ - r_-)} \right|^2 \right] \le \frac{K}{N^2}. $$

First, note that

$$ Q^N L^{N, \bullet-} = -pr_-^N \ell^N + Q^N (L^{N, \bullet-} + pr_-^N 1_N). $$

Hence, it follows that

$$ \begin{aligned} \overline{Q^N L^{N, \bullet-}} &= -pr_-^N \overline{\ell^N} + N^{-1} \sum_{i=1}^N \sum_{j=1}^N Q^N(i,j) (L_j^{N, \bullet-} + pr_-^N) \\ &= -pr_-^N \overline{\ell^N} + N^{-1} \sum_{j=1}^N (L_j^{N, \bullet-} + pr_-^N) c_j^N \\ &= -pr_-^N \overline{\ell^N} + (1 - \lambda)^{-1} \xi^{N,-}, \end{aligned} $$

where $\xi^{N,-} = (1 - \lambda)N^{-1} \langle L^{N, \bullet-} + pr_-^N 1_N, c^N \rangle$ is already defined in Step 6 of the proof of Lemma F.4. From there, we know that $\mathrm{IE}[|\xi^{N,-}|^2] \le KN^{-2}$. As a consequence, it follows that

$$ \mathrm{IE} \left[ \left| \overline{Q^N L^{N, \bullet-}} + \frac{pr_-}{1 - (1 - \lambda)p(r_+ - r_-)} \right|^2 \right] \le K \left( \mathrm{IE} \left[ \left| \overline{\ell^N} - \frac{1}{1 - (1 - \lambda)p(r_+ - r_-)} \right|^2 \right] + N^{-2} \right) \le KN^{-2}, $$

where in the last inequality we used Equation (90).

<figure><img src="image_18.png" /></figure>

PROOF OF INEQUALITY (36). Expanding the scalar product $v_\infty^N = \|m^N - \overline{m^N} 1_N\|_2^2$ and using Lemma F.6, we have

$$ \mathrm{IE} \left[ \left| \|m^N - \overline{m^N} 1_N\|_2^2 - \left( \mu^2 \|\tilde{\ell}^N\|_2^2 - 2\mu \langle \tilde{\ell}^N, \tilde{\ell}^N, \bullet- \rangle + \|\tilde{\ell}^N, \bullet-\|_2^2 \right) \right| \right] \le \frac{K}{N}. $$

Yet, by the polarization identity, $-2 \langle \tilde{\ell}^N, \tilde{\ell}^N, \bullet- \rangle = (\|\tilde{\ell}^N, \bullet+\|_2^2 - \|\tilde{\ell}^N\|_2^2 - \|\tilde{\ell}^N, \bullet-\|_2^2)$, so that

$$ \mathrm{IE} \left[ \left| \|m^N - \overline{m^N} 1_N\|_2^2 - \left( (\mu^2 - \mu) \|\tilde{\ell}^N\|_2^2 + \mu \|\tilde{\ell}^N, \bullet+\|_2^2 + (1 - \mu) \|\tilde{\ell}^N, \bullet-\|_2^2 \right) \right| \right] \le \frac{K}{N}, $$

and we conclude this step thanks to Equations (88) and (89) as soon as we check the simplification:

$$ \begin{aligned} I &= (\mu^2 - \mu) + \mu [r_+ + (1 - \lambda)pr_+r_- (2 + (1 - \lambda)p)] \\ &\quad + (1 - \mu) [r_- + (1 - \lambda)pr_+r_- (-2 + (1 - \lambda)p)] \\ &= \mu^2 + \mu(-1 + r_+ - r_-) + r_- \end{aligned} $$
$$ \begin{aligned} &+(1-\lambda)pr_{+}r_{-}[\mu(2+(1-\lambda)p+2-(1-\lambda)p)-2+(1-\lambda)p] \\ &=\mu^2-2\mu r_{-}+r_{-}+(1-\lambda)pr_{+}r_{-}(4\mu-2+(1-\lambda)p) \\ &= (\mu+(1-\lambda)pr_{-})^2+r_{-}(1-(1-\lambda)p(r_{+}-r_{-}))(1-2\mu-(1-\lambda)p) \end{aligned} $$

and the identity

$$ \frac{1-2\mu-(1-\lambda)p}{1-(1-\lambda)p(r_{+}-r_{-})} = 1-2m. $$

<figure><img src="image_19.png" /></figure>

PROOF OF INEQUALITY (37). The proof is divided in the 5 steps below.

Step 1. Here we prove that

$$ (103) \quad \mathrm{IE} \left[ \left| \frac{1}{N} \sum_{i=1}^{N} \ell_i^N (c_i^N)^2 - \frac{1+4(1-\lambda)^2 p^2 r_- r_+}{(1-(1-\lambda)p(r_+-r_-))^3} \right|^2 \right] \le \frac{K}{N^2}. $$

Notice that $\sum_{i=1}^N \ell_i^N (c_i^N)^2 = \sum_{a \in \{-,+\}} \langle \ell^{N,a\bullet}, (c^{N,a\bullet})^2 \rangle$ and $(c^{N,a\bullet})^2 = (\tilde{c}^{N,a\bullet})^2 + 2(r_a^N)^{-1} \overline{c^{N,a\bullet} \tilde{c}^{N,a\bullet}} + ((r_a^N)^{-1} \overline{c^{N,a\bullet}})^2 1_{\mathcal{P}_a}$ for each $a \in \{-,+\}$. We argue that Lemma H.2 can be applied to

$$ \left\{ \begin{aligned} u_1^N &= 0, & u_2^N &= \tilde{\ell}^{N,a\bullet}, & u_3^N &= (r_a^N)^{-1} \overline{\ell^{N,a\bullet}}, \\ u_3^\infty &= (1-(1-\lambda)p(r_+-r_-))^{-1}, & & & \\ v_1^N &= (\tilde{c}^{N,a\bullet})^2, & v_2^N &= (r_a^N)^{-1} \overline{c^{N,a\bullet} \tilde{c}^{N,a\bullet}}, & v_3^N &= ((r_a^N)^{-1} \overline{c^{N,a\bullet}})^2, \\ v_3^\infty &= \left( \frac{[1+(2a)(1-\lambda)p(1-r_a)]}{1-(1-\lambda)p(r_+-r_-)} \right)^2. & & & \end{aligned} \right. $$

Indeed, the assumptions are satisfied thanks to Equations (84), (85), (87), the fact that $\tilde{\ell}^{N,a\bullet}$ and $\tilde{c}^{N,a\bullet}$ are centered (for instance $\tilde{\ell}^{N,a\bullet} = 0$) and the following argument: using the fact that $\overline{c^{N,a\bullet}} \le 1$ and the previous step, we have

$$ \mathrm{IE} \left[ |v_3^N - v_3^\infty|^2 \right] \le K \mathrm{IE} \left[ \left| \overline{c^{N,a\bullet}} - r_a \frac{[1+(2a)(1-\lambda)p(1-r_a)]}{1-(1-\lambda)p(r_+-r_-)} \right|^2 \right] \le \frac{K}{N^2}. $$

The conclusion of Lemma H.2 is

$$ \mathrm{IE} \left[ \left| \frac{1}{N} \langle \ell^{N,a\bullet}, (c^{N,a\bullet})^2 \rangle - r_a \frac{[1+(2a)(1-\lambda)p(1-r_a)]^2}{(1-(1-\lambda)p(r_+-r_-))^3} \right|^2 \right] \le \frac{K}{N^2}. $$

Finally, the conclusion of this step follows from the fact that

$$ (104) \quad \sum_{a \in \{-,+\}} r_a [1+(2a)(1-\lambda)p(1-r_a)]^2 = 1+4(1-\lambda)^2 p^2 r_- r_+ . $$

Step 2. Here we prove that

$$ (105) \quad \mathrm{IE} \left[ \left| \frac{1}{N} \sum_{i=1}^{N} (\ell_i^N c_i^N)^2 - \frac{1+4(1-\lambda)^2 p^2 r_- r_+}{(1-(1-\lambda)p(r_+-r_-))^4} \right|^2 \right] \le \frac{K}{N^2}. $$
Notice that $\sum_{i=1}^{N} (\ell_i^N c_i^N)^2 = \sum_{a \in \{-,+\}} \langle (\ell^{N,a\bullet})^2, (c^{N,a\bullet})^2 \rangle$. We argue that Lemma H.2 can be applied to

$$
\left\{
\begin{aligned}
u_1^N &= (\tilde{\ell}^{N,a\bullet})^2, & u_2^N &= (r_a^N)^{-1} \overline{\ell^{N,a\bullet} \tilde{\ell}^{N,a\bullet}}, & u_3^N &= ((r_a^N)^{-1} \overline{\ell^{N,a\bullet}})^2, \\
u_3^\infty &= (1 - (1-\lambda)p(r_+ - r_-))^{-2}, & & & \\
v_1^N &= (\tilde{c}^{N,a\bullet})^2, & v_2^N &= (r_a^N)^{-1} \overline{c^{N,a\bullet} \tilde{c}^{N,a\bullet}}, & v_3^N &= ((r_a^N)^{-1} \overline{c^{N,a\bullet}})^2, \\
v_3^\infty &= \left( \frac{[1+(2a)(1-\lambda)p(1-r_a)]}{1-(1-\lambda)p(r_+-r_-)} \right)^2.
\end{aligned}
\right.
$$

