Large Deviations for the Annealed Ising Model on Inhomogeneous Random Graphs: Spins and Degrees

Abstract

We prove a large deviations principle for the total spin and the number of edges under the annealed Ising measure on generalized random graphs. We also give detailed results on how the annealing over the Ising model changes the degrees of the vertices in the graph and show how it gives rise to interesting correlated random graphs.

Appendix: LDP for the total spin using combinatorial arguments

In this appendix, we obtain the large deviation function of the total spin in the rank-1 inhomogeneous Curie–Weiss model (and thus in the annealed Ising model) by employing direct combinatorial arguments. We will restrict to the finite-type setting in which, roughly, there is a finite set of values for $w_i$’s. More precisely, we define this setting as follows:

Condition A.1

(Finite-type setting) The vertex weight sequences $\varvec{w} = (w_i)_{i \in [n]}$ satisfy the following conditions:

• (a) There exists a $K\in \mathbb {N}$ and a set of positive numbers $\mathtt{A}=\{a_1,a_2,\ldots , a_K\}$, with $a_1<a_2<\ldots < a_K$, such that $w_i\in \mathtt{A}$ for all $i\in [n]$;

• (b) Denoting by $\hat{n}_k (n)$ the number of weights $(w_i)_{i \in [n]}$ such that $w_i = a_k$, then the following limits exist

  $$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\hat{n}_k (n)}{n} = p_k\, ,\quad k=1,\ldots ,K, \end{aligned}$$

  (obviously $p=(p_1,\ldots ,p_K)$ is a probability vector). We define also $\hat{p}_k(n):=\frac{\hat{n}_k (n)}{n}$ and $e_k(n):=\hat{p}_k(n)-p_k$.

Hereafter, for the sake of notation we drop n from the notation of $\hat{n}_k (n)$, $\hat{p}_k(n)$, $e_k(n)$.

In this finite-type setting, the previous Condition A.1 is equivalent to Condition 1.1 in which $W_n$ is the uniformly chosen weight with

$$\begin{aligned} \mathbb E[W_n]= \sum _{k=1}^K a_k \hat{p}_k\, ,\quad \mathbb E[W^2_n]= \sum _{k=1}^K a^2_k \hat{p}_k\, , \end{aligned}$$

and W is the limit weight assuming values $a_k$ with probability $p_k$, so that

$$\begin{aligned} \mathbb E[W] =\sum _{k=1}^K a_k p_k\, , \quad \mathbb E[W^2] =\sum _{k=1}^K a^2_k p_k\, . \end{aligned}$$

(A.1)

Assuming Condition A.1, we consider the Hamiltonian (1.22) and defining

$$\begin{aligned} m_{n}=\frac{1}{n} \sum _{i\in [n]} \sigma _{i},\qquad \quad m^{\scriptscriptstyle (w)}_{n}=\frac{1}{n} \sum _{i\in [n]} w_{i}\sigma _{i}, \end{aligned}$$

(A.2)

we rewrite

$$\begin{aligned} H^{{\scriptscriptstyle {\mathrm {ICW}}}}_{n}(\sigma )=\tilde{\beta }\frac{n}{2\mathbb E[W]} ({{m^{\scriptscriptstyle (w)}_{n}}})^2 + nB\ m_{n}\, . \end{aligned}$$

(A.3)

In the theorem below, we write $\lfloor x \rfloor$ for the integer part of $x>0$.

Theorem A.2

(LDPs for the total spin in the finite-type ICW model) In the inhomogeneous Curie–Weiss model defined by (1.22), and assuming the finite-type setting in Condition A.1, the total spin $S_n$ satisfies that for $m\in (-1,1)$, with $\mathcal{A}= \left[ \frac{1}{2}(1+m)\, a_1 , \frac{1}{2}(1+m)\, a_K \right]$,

$$\begin{aligned}&\lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb {P}_{{{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n}(S_{n}= \lfloor m\, n \rfloor )\nonumber \\&\qquad \qquad = - \inf _{x\in \mathcal{A}} \left[ -\frac{\tilde{\beta }}{2} \mathbb E[W] - \frac{2\tilde{\beta }}{\mathbb E[W]} x^2 + 2 \tilde{\beta }x - B\, m + {\mathtt{I}_m}(x) +\psi ^{\scriptscriptstyle {\mathrm {ICW}}} (\tilde{\beta }, B) \right] , \end{aligned}$$

(A.4)

where $\psi ^{\scriptscriptstyle {\mathrm {ICW}}} (\tilde{\beta }, B)$ is the pressure of the model and where

$$\begin{aligned} \mathtt{I}_m(x)&= \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) + \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) \right] , \end{aligned}$$

