Introduction
A multi-state spin system is composed of a set V = { 1 , … , n } V = \{1,\ldots,n\} V = { 1 , … , n } i i i S v S_v S v S : = S 1 × ⋯ × S n \mathcal{S} := S_1 \times\cdots\times S_n S := S 1 × ⋯ × S n μ \mu μ
Given such a μ \mu μ μ \mu μ μ \mu μ n n n μ \mu μ
The natural Markov chain is the Glauber dynamics: Given a state x ∈ S \mathbf{x} \in\mathcal{S} x ∈ S 1 ≤ i ≤ n 1 \le i \le n 1 ≤ i ≤ n i i i μ \mu μ x − i \mathbf{x}_{-i} x − i x ∈ S i x \in S_i x ∈ S i x x x
μ ( x 1 , … , x i − 1 , x , x i + 1 , … , x n ) ∑ x ′ ∈ S i μ ( x 1 , … , x i − 1 , x ′ , x i + 1 , … , x n ) \frac{\mu(x_1,\ldots,x_{i-1},x,x_{i+1},\ldots,x_n)}{\sum_{x' \in S_i}\mu(x_1,\ldots,x_{i-1},x',x_{i+1},\ldots,x_n)} ∑ x ′ ∈ S i μ ( x 1 , … , x i − 1 , x ′ , x i + 1 , … , x n ) μ ( x 1 , … , x i − 1 , x , x i + 1 , … , x n ) One fundamental question is how long it takes for this chain to mix. Let 1 n P \frac{1}{n}\mathcal{P} n 1 P 1 / n 1/n 1/ n P \mathcal{P} P
max x ∈ S min { t : ∥ 1 x T ( 1 n P ) t − μ ∥ T V ≤ 1 4 } . \max_{x\in S}\min\left\{t:\left\|\mathbf{1}_x^{\mathsf{T}}\left(\frac{1}{n}\mathcal{P}\right)^t-\mu\right\|_{\mathrm{TV}}\le\frac{1}{4}\right\}. x ∈ S max min { t : 1 x T ( n 1 P ) t − μ TV ≤ 4 1 } . To bound the mixing time of 1 n P \frac{1}{n}\mathcal{P} n 1 P
Theorem 1.1. Suppose that for any function f : S → R f:S\to\mathbb{R} f : S → R E μ f = 0 \mathbb{E}_{\mu}f=0 E μ f = 0
⟨ L f , f ⟩ μ ≥ ε ⟨ f , f ⟩ μ . \langle\mathcal{L}f,f\rangle_{\mu}\ge\varepsilon\langle f,f\rangle_{\mu}. ⟨ L f , f ⟩ μ ≥ ε ⟨ f , f ⟩ μ . where L = n I − P \mathcal{L}=nI-\mathcal{P} L = n I − P f , g : S → R f,g:S\to\mathbb{R} f , g : S → R ⟨ f , g ⟩ μ = E X ∼ μ [ f ( X ) g ( X ) ] \langle f,g\rangle_{\mu}=\mathbb{E}_{X\sim\mu}[f(X)g(X)] ⟨ f , g ⟩ μ = E X ∼ μ [ f ( X ) g ( X )] 1 n P \frac{1}{n}\mathcal{P} n 1 P
O ( n ⋅ log ( 1 μ min ) ε ) . O\left(\frac{n\cdot\log\left(\frac{1}{\mu_{\min}}\right)}{\varepsilon}\right). O ε n ⋅ log ( μ m i n 1 ) . where μ min = min x μ ( x ) \mu_{\min}=\min_x\mu(x) μ m i n = min x μ ( x )
Over the last few decades, several techniques have been developed to bound the mixing time of such a chain such as (path) coupling, canonical paths, and spectral independence. Of particular interest to us is a family of local to global theorems that often go by the name of trickledown theorems (or more classically by the name of Dobrushin’s condition or path coupling).
In such techniques given two arbitrary particles i , j i,j i , j worst case interaction between i , j i,j i , j j j j i i i j j j i i i [18 , 14 , 15 ] . More modern techniques show that it is enough to bound the worst case “spectral influence” of j j j i i i X i , X j X_i,X_j X i , X j R ( S i ∪ S j ) × ( S i ∪ S j ) \mathbb{R}^{(S_i\cup S_j)\times(S_i\cup S_j)} R ( S i ∪ S j ) × ( S i ∪ S j ) P [ X i ∣ X j = s ] \mathbb{P}[X_i\mid X_j=s] P [ X i ∣ X j = s ] s s s j j j
Let λ i , j \lambda_{i,j} λ i , j j j j i i i Λ \Lambda Λ ∥ Λ ∥ o p < 1 \|\Lambda\|_{\mathrm{op}}<1 ∥Λ ∥ op < 1 [22 ] .
The aforementioned technique has found many applications e.g., in studying graph coloring or multi-state hard core model or even in studying graphical models [18 , 19 , 22 , 6 , 23 ] . Although it is very powerful, it comes short of analyzing many applications of the natural Glauber dynamics. Ideally, one would like an average local-to-global theorem which allows for an “average spectral influence” of particle j j j i i i
Rank-1-perturbed trickledown theorem: Given the n n n f : S 1 × ⋯ × S n → R f:S_1\times\cdots\times S_n\to\mathbb{R} f : S 1 × ⋯ × S n → R P i f P_if P i f f f f S i S_i S i
P i f ( x 1 , … , x n ) = E μ [ f ( X 1 , … , X n ) ∣ X j = x j , ∀ j ≠ i ] . P_if(x_1,\ldots,x_n)=\mathbb{E}_{\mu}[f(X_1,\ldots,X_n)\mid X_j=x_j,\forall j\ne i]. P i f ( x 1 , … , x n ) = E μ [ f ( X 1 , … , X n ) ∣ X j = x j , ∀ j = i ] . Let P = ∑ i P i \mathcal{P} = \sum_i P_i P = ∑ i P i 1 n P \frac{1}{n}\mathcal{P} n 1 P L i = I − P i L_i = I - P_i L i = I − P i L = ∑ i L i \mathcal{L} = \sum_i L_i L = ∑ i L i P i P_i P i f f f P i 2 f = P i f P_i^2 f = P_i f P i 2 f = P i f L i L_i L i
⟨ f , g ⟩ μ = E X ∼ μ [ f ( X ) g ( X ) ] . \langle f,g\rangle_\mu= \mathbb{E}_{X\sim\mu}[f(X)g(X)]. ⟨ f , g ⟩ μ = E X ∼ μ [ f ( X ) g ( X )] . It is not hard to see that P i , L i P_i,L_i P i , L i f , g f,g f , g
⟨ f , P i g ⟩ = E μ [ f E μ [ g ∣ X − i ] ] = E μ [ E μ [ f ∣ X − i ] E μ [ g ∣ X − i ] ] = E μ [ E μ [ f ∣ X − i ] g ] = ⟨ P i f , g ⟩ \langle f,P_i g\rangle= \mathbb{E}_\mu[f\mathbb{E}_\mu[g\mid X_{-i}]] = \mathbb{E}_\mu[\mathbb{E}_\mu[f\mid X_{-i}]\mathbb{E}_\mu[g\mid X_{-i}]] = \mathbb{E}_\mu[\mathbb{E}_\mu[f\mid X_{-i}]g] = \langle P_i f,g\rangle ⟨ f , P i g ⟩ = E μ [ f E μ [ g ∣ X − i ]] = E μ [ E μ [ f ∣ X − i ] E μ [ g ∣ X − i ]] = E μ [ E μ [ f ∣ X − i ] g ] = ⟨ P i f , g ⟩ and since L i = I − P i L_i = I - P_i L i = I − P i L i L_i L i
E ( f ) : = E μ [ f L f ] = ∑ i E μ [ f L i f ] = ∑ i E μ [ L i f L i f ] = ∑ i E μ [ ( L i f ) 2 ] . (1) \mathcal{E}(f) := \mathbb{E}_\mu[f\mathcal{L}f] = \sum_i \mathbb{E}_\mu[fL_i f] = \sum_i \mathbb{E}_\mu[L_i fL_i f] = \sum_i \mathbb{E}_\mu[(L_i f)^2]. \tag*{(1)} E ( f ) := E μ [ f L f ] = i ∑ E μ [ f L i f ] = i ∑ E μ [ L i f L i f ] = i ∑ E μ [( L i f ) 2 ] . (1) Fact 1.2. If for any function f : S → R f:S\to\mathbb{R} f : S → R
⟨ L f , L f ⟩ ≥ ε ⟨ L f , f ⟩ = ε ⋅ E ( f ) , \langle\mathcal{L}f,\mathcal{L}f\rangle\ge\varepsilon\langle\mathcal{L}f,f\rangle= \varepsilon\cdot\mathcal{E}(f), ⟨ L f , L f ⟩ ≥ ε ⟨ L f , f ⟩ = ε ⋅ E ( f ) , then the spectral gap of the Glauber dynamics 1 n P \frac{1}{n}\mathcal{P} n 1 P ε / n \varepsilon/n ε / n
Proof. Given an arbitrary function g g g L ⪰ 0 \mathcal{L}\succeq0 L ⪰ 0 f = L − 1 / 2 g f = \mathcal{L}^{-1/2}g f = L − 1/2 g L \mathcal{L} L L \mathcal{L} L 0 0 0
⟨ L g , g ⟩ = ⟨ L f , L f ⟩ ≥ ε ⋅ ⟨ L f , f ⟩ = ⟨ g , g ⟩ , \langle\mathcal{L}g,g\rangle= \langle\mathcal{L}f,\mathcal{L}f\rangle\ge\varepsilon\cdot\langle\mathcal{L}f,f\rangle= \langle g,g\rangle, ⟨ L g , g ⟩ = ⟨ L f , L f ⟩ ≥ ε ⋅ ⟨ L f , f ⟩ = ⟨ g , g ⟩ , where all the inner-products are with respect to μ \mu μ
We emphasize that the above machinery of using the “averaged curvature condition” ([? ], Equation 1.16.6, Proposition 4.8.3) was recently employed by Göbel, Jenssen, Michelen, Pappik, Perkins and Schiller [21 ] to re-prove a recent trickledown theorem of Chen, Chen, Chen, Yin, and Zhang [10 ] . Their proof was a motivation for us to consider this technique for the SK model.
Definition 1.3. We say P \mathcal{P} P { m ( X ) } ∈ R n \{m(X)\}\in\mathbb{R}^n { m ( X )} ∈ R n X ∈ S X\in S X ∈ S f : S → R f:S\to\mathbb{R} f : S → R 1 ≤ i < j ≤ n 1\le i<j\le n 1 ≤ i < j ≤ n x − i − j x_{-i-j} x − i − j i , j i,j i , j
E X ∼ μ [ L i f ( X ) L j f ( X ) ∣ x − i − j ] ≥ − a i , j E [ L i f ( X ) L j f ( X ) m i ( X ) m j ( X ) ∣ x − i − j ] \mathbb{E}_{X\sim\mu}[L_i f(X)L_j f(X)\mid x_{-i-j}] \ge-a_{i,j}\mathbb{E}[L_i f(X)L_j f(X)m_i(X)m_j(X)\mid x_{-i-j}] E X ∼ μ [ L i f ( X ) L j f ( X ) ∣ x − i − j ] ≥ − a i , j E [ L i f ( X ) L j f ( X ) m i ( X ) m j ( X ) ∣ x − i − j ] − b i , j E [ L i f ( X ) 2 + L j f ( X ) 2 ∣ x − i − j ] . (2) -b_{i,j}\mathbb{E}[L_i f(X)^2+L_j f(X)^2\mid x_{-i-j}]. \tag*{(2)} − b i , j E [ L i f ( X ) 2 + L j f ( X ) 2 ∣ x − i − j ] . (2) The following abstracts our main technique.
Proposition 1.4. Suppose P \mathcal{P} P m ( x ) \mathbf{m}(\mathbf{x}) m ( x ) { a i , j , b i , j } 1 ≤ i , j ≤ n \{a_{i,j}, b_{i,j}\}_{1\leq i,j\leq n} { a i , j , b i , j } 1 ≤ i , j ≤ n i i i
∑ j b i , j E [ L i f ( X ) 2 ] + λ max ( A ) E [ m i ( X ) 2 L i f ( X ) 2 ] ≤ ( 1 − ε ) E [ L i f ( X ) 2 ] , (3) \sum_j b_{i,j}\mathbb{E}[L_i f(X)^2] + \lambda_{\max}(A)\mathbb{E}[m_i(X)^2 L_i f(X)^2] \leq(1-\varepsilon)\mathbb{E}[L_i f(X)^2], \tag*{(3)} j ∑ b i , j E [ L i f ( X ) 2 ] + λ m a x ( A ) E [ m i ( X ) 2 L i f ( X ) 2 ] ≤ ( 1 − ε ) E [ L i f ( X ) 2 ] , (3) where A ∈ R n × n A \in\mathbb{R}^{n\times n} A ∈ R n × n a i , j a_{i,j} a i , j ⟨ L f , L f ⟩ ≥ ε ⋅ E ( f ) \langle Lf,Lf\rangle\geq\varepsilon\cdot\mathcal{E}(f) ⟨ L f , L f ⟩ ≥ ε ⋅ E ( f )
Proof. Let w i ( X ) = L i f ( X ) m i ( X ) w_i(X)=L_i f(X)m_i(X) w i ( X ) = L i f ( X ) m i ( X ) ∑ X ∈ S : X − i − j = σ − i − j μ ( X ) \sum_{X\in S:X_{-i-j}=\sigma_{-i-j}}\mu(X) ∑ X ∈ S : X − i − j = σ − i − j μ ( X ) i , j i,j i , j
⟨ L f , L f ⟩ ≥ − E [ w w ⊤ ] ∙ A + ∑ i ( 1 − ∑ j b i , j ) E [ L i f ( X ) 2 ] ≥ − λ max ( A ) tr ( E [ w w ⊤ ] ) + ∑ i ( 1 − ∑ j b i , j ) E [ L i f ( X ) 2 ] = − λ max ( A ) ⋅ E [ ∥ w ∥ 2 ] + ∑ i ( 1 − ∑ j b i , j ) E [ L i f ( X ) 2 ] = ∑ i ( ( 1 − ∑ j b i , j ) E [ L i f ( X ) 2 ] − λ max ( A ) E [ m i ( X ) 2 L i f ( X ) 2 ] ) ≥ ε ⋅ E ( f ) , \begin{aligned}
\langle Lf,Lf\rangle&\geq-\mathbb{E}[ww^{\top}]\bullet A+\sum_i\left(1-\sum_j b_{i,j}\right)\mathbb{E}[L_i f(X)^2]\\
&\geq-\lambda_{\max}(A)\operatorname{tr}(\mathbb{E}[ww^{\top}])+\sum_i\left(1-\sum_j b_{i,j}\right)\mathbb{E}[L_i f(X)^2]\\
&=-\lambda_{\max}(A)\cdot\mathbb{E}[\lVert w\rVert^2]+\sum_i\left(1-\sum_j b_{i,j}\right)\mathbb{E}[L_i f(X)^2]\\
&=\sum_i\left(\left(1-\sum_j b_{i,j}\right)\mathbb{E}[L_i f(X)^2]-\lambda_{\max}(A)\mathbb{E}[m_i(X)^2L_i f(X)^2]\right)\\
&\geq\varepsilon\cdot\mathcal{E}(f),
\end{aligned} ⟨ L f , L f ⟩ ≥ − E [ w w ⊤ ] ∙ A + i ∑ ( 1 − j ∑ b i , j ) E [ L i f ( X ) 2 ] ≥ − λ m a x ( A ) tr ( E [ w w ⊤ ]) + i ∑ ( 1 − j ∑ b i , j ) E [ L i f ( X ) 2 ] = − λ m a x ( A ) ⋅ E [∥ w ∥ 2 ] + i ∑ ( 1 − j ∑ b i , j ) E [ L i f ( X ) 2 ] = i ∑ ( ( 1 − j ∑ b i , j ) E [ L i f ( X ) 2 ] − λ m a x ( A ) E [ m i ( X ) 2 L i f ( X ) 2 ] ) ≥ ε ⋅ E ( f ) , where the second inequality follows by Lemma 2.1.
To put the above statement in context, suppose that we let a i , j = 0 a_{i,j}=0 a i , j = 0 b i , j b_{i,j} b i , j j j j i i i ∑ j b i , j ≤ 1 − ε \sum_j b_{i,j}\leq1-\varepsilon ∑ j b i , j ≤ 1 − ε i i i ε / n \varepsilon/n ε / n i , j i,j i , j m ( X ) m ( X ) ⊤ \mathbf{m}(X)\mathbf{m}(X)^{\top} m ( X ) m ( X ) ⊤ i i i
Application: The Sherrington-Kirkpatrick (SK) model
Given a 2-state spin system with n n n S i = { − 1 , + 1 } S_i=\{-1,+1\} S i = { − 1 , + 1 } i i i J J J μ J \mu_J μ J x ∈ { ± 1 } n \mathbf{x}\in\{\pm1\}^n x ∈ { ± 1 } n
μ J ( x ) = 1 Z n exp ( 1 2 x ⊤ J x ) , (4) \mu_J(\mathbf{x})=\frac{1}{Z_n}\exp\left(\frac{1}{2}\mathbf{x}^{\top}J\mathbf{x}\right), \tag*{(4)} μ J ( x ) = Z n 1 exp ( 2 1 x ⊤ J x ) , (4) where Z n : = ∑ x exp ( x ⊤ J x ) Z_n:=\sum_{\mathbf{x}}\exp(\mathbf{x}^{\top}J\mathbf{x}) Z n := ∑ x exp ( x ⊤ J x ) J J J J J J β 2 n \frac{\beta^2}{n} n β 2 average case problem in approximate counting and sampling with the hope that the techniques developed here can be extended more generally to sampling problems over random constraint satisfaction problems.
The following is our main theorem.
Theorem 1.5. Let L J = ∑ i L i = ∑ i ( I − P i ) \mathcal{L}_J = \sum_i L_i = \sum_i(I-P_i) L J = ∑ i L i = ∑ i ( I − P i ) ε ≥ 5 ⋅ 10 − 5 \varepsilon\ge5 \cdot10^{-5} ε ≥ 5 ⋅ 1 0 − 5 β < 1 2 + ε \beta< \frac{1}{2}+\varepsilon β < 2 1 + ε n > n 0 n > n_0 n > n 0 J J J
⟨ L J f , L J f ⟩ μ ≥ C ⋅ ⟨ L J f , f ⟩ μ \langle\mathcal{L}_J f,\mathcal{L}_J f\rangle_\mu\ge C \cdot\langle\mathcal{L}_J f,f\rangle_\mu ⟨ L J f , L J f ⟩ μ ≥ C ⋅ ⟨ L J f , f ⟩ μ where C C C ε \varepsilon ε
Now, using Fact 1.2 we have the following immediate consequence.
Corollary 1.6. There exists an absolute constant ε > 0 \varepsilon> 0 ε > 0 β < 1 / 2 + ε \beta< 1/2+\varepsilon β < 1/2 + ε spectral gap of 1 n P J \frac{1}{n}P_J n 1 P J Ω ( 1 / n ) \Omega(1/n) Ω ( 1/ n ) Glauber dynamics for the SK model mixes in time O ( n 2 ) O(n^2) O ( n 2 )
Proof. We will show that with probability 1 − o n ( 1 ) 1-o_n(1) 1 − o n ( 1 )
μ min : = min x ∈ { ± 1 } n μ J ( x ) ≥ exp ( − ( ln 2 + 4 β ) n ) . (5) \mu_{\min} := \min_{x\in\{\pm1\}^n}\mu_J(x) \ge\exp\left(-(\ln2+4\beta)n\right). \tag*{(5)} μ m i n := x ∈ { ± 1 } n min μ J ( x ) ≥ exp ( − ( ln 2 + 4 β ) n ) . (5) With (5), it follows from Theorem 1.1 that the mixing time is O ( log ( 1 / μ min ) / Ω ( 1 / n ) ) = O ( n 2 ) O(\log(1/\mu_{\min})/\Omega(1/n))=O(n^2) O ( log ( 1/ μ m i n ) /Ω ( 1/ n )) = O ( n 2 )
It remains to prove (5). By (4), a step towards this is to show that with probability 1 − o n ( 1 ) 1-o_n(1) 1 − o n ( 1 ) x ∈ { ± 1 } n x\in\{\pm1\}^n x ∈ { ± 1 } n
∣ x T J x ∣ ≤ 2 β n . (6) \lvert x^{\mathsf{T}}Jx\rvert\le2\beta n. \tag*{(6)} ∣ x T J x ∣ ≤ 2 β n . (6) Assuming that we are on the high probability event for which (6) is true, we can bound Z n = ∑ x ∈ { ± 1 } n exp ( 1 2 x T J x ) ≤ 2 n exp ( β n ) Z_n=\sum_{x\in\{\pm1\}^n}\exp\left(\frac{1}{2}x^{\mathsf{T}}Jx\right)\le2^n\exp(\beta n) Z n = ∑ x ∈ { ± 1 } n exp ( 2 1 x T J x ) ≤ 2 n exp ( β n )
min x ∈ { ± 1 } n μ J ( x ) ≥ exp ( − β n ) 2 n exp ( β n ) = exp ( − ( ln 2 + 2 β ) n ) . \min_{x\in\{\pm1\}^n}\mu_J(x)\ge\frac{\exp(-\beta n)}{2^n\exp(\beta n)}=\exp\left(-(\ln2+2\beta)n\right). x ∈ { ± 1 } n min μ J ( x ) ≥ 2 n exp ( β n ) exp ( − β n ) = exp ( − ( ln 2 + 2 β ) n ) . Finally, it remains to prove (6). We fix x ∈ { ± 1 } n x\in\{\pm1\}^n x ∈ { ± 1 } n Var ( x T J x ) = 4 β 2 n ∑ i < j Var ( g i j x i x j ) = 4 β 2 n ( n 2 ) = 2 β 2 ( n − 1 ) \operatorname{Var}(x^{\mathsf{T}}Jx)=\frac{4\beta^2}{n}\sum_{i<j}\operatorname{Var}(g_{ij}x_ix_j)=\frac{4\beta^2}{n}\binom{n}{2}=2\beta^2(n-1) Var ( x T J x ) = n 4 β 2 ∑ i < j Var ( g ij x i x j ) = n 4 β 2 ( 2 n ) = 2 β 2 ( n − 1 ) g i j ∼ N ( 0 , 1 ) g_{ij}\sim\mathcal{N}(0,1) g ij ∼ N ( 0 , 1 ) i , j ∈ [ n ] i,j\in[n] i , j ∈ [ n ] x T J x ∼ N ( 0 , 2 β 2 ( n − 1 ) ) x^{\mathsf{T}}Jx\sim\mathcal{N}(0,2\beta^2(n-1)) x T J x ∼ N ( 0 , 2 β 2 ( n − 1 )) x T J x β 2 ( n − 1 ) \frac{x^{\mathsf{T}}Jx}{\beta\sqrt{2(n-1)}} β 2 ( n − 1 ) x T J x
P [ ∣ x T J x ∣ > 2 β n ] ≤ P [ ∣ x T J x ∣ β 2 ( n − 1 ) > 2 n ] ≤ 2 exp ( − ( 2 n ) 2 2 ) ≤ 2 e − n . \mathbb{P}\left[\lvert x^{\mathsf{T}}Jx\rvert>2\beta n\right]\le\mathbb{P}\left[\frac{\lvert x^{\mathsf{T}}Jx\rvert}{\beta\sqrt{2(n-1)}}>\sqrt{2n}\right]\le2\exp\left(-\frac{(\sqrt{2n})^2}{2}\right)\le2e^{-n}. P [ ∣ x T J x ∣ > 2 β n ] ≤ P [ β 2 ( n − 1 ) ∣ x T J x ∣ > 2 n ] ≤ 2 exp ( − 2 ( 2 n ) 2 ) ≤ 2 e − n . Taking a union bound over all x ∈ { ± 1 } n x\in\{\pm1\}^n x ∈ { ± 1 } n
P ( max x ∈ { ± 1 } n ∣ x T J x ∣ > 2 n ) ≤ 2 n + 1 exp ( − n ) = o n ( 1 ) . \mathbb{P}\left(\max_{x\in\{\pm1\}^n}\lvert x^{\mathsf{T}}Jx\rvert>2n\right)\le2^{n+1}\exp(-n)=o_n(1). P ( x ∈ { ± 1 } n max ∣ x T J x ∣ > 2 n ) ≤ 2 n + 1 exp ( − n ) = o n ( 1 ) . This shows (6).
