The Multi-species Mean-Field Spin-Glass on the Nishimori Line

Abstract

In this paper we study a multi-species disordered model on the Nishimori line. The typical properties of this line, a set of identities and inequalities among correlation functions, allow us to prove the replica symmetry i.e. the concentration of the order parameter. When the interaction structure is elliptic we rigorously compute the exact solution of the model in terms of a finite-dimensional variational principle and we study its properties.

Appendices

A Appendix: Proof of the Concentration Lemma

Proof of Lemma 3

Let us split the proof into three steps for the sake of clarity. As anticipated, it is convenient to split the total fluctuation of ${\mathcal {L}}_r$ into two parts, thus proving that:

$$\begin{aligned}&{\mathbb {E}}_{\varvec{\epsilon }}{\mathbb {E}} \Big \langle \left( {\mathcal {L}}_r-\langle {\mathcal {L}}_r\rangle _{N,t}^{(\epsilon )}\right) ^2\Big \rangle _{N,t}^{(\epsilon )}\longrightarrow 0\quad \text {as }N\rightarrow \infty \end{aligned}$$

(A.1)

$$\begin{aligned}&{\mathbb {E}}_{\varvec{\epsilon }}{\mathbb {E}} \left( \langle {\mathcal {L}}_r\rangle _{N,t}^{(\epsilon )}-{\mathbb {E}}\langle {\mathcal {L}}_r\rangle _{N,t}^{(\epsilon )}\right) ^2\longrightarrow 0\quad \text {as }N\rightarrow \infty \;. \end{aligned}$$

(A.2)

From this moment on, we neglect sub and superscripts in the Gibbs brackets as well as t-dependencies. We start by proving the inequality (45).

Proof of inequality (45): To begin with, we compute:

$$\begin{aligned}&{\mathbb {E}}[\langle {\mathcal {L}}_r\rangle ]=\frac{1}{N_r}\sum _{i\in \varLambda _r}{\mathbb {E}}\Big \langle \sigma _i+\frac{J_i^r\sigma _i}{2\sqrt{Q_{\epsilon , r}}} \Big \rangle ={\mathbb {E}}\langle m_r\rangle +\frac{1}{2}{\mathbb {E}}[1-\langle m_r\rangle ]=\frac{1}{2}{\mathbb {E}}[1+\langle m_r\rangle ] \end{aligned}$$

(A.3)

where integration by parts has been used.

Then, we proceed with:

$$\begin{aligned} {\mathbb {E}}\langle {\mathcal {L}}_r^2\rangle&=\frac{1}{N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}}\Big \langle \sigma _i\sigma _j+\frac{J_i^r\sigma _i\sigma _j}{\sqrt{Q_{\epsilon ,r}}}+\frac{J_i^rJ_j^r\sigma _i\sigma _j}{4Q_{\epsilon ,r}} \Big \rangle =\underbrace{{\mathbb {E}}\langle m_r^2\rangle }_{R_1}+\underbrace{ \frac{1}{N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}}\Big \langle \frac{J_i^r\sigma _i\sigma _j}{\sqrt{Q_{\epsilon ,r}}}\Big \rangle }_{R_2}+\nonumber \\&\quad + \underbrace{ \frac{1}{N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}}\Big \langle \frac{J_i^rJ_j^r\sigma _i\sigma _j}{4Q_{\epsilon ,r}}\Big \rangle }_{R_3}\,. \end{aligned}$$

(A.4)

We treat the three terms $R_1$, $R_2$ and $R_3$ separately with repeated integrations by parts.

$$\begin{aligned} R_2=\frac{1}{N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}} \left[ \langle \sigma _j\rangle -\langle \sigma _i\sigma _j\rangle \langle \sigma _i\rangle \right] ={\mathbb {E}}\langle m_r\rangle -{\mathbb {E}}\langle m_r\rangle ^2 \end{aligned}$$

(A.5)

where we have used the Nishimori identity (16).