(A.5)

with $\lambda _1=\lambda _1(x,m)$, $\lambda _2=\lambda _2(x,m)$ defined implicitly by

$$\begin{aligned} \left\{ \begin{array}{l} \mathbb E\left[ \frac{\displaystyle {\mathrm e}^{\lambda _1 W +\lambda _2}}{\displaystyle 1+ {\mathrm e}^{\lambda _1 W +\lambda _2}} \right] = \frac{\displaystyle 1+m}{\displaystyle 2},\\ \mathbb E\left[ W \frac{\displaystyle {\mathrm e}^{\lambda _1 W +\lambda _2}}{ \displaystyle 1+ {\mathrm e}^{\lambda _1 W +\lambda _2}} \right] = {\displaystyle x}. \end{array} \right. \end{aligned}$$

(A.6)

Remark A.3

The expression for the large deviation rate function of the total spin in the Theorem A.2 coincides with the one that is obtained from Theorem 1.6 by application of the contraction principle and the relation between the annealed Ising model and the inhomogeneous Curie–Weiss model. Indeed, recalling that the annealed measure ${\mu }^\mathrm{{an}}_{n}$ at inverse temperature $\beta$ is close to the Boltzmann–Gibbs measure ${{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n$ of the inhomogeneous Curie–Weiss model at inverse temperature $\tilde{\beta }=\sinh (\beta )$ (in the sense of Eq. (2.7)) and by using $\psi ^{\scriptscriptstyle {\mathrm {ICW}}}(\tilde{\beta },B) = - \alpha (\beta )+ \psi ^\mathrm {an}(\beta ,B)$, one finds that the large deviation function of the total spin in the inhomogeneous Curie–Weiss model obtained from (2.4) reads

$$\begin{aligned} I(m) = \inf _{x_2} \left[ I(m,x_2) - \frac{\tilde{\beta }}{2\mathbb E[W]}x_2^{2}- B m - \log (2) + \psi ^{\scriptscriptstyle {\mathrm {ICW}}}(\tilde{\beta },B) \right] . \end{aligned}$$

(A.7)

To see that (A.7) is equal to the r.h.s of (A.4) one employs the substitution $x_2 = 2x - \mathbb {E}(W)$. In doing so clearly the energetic contribution are equal, since

$$\begin{aligned} -\frac{\tilde{\beta }}{2\mathbb E[W]}x_2^{2} = -\frac{\tilde{\beta }}{2} \mathbb E[W] - \frac{2\tilde{\beta }}{\mathbb E[W]} x^2 + 2 \tilde{\beta }x. \end{aligned}$$

It remains to prove that

$$\begin{aligned} I(m,x_2) - \log (2) = {\mathtt{I}_m}(x). \end{aligned}$$

This can be shown by changing the spin variables $\sigma _i$ to the variables $y_i=\frac{1}{2} (\sigma _i +1)$ and introducing

$$\begin{aligned} \hat{S}_{n}=\sum _{i\in [n]} y_{i},\quad \hat{S}^{\scriptscriptstyle (w)}_{n}=\sum _{i\in [n]} w_{i}y_{i}. \end{aligned}$$

Observe that

$$\begin{aligned} {S}_{n}=2\hat{S}_{n}-n,\quad {S}^{\scriptscriptstyle (w)}_{n}=2 \hat{S}^{\scriptscriptstyle (w)}_{n}-n\,\mathbb E[W_{n}], \end{aligned}$$

so that we can write

$$\begin{aligned} \mathbb E_{P_{n}} [\exp ( t_1 S_{n}+ t_2 S^{\scriptscriptstyle (w)}_{n})] = \exp (-n(t_1+ t_2 \mathbb E[W_{n}]))\, \mathbb E_{P_{n}} [\exp (2 t_1 \hat{S}_{n}+2 t_2 \hat{S}^{\scriptscriptstyle (w)}_{n})]. \end{aligned}$$

Since

$$\begin{aligned} \mathbb E_{P_{n}} [\exp ( 2t_1 \hat{S}_{n}+ 2t_2 \hat{S}^{\scriptscriptstyle (w)}_{n})] = \mathbb E_{P_{n}} [ \Pi _{i\in [n]} \exp (2 t_1 + 2 w_i t_2)y_i ] = \frac{1}{2^{n}} \Pi _{i\in [n]} (1+{\mathrm e}^{2(t_1+w_i t_2)}), \end{aligned}$$

we obtain that the moment generating function of $(S_{n},{S}^{\scriptscriptstyle (w)}_{n})$ w.r.t. the product measure (2.2) can be expressed as

$$\begin{aligned} {c}_{n}(\mathbf{t})= -\log 2 - (t_1+ t_2 \mathbb E[W_{n}])+\mathbb E[\log (1+{\mathrm e}^{2(t_1+W_n t_2)}) ]. \end{aligned}$$