Concurrent work. This work is done concurrently and independently of the work of Wang [30 ] who obtains related results based on the Bochner framework, although our detailed proof techniques are different.
Acknowledgements The authors have used AI systems for many heavy computations involved in this project especially in proving Theorem 3.1 and also for vibe-coding Mathematica to solve for the parameters in the proof of Theorem 1.5. The first author would like to thank Leon Schiller for discussing with him their paper [21 ] . This paper is the main inspiration for our work. The first author would also like to thank Thomas Rothvoss, Victor Reis, Arvin Sahami, Dan Mikulincer, Ramon van Handel and Santosh Vempala for discussions on stochastic localization. The second author would like to thank Tselil Schramm and her lab for fruitful discussions on spin systems. The last author’s research is supported by an NSF grant CCF-2203541, a Simons Investigator Award 928589, and a Lazowska Endowed Professorship in Computer Science & Engineering.
Previous Work
Physicists [28 ] , [25 ] predicted that Glauber dynamics for the SK model would converge fast for all β < 1 \beta< 1 β < 1 [5 ] proved a log-Sobolev inequality for μ J , h \mu_{J,h} μ J , h J J J ∥ J ∥ o p < 1 \lVert J\rVert_{\mathrm{op}} < 1 ∥ J ∥ op < 1
A few years later Eldan, Koehler and Zeitouni [20 ] proved that Glauber dynamics for μ J , h \mu_{J,h} μ J , h J J J C n \frac{C}{n} n C C > 0 C > 0 C > 0 O ( n 2 ) O(n^2) O ( n 2 ) β < 1 4 \beta< \frac{1}{4} β < 4 1 ∥ J ∥ o p < 1 \lVert J\rVert_{\mathrm{op}} < 1 ∥ J ∥ op < 1 β J \beta J β J J J J 1 n \frac{1}{n} n 1 β > 1 \beta> 1 β > 1
Anari, Jain, Koehler, Pham and Vuong [1 ] proved a modified log-Sobolev inequality for μ J , h \mu_{J,h} μ J , h [20 ] . Consequently they obtained an optimal O ( n log n ) O(n\log n) O ( n log n ) β < 1 4 \beta< \frac{1}{4} β < 4 1 [11 ] reproved this result using an entropy-annealing theorem and the log-Sobolev inequality result by Bauerschmidt and Bodineau [5 ] .
Later Anari, Koehler and Vuong [2 ] managed to extend trickledown theorems [26 ] , [3 ] to the stochastic localization machinery, and used it to show that if J J J ∥ J ∥ o p < 4 β \lVert J\rVert_{\mathrm{op}} < 4\beta ∥ J ∥ op < 4 β J J J 2 β 2\beta 2 β μ J , h \mu_{J,h} μ J , h β \beta β 0.295 0.295 0.295
On a separate line of work, El Alaoui, Montanari and Sellke [4 ] gave an efficient statistical algorithm to sample from the SK model for β \beta β 1 2 \frac{1}{2} 2 1 β < 1 \beta< 1 β < 1 [12 ] . So, in a sense these results provide a substantial indication that polynomial mixing bound for Glauber dynamics should be reachable.
Very recently, Davies, Lee, Sandhu and Shi [17 ] proved a weak-Poincare inequality for Glauber dynamics for SK model for β < 1 2 \beta< \frac{1}{2} β < 2 1
Proof overview
By Fact 1.2, to bound the spectral gap of L \mathcal{L} L
⟨ L f , L f ⟩ E ( f ) = ∑ i E [ ( L i f ) 2 ] + ∑ i ≠ j E [ ( L i f ) ( L j f ) ] E ( f ) = 1 + ∑ i ≠ j E [ ( L i f ) ( L j f ) ] E ( f ) . (7) \frac{\langle\mathcal{L}f,\mathcal{L}f\rangle}{\mathcal{E}(f)} = \frac{\sum_i \mathbb{E}[(L_i f)^2] + \sum_{i\ne j} \mathbb{E}[(L_i f)(L_j f)]}{\mathcal{E}(f)} = 1 + \frac{\sum_{i\ne j} \mathbb{E}[(L_i f)(L_j f)]}{\mathcal{E}(f)}. \tag*{(7)} E ( f ) ⟨ L f , L f ⟩ = E ( f ) ∑ i E [( L i f ) 2 ] + ∑ i = j E [( L i f ) ( L j f )] = 1 + E ( f ) ∑ i = j E [( L i f ) ( L j f )] . (7) The term E [ ( L i f ) ( L j f ) ] \mathbb{E}[(L_i f)(L_j f)] E [( L i f ) ( L j f )] x i , x j x_i,x_j x i , x j x − i − j x_{-i-j} x − i − j μ J i j , a , b ( x i , x j ) = exp ( J i j x i x j + a x i + b x j ) \mu_{J_{ij},a,b}(x_i,x_j) = \exp(J_{ij}x_i x_j + ax_i + bx_j) μ J ij , a , b ( x i , x j ) = exp ( J ij x i x j + a x i + b x j )
a = b = 0 a=b=0 a = b = 0 i , j i,j i , j J i j J_{ij} J ij x i , x j x_i,x_j x i , x j μ J i j ( x i , x j ) = exp ( J i j x i x j ) \mu_{J_{ij}}(x_i,x_j) = \exp(J_{ij}x_i x_j) μ J ij ( x i , x j ) = exp ( J ij x i x j )
For any function f : { ± 1 } 2 f:\{\pm1\}^2 f : { ± 1 } 2 μ \mu μ m i ( x ) = x i m_i(x)=x_i m i ( x ) = x i a i , j = J i , j a_{i,j}=J_{i,j} a i , j = J i , j b i , j = ∣ J i j ∣ 3 / 6 b_{i,j} = \lvert J_{ij}\rvert^3/6 b i , j = ∣ J ij ∣ 3 /6
E μ J i j [ ( L i f ) ( L j f ) ∣ X − i − j ] ≥ − J i j E μ J i j [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − ∣ J i j ∣ 3 6 E μ J i j [ ( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] . (8) \mathbb{E}_{\mu_{J_{ij}}}[(L_i f)(L_j f)\mid X_{-i-j}] \ge-J_{ij}\mathbb{E}_{\mu_{J_{ij}}}[X_iX_j(L_i f)(L_j f)\mid X_{-i-j}] - \frac{\lvert J_{ij}\rvert^3}{6}\mathbb{E}_{\mu_{J_{ij}}}[(L_i f)^2+(L_j f)^2\mid X_{-i-j}]. \tag*{(8)} E μ J ij [( L i f ) ( L j f ) ∣ X − i − j ] ≥ − J ij E μ J ij [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − 6 ∣ J ij ∣ 3 E μ J ij [( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] . (8) Assuming (7), we are in a position to apply Proposition 1.4.
Using standard facts about GOE random matrices, we have 2 ∑ j ≠ i b i j ≤ o ( 1 ) 2\sum_{j\ne i} b_{ij} \le o(1) 2 ∑ j = i b ij ≤ o ( 1 ) i ∈ [ n ] i\in[n] i ∈ [ n ] ε = 1 − ∥ J ∥ o p − o ( 1 ) \varepsilon= 1-\lVert J\rVert_{\mathrm{op}}-o(1) ε = 1 − ∥ J ∥ op − o ( 1 )
( 1 6 ∑ j ∣ J i j ∣ 3 ) ⋅ E [ L i f ( X ) 2 ] + λ max ( J ) E [ X i 2 L i f ( X ) 2 ] ≤ ( 1 − λ max ( J ) − o ( 1 ) ) E [ L i f ( X ) 2 ] . (9) \left(\frac{1}{6}\sum_j \lvert J_{ij}\rvert^3\right)\cdot\mathbb{E}[L_i f(X)^2] + \lambda_{\max}(J)\mathbb{E}[X_i^2L_i f(X)^2] \le\left(1-\lambda_{\max}(J)-o(1)\right)\mathbb{E}[L_i f(X)^2]. \tag*{(9)} ( 6 1 j ∑ ∣ J ij ∣ 3 ) ⋅ E [ L i f ( X ) 2 ] + λ m a x ( J ) E [ X i 2 L i f ( X ) 2 ] ≤ ( 1 − λ m a x ( J ) − o ( 1 ) ) E [ L i f ( X ) 2 ] . (9) Therefore, since λ max ( J ) → 2 β + o ( 1 ) \lambda_{\max}(J)\to2\beta+o(1) λ m a x ( J ) → 2 β + o ( 1 ) β < 1 / 2 − ε \beta< 1/2-\varepsilon β < 1/2 − ε L \mathcal{L} L ≥ ε \ge\varepsilon ≥ ε
Proof of (7). For simplicity, we drop the pinned variables x − i − j x_{-i-j} x − i − j
We observe that exp ( J i j x i x j ) = cosh J i j ( 1 + x i x j J ‾ i j ) \exp(J_{ij}x_i x_j)=\cosh J_{ij}(1+x_i x_j\overline{J}_{ij}) exp ( J ij x i x j ) = cosh J ij ( 1 + x i x j J ij ) J ‾ i j = tanh J i j \overline{J}_{ij}=\tanh J_{ij} J ij = tanh J ij μ J ( x i , x j ) = ( 1 + x i x j J ‾ i j ) μ 0 ( x i , x j ) \mu_J(x_i,x_j)=(1+x_i x_j\overline{J}_{ij})\mu_0(x_i,x_j) μ J ( x i , x j ) = ( 1 + x i x j J ij ) μ 0 ( x i , x j )
E μ J i j [ ( L i f ) ( L j f ) ] = E μ 0 [ ( 1 + X i X j J ‾ ) ( L i f ) ( L j f ) ] . (10) \mathbb{E}_{\mu_{J_{ij}}}[(L_i f)(L_j f)] = \mathbb{E}_{\mu_0}[(1+X_iX_j\overline{J})(L_i f)(L_j f)]. \tag*{(10)} E μ J ij [( L i f ) ( L j f )] = E μ 0 [( 1 + X i X j J ) ( L i f ) ( L j f )] . (10) We don’t know a direct lower-bound on this quantity.
Instead, first we use μ J ˉ i j ( X i = x i ∣ X j = x j ) = 1 + J ˉ i j x i x j 2 \mu_{\bar{J}_{ij}}(X_i=x_i\mid X_j=x_j)=\frac{1+\bar{J}_{ij}x_ix_j}{2} μ J ˉ ij ( X i = x i ∣ X j = x j ) = 2 1 + J ˉ ij x i x j
L i f ( x i , x j ) = f ( x i , x j ) − E X i [ f ( X i , x j ) ∣ x j ] = f ( x i , x j ) − ( 1 + J ˉ i j x i x j 2 ⋅ f ( x i , x j ) + 1 − J ˉ i j x i x j 2 ⋅ f ( − x i , x j ) ) = 1 − J ˉ i j x i x j 2 ( f ( x i , x j ) − f ( − x i , x j ) ) = ( 1 − J ˉ i j x i x j ) L i 0 f ( x i , x j ) (11) \begin{aligned}
L_i f(x_i,x_j)=f(x_i,x_j)-\mathbb{E}_{X_i}[f(X_i,x_j)\mid x_j] \\
&=\ f(x_i,x_j)-\left(\frac{1+\bar{J}_{ij}x_ix_j}{2}\cdot f(x_i,x_j)+\frac{1-\bar{J}_{ij}x_ix_j}{2}\cdot f(-x_i,x_j)\right) \\
&=\frac{1-\bar{J}_{ij}x_ix_j}{2}\left(f(x_i,x_j)-f(-x_i,x_j)\right)=(1-\bar{J}_{ij}x_ix_j)L_i^0f(x_i,x_j) \tag*{(11)}
\end{aligned} L i f ( x i , x j ) = f ( x i , x j ) − E X i [ f ( X i , x j ) ∣ x j ] = f ( x i , x j ) − ( 2 1 + J ˉ ij x i x j ⋅ f ( x i , x j ) + 2 1 − J ˉ ij x i x j ⋅ f ( − x i , x j ) ) = 2 1 − J ˉ ij x i x j ( f ( x i , x j ) − f ( − x i , x j ) ) = ( 1 − J ˉ ij x i x j ) L i 0 f ( x i , x j ) (11) where L i 0 f ( x i , x j ) = f ( x i , x j , x − i − j ) − f ( − x i , x j , x − i − j ) 2 L_i^0f(x_i,x_j)=\frac{f(x_i,x_j,x_{-i-j})-f(-x_i,x_j,x_{-i-j})}{2} L i 0 f ( x i , x j ) = 2 f ( x i , x j , x − i − j ) − f ( − x i , x j , x − i − j ) μ 0 \mu_0 μ 0 J ˉ i j = 0 \bar{J}_{ij}=0 J ˉ ij = 0 i i i j j j L j f ( x i , x j ) = ( 1 − J ˉ i j x i x j ) L j 0 f ( x i , x j ) L_jf(x_i,x_j)=(1-\bar{J}_{ij}x_ix_j)L_j^0f(x_i,x_j) L j f ( x i , x j ) = ( 1 − J ˉ ij x i x j ) L j 0 f ( x i , x j )
E μ J ˉ i j [ ( L i f ) ( L j f ) ] = E μ 0 [ ( 1 + X i X j J ˉ i j ) ( 1 − X i X j J ˉ i j ) 2 ( L i 0 f ) ( L j 0 f ) ] . \mathbb{E}_{\mu_{\bar{J}_{ij}}}[(L_if)(L_jf)]=\mathbb{E}_{\mu_0}[(1+X_iX_j\bar{J}_{ij})(1-X_iX_j\bar{J}_{ij})^2(L_i^0f)(L_j^0f)]. E μ J ˉ ij [( L i f ) ( L j f )] = E μ 0 [( 1 + X i X j J ˉ ij ) ( 1 − X i X j J ˉ ij ) 2 ( L i 0 f ) ( L j 0 f )] . Now, notice if we introduce a 1 + X i X j J ˉ i j 1+X_iX_j\bar{J}_{ij} 1 + X i X j J ˉ ij ( 1 − J ˉ i j 2 ) 2 E [ ( L i 0 f ) ( L j 0 f ) ] (1-\bar{J}_{ij}^2)^2\mathbb{E}[(L_i^0f)(L_j^0f)] ( 1 − J ˉ ij 2 ) 2 E [( L i 0 f ) ( L j 0 f )]
E μ J ˉ i j [ ( 1 + J ˉ i j X i X j ) ( L i f ) ( L j f ) ] ≥ independence x i 2 = x j 2 = 1 E μ 0 [ ( 1 − J ˉ i j 2 ) ( L i , 0 f ) ] E μ 0 [ ( 1 − J ˉ i j 2 ) ( L j , 0 f ) ] ≥ independence 0. \mathbb{E}_{\mu_{\bar{J}_{ij}}}\left[(1+\bar{J}_{ij}X_iX_j)(L_if)(L_jf)\right]
\overset{x_i^2=x_j^2=1}{\underset{\text{independence}}{\geq}}
\mathbb{E}_{\mu_0}\left[(1-\bar{J}_{ij}^2)(L_{i,0}f)\right]\mathbb{E}_{\mu_0}\left[(1-\bar{J}_{ij}^2)(L_{j,0}f)\right]
\overset{\text{independence}}{\geq}0. E μ J ˉ ij [ ( 1 + J ˉ ij X i X j ) ( L i f ) ( L j f ) ] independence ≥ x i 2 = x j 2 = 1 E μ 0 [ ( 1 − J ˉ ij 2 ) ( L i , 0 f ) ] E μ 0 [ ( 1 − J ˉ ij 2 ) ( L j , 0 f ) ] ≥ independence 0. Having figured out the right rank-1 shift, we have
E μ J i j [ ( L i f ) ( L j f ) ∣ X − i − j ] \mathbb{E}_{\mu_{J_{ij}}}[(L_if)(L_jf)\mid X_{-i-j}] E μ J ij [( L i f ) ( L j f ) ∣ X − i − j ] ≥ AM-GM − J i j E μ J i j [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − ∣ J i j − J ˉ i j ∣ 2 E μ J i j [ ( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] ≥ − J i j E μ J i j [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − ∣ J i j ∣ 3 6 E μ J i j [ ( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] , \begin{aligned}
\overset{\text{AM-GM}}{\geq}
-J_{ij}\mathbb{E}_{\mu_{J_{ij}}}[X_iX_j(L_if)(L_jf)\mid X_{-i-j}]
-\frac{|J_{ij}-\bar{J}_{ij}|}{2}\mathbb{E}_{\mu_{J_{ij}}}[(L_if)^2+(L_jf)^2\mid X_{-i-j}] \\
&\geq
-J_{ij}\mathbb{E}_{\mu_{J_{ij}}}[X_iX_j(L_if)(L_jf)\mid X_{-i-j}]
-\frac{|J_{ij}|^3}{6}\mathbb{E}_{\mu_{J_{ij}}}[(L_if)^2+(L_jf)^2\mid X_{-i-j}],
\end{aligned} ≥ AM-GM − J ij E μ J ij [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − 2 ∣ J ij − J ˉ ij ∣ E μ J ij [( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] ≥ − J ij E μ J ij [ X i X j ( L i f ) ( L j f ) ∣ X − i − j ] − 6 ∣ J ij ∣ 3 E μ J ij [( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] , showing (7).