Indeed, the assumptions are satisfied thanks to Equations (84), (85), (87), the fact that $\tilde{\ell}^{N,a\bullet}$ and $\tilde{c}^{N,a\bullet}$ are centered (for instance $\tilde{\ell}^{N,a\bullet} = 0$) and the fact that $\mathbb{E} [|u_3^N - u_3^\infty|^2] \le KN^{-2}$ can be proved like we did in the previous step for $v_3^N$

The conclusion of Lemma H.2 is

$$ \mathbb{E} \left[ \left| \frac{1}{N} \langle (\ell^{N,a\bullet})^2, (c^{N,a\bullet})^2 \rangle - r_a \frac{[1+(2a)(1-\lambda)p(1-r_a)]^2}{(1-(1-\lambda)p(r_+-r_-))^4} \right|^2 \right] \le \frac{K}{N^2}. $$

Finally, the conclusion of this step follows one again from Equation (104).

Step 3. In this step, we show the following result:

$$ (106) \qquad N^{-1} \langle (c^N)^2, m^N \rangle = (\mu + (1-\lambda)pr_-^N) N^{-1} \langle (c^N)^2, \ell^N \rangle + \xi^{N,(2),-}, $$

where $\mathbb{E} [(\xi^{N,(2),-})^2] \le KN^{-2}$ for some constant $K > 0$.

On the one hand, recall that $m^N = \mu\ell^N - (1-\lambda)Q^N L^{N,\bullet-} = (\mu + (1-\lambda)pr_-^N)\ell^N - (1-\lambda)Q^N (L^{N,\bullet-} + pr_-^N 1_N)$. On the other hand, remark that $(c^N)^2 = (c^{N,+\bullet})^2 + (c^{N,-\bullet})^2$. Hence, (106) is satisfied with $\xi^{N,(2),-} = -(1-\lambda)N^{-1}(\xi_+^{N,(2),-} + \xi_-^{N,(2),-})$ where for each $e \in \{-,+\}$,

$$ \xi_e^{N,(2),-} = \langle (Q^N)[L^{N,\bullet-} + pr_-^N 1_N], (c^{N,e\bullet})^2 \rangle = \langle L^{N,\bullet-} + pr_-^N 1_N, (Q^N)^\top (c^{N,e\bullet})^2 \rangle. $$

Therefore, it remains to prove that $\mathbb{E} [(\xi_e^{N,(2),-})^2] \le K$ for some constant $K > 0$. Remind that $(c^{N,e\bullet})^2 = (\tilde{c}^{N,e\bullet})^2 + 2(r_e^N)^{-1} \overline{c^{N,e\bullet} \tilde{c}^{N,e\bullet}} + ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 1_{P_e}$ for each $e \in \{-,+\}$. Then, Lemma H.1 can be applied with $V^N = L^{N,\bullet-} + pr_-^N 1_N$, $v_1^N = (Q^N)^\top (\tilde{c}^{N,e\bullet})^2$, $v_2^N = 2(r_e^N)^{-1} c^{N,e\bullet} (Q^N)^\top \tilde{c}^{N,e\bullet}$ and $v_3^N = ((r_e^N)^{-1} c^{N,e\bullet})^2 (Q^N)^\top 1_{P_e} = ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 c^{N,e\bullet}$. Assumption (i) is satisfied thanks to Equation (82) and the fact that $\|L^{N,\bullet-}\|_\infty \le 1$, assumption (ii) is satisfied thanks to Equations (87), the facts that $\|(\tilde{c}^{N,e\bullet})^2\|_1 = \|\tilde{c}^{N,e\bullet}\|_2^2$ and $\max\{\|(Q^N)^\top\|_2, \|(Q^N)^\top\|_1\} \le \lambda^{-1}$, assumption (iii) is satisfied because $\langle V^N, v_3^N \rangle = ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 \langle L^{N,\bullet-} + pr_-^N 1_N, c^{N,e\bullet} \rangle = ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 \xi_e^{N,-}$ which is defined and controlled in Step 7.

$$ \text{Hence, } \mathbb{E}[(\xi_e^{N,(2),-})^2] \le K \text{ which in turn implies that } \mathbb{E}[(\xi^{N,(2),-})^2] \le KN^{-2}. $$

Step 4. In this step, we show the following result:

$$ (107) \qquad N^{-1} \langle (c^N)^2, (m^N)^2 \rangle = (\mu + (1-\lambda)pr_-^N)^2 N^{-1} \langle (c^N)^2, (\ell^N)^2 \rangle + \xi^{N,(3),-} + \xi^{N,(2,2),-}, $$

where $\xi^{N,(3),-}$ and $\xi^{N,(2,2),-}$ satisfy $\mathbb{E} [(\xi^{N,(3),-})^2 + (\xi^{N,(2,2),-})^2] \le KN^{-2}$ for some constant $K > 0$.
On the one hand, from the identity $m^N = (\mu + (1 - \lambda)pr_{-}^N)\ell^N - (1 - \lambda)Q^N(L^{N,\bullet} + pr_{-}^N 1_N)$, one can check that

$$ (m^N)^2 = (\mu + (1 - \lambda)pr_{-}^N)^2 (\ell^N)^2 - 2(\mu + (1 - \lambda)pr_{-}^N)(1 - \lambda)\ell^N \odot Q^N(L^{N,\bullet} + pr_{-}^N 1_N) + (1 - \lambda)^2(Q^N(L^{N,\bullet} + pr_{-}^N 1_N))^2. $$

On the other hand, remind that $(c^N)^2 = (c^{N,+\bullet})^2 + (c^{N,-\bullet})^2$ and $\ell^N = \ell^{N,+\bullet} + \ell^{N,-\bullet}$. Hence, Equation (107) is satisfied with $\xi^{N,(3),-} = -2(\mu + (1 - \lambda)pr_{-}^N)(1 - \lambda)N^{-1}(\xi_{+}^{N,(3),-} + \xi_{-}^{N,(3),-})$ where, for each $e \in \{-,+\}$,

$$ \xi_{e}^{N,(3),-} = \langle (Q^N)^{\top} [(c^{N,e\bullet})^2 \odot \ell^{N,e\bullet}], L^{N,\bullet} + pr_{-}^{N} 1_{N} \rangle, $$

and $\xi^{N,(2,2),-} = (1 - \lambda)^2 N^{-1}(\xi_{+}^{N,(2,2),-} + \xi_{-}^{N,(2,2),-})$ where

$$ \xi_{e}^{N,(2,2),-} = \langle (Q^N)^{\top} [(c^{N,e\bullet})^2 \odot (Q^N(L^{N,\bullet} + pr_{-}^{N} 1_{N}))], L^{N,\bullet} + pr_{-}^{N} 1_{N} \rangle. $$

Let us first check that $\mathbb{E}[(\xi_{e}^{N,(3),-})^2] \le K$.

Remind that $\ell^{N,e\bullet} = \tilde{\ell}^{N,e\bullet} + (r_e^N)^{-1}\overline{\ell^{N,e\bullet}}1_{P_e}$ for each $e \in \{-,+\}$. Then, Lemma H.1 can be applied with $V^N = L^{N,\bullet} + pr_{-}^N 1_N$, $v_1^N = 0$, $v_2^N = (Q^N)^{\top}[(c^{N,e\bullet})^2 \odot \ell^{N,e\bullet}]$ and $v_3^N = (r_e^N)^{-1}\overline{\ell^{N,e\bullet}}(Q^N)^{\top}(c^{N,e\bullet})^2$. Assumption (i) is satisfied thanks to Equation (82) and the fact that $\|L^{N,\bullet}\|_{\infty} \le 1$, assumption (ii) is satisfied thanks to Equations (84), (87), the facts that $\|(Q^N)^{\top}\|_2 \le \lambda^{-1}$ and $\|(c^{N,e\bullet})^2 \odot \tilde{\ell}^{N,e\bullet}\|_{\infty} \le \|c^{N,e\bullet}\|_{\infty}^2 \|\tilde{\ell}^{N,e\bullet}\|_2$, assumption (iii) is satisfied because $\langle V^N, v_3^N \rangle = (r_e^N)^{-1}\overline{\ell^{N,e\bullet}}\langle L^{N,\bullet} + pr_{-}^N 1_N, (Q^N)^{\top}(c^{N,e\bullet})^2 \rangle = (r_e^N)^{-1}\overline{\ell^{N,e\bullet}}\xi_{e}^{N,(2),-}$ which is defined and controlled in Step 3. Hence, $\mathbb{E}[(\xi_{e}^{N,(3),-})^2] \le K$.