Thus, arguing as in the proof of Theorem 1.6, we obtain that the limit of ${c}_{n}(\mathbf{t})$ exists and equals

$$\begin{aligned} {c}(\mathbf{t})= -\log 2 - (t_1+ t_2 \mathbb E[W])+\mathbb E[\log (1+{\mathrm e}^{2(t_1+W t_2)}) ]. \end{aligned}$$

By applying the Gärtner–Ellis theorem we get the expression for the rate function

$$\begin{aligned} {I}(x_1,x_2)= \sup _{(t_1,t_2)} \left( t_1 x_1 + t_2 x_2 +\log 2 + (t_1+ t_2 \mathbb E[W])-\mathbb E[ \log (1+{\mathrm e}^{2(t_1+W t_2)})]\right) . \end{aligned}$$

The stationarity conditions read as

$$\begin{aligned} \left\{ \begin{array}{l} \mathbb E\left[ \frac{\displaystyle {\mathrm e}^{2(t_1 + W t_2)}}{\displaystyle 1+ {\mathrm e}^{2(t_1 +W t_2)}} \right] = \frac{\displaystyle 1+x_1}{\displaystyle 2},\\ \\ \mathbb E\left[ W \frac{\displaystyle {\mathrm e}^{2(t_1+ W t_2)}}{ \displaystyle 1+ {\mathrm e}^{2(t_1 + W t_2)}} \right] = {\frac{\displaystyle \mathbb E[W] +x_2}{\displaystyle 2}}. \end{array} \right. \end{aligned}$$

(A.8)

Since $x_1$ represents the magnetization m and $x_2$ represents the weighted magnetization $m^{\scriptscriptstyle (w)}$, and using again the substitution $x = (x_2 + \mathbb {E}(W))/2$ we obtain that (A.8) is identical to (A.6) provided that $\lambda _1$ is identified with $2 t_2$ and $\lambda _2$ with $2 t_1$.

Proof of Theorem A.2

Given a configuration $\sigma$, we denote by $n_+$ and $n_-$ the number of its positive resp. negative spins. We can group the spins in $\sigma$ according to either $m_n$ or $n_+$, since these quantities are related by $n_+=n(1+m_{n})/2$. We can also identify each configuration of spins $\sigma$ with the set $I_+\subset [n]$ of vertices in which $\sigma _i=1$, obviously the cardinality of this set is $|I_+|=n_+$.

Given any $I_+ \subset [n]$, we define

$$\begin{aligned} q_k=\frac{1}{n}\# \{ i\in I_+ \mid w_i = a_k \}, \qquad {k\in [K]} \end{aligned}$$

(A.9)

to be the frequency of type $a_k$ in $I_+$. Then

$$\begin{aligned} |I_+|=n_+= n\sum _{k=1}^K q_k, \end{aligned}$$

(A.10)

and

$$\begin{aligned} r_n^{\scriptscriptstyle (w)}:=\frac{1}{n} \sum _{i \in I_+} w_i = \sum _{k=1}^K a_k q_k \equiv a \cdot q. \end{aligned}$$

(A.11)

Moreover, given n, we define the set

$$\begin{aligned} \mathcal{Q}_n:=\left\{ \left( \frac{\ell _1}{n},\ldots ,\frac{\ell _K}{n} \right) \mid \ell _k\in \mathbb {N},\, \ell _k \le \hat{n}_k (n),\, k=1,\ldots , K \right\} \end{aligned}$$

of the possible frequency vectors $q=(q_1,q_2,\ldots , q_K)$.

Exponential estimate for the conditional probability of $q=(q_1,q_2,\ldots , q_K)$. We start by counting the number of sets $I_+$ with $\lfloor n(1+m)/2\rfloor$ elements and a given $q=(q_1,q_2,\ldots , q_K) \in \mathcal{Q}_n$ that satisfies the condition $n\sum _{k=1}^K q_k= \lfloor n(\frac{1+m}{2})\rfloor =|I_+|=n_+$. For any $k=1,\ldots , K$, in [n] there are $n\hat{p}_k$ sites corresponding to $a_k$, and we choose $nq_k$ out of them to form $I_+$. On the other hand, there are ${n\atopwithdelims ()n_+}\equiv {n\atopwithdelims ()\lfloor n( {1+m})/{2} \rfloor }$ possible ways to form a set $I_+$ with $n_+$ elements. Thus, the conditional distribution of $q=(q_1,q_2,\ldots , q_K)$ given m is multi-hypergeometric, i.e.,