General Case: a , b a,b a , b Now, we consider the more general case for μ J i j , a , b \mu_{J_{ij},a,b} μ J ij , a , b a = ∑ k ≠ i , j J i k x k a=\sum_{k\ne i,j}J_{ik}x_k a = ∑ k = i , j J ik x k b = ∑ k ≠ i , j J j k x k b=\sum_{k\ne i,j}J_{jk}x_k b = ∑ k = i , j J j k x k i i i j j j x − i − j \mathbf{x}_{-i-j} x − i − j E μ J i j , a , b [ L i f ⋅ L j f ] \mathbb{E}_{\mu_{J_{ij},a,b}}[L_if\cdot L_jf] E μ J ij , a , b [ L i f ⋅ L j f ]
E [ ( L i f ) ( L j f ) ] + J i j ⋅ E [ m i ( x i , x j ) m j ( x i , x j ) ( L i f ) ( L j f ) ] + ( λ J i j ) 2 2 ⋅ E [ ( L i f ) 2 + ( L j f ) 2 ] + d J i j 2 2 E [ X ‾ i ( x j ) 2 ⋅ L i f ( x i , x j ) 2 + X ‾ j ( x i ) 2 ⋅ L j f ( x i , x j ) 2 ] ≥ 0. (12) \begin{aligned}
\mathbb{E}[(L_if)(L_jf)]+J_{ij}\cdot\mathbb{E}[m_i(x_i,x_j)m_j(x_i,x_j)(L_if)(L_jf)]+\frac{(\lambda J_{ij})^2}{2}\cdot\mathbb{E}[(L_if)^2+(L_jf)^2] \\
&+\frac{dJ_{ij}^2}{2}\mathbb{E}\left[\overline{X}_i(x_j)^2\cdot L_if(x_i,x_j)^2+\overline{X}_j(x_i)^2\cdot L_jf(x_i,x_j)^2\right]\geq0. \tag*{(12)}
\end{aligned} E [( L i f ) ( L j f )] + J ij ⋅ E [ m i ( x i , x j ) m j ( x i , x j ) ( L i f ) ( L j f )] + 2 ( λ J ij ) 2 ⋅ E [( L i f ) 2 + ( L j f ) 2 ] + 2 d J ij 2 E [ X i ( x j ) 2 ⋅ L i f ( x i , x j ) 2 + X j ( x i ) 2 ⋅ L j f ( x i , x j ) 2 ] ≥ 0. (12) where X ‾ i \overline{X}_i X i X i X_i X i x − i \mathbf{x}_{-i} x − i λ = o ( 1 ) \lambda=o(1) λ = o ( 1 ) J i j J_{ij} J ij d , s d,s d , s m ( x ) \mathbf{m}(x) m ( x ) m i ( x ) = x i + s X ‾ i m_i(x)=x_i+s\overline{X}_i m i ( x ) = x i + s X i
In this case we write
μ J i j , a , b ( x i , x j ) ( 1 + J ‾ i j X ‾ i , 0 X ‾ j , 0 ) = Lemma 2.5 μ 0 , a , b ( x i , x j ) ( 1 + J ‾ i j x i x j ) , \mu_{J_{ij},a,b}(x_i,x_j)\left(1+\overline{J}_{ij}\overline{X}_{i,0}\overline{X}_{j,0}\right) \overset{\text{Lemma 2.5}}{=} \mu_{0,a,b}(x_i,x_j)\left(1+\overline{J}_{ij}x_i x_j\right), μ J ij , a , b ( x i , x j ) ( 1 + J ij X i , 0 X j , 0 ) = Lemma 2.5 μ 0 , a , b ( x i , x j ) ( 1 + J ij x i x j ) , where X ‾ i , 0 = E μ 0 , a , b [ X i ] = tanh a \overline{X}_{i,0}=\mathbb{E}_{\mu_{0,a,b}}[X_i]=\tanh a X i , 0 = E μ 0 , a , b [ X i ] = tanh a X ‾ i , 0 = 0 \overline{X}_{i,0}=0 X i , 0 = 0 a = 0 a=0 a = 0
( 1 + J ‾ i j X ‾ i , 0 x j ) L i f ( x i , x j ) = ( 1 − J ‾ i j x i x j ) L i 0 f ( x i , x j ) . \left(1+\overline{J}_{ij}\overline{X}_{i,0}x_j\right)L_i f(x_i,x_j)=\left(1-\overline{J}_{ij}x_i x_j\right)L_i^0 f(x_i,x_j). ( 1 + J ij X i , 0 x j ) L i f ( x i , x j ) = ( 1 − J ij x i x j ) L i 0 f ( x i , x j ) . Now, we run a calculation similar to (9), which gives
E μ J i j , a , b [ ( L i f ) ( L j f ) ] = E μ 0 , a , b [ ( 1 − J ‾ i j 2 ) 1 − J ‾ i j X i X j ( 1 + J ‾ i j X ‾ i , 0 X ‾ j , 0 ) ( 1 + J ‾ i j X ‾ i , 0 X j ) ( 1 + J ‾ i j X ‾ j , 0 X i ) ( L i 0 f ) ( L j 0 f ) ] . \mathbb{E}_{\mu_{J_{ij},a,b}}\left[(L_i f)(L_j f)\right]=\mathbb{E}_{\mu_{0,a,b}}\left[\left(1-\overline{J}_{ij}^{\,2}\right)\frac{1-\overline{J}_{ij}X_iX_j}{\left(1+\overline{J}_{ij}\overline{X}_{i,0}\overline{X}_{j,0}\right)\left(1+\overline{J}_{ij}\overline{X}_{i,0}X_j\right)\left(1+\overline{J}_{ij}\overline{X}_{j,0}X_i\right)}(L_i^0f)(L_j^0f)\right]. E μ J ij , a , b [ ( L i f ) ( L j f ) ] = E μ 0 , a , b [ ( 1 − J ij 2 ) ( 1 + J ij X i , 0 X j , 0 ) ( 1 + J ij X i , 0 X j ) ( 1 + J ij X j , 0 X i ) 1 − J ij X i X j ( L i 0 f ) ( L j 0 f ) ] . This suggests that we should multiply the LHS with ( 1 + J ‾ i j X i X j ) ( 1 + J ‾ i j X ‾ i , 0 X ‾ j , 0 ) ( 1 + J ‾ i j X ‾ i , 0 X j ) ( 1 + J ‾ i j X ‾ j , 0 X i ) \left(1+\overline{J}_{ij}X_iX_j\right)\left(1+\overline{J}_{ij}\overline{X}_{i,0}\overline{X}_{j,0}\right)\left(1+\overline{J}_{ij}\overline{X}_{i,0}X_j\right)\left(1+\overline{J}_{ij}\overline{X}_{j,0}X_i\right) ( 1 + J ij X i X j ) ( 1 + J ij X i , 0 X j , 0 ) ( 1 + J ij X i , 0 X j ) ( 1 + J ij X j , 0 X i )
E μ J i j , a , b [ ( 1 + ( X i + X ‾ i ) ( X j + X ‾ j ) + O ( J ‾ i j 2 ) ) ( L i f ) ( L j f ) ] ≥ ∣ J ‾ i j ∣ ≤ 1 / 2 E μ J i j , a , b [ ( 1 + J ‾ i j X i X j ) ( 1 + J ‾ i j X ‾ i , 0 X ‾ j , 0 ) ( 1 + J ‾ i j X ‾ i , 0 X j ) ( 1 + J ‾ i j X ‾ j , 0 X i ) ( L i f ) ( L j f ) ] = ( 1 − J ‾ i j 2 ) 2 E μ 0 [ L i 0 f ] E μ 0 [ L j 0 f ] ≥ 0 , \begin{aligned}
\mathbb{E}_{\mu_{J_{ij},a,b}}\left[\left(1+(X_i+\overline{X}_i)(X_j+\overline{X}_j)+O(\overline{J}_{ij}^{\,2})\right)(L_i f)(L_j f)\right]
&\underset{|\overline{J}_{ij}|\leq1/2}{\geq}
\mathbb{E}_{\mu_{J_{ij},a,b}}\left[\left(1+\overline{J}_{ij}X_iX_j\right)\left(1+\overline{J}_{ij}\overline{X}_{i,0}\overline{X}_{j,0}\right)\left(1+\overline{J}_{ij}\overline{X}_{i,0}X_j\right)\left(1+\overline{J}_{ij}\overline{X}_{j,0}X_i\right)(L_i f)(L_j f)\right]\\
&=\left(1-\overline{J}_{ij}^{\,2}\right)^2\mathbb{E}_{\mu_0}[L_i^0f]\mathbb{E}_{\mu_0}[L_j^0f]\geq0,
\end{aligned} E μ J ij , a , b [ ( 1 + ( X i + X i ) ( X j + X j ) + O ( J ij 2 ) ) ( L i f ) ( L j f ) ] ∣ J ij ∣ ≤ 1/2 ≥ E μ J ij , a , b [ ( 1 + J ij X i X j ) ( 1 + J ij X i , 0 X j , 0 ) ( 1 + J ij X i , 0 X j ) ( 1 + J ij X j , 0 X i ) ( L i f ) ( L j f ) ] = ( 1 − J ij 2 ) 2 E μ 0 [ L i 0 f ] E μ 0 [ L j 0 f ] ≥ 0 , where the last inequality similarly follows from μ 0 \mu_0 μ 0
Using AM-GM the above rearranges to the following analogous inequality as (7)
E μ J i j , a , b [ ( L i f ) ( L j f ) ∣ X − i − j ] ≥ − J ‾ i j E μ J i j [ ( X i + X ‾ i ( X j ) ) L i f ⋅ ( X j + X ‾ j ( X i ) ) L j f ∣ X − i − j ] − O ( J ‾ i j 2 ) E μ J i j [ ( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] . \mathbb{E}_{\mu_{J_{ij},a,b}}\left[(L_i f)(L_j f)\mid X_{-i-j}\right]\geq-\overline{J}_{ij}\mathbb{E}_{\mu_{J_{ij}}}\left[(X_i+\overline{X}_i(X_j))L_i f\cdot(X_j+\overline{X}_j(X_i))L_j f\mid X_{-i-j}\right]-O(\overline{J}_{ij}^{\,2})\mathbb{E}_{\mu_{J_{ij}}}\left[(L_i f)^2+(L_j f)^2\mid X_{-i-j}\right]. E μ J ij , a , b [ ( L i f ) ( L j f ) ∣ X − i − j ] ≥ − J ij E μ J ij [ ( X i + X i ( X j )) L i f ⋅ ( X j + X j ( X i )) L j f ∣ X − i − j ] − O ( J ij 2 ) E μ J ij [ ( L i f ) 2 + ( L j f ) 2 ∣ X − i − j ] . In comparison to the earlier computations in (7), the error terms here are of order O ( J ‾ i j 2 ) O(\overline{J}_{ij}^{\,2}) O ( J ij 2 ) O ( J ‾ i j 3 ) O(\overline{J}_{ij}^{\,3}) O ( J ij 3 ) ∑ j ≠ i J ‾ i j 2 = β 2 + o ( 1 ) \sum_{j\ne i}\overline{J}_{ij}^{\,2}=\beta^2+o(1) ∑ j = i J ij 2 = β 2 + o ( 1 ) i ∈ [ n ] i\in[n] i ∈ [ n ] ∑ j ≠ i ∣ J ‾ i j ∣ 3 = o ( 1 ) \sum_{j\ne i}|\overline{J}_{ij}|^3=o(1) ∑ j = i ∣ J ij ∣ 3 = o ( 1 ) i ∈ [ n ] i\in[n] i ∈ [ n ] O ( ∣ J ‾ i j ∣ 3 ) O(|\overline{J}_{ij}|^3) O ( ∣ J ij ∣ 3 )
For this reason, we work with a slightly different rank one term in Lemma 3.1, where we take m i ( x ) = x i + s x ‾ i m_i(x)=x_i+s\overline{x}_i m i ( x ) = x i + s x i s < 1 s<1 s < 1 x i + x ‾ i x_i+\overline{x}_i x i + x i ( 1 − s ) x ‾ i (1-s)\overline{x}_i ( 1 − s ) x i O ( J ‾ i j 2 ) O(\overline{J}_{ij}^{\,2}) O ( J ij 2 )
β < 1 / 2 \beta<1/2 β < 1/2 i , j i,j i , j
⟨ L f , L f ⟩ ≥ ( 1 − ∥ J ∥ op − o n ( 1 ) ) E ( f ) + s ( 2 − s ) ∥ J ∥ op ∑ i ∣ X ‾ i ∣ 2 ( L i f ) 2 ⏟ (A) + E [ w ⊤ ( ∥ J ∥ op I − J ) w ] ⏟ (B) . \langle Lf,Lf\rangle\geq\left(1-\lVert J\rVert_{\operatorname{op}}-o_n(1)\right)\mathcal{E}(f)+\underbrace{s(2-s)\lVert J\rVert_{\operatorname{op}}\sum_i|\overline{X}_i|^2(L_i f)^2}_{\text{(A)}}+\underbrace{\mathbb{E}\left[w^\top\left(\lVert J\rVert_{\operatorname{op}}I-J\right)w\right]}_{\text{(B)}}. ⟨ L f , L f ⟩ ≥ ( 1 − ∥ J ∥ op − o n ( 1 ) ) E ( f ) + (A) s ( 2 − s ) ∥ J ∥ op i ∑ ∣ X i ∣ 2 ( L i f ) 2 + (B) E [ w ⊤ ( ∥ J ∥ op I − J ) w ] . At this point, we note that A , B ≥ 0 A,B \ge0 A , B ≥ 0 ∥ J ∥ op → 2 β + o n ( 1 ) \lVert J\rVert_{\operatorname{op}} \to2\beta+ o_n(1) ∥ J ∥ op → 2 β + o n ( 1 ) ε / n \varepsilon/n ε / n β < 1 2 − ε \beta< \frac{1}{2}-\varepsilon β < 2 1 − ε β < 1 2 − ε \beta< \frac{1}{2}-\varepsilon β < 2 1 − ε J J J ∥ J ∥ op = 2 β \lVert J\rVert_{\operatorname{op}}=2\beta ∥ J ∥ op = 2 β J i j = β / n J_{ij}=\beta/n J ij = β / n i ≠ j i\ne j i = j J i i = 0 J_{ii}=0 J ii = 0 i ∈ [ n ] i\in[n] i ∈ [ n ] β = 1 / 2 \beta=1/2 β = 1/2 Ω ( 1 / n ) \Omega(1/n) Ω ( 1/ n ) [16 , 24 ] 1 β > 1 / 2 \beta>1/2 β > 1/2 exp ( − Ω ( n ) ) \exp(-\Omega(n)) exp ( − Ω ( n ))
β < 1 / 2 + ε \beta< 1/2+\varepsilon β < 1/2 + ε β ∈ [ 1 / 2 − ε , 1 / 2 + ε ) \beta\in[1/2-\varepsilon,1/2+\varepsilon) β ∈ [ 1/2 − ε , 1/2 + ε ) ∥ J ∥ op → 2 β + o n ( 1 ) \lVert J\rVert_{\operatorname{op}}\to2\beta+o_n(1) ∥ J ∥ op → 2 β + o n ( 1 ) A ≳ E ( f ) A\gtrsim\mathcal{E}(f) A ≳ E ( f ) B ≳ E ( f ) B\gtrsim\mathcal{E}(f) B ≳ E ( f )
To that end, we use two properties of GOE random matrices. First, that their eigenvectors are delocalized, namely that the mass of the eigenvector is spread across all coordinates i ∈ [ n ] i\in[n] i ∈ [ n ] x ∈ { ± 1 } n x\in\{\pm1\}^{n} x ∈ { ± 1 } n ( 1 − δ ) (1-\delta) ( 1 − δ ) δ \delta δ i ∈ [ n ] i\in[n] i ∈ [ n ] X ‾ i = tanh ( ∑ k ≠ i J i k x k ) \overline{X}_i=\tanh\left(\sum_{k\ne i}J_{ik}x_k\right) X i = tanh ( ∑ k = i J ik x k ) H > H_> H > X ‾ i \overline{X}_i X i i i i
As such, we consider the following two cases:
(ii) If f f f L i f L_i f L i f i ∈ H > i\in H_> i ∈ H > w i ≈ L i f w_i\approx L_i f w i ≈ L i f w w w δ n \delta n δ n i ∉ H > i\notin H_> i ∈ / H > w w w J J J
B ≳ ∥ w ∥ 2 2 ≳ ∑ i = 1 n ( L i f ) 2 ≳ 2 ε ⋅ E ( f ) ( 1 ) . B \gtrsim\lVert w\rVert_2^2 \gtrsim\sum_{i=1}^{n}(L_i f)^2 \gtrsim2\varepsilon\cdot\mathcal{E}(f)_{(1)}. B ≳ ∥ w ∥ 2 2 ≳ i = 1 ∑ n ( L i f ) 2 ≳ 2 ε ⋅ E ( f ) ( 1 ) . (i) Suppose B ≈ 0 B\approx0 B ≈ 0 w w w J J J w w w w i = ( X i + X ‾ i ) ⋅ L i f w_i=(X_i+\overline{X}_i)\cdot L_i f w i = ( X i + X i ) ⋅ L i f L i f L_i f L i f
A = s ( 2 − s ) ∥ J ∥ op ∑ i ∣ X ‾ i ∣ 2 ( L i f ) 2 ≳ ( 1 n ∑ i ∈ H > ∣ X ‾ i ∣ 2 ) ∑ i ( L i f ) 2 ≳ 2 ε ⋅ E ( f ) ( 1 ) . A=s(2-s)\lVert J\rVert_{\operatorname{op}}\sum_i\lvert\overline{X}_i\rvert^2(L_i f)^2
\gtrsim\left(\frac{1}{n}\sum_{i\in H_>}\lvert\overline{X}_i\rvert^2\right)\sum_i(L_i f)^2
\gtrsim2\varepsilon\cdot\mathcal{E}(f)_{(1)}. A = s ( 2 − s ) ∥ J ∥ op i ∑ ∣ X i ∣ 2 ( L i f ) 2 ≳ ( n 1 i ∈ H > ∑ ∣ X i ∣ 2 ) i ∑ ( L i f ) 2 ≳ 2 ε ⋅ E ( f ) ( 1 ) . We refer the reader to Section 3 where we carry out this dichotomy to prove a spectral gap for L \mathcal{L} L β ≤ 1 2 + 5 ⋅ 10 − 5 \beta\le\frac{1}{2}+5\cdot10^{-5} β ≤ 2 1 + 5 ⋅ 1 0 − 5 L \mathcal{L} L β \beta β
Preliminaries
Notation and basic facts
We will use capital letters to denote random variables and bold letters to denote vectors. For a vector x ∈ R n \mathbf{x} \in\mathbb{R}^n x ∈ R n
∥ x ∥ ≤ m 2 : = min S ⊆ [ n ] , ∣ S ∣ = m ∑ i ∈ S x i 2 \|\mathbf{x}\|_{\leq m}^{2} := \min_{\substack{S \subseteq[n], |S|=m}} \sum_{i \in S} x_i^2 ∥ x ∥ ≤ m 2 := S ⊆ [ n ] , ∣ S ∣ = m min i ∈ S ∑ x i 2 We say that the matrix J ∈ R n × n J \in\mathbb{R}^{n \times n} J ∈ R n × n β \beta β J J J J i j ∼ i . i . d . N ( 0 , β 2 / n ) J_{ij} \overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0,\beta^2/n) J ij ∼ i.i.d. N ( 0 , β 2 / n ) i ≠ j i \ne j i = j J i i = 0 J_{ii}=0 J ii = 0 i ∈ [ n ] i \in[n] i ∈ [ n ] Φ ( x ) \Phi(x) Φ ( x )
Lemma 2.1. Let C ⪰ 0 C \succeq0 C ⪰ 0 A ∈ R n × n A \in\mathbb{R}^{n \times n} A ∈ R n × n
A ∙ C ≤ λ max ( A ) Tr ( C ) . A \mathbin{\bullet} C \leq\lambda_{\max}(A) \operatorname{Tr}(C). A ∙ C ≤ λ m a x ( A ) Tr ( C ) . Proof. Since λ max ( A ) I − A ⪰ 0 \lambda_{\max}(A)I-A \succeq0 λ m a x ( A ) I − A ⪰ 0 C 1 / 2 ( λ max ( A ) I − A ) C 1 / 2 ⪰ 0 C^{1/2}(\lambda_{\max}(A)I-A)C^{1/2} \succeq0 C 1/2 ( λ m a x ( A ) I − A ) C 1/2 ⪰ 0
0 ≤ Tr ( C 1 / 2 ( λ max ( A ) I − A ) C 1 / 2 ) = Tr ( ( λ max ( A ) I − A ) C ) = λ max ( A ) Tr ( C ) − A ∙ C . 0 \leq\operatorname{Tr}\left(C^{1/2}(\lambda_{\max}(A)I-A)C^{1/2}\right)
= \operatorname{Tr}((\lambda_{\max}(A)I-A)C)
= \lambda_{\max}(A)\operatorname{Tr}(C)-A \mathbin{\bullet} C. 0 ≤ Tr ( C 1/2 ( λ m a x ( A ) I − A ) C 1/2 ) = Tr (( λ m a x ( A ) I − A ) C ) = λ m a x ( A ) Tr ( C ) − A ∙ C . Fact 2.2. Let X ∈ { − 1 , 1 } X \in\{-1,1\} X ∈ { − 1 , 1 } X ‾ \overline{X} X f : { − 1 , 1 } → R f:\{-1,1\}\to\mathbb{R} f : { − 1 , 1 } → R L L L L f ( x ) = f ( x ) − E X f ( X ) Lf(x)=f(x)-\mathbb{E}_{X}f(X) L f ( x ) = f ( x ) − E X f ( X ) κ \kappa κ f f f
L f ( X ) = κ ⋅ ( X − X ‾ ) . Lf(X)=\kappa\mathbin{\cdot}(X-\overline{X}). L f ( X ) = κ ⋅ ( X − X ) . Proof. In fact, we claim that we may take
κ = f ( 1 ) − f ( − 1 ) 2 . \kappa=\frac{f(1)-f(-1)}{2}. κ = 2 f ( 1 ) − f ( − 1 ) . We also define c = κ + f ( − 1 ) c=\kappa+f(-1) c = κ + f ( − 1 ) f ( x ) = c + κ x f(x)=c+\kappa x f ( x ) = c + κ x x ∈ { − 1 , 1 } x\in\{-1,1\} x ∈ { − 1 , 1 }
L f ( X ) = f ( X ) − E [ f ( X ) ] = ( c + κ X ) − ( c + κ X ‾ ) = κ ( X − X ‾ ) . Lf(X)=f(X)-\mathbb{E}[f(X)]=(c+\kappa X)-(c+\kappa\overline{X})=\kappa(X-\overline{X}). L f ( X ) = f ( X ) − E [ f ( X )] = ( c + κ X ) − ( c + κ X ) = κ ( X − X ) . Ising measure
In this section, we collect some basic facts about the two-state Ising measure: let J , a , b ∈ R J,a,b\in\mathbb{R} J , a , b ∈ R μ J , a , b ( x , y ) = 1 Z exp ( J x y + a x + b y ) \mu_{J,a,b}(x,y)=\frac{1}{Z}\exp(Jxy+ax+by) μ J , a , b ( x , y ) = Z 1 exp ( J x y + a x + b y ) x , y ∈ { − 1 , 1 } x,y\in\{-1,1\} x , y ∈ { − 1 , 1 }
At J = 0 J=0 J = 0
μ 0 , a , b ( x , y ) = exp ( a x ) 2 cosh a ⋅ exp ( b y ) 2 cosh b . \mu_{0,a,b}(x,y)=\frac{\exp(ax)}{2\cosh a}\cdot\frac{\exp(by)}{2\cosh b}. μ 0 , a , b ( x , y ) = 2 cosh a exp ( a x ) ⋅ 2 cosh b exp ( b y ) . Because of this factorization, it follows that
E μ 0 , a , b [ X ] = tanh a = : X ‾ 0 , E μ 0 , a , b [ Y ] = tanh b = : Y ‾ 0 and E μ 0 , a , b [ X Y ] = X ‾ 0 Y ‾ 0 . (13) \mathbb{E}_{\mu_{0,a,b}}[X]=\tanh a=:\overline{X}_0,\qquad
\mathbb{E}_{\mu_{0,a,b}}[Y]=\tanh b=:\overline{Y}_0
\quad\text{and}\quad
\mathbb{E}_{\mu_{0,a,b}}[XY]=\overline{X}_0\overline{Y}_0. \tag*{(13)} E μ 0 , a , b [ X ] = tanh a =: X 0 , E μ 0 , a , b [ Y ] = tanh b =: Y 0 and E μ 0 , a , b [ X Y ] = X 0 Y 0 . (13) For simplicity, we define J ‾ = tanh J \overline{J}=\tanh J J = tanh J
σ X 0 = 1 − X ‾ 0 2 and σ Y 0 = 1 − Y ‾ 0 2 . \sigma_{X_0}=\sqrt{1-\overline{X}_0^2}
\quad\text{and}\quad
\sigma_{Y_0}=\sqrt{1-\overline{Y}_0^2}. σ X 0 = 1 − X 0 2 and σ Y 0 = 1 − Y 0 2 . We also define
X ‾ ( y ) = E [ X ∣ y ] and Y ‾ ( x ) = E [ Y ∣ x ] . \overline{X}(y)=\mathbb{E}[X\mid y]\quad\text{and}\quad
\overline{Y}(x)=\mathbb{E}[Y\mid x]. X ( y ) = E [ X ∣ y ] and Y ( x ) = E [ Y ∣ x ] . Lemma 2.3. For μ = μ J , a , b \mu=\mu_{J,a,b} μ = μ J , a , b