Now, let us check that $\mathbb{E}[(\xi_{e}^{N,(2,2),-})^2] \le K$.

Remind that $(c^{N,e\bullet})^2 = (\tilde{c}^{N,e\bullet})^2 + 2(r_e^N)^{-1}\overline{c^{N,e\bullet}}\tilde{c}^{N,e\bullet} + ((r_e^N)^{-1}\overline{c^{N,e\bullet}})^21_{P_e}$ for each $e \in \{-,+\}$. Then, Lemma H.1 can be applied with $V^N = L^{N,\bullet} + pr_{-}^N 1_N$,

$$ v_1^N = (Q^N)^{\top} [(\tilde{c}^{N,e\bullet})^2 \odot (Q^N V^N)], $$

$$ v_2^N = 2(r_e^N)^{-1} \overline{c^{N,e\bullet}} (Q^N)^{\top} [\tilde{c}^{N,e\bullet} \odot (Q^N V^N)], $$

and

$$ v_3^N = ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 (Q^N)^{\top} [1_{P_e} \odot (Q^N V^N)]. $$

Assumption (i) is satisfied thanks to Equation (82) and the fact that $\|L^{N,\bullet}\|_{\infty} \le 1$, assumption (ii) is satisfied thanks to Equations (84), (87), the facts that $\|(\tilde{c}^{N,e\bullet})^2\|_1 = \|\tilde{c}^{N,e\bullet}\|_2^2$, $\max\{\|(Q^N)^{\top}\|_2, \|(Q^N)^{\top}\|_1, \|Q^N\|_{\infty}\} \le \lambda^{-1}$ and $\|V^N\|_{\infty} \le 2$, assumption (iii) is satisfied thanks to Equation (82) because

$$ \begin{aligned} |\langle V^N, v_3^N \rangle| &= ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 |\langle V^N, (Q^N)^{\top} [1_{P_e} \odot (Q^N V^N)] \rangle| \\ &\le ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 \|V^N\|_2 \|(Q^N)^{\top}\|_2 \|1_{P_e} \odot (Q^N V^N)\|_2 \\ &\le ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 \|V^N\|_2 \|(Q^N)^{\top}\|_2 \|Q^N V^N\|_2 \\ &\le ((r_e^N)^{-1} \overline{c^{N,e\bullet}})^2 \|(Q^N)^{\top}\|_2 \|(Q^N)^{\top}\|_2 \|V^N\|_2^2. \end{aligned} $$

Hence, $\mathbb{E}[(\xi_{e}^{N,(2,2),-})^2] \le K$ which ends the step.
Step 5. Here we prove Equation (37).

First remark that $N^{-1} \sum_{i=1}^{N} (c_i^N)^2 (m_i^N - (m_i^N)^2) = N^{-1} \langle (c^N)^2, m^N - (m^N)^2 \rangle$. Then, combining Equations (106) and (103), we know that $N^{-1} \langle (c^N)^2, m^N \rangle$ converges to (remind the definition of $m$ in (8))

$$ (\mu + (1 - \lambda)pr_{-}) \frac{1 + 4(1 - \lambda)^2 p^2 r_{+}r_{-}}{(1 - (1 - \lambda)p(r_{+} - r_{-}))^3} = m \frac{1 + 4(1 - \lambda)^2 p^2 r_{+}r_{-}}{(1 - (1 - \lambda)p(r_{+} - r_{-}))^2}. $$

Similarly, combining Equations (107) and (105), we know that $N^{-1} \langle (c^N)^2, (m^N)^2 \rangle$ converges to

$$ (\mu + (1 - \lambda)pr_{-})^2 \frac{1 + 4(1 - \lambda)^2 p^2 r_{+}r_{-}}{(1 - (1 - \lambda)p(r_{+} - r_{-}))^4} = m^2 \frac{1 + 4(1 - \lambda)^2 p^2 r_{+}r_{-}}{(1 - (1 - \lambda)p(r_{+} - r_{-}))^2}. $$

Finally, one ends up with the definition of $w$ by summing these two limits, which ends this step. $\square$

## APPENDIX G: INVERSION OF Ψ

The main objective of this section is to prove Proposition 2.2 and in particular provide the expressions of $\Phi^{(+)}$ and $\Phi^{(-)}$. As it appears below, the two functions $\Phi^{(+)}$ and $\Phi^{(-)}$ are related with the two roots $d^{(+)}$ and $d^{(-)}$ of a quadratic equation. In turn, these roots are related with the function $D$ defined in Equation (9).

Here is a collection of preliminary results on the functions $D$ and $\Psi$.

**PROPOSITION G.1.** For all $(\mu, \lambda, p) \in \Lambda$,

1. if $r_+ < 1/2$, then $1 < D(\lambda, p) < 2r_-$;

2. if $r_+ = 1/2$, then $D(\lambda, p) = 1$;

3. if $r_+ > 1/2$, then $2r_- < D(\lambda, p) < 1$.

Whatever the value of $r_+$, the image $\Psi(\Lambda)$ is included in $(0, 1) \times (0, \infty)^2$.

Finally, if $r_+ = 1/2$, then $\Psi_3(\mu, \lambda, p) > \Psi_1(\mu, \lambda, p)[1 - \Psi_1(\mu, \lambda, p)]$, for all $(\mu, \lambda, p) \in \Lambda$.

**PROOF.** The statements regarding the function $D$ are obvious.

Let $r_+ \in [0, 1]$ and $(\mu, \lambda, p) \in \Lambda$. For all $k = 1, 2, 3$, the fact that $\Psi_k(\mu, \lambda, p) \in (0, \infty)$ is obvious. It only remains to prove that $\Psi_1(\mu, \lambda, p) < 1$. This follows from

$$ D(\lambda, p) - (\mu + (1 - \lambda)pr_{-}) = 1 - \mu - (1 - \lambda)pr_{+} > 1 - \lambda - (1 - \lambda)r_{+} = (1 - \lambda)pr_{-} \geq 0. $$

Finally, if $r_+ = 1/2$, we have $D(\lambda, p) = 1$ and the statement follows from the fact that $1 + 4(1 - \lambda)^2 p^2 r_+ r_- > 1$. $\square$

From now on, the objective is to invert the function $\Psi$ on the set of admissible parameters $\Lambda$. In view of Proposition G.1, it suffices to find this inverse function on the set

$$ (108) \qquad \mathcal{M} = \begin{cases} (0, 1) \times (0, \infty)^2, & \text{if } r_+ \neq 1/2; \\ \{(m, v, w) \in (0, 1) \times (0, \infty)^2 : w > m(1-m)\}, & \text{if } r_+ = 1/2. \end{cases} $$

Hence, from now on, $(m, v, w)$ will denote an arbitrary vector in $\mathcal{M}$ (in particular, it is not related to $(\mu, \lambda, p)$). Remind the function $\kappa$ defined in Equation 11 and let us define $d^{(+)}, d^{(-)} : (0, 1) \times (0, \infty) \to \mathbb{R}$ by, for $a \in \{-, +\}$,

$$ (109) \qquad d^{(a)}(m, w) = \begin{cases} \frac{4r_+r_- + a\sqrt{(4r_+r_-)^2 - 4r_+r_- + \kappa(m,w)}}{4r_+r_- - \kappa(m,w)}, & \text{if } \kappa(m, w) \neq 4r_+r_-; \\ (8r_+r_-)^{-1}, & \text{if } \kappa(m, w) = 4r_+r_-; \end{cases} $$
If $(\mu, \lambda, p)$ and $(m, v, w)$ are related through (8) then the value $d^{(a)}(m, w)$ is a candidate to be equal to $D(\lambda, p)$. The values $d^{(+)}(m, w)$ and $d^{(-)}(m, w)$ are the roots of the quadratic equation

$$ (110) \qquad [4r_+r_- - \kappa(m, w)]u^2 - (8r_+r_-)u + 1 = 0. $$

On the one hand, $d^{(-)}$ is positive and one can check that it is $C^1$ (even when $\kappa(m, w) \to 4r_+r_-$. On the other hand, $d^{(+)}$ may be negative and goes to infinity when $\kappa(m, w) \to 4r_+r_-$. In view of this remark and Proposition G.1, $d^{(-)}(m, w)$ is expected to be the good candidate to be equal to $D(\lambda, p)$. This is true most of the time but not for the whole range of parameters (see Proposition G.3 below).