$$\begin{aligned} \mathbb P_n(q_1,q_2,\ldots , q_K \mid m)=\frac{\displaystyle \prod _{k=1}^K {n\hat{p}_k\atopwithdelims ()nq_k}}{\displaystyle {n\atopwithdelims ()\lfloor n(\frac{1+m}{2}) \rfloor }}\, 1\mathrm l_{\{ q\in \mathcal{D}_n, \sum _{k=1}^K q_k= \frac{1}{n} \left\lfloor n \left( \frac{1+m}{2} \right) \right\rfloor \}}. \end{aligned}$$

(A.12)

The asymptotic behavior of this probability as $n\rightarrow \infty$, can be obtained by using the Stirling’s approximation $n! = {\mathrm e}^{-n}n^n \sqrt{2\pi n}(1+o(1))$ to estimate of the binomial coefficient as

$$\begin{aligned} {nb \atopwithdelims ()na} = {\mathrm e}^{n[ b \log b - a \log a - (b-a) \log (b-a)]}\cdot \frac{\sqrt{b}(1+o(1))}{\sqrt{a}\sqrt{b-a} \sqrt{2\pi n}}, \end{aligned}$$

where $0<a<b$. Then, generalizing the previous formula to a set of variables $a_k<b_k,\, k=1,\ldots , K$, we obtain

$$\begin{aligned} \prod _{k=1}^K {nb_k\atopwithdelims ()na_k} = {\mathscr {C}}_n (a_k,b_k)(1+o(1)), \end{aligned}$$

(A.13)

where

$$\begin{aligned} {\mathscr {C}}_n ({ a},{ b}): = c_1\, (2\pi n)^{- K/2} \exp \left( n\sum _{k=1}^K [ b_k \log b_k - a_k \log a_k -(b_k - a_k)\log (b_k - a_k)] \right) , \end{aligned}$$

(A.14)

with $c_1= \prod _{k=1}^K \sqrt{\frac{b_k}{a_k(b_k-a_k)}}$ and the function is defined on the set

$$\begin{aligned} \{ (a_1,\ldots , a_K,b_1,\ldots , b_K) \in \mathbb R^{2K} |\, 0<a_k<b_k,\, k=1, \ldots , K\}. \end{aligned}$$

We now compute the asymptotics of the numerator in (A.12). Recalling that $\hat{p}_k=p_k+ e_k$ and Taylor expanding the sum in (A.14) and $c_1$ as a function of $b_k$’s, we obtain

$$\begin{aligned} \prod _{k=1}^K {n(p_k+ e_k)\atopwithdelims ()nq_k} =&\, {\mathscr {C}}_n ({ q},{ p})\, [1+\sum _{k=1}^K c^{\scriptscriptstyle (1)}_k e_k + \sum _{k=1}^K O( e^2_k)]\nonumber \\&\times \exp \left( n \sum _{k=1}^K c^{\scriptscriptstyle (2)}_k e_k(1+O(e_k))\right) (1+o(1)), \end{aligned}$$

(A.15)

for some constants $c^{\scriptscriptstyle (1)}_k$ and $c^{\scriptscriptstyle (2)}_k$. The second factor in the r.h.s. comes from the substitution $p_k \rightarrow p_k+e_k$ in the factor $c_1$ of (A.14), and the third form the sum in the same equation. From Condition A.1 we have that these terms are both $(1+o(1))$. Then, we conclude that the numerator in (A.12) is

$$\begin{aligned} \prod _{k=1}^K {n\hat{p}_k\atopwithdelims ()nq_k} = {\mathscr {C}}_n ({ q},{ p}) (1+o(1)). \end{aligned}$$

(A.16)

We can deal with the denominator in (A.12) in a similar fashion, obtaining:

$$\begin{aligned}&{n\atopwithdelims ()\lfloor n\big (\frac{1+m}{2}\big ) \rfloor } =c_2\, s_n (2\pi n)^{-\frac{1}{2}} \nonumber \\&\quad \exp \left( n\left[ - \frac{1+m}{2} \log \Big ( \frac{1+m}{2}\Big ) - \frac{1-m}{2} \log \Big ( \frac{1-m}{2}\Big ) \right] \right) (1+o(1)), \end{aligned}$$