μ ( x , y ) = ( 1 + X ‾ 0 x ) ( 1 + Y ‾ 0 y ) ( 1 + J ‾ x y ) 4 ( 1 + J ‾ X ‾ 0 Y ‾ 0 ) , X ‾ ( y ) = X ‾ 0 + J ‾ y 1 + X ‾ 0 J ‾ y and Y ‾ ( x ) = Y ‾ 0 + J ‾ x 1 + Y ‾ 0 J ‾ x . \mu(x,y)=\frac{(1+\overline{X}_0x)(1+\overline{Y}_0y)(1+\overline{J}xy)}{4(1+\overline{J}\overline{X}_0\overline{Y}_0)},\qquad\overline{X}(y)=\frac{\overline{X}_0+\overline{J}y}{1+\overline{X}_0\overline{J}y}\qquad\text{and}\qquad\overline{Y}(x)=\frac{\overline{Y}_0+\overline{J}x}{1+\overline{Y}_0\overline{J}x}. μ ( x , y ) = 4 ( 1 + J X 0 Y 0 ) ( 1 + X 0 x ) ( 1 + Y 0 y ) ( 1 + J x y ) , X ( y ) = 1 + X 0 J y X 0 + J y and Y ( x ) = 1 + Y 0 J x Y 0 + J x . Proof. First, for x ∈ { ± 1 } x\in\{\pm1\} x ∈ { ± 1 }
exp ( c x ) = cosh ( c ) + x sinh ( c ) = cosh ( c ) ( 1 + x tanh ( c ) ) . (14) \exp(cx)=\cosh(c)+x\sinh(c)=\cosh(c)(1+x\tanh(c)). \tag*{(14)} exp ( c x ) = cosh ( c ) + x sinh ( c ) = cosh ( c ) ( 1 + x tanh ( c )) . (14) Recall that X ‾ 0 = tanh ( a ) \overline{X}_0=\tanh(a) X 0 = tanh ( a ) Y ‾ 0 = tanh ( b ) \overline{Y}_0=\tanh(b) Y 0 = tanh ( b )
exp ( a x ) = cosh ( a ) ( 1 + X ‾ 0 x ) , exp ( b y ) = cosh ( b ) ( 1 + Y ‾ 0 y ) and exp ( J ‾ x y ) = cosh ( J ‾ ) ( 1 + J ‾ x y ) . \exp(ax)=\cosh(a)(1+\overline{X}_0x),\qquad\exp(by)=\cosh(b)(1+\overline{Y}_0y)\qquad\text{and}\qquad\exp(\overline{J}xy)=\cosh(\overline{J})(1+\overline{J}xy). exp ( a x ) = cosh ( a ) ( 1 + X 0 x ) , exp ( b y ) = cosh ( b ) ( 1 + Y 0 y ) and exp ( J x y ) = cosh ( J ) ( 1 + J x y ) . The numerator of μ ( x , y ) \mu(x,y) μ ( x , y ) cosh ( a ) cosh ( b ) cosh ( J ‾ ) ( 1 + X ‾ 0 x ) ( 1 + Y ‾ 0 y ) ( 1 + J ‾ x y ) \cosh(a)\cosh(b)\cosh(\overline{J})(1+\overline{X}_0x)(1+\overline{Y}_0y)(1+\overline{J}xy) cosh ( a ) cosh ( b ) cosh ( J ) ( 1 + X 0 x ) ( 1 + Y 0 y ) ( 1 + J x y )
∑ x , y ∈ { ± 1 } ( 1 + X ‾ 0 x ) ( 1 + Y ‾ 0 y ) ( 1 + J ‾ x y ) = 4 ( 1 + J ‾ X ‾ 0 Y ‾ 0 ) , \sum_{x,y\in\{\pm1\}}(1+\overline{X}_0x)(1+\overline{Y}_0y)(1+\overline{J}xy)=4(1+\overline{J}\overline{X}_0\overline{Y}_0), x , y ∈ { ± 1 } ∑ ( 1 + X 0 x ) ( 1 + Y 0 y ) ( 1 + J x y ) = 4 ( 1 + J X 0 Y 0 ) , it follows that
μ ( x , y ) = ( 1 + X ‾ 0 x ) ( 1 + Y ‾ 0 y ) ( 1 + J ‾ x y ) 4 ( 1 + J ‾ X ‾ 0 Y ‾ 0 ) . \mu(x,y)=\frac{(1+\overline{X}_0x)(1+\overline{Y}_0y)(1+\overline{J}xy)}{4(1+\overline{J}\overline{X}_0\overline{Y}_0)}. μ ( x , y ) = 4 ( 1 + J X 0 Y 0 ) ( 1 + X 0 x ) ( 1 + Y 0 y ) ( 1 + J x y ) . Next, we also have
∑ x ∈ { ± 1 } ( 1 + X ‾ 0 x ) ( 1 + J ‾ x y ) = 2 ( 1 + X ‾ 0 J ‾ y ) , \sum_{x\in\{\pm1\}}(1+\overline{X}_0x)(1+\overline{J}xy)=2(1+\overline{X}_0\overline{J}y), x ∈ { ± 1 } ∑ ( 1 + X 0 x ) ( 1 + J x y ) = 2 ( 1 + X 0 J y ) , so that
X ‾ ( y ) = ∑ x ∈ { ± 1 } ( 1 + X ‾ 0 x ) ( 1 + J ‾ x y ) 2 ( 1 + X ‾ 0 J ‾ y ) ⋅ x = X ‾ 0 + J ‾ y 1 + X ‾ 0 J ‾ y . \overline{X}(y)=\sum_{x\in\{\pm1\}}\frac{(1+\overline{X}_0x)(1+\overline{J}xy)}{2(1+\overline{X}_0\overline{J}y)}\cdot x=\frac{\overline{X}_0+\overline{J}y}{1+\overline{X}_0\overline{J}y}. X ( y ) = x ∈ { ± 1 } ∑ 2 ( 1 + X 0 J y ) ( 1 + X 0 x ) ( 1 + J x y ) ⋅ x = 1 + X 0 J y X 0 + J y . Symmetrically we get the result for Y ‾ ( x ) \overline{Y}(x) Y ( x )
Taylor expansion of two-state Ising model around zero interaction
We do a Taylor expansion of the two-state spin system for the Ising measure and various functions of the underlying spins, when the interaction J ‾ \overline{J} J 0 0 0
Lemma 2.4. For all x ∈ { ± 1 } x\in\{\pm1\} x ∈ { ± 1 } y ∈ { ± 1 } y\in\{\pm1\} y ∈ { ± 1 } ∣ J ‾ ∣ ≤ 1 2 |\overline{J}|\le\frac{1}{2} ∣ J ∣ ≤ 2 1 X ‾ ( y ) = X ‾ 0 + J ‾ σ X ‾ 0 2 y + O ( ∣ J ‾ ∣ 2 ) \overline{X}(y)=\overline{X}_0+\overline{J}\sigma^2_{\overline{X}_0}y+O(|\overline{J}|^2) X ( y ) = X 0 + J σ X 0 2 y + O ( ∣ J ∣ 2 ) Y ‾ ( x ) = Y ‾ 0 + J ‾ σ Y ‾ 0 2 x ± O ( ∣ J ‾ ∣ 2 ) \overline{Y}(x)=\overline{Y}_0+\overline{J}\sigma^2_{\overline{Y}_0}x\pm O(|\overline{J}|^2) Y ( x ) = Y 0 + J σ Y 0 2 x ± O ( ∣ J ∣ 2 )
Proof. By the expression that we have computed in Claim 2.3, it follows that
∣ X ‾ ( y ) − X ‾ 0 − J ‾ σ X ‾ 0 2 y ∣ = ∣ X ‾ 0 + J ‾ y 1 + J ‾ X ‾ 0 y − X ‾ 0 − J ‾ ( 1 − X ‾ 0 2 ) y ∣ = ∣ J ‾ y ( 1 − X ‾ 0 2 ) ( 1 − ( 1 + J ‾ X ‾ 0 y ) ) 1 + J ‾ X ‾ 0 y ∣ = ∣ J ‾ 2 y 2 ( 1 − X ‾ 0 2 ) X ‾ 0 1 + J ‾ X ‾ 0 y ∣ ≤ ∣ y ∣ , ∣ X ‾ 0 ∣ ≤ 1 , J ‾ ≤ 1 / 2 2 J ‾ 2 \begin{aligned}
\left|\overline{X}(y)-\overline{X}_0-\overline{J}\sigma^2_{\overline{X}_0}y\right|=\left|\frac{\overline{X}_0+\overline{J}y}{1+\overline{J}\overline{X}_0y}-\overline{X}_0-\overline{J}(1-\overline{X}_0^2)y\right| \\
&=\left|\frac{\overline{J}y(1-\overline{X}_0^2)(1-(1+\overline{J}\overline{X}_0y))}{1+\overline{J}\overline{X}_0y}\right| \\
&=\left|\frac{\overline{J}^2y^2(1-\overline{X}_0^2)\overline{X}_0}{1+\overline{J}\overline{X}_0y}\right|\underset{|y|,|\overline{X}_0|\le1,\overline{J}\le1/2}{\le}2\overline{J}^2
\end{aligned} X ( y ) − X 0 − J σ X 0 2 y = 1 + J X 0 y X 0 + J y − X 0 − J ( 1 − X 0 2 ) y = 1 + J X 0 y J y ( 1 − X 0 2 ) ( 1 − ( 1 + J X 0 y )) = 1 + J X 0 y J 2 y 2 ( 1 − X 0 2 ) X 0 ∣ y ∣ , ∣ X 0 ∣ ≤ 1 , J ≤ 1/2 ≤ 2 J 2 for small enough ∣ J ˉ ∣ |\bar{J}| ∣ J ˉ ∣ X ‾ ( y ) = X ‾ 0 + J ˉ σ X 0 2 y + O ( ∣ J ˉ ∣ 2 ) \overline{X}(y)=\overline{X}_0+\bar{J}\sigma_{X_0}^2y+O(|\bar{J}|^2) X ( y ) = X 0 + J ˉ σ X 0 2 y + O ( ∣ J ˉ ∣ 2 ) Y ‾ ( x ) \overline{Y}(x) Y ( x )
Lemma 2.5. For ∣ J ˉ ∣ ≤ 1 / 2 |\bar{J}|\le1/2 ∣ J ˉ ∣ ≤ 1/2
μ J , a , b ( x , y ) μ 0 , a , b ( x , y ) = 1 + J ˉ x y 1 + J ˉ X ‾ 0 Y ‾ 0 = 1 + J ˉ ( x y − X ‾ 0 Y ‾ 0 ) + μ J , a , b R ( x , y ) \frac{\mu_{J,a,b}(x,y)}{\mu_{0,a,b}(x,y)}=\frac{1+\bar{J}xy}{1+\bar{J}\overline{X}_0\overline{Y}_0}=1+\bar{J}(xy-\overline{X}_0\overline{Y}_0)+\mu^R_{J,a,b}(x,y) μ 0 , a , b ( x , y ) μ J , a , b ( x , y ) = 1 + J ˉ X 0 Y 0 1 + J ˉ x y = 1 + J ˉ ( x y − X 0 Y 0 ) + μ J , a , b R ( x , y ) where ∣ μ J , a , b R ( x , y ) ∣ = O ( J ˉ 2 ) |\mu^R_{J,a,b}(x,y)|=O(\bar{J}^2) ∣ μ J , a , b R ( x , y ) ∣ = O ( J ˉ 2 ) x , y ∈ { ± 1 } x,y\in\{\pm1\} x , y ∈ { ± 1 }
Proof. We first prove the first equality. By an analysis similar to (13), we have
exp ( J ˉ x y ) = ( cosh J ) ⋅ ( 1 + J ˉ x y ) . \exp(\bar{J}xy)=(\cosh J)\cdot(1+\bar{J}xy). exp ( J ˉ x y ) = ( cosh J ) ⋅ ( 1 + J ˉ x y ) . In particular, this means that we may write
Z J , a , b = cosh J ∑ x , y ∈ { ± 1 } exp ( a x + b y ) ( 1 + J ˉ x y ) = ( cosh J ) Z 0 , a , b E μ 0 , a , b ( 1 + J ˉ x y ) = ( cosh J ) Z 0 , a , b ( 1 + J ˉ X ‾ 0 Y ‾ 0 ) . \begin{aligned}
Z_{J,a,b}&=\cosh J\sum_{x,y\in\{\pm1\}}\exp(ax+by)(1+\bar{J}xy)\\
&=(\cosh J)Z_{0,a,b}\mathbb{E}_{\mu_{0,a,b}}(1+\bar{J}xy)\\
&=(\cosh J)Z_{0,a,b}(1+\bar{J}\overline{X}_0\overline{Y}_0).
\end{aligned} Z J , a , b = cosh J x , y ∈ { ± 1 } ∑ exp ( a x + b y ) ( 1 + J ˉ x y ) = ( cosh J ) Z 0 , a , b E μ 0 , a , b ( 1 + J ˉ x y ) = ( cosh J ) Z 0 , a , b ( 1 + J ˉ X 0 Y 0 ) . Consequently, it follows that
μ J , a , b ( x , y ) = exp ( J ˉ x y + a x + b y ) Z J , a , b = ( cosh J ) ( 1 + J ˉ x y ) exp ( a x + b y ) Z 0 , a , b ( cosh J ) ( 1 + J ˉ X ‾ 0 Y ‾ 0 ) = μ 0 , a , b ( x , y ) ( 1 + J ˉ x y ) 1 + J ˉ X ‾ 0 Y ‾ 0 . \mu_{J,a,b}(x,y)=\frac{\exp(\bar{J}xy+ax+by)}{Z_{J,a,b}}=\frac{(\cosh J)(1+\bar{J}xy)\exp(ax+by)}{Z_{0,a,b}(\cosh J)(1+\bar{J}\overline{X}_0\overline{Y}_0)}=\frac{\mu_{0,a,b}(x,y)(1+\bar{J}xy)}{1+\bar{J}\overline{X}_0\overline{Y}_0}. μ J , a , b ( x , y ) = Z J , a , b exp ( J ˉ x y + a x + b y ) = Z 0 , a , b ( cosh J ) ( 1 + J ˉ X 0 Y 0 ) ( cosh J ) ( 1 + J ˉ x y ) exp ( a x + b y ) = 1 + J ˉ X 0 Y 0 μ 0 , a , b ( x , y ) ( 1 + J ˉ x y ) . Rearranging this proves the first equality of the lemma. Next, we prove the second equality. To that end, we note that
∣ μ J , a , b ( x , y ) μ 0 , a , b ( x , y ) − 1 ∣ = ∣ 1 + J ˉ x y 1 + J ˉ X ‾ 0 Y ‾ 0 − 1 ∣ = ∣ J ˉ ( x y − X ‾ 0 Y ‾ 0 ) 1 + J ˉ X ‾ 0 Y ‾ 0 ∣ \left|\frac{\mu_{J,a,b}(x,y)}{\mu_{0,a,b}(x,y)}-1\right|=\left|\frac{1+\bar{J}xy}{1+\bar{J}\overline{X}_0\overline{Y}_0}-1\right|=\left|\frac{\bar{J}(xy-\overline{X}_0\overline{Y}_0)}{1+\bar{J}\overline{X}_0\overline{Y}_0}\right| μ 0 , a , b ( x , y ) μ J , a , b ( x , y ) − 1 = 1 + J ˉ X 0 Y 0 1 + J ˉ x y − 1 = 1 + J ˉ X 0 Y 0 J ˉ ( x y − X 0 Y 0 ) and so
∣ μ J , a , b R ( x , y ) ∣ : = ∣ μ J , a , b ( x , y ) μ 0 , a , b ( x , y ) − 1 − J ˉ ( x y − X ‾ 0 Y ‾ 0 ) ∣ = J ˉ 2 ∣ ( x y − X ‾ 0 Y ‾ 0 ) X ‾ 0 Y ‾ 0 1 + J ˉ X ‾ 0 Y ‾ 0 ∣ ≤ 4 J ˉ 2 |\mu^R_{J,a,b}(x,y)|:=\left|\frac{\mu_{J,a,b}(x,y)}{\mu_{0,a,b}(x,y)}-1-\bar{J}(xy-\overline{X}_0\overline{Y}_0)\right|=\bar{J}^2\left|(xy-\overline{X}_0\overline{Y}_0)\frac{\overline{X}_0\overline{Y}_0}{1+\bar{J}\overline{X}_0\overline{Y}_0}\right|\le4\bar{J}^2 ∣ μ J , a , b R ( x , y ) ∣ := μ 0 , a , b ( x , y ) μ J , a , b ( x , y ) − 1 − J ˉ ( x y − X 0 Y 0 ) = J ˉ 2 ( x y − X 0 Y 0 ) 1 + J ˉ X 0 Y 0 X 0 Y 0 ≤ 4 J ˉ 2 which rearranges to the second equality.
Lemma 2.6 (Master Lemma). Let F : R × { ± 1 } 2 → R F:\mathbb{R}\times\{\pm1\}^2\to\mathbb{R} F : R × { ± 1 } 2 → R
F ( J ˉ , x , y ) = F ( 0 , x , y ) + J ˉ F ˙ ( 0 , x , y ) + J ˉ 2 F ¨ ( J ˉ ∗ ( x , y ) , x , y ) (15) F(\bar{J},x,y)=F(0,x,y)+\bar{J}\dot{F}(0,x,y)+\bar{J}^2\ddot{F}(\bar{J}^*(x,y),x,y) \tag*{(15)} F ( J ˉ , x , y ) = F ( 0 , x , y ) + J ˉ F ˙ ( 0 , x , y ) + J ˉ 2 F ¨ ( J ˉ ∗ ( x , y ) , x , y ) (15) for some J ˉ ∗ ( x , y ) \bar{J}^*(x,y) J ˉ ∗ ( x , y ) 0 0 0 J ˉ \bar{J} J ˉ F ˙ ( 0 , x , y ) = d d J ˉ F ( J ˉ , x , y ) ∣ J ˉ = 0 \dot{F}(0,x,y)=\left.\frac{d}{d\bar{J}}F(\bar{J},x,y)\right|_{\bar{J}=0} F ˙ ( 0 , x , y ) = d J ˉ d F ( J ˉ , x , y ) J ˉ = 0 F ¨ ( J ˉ ∗ ( x , y ) , x , y ) = d 2 d J ˉ 2 F ( J ˉ , x , y ) ∣ J ˉ = J ˉ ∗ ( x , y ) \ddot{F}(\bar{J}^*(x,y),x,y)=\left.\frac{d^2}{d\bar{J}^2}F(\bar{J},x,y)\right|_{\bar{J}=\bar{J}^*(x,y)} F ¨ ( J ˉ ∗ ( x , y ) , x , y ) = d J ˉ 2 d 2 F ( J ˉ , x , y ) J ˉ = J ˉ ∗ ( x , y ) C > 0 C>0 C > 0 x , y ∈ { ± 1 } x,y\in\{\pm1\} x , y ∈ { ± 1 } ∣ J ˉ ∣ ≤ 1 / 2 |\bar{J}|\le1/2 ∣ J ˉ ∣ ≤ 1/2
E μ 0 , a , b [ ∣ F ( 0 , x , y ) ∣ ] , E μ 0 , a , b [ ∣ F ˙ ( 0 , x , y ) ∣ ] , E μ 0 , a , b [ ∣ F ¨ ( J ˉ ∗ ( x , y ) , x , y ) ∣ ] ≤ C (16) \mathbb{E}_{\mu_{0,a,b}}[|F(0,x,y)|],\ \mathbb{E}_{\mu_{0,a,b}}[|\dot{F}(0,x,y)|],\ \mathbb{E}_{\mu_{0,a,b}}[|\ddot{F}(\bar{J}^*(x,y),x,y)|]\le C \tag*{(16)} E μ 0 , a , b [ ∣ F ( 0 , x , y ) ∣ ] , E μ 0 , a , b [ ∣ F ˙ ( 0 , x , y ) ∣ ] , E μ 0 , a , b [ ∣ F ¨ ( J ˉ ∗ ( x , y ) , x , y ) ∣ ] ≤ C (16) then for ∣ J ˉ ∣ ≤ 1 2 \lvert\bar{J}\rvert\le\frac{1}{2} ∣ J ˉ ∣ ≤ 2 1
E μ J ˉ , a , b [ F ( J ˉ , x , y ) ] = E μ 0 , a , b [ F ( 0 , x , y ) ] + J ˉ ( E μ 0 , a , b [ F ˙ ( 0 , x , y ) ] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 , x , y ) ] ) + O ( C ⋅ J ˉ 2 ) . \mathbb{E}_{\mu_{\bar{J},a,b}}[F(\bar{J},x,y)] = \mathbb{E}_{\mu_{0,a,b}}[F(0,x,y)] + \bar{J}\left(\mathbb{E}_{\mu_{0,a,b}}[\dot{F}(0,x,y)]+\mathbb{E}_{\mu_{0,a,b}}[(XY-\bar{X}_{0}\bar{Y}_{0})F(0,x,y)]\right)+O(C\cdot\bar{J}^{2}). E μ J ˉ , a , b [ F ( J ˉ , x , y )] = E μ 0 , a , b [ F ( 0 , x , y )] + J ˉ ( E μ 0 , a , b [ F ˙ ( 0 , x , y )] + E μ 0 , a , b [( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 , x , y )] ) + O ( C ⋅ J ˉ 2 ) . Proof. We begin by remarking that the form of F F F F ( J ˉ ) F(\bar{J}) F ( J ˉ ) F ( J ˉ , x , y ) F(\bar{J},x,y) F ( J ˉ , x , y )
E μ J ˉ , a , b [ F ( J ˉ ) ] = E μ 0 , a , b [ μ J ˉ , a , b ( x , y ) μ 0 , a , b ( x , y ) ( F ( 0 ) + J ˉ F ˙ ( 0 ) + J ˉ 2 F ¨ ( J ˉ ∗ ) ) ] = E μ 0 , a , b [ ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + μ J ˉ , a , b R ( x , y ) ) ( F ( 0 ) + J ˉ F ˙ ( 0 ) + J ˉ 2 F ¨ ( J ˉ ∗ ) ) ] . \begin{aligned}
\mathbb{E}_{\mu_{\bar{J},a,b}}[F(\bar{J})]=\mathbb{E}_{\mu_{0,a,b}}\left[\frac{\mu_{\bar{J},a,b}(x,y)}{\mu_{0,a,b}(x,y)}\left(F(0)+\bar{J}\dot{F}(0)+\bar{J}^{2}\ddot{F}(\bar{J}^{*})\right)\right] \\
&=\mathbb{E}_{\mu_{0,a,b}}\left[\left(1+\bar{J}(xy-\bar{X}_{0}\bar{Y}_{0})+\mu^{R}_{\bar{J},a,b}(x,y)\right)\left(F(0)+\bar{J}\dot{F}(0)+\bar{J}^{2}\ddot{F}(\bar{J}^{*})\right)\right].
\end{aligned} E μ J ˉ , a , b [ F ( J ˉ )] = E μ 0 , a , b [ μ 0 , a , b ( x , y ) μ J ˉ , a , b ( x , y ) ( F ( 0 ) + J ˉ F ˙ ( 0 ) + J ˉ 2 F ¨ ( J ˉ ∗ ) ) ] = E μ 0 , a , b [ ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + μ J ˉ , a , b R ( x , y ) ) ( F ( 0 ) + J ˉ F ˙ ( 0 ) + J ˉ 2 F ¨ ( J ˉ ∗ ) ) ] . Lemma 2.5
The constant term of the above is E μ 0 , a , b [ F ( 0 ) ] \mathbb{E}_{\mu_{0,a,b}}[F(0)] E μ 0 , a , b [ F ( 0 )] J ˉ \bar{J} J ˉ
J ˉ ( E μ 0 , a , b [ F ˙ ( 0 ) ] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 ) ] ) . \bar{J}\left(\mathbb{E}_{\mu_{0,a,b}}[\dot{F}(0)]+\mathbb{E}_{\mu_{0,a,b}}[(XY-\bar{X}_{0}\bar{Y}_{0})F(0)]\right). J ˉ ( E μ 0 , a , b [ F ˙ ( 0 )] + E μ 0 , a , b [( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 )] ) . The quadratic term in J ˉ \bar{J} J ˉ
E μ 0 , a , b [ μ J ˉ , a , b R ⋅ F ( 0 ) ] + J ˉ 2 E μ 0 , a , b [ ( x y − X ˉ 0 Y ˉ 0 ) F ˙ ( 0 ) ] + E μ 0 , a , b [ ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + O ( J ˉ 2 ) ) J ˉ 2 F ¨ ( J ˉ ∗ ) ] \mathbb{E}_{\mu_{0,a,b}}[\mu^{R}_{\bar{J},a,b}\cdot F(0)]+\bar{J}^{2}\mathbb{E}_{\mu_{0,a,b}}\left[(xy-\bar{X}_{0}\bar{Y}_{0})\dot{F}(0)\right]+\mathbb{E}_{\mu_{0,a,b}}\left[\left(1+\bar{J}(xy-\bar{X}_{0}\bar{Y}_{0})+O(\bar{J}^{2})\right)\bar{J}^{2}\ddot{F}(\bar{J}^{*})\right] E μ 0 , a , b [ μ J ˉ , a , b R ⋅ F ( 0 )] + J ˉ 2 E μ 0 , a , b [ ( x y − X ˉ 0 Y ˉ 0 ) F ˙ ( 0 ) ] + E μ 0 , a , b [ ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + O ( J ˉ 2 ) ) J ˉ 2 F ¨ ( J ˉ ∗ ) ] ≤ ∣ x ∣ , ∣ y ∣ , ∣ X ˉ 0 ∣ , ∣ Y ˉ 0 ∣ ≤ 1 O ( J ˉ 2 ) E μ 0 , a , b [ ∣ F ( 0 ) ∣ ] + 2 J ˉ 2 E μ 0 , a , b [ ∣ F ˙ ( 0 ) ∣ ] + ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) J ˉ 2 E μ 0 , a , b [ ∣ F ¨ ( J ˉ ∗ ) ∣ ] \leq_{\lvert x\rvert,\lvert y\rvert,\lvert\bar{X}_{0}\rvert,\lvert\bar{Y}_{0}\rvert\leq1} O(\bar{J}^{2})\mathbb{E}_{\mu_{0,a,b}}[\lvert F(0)\rvert]+2\bar{J}^{2}\mathbb{E}_{\mu_{0,a,b}}[\lvert\dot{F}(0)\rvert]+(1+2\bar{J}+O(\bar{J}^{2}))\bar{J}^{2}\mathbb{E}_{\mu_{0,a,b}}[\lvert\ddot{F}(\bar{J}^{*})\rvert] ≤ ∣ x ∣ , ∣ y ∣ , ∣ X ˉ 0 ∣ , ∣ Y ˉ 0 ∣ ≤ 1 O ( J ˉ 2 ) E μ 0 , a , b [∣ F ( 0 )∣] + 2 J ˉ 2 E μ 0 , a , b [∣ F ˙ ( 0 )∣] + ( 1 + 2 J ˉ + O ( J ˉ 2 )) J ˉ 2 E μ 0 , a , b [∣ F ¨ ( J ˉ ∗ )∣] Lemma 2.5: ∣ μ J ˉ , a , b R ∣ ≤ O ( J ˉ 2 ) \lvert\mu^{R}_{\bar{J},a,b}\rvert\leq O(\bar{J}^{2}) ∣ μ J ˉ , a , b R ∣ ≤ O ( J ˉ 2 )
≤ J ˉ , ∣ J ˉ ∣ ≤ 1 2 O ( C ⋅ J ˉ 2 ) . (17) \leq_{\bar{J},\lvert\bar{J}\rvert\leq\frac{1}{2}} O(C\cdot\bar{J}^{2}). \tag*{(17)} ≤ J ˉ , ∣ J ˉ ∣ ≤ 2 1 O ( C ⋅ J ˉ 2 ) . (17) GOE random matrices
Proposition 2.7 ([9 ] ). Let β > 0 \beta>0 β > 0 J J J β \beta β ∥ J ∥ o p → 2 β \lVert J\rVert_{\mathrm{op}}\to2\beta ∥ J ∥ op → 2 β
Lemma 2.8. Let J ∈ R n × n J\in\mathbb{R}^{n\times n} J ∈ R n × n β \beta β ∑ j J i j 2 → β 2 \sum_j J_{ij}^{2}\to\beta^{2} ∑ j J ij 2 → β 2
Proof. Observe that J i 1 , … , J i n J_{i1},\ldots,J_{in} J i 1 , … , J in β 2 / n \beta^{2}/n β 2 / n ∑ j J i j 2 \sum_j J_{ij}^{2} ∑ j J ij 2 β 2 \beta^{2} β 2
Lemma 2.9. Let β > 0 \beta>0 β > 0 J J J β \beta β
max 1 ≤ i , j ≤ n ∣ J i j ∣ → 0 \max_{1\leq i,j\leq n}\lvert J_{ij}\rvert\to0 1 ≤ i , j ≤ n max ∣ J ij ∣ → 0 in probability.