From now on, let $a \in \{-, +\}$. Let us define $\phi_1^{(a)}: (0, 1) \times (0, \infty) \to (0, \infty)$ by

$$ (111) \qquad \phi_1^{(a)}(m, w) = \begin{cases} w[m(1-m)]^{-1} - 1, & \text{if } r_+ = 1/2; \\ [1 - d^{(a)}(m, w)]^2(r_+ - r_-)^{-2}, & \text{else,} \end{cases} $$

and $\phi_2^{(a)}: M \to (1, \infty)$ by

$$ (112) \qquad \phi_2^{(a)}(m, v, w) = 1 + \frac{v}{[(m - r_-)^2 + r_+r_-]\phi_1^{(a)}(m, w)}. $$

If $(\mu, \lambda, p)$ and $(m, v, w)$ are related through (8) then $\phi_1^{(a)}(m, w)$ (respectively $\phi_2^{(a)}(m, v, w)$) are two candidates to be equal to $(1 - \lambda)^2 p^2$ (resp. $p^{-1}$). Then, for $k = 1, 2, 3$, let $\Phi_k^{(a)}: M \to \mathbb{R}$ be defined by,

$$ (113) \qquad \left\{ \begin{aligned} \Phi_1^{(a)}(m, v, w) &= m \left( 1 - (r_+ - r_-) \sqrt{\phi_1^{(a)}(m, w)} \right) - r_- \sqrt{\phi_1^{(a)}(m, w)}, \\ \Phi_2^{(a)}(m, v, w) &= 1 - \phi_2^{(a)}(m, v, w) \sqrt{\phi_1^{(a)}(m, w)}, \\ \Phi_3^{(a)}(m, v, w) &= (\phi_2^{(a)}(m, v, w))^{-1}. \end{aligned} \right. $$

Finally, let $\Phi^{(a)}: M \to \mathbb{R}^3$ be defined by the three coordinate functions above. Let us remark that all the functions involved in the definition of $\Phi^{(a)}$ are obviously $C^\infty$ except $d^{(a)}$. Nevertheless, it is easy to check that $d^{(-)}$ is (at least) $C^1$ when $\kappa(m, w) \to 4r_+r_-$. In turn, $\Phi^{(-)}$ is regular.

**PROPOSITION G.2.** The function $\Phi^{(-)}$ is $C^1$.

Finally, the functions $\Phi^{(+)}$ and $\Phi^{(-)}$ are related with $\Psi$ in the following sense.

**PROPOSITION G.3.** Whatever the value of $r_+$, the following results hold.

1. For all $(\mu, \lambda, p) \in \Lambda$, $(\mu, \lambda, p) \in \{\Phi^{(+)} \circ \Psi(\mu, \lambda, p), \Phi^{(-)} \circ \Psi(\mu, \lambda, p)\}$,

2. moreover, if $r_+ \ge 1/2$ or

$$ \begin{gathered} \kappa(\Psi_1(\mu, \lambda, p), \Psi_3(\mu, \lambda, p)) \ge 4r_+r_-, \\ \text{then } (\mu, \lambda, p) = \Phi^{(-)} \circ \Psi(\mu, \lambda, p). \end{gathered} $$

3. Let $(m, v, w) \in M$ and $a \in \{-, +\}$. If $\Phi^{(a)}(m, v, w) \in \Lambda$ and

$$ \begin{gathered} r_+ = 1/2 \text{ or } \operatorname{sgn}(1 - d^{(a)}(m, w)) = \operatorname{sgn}(r_+ - r_-), \\ \text{then } \Psi \circ \Phi^{(a)}(m, v, w) = (m, v, w). \end{gathered} $$

**REMARK G.4.** Remind that $d^{(a)}(m, w)$ is a candidate for $D(\lambda, p)$. In that regard, the sign condition appearing in Item 2 above is consistent with Proposition G.1.

**PROOF.**
*Proof of 1.* Let $(\mu, \lambda, p) \in \Lambda$. In the following, we shorten the notation $\Psi_k = \Psi_k(\mu, \lambda, p)$. Finally, we denote $\Psi = (\Psi_1, \Psi_2, \Psi_3) \in \mathbb{R}^3$.

Let us first consider the case $r_+ = 1/2$. In that case, the $\phi$ and $\Phi$ functions do not depend on the superscript $a \in \{-, +\}$ and we omit it in the following. Furthermore, $D(\lambda, p) = 1$ and $\Psi$ reduces to

$$ 
\begin{cases} 
\Psi_1 = \mu + (1 - \lambda)p/2 \\ 
\Psi_2 = (1 - \lambda)^2 p(1 - p)[(\Psi_1 - 1/2)^2 + 1/4] \\ 
\Psi_3 = \Psi_1[1 - \Psi_1][1 + (1 - \lambda)^2 p^2]. 
\end{cases} 
$$

Hence, it is easy to check that $\phi_1(\Psi_1, \Psi_3) = (1 - \lambda)^2 p^2$ and

$$ \phi_2(\Psi) = 1 + \frac{(1 - \lambda)^2 p(1 - p)[(\Psi_1 - 1/2)^2 + 1/4]}{[(\Psi_1 - 1/2)^2 + 1/4](1 - \lambda)^2 p^2} = 1 + \frac{1 - p}{p} = \frac{1}{p}. $$

Then, we conclude by

$$ 
\begin{cases} 
\Phi_1^{(a)}(\Psi) = \Psi_1 - (1 - \lambda)p/2 = \mu, \\ 
\Phi_2^{(a)}(\Psi) = 1 - \frac{1}{p}(1 - \lambda)p = \lambda, \\ 
\Phi_3^{(a)}(\Psi) = (1/p)^{-1} = p. 
\end{cases} 
$$

Let us then consider the case $r_+ \neq 1/2$. First, we remark that

$$ \kappa(\Psi_1, \Psi_3) = (r_+ - r_-)^2 \frac{1 + 4(1 - \lambda)^2 p^2 r_+ r_-}{D(\lambda, p)^2}. $$

Then, substituting $(1 - \lambda)p = (1 - D(\lambda, p))/(r_+ - r_-)$ into the equation above gives that

$$ \kappa(\Psi_1, \Psi_3) = \frac{(r_+ - r_-)^2 + 4r_+r_- (1 - D(\lambda, p))^2}{D(\lambda, p)^2}. $$

Then, using the fact that $(r_+ - r_-)^2 = (r_+ + r_-)^2 - 4r_+r_- = 1 - 4r_+r_-$, it follows that

$$ \kappa(\Psi_1, \Psi_3)D(\lambda, p)^2 = 1 + 4r_+r_-[D(\lambda, p)^2 - 2D(\lambda, p)], $$

which means that $D(\lambda, p)$ is a solution of the quadratic equation

$$ [4r_+r_- - \kappa(\Psi_1, \Psi_3)]X^2 - (8r_+r_-)X + 1 = 0. $$

By definition of $d^{(+)}$ and $d^{(-)}$, we necessarily have $D(\lambda, p) \in \{d^{(+)}(\Psi_1, \Psi_3), d^{(-)}(\Psi_1, \Psi_3)\}$. In turn, since $(1 - D(\lambda, p))^2(r_+ - r_-)^{-2} = (1 - \lambda)^2 p^2$, we have

$$ (1 - \lambda)^2 p^2 \in \{\phi_1^{(+)}(\Psi_1, \Psi_2), \phi_1^{(-)}(\Psi_1, \Psi_2)\}, $$

and the proof is concluded in the same manner as the case $r_+ = 1/2$.

*Proof of 2.* Let us first remark that, if $r_+ = 1/2$ or $\kappa(\Psi_1, \Psi_3) = 4r_+r_-$, then $\Phi^{(+)} = \Phi^{(-)}$ and the result is trivial.

On the one hand, if $\kappa(\Psi_1, \Psi_3) > 4r_+r_-$ then the root $d^{(+)}(\Psi_1, \Psi_3)$ is negative which implies that $D(\lambda, p) = d^{(-)}(\Psi_1, \Psi_3)$. On the other hand, if $r_+ > 1/2$ and $\kappa(\Psi_1, \Psi_3) < 4r_+r_-$ then

$$ d^{(+)}(\Psi_1, \Psi_3) \geq \frac{4r_+r_-}{4r_+r_- - \kappa(\Psi_1, \Psi_3)} > 1 $$

Yet, we know that $D(\lambda, p) < 1$ by Proposition G.1 which implies that $D(\lambda, p) = d^{(-)}(\Psi_1, \Psi_3)$.