(A.17)

with $c_2= \frac{2}{\sqrt{1-m^2}}$ and $s_n=s_n(m)=\exp [ ( n\frac{1+ m}{2} - \lfloor n\frac{1+ m}{2} \rfloor ) \log (\frac{1-m^2}{4})]$. By plugging this estimate and (A.13) in (A.12), we finally obtain that

$$\begin{aligned} \mathbb P_n(q_1,q_2,\ldots , q_K \mid m) = \frac{c_1}{c_2}\, s_n\, (2\pi n)^{\frac{1-K}{2}} {\mathrm e}^{ng(q,m)} (1+o(1)), \end{aligned}$$

(A.18)

with

$$\begin{aligned} g(q,m)= \left\{ \begin{array}{ll} h(q,m), &{} \text{ if }\; \; q_k \le p_k,\; \sum _k q_k = \frac{1}{n} \lfloor n \big (\frac{1+m}{2}\big ) \rfloor ,\\ -\infty , &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(A.19)

and

$$\begin{aligned} h(q,m)= & {} \frac{1+m}{2} \log \Big ( \frac{1+m}{2}\Big ) + \frac{1-m}{2} \log \Big ( \frac{1-m}{2}\Big )\nonumber \\&+\sum _{k=1}^K [ p_k \log p_k - q_k \log q_k -(p_k - q_k)\log (p_k - q_k)] . \end{aligned}$$

(A.20)

Exponential estimate for the conditional probability of $r_n^{\scriptscriptstyle (w)}$. Let use introduce

$$\begin{aligned} \mathcal{A}_n(m)= \left\{ q_1 a_1+\cdots +q_K a_K\, |\, q \in \mathcal{Q}_n,\, \sum _k q_k = \frac{1}{n} \left\lfloor n \left( \frac{1+m}{2} \right) \right\rfloor \right\} , \end{aligned}$$

which are the sets of values of $r_n^{\scriptscriptstyle (w)}=\frac{1}{n} \sum _{i \in I_+} w_i$ corresponding to subsets $I_+ \subset [n]$ with $\lfloor n(1+m)/2\rfloor$ elements. We have that

$$\begin{aligned} \frac{1}{2}(1+m)\, a_1 - \frac{\rho _n(m)}{n} \le \inf \mathcal{A}_n(m), \qquad \sup \mathcal{A}_n(m) \le \frac{1}{2}(1+m)\, a_K - \frac{\rho _n(m)}{n}, \end{aligned}$$

(A.21)

with $\rho _n(m)=n (1+m)/2- \lfloor n (1+m)/2\rfloor$. Obviously $\frac{\rho _n(m)}{n}=O(n^{-1})$, since $0\le \rho _n(m)<1$. Moreover, by (A.21) and the fact that $\inf \mathcal{A}_n(m)\ge a_1 \sum _k q_k$ and $\sup \mathcal{A}_n(m)\le a_K \sum _k q_k$,

$$\begin{aligned} \inf \mathcal{A}_n(m) \rightarrow \frac{1}{2}(1+m)\, a_1, \quad \quad \sup \mathcal{A}_n(m) \rightarrow \frac{1}{2}(1+m)\, a_K \end{aligned}$$

(A.22)

as $n \rightarrow \infty$. The previous remark implies that $r^{\scriptscriptstyle (w)}_n$ is close to some $x\in \big [ \frac{1}{2}(1+m)\, a_1 , \frac{1}{2}(1+m)\, a_K \big ]$ for large n. Therefore, we claim that

$$\begin{aligned}&\lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb P_{\pi _n} \left( \sum _{i \in I_+} w_i = n\, x ~\bigg |~ |I_+ | = \left\lfloor n(1+m)/2\right\rfloor \right) \nonumber \\&\qquad \qquad \qquad = \left\{ \begin{array}{ll} \mathtt{S}_m(x), &{} x\in \left[ \frac{1}{2}(1+m)\, a_1 , \frac{1}{2}(1+m)\, a_K \right] ,\\ -\infty , &{} \text{ otherwise }, \end{array} \right. \end{aligned}$$