Proof. Fix t > 0 t>0 t > 0
P ( ∣ J i j ∣ > t ) = P Z ∼ N ( 0 , 1 ) ( ∣ Z ∣ > t n β ) ≤ 2 exp ( − n t 2 2 β 2 ) . \mathbb{P}(\lvert J_{ij}\rvert>t)=\mathbb{P}_{Z\sim\mathcal{N}(0,1)}\left(\lvert Z\rvert>\frac{t\sqrt{n}}{\beta}\right)\leq2\exp\left(-\frac{nt^{2}}{2\beta^{2}}\right). P (∣ J ij ∣ > t ) = P Z ∼ N ( 0 , 1 ) ( ∣ Z ∣ > β t n ) ≤ 2 exp ( − 2 β 2 n t 2 ) . By taking a union bound over the n ( n − 1 ) 2 \frac{n(n-1)}{2} 2 n ( n − 1 )
P ( max 1 ≤ i , j ≤ n ∣ J i j ∣ > t ) = P ( max 1 ≤ i < j ≤ n ∣ J i j ∣ > t ) ≤ ∑ 1 ≤ i < j ≤ n P ( ∣ J i j ∣ > t ) ≤ n ( n − 1 ) 2 exp ( − n t 2 2 β 2 ) , \mathbb{P}\left(\max_{1\leq i,j\leq n}\lvert J_{ij}\rvert>t\right)=\mathbb{P}\left(\max_{1\leq i<j\leq n}\lvert J_{ij}\rvert>t\right)\leq\sum_{1\leq i<j\leq n}\mathbb{P}(\lvert J_{ij}\rvert>t)\leq\frac{n(n-1)}{2}\exp\left(-\frac{nt^{2}}{2\beta^{2}}\right), P ( 1 ≤ i , j ≤ n max ∣ J ij ∣ > t ) = P ( 1 ≤ i < j ≤ n max ∣ J ij ∣ > t ) ≤ 1 ≤ i < j ≤ n ∑ P (∣ J ij ∣ > t ) ≤ 2 n ( n − 1 ) exp ( − 2 β 2 n t 2 ) , which converges to zero as n → ∞ n\to\infty n → ∞
Lemma 2.10 (GOE eigenvector delocalization). Let δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 ) ρ , β > 0 \rho,\beta> 0 ρ , β > 0 J J J β \beta β J J J z ∈ R n z \in\mathbb{R}^{n} z ∈ R n
z T ( 2 β I − J ) z ≥ π β ρ 6.1 ( β + ρ ) 3 ⋅ ∥ z ∥ 2 2 − ρ ∥ z ∥ ≤ δ n 2 . \mathbf{z}^{\mathsf{T}}(2\beta I-J)\mathbf{z} \ge\frac{\pi\beta\rho}{6.1(\beta+\rho)^{3}} \cdot\lVert z\rVert_{2}^{2}-\rho\lVert z\rVert_{\leq\delta n}^{2}. z T ( 2 β I − J ) z ≥ 6.1 ( β + ρ ) 3 π β ρ ⋅ ∥ z ∥ 2 2 − ρ ∥ z ∥ ≤ δ n 2 . One way of interpreting the above statement is that if z z z J J J λ max ( J ) ≈ 2 β \lambda_{\max}(J) \approx2\beta λ m a x ( J ) ≈ 2 β z z z δ n \delta n δ n
Proof of Lemma 2.10. For z ∈ S n − 1 \mathbf{z} \in\mathbb{S}^{n-1} z ∈ S n − 1
X z = z T J z and Y z = 2 β n g T z where g ∼ N ( 0 , I n ) . X_{\mathbf{z}}=\mathbf{z}^{\mathsf{T}}J\mathbf{z}\quad\text{and}\quad Y_{\mathbf{z}}=\frac{2\beta}{\sqrt{n}}\mathbf{g}^{\mathsf{T}}\mathbf{z}\quad\text{where }\mathbf{g}\sim\mathcal{N}(0,I_n). X z = z T J z and Y z = n 2 β g T z where g ∼ N ( 0 , I n ) . We note that for z , z ~ ∈ S n − 1 \mathbf{z},\widetilde{\mathbf{z}}\in\mathbb{S}^{n-1} z , z ∈ S n − 1
E [ ( X z − X z ~ ) 2 ] = 4 E [ ( ∑ i < j J i j ( z i z j − z ~ i z ~ j ) ) 2 ] = 4 β 2 n ∑ i < j ( z i z j − z ~ i z ~ j ) 2 ≤ 2 β 2 n ∥ z z T − z ~ z ~ T ∥ F 2 = 4 β 2 n [ ( 1 − ( z T z ~ ) 2 ) ] ≤ 1 − x 2 ≤ 2 ( 1 − x ) 4 β 2 n [ 2 ( 1 − z T z ~ ) ] = 4 β 2 n ∥ z − z ~ ∥ 2 2 = E [ ( Y z − Y z ~ ) 2 ] . \begin{aligned}
\mathbb{E}\left[(X_{\mathbf{z}}-X_{\widetilde{\mathbf{z}}})^{2}\right]
=4\mathbb{E}\left[\left(\sum_{i<j}J_{ij}(z_i z_j-\widetilde{z}_i\widetilde{z}_j)\right)^{2}\right] \\
&=\frac{4\beta^{2}}{n}\sum_{i<j}(z_i z_j-\widetilde{z}_i\widetilde{z}_j)^{2}
\leq\frac{2\beta^{2}}{n}\left\|\mathbf{z}\mathbf{z}^{\mathsf{T}}-\widetilde{\mathbf{z}}\widetilde{\mathbf{z}}^{\mathsf{T}}\right\|_{F}^{2} \\
&=\frac{4\beta^{2}}{n}\left[(1-(\mathbf{z}^{\mathsf{T}}\widetilde{\mathbf{z}})^{2})\right]
\underset{1-x^{2}\leq2(1-x)}{\leq}\frac{4\beta^{2}}{n}\left[2(1-\mathbf{z}^{\mathsf{T}}\widetilde{\mathbf{z}})\right] \\
&=\frac{4\beta^{2}}{n}\lVert\mathbf{z}-\widetilde{\mathbf{z}}\rVert_{2}^{2}
=\mathbb{E}\left[(Y_{\mathbf{z}}-Y_{\widetilde{\mathbf{z}}})^{2}\right].
\end{aligned} E [ ( X z − X z ) 2 ] = 4 E ( i < j ∑ J ij ( z i z j − z i z j ) ) 2 = n 4 β 2 i < j ∑ ( z i z j − z i z j ) 2 ≤ n 2 β 2 z z T − z z T F 2 = n 4 β 2 [ ( 1 − ( z T z ) 2 ) ] 1 − x 2 ≤ 2 ( 1 − x ) ≤ n 4 β 2 [ 2 ( 1 − z T z ) ] = n 4 β 2 ∥ z − z ∥ 2 2 = E [ ( Y z − Y z ) 2 ] . Theorem 2.11 (Variant of Sudakov-Fernique comparison inequality [29 ] ). Let I \mathcal{I} I { X i } i ∈ I \{X_i\}_{i\in\mathcal{I}} { X i } i ∈ I { Y i } i ∈ I \{Y_i\}_{i\in\mathcal{I}} { Y i } i ∈ I
E ( X i − X j ) 2 ≤ E ( Y i − Y j ) 2 ∀ i , j ∈ I . \mathbb{E}(X_i-X_j)^2\leq\mathbb{E}(Y_i-Y_j)^2\quad\forall i,j\in\mathcal{I}. E ( X i − X j ) 2 ≤ E ( Y i − Y j ) 2 ∀ i , j ∈ I . Let { m i } i ∈ I \{m_i\}_{i\in\mathcal{I}} { m i } i ∈ I m i ∈ R m_i\in\mathbb{R} m i ∈ R i ∈ I i\in\mathcal{I} i ∈ I
E sup i ( X i + m i ) ≤ E sup i ( Y i + m i ) ∀ i ∈ I . \mathbb{E}\sup_{i}(X_i+m_i)\leq\mathbb{E}\sup_{i}(Y_i+m_i)\quad\forall i\in\mathcal{I}. E i sup ( X i + m i ) ≤ E i sup ( Y i + m i ) ∀ i ∈ I . Applying above to { X z } z ∈ S n − 1 \{X_{\mathbf{z}}\}_{\mathbf{z}\in\mathbb{S}^{n-1}} { X z } z ∈ S n − 1 { Y z } z ∈ S n − 1 \{Y_{\mathbf{z}}\}_{\mathbf{z}\in\mathbb{S}^{n-1}} { Y z } z ∈ S n − 1 m z = − ρ ∥ z ∥ ≤ δ n 2 m_{\mathbf{z}}=-\rho\lVert\mathbf{z}\rVert_{\leq\delta n}^{2} m z = − ρ ∥ z ∥ ≤ δ n 2 z ∈ S n − 1 \mathbf{z}\in\mathbb{S}^{n-1} z ∈ S n − 1
E sup z ∈ S n − 1 { X z − ρ ∥ z ∥ ≤ δ n 2 } ≤ E sup z ∈ S n − 1 { 2 β n g T z − ρ ∥ z ∥ ≤ δ n 2 } . \mathbb{E}\sup_{\mathbf{z}\in\mathbb{S}^{n-1}}\left\{X_{\mathbf{z}}-\rho\lVert\mathbf{z}\rVert_{\leq\delta n}^{2}\right\}
\leq
\mathbb{E}\sup_{\mathbf{z}\in\mathbb{S}^{n-1}}\left\{\frac{2\beta}{\sqrt{n}}\mathbf{g}^{\mathsf{T}}\mathbf{z}-\rho\lVert\mathbf{z}\rVert_{\leq\delta n}^{2}\right\}. E z ∈ S n − 1 sup { X z − ρ ∥ z ∥ ≤ δ n 2 } ≤ E z ∈ S n − 1 sup { n 2 β g T z − ρ ∥ z ∥ ≤ δ n 2 } . We note that S n − 1 \mathbb{S}^{n-1} S n − 1 S n − 1 \mathbb{S}^{n-1} S n − 1
Claim 2.12.
E sup z ∈ S n − 1 { 2 β n g T z − ρ ∥ z ∥ ≤ δ n 2 } ≤ 2 β − π β ρ 120 ( β + ρ ) δ 3 \mathbb{E}\sup_{\mathbf{z}\in\mathbb{S}^{n-1}}\left\{\frac{2\beta}{\sqrt{n}}\mathbf{g}^{\mathsf{T}}\mathbf{z}-\rho\lVert\mathbf{z}\rVert_{\leq\delta n}^{2}\right\}
\leq2\beta-\frac{\pi\beta\rho}{120(\beta+\rho)}\delta^{3} E z ∈ S n − 1 sup { n 2 β g T z − ρ ∥ z ∥ ≤ δ n 2 } ≤ 2 β − 120 ( β + ρ ) π β ρ δ 3 Proof. Let g ˉ = g / n \bar{\mathbf{g}} = \mathbf{g}/\sqrt{n} g ˉ = g / n ∥ z ∥ = 1 \|\mathbf{z}\| = 1 ∥ z ∥ = 1
2 β g ˉ T z − ρ ∥ z ∥ ≤ δ n 2 = β ( 1 + ∥ g ˉ ∥ 2 ) − ( β ∥ g ˉ − z ∥ 2 + ρ ∥ z ∥ ≤ δ n 2 ) ≤ β ( 1 + ∥ g ˉ ∥ 2 ) − β ρ β + ρ ∥ g ˉ ∥ ≤ δ n 2 (18) 2\beta\bar{\mathbf{g}}^{\mathsf{T}}\mathbf{z} - \rho\|\mathbf{z}\|^{2}_{\le\delta n} = \beta(1+\|\bar{\mathbf{g}}\|^{2}) - \left(\beta\|\bar{\mathbf{g}}-\mathbf{z}\|^{2}+\rho\|\mathbf{z}\|^{2}_{\le\delta n}\right) \le\beta(1+\|\bar{\mathbf{g}}\|^{2})-\frac{\beta\rho}{\beta+\rho}\|\bar{\mathbf{g}}\|^{2}_{\le\delta n} \tag*{(18)} 2 β g ˉ T z − ρ ∥ z ∥ ≤ δ n 2 = β ( 1 + ∥ g ˉ ∥ 2 ) − ( β ∥ g ˉ − z ∥ 2 + ρ ∥ z ∥ ≤ δ n 2 ) ≤ β ( 1 + ∥ g ˉ ∥ 2 ) − β + ρ β ρ ∥ g ˉ ∥ ≤ δ n 2 (18) To see the inequality, let S S S ⌊ δ n ⌋ \lfloor\delta n\rfloor ⌊ δ n ⌋ z \mathbf{z} z
∥ g ˉ ∥ ≤ δ n = ∑ i ∈ S g ˉ i 2 ≤ triangle ineq ∑ i ∈ S ( g ˉ i − z i ) 2 + ∑ i ∈ S z i 2 ≤ ∥ g ˉ − z ∥ + ∥ z ∥ ≤ δ n \|\bar{\mathbf{g}}\|_{\le\delta n}=\sqrt{\sum_{i\in S}\bar{g}_{i}^{2}}\le_{\text{triangle ineq}}\sqrt{\sum_{i\in S}(\bar{g}_{i}-z_{i})^{2}}+\sqrt{\sum_{i\in S}z_{i}^{2}}\le\|\bar{\mathbf{g}}-\mathbf{z}\|+\|\mathbf{z}\|_{\le\delta n} ∥ g ˉ ∥ ≤ δ n = i ∈ S ∑ g ˉ i 2 ≤ triangle ineq i ∈ S ∑ ( g ˉ i − z i ) 2 + i ∈ S ∑ z i 2 ≤ ∥ g ˉ − z ∥ + ∥ z ∥ ≤ δ n Therefore, by the Cauchy-Schwarz inequality,
∥ g ˉ ∥ ≤ δ n 2 ≤ ( 1 β + 1 ρ ) ( β ∥ g ˉ − z ∥ 2 + ρ ∥ z ∥ ≤ δ n 2 ) \|\bar{\mathbf{g}}\|_{\le\delta n}^{2}\le\left(\frac{1}{\beta}+\frac{1}{\rho}\right)\left(\beta\|\bar{\mathbf{g}}-\mathbf{z}\|^{2}+\rho\|\mathbf{z}\|_{\le\delta n}^{2}\right) ∥ g ˉ ∥ ≤ δ n 2 ≤ ( β 1 + ρ 1 ) ( β ∥ g ˉ − z ∥ 2 + ρ ∥ z ∥ ≤ δ n 2 ) Having (16) and using E ∥ g ˉ ∥ 2 = 1 \mathbb{E}\|\bar{\mathbf{g}}\|^{2}=1 E ∥ g ˉ ∥ 2 = 1
E sup z ∈ S n − 1 { 2 β n g T z − ρ ∥ z ∥ ≤ δ n 2 } ≤ 2 β − β ρ β + ρ E ∥ g ∥ ≤ δ n 2 n \mathbb{E}\sup_{\mathbf{z}\in\mathbb{S}^{n-1}}\left\{\frac{2\beta}{\sqrt{n}}\mathbf{g}^{\mathsf{T}}\mathbf{z}-\rho\|\mathbf{z}\|_{\le\delta n}^{2}\right\}\le2\beta-\frac{\beta\rho}{\beta+\rho}\frac{\mathbb{E}\|\mathbf{g}\|_{\le\delta n}^{2}}{n} E z ∈ S n − 1 sup { n 2 β g T z − ρ ∥ z ∥ ≤ δ n 2 } ≤ 2 β − β + ρ β ρ n E ∥ g ∥ ≤ δ n 2 So, to finish the proof it is enough to show that E ∥ g ∥ ≤ δ n 2 / n ≥ π δ 3 / 6 \mathbb{E}\|\mathbf{g}\|_{\le\delta n}^{2}/n\ge\pi\delta^{3}/6 E ∥ g ∥ ≤ δ n 2 / n ≥ π δ 3 /6 t > 0 t>0 t > 0
E ∥ g ∥ ≤ δ n 2 n ≥ E min S ⊂ [ n ] : ∣ S ∣ = ⌊ δ n ⌋ ∑ i ∈ S min { g i 2 , t 2 } n ≥ E min S : ∣ S ∣ = ⌊ δ n ⌋ { δ n t 2 − ∑ i = 1 n ( t 2 − g i 2 ) + } n = δ t 2 − E g ∼ N ( 0 , 1 ) ( t 2 − g 2 ) + = δ t 2 − ∫ − t t ( t 2 − x 2 ) exp ( − x 2 / 2 ) 2 π d x \begin{aligned}
\frac{\mathbb{E}\|\mathbf{g}\|_{\le\delta n}^{2}}{n}\ge\frac{\mathbb{E}\min_{S\subset[n]:|S|=\lfloor\delta n\rfloor}\sum_{i\in S}\min\{g_{i}^{2},t^{2}\}}{n} \\
&\ge\frac{\mathbb{E}\min_{S:|S|=\lfloor\delta n\rfloor}\left\{\delta n t^{2}-\sum_{i=1}^{n}(t^{2}-g_{i}^{2})_{+}\right\}}{n} \\
&=\delta t^{2}-\mathbb{E}_{g\sim\mathcal{N}(0,1)}(t^{2}-g^{2})_{+}=\delta t^{2}-\int_{-t}^{t}(t^{2}-x^{2})\frac{\exp(-x^{2}/2)}{\sqrt{2\pi}}\,dx
\end{aligned} n E ∥ g ∥ ≤ δ n 2 ≥ n E min S ⊂ [ n ] : ∣ S ∣ = ⌊ δ n ⌋ ∑ i ∈ S min { g i 2 , t 2 } ≥ n E min S : ∣ S ∣ = ⌊ δ n ⌋ { δ n t 2 − ∑ i = 1 n ( t 2 − g i 2 ) + } = δ t 2 − E g ∼ N ( 0 , 1 ) ( t 2 − g 2 ) + = δ t 2 − ∫ − t t ( t 2 − x 2 ) 2 π exp ( − x 2 /2 ) d x ≥ e − x 2 / 2 ≤ 1 δ t 2 − 4 t 3 3 2 π = t = π / 2 δ π 6 ⋅ δ 3 . \ge_{e^{-x^{2}/2}\le1}\delta t^{2}-\frac{4t^{3}}{3\sqrt{2\pi}}\mathrel{\mathop=^{t=\sqrt{\pi/2}\delta}}\frac{\pi}{6}\cdot\delta^{3}. ≥ e − x 2 /2 ≤ 1 δ t 2 − 3 2 π 4 t 3 = t = π /2 δ 6 π ⋅ δ 3 . where we write ( a ) + = max { a , 0 } (a)_{+}=\max\{a,0\} ( a ) + = max { a , 0 }
Next, we use the following theorem to prove that the LHS is concentrated around its expectation.