As a summary, we have $D(\lambda, p) = d^{(-)}(\Psi_1, \Psi_3)$ in any case. In turn, it implies that $(1 - \lambda)^2 p^2 = \phi_1^{(-)}(\Psi_1, \Psi_3)$ and the proof is concluded as above.
Proof of 3. Let $(m, v, w) \in \mathcal{M}$, and $a \in \{-, +\}$ such that $\Phi^{(a)}(m, v, w) \in \Lambda$. In the following, we shorten the notation $d^{(a)} = d^{(a)}(m, w)$, $\phi_1^{(a)} = \phi_1^{(a)}(m, w)$, $\phi_2^{(a)} = \phi_2^{(a)}(m, v, w)$ and $\Phi_k^{(a)} = \Phi_k^{(a)}(m, v, w)$ for $k = 1, 2, 3$. Finally, we denote $\Phi^{(a)} = (\Phi_1^{(a)}, \Phi_2^{(a)}, \Phi_3^{(a)}) \in \Lambda$.

Without any condition, it is easy to check that $D(\Phi_2^{(a)}, \Phi_3^{(a)}) = 1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}}$. In particular, it gives

$$ \Psi_1(\Phi^{(a)}) = \frac{\Phi_1^{(a)} + (1 - \Phi_2^{(a)})\Phi_3^{(a)}r_-}{1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}}} \\ = \frac{m\left[1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}}\right] - r_-\sqrt{\phi_1^{(a)}} + r_+\sqrt{\phi_1^{(a)}}}{1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}}} = m, $$

and

$$ \Psi_2(\Phi^{(a)}) = (1 - \Phi_2^{(a)})^2 \Phi_3^{(a)} (1 - \Phi_3^{(a)}) [(m - r_-)^2 + r_+ r_-] \\ = \phi_1^{(a)} [\phi_2^{(a)} - 1] [(m - r_-)^2 + r_+ r_-] = v. $$

For the last coordinate, we use the condition stated in the Proposition.

Let us first consider the case $r_+ = 1/2$. In that case, the $\phi$ and $\Phi$ functions do not depend on the superscript $a \in \{-, +\}$ and we omit it in the following. First, remark that $D(\Phi_2^{(a)}, \Phi_3^{(a)}) = 1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}} = 1$ in that case, so that

$$ \Psi_3(\Phi^{(a)}) = m(1-m)[1 + (1 - \Phi_2^{(a)})^2(\Phi_3^{(a)})^2] \\ = m(1-m)\left[1 + \frac{w}{m(1-m)} - 1\right] = w. $$

Let us then consider the case $r_+ \neq 1/2$ and assume the sign condition: $\operatorname{sgn}(1 - d^{(a)}) = \operatorname{sgn}(r_+ - r_-)$. This condition implies that $D(\Phi_2^{(a)}, \Phi_3^{(a)}) = 1 - (r_+ - r_-)\sqrt{\phi_1^{(a)}} = d^{(a)}$. Hence,

$$ \Psi_3(\Phi^{(a)}) = m(1-m)[1 + 4r_+r_- (1 - \Phi_2^{(a)})^2 (\Phi_3^{(a)})^2] / (d^{(a)})^2 \\ = m(1-m) \left[ 1 + 4r_+r_- \frac{(1-d^{(a)})^2}{(r_+ - r_-)^2} \right] / (d^{(a)})^2. $$

Yet,

$$ 4r_+r_- (1 - d^{(a)})^2 = 4r_+r_- - 1 + (1 - 8r_+r_-d^{(a)} + 4r_+r_- (d^{(a)})^2) = 4r_+r_- - 1 + \kappa(m, w)(d^{(a)})^2 $$

because $d^{(a)}$ solves the quadratic equation (110). Finally, using the definition of $\kappa$ and the fact that $1 - 4r_+r_- = (r_+ - r_-)^2$ we get that $\Psi_3(\Phi^{(a)}) = w$ which concludes the proof. $\square$

## APPENDIX H: AUXILIARY RESULTS

Here are two technical lemmas used throughout the proof of Proposition 4.2.

**LEMMA H.1.** Let $(V^N)_N, (v^N)_N$ be two sequences of random vectors such that for all $N > 0$, $V^N, v^N \in \mathbb{R}^N$. Assume that there exists a constant $K$ such that, for all $N > 0$, $v^N$ can be written as $v^N = v_1^N + v_2^N + v_3^N$, where $v_1^N, v_2^N, v_3^N \in \mathbb{R}^N$, and
(i) $\mathbb{E} [\|V^N\|_2^4] + \|V^N\|_\infty^2 \le K$ almost surely,

(ii) $\mathbb{E} [\|v_1^N\|_1^2 + \|v_2^N\|_2^4]^2 \le K$,

(iii) $\mathbb{E} [\langle V^N, v_3^N \rangle^2] \le K$.

Then, for all $N > 0$,

$$ \mathbb{E} [\langle V^N, v^N \rangle^2] \le 3(K^2 + 2K). $$

PROOF. First, by Holder inequality,

$$ \mathbb{E} [\langle V^N, v_1^N \rangle^2] \le \mathbb{E} [\|V^N\|_\infty^2 \|v_1^N\|_1^2] \le K \mathbb{E} [\|v_1^N\|_1^2] \le K^2. $$

Secondly, using Cauchy Schwarz inequality twice, we have

$$ \mathbb{E} [\langle V^N, v_2^N \rangle^2] \le \mathbb{E} [\|V^N\|_2^2 \|v_2^N\|_2^2] \le (\mathbb{E} [\|V^N\|_2^4] \mathbb{E} [\|v_2^N\|_2^4])^{1/2} \le K. $$

And, $\mathbb{E} [\langle V^N, v_3^N \rangle^2] \le K$ by assumption. Finally, we conclude by combining those three inequalities thanks to the convexity of the square function. $\square$

LEMMA H.2. Let $a \in \{-, +\}$. Let $(u^N)_N$, $(v^N)_N$ be two sequences of random vectors such that for all $N > 0$, $u^N, v^N \in \mathbb{R}^N$ are supported in $P_a$. Assume that there exist two random variables $u_3^\infty, v_3^\infty \in \mathbb{R}$ and a constant $K$ such that, for all $N > 0$, $u^N = u_1^N + u_2^N + u_3^N 1_{P_a}$ and $v^N = v_1^N + v_2^N + v_3^N 1_{P_a}$, where $u_1^N, u_2^N, v_1^N, v_2^N \in \mathbb{R}^N$ are random vectors supported in $P_a$ and $u_3^N, v_3^N \in \mathbb{R}$ are random variables, and

$$ \bullet \mathbb{E} [\|u_1^N\|_1^2 + \|u_2^N\|_2^4 + N^2(\overline{u_2^N})^2] + \|u_1^N + u_2^N\|_\infty + |u_3^N| + |u_3^\infty| \le K \text{ almost surely,} $$

$$ \bullet \mathbb{E} [\|v_1^N\|_1^2 + \|v_2^N\|_2^4 + N^2(\overline{v_2^N})^2] + \|v_1^N + v_2^N\|_\infty + |v_3^N| + |v_3^\infty| \le K \text{ almost surely,} $$

$$ \bullet \mathbb{E} [|u_3^N - u_3^\infty|^2 + |v_3^N - v_3^\infty|^2] \le KN^{-2}. $$

Then, there exists another constant $K$ (independent of $N$) such that, for all $N > 0$,

$$ \mathbb{E} \left[ \left| \frac{1}{N} \langle u^N, v^N \rangle - r_a v_3^\infty u_3^\infty \right|^2 \right] \le KN^{-2}. $$

PROOF. Throughout the proof, the constant $K$ may change from line to line and even within the same line. Since $u^N$ and $v^N$ are decomposed into three vectors each, there are naturally nine contributions to the scalar product $\langle u^N, v^N \rangle$.

Let us first prove that the limit term $r_a v_3^\infty u_3^\infty$ comes from the contribution of the constant parts, i.e. $N^{-1}\langle u_3^N 1_{P_a}, v_3^N 1_{P_a} \rangle = r_a^N u_3^N v_3^\infty$. Indeed,

$$ \begin{aligned} |r_a^N u_3^N v_3^N - r_a v_3^\infty u_3^\infty| &\le |r_a^N u_3^N| |v_3^N - v_3^\infty| + |r_a^N v_3^\infty| |u_3^N - u_3^\infty| + |u_3^N v_3^\infty| |r_a^N - r_a| \\ &\le K (|u_3^N - u_3^\infty| + |v_3^N - v_3^\infty|) + K^2 |r_a^N - r_a|, \end{aligned} $$

so that $\mathbb{E} [|N^{-1}\langle u_3^N 1_{P_a}, v_3^N 1_{P_a} \rangle - r_a v_3^\infty u_3^\infty|^2] \le KN^{-2}$ thanks to the assumptions on $u_3^N$, $v_3^N$ and the sequence of fractions $r_a^N$.