(A.23)

where $\mathtt{S}_m(x)$ has to be computed. To this end, we observe now that the probability in (A.23) can be written as

$$\begin{aligned} \mathbb P_{\pi _n} \left( \sum _{i \in I_+} w_i = n\, x ~\bigg |~ |I_+ | = \left\lfloor n{(1+m)}{/2} \right\rfloor \right) = \sum \limits _{\begin{array}{*{20}{c}} {q \in {Q_n}{} }\\ {{} a\cdot q = x} \end{array}} \mathbb P_n (q_1,q_2,\ldots , q_K \mid m) , \end{aligned}$$

where the sum is extended to those k-tuples $q \in \mathcal{Q}_n$ for which the event $\sum _{i \in I_+} w_i =n\, x$ is realized. In the previous sum the term that corresponds to the larger value of the exponent g(q, m) in (A.18) controls the behavior in the limit, the remaining terms being sub-leading. The quantity depending on m in the definition of h(q, m), see (A.20), is negative and the sum on k is positive, while h(q, m) is negative in the range defined in the first line of (A.19). Thus, defining

$$\begin{aligned} \tilde{h}(q_1,q_2,\ldots , q_K)= \sum _{k=1}^K [ p_k \log p_k - q_k \log q_k -(p_k - q_k)\log (p_k - q_k)] , \end{aligned}$$

we have to find

$$\begin{aligned} \mathscr {S}_n=\sup _{\begin{array}{*{20}{c}} {a\cdot q = x,}\\ {\sum \nolimits _k {{q_k}} = \frac{1}{n} \lfloor n(m + 1)/2 \rfloor } \end{array}} \tilde{h}(q_1,q_2,\ldots , q_K). \end{aligned}$$

In the previous equation, the notation $\mathscr {S}_n$ emphasizes the fact that due to the constraints, the sup depends on n. As a consequence, the optimization point $q^*=(q_1^*,\ldots , q_k^*)$ will depend on n. In order to find $q^*$ we introduce the multipliers $\lambda _1$ and $\lambda _2$ conjugate to x and m, and write the Lagrangian function as

$$\begin{aligned} L(q_1,q_2,\ldots , q_K; \lambda _1,\lambda _2)= \tilde{h}(q_1,q_2,\ldots , q_K) + \lambda _1 \left( \sum _{k=1}^K a_k q_k -x\right) + \lambda _2 \left( \sum _k q_k-\widetilde{m}_n\right) , \end{aligned}$$

where we set

$$\begin{aligned} \widetilde{m}_n:= \frac{1}{n} \lfloor n (m+1)/2 \rfloor = \frac{1+m}{2} -\frac{\rho _n(m)}{n}= \frac{1+m}{2} +O(n^{-1}). \end{aligned}$$

By imposing that $\partial L/\partial q_k=0,\, k=1,\ldots , K$, we obtain that the stationarity point $q^*(n)=(q_1^*(n),\ldots , q_k^*(n))$ of the function $\tilde{h}$ satisfies

$$\begin{aligned} q^*_k(n)=\frac{p_k {\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}, \qquad k\in [K], \end{aligned}$$

with $\lambda _1(n)=\lambda _1(x,\widetilde{m}_n)$, $\lambda _2(n)=\lambda _2(x,\widetilde{m}_n)$. By introducing the notation

$$\begin{aligned} u_k(n)= \frac{q^*_k(n)}{p_k}=\frac{{\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}\, , \end{aligned}$$

we write

$$\begin{aligned} \mathscr {S}_n= & {} \tilde{h}(q^*_1(n),q^*_2(n),\ldots , q^*_K(n))\nonumber \\= & {} -\sum _{k=1}^K p_k [u_k(n) \log u_k(n) + (1-u_k(n)) \log (1-u_k(n)) ]\nonumber \\= & {} \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)}}{1+{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)} } \log \left( \frac{{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)}}{1+{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)} } \right) + \frac{1}{1+{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)} }\right. \nonumber \\&\left. \times \,\log \left( \frac{1}{1+{\mathrm e}^{\lambda _1(n) W + \lambda _2(n)} } \right) \right] . \end{aligned}$$

(A.24)

The relation between the multipliers $\lambda _1$, $\lambda _2$ and the parameters x, $\widetilde{m}_n$ can be made explicit by recalling that, since the probability vector $(p_1,\ldots , p_K)$ is the distribution of W, and by (A.11),

$$\begin{aligned} x= \sum _{k=1}^K a_k p_k \frac{{\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}} = \mathbb E\left[ W \frac{{\mathrm e}^{\lambda _1(n) W +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) W +\lambda _2(n)}} \right] , \end{aligned}$$

(A.25)

and, from (A.10),

$$\begin{aligned} \widetilde{m}_n=\frac{1+m}{2}+O(n^{-1})= \sum _{k=1}^K p_k \frac{{\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) a_k +\lambda _2(n)}} = \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1 (n)W +\lambda _2(n)}}{ 1+ {\mathrm e}^{\lambda _1(n) W +\lambda _2(n)}} \right] , \end{aligned}$$