Theorem 2.13 (Borell-TIS inequality [8 , 13 ] ; see [27 ] for non-centered version). Let { X i } i ∈ I \{X_i\}_{i\in\mathcal{I}} { X i } i ∈ I i ∈ I i\in\mathcal{I} i ∈ I Var X i ≤ σ 2 \operatorname{Var}X_i\le\sigma^{2} Var X i ≤ σ 2
P ( max i ∈ I X i − E [ max i ∈ I X i ] ≥ t ) ≤ exp ( − t 2 8 σ 2 ) . \mathbb{P}\left(\max_{i\in\mathcal{I}}X_i-\mathbb{E}\left[\max_{i\in\mathcal{I}}X_i\right]\ge t\right)\le\exp\left(-\frac{t^{2}}{8\sigma^{2}}\right). P ( i ∈ I max X i − E [ i ∈ I max X i ] ≥ t ) ≤ exp ( − 8 σ 2 t 2 ) . For convenience let X z ′ = X z − ρ ∥ z ∥ ≤ δ n 2 X'_{\mathbf{z}}=X_{\mathbf{z}}-\rho\|\mathbf{z}\|_{\le\delta n}^{2} X z ′ = X z − ρ ∥ z ∥ ≤ δ n 2
Var ( X z ′ ) = E X z 2 = E [ ( ∑ i < j 2 J i j z i z j ) 2 ] = 2 β n ∑ i , j 2 z i 2 z j 2 ≤ 2 β n ∥ z ∥ 4 = 2 β n . (19) \operatorname{Var}(X'_{\mathbf{z}})=\mathbb{E}X_{\mathbf{z}}^{2}=\mathbb{E}\left[\left(\sum_{i<j}2J_{ij}z_{i}z_{j}\right)^{2}\right]=\frac{2\beta}{n}\sum_{i,j}2z_{i}^{2}z_{j}^{2}\le\frac{2\beta}{n}\|\mathbf{z}\|^{4}=\frac{2\beta}{n}. \tag*{(19)} Var ( X z ′ ) = E X z 2 = E ( i < j ∑ 2 J ij z i z j ) 2 = n 2 β i , j ∑ 2 z i 2 z j 2 ≤ n 2 β ∥ z ∥ 4 = n 2 β . (19) Consequently, this implies that
P ( sup z ∈ S n − 1 { X z ′ } ≥ 2 β − π β ρ 6.1 ( β + ρ ) δ 3 ) ≤ P ( sup z ∈ S n − 1 { X z ′ } − E sup z ∈ S n − 1 { X z ′ } ≥ π β ρ 36.6 ( β + ρ ) δ 3 ) ≤ exp ( − ( π β ρ 14520 ( β + ρ ) δ 3 ) 2 n 36.6 β ) (20) \begin{aligned}
\mathbb{P}\left(\sup_{z\in\mathbb{S}^{n-1}}\left\{X'_z\right\}\ge2\beta-\frac{\pi\beta\rho}{6.1(\beta+\rho)}\delta^3\right) \\
&\le\mathbb{P}\left(\sup_{z\in\mathbb{S}^{n-1}}\left\{X'_z\right\}-\mathbb{E}\sup_{z\in\mathbb{S}^{n-1}}\left\{X'_z\right\}\ge\frac{\pi\beta\rho}{36.6(\beta+\rho)}\delta^3\right) \\
&\le\exp\left(-\left(\frac{\pi\beta\rho}{14520(\beta+\rho)}\delta^3\right)^2\frac{n}{36.6\beta}\right) \tag*{(20)}
\end{aligned} P ( z ∈ S n − 1 sup { X z ′ } ≥ 2 β − 6.1 ( β + ρ ) π β ρ δ 3 ) ≤ P ( z ∈ S n − 1 sup { X z ′ } − E z ∈ S n − 1 sup { X z ′ } ≥ 36.6 ( β + ρ ) π β ρ δ 3 ) ≤ exp ( − ( 14520 ( β + ρ ) π β ρ δ 3 ) 2 36.6 β n ) (20) Consequently, it follows that with probability 1 − o n ( 1 ) 1-o_n(1) 1 − o n ( 1 )
z T ( 2 β I − J ) z = 2 β − X z ≥ π β ρ 6.1 ( β + ρ ) δ 3 − ρ ∥ z ∥ ≤ δ n 2 . \mathbf{z}^{\mathsf{T}}(2\beta I-J)\mathbf{z}=2\beta-X_{\mathbf{z}}\ge\frac{\pi\beta\rho}{6.1(\beta+\rho)}\delta^3-\rho\|\mathbf{z}\|^2_{\leq\delta n}. z T ( 2 β I − J ) z = 2 β − X z ≥ 6.1 ( β + ρ ) π β ρ δ 3 − ρ ∥ z ∥ ≤ δ n 2 . with probability 1 − o n ( 1 ) 1-o_n(1) 1 − o n ( 1 )
In the context of Sherrington-Kirkpatrick μ J , h \mu_{J,\mathbf{h}} μ J , h h ∈ R n \mathbf{h}\in\mathbb{R}^n h ∈ R n H i ( x ) = h i + ∑ j ≠ i J i j x j H_i(\mathbf{x})=h_i+\sum_{j\ne i}J_{ij}x_j H i ( x ) = h i + ∑ j = i J ij x j i i i
Lemma 2.14 (Typical large local fields). Let β > 0 \beta>0 β > 0 and J ∈ R n × n J\in\mathbb{R}^{n\times n} J ∈ R n × n be β \beta β GOE. Suppose 0 < δ ′ < δ < 1 0<\delta'<\delta<1 0 < δ ′ < δ < 1 is such that
D K L ( Ber ( δ ) ∥ Ber ( δ ′ ) ) > ln 2. (21) D_{\mathrm{KL}}\bigl(\operatorname{Ber}(\delta)\,\|\,\operatorname{Ber}(\delta')\bigr)>\ln2. \tag*{(21)} D KL ( Ber ( δ ) ∥ Ber ( δ ′ ) ) > ln 2. (21) Let h ∈ R n \mathbf{h}\in\mathbb{R}^n h ∈ R n be arbitrary. With high probability, for all x ∈ { ± 1 } n \mathbf{x}\in\{\pm1\}^n x ∈ { ± 1 } n we have
∣ { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } ∣ < δ n . \left|\left\{i:|H_i(\mathbf{x})|\leq\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\right|<\delta n. { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } < δ n . Proof. Fix a vector x ∈ { ± 1 } n \mathbf{x}\in\{\pm1\}^n x ∈ { ± 1 } n
P [ ∣ { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } ∣ ≥ δ n ] ≤ exp ( − n D K L ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) (22) \mathbb{P}\left[\left|\left\{i:|H_i(\mathbf{x})|\leq\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\right|\geq\delta n\right]\leq\exp\left(-nD_{\mathrm{KL}}\left(\operatorname{Ber}(\delta)\,\|\,\operatorname{Ber}(\delta'+o_n(1))\right)\right) \tag*{(22)} P [ { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } ≥ δ n ] ≤ exp ( − n D KL ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 )) ) ) (22) Now running a union bound over all 2 n 2^n 2 n x \mathbf{x} x
P [ ∃ x : ∣ { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } ∣ ≥ δ n ] ≤ 2 n exp ( − n D K L ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) ≤ o ( 1 ) , \mathbb{P}\left[\exists\mathbf{x}:\left|\left\{i:|H_i(\mathbf{x})|\leq\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\right|\geq\delta n\right]\leq2^n\exp\left(-nD_{\mathrm{KL}}\left(\operatorname{Ber}(\delta)\,\|\,\operatorname{Ber}(\delta'+o_n(1))\right)\right)\leq o(1), P [ ∃ x : { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } ≥ δ n ] ≤ 2 n exp ( − n D KL ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 )) ) ) ≤ o ( 1 ) , where the last inequality uses the assumption that D K L ( Ber ( δ ) ∥ Ber ( δ ′ ) ) > ln 2 D_{\mathrm{KL}}(\operatorname{Ber}(\delta)\,\|\,\operatorname{Ber}(\delta'))>\ln2 D KL ( Ber ( δ ) ∥ Ber ( δ ′ )) > ln 2
It remains to prove (22) . The covariance matrix of ( H 1 ( x ) , … , H n ( x ) ) (H_1(\mathbf{x}),\ldots,H_n(\mathbf{x})) ( H 1 ( x ) , … , H n ( x ))
Cov ( H 1 ( x ) , … , H n ( x ) ) = β 2 n ( ( n − 2 ) I n + x x T ) . \operatorname{Cov}(H_1(\mathbf{x}),\ldots,H_n(\mathbf{x}))=\frac{\beta^2}{n}\left((n-2)I_n+\mathbf{x}\mathbf{x}^{\mathsf{T}}\right). Cov ( H 1 ( x ) , … , H n ( x )) = n β 2 ( ( n − 2 ) I n + x x T ) . Note that E [ H i ( x ) 2 ] = β 2 ( n − 1 ) / n = β 2 ( n − 2 ) / n + x i 2 \mathbb{E}[H_i(\mathbf{x})^2]=\beta^2(n-1)/n=\beta^2(n-2)/n+x_i^2 E [ H i ( x ) 2 ] = β 2 ( n − 1 ) / n = β 2 ( n − 2 ) / n + x i 2 Z 0 , Z 1 , … , Z n Z_0,Z_1,\ldots,Z_n Z 0 , Z 1 , … , Z n N ( 0 , 1 ) \mathcal{N}(0,1) N ( 0 , 1 ) ( H 1 ( x ) , … , H n ( x ) ) (H_1(\mathbf{x}),\ldots,H_n(\mathbf{x})) ( H 1 ( x ) , … , H n ( x ))
H i ′ ( x ) = ( h i + β n − 2 n Z i + β n x i Z 0 ) i ∈ [ n ] . H'_i(\mathbf{x})=\left(h_i+\beta\sqrt{\frac{n-2}{n}}Z_i+\frac{\beta}{\sqrt{n}}x_iZ_0\right)_{i\in[n]}. H i ′ ( x ) = ( h i + β n n − 2 Z i + n β x i Z 0 ) i ∈ [ n ] . So, to prove (22) it is enough to show that
P [ ∣ { i : ∣ H i ′ ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } ∣ ≥ δ n ] ≤ exp ( − n D K L ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) . (23) \mathbb{P}\left[\left|\left\{i:\left|H'_i\right|\le\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\right|\ge\delta n\right]\le\exp\left(-nD_{\mathrm{KL}}\left(\operatorname{Ber}(\delta)\mathbin\Vert\operatorname{Ber}\left(\delta'+o_n(1)\right)\right)\right). \tag*{(23)} P [ { i : ∣ H i ′ ∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } ≥ δ n ] ≤ exp ( − n D KL ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) . (23) For Z i , Z 0 ∼ N ( 0 , 1 ) Z_i,Z_0\sim\mathcal{N}(0,1) Z i , Z 0 ∼ N ( 0 , 1 ) β ( n − 2 ) / n + β / n x i Z 0 \beta\sqrt{(n-2)/n}+\beta/\sqrt{n}x_iZ_0 β ( n − 2 ) / n + β / n x i Z 0
h i ↦ P [ ∣ h i + β n − 2 n Z i + β n x i Z 0 ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) ] h_i\mapsto\mathbb{P}\left[\left|h_i+\beta\sqrt{\frac{n-2}{n}}Z_i+\frac{\beta}{\sqrt{n}}x_iZ_0\right|\le\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right] h i ↦ P [ h i + β n n − 2 Z i + n β x i Z 0 ≤ β Φ − 1 ( 2 1 + δ ′ ) ] is nonincreasing for h i ≥ 0 h_i\ge0 h i ≥ 0 Z 0 Z_0 Z 0 H i ′ ( x ) H'_i(x) H i ′ ( x ) Z 0 Z_0 Z 0 i i i h i = 0 h_i=0 h i = 0 Z 0 = 0 Z_0=0 Z 0 = 0 H i ′ ′ = β n − 2 n Z i H''_i=\beta\sqrt{\frac{n-2}{n}}Z_i H i ′′ = β n n − 2 Z i x x x (23) it is enough to show that
P [ ∣ { i : ∣ H i ′ ′ ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } ∣ ≥ δ n ] ≤ exp ( − n D K L ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) . (24) \mathbb{P}\left[\left|\left\{i:\left|H''_i\right|\le\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\right|\ge\delta n\right]\le\exp\left(-nD_{\mathrm{KL}}\left(\operatorname{Ber}(\delta)\mathbin\Vert\operatorname{Ber}\left(\delta'+o_n(1)\right)\right)\right). \tag*{(24)} P [ { i : ∣ H i ′′ ∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } ≥ δ n ] ≤ exp ( − n D KL ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) . (24) Let Y i Y_i Y i H i ′ ′ > β Φ − 1 ( 1 + δ ′ 2 ) H''_i>\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right) H i ′′ > β Φ − 1 ( 2 1 + δ ′ ) i ∈ [ n ] i\in[n] i ∈ [ n ]
P ( Y i ) ≤ 2 Φ ( Φ − 1 ( 1 + δ ′ 2 ) n n − 2 ) − 1 = δ ′ + o n ( 1 ) . \mathbb{P}(Y_i)\le2\Phi\left(\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\sqrt{\frac{n}{n-2}}\right)-1=\delta'+o_n(1). P ( Y i ) ≤ 2Φ ( Φ − 1 ( 2 1 + δ ′ ) n − 2 n ) − 1 = δ ′ + o n ( 1 ) . It follows by the Chernoff bound that
P [ ∑ i = 1 n Y i ≥ δ n ] ≤ exp ( − n D K L ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) δ ′ < δ . \mathbb{P}\left[\sum_{i=1}^{n}Y_i\ge\delta n\right]\le\exp\left(-nD_{\mathrm{KL}}\left(\operatorname{Ber}(\delta)\mathbin\Vert\operatorname{Ber}\left(\delta'+o_n(1)\right)\right)\right)_{\delta'<\delta}. P [ i = 1 ∑ n Y i ≥ δ n ] ≤ exp ( − n D KL ( Ber ( δ ) ∥ Ber ( δ ′ + o n ( 1 ) ) ) ) δ ′ < δ . This proves (24) which concludes the proof.
Proof of Theorem 1.5
In this section we prove Theorem 1.5. All expectations in this section are with respect to μ J \mu_J μ J f : { ± 1 } n → R f:\{\pm1\}^n\to\mathbb{R} f : { ± 1 } n → R
First, we observe that
⟨ L J f , L J f ⟩ = ∑ i E [ ( L i f ) 2 ] + ∑ i ≠ j E [ L i f L j f ] = ∑ i E [ ( L i f ) 2 ] + ∑ i ≠ j E X − i − j [ E X − i − j [ L i f L j f ] ] . (25) \langle L_Jf,L_Jf\rangle=\sum_i\mathbb{E}[(L_if)^2]+\sum_{i\ne j}\mathbb{E}[L_ifL_jf]=\sum_i\mathbb{E}[(L_if)^2]+\sum_{i\ne j}\mathbb{E}_{X_{-i-j}}\left[\mathbb{E}_{X_{-i-j}}[L_ifL_jf]\right]. \tag*{(25)} ⟨ L J f , L J f ⟩ = i ∑ E [( L i f ) 2 ] + i = j ∑ E [ L i f L j f ] = i ∑ E [( L i f ) 2 ] + i = j ∑ E X − i − j [ E X − i − j [ L i f L j f ] ] . (25) where E X − i − j \mathbb{E}_{X_{-i-j}} E X − i − j μ X − i − j \mu_{X_{-i-j}} μ X − i − j ( X i , X j ) (X_i,X_j) ( X i , X j ) X − i − j X_{-i-j} X − i − j x ∈ { ± 1 } n x\in\{\pm1\}^n x ∈ { ± 1 } n
μ x − i , − j ( x i , x j ) = μ ( x ) ∑ X i , X j ∈ { ± 1 } μ ( x − i − j , X i , X j ) ∝ exp ( 1 2 J i j x i x j + 1 2 x i ∑ k ≠ i , j J i k x k + 1 2 x j ∑ k ≠ i , j J j k x k ) . \mu_{x_{-i,-j}}(x_i,x_j)=\frac{\mu(x)}{\sum_{X_i,X_j\in\{\pm1\}}\mu(x_{-i-j},X_i,X_j)}\propto\exp\left(\frac{1}{2}J_{ij}x_ix_j+\frac{1}{2}x_i\sum_{k\ne i,j}J_{ik}x_k+\frac{1}{2}x_j\sum_{k\ne i,j}J_{jk}x_k\right). μ x − i , − j ( x i , x j ) = ∑ X i , X j ∈ { ± 1 } μ ( x − i − j , X i , X j ) μ ( x ) ∝ exp 2 1 J ij x i x j + 2 1 x i k = i , j ∑ J ik x k + 2 1 x j k = i , j ∑ J j k x k . The following theorem is our local rank-1-perturbation lemma; here, we rename X i , X j X_i,X_j X i , X j X , Y X,Y X , Y L X , L Y L_X,L_Y L X , L Y E [ X ‾ ( y ) 2 ⋅ L X f ( x , y ) 2 + Y ‾ ( x ) 2 ⋅ L Y f ( x , y ) 2 ] \mathbb{E}[\overline{X}(y)^2\cdot L_Xf(x,y)^2+\overline{Y}(x)^2\cdot L_Yf(x,y)^2] E [ X ( y ) 2 ⋅ L X f ( x , y ) 2 + Y ( x ) 2 ⋅ L Y f ( x , y ) 2 ]
Theorem 3.1 (Main Technical Lemma). Let d , s ≤ 1 d,s\leq1 d , s ≤ 1 λ > 0 \lambda>0 λ > 0
d ≥ max { − ( s 2 + 2 s − 1 ) , 2 s 2 } (26) d\geq\max\left\{-(s^2+2s-1),2s^2\right\} \tag*{(26)} d ≥ max { − ( s 2 + 2 s − 1 ) , 2 s 2 } (26) Then there exists j ( d , λ , s ) j(d,\lambda,s) j ( d , λ , s ) J J J ∣ J ∣ ≤ j ( d , λ , s ) |J|\leq j(d,\lambda,s) ∣ J ∣ ≤ j ( d , λ , s ) a , b ∈ R a,b\in\mathbb{R} a , b ∈ R
For all f : { ± 1 } 2 → R f:\{\pm1\}^2\to\mathbb{R} f : { ± 1 } 2 → R
E [ ( L X f ) ( L Y f ) ] + J ⋅ E [ M ( L X f ) ( L Y f ) ] + ( λ J ) 2 2 ⋅ E [ ( L X f ) 2 + ( L Y f ) 2 ] + d J 2 2 E [ X ‾ ( y ) 2 ⋅ L X f ( x , y ) 2 + Y ‾ ( x ) 2 ⋅ L Y f ( x , y ) 2 ] ≥ 0. \mathbb{E}[(L_Xf)(L_Yf)]+J\cdot\mathbb{E}[M(L_Xf)(L_Yf)]+\frac{(\lambda J)^2}{2}\cdot\mathbb{E}[(L_Xf)^2+(L_Yf)^2]
+\frac{dJ^2}{2}\mathbb{E}\left[\overline{X}(y)^2\cdot L_Xf(x,y)^2+\overline{Y}(x)^2\cdot L_Yf(x,y)^2\right]\geq0. E [( L X f ) ( L Y f )] + J ⋅ E [ M ( L X f ) ( L Y f )] + 2 ( λ J ) 2 ⋅ E [( L X f ) 2 + ( L Y f ) 2 ] + 2 d J 2 E [ X ( y ) 2 ⋅ L X f ( x , y ) 2 + Y ( x ) 2 ⋅ L Y f ( x , y ) 2 ] ≥ 0. where M ( x , y ) = ( x + s X ‾ ) ( y + s Y ‾ ) M(x,y)=(x+s\overline{X})(y+s\overline{Y}) M ( x , y ) = ( x + s X ) ( y + s Y )
We defer the proof of Theorem 3.1 to Section 4. But we would like to point out that the main advantage of the above statement is that it has no dependency on the external fields a , b a,b a , b μ J , a , b \mu_{J,a,b} μ J , a , b
We will later choose ∣ d ∣ , ∣ s ∣ ≤ 1 |d|,|s|\leq1 ∣ d ∣ , ∣ s ∣ ≤ 1 J J J j ( d , λ , s ) j(d,\lambda,s) j ( d , λ , s ) μ x − i − j \mu_{x-i-j} μ x − i − j
E x − i − j [ ( L i f ) ( L j f ) ] + J i j ⋅ E x − i − j [ M i j ( L i f ) ( L j f ) ] + ( λ J i j ) 2 2 ⋅ E x − i − j [ ( L i f ) 2 + ( L j f ) 2 ] + d J i j 2 2 E x − i − j [ X ‾ i 2 ⋅ ( L i f ) 2 + X ‾ j 2 ⋅ ( L j f ) 2 ] ≥ 0. (27) \mathbb{E}_{x-i-j}[(L_if)(L_jf)]+J_{ij}\cdot\mathbb{E}_{x-i-j}[M_{ij}(L_if)(L_jf)]+\frac{(\lambda J_{ij})^2}{2}\cdot\mathbb{E}_{x-i-j}[(L_if)^2+(L_jf)^2]
+\frac{dJ_{ij}^2}{2}\mathbb{E}_{x-i-j}\left[\overline{X}_i^2\cdot(L_if)^2+\overline{X}_j^2\cdot(L_jf)^2\right]\geq0. \tag*{(27)} E x − i − j [( L i f ) ( L j f )] + J ij ⋅ E x − i − j [ M ij ( L i f ) ( L j f )] + 2 ( λ J ij ) 2 ⋅ E x − i − j [( L i f ) 2 + ( L j f ) 2 ] + 2 d J ij 2 E x − i − j [ X i 2 ⋅ ( L i f ) 2 + X j 2 ⋅ ( L j f ) 2 ] ≥ 0. (27) For 1 ≤ i ≤ n 1\leq i\leq n 1 ≤ i ≤ n ∥ J i ∥ 2 = ∑ j J i , j 2 \lVert J_i\rVert^2=\sum_j J_{i,j}^2 ∥ J i ∥ 2 = ∑ j J i , j 2 μ − i , − j ( X − i , − j ) \mu_{-i,-j}(X_{-i,-j}) μ − i , − j ( X − i , − j ) i , j ∈ [ n ] i,j\in[n] i , j ∈ [ n ]
∑ i ≠ j E [ ( L i f ) ( L j f ) ] ≥ − ∑ i ≠ j J i j E [ M i j ( L i f ) ( L j f ) ] − λ 2 ∑ i ∥ J i ∥ 2 E [ ( L i f ) 2 ] − 2 d ∑ i ∥ J i ∥ 2 E [ X ‾ i 2 L i f 2 ] ≥ − ∑ i ≠ j J i j E [ M i j ( L i f ) ( L j f ) ] − λ 2 ( β 2 + o ( 1 ) ) E ( f ) − d ( β 2 + o ( 1 ) ) ∑ i E [ X ‾ i 2 L i f 2 ] = − ∑ i ≠ j J i j E [ M i j ( L i f ) ( L j f ) ] − λ 2 ( β 2 + o ( 1 ) ) E ( f ) − d ( β 2 + o ( 1 ) ) D ( f ) . (28) \begin{aligned}
\sum_{i\ne j}\mathbb{E}[(L_if)(L_jf)]\geq-\sum_{i\ne j}J_{ij}\mathbb{E}[M_{ij}(L_if)(L_jf)]-\lambda^2\sum_i\lVert J_i\rVert^2\mathbb{E}[(L_if)^2]-2d\sum_i\lVert J_i\rVert^2\mathbb{E}[\overline{X}_i^2L_i f^2] \\
&\geq-\sum_{i\ne j}J_{ij}\mathbb{E}[M_{ij}(L_if)(L_jf)]-\lambda^2(\beta^2+o(1))\mathcal{E}(f)-d(\beta^2+o(1))\sum_i\mathbb{E}[\overline{X}_i^2L_i f^2] \\
&=-\sum_{i\ne j}J_{ij}\mathbb{E}[M_{ij}(L_if)(L_jf)]-\lambda^2(\beta^2+o(1))\mathcal{E}(f)-d(\beta^2+o(1))\mathcal{D}(f). \tag*{(28)}
\end{aligned} i = j ∑ E [( L i f ) ( L j f )] ≥ − i = j ∑ J ij E [ M ij ( L i f ) ( L j f )] − λ 2 i ∑ ∥ J i ∥ 2 E [( L i f ) 2 ] − 2 d i ∑ ∥ J i ∥ 2 E [ X i 2 L i f 2 ] ≥ − i = j ∑ J ij E [ M ij ( L i f ) ( L j f )] − λ 2 ( β 2 + o ( 1 )) E ( f ) − d ( β 2 + o ( 1 )) i ∑ E [ X i 2 L i f 2 ] = − i = j ∑ J ij E [ M ij ( L i f ) ( L j f )] − λ 2 ( β 2 + o ( 1 )) E ( f ) − d ( β 2 + o ( 1 )) D ( f ) . (28) We note that we lose a factor of 2 for the two terms in the RHS of the first inequality because we sum (25) over all un-ordered pairs i , j i,j i , j
Lemma 3.2. Let J ~ = ∥ J ∥ op I − J \widetilde{J}=\lVert J\rVert_{\operatorname{op}}I-J J = ∥ J ∥ op I − J w : { ± 1 } n → R n \mathbf{w}:\{\pm1\}^{n}\to\mathbb{R}^{n} w : { ± 1 } n → R n 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n w i : { ± 1 } n → R w_i:\{\pm1\}^{n}\to\mathbb{R} w i : { ± 1 } n → R
w i ( x 1 , … , x n ) = ( x i + s X ‾ i ( x − i ) ) ⋅ L i f ( x ) . w_i(x_1,\ldots,x_n)=(x_i+s\overline{X}_i(\mathbf{x}_{-i}))\cdot L_i f(\mathbf{x}). w i ( x 1 , … , x n ) = ( x i + s X i ( x − i )) ⋅ L i f ( x ) . Then,
∑ i ≠ j J i j E [ M i j ( L i f ) ( L j f ) ] = ∥ J ∥ op ( E ( f ) − s ( 2 − s ) D ( f ) ) + E [ w T J ~ w ] . \sum_{i\ne j}J_{ij}\mathbb{E}[M_{ij}(L_i f)(L_j f)]=\lVert J\rVert_{\operatorname{op}}\left(\mathcal{E}(f)-s(2-s)\mathcal{D}(f)\right)+\mathbb{E}[\mathbf{w}^{\mathsf{T}}\widetilde{J}\mathbf{w}]. i = j ∑ J ij E [ M ij ( L i f ) ( L j f )] = ∥ J ∥ op ( E ( f ) − s ( 2 − s ) D ( f ) ) + E [ w T J w ] . Proof. Given x ∈ { ± 1 } n \mathbf{x}\in\{\pm1\}^{n} x ∈ { ± 1 } n W ( x ) = w w T W(\mathbf{x})=\mathbf{w}\mathbf{w}^{\mathsf{T}} W ( x ) = w w T
∑ i ≠ j M i j ( x ) J i j L i f ( x ) L j f ( x ) = J ∙ W ( x ) , \sum_{i\ne j}M_{ij}(\mathbf{x})J_{ij}L_i f(\mathbf{x})L_j f(\mathbf{x})=J\mathbin{\bullet}W(\mathbf{x}), i = j ∑ M ij ( x ) J ij L i f ( x ) L j f ( x ) = J ∙ W ( x ) , where we are crucially using that the matrix M M M M i j = ( x i + s X ‾ i ( x − i ) ) ( x j + s X ‾ j ( x − j ) ) M_{ij}=(x_i+s\overline{X}_i(\mathbf{x}_{-i}))(x_j+s\overline{X}_j(\mathbf{x}_{-j})) M ij = ( x i + s X i ( x − i )) ( x j + s X j ( x − j )) W = E x ∼ μ J W ( x ) W=\mathbb{E}_{\mathbf{x}\sim\mu_J}W(\mathbf{x}) W = E x ∼ μ J W ( x )
∑ i ≠ j J i j E [ M i j ( L i f ) ( L j f ) ] = J ∙ E [ W ( x ) ] = ( ∥ J ∥ op I − J ~ ) ∙ W = ∥ J ∥ op ( I ∙ W ) − J ~ ∙ W = ∥ J ∥ op E μ β [ ∥ w ∥ 2 2 ] − E [ w T J ~ w ] . \begin{aligned}
\sum_{i\ne j}J_{ij}\mathbb{E}[M_{ij}(L_i f)(L_j f)]
&=J\mathbin{\bullet}\mathbb{E}[W(\mathbf{x})]
=(\lVert J\rVert_{\operatorname{op}}I-\widetilde{J})\mathbin{\bullet}W \\
&=\lVert J\rVert_{\operatorname{op}}(I\mathbin{\bullet}W)-\widetilde{J}\mathbin{\bullet}W \\
&=\lVert J\rVert_{\operatorname{op}}\mathbb{E}_{\mu_\beta}[\lVert\mathbf{w}\rVert_2^2]-\mathbb{E}[\mathbf{w}^{\mathsf{T}}\widetilde{J}\mathbf{w}].