$$\begin{aligned} R_3&=\frac{1}{4N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}}\left[ \frac{\delta _{ij}\overbrace{\langle \sigma _i\sigma _j\rangle }^{=1}}{Q_{\epsilon ,r}} +J_j^r\frac{ (\langle \sigma _j\rangle -\langle \sigma _i\rangle \langle \sigma _i\sigma _j\rangle )}{\sqrt{Q_{\epsilon ,r}}}\right] = \frac{1}{4N_rQ_{\epsilon ,r}}\nonumber \\&\quad +\frac{1}{4N_r^2}\sum _{i,j\in \varLambda _r}{\mathbb {E}}\left[ 1-\langle \sigma _j\rangle ^2-\langle \sigma _i\rangle (\langle \sigma _i\rangle -\langle \sigma _j\rangle \langle \sigma _i\sigma _j\rangle )-\langle \sigma _i\sigma _j\rangle (\langle \sigma _i\sigma _j\rangle -\langle \sigma _i\rangle \langle \sigma _j\rangle ) \right] \nonumber \\&= \frac{1}{4N_rQ_{\epsilon ,r}}+\frac{1}{4}-\frac{1}{2}{\mathbb {E}}\langle m_r\rangle +\frac{1}{2}{\mathbb {E}}\langle m_r\rangle ^2-\frac{1}{4}{\mathbb {E}}\langle m_r^2\rangle \,. \end{aligned}$$

(A.6)

Hence:

$$\begin{aligned} R_1+R_2+R_3=\frac{1}{4}+\frac{1}{4N_rQ_{\epsilon ,r}}+\frac{3}{4}{\mathbb {E}}\langle m_r^2\rangle +\frac{1}{2}{\mathbb {E}}\langle m_r\rangle -\frac{1}{2}{\mathbb {E}}\langle m_r\rangle ^2\,. \end{aligned}$$

(A.7)

Summing up all the contributions:

$$\begin{aligned}&{\mathbb {E}}\langle {\mathcal {L}}_r^2\rangle -({\mathbb {E}}\langle {\mathcal {L}}_r\rangle )^2=\frac{1}{4N_rQ_{\epsilon ,r}}+\frac{3}{4}{\mathbb {E}}\langle m_r^2\rangle -\frac{1}{2}{\mathbb {E}}\langle m_r\rangle ^2-\frac{1}{4}({\mathbb {E}}\langle m_r\rangle )^2\nonumber \\&\quad =\frac{1}{4N_rQ_{\epsilon ,r}}+\frac{1}{4}({\mathbb {E}}\langle m_r^2\rangle -({\mathbb {E}}\langle m_r\rangle )^2)+\frac{1}{2}({\mathbb {E}}\langle m_r^2\rangle -{\mathbb {E}}\langle m_r\rangle ^2)\ge \frac{1}{4}{\mathbb {E}}\Big \langle (m_r-{\mathbb {E}}\langle m_r\rangle )^2\Big \rangle \,.\nonumber \\ \end{aligned}$$

(A.8)

Proof of (A.1): Notice that:

(A.9)

$$\begin{aligned}&\frac{\partial ^2 {\bar{p}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}^2}=\alpha _r N_r{\mathbb {E}}\langle ({\mathcal {L}}_r-\langle {\mathcal {L}}_r\rangle )^2\rangle -\frac{1}{4NQ_{\epsilon ,r}^{3/2}}\sum _{i\in \varLambda _r}{\mathbb {E}}\langle J_i^r\sigma _i\rangle \,. \end{aligned}$$

(A.10)

From the last one, after an integration by parts and using the regularity of the map $\epsilon \longmapsto {\mathbf {Q}}_\epsilon (\cdot )$ and Lemma 2 we get:

$$\begin{aligned}&{\mathbb {E}}_{\varvec{\epsilon }}{\mathbb {E}}\langle ({\mathcal {L}}_r-\langle {\mathcal {L}}_r\rangle )^2\rangle \le \frac{1}{N_r\alpha _rs_N^K}\prod _{s=1}^K\int _{Q_{s_N,s}}^{Q_{2s_N,s}}dQ_{\epsilon ,s}\frac{\partial ^2 {\bar{p}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}^2}+{\mathbb {E}}_{\varvec{\epsilon }}\frac{1}{4N_r\epsilon _r}{\mathbb {E}}[1-\langle m_r\rangle ]\nonumber \\&\quad \le \frac{1}{N_r\alpha _rs_N^K}\prod _{s\ne r,1}^K\int _{Q_{s_N,s}}^{Q_{2s_N,s}}dQ_{\epsilon ,s}\left[ \left. \frac{\partial {\bar{p}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}}\right| _{Q_{2s_N,r}}-\left. \frac{\partial {\bar{p}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}}\right| _{Q_{s_N,r}}\right] +\frac{\log 2}{4N_rs_N}\nonumber \\&\quad \le \frac{2K_r(\varDelta )}{N_rs_N^K}+\frac{\log 2}{4N_rs_N}={\mathcal {O}}\left( \frac{1}{N_rs_N^K} \right) \longrightarrow 0 \end{aligned}$$

(A.11)

where:

$$\begin{aligned} \prod _{s\ne r,1}^K(Q_{2s_N,s}-Q_{s_N,s})\le K_r(\varDelta )\,. \end{aligned}$$

(A.12)

Proof of (A.2): Let $p_{N,\epsilon }$ be the random interpolating pressure, such that ${\mathbb {E}}p_{N,\epsilon }={\bar{p}}_{N,\epsilon }$. Define:

$$\begin{aligned}&\hat{p}_{N,\epsilon }=p_{N,\epsilon }-\alpha _r\sqrt{Q_{\epsilon ,r}}\sum _{i\in \varLambda _r}\frac{|J_i^r|}{N_r},\;\quad \hat{{\bar{p}}}_{N,\epsilon }={\mathbb {E}}\hat{p}_{N,\epsilon } \end{aligned}$$

(A.13)

$$\begin{aligned}&\frac{\partial ^2 \hat{p}_{N,\epsilon }}{\partial Q_{\epsilon ,r}^2}=\alpha _r N_r\langle ({\mathcal {L}}_r-\langle {\mathcal {L}}_r\rangle )^2\rangle +\frac{\alpha _r}{4Q_{\epsilon ,r}^{3/2}}\sum _{i\in \varLambda _r}\frac{|J_i^r|- J_i^r\langle \sigma _i\rangle }{N_r} \ge 0\,. \end{aligned}$$

(A.14)

Let us evaluate:

$$\begin{aligned} \left| \frac{\partial \hat{p}_{N,\epsilon }}{\partial Q_{\epsilon ,r}}-\frac{\partial \hat{{\bar{p}}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}} \right| \ge \alpha _r\left| \langle {\mathcal {L}}_r\rangle -{\mathbb {E}}\langle {\mathcal {L}}_r\rangle \right| -\frac{\alpha _r|A_r|}{2\sqrt{Q_{\epsilon ,r}}} \end{aligned}$$

(A.15)

where:

$$\begin{aligned} A_r:=\frac{1}{N_r}\sum _{i\in \varLambda _r}[|J_i^r|-{\mathbb {E}}|J_i^r|]. \end{aligned}$$

(A.16)

Thanks to the independence of the $J_i^r$ it is immediate to verify that $\exists \, a\ge 0$ s.t.:

$$\begin{aligned} {\mathbb {E}}[A_r^2]\le \frac{a}{N_r}\,. \end{aligned}$$

(A.17)

Using Lemma 3.2 in [25], with notations used in [8]:

$$\begin{aligned}&\left| \frac{\partial \hat{p}_{N,\epsilon }}{\partial Q_{\epsilon ,r}}-\frac{\partial \hat{{\bar{p}}}_{N,\epsilon }}{\partial Q_{\epsilon ,r}} \right| \le \frac{1}{\delta }\sum _{u\in \{Q_{\epsilon ,r}+\delta ,Q_{\epsilon ,r},Q_{\epsilon ,r}-\delta \}}[|\hat{p}_{N,\epsilon }-\hat{{\bar{p}}}_{N,\epsilon }|+\alpha _r\sqrt{u}|A_r|]\nonumber \\&\quad +\;C_\delta ^+(Q_{\epsilon ,r})+C_\delta ^-(Q_{\epsilon ,r}) \end{aligned}$$