(A.26)

where (A.11) and (A.10) have been used. By taking the limit of (A.25) and (A.26) as $n\rightarrow \infty$, we see that $\lambda _1(n)$ and $\lambda _2(n)$ converge to $\lambda _1$ and $\lambda _2$ that solve

$$\begin{aligned} x = \sum _{k=1}^K a_k p_k \frac{{\mathrm e}^{\lambda _1 a_k +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1 a_k +\lambda _2}} = \mathbb E\left[ W \frac{{\mathrm e}^{\lambda _1 W +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1 W +\lambda _2}} \right] , \end{aligned}$$

and

$$\begin{aligned} \frac{1+m}{2}= \sum _{k=1}^K p_k \frac{{\mathrm e}^{\lambda _1 a_k +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1 a_k +\lambda _2}} = \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1 W +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1 W +\lambda _2}} \right] , \end{aligned}$$

that is (A.6). From this fact it follows that in the same limit $n \rightarrow \infty$,

$$\begin{aligned} q^*_k(n) \rightarrow q^*_k =\frac{p_k {\mathrm e}^{\lambda _1 a_k +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1a_k +\lambda _2}}\, \quad \text{ and }\quad u_k(n)\rightarrow u_k=\frac{{\mathrm e}^{\lambda _1 a_k +\lambda _2}}{ 1+ {\mathrm e}^{\lambda _1 a_k +\lambda _2}}\,\nonumber \\ \end{aligned}$$

and, thus,

$$\begin{aligned} \mathscr {S}_n\rightarrow \mathscr {S}:= \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) + \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) \right] \nonumber \\ \end{aligned}$$

(A.27)

Then, from (A.18), (A.20), and the previous display, we obtain the limit in (A.23) with

$$\begin{aligned} \mathtt{S}_m(x)&= \frac{1+m}{2} \log \left( \frac{1+m}{2}\right) + \frac{1-m}{2} \log \left( \frac{1-m}{2}\right) \nonumber \\&\qquad + \mathbb E\left[ \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{{\mathrm e}^{\lambda _1 W + \lambda _2}}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) + \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \log \left( \frac{1}{1+{\mathrm e}^{\lambda _1 W + \lambda _2} } \right) \right] . \end{aligned}$$

(A.28)

Moment generating function of the Hamiltonian $H^{{\scriptscriptstyle {\mathrm {ICW}}}}_{n}(\sigma )$. Our next step to compute the cumulant generating function of the Hamiltonian (A.3) that we rewrite as a function of

$$\begin{aligned} {r^{\scriptscriptstyle (w)}_{n}}= \frac{1}{n} \sum _{i\in I_+} w_{i}, \end{aligned}$$

for which we have proven (A.23). In this way we obtain

$$\begin{aligned} H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}( {r^{\scriptscriptstyle (w)}_{n}},m_n)= n\left[ \frac{ 2 \tilde{\beta }}{\mathbb E[W]} ({r^{\scriptscriptstyle (w)}_{n}})^2-2\tilde{\beta }\frac{\mathbb E{[W_{n}]}}{\mathbb E[W]} {r^{\scriptscriptstyle (w)}_{n}}+ \frac{\tilde{\beta }}{2} \frac{\mathbb E{[W_{n}]}^2}{\mathbb E[W]} + B m_n\right] . \end{aligned}$$

Now, writing $\mathbb E[W_n] = \mathbb E[W] + \epsilon _n$ and defining

$$\begin{aligned} h^{\scriptscriptstyle {\mathrm {ICW}}}_{n}( {r^{\scriptscriptstyle (w)}_{n}},m_n):= \frac{2 \tilde{\beta }}{\mathbb E[W]} ({r^{\scriptscriptstyle (w)}_{n}})^2-2\tilde{\beta }{r^{\scriptscriptstyle (w)}_{n}}+ \frac{\tilde{\beta }}{2} \mathbb E{[W]} + B m_n , \end{aligned}$$

we have

$$\begin{aligned} H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}( {r^{\scriptscriptstyle (w)}_{n}},m_n)= n\, h^{\scriptscriptstyle {\mathrm {ICW}}}_{n}( {r^{\scriptscriptstyle (w)}_{n}},m_n) + n\, \epsilon _n \left[ - \frac{2 \tilde{\beta }}{\mathbb E[W]} {r^{\scriptscriptstyle (w)}_{n}}+\tilde{\beta }+ \frac{\tilde{\beta }}{2 \mathbb E[W]} \varepsilon _n\right] . \end{aligned}$$