Then, let us prove that the other contributions to the scalar product are negligible. We have,

$$ \mathbb{E} [\langle u_1^N, v^N \rangle^2] \le \mathbb{E} [\|u_1^N\|_1^2 \|v^N\|_\infty^2] \le K \mathbb{E} [\|u_1^N\|_1^2] \le K, $$

and similarly $\mathbb{E} [\langle u^N, v_1^N \rangle^2] \le K$. Applying Cauchy-Schwarz inequality twice, we have

$$ \mathbb{E} [\langle u_2^N, v_2^N \rangle^2] \le \mathbb{E} [\|u_2^N\|_2^2 \|v_2^N\|_2^2] \le (\mathbb{E} [\|u_2^N\|_2^4] \mathbb{E} [\|v_2^N\|_2^4])^{1/2} \le K. $$
Finally,

$$ \mathbb{E} \left[ \left\langle u_2^N, v_3^N 1_{\mathcal{P}_a} \right\rangle^2 \right] \leq K \mathbb{E} \left[ \left\langle u_2^N, 1_{\mathcal{P}_a} \right\rangle^2 \right] = K \mathbb{E} \left[ N^2 (\overline{u_2^N})^2 \right] \leq K, $$

and similarly $\mathbb{E} \left[ \left\langle u_3^N 1_{\mathcal{P}_a}, v_2^N \right\rangle^2 \right] \leq K$. Finally, we conclude by combining all those inequalities thanks to the convexity of the square function. □

## APPENDIX I: A NOTE ON LOWER BOUNDS

The goal of this section is to discuss the optimality of our estimation rate. To that end, we analyze two simple and related statistical settings (both inspired by the one considered in Delattre and Fournier (2016)) where all the computations can be done more transparently. The first statistical setting is discussed in Section I.1 and the second one in Section I.2. In the end of that subsection, we also present some concluding remarks connecting the results proved. Finally, in Section I.3, we state and prove a Gaussian approximation used in Section I.1.

Throughout the section, we write $\theta$ to denote a random variable distributed as $\text{Bin}(N, p)$ where $0 < p_{\min} \le p \le p_{\max} < 1$ is an unknown parameter. Let $\kappa \in (0, p_{\max}/2)$ be a known value and define $\gamma(p) = \kappa/p$ for all $p_{\min} \le p \le p_{\max}$. Observe that $\gamma(p) < 1/2$ for all values of $p$. In what follows, we denote $m = 1/2 + \kappa$.

**I.1. Statistical setting 1 and problem formulation.** Let $\theta \sim \text{Bin}(N, p)$ and consider a discrete random variable $B$ taking values in $\{0, \dots, T\}$ such that $B|\theta \sim \text{Bin}(T, 1/2 + \gamma(p)\theta/N)$. Note $1/2 + \gamma(p)\theta/N < 1$ for all $p$ and all realizations of $\theta$, since $\gamma(p) < 1/2$. In particular, the conditional distribution of $B|\theta$ is well-defined no matter the realization of $\theta$.

Suppose we observe $N$ independent copies $B_1, \dots, B_N$ of the random variable $B$ as defined above. Given these observations, we want to find an unbiased estimator for the parameter $1/p$ whose variance is of order $(N^{1/2}/T + 1/N^{1/2})^2$.

In what follows, we write $\mathbb{P}_p$ to denote the probability distribution of $B_1, \dots, B_N$ associated to the choice of $p$. The expectation and variance computed with respect to $\mathbb{P}_p$ are denoted $\mathbb{E}_p$ and $\text{Var}_p$ respectively. One can easily check that, for all values of $p$, the mean of each random variable $B_i$ under $\mathbb{P}_p$ is a known quantity: $\mathbb{E}_p[B_i] = \mathbb{E}_p[B] = Tm$.

**REMARK I.1.** Note that the goal is to determine an unbiased estimator for the parameter $1/p$ and not for the parameter $p$ itself. In Section I.2, we explain why the parameter $1/p$ is a “natural” parameter to be estimated in this setting.

Note also that the random variables $B_1, \dots, B_N$ can be seen as an oversimplified version of the random variables $\sum_{t=1}^T X_{1,t}, \dots, \sum_{t=1}^T X_{N,t}$ in which we drop the temporal and spatial dependence between the random variables $X_{i,t}$ given the realization of the environment and consider $\mu = 1/2$ and $\mathcal{P}_- = \emptyset$ (no inhibition). The parameter $\gamma$ plays the role of the parameter $(1-\lambda)$ in the original model. In particular, it is implicitly assumed that the baseline parameter $\mu$ is known and equals to $1/2$, making the estimation problem even simpler.

Recall that $m = 1/2 + \kappa$, where $\kappa$ is a known quantity. Given the random variables $B_1, \dots, B_N$, define

$$ \hat{V} = \frac{1}{N} \sum_{i=1}^{N} (B_i - Tm)^2, $$
and consider the following estimator for $1/p$:

$$ (114) \qquad \hat{S} = \frac{N}{T(T-1)\kappa^2} (\hat{V} - Tm(1-m)) + 1. $$

The estimator $\hat{S}$ is unbiased (see the proposition below) and follows from the method of moments. We expect that its variance gives the optimal rate of convergence within the class of unbiased estimators of $1/p$. However, we do not have a proof of such result at the moment. The goal of this subsection is to show the following result.

**PROPOSITION I.2.** For the estimator $\hat{S}$ defined in (114) the following results hold:

1. $\hat{S}$ is an unbiased estimator of $1/p$: for all $p$,

$$ \mathbb{E}_p[\hat{S}] = 1/p. $$

2. For all $p$,

$$ \mathrm{Var}_p(\hat{S}) = \frac{2}{\kappa^4} \left( \frac{m(1-m)N^{1/2}}{T} + \frac{\gamma^2 p(1-p)}{N^{1/2}} \right)^2 (1+o(1)). $$

**PROOF.** We start proving Item 1. First, use the decomposition

$$ \mathrm{Var}_p(B) = \mathbb{E}_p[\mathrm{Var}_p(B|\theta)] + \mathrm{Var}_p[\mathbb{E}_p(B|\theta)] $$

to check that

$$ (115) \qquad \mathrm{Var}_p(B) = Tm(1-m) + \frac{T(T-1)}{N} \gamma^2 p(1-p). $$

Then, to conclude the proof, combine the above identity with the fact that

$$ \mathbb{E}_p(\hat{S}) = \frac{N}{T(T-1)(\gamma p)^2} (\mathrm{Var}_p(B) - Tm(1-m)) + 1. $$

Next, we prove Item 2. First, observe that

$$ (116) \qquad \mathrm{Var}_p(\hat{S}) = \frac{N}{T^2(T-1)^2(\gamma p)^4} \mathrm{Var}_p((B-Tm)^2). $$

Next, write $\mathrm{Var}_p((B-Tm)^2) = \mathbb{E}_p[(B-Tm)^4] - \mathrm{Var}_p^2(B)$ and use (115) to show that

$$ (117) \qquad \frac{N}{T^2(T-1)^2} \mathrm{Var}_p^2(B) = \left( \frac{N^{1/2}}{(T-1)} m(1-m) + \frac{\gamma^2 p(1-p)}{N^{1/2}} \right)^2. $$

To conclude the proof, we approximate the term $\mathbb{E}_p[(B-Tm)^4]$. To that end, we use the Gaussian approximation provided in Lemma I.5 below. Using this approximation, we can write $B/T - m = G + \epsilon$ so that

$$ (118) \qquad \mathbb{E}_p[(B-Tm)^4] = T^4 \left( \mathbb{E}_p[G^4] + \sum_{k=1}^4 \binom{4}{k} \mathbb{E}_p[G^{4-k}\epsilon^k] \right). $$

Now, since $G$ is a centered Gaussian random variable with variance $m(1-m)T^{-1} + \gamma^2 p(1-p)N^{-1}$, we obtain that

$$ (119) \qquad \mathbb{E}_p[G^4] = 3 \left( \frac{m(1-m)}{T} + \frac{\gamma^2 p(1-p)}{N} \right)^2. $$
Moreover, using Cauchy-Schwarz inequality, the estimate (121), and the fact that G is a centered Gaussian random variable, one can show that there exists a constant $C_k$ depending only on $k$ such that

$$ \mathbb{E}_p[G^{4-k} \epsilon^k] \le C_k \left( \frac{m(1-m)}{T} + \frac{\gamma^2 p(1-p)}{N} \right)^{(4-k)} \times \left( \frac{1}{T^{k/2} N^{k/4}} + \frac{\log^k(T)}{T^k} + \frac{\log^k(N)}{N^k} \right). $$