(A.18)

with:

$$\begin{aligned}&C_\delta ^\pm (Q_{\epsilon ,r})=|\hat{{\bar{p}}}'_{N,\epsilon }(Q_{\epsilon ,r}\pm \delta )-\hat{{\bar{p}}}'_{N,\epsilon }(Q_{\epsilon ,r})| \end{aligned}$$

(A.19)

$$\begin{aligned}&|\hat{{\bar{p}}}'_{N,\epsilon }|=\left| \frac{\alpha _r}{2}{\mathbb {E}}[1+\langle m_r\rangle ]-\frac{\alpha _r{\mathbb {E}}|J_1^r|}{2\sqrt{Q_{\epsilon ,r}}}\right| \le \alpha _r\left( 1+\frac{C}{2\sqrt{s_N}}\right) \end{aligned}$$

(A.20)

$$\begin{aligned}&C_\delta ^\pm (Q_{\epsilon ,r})\le \alpha _r\left( 2+\frac{C}{\sqrt{s_N}}\right) \end{aligned}$$

(A.21)

where for simplicity we have kept the dependence on $Q_{\epsilon ,r}$ only, $\hat{{\bar{p}}}'_{N,\epsilon }$ is the derivative w.r.t. it and $\delta >0$. Notice that $\delta$ will be chosen strictly smaller than $s_N$, so that $Q_{\epsilon ,r}-\delta \ge \epsilon -\delta \ge s_N-\delta >0$.

Then, using (A.15), (A.16) and (A.18), and thanks to the fact that $(\sum _{i=1}^p\nu _i)^2\le p\sum _{i=1}^p\nu _i^2$, we get:

$$\begin{aligned} \frac{\alpha ^2_r}{9}\left| \langle {\mathcal {L}}_r\rangle -{\mathbb {E}}\langle {\mathcal {L}}_r\rangle \right| ^2&\le \frac{1}{\delta ^2}\sum _{u\in \{Q_{\epsilon ,r}+\delta ,Q_{\epsilon ,r},Q_{\epsilon ,r}-\delta \}}[|\hat{p}_{N,\epsilon }-\hat{{\bar{p}}}_{N,\epsilon }|^2+\alpha _r^2u|A_r|^2]\nonumber \\&\quad +C_\delta ^+(Q_{\epsilon ,r})^2+C_\delta ^-(Q_{\epsilon ,r})^2+\frac{\alpha _r^2A_r^2}{4\epsilon _r}\,. \end{aligned}$$

(A.22)

We first evaluate the two terms containing $C_\delta ^\pm$:

$$\begin{aligned}&{\mathbb {E}}_{\varvec{\epsilon }}[C_\delta ^+(Q_{\epsilon ,r})^2+C_\delta ^-(Q_{\epsilon ,r})^2]\le 2\alpha _r\left( 2+\frac{C}{\sqrt{s_N}} \right) {\mathbb {E}}_{\varvec{\epsilon }}[C_\delta ^+(Q_{\epsilon ,r})+C_\delta ^-(Q_{\epsilon ,r})] \nonumber \\&\quad \le \frac{2\alpha _r}{s_N^K}\left( 2+\frac{C}{\sqrt{s_N}} \right) \prod _{s=1}^K\int _{Q_{s_N,s}}^{Q_{2s_N,s}}dQ_{\epsilon ,s}[\hat{{\bar{p}}}'_{N,\epsilon }(Q_{\epsilon ,r}+\delta )-\hat{{\bar{p}}}'_{N,\epsilon }(Q_{\epsilon ,r}-\delta )]\nonumber \\&\quad = \frac{2\alpha _r}{s_N^K}\left( 2+\frac{C}{\sqrt{s_N}} \right) \prod _{s\ne r,1}^K \int _{Q_{s_N,s}}^{Q_{2s_N,s}}dQ_{\epsilon ,s}[\hat{{\bar{p}}}_{N,\epsilon }(Q_{2s_N,r}+\delta )-\hat{{\bar{p}}}_{N,\epsilon }(Q_{2s_N,r}-\delta )+\nonumber \\&\quad -\hat{{\bar{p}}}_{N,\epsilon }(Q_{s_N,r}+\delta )+\hat{{\bar{p}}}_{N,\epsilon }(Q_{s_N,r}-\delta )]\le \frac{8\alpha _r^2 K_r(\varDelta )}{s_N^K}\delta \left( 2+\frac{C}{\sqrt{s_N}} \right) ^2\,. \end{aligned}$$