Since Condition A.1 implies that $\varepsilon _n=o(1)$ the last addend in the previous display is o(n). Now we can finally write the cumulant generating function and apply Varadhan’s lemma to compute

$$\begin{aligned}&\lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb E_{\pi _n}\left[ {\mathrm e}^{H^{{\scriptscriptstyle {\mathrm {ICW}}}}_{n}({r^{\scriptscriptstyle (w)}_{n}}, m_n)} \mid m_{n}=m\right] \nonumber \\&\qquad \qquad \qquad =\lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb E_{\pi _n}\left[ {\mathrm e}^{ n\, [h^{\scriptscriptstyle {\mathrm {ICW}}}_{n}( {r^{\scriptscriptstyle (w)}_{n}},m_n) +o(1)]} \mid m_{n}=m\ \right] \nonumber \\&\qquad \qquad \qquad = \frac{\tilde{\beta }}{2} \mathbb E[W]+ B\, m + \sup _{x\in \mathcal{A}} \left[ \frac{2 \tilde{\beta }}{\mathbb E[W]} x^2 -2 \tilde{\beta }x - \mathtt{S}_m(x) \right] , \end{aligned}$$

(A.29)

where the large deviation property (A.23) has been used and $\mathcal{A}= \big [ \frac{1}{2}(1+m)\, a_1 , \frac{1}{2}(1+m)\, a_K \big ]$. We can now move to the final step.

Asymptotic behavior of $\mathbb {P}_{{{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n}(m_{n}=m)$. Let us observe that, since the conditional average on the left hand side to the previous display is computed with respect to the uniform measure $\pi _n(\sigma )=2^{-n}$ on the spins $\sigma$,

$$\begin{aligned} \mathbb E_{\pi _n} \left[ {\mathrm e}^{H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}(\sigma )} | m_{n}=m \right]= & {} \frac{\sum _\sigma 1\mathrm l_{\{m_{n}(\sigma )=m\}} {\mathrm e}^{H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}(\sigma )} \pi _n(\sigma )}{\mathbb {P}_{\pi _n}(m_{n}=m)}\\= & {} \frac{ 2^{-n}\,Z^{\scriptscriptstyle {\mathrm {ICW}}}_{n}}{\mathbb {P}_{\pi _n}(m_{n}=m) }\,\mathbb {P}_{{{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n}(m_{n}=m), \end{aligned}$$

with $Z^{\scriptscriptstyle {\mathrm {ICW}}}_{n}=\sum _\sigma {\mathrm e}^{H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}(\sigma )}$, the partition function of the ICW model. Thus,

$$\begin{aligned} \frac{1}{n} \log \mathbb {P}_{{{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n}(m_{n}=m)= & {} \frac{1}{n} \log \mathbb E_{\pi _n} \left[ {\mathrm e}^{H^{\scriptscriptstyle {\mathrm {ICW}}}_{n}(\sigma )} | m_{n}=m \right] + \frac{1}{n} \log {\mathbb {P}_{\pi _n}(m_{n}=m) }\\&\quad -\,\frac{1}{n} \log Z^{\scriptscriptstyle {\mathrm {ICW}}}_{n}\, + \log 2. \end{aligned}$$

Since ${\mathbb {P}_{\pi _n}(m_{n}=m) } =2^{-n} {n\atopwithdelims ()n(\frac{1+m}{2}) }$, by (A.17),

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \log {\mathbb {P}_{\pi _n}(m_{n}=m) } = -\log 2 - \frac{1+m}{2} \log \Big ( \frac{1+m}{2}\Big ) - \frac{1-m}{2} \log \Big ( \frac{1-m}{2}\Big ), \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \log Z^{\scriptscriptstyle {\mathrm {ICW}}}_{n}= \psi ^{\scriptscriptstyle {\mathrm {ICW}}} (\tilde{\beta }, B), \end{aligned}$$

is the pressure of the Inhomogeneous Curie–Weiss model [13]. Thus, by (A.29),

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \log \mathbb {P}_{{{\mu }}^{\scriptscriptstyle {\mathrm {ICW}}}_n}(m_{n}=m)&\,= \frac{\tilde{\beta }}{2} \mathbb E[W]+ B\, m + \sup _{x\in \mathcal{A}} \left[ \frac{2 \tilde{\beta }}{\mathbb E[W]} x^2 -2 \tilde{\beta }x - \mathtt{S}_m(x) \right] \\ {}&\quad -\psi ^{\scriptscriptstyle {\mathrm {ICW}}} (\tilde{\beta }, B) \nonumber \\&\quad - \frac{1+m}{2} \log \Big ( \frac{1+m}{2}\Big ) - \frac{1-m}{2} \log \Big ( \frac{1-m}{2}\Big ), \end{aligned}$$

from which, recalling (A.28), we obtain (A.4) and (A.5). $\square$