Combining the above inequality with Jensen inequality, we then conclude that

$$ N \mathbb{E}_p[G^{4-k} \epsilon^k] \le \tilde{C}_k \left( \frac{N^{(4-k)/4}}{T^{4-k/2}} + \frac{N \log^k(T)}{T^4} + \frac{\log^k(N)}{N^{(k-1)T^{(4-k)}}} + \frac{1}{T^{k/2} N^{3-3k/4}} + \frac{\log^k(T)}{N^{(3-k)T^k}} + \frac{\log^k(N)}{N^3} \right), $$

where $\tilde{C}_k$ is a constant depending only on $k$. Since $\frac{N^{(4-k)/4} T^2}{T^{4-k/2} N} = \frac{1}{T^{2-k/2} N^{k/4}} \to 0$ and $\frac{N \log^k(T) T^2}{T^4 N} = \frac{\log^k(T)}{T^2} \to 0$ as $N, T \to \infty$, one can deduce from the above inequality that

$$ (120) \quad \max_{k=1,\dots,4} N \mathbb{E}_p[G^{4-k} \epsilon^k] = o \left( \left( \frac{m(1-m)N^{1/2}}{T} + \frac{\gamma^2 p(1-p)}{N^{1/2}} \right)^2 \right). $$

Combining (116), (117), (118), (119) and (120), the result follows. □

<figure><img src="image_20.png" /></figure>

**I.2. Statistical setting 2 and problem formulation.** As in the previous subsection, let $\theta \sim \text{Bin}(N, p)$. Here, we assume that the random variable B takes values in $\{0, 1, 2\}$ and is such that $B|\theta \sim \text{Bin}(2, 1/2 + \gamma(p)\theta/N)$.

Suppose that $T$ is even and that we observe $n := (NT)/2$ independent copies $B_1, \dots, B_n$ of the random variable $B$ as defined above. We want to show that the variance of any unbiased estimator of the parameter $1/p$ is larger than $K(N/T)$ for some positive universal constant $K$. Hence, if $N/T$ remains bounded away from 0, then no unbiased estimator of the parameter $1/p$ is consistent.

With a slight abuse of notation, we also write $\mathbb{P}_p$ to denote the probability distribution of $B_1, \dots, B_n$ associated to the choice of $p$. The expectation and variance computed with respect to $\mathbb{P}_p$ are denoted $\mathbb{E}_p$ and $\text{Var}_p$ respectively. In this setting, observe that $\mathbb{E}_p[B_i] = 2(1/2 + \kappa) = 2m$ for all values of $p$.

**REMARK I.3.** In this setting, the random variables $B_1, \dots, B_n$ are distributed as the random variables $(X_{1,1} + X_{1,2}), (X_{1,3} + X_{1,4}), \dots, (X_{1,T-1} + X_{1,T}), \dots, (X_{N,1} + X_{N,2}), (X_{N,3} + X_{N,4}), \dots, (X_{N,T-1} + X_{N,T})$ under the assumptions that: 1) $\mu = 1/2$ and $\mathcal{P}_- = \emptyset$ (no inhibition), 2) for all $1 \le i \le N$ and $1 \le t \le T/2$, the random variables $X_{i,2t-1}$ and $X_{i,2t}$ are conditionally independent given the realization of the environment and 3) all the other dependencies are dropped. Note that the random variables $X_{i,t}$ are less dependent under this set of assumptions than under the assumptions made in the statistical setting 1.

While in our original problem, the “natural” parameter to estimate is $p$, in this statistical setting the “natural” parameter is $1/p$ (as well as in the statistical setting 1). We can see this
by checking that the probability mass function $f_p$ of each random variable $B_i$ under $\mathbb{P}_p$ can be written as

$$f_p(b) = \binom{2}{b} \left[ m^b (1-m)^{(2-b)} + \frac{\kappa^2}{N} \left( \frac{1}{p} - 1 \right) s(b) \right], b \in \{0, 1, 2\},$$

where $s(b) = 1$ if $b \in \{0, 2\}$ and $s(b) = -1$ if $b = 1$. The expression above for $f_p(b)$ follows from the fact that

$$f_p(b) = \mathbb{E}_p \left[ \binom{2}{b} \left( \frac{1}{2} + \frac{\gamma(p)\theta}{N} \right)^b \left( \frac{1}{2} - \frac{\gamma(p)\theta}{N} \right)^{2-b} \right]$$

and simple calculations involving binomial random variables.

To alleviate the notation, we parametrize the model by $\eta = 1/p$ in what follows. In particular, we write $\mathbb{P}_\eta = \mathbb{P}_{1/p}$, $\mathbb{E}_\eta = \mathbb{E}_{1/p}$ and $\text{Var}_\eta = \text{Var}_{1/p}$. Therefore, we assume that we observe $n$ independent copies $B_1, \dots, B_n$ of a random variable $B$ taking values in $\{0, 1, 2\}$ having as probability mass function $g_\eta(b) = f_{1/\eta}(b)$. We want to show that no unbiased estimator of $\eta$ based on these observations is consistent whenever $N/T$ remains bounded away from 0. This is the content of the proposition below.

**PROPOSITION I.4.** For all $N$ large enough, we have for any unbiased estimator $\hat{\eta}(B_1, \dots, B_n)$ of $\eta$ that

$$\mathrm{Var}_{\eta}(\hat{\eta}(B_1, \dots, B_n)) \geq \frac{NC^2}{T\kappa^4},$$

where $C := \min\{m^2, m(1-m), (1-m)^2\} > 0$. In particular, if $N/T$ remains bounded away from 0, then no unbiased estimator of $\eta$ is consistent.

**PROOF.** First, verify that the Fisher information of a single observation is given by

$$I_1(\eta) := -\mathbb{E}_{\eta} \left[ \frac{\partial^2 \log g_{\eta}(B)}{\partial \eta^2} \right] = \frac{\kappa^4}{N^2} \mathbb{E}_{\eta} \left[ \frac{1}{(m^B(1-m)^{(2-B)} + \frac{\kappa^2}{N}(\eta-1)s(B))^2} \right].$$

Then, use Corollary 5.9 of Lehmann and Casella (1998) to deduce that the Fisher information of $n = NT/2$ i.i.d observations is given by

$$I_n(\eta) = nI_1(\eta) = \frac{T\kappa^4}{2N} \mathbb{E}_{\eta} \left[ \frac{1}{(m^B(1-m)^{(2-B)} + \frac{\kappa^2}{N}(\eta-1)s(B))^2} \right].$$

Next, observe that

$$(m^B(1-m)^{(2-B)} + \frac{\kappa^2}{N}(\eta-1)s(B))^2 \\ \geq m^B(1-m)^{2-B} \left(m^B(1-m)^{2-B} + \frac{2\kappa^2(\eta-1)}{Ns(B)}\right)$$

and that $(\eta - 1)/Ns(B) \ge 0$ for $B = 0, 2$. Combining these inequalities with the definition of $C = \min\{m^2, m(1-m), (1-m)^2\}$, we deduce that

$$(m^B(1-m)^{(2-B)} + \frac{\kappa^2}{N}(\eta-1)s(B))^2 \geq C \left(C - \frac{2\kappa^2(\eta-1)}{N}\right),$$

from which it follows that for $N \ge 4(\eta - 1)\kappa^2/C$,

$$(m^B(1-m)^{(2-B)} + \frac{\kappa^2}{N}(\eta-1)s(B))^2 \geq C^2/2,$$
which, in turn, implies that

$$I_n(\eta) \le \frac{T\kappa^4}{NC^2},$$

as long as $N \ge 4(\eta - 1)\kappa^2/C$. By Assumption 1 (and restricting the range of $p$ if necessary), we have that $(\eta - 1) \le M$ for some suitable constant $M$ so that for all $N \ge 4M\kappa^2/C$, we have that

$$I_n(\eta) \le \frac{T\kappa^4}{NC^2}.$$

Therefore, by Cramér-Rao lower bound (see Theorem 5.10 of Lehmann and Casella (1998)), it then follows that for all unbiased estimators $\hat{\eta}(B_1, \dots, B_n)$ of $\eta$,

$$\mathrm{Var}_{\eta}(\hat{\eta}(B_1, \dots, B_n)) \ge (I_n(\eta))^{-1} = \frac{NC^2}{T\kappa^4},$$

and the result follows. □

*Concluding remarks.* Both statistical settings 1 and 2 consider a statistical model with a hidden layer described by the random variable $\theta$, similar as in the original model. As indicated in Remarks I.1 and I.3, the laws of the statistical models considered in these settings are related to the law of the original model under different sets of assumptions dropping many dependencies. Moreover, more dependencies of the original model are preserved in setting 1.