\end{aligned} i = j ∑ J ij E [ M ij ( L i f ) ( L j f )] = J ∙ E [ W ( x )] = (∥ J ∥ op I − J ) ∙ W = ∥ J ∥ op ( I ∙ W ) − J ∙ W = ∥ J ∥ op E μ β [∥ w ∥ 2 2 ] − E [ w T J w ] . Therefore, to finish the proof it is enough to show that E μ β [ ∥ w ∥ 2 2 ] = E ( f ) − s ( 2 − s ) D ( f ) \mathbb{E}_{\mu_\beta}[\lVert\mathbf{w}\rVert_2^2]=\mathcal{E}(f)-s(2-s)\mathcal{D}(f) E μ β [∥ w ∥ 2 2 ] = E ( f ) − s ( 2 − s ) D ( f ) g ≡ 1 g\equiv1 g ≡ 1 i ∈ [ n ] i\in[n] i ∈ [ n ]
Claim 3.3. Let w \mathbf{w} w g : [ − 1 , 1 ] → R g:[-1,1]\to\mathbb{R} g : [ − 1 , 1 ] → R i ∈ [ n ] i\in[n] i ∈ [ n ]
E [ g ( X ‾ i ) w i ( X ) 2 ] = E [ g ( X ‾ i ) ( 1 − s ( 2 − s ) X ‾ i 2 ) ( L i f ) 2 ] . \mathbb{E}\left[g(\overline{X}_i)w_i(\mathbf{X})^2\right]=\mathbb{E}\left[g(\overline{X}_i)\left(1-s(2-s)\overline{X}_i^2\right)(L_i f)^2\right]. E [ g ( X i ) w i ( X ) 2 ] = E [ g ( X i ) ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ] . Proof. By Fact 2.2, it follows that there exists some function κ ( x − i ) \kappa(\mathbf{x}_{-i}) κ ( x − i ) f f f
L i f ( x ) = κ ( x − i ) ( x i − X ‾ i ( x − i ) ) . L_i f(\mathbf{x})=\kappa(\mathbf{x}_{-i})(x_i-\overline{X}_i(\mathbf{x}_{-i})). L i f ( x ) = κ ( x − i ) ( x i − X i ( x − i )) . Consequently, it follows that
w i ( x ) = ( x i + s X ‾ i ( x − i ) ) L i f ( x ) = κ ( x − i ) ( 1 − s X ‾ i ( x − i ) 2 + ( s − 1 ) X ‾ i ( x − i ) x i ) . w_i(\mathbf{x})=(x_i+s\overline{X}_i(\mathbf{x}_{-i}))L_i f(\mathbf{x})=\kappa(\mathbf{x}_{-i})\left(1-s\overline{X}_i(\mathbf{x}_{-i})^2+(s-1)\overline{X}_i(\mathbf{x}_{-i})x_i\right). w i ( x ) = ( x i + s X i ( x − i )) L i f ( x ) = κ ( x − i ) ( 1 − s X i ( x − i ) 2 + ( s − 1 ) X i ( x − i ) x i ) . That is, we have
E [ g ( X ‾ i ) w i 2 ∣ X − i = x − i ] = g ( X ‾ i ) κ ( x − i ) 2 E [ ( 1 − s X ‾ i ( x − i ) 2 + ( s − 1 ) X ‾ i ( x − i ) X i ) 2 | X − i = x − i ] = g ( X ‾ i ) κ ( x − i ) 2 ( 1 − X ‾ i ( x − i ) 2 ) ( 1 − s ( 2 − s ) X ‾ i ( x − i ) 2 ) = E [ g ( X ‾ i ) ( L i f ) 2 ∣ X − i = x − i ] ( 1 − s ( 2 − s ) X ‾ i ( x − i ) 2 ) = E [ g ( X ‾ i ) ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ∣ X − i = x − i ] . \begin{aligned}
\mathbb{E}[g(\overline{X}_i)w_i^2\mid X_{-i}=x_{-i}]
&=g(\overline{X}_i)\kappa(x_{-i})^2\mathbb{E}\left[\left(1-s\overline{X}_i(x_{-i})^2+(s-1)\overline{X}_i(x_{-i})X_i\right)^2\middle|X_{-i}=x_{-i}\right]\\
&=g(\overline{X}_i)\kappa(x_{-i})^2(1-\overline{X}_i(x_{-i})^2)(1-s(2-s)\overline{X}_i(x_{-i})^2)\\
&=\mathbb{E}[g(\overline{X}_i)(L_i f)^2\mid X_{-i}=x_{-i}](1-s(2-s)\overline{X}_i(x_{-i})^2)\\
&=\mathbb{E}[g(\overline{X}_i)(1-s(2-s)X_i^2)(L_i f)^2\mid X_{-i}=x_{-i}].
\end{aligned} E [ g ( X i ) w i 2 ∣ X − i = x − i ] = g ( X i ) κ ( x − i ) 2 E [ ( 1 − s X i ( x − i ) 2 + ( s − 1 ) X i ( x − i ) X i ) 2 X − i = x − i ] = g ( X i ) κ ( x − i ) 2 ( 1 − X i ( x − i ) 2 ) ( 1 − s ( 2 − s ) X i ( x − i ) 2 ) = E [ g ( X i ) ( L i f ) 2 ∣ X − i = x − i ] ( 1 − s ( 2 − s ) X i ( x − i ) 2 ) = E [ g ( X i ) ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ∣ X − i = x − i ] . Taking another expectation and using the tower property gives the desired conclusion.
Putting equations (25) and (28) and Lemma 3.2 together we obtain
⟨ L J f , L J f ⟩ = E ( f ) + ∑ i ≠ j E [ ( L i f ) ( L j f ) ] ≥ ( 1 − ∥ J ∥ op − λ 2 β 2 − o ( 1 ) ) E ( f ) + E [ w T J ~ w ] + ( s ( 2 − s ) ∥ J ∥ op − d β 2 − o ( 1 ) ) D ( f ) , (29) \begin{aligned}
\langle\mathcal{L}_J f,\mathcal{L}_J f\rangle= \mathcal{E}(f) + \sum_{i\ne j} \mathbb{E}[(L_i f)(L_j f)] \\
&\ge\left(1-\lVert J\rVert_{\operatorname{op}}-\lambda^2\beta^2-o(1)\right)\mathcal{E}(f)+\mathbb{E}[\mathbf{w}^{\mathsf{T}}\widetilde{J}\mathbf{w}]+\left(s(2-s)\lVert J\rVert_{\operatorname{op}}-d\beta^2-o(1)\right)\mathcal{D}(f), \tag*{(29)}
\end{aligned} ⟨ L J f , L J f ⟩ = E ( f ) + i = j ∑ E [( L i f ) ( L j f )] ≥ ( 1 − ∥ J ∥ op − λ 2 β 2 − o ( 1 ) ) E ( f ) + E [ w T J w ] + ( s ( 2 − s ) ∥ J ∥ op − d β 2 − o ( 1 ) ) D ( f ) , (29) where w : { ± 1 } n → R n \mathbf{w}:\{\pm1\}^{n}\to\mathbb{R}^{n} w : { ± 1 } n → R n ε ′ > 0 \varepsilon'>0 ε ′ > 0
Case 1: β ∈ ( 0 , 1 2 − ε ) \beta\in\left(0,\frac{1}{2}-\varepsilon\right) β ∈ ( 0 , 2 1 − ε ) J ~ ⪰ 0 \widetilde{J}\succeq0 J ⪰ 0 D ( f ) ≥ 0 \mathcal{D}(f)\ge0 D ( f ) ≥ 0 d , s d,s d , s s ( 2 − s ) ( 2 β ) − d β 2 > 0 s(2-s)(2\beta)-d\beta^2>0 s ( 2 − s ) ( 2 β ) − d β 2 > 0 (26) , then we have for large enough n n n
⟨ L J f , L J f ⟩ ≥ ( 1 − 2 β − λ 2 β 2 − o ( 1 ) ) E ( f ) > β ≤ ( 1 − ε ) / 2 , λ sufficiently small ε ⋅ E ( f ) . \langle\mathcal{L}_J f,\mathcal{L}_J f\rangle\ge\left(1-2\beta-\lambda^2\beta^2-o(1)\right)\mathcal{E}(f) \mathrel{\underset{\beta\le(1-\varepsilon)/2,\ \lambda\ \text{sufficiently small}}{>}} \varepsilon\cdot\mathcal{E}(f). ⟨ L J f , L J f ⟩ ≥ ( 1 − 2 β − λ 2 β 2 − o ( 1 ) ) E ( f ) β ≤ ( 1 − ε ) /2 , λ sufficiently small > ε ⋅ E ( f ) . It can be checked for instance that s = 1 3 s=\frac{1}{3} s = 3 1 d = 2 9 d=\frac{2}{9} d = 9 2 s ( 2 − s ) ( 2 β ) − d β 2 > 0 s(2-s)(2\beta)-d\beta^2>0 s ( 2 − s ) ( 2 β ) − d β 2 > 0 (26) . This concludes the proof of Theorem 1.5 in this case.
Case 2: 1 2 − ε ≤ β ≤ 1 2 + ε \frac{1}{2}-\varepsilon\le\beta\le\frac{1}{2}+\varepsilon 2 1 − ε ≤ β ≤ 2 1 + ε β ≤ 1 2 + ε \beta\le\frac{1}{2}+\varepsilon β ≤ 2 1 + ε
G ( f ) > ( 2 ε + 9 ⋅ 10 − 6 ) ⋅ E ( f ) . (30) \mathcal{G}(f)>(2\varepsilon+9\cdot10^{-6})\cdot\mathcal{E}(f). \tag*{(30)} G ( f ) > ( 2 ε + 9 ⋅ 1 0 − 6 ) ⋅ E ( f ) . (30) We first show how to complete the proof assuming (30) . By Proposition 2.7, with high probability we have ∥ J ∥ op = 2 β ± o ( 1 ) \lVert J\rVert_{\operatorname{op}}=2\beta\pm o(1) ∥ J ∥ op = 2 β ± o ( 1 )
E [ ( L J f ) 2 ] ≥ ( 1 − 2 β − λ 2 β 2 + 2 ε + 9 ⋅ 10 − 6 ) ⋅ E ( f ) ≥ ( 9 ⋅ 10 − 6 − λ 2 β 2 ) ⋅ E ( f ) , \mathbb{E}\left[(\mathcal{L}_Jf)^2\right]\ge\left(1-2\beta-\lambda^2\beta^2+2\varepsilon+9\cdot10^{-6}\right)\cdot\mathcal{E}(f)\ge\left(9\cdot10^{-6}-\lambda^2\beta^2\right)\cdot\mathcal{E}(f), E [ ( L J f ) 2 ] ≥ ( 1 − 2 β − λ 2 β 2 + 2 ε + 9 ⋅ 1 0 − 6 ) ⋅ E ( f ) ≥ ( 9 ⋅ 1 0 − 6 − λ 2 β 2 ) ⋅ E ( f ) , which would complete the proof of Theorem 1.5 in this case by taking λ \lambda λ
It remains to justify (30) . For the remainder of the proof, we work in the high probability event that Proposition 2.7, Lemma 2.10 and Lemma 2.14 hold. We will choose 0 < δ ′ < δ 0<\delta'<\delta 0 < δ ′ < δ D K L ( Ber ( δ ) ∥ Ber ( δ ′ ) ) > ln 2 D_{\mathrm{KL}}(\operatorname{Ber}(\delta)\mathbin\Vert\operatorname{Ber}(\delta'))>\ln2 D KL ( Ber ( δ ) ∥ Ber ( δ ′ )) > ln 2
Fix x ∈ { ± 1 } n \mathbf{x}\in\{\pm1\}^{n} x ∈ { ± 1 } n
H ≤ ( x ) = { i : ∣ H i ( x ) ∣ ≤ β Φ − 1 ( 1 + δ ′ 2 ) } , H_{\le}(\mathbf{x})=\left\{i: \lvert H_i(\mathbf{x})\rvert\le\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}, H ≤ ( x ) = { i : ∣ H i ( x )∣ ≤ β Φ − 1 ( 2 1 + δ ′ ) } , and we define H > ( x ) = [ n ] ∖ H ≤ ( x ) H_{>}(\mathbf{x})=[n]\setminus H_{\le}(\mathbf{x}) H > ( x ) = [ n ] ∖ H ≤ ( x )
By Lemma 2.14, ∣ H ≤ ( x ) ∣ ≤ δ n \lvert H_{\le}(\mathbf{x})\rvert\le\delta n ∣ H ≤ ( x )∣ ≤ δ n
∥ w ( x ) ∥ 2 , ≤ ( 1 − δ ) n 2 ≤ ∑ i ∈ H > ( x ) w i ( x ) 2 , (31) \lVert\mathbf{w}(\mathbf{x})\rVert_{2,\le(1-\delta)n}^{2}\le\sum_{i\in H_{>}(\mathbf{x})}w_i(\mathbf{x})^2, \tag*{(31)} ∥ w ( x ) ∥ 2 , ≤ ( 1 − δ ) n 2 ≤ i ∈ H > ( x ) ∑ w i ( x ) 2 , (31) Next, by Lemma 2.10 with β \beta β ρ > 0 \rho>0 ρ > 0 G ( f ) \mathcal{G}(f) G ( f )
w T ( x ) J ~ w ( x ) ≥ ( c − o ( 1 ) ) ∥ w ( x ) ∥ 2 2 − ρ ∥ w ( x ) ∥ 2 , ≤ ( 1 − δ ) n 2 ≥ ( 28 ) ( c − o ( 1 ) ) ∥ w ( x ) ∥ 2 2 − ρ ∑ i ∈ H > ( x ) w i ( x ) 2 , \mathbf{w}^{\mathsf{T}}(\mathbf{x})\widetilde{J}\mathbf{w}(\mathbf{x})\ge(c-o(1))\lVert\mathbf{w}(\mathbf{x})\rVert_2^2-\rho\lVert\mathbf{w}(\mathbf{x})\rVert_{2,\le(1-\delta)n}^2\underset{(28)}{\ge}(c-o(1))\lVert\mathbf{w}(\mathbf{x})\rVert_2^2-\rho\sum_{i\in H_{>}(\mathbf{x})}w_i(\mathbf{x})^2, w T ( x ) J w ( x ) ≥ ( c − o ( 1 )) ∥ w ( x ) ∥ 2 2 − ρ ∥ w ( x ) ∥ 2 , ≤ ( 1 − δ ) n 2 ( 28 ) ≥ ( c − o ( 1 )) ∥ w ( x ) ∥ 2 2 − ρ i ∈ H > ( x ) ∑ w i ( x ) 2 , where c = c ( β , ρ , δ ) = π β ρ 6.1 ( β + ρ ) ( 1 − δ ) 3 c=c(\beta,\rho,\delta)=\frac{\pi\beta\rho}{6.1(\beta+\rho)}(1-\delta)^3 c = c ( β , ρ , δ ) = 6.1 ( β + ρ ) π β ρ ( 1 − δ ) 3
Now, taking expectations in the above, we obtain
E [ w T J ~ w ] ≥ ( c − o ( 1 ) ) E [ ∥ w ∥ 2 2 ] − ρ E [ ∑ i ∈ H > ( x ) w i 2 ] . (32) \mathbb{E}[\mathbf{w}^{\mathsf{T}}\widetilde{J}\mathbf{w}] \ge(c-o(1))\mathbb{E}[\lVert\mathbf{w}\rVert_{2}^{2}]-\rho\mathbb{E}\left[\sum_{i\in H_{>}(x)}w_i^{2}\right]. \tag*{(32)} E [ w T J w ] ≥ ( c − o ( 1 )) E [∥ w ∥ 2 2 ] − ρ E i ∈ H > ( x ) ∑ w i 2 . (32) We write,
G ( f ) = E [ w T J ~ w ] + ( s ( 2 − s ) ∥ J ∥ op − d β 2 − o ( 1 ) ) D ( f ) G(f)=\mathbb{E}[\mathbf{w}^{\mathsf{T}}\widetilde{J}\mathbf{w}]+\left(s(2-s)\lVert J\rVert_{\operatorname{op}}-d\beta^{2}-o(1)\right)D(f) G ( f ) = E [ w T J w ] + ( s ( 2 − s ) ∥ J ∥ op − d β 2 − o ( 1 ) ) D ( f ) ≥ ( c − o ( 1 ) ) E [ ∥ w ∥ 2 2 ] − ρ E [ ∑ i ( 1 { i ∈ H > ( x ) } ⋅ w i 2 ) ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) ∑ i E [ X ‾ i 2 ( L i f ) 2 ] \begin{aligned}
&\ge(c-o(1))\mathbb{E}[\lVert\mathbf{w}\rVert_{2}^{2}]-\rho\mathbb{E}\left[\sum_i\left(\mathbf{1}\{i\in H_{>}(x)\}\cdot w_i^{2}\right)\right]\\
&\quad+\left(2s(2-s)\beta-\beta^{2}d-o(1)\right)\sum_i\mathbb{E}\left[\overline{X}_i^{2}(L_i f)^{2}\right]
\end{aligned} ≥ ( c − o ( 1 )) E [∥ w ∥ 2 2 ] − ρ E [ i ∑ ( 1 { i ∈ H > ( x )} ⋅ w i 2 ) ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) i ∑ E [ X i 2 ( L i f ) 2 ] Claim 3.3
≥ ( c − o ( 1 ) ) E [ ∥ w ∥ 2 2 ] − ρ ∑ i E [ 1 { H i ( x ) > β Φ − 1 ( 1 + δ ′ 2 ) } ( 1 − s ( 2 − s ) X ‾ i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) ∑ i E [ X ‾ i 2 ( L i f ) 2 ] . \begin{aligned}
&\ge(c-o(1))\mathbb{E}[\lVert\mathbf{w}\rVert_{2}^{2}]-\rho\sum_i\mathbb{E}\left[\mathbf{1}\left\{H_i(x)>\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right\}\left(1-s(2-s)\overline{X}_i^{2}\right)(L_i f)^{2}\right]\\
&\quad+\left(2s(2-s)\beta-\beta^{2}d-o(1)\right)\sum_i\mathbb{E}\left[\overline{X}_i^{2}(L_i f)^{2}\right].
\end{aligned} ≥ ( c − o ( 1 )) E [∥ w ∥ 2 2 ] − ρ i ∑ E [ 1 { H i ( x ) > β Φ − 1 ( 2 1 + δ ′ ) } ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) i ∑ E [ X i 2 ( L i f ) 2 ] . using X ‾ i ( x ) = tanh H i ( x ) \overline{X}_i(x)=\tanh H_i(x) X i ( x ) = tanh H i ( x ) δ ‾ = tanh ( β Φ − 1 ( 1 + δ ′ 2 ) ) \overline{\delta}=\tanh\left(\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right) δ = tanh ( β Φ − 1 ( 2 1 + δ ′ ) ) tanh ( x ) \tanh(x) tanh ( x ) x > 0 x>0 x > 0
≥ ( c − o ( 1 ) ) E [ ∥ w ∥ 2 2 ] − ρ ∑ i E [ 1 { X ‾ i > δ ‾ } ( 1 − s ( 2 − s ) X ‾ i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) ∑ i E [ X ‾ i 2 ( L i f ) 2 ] . \begin{aligned}
&\ge(c-o(1))\mathbb{E}[\lVert\mathbf{w}\rVert_{2}^{2}]-\rho\sum_i\mathbb{E}\left[\mathbf{1}\{\overline{X}_i>\overline{\delta}\}\left(1-s(2-s)\overline{X}_i^{2}\right)(L_i f)^{2}\right]\\
&\quad+\left(2s(2-s)\beta-\beta^{2}d-o(1)\right)\sum_i\mathbb{E}\left[\overline{X}_i^{2}(L_i f)^{2}\right].