(A.23)

Taking the expectation ${\mathbb {E}}_{\varvec{\epsilon }}{\mathbb {E}}$ in (A.22), and defining $W_r$ s.t. $Q_{\epsilon ,r}\le W_r$, we get:

$$\begin{aligned}&\frac{\alpha ^2_r}{9}{\mathbb {E}}_{\varvec{\epsilon }}{\mathbb {E}}\left| \langle {\mathcal {L}}_r\rangle -{\mathbb {E}}\langle {\mathcal {L}}_r\rangle \right| ^2\le \frac{3}{\delta ^2}\left[ \frac{S}{N}+\frac{\alpha _r^2W_ra}{N_r}\right] \nonumber \\&\quad +\frac{8\alpha _r^2 K_r(\varDelta )}{s_N^K}\delta \left( 2+\frac{C}{\sqrt{s_N}} \right) ^{2}+\frac{\alpha _r^2a\log 2}{4N_rs_N}\,. \end{aligned}$$

(A.24)

We can make the r.h.s. vanish by choosing for example: $\delta =s_N^{2K/3}N^{-1/3}$. The choice $s_N\propto N^{-1/16K}$ makes the r.h.s. (A.24) behave like ${\mathcal {O}}(N^{-1/4})$. $\square$

B Appendix: the SK Case

In the case $K=1$ the equation (46) reduces to:

$$\begin{aligned} \lim _{N\rightarrow \infty }{\bar{p}}_N(\mu ,h)=\sup _{x\in {\mathbb {R}}_{\ge 0}}\left\{ \mu \frac{(1-x)^2}{4}-\frac{\mu x^2}{2}+\psi (\mu x+h) \right\} \end{aligned}$$

(B.1)

while (48) simply becomes:

$$\begin{aligned} x={\mathbb {E}}_z\tanh \left( z\sqrt{\mu x+h}+\mu x+h\right) :=T(x;\mu ,h)\;. \end{aligned}$$

(B.2)

We collect the main results on this model in the following proposition.

Proposition 4

Define:

$$\begin{aligned} {\bar{p}}_{var}(x;\mu ,h)=\mu \frac{(1-x)^2}{4}-\frac{\mu x^2}{2}+\psi (\mu x+h)\;. \end{aligned}$$

(B.3)

The following hold:

• 1. if $\mu <1$ then ${\bar{p}}_{var}$ is concave in x. Equivalently if $\mu <1$ then $T(x;\mu ,h)$ is a contraction, and if further $h=0$ then $x=0$ is its fixed point;

• 2. the stable solution of the consistency equation (B.2) is continuous at $(\mu ,h)=(1,0)$:

  $$\begin{aligned} \lim _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)=0={\bar{x}}(1,0)\;; \end{aligned}$$

  (B.4)

• 3. for fixed $h=0$, the magnetization goes to 0 linearly with $\mu -1$ as $\mu \rightarrow 1_+$, more precisely:

  $$\begin{aligned} {\bar{x}}=(1+o(1))\frac{\mu -1}{\mu ^2} \end{aligned}$$

  (B.5)

  where o(1) goes to 0 when $\mu \rightarrow 1_+$. Therefore the critical exponent $\beta$ (in the Landau classification) is 1, which means that the derivative of the magnetization w.r.t. $\mu$ does not diverge at the critical point, it only jumps from 0 to 1 and then decreases;

• 4. Along the line $(\mu ,\lambda (\mu -1)),\;\lambda >0$ in the plane $(\mu ,h)$ the magnetization goes to 0 as follows:

  $$\begin{aligned} {\bar{x}}=\sqrt{\frac{\lambda (\mu -1)}{\mu ^2}}(1+o(1))\; \end{aligned}$$

  (B.6)

  when $\mu \rightarrow 1_+$, therefore with a critical exponent 1/2;

• 5. For fixed $\mu =1$ and $h\rightarrow 0_+$ the magnetization behaves as:

  $$\begin{aligned} {\bar{x}}^2=h(1+o(1)) \end{aligned}$$

  (B.7)

  where $o(1)\rightarrow 0$ when $h\rightarrow 0_+$. Therefore we have a critical exponent $\delta =2$ (according to Landau’s classification).