Both Propositions I.2 and I.4 highlight the interplay between the dimension of the original model (number $N$ of observed nodes) and the sample size (the number $T$ of time units observed per node). In the regime $N^2 \gg T$ (which implies that $N/T \gg (N^{1/2}/T + 1/N^{1/2})^2$), the results of Proposition I.2 and I.4 combined indicate that estimating the model parameter becomes easier as more dependencies of the original model are preserved. This partially explains why the lower bound in Proposition I.4 suggests that the condition $T \gg N$ is necessary for our original estimation problem, while Item 2 of Proposition I.2 indicates that the weaker condition $T \gg N^{1/2}$ might suffice for consistently estimating the model parameters. Proving such a result in the original model remains an interesting open question.

**I.3. Auxiliary results.** In this section, we state and prove the Gaussian approximation used to prove Item 2 of Proposition I.2. This result follows almost immediately from the KMT approximation theorem (see for example Komlós, Major and Tusnády (1975)).

**LEMMA I.5.** Let $B$ be a discrete random variable taking values in $\{0, \dots, T\}$ such that $B|\theta \sim \mathrm{Bin}(T, 1/2 + \gamma\theta/N)$ where $\theta \sim \mathrm{Bin}(N, p)$, $0 < p_{\min} \le p \le p_{\max} < 1$ and $0 < \gamma_{\min} \le \gamma \le \gamma_{\max} < 1/2$. Denote $m = 1/2 + \gamma p$. There exists a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ on which almost surely

$$\frac{B}{T} - m = G + \epsilon,$$

where $G$ is a centered Gaussian random variable with variance $m(1 - m)T^{-1} + \gamma^2 p(1 - p)N^{-1}$ and the error term $\epsilon$ satisfies for all integer $\alpha \ge 1$,

$$(121) \qquad \mathrm{E}[|\epsilon|^\alpha] \le K_\alpha \left( \frac{1}{T^{\alpha/2}N^{\alpha/4}} + \left(\frac{\log(T)}{T}\right)^\alpha + \left(\frac{\log(N)}{N}\right)^\alpha \right),$$

for some positive constant $K_\alpha$ depending only $\alpha$.
PROOF. Throughout the proof, we assume we work on a probability space $(\Omega, \mathcal{F}, \mathcal{P})$ rich enough so that all the processes below are well-defined.

Let $(U_t)_{1\le t\le T}$ and $(V_j)_{j\in[N]}$ be two independent collections of i.i.d random variables uniformly distributed on $(0, 1)$. For each $u \in [0, 1]$, define

$$F_T(u) = \frac{1}{T} \sum_{t=1}^{T} 1_{U_t \le u} \quad \text{and} \quad F_N(u) = \frac{1}{N} \sum_{j=1}^{N} 1_{V_j \le u},$$

and

$$\alpha_T(u) = \sqrt{T}(F_T(u) - u) \text{ and } \alpha_N(u) = \sqrt{N}(F_N(u) - u).$$

With this notation, one can check that

$$(122) \qquad \frac{B}{T} = m + \frac{1}{\sqrt{T}} \alpha_T(1/2 + \gamma\theta/N) + \frac{\gamma}{\sqrt{N}} \alpha_N(p).$$

Next, consider two independent Brownian Bridges $(B_T(t))_{0\le u\le 1}$ and $(B_N(t))_{0\le u\le 1}$ and define for each $u \in [0, 1]$,

$$(123) \qquad \epsilon_T(u) = \alpha_T(u) - B_T(u) \text{ and } \epsilon_N(u) = \alpha_N(u) - B_N(u).$$

We also suppose that the first Brownian Bridge $(B_T(t))_{0\le t\le 1}$ is independent of $\theta$ as well. Furthermore, we follow the construction used in the KMT approximation (see Komlós, Major and Tusnády (1975)) in such a way that we can also assume that for $z > 0$,

$$(124) \qquad \mathbb{P} \left( \sup_{0 \le u \le 1} |\epsilon_T(u)| \ge \frac{1}{T^{1/2}} (a \log(T) + z) \right) \le be^{-cz},$$

and

$$\mathbb{P} \left( \sup_{0 \le u \le 1} |\epsilon_N(u)| \ge \frac{1}{N^{1/2}} (a \log(N) + z) \right) \le be^{-cz},$$

for some positive universal constants $a, b$, and $c$. Using that $B_N(u) \sim N(0, u(1-u))$ for each $u \in [0, 1]$, we immediately see that

$$\frac{\gamma}{\sqrt{N}} B_N(p) \sim N \left( 0, \frac{\gamma^2 p(1-p)}{N} \right).$$

By similar arguments, we also have that $\frac{1}{\sqrt{T}} B_T(m) \sim N(0, \frac{m(1-m)}{T})$, so that it follows from the independence between $B_T(p)$ and $B_N(p)$ that

$$(125) \qquad \frac{1}{\sqrt{T}} B_T(m) + \frac{\gamma}{\sqrt{N}} B_N(p) \sim N \left( 0, \frac{m(1-m)}{T} + \frac{\gamma^2 p(1-p)}{N} \right).$$

Combining identities (122), (123) and (125), we then deduce that

$$\frac{B}{T} - m = G + \epsilon,$$

where

$$G = \frac{1}{\sqrt{T}} B_T(m) + \frac{\gamma}{\sqrt{N}} B_N(p)$$

is a centered Gaussian random variable with variance $m(1-m)T^{-1} + \gamma^2 p(1-p)N^{-1}$ and

$$(126) \qquad \epsilon = \frac{1}{\sqrt{T}} \left( (B_T(1/2 + \gamma\theta/N) - B_T(m)) + \epsilon_T(1/2 + \gamma\theta/N) \right) + \frac{\gamma}{\sqrt{N}} \epsilon_N(p).$$
Next, we will prove that the error term $\epsilon$ defined in (126) satisfies the bound stated in (121). By combining the tail integral formula for the moments with (124), one can show that for any integer $\alpha \ge 1$

$$ (127) \qquad \mathbb{E} \left[ \left( \sup_{0 \le u \le 1} |\epsilon_T(u)| \right)^\alpha \right] \le K \frac{\log^\alpha(T)}{T^{\alpha/2}}, $$

for some constant $K$ depending only on $a, b, c$ and $\alpha$. Similarly, one can also show that for any integer $\alpha \ge 1$

$$ (128) \qquad \mathbb{E} \left[ \left( \sup_{0 \le u \le 1} |\epsilon_N(u)| \right)^\alpha \right] \le K \frac{\log^\alpha(N)}{N^{\alpha/2}}, $$

for some constant $K$ depending only on $a, b, c$ and $\alpha$. Moreover, by the independence between $(B_T(u))_{0 \le u \le 1}$ and $\theta$, we can write

$$ \mathbb{E}[(B_{i,T}(1/2 + \gamma\theta/N) - B_{i,T}(m))^\alpha] = \mathbb{E}[\varphi_\alpha(1/2 + \gamma\theta/N)], $$

where $\varphi_\alpha(q) = \mathbb{E}[(B_T(q) - B_T(m))^\alpha]$, $q \in [0, 1]$. Next, using properties of a Brownian Bridge, we can check that

$$ \varphi_\alpha(q) \le 2^{\alpha-1} |q-m|^{\alpha/2} \mathbb{E}[|N(0,1)|^\alpha] (1 - |q-m|^{\alpha/2}) $$

so that denoting $K_\alpha = 2^{\alpha-1} \mathbb{E}[|N(0,1)|^\alpha]$, we obtain that

$$ \mathbb{E}[(B_T(1/2 + \gamma\theta/N) - B_T(m))^\alpha] \le K_\alpha \gamma \mathbb{E}[|\theta/N - p|^{\alpha/2}], $$

where we have also used that $0 \le |q-m| \le 1$ which implies that $(1 - |q-m|^{\alpha/2}) \le 1$. By applying Lemma F.7 of the main text, we then deduce that

$$ (129) \qquad \mathbb{E}[(B_T(1/2 + \gamma\theta/N) - B_T(m))^\alpha] \le K N^{-\alpha/4} $$

for $K = K_\alpha \gamma_{max}^\alpha C_\alpha$, where $C_\alpha$ is a constant depending only on $\alpha$. Combining (127), (128), (129) with Jensen inequality, we then obtain that

$$ (130) \qquad \mathbb{E}[|\epsilon|^\alpha] \le K \left( \frac{1}{T^{\alpha/2} N^{\alpha/4}} + \left( \frac{\log(T)}{T} \right)^\alpha + \left( \frac{\log(N)}{N} \right)^\alpha \right), $$

for some constant $K$ depending only on $a, b, c, \gamma_{max}$ and $\alpha$.

□