Proof

1. The fist assertion follows immediately from (56), since ${\hat{\alpha }}\equiv 1$ and $\varDelta \equiv \mu$. Then, by (51):

$$\begin{aligned} \frac{dT}{dx}(x;\mu ,h)=\mu {\mathbb {E}}_z\left[ \left( 1-\tanh ^2\left( z\sqrt{\mu x+h}+\mu x+h\right) \right) ^2\right] \le \mu <1 \end{aligned}$$

that implies T is a contraction. It is easy to see that if $h=0$ then $x=0$ is a solution of the fixed point equation which must be unique by Banach’s fixed point theorem.

2. Using continuity and monotonicity of T (see (51)):

$$\begin{aligned}&\limsup _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)=T(\limsup _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h);1,0)\\&\liminf _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)=T(\liminf _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h);1,0) \end{aligned}$$

hence both $\limsup _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)$ and $\liminf _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)$ satisfy the consistency equation:

$$\begin{aligned} m={\mathbb {E}}_z\tanh (z\sqrt{m}+m) \end{aligned}$$

whose solution $m=0$ is unique, since the derivative of T(m; 1, 0) is $\le 1$ and equality holds only at $m=0$. We conclude that there exists

$$\begin{aligned} \lim _{(\mu ,h)\rightarrow (1,0)}{\bar{x}}(\mu ,h)=0={\bar{x}}(1,0)\;. \end{aligned}$$

(B.8)

3. We first notice that

$$\begin{aligned} {\mathbb {E}}\tanh ^2\left( z\sqrt{Q}+Q\right) ={\mathbb {E}}\tanh \left( z\sqrt{Q}+Q\right) \,,\quad Q\ge 0\,,z\sim {\mathcal {N}}(0,1) \end{aligned}$$

(B.9)

which simply follows from the third relation in (51) and the identity (14). Indeed, the quantity in (B.9) is nothing but the quenched average magnetization of a free system. Now, by computing the first and second derivatives of the map $T(x;\mu ,0)$ and using (B.9) we get:

$$\begin{aligned}&T'(0;\mu ,0)=\mu \,,\quad T''(0;\mu ,0)=-2\mu ^2\\&{\bar{x}}=\mu {\bar{x}}-\mu ^2 {\bar{x}}^2(1+o(1))\quad \Rightarrow \quad {\bar{x}}=(1+o(1))\left( \frac{\mu -1}{\mu ^2}\right) \;, \end{aligned}$$

which implies that, in proximity of $\mu =1$, the magnetization goes to 0 with a critical exponent $\beta =1$ (not to be confused with inverse absolute temperature) and with slope 1.

4. An analogous expansion of T yields:

$$\begin{aligned} {\bar{x}}=T({\bar{x}};\mu ,\lambda (\mu -1))=\mu {\bar{x}}+\lambda (\mu -1)-(\mu ^2{\bar{x}}^2+o(\mu -1))(1+o(1)) \end{aligned}$$

which in turn entails:

$$\begin{aligned} {\bar{x}}^2=\frac{\lambda (\mu -1)}{\mu ^2}(1+o(1))\;. \end{aligned}$$

5. Here by o(1) we mean a quantity that approaches 0 as $h\rightarrow 0_+$. As in the previous steps:

$$\begin{aligned} {\bar{x}}=T({\bar{x}};1,h)={\bar{x}}+h-({\bar{x}}^2+o(h))(1+o(1))\;, \end{aligned}$$

then we get:

$$\begin{aligned} {\bar{x}}^2=h(1+o(1))\quad \Rightarrow \quad \delta =2\;. \end{aligned}$$

$\square$