\end{aligned} ≥ ( c − o ( 1 )) E [∥ w ∥ 2 2 ] − ρ i ∑ E [ 1 { X i > δ } ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) i ∑ E [ X i 2 ( L i f ) 2 ] . Claim 3.3
≥ ∑ i ( c − o ( 1 ) ) E [ ( 1 − s ( 2 − s ) X ‾ i 2 ) ( L i f ) 2 ] − ρ ∑ i E [ 1 { X ‾ i > δ ‾ } ( 1 − s ( 2 − s ) X ‾ i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) ∑ i E [ X ‾ i 2 ( L i f ) 2 ] > ( 2 ε + 9 ⋅ 10 − 6 ) ⋅ E ( f ) . (33) \begin{aligned}
&\ge\sum_i(c-o(1))\mathbb{E}\left[(1-s(2-s)\overline{X}_i^{2})(L_i f)^{2}\right]\\
&\quad-\rho\sum_i\mathbb{E}\left[\mathbf{1}\{\overline{X}_i>\overline{\delta}\}(1-s(2-s)\overline{X}_i^{2})(L_i f)^{2}\right]\\
&\quad+\left(2s(2-s)\beta-\beta^{2}d-o(1)\right)\sum_i\mathbb{E}\left[\overline{X}_i^{2}(L_i f)^{2}\right] \\
&> (2\varepsilon+9\cdot10^{-6})\cdot\mathcal{E}(f). \tag*{(33)}
\end{aligned} ≥ i ∑ ( c − o ( 1 )) E [ ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ] − ρ i ∑ E [ 1 { X i > δ } ( 1 − s ( 2 − s ) X i 2 ) ( L i f ) 2 ] + ( 2 s ( 2 − s ) β − β 2 d − o ( 1 ) ) i ∑ E [ X i 2 ( L i f ) 2 ] > ( 2 ε + 9 ⋅ 1 0 − 6 ) ⋅ E ( f ) . (33) To see the last inequality we set ε = 5 ⋅ 10 − 5 \varepsilon=5\cdot10^{-5} ε = 5 ⋅ 1 0 − 5 s = 2 3 s=\frac{2}{3} s = 3 2 d = 8 9 d=\frac{8}{9} d = 9 8 δ = 0.642 \delta=0.642 δ = 0.642 δ ′ = 0.13319 \delta'=0.13319 δ ′ = 0.13319 ρ = 0.004693 \rho=0.004693 ρ = 0.004693
First, we observe that max { − ( s 2 + 2 s − 1 ) , 2 s 2 } = max { − 7 9 , 8 9 } = d \max\{-(s^{2}+2s-1),2s^{2}\}=\max\{-\frac{7}{9},\frac{8}{9}\}=d max { − ( s 2 + 2 s − 1 ) , 2 s 2 } = max { − 9 7 , 9 8 } = d D K L ( Ber ( 0.642 ) ∥ Ber ( 0.13319 ) ) > 0.69317 > ln 2 D_{\mathrm{KL}}(\operatorname{Ber}(0.642)\mathbin{\|}\operatorname{Ber}(0.13319))>0.69317>\ln2 D KL ( Ber ( 0.642 ) ∥ Ber ( 0.13319 )) > 0.69317 > ln 2
2 s ( 2 − s ) β − d β 2 = 8 9 ( 2 β − β 2 ) ≥ monotonic in β 8 9 ( 2 ( 1 2 − ε ) − ( 1 2 − ε ) 2 ) = 0.66662222. (34) 2s(2-s)\beta-d\beta^{2}=\frac{8}{9}(2\beta-\beta^{2})\underset{\text{monotonic in }\beta}{\ge}\frac{8}{9}\left(2\left(\frac{1}{2}-\varepsilon\right)-\left(\frac{1}{2}-\varepsilon\right)^{2}\right)=0.66662222. \tag*{(34)} 2 s ( 2 − s ) β − d β 2 = 9 8 ( 2 β − β 2 ) monotonic in β ≥ 9 8 ( 2 ( 2 1 − ε ) − ( 2 1 − ε ) 2 ) = 0.66662222. (34) 0.004693 = ρ > c ( β , ρ , δ ) > monotonic in β π ⋅ 0.49995 ⋅ 0.004693 6.1 ( 0.49995 + 0.004693 ) ( 0.358 ) 3 > 0.0001098 > 2 ε + 9 ⋅ 10 − 6 . (35) 0.004693=\rho>c(\beta,\rho,\delta)\underset{\text{monotonic in }\beta}{>}\frac{\pi\cdot0.49995\cdot0.004693}{6.1(0.49995+0.004693)}(0.358)^{3}>0.0001098>2\varepsilon+9\cdot10^{-6}. \tag*{(35)} 0.004693 = ρ > c ( β , ρ , δ ) monotonic in β > 6.1 ( 0.49995 + 0.004693 ) π ⋅ 0.49995 ⋅ 0.004693 ( 0.358 ) 3 > 0.0001098 > 2 ε + 9 ⋅ 1 0 − 6 . (35) δ ‾ 2 = tanh 2 ( β Φ − 1 ( 1 + δ ′ 2 ) ) ≥ tanh is monotonic tanh 2 ( 0.49995 ⋅ Φ − 1 ( 0.566595 ) ) > 0.00699758. (36) \overline{\delta}^{2}=\tanh^{2}\left(\beta\Phi^{-1}\left(\frac{1+\delta'}{2}\right)\right)\underset{\tanh\text{ is monotonic}}{\ge}\tanh^{2}\left(0.49995\cdot\Phi^{-1}(0.566595)\right)>0.00699758. \tag*{(36)} δ 2 = tanh 2 ( β Φ − 1 ( 2 1 + δ ′ ) ) t a n h is monotonic ≥ tanh 2 ( 0.49995 ⋅ Φ − 1 ( 0.566595 ) ) > 0.00699758. (36) For i ∈ [ n ] i\in[n] i ∈ [ n ] ∣ X ‾ i ∣ ≤ δ ‾ \lvert\overline{X}_i\rvert\le\overline{\delta} ∣ X i ∣ ≤ δ ( L i f ) 2 (L_i f)^{2} ( L i f ) 2 (33) is
c + ( 2 s ( 2 − s ) β − β 2 d − c s ( 2 − s ) ) ∣ X ‾ i ∣ 2 ≥ (31) c + ( 0.66662222 − 8 9 ⋅ 0.004693 ) ∣ X ‾ i ∣ 2 c+\left(2s(2-s)\beta-\beta^{2}d-cs(2-s)\right)\lvert\overline{X}_i\rvert^{2}\underset{\text{(31)}}{\ge}c+\left(0.66662222-\frac{8}{9}\cdot0.004693\right)\lvert\overline{X}_i\rvert^{2} c + ( 2 s ( 2 − s ) β − β 2 d − cs ( 2 − s ) ) ∣ X i ∣ 2 (31) ≥ c + ( 0.66662222 − 9 8 ⋅ 0.004693 ) ∣ X i ∣ 2 ≥ (32) c ≥ > 0 0.00010986 > 2 ε + 9 ⋅ 10 − 6 . (37) \underset{\text{(32)}}{\ge}c\underset{>0}{\ge}0.00010986>2\varepsilon+9\cdot10^{-6}. \tag*{(37)} (32) ≥ c > 0 ≥ 0.00010986 > 2 ε + 9 ⋅ 1 0 − 6 . (37) For i ∈ [ n ] i \in[n] i ∈ [ n ] ∣ X ‾ i ∣ > δ ˉ \lvert\overline{X}_i\rvert> \bar{\delta} ∣ X i ∣ > δ ˉ ( L i f ) 2 (L_i f)^2 ( L i f ) 2
( c − ρ ) + ( 2 s ( 2 − s ) β − β 2 d − s ( 2 − s ) ( c − ρ ) ) ∣ X ‾ i ∣ 2 = ( c − ρ ) ( 1 − 8 9 X ‾ i 2 ) + ( 2 s ( 2 − s ) β − β 2 d ) X ‾ i 2 (38) (c-\rho) + \left(2s(2-s)\beta-\beta^2d-s(2-s)(c-\rho)\right)\lvert\overline{X}_i\rvert^2
= (c-\rho)\left(1-\frac{8}{9}\overline{X}_i^2\right) + \left(2s(2-s)\beta-\beta^2d\right)\overline{X}_i^2 \tag*{(38)} ( c − ρ ) + ( 2 s ( 2 − s ) β − β 2 d − s ( 2 − s ) ( c − ρ ) ) ∣ X i ∣ 2 = ( c − ρ ) ( 1 − 9 8 X i 2 ) + ( 2 s ( 2 − s ) β − β 2 d ) X i 2 (38) ≥ − 0.00458314 + 0.67069612 ⋅ ∣ X ‾ i ∣ 2 > 0.000110109 > 2 ε + 9 ⋅ 10 − 6 . (39) \ge-0.00458314 + 0.67069612\cdot\lvert\overline{X}_i\rvert^2 > 0.000110109 > 2\varepsilon+ 9\cdot10^{-6}. \tag*{(39)} ≥ − 0.00458314 + 0.67069612 ⋅ ∣ X i ∣ 2 > 0.000110109 > 2 ε + 9 ⋅ 1 0 − 6 . (39) Summing up (34) and (35) over all i ∈ [ n ] i \in[n] i ∈ [ n ] ∑ i E [ ( L i f ) 2 ] = E ( f ) \sum_i \mathbb{E}[(L_i f)^2]=\mathcal{E}(f) ∑ i E [( L i f ) 2 ] = E ( f ) G ( f ) > ( 2 ε + 9 ⋅ 10 − 6 ) ⋅ E ( f ) \mathcal{G}(f)>(2\varepsilon+9\cdot10^{-6})\cdot\mathcal{E}(f) G ( f ) > ( 2 ε + 9 ⋅ 1 0 − 6 ) ⋅ E ( f )
Proof of the local inequality
In this section, we prove Theorem 3.1. We begin with decomposing a given function f : { ± 1 } 2 → R f:\{\pm1\}^2\to\mathbb{R} f : { ± 1 } 2 → R μ 0 , a , b \mu_{0,a,b} μ 0 , a , b x x x y y y
For the random variable X X X Y Y Y
φ ( X ) = X − X ‾ 0 σ X 0 , \varphi(X)=\frac{X-\overline{X}_0}{\sigma_{X_0}}, φ ( X ) = σ X 0 X − X 0 , where X ‾ 0 \overline{X}_0 X 0 X X X σ X 0 \sigma_{X_0} σ X 0 X X X X X X Y Y Y
Note that by definition E μ 0 , a , b φ ( X ) = E μ 0 , a , b φ ( Y ) = 0 \mathbb{E}_{\mu_{0,a,b}}\varphi(X)=\mathbb{E}_{\mu_{0,a,b}}\varphi(Y)=0 E μ 0 , a , b φ ( X ) = E μ 0 , a , b φ ( Y ) = 0 E μ 0 , a , b [ φ ( X ) φ ( Y ) ] = 0 \mathbb{E}_{\mu_{0,a,b}}[\varphi(X)\varphi(Y)]=0 E μ 0 , a , b [ φ ( X ) φ ( Y )] = 0 1 , φ ( X ) , φ ( Y ) , φ ( X ) φ ( Y ) 1,\varphi(X),\varphi(Y),\varphi(X)\varphi(Y) 1 , φ ( X ) , φ ( Y ) , φ ( X ) φ ( Y ) μ 0 , a , b \mu_{0,a,b} μ 0 , a , b f : { ± 1 } 2 → R f:\{\pm1\}^2\to\mathbb{R} f : { ± 1 } 2 → R
f ( x , y ) = f 0 + α φ ( x ) + β φ ( y ) + γ φ ( x ) ⋅ φ ( y ) . f(x,y)=f_0+\alpha\varphi(x)+\beta\varphi(y)+\gamma\varphi(x)\cdot\varphi(y). f ( x , y ) = f 0 + α φ ( x ) + β φ ( y ) + γ φ ( x ) ⋅ φ ( y ) . Recall that L X f = f − E [ f ∣ Y ] L_Xf=f-\mathbb{E}[f\mid Y] L X f = f − E [ f ∣ Y ] L Y f = f − E [ f ∣ X ] L_Yf=f-\mathbb{E}[f\mid X] L Y f = f − E [ f ∣ X ] f f f L X , L Y L_X,L_Y L X , L Y f f f
Lemma 4.1 (Explicit description of A A A For a function f : { ± 1 } 2 → R f:\{\pm1\}^2\to\mathbb{R} f : { ± 1 } 2 → R with f = f 0 + α φ ( x ) + β φ ( y ) + γ φ ( x ) φ ( y ) f=f_0+\alpha\varphi(x)+\beta\varphi(y)+\gamma\varphi(x)\varphi(y) f = f 0 + α φ ( x ) + β φ ( y ) + γ φ ( x ) φ ( y ) let A ∈ R 3 × 3 A\in\mathbb{R}^{3\times3} A ∈ R 3 × 3 be such that
\begin{bmatrix}\alpha& \beta& \gamma\end{bmatrix}
A
\begin{bmatrix}\alpha& \beta& \gamma\end{bmatrix}^{\mathsf{T}}
=\mathbb{E}[(L_Xf)(L_Yf)]+J\cdot\mathbb{E}[M(L_Xf)(L_Yf)]
+\frac{(\lambda J)^2}{2}\cdot\mathbb{E}[(L_Xf)^2+(L_Yf)^2]
+\frac{dJ^2}{2}\mathbb{E}[\overline{X}(y)^2\cdot L_Xf^2+\overline{Y}(x)^2\cdot L_Yf^2]. \tag*{(40)}\end{bmatrix}
Suppose also that ∣ J ∣ ≤ 1 2 \lvert J\rvert\le\frac{1}{2} ∣ J ∣ ≤ 2 1 Then we can write, A = diag ( J , J , 1 ) ( A 0 + E ) diag ( J , J , 1 ) A=\operatorname{diag}(J,J,1)(A_0+E)\operatorname{diag}(J,J,1) A = diag ( J , J , 1 ) ( A 0 + E ) diag ( J , J , 1 ) where ∥ E ∥ op ≤ O ( ∣ J ∣ ) \lVert E\rVert_{\operatorname{op}}\le O(\lvert J\rvert) ∥ E ∥ op ≤ O (∣ J ∣) and
A 0 = ( 1 2 ( λ 2 + d X ‾ 0 2 ) − 1 2 ( s 2 + 2 s − 1 ) X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 − 1 2 ( 1 − s ) X ‾ 0 σ Y 0 − 1 2 ( s 2 + 2 s − 1 ) X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 1 2 ( λ 2 + d Y ‾ 0 2 ) − 1 2 ( 1 − s ) Y ‾ 0 σ X 0 − 1 2 ( 1 − s ) X ‾ 0 σ Y 0 − 1 2 ( 1 − s ) Y ‾ 0 σ X 0 1 ) . A_0=
\begin{pmatrix}
\frac{1}{2}(\lambda^2+d\overline{X}_0^2) & -\frac{1}{2}(s^2+2s-1)\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0} & -\frac{1}{2}(1-s)\overline{X}_0\sigma_{Y_0}\\
-\frac{1}{2}(s^2+2s-1)\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0} & \frac{1}{2}(\lambda^2+d\overline{Y}_0^2) & -\frac{1}{2}(1-s)\overline{Y}_0\sigma_{X_0}\\
-\frac{1}{2}(1-s)\overline{X}_0\sigma_{Y_0} & -\frac{1}{2}(1-s)\overline{Y}_0\sigma_{X_0} & 1
\end{pmatrix}. A 0 = 2 1 ( λ 2 + d X 0 2 ) − 2 1 ( s 2 + 2 s − 1 ) X 0 Y 0 σ X 0 σ Y 0 − 2 1 ( 1 − s ) X 0 σ Y 0 − 2 1 ( s 2 + 2 s − 1 ) X 0 Y 0 σ X 0 σ Y 0 2 1 ( λ 2 + d Y 0 2 ) − 2 1 ( 1 − s ) Y 0 σ X 0 − 2 1 ( 1 − s ) X 0 σ Y 0 − 2 1 ( 1 − s ) Y 0 σ X 0 1 . We defer the proof of Lemma 4.1 to the end of this section. First we give a proof Theorem 3.1 assuming the above lemma.
Proof of Theorem 3.1. We will see that entries of A 0 A_0 A 0 A A A
Since ∥ E ∥ o p ≤ O ( ∣ J ∣ ) \lVert E\rVert_{\mathrm{op}} \le O(\lvert J\rvert) ∥ E ∥ op ≤ O (∣ J ∣) A A A c > 0 c > 0 c > 0 d d d s s s λ \lambda λ A 0 ⪰ c I A_0 \succeq cI A 0 ⪰ c I j ( d , λ , s ) = O ( c ) j(d,\lambda,s) = O(c) j ( d , λ , s ) = O ( c ) J ≤ j ( d , λ , s ) J \le j(d,\lambda,s) J ≤ j ( d , λ , s ) ∥ E ∥ o p ≤ c \lVert E\rVert_{\mathrm{op}} \le c ∥ E ∥ op ≤ c
As a first step, we will show that
[ α β γ ] A 0 [ α β γ ] T ≥ λ 2 2 ( α 2 + β 2 ) + ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 . (41) [\alpha\ \beta\ \gamma]A_0[\alpha\ \beta\ \gamma]^{\mathsf T} \ge\frac{\lambda^2}{2}(\alpha^2+\beta^2)+\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2. \tag*{(41)} [ α β γ ] A 0 [ α β γ ] T ≥ 2 λ 2 ( α 2 + β 2 ) + ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 . (41) Assuming this, we note that
α 2 + β 2 + γ 2 ≤ 2 λ 2 ⋅ ( λ 2 2 ( α 2 + β 2 ) + ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 ) + ( 1 ⋅ ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) + ( 1 − s ) X ‾ 0 σ Y 0 2 ⋅ α + ( 1 − s ) Y ‾ 0 σ X 0 2 ⋅ β ) 2 ≤ 2 λ 2 ⋅ ( λ 2 2 ( α 2 + β 2 ) + ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 ) + ( 1 + ( 1 − s ) 2 2 λ 2 + ( 1 − s ) 2 2 λ 2 ) ( ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 + λ 2 2 ( α 2 + β 2 ) ) = ( 2 λ 2 + 1 + ( 1 − s ) 2 λ 2 ) ( ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 + λ 2 2 ( α 2 + β 2 ) ) ≤ ( 2 λ 2 + 1 + ( 1 − s ) 2 λ 2 ) [ α β γ ] A 0 [ α β γ ] T . \begin{aligned}
\alpha^2+\beta^2+\gamma^2
&\le\frac{2}{\lambda^2}\cdot\left(\frac{\lambda^2}{2}(\alpha^2+\beta^2)+\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2\right)\\
&\quad+\left(1\cdot\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)+\frac{(1-s)\overline{X}_0\sigma_{Y_0}}{2}\cdot\alpha+\frac{(1-s)\overline{Y}_0\sigma_{X_0}}{2}\cdot\beta\right)^2\\
&\le\frac{2}{\lambda^2}\cdot\left(\frac{\lambda^2}{2}(\alpha^2+\beta^2)+\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2\right)\\
&\quad+\left(1+\frac{(1-s)^2}{2\lambda^2}+\frac{(1-s)^2}{2\lambda^2}\right)\left(\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2+\frac{\lambda^2}{2}(\alpha^2+\beta^2)\right)\\
&=\left(\frac{2}{\lambda^2}+1+\frac{(1-s)^2}{\lambda^2}\right)\left(\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2+\frac{\lambda^2}{2}(\alpha^2+\beta^2)\right)\\
&\le\left(\frac{2}{\lambda^2}+1+\frac{(1-s)^2}{\lambda^2}\right)[\alpha\ \beta\ \gamma]A_0[\alpha\ \beta\ \gamma]^{\mathsf T}.
\end{aligned} α 2 + β 2 + γ 2 ≤ λ 2 2 ⋅ ( 2 λ 2 ( α 2 + β 2 ) + ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 ) + ( 1 ⋅ ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) + 2 ( 1 − s ) X 0 σ Y 0 ⋅ α + 2 ( 1 − s ) Y 0 σ X 0 ⋅ β ) 2 ≤ λ 2 2 ⋅ ( 2 λ 2 ( α 2 + β 2 ) + ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 ) + ( 1 + 2 λ 2 ( 1 − s ) 2 + 2 λ 2 ( 1 − s ) 2 ) ( ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 + 2 λ 2 ( α 2 + β 2 ) ) = ( λ 2 2 + 1 + λ 2 ( 1 − s ) 2 ) ( ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 + 2 λ 2 ( α 2 + β 2 ) ) ≤ ( λ 2 2 + 1 + λ 2 ( 1 − s ) 2 ) [ α β γ ] A 0 [ α β γ ] T . Cauchy-Schwarz σ X 0 , σ Y 0 , X ‾ 0 , Y ‾ 0 ≤ 1 \sigma_{X_0},\sigma_{Y_0},\overline{X}_0,\overline{Y}_0\le1 σ X 0 , σ Y 0 , X 0 , Y 0 ≤ 1
Consequently, this shows that for c = ( 2 λ 2 + 1 + ( 1 − s ) 2 λ 2 ) − 1 c=\left(\frac{2}{\lambda^2}+1+\frac{(1-s)^2}{\lambda^2}\right)^{-1} c = ( λ 2 2 + 1 + λ 2 ( 1 − s ) 2 ) − 1 A 0 ⪰ c I A_0\succeq cI A 0 ⪰ c I
In the rest of the proof we show (37). Expanding the quadratic form gives
[ α β γ ] A 0 [ α β γ ] T = λ 2 2 ( α 2 + β 2 ) + d 2 ( X ‾ 0 2 α 2 + Y ‾ 0 2 β 2 ) − ( s 2 + 2 s − 1 ) X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 α β − ( 1 − s ) γ ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) + γ 2 = λ 2 + d X ‾ 0 2 Y ‾ 0 2 2 ( α 2 + β 2 ) + d + ( s 2 + 2 s − 1 ) 4 ( X ‾ 0 σ Y 0 α − Y ‾ 0 σ X 0 β ) 2 + d − ( s 2 + 2 s − 1 ) − ( 1 − s ) 2 4 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) 2 + ( γ − 1 − s 2 ( X ‾ 0 σ Y 0 α + Y ‾ 0 σ X 0 β ) ) 2 , \begin{aligned}
[\alpha\ \beta\ \gamma]A_0[\alpha\ \beta\ \gamma]^{\mathsf T}
&=\frac{\lambda^2}{2}(\alpha^2+\beta^2)+\frac{d}{2}(\overline{X}_0^2\alpha^2+\overline{Y}_0^2\beta^2)-(s^2+2s-1)\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0}\alpha\beta\\
&\quad-(1-s)\gamma(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)+\gamma^2\\
&=\frac{\lambda^2+d\overline{X}_0^2\overline{Y}_0^2}{2}(\alpha^2+\beta^2)+\frac{d+(s^2+2s-1)}{4}(\overline{X}_0\sigma_{Y_0}\alpha-\overline{Y}_0\sigma_{X_0}\beta)^2\\
&\quad+\frac{d-(s^2+2s-1)-(1-s)^2}{4}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)^2\\
&\quad+\left(\gamma-\frac{1-s}{2}(\overline{X}_0\sigma_{Y_0}\alpha+\overline{Y}_0\sigma_{X_0}\beta)\right)^2,
\end{aligned} [ α β γ ] A 0 [ α β γ ] T = 2 λ 2 ( α 2 + β 2 ) + 2 d ( X 0 2 α 2 + Y 0 2 β 2 ) − ( s 2 + 2 s − 1 ) X 0 Y 0 σ X 0 σ Y 0 α β − ( 1 − s ) γ ( X 0 σ Y 0 α + Y 0 σ X 0 β ) + γ 2 = 2 λ 2 + d X 0 2 Y 0 2 ( α 2 + β 2 ) + 4 d + ( s 2 + 2 s − 1 ) ( X 0 σ Y 0 α − Y 0 σ X 0 β ) 2 + 4 d − ( s 2 + 2 s − 1 ) − ( 1 − s ) 2 ( X 0 σ Y 0 α + Y 0 σ X 0 β ) 2 + ( γ − 2 1 − s ( X 0 σ Y 0 α + Y 0 σ X 0 β ) ) 2 , where X ‾ 0 = X ‾ 0 2 Y ‾ 0 2 + X ‾ 0 2 σ Y 0 2 \overline{X}_0=\overline{X}_0^2\overline{Y}_0^2+\overline{X}_0^2\sigma_{Y_0}^2 X 0 = X 0 2 Y 0 2 + X 0 2 σ Y 0 2
To get (37) we just need to choose s , d s,d s , d d ≥ max { − ( s 2 + 2 s − 1 ) , 2 s 2 } d\ge\max\{-(s^2+2s-1),2s^2\} d ≥ max { − ( s 2 + 2 s − 1 ) , 2 s 2 }
In the remainder of this section, we prove Lemma 4.1.
To prove the lemma we first need to calculate the action of the L X , L Y L_X,L_Y L X , L Y φ ( X ) , φ ( Y ) \varphi(X),\varphi(Y) φ ( X ) , φ ( Y ) φ ( X ) φ ( Y ) \varphi(X)\varphi(Y) φ ( X ) φ ( Y )
L Y φ ( X ) = φ ( X ) − E [ φ ( X ) ∣ x ] = 0 and L X φ ( Y ) = 0. (42) L_Y\varphi(X)=\varphi(X)-\mathbb{E}[\varphi(X)\mid x]=0 \quad\text{and}\quad L_X\varphi(Y)=0. \tag*{(42)} L Y φ ( X ) = φ ( X ) − E [ φ ( X ) ∣ x ] = 0 and L X φ ( Y ) = 0. (42) Second,
L X φ ( X ) = φ ( X ) − E [ φ ( X ) ∣ y ] = X − X ‾ 0 σ X 0 − X ‾ ( y ) − X ‾ 0 σ X 0 = X − X ‾ ( y ) σ X 0 = : R X ( y ) , L Y φ ( Y ) = : R Y ( x ) . (43) L_X\varphi(X)=\varphi(X)-\mathbb{E}[\varphi(X)\mid y]=\frac{X-\overline{X}_0}{\sigma_{X_0}}-\frac{\overline{X}(y)-\overline{X}_0}{\sigma_{X_0}}=\frac{X-\overline{X}(y)}{\sigma_{X_0}}=:R_X(y), \quad L_Y\varphi(Y)=:R_Y(x). \tag*{(43)} L X φ ( X ) = φ ( X ) − E [ φ ( X ) ∣ y ] = σ X 0 X − X 0 − σ X 0 X ( y ) − X 0 = σ X 0 X − X ( y ) =: R X ( y ) , L Y φ ( Y ) =: R Y ( x ) . (43) Lastly,
L X ( φ ( X ) φ ( Y ) ) = φ ( X ) φ ( Y ) − E [ φ ( X ) φ ( Y ) ∣ y ] = φ ( Y ) L X φ ( X ) = φ ( Y ) ⋅ R X , (44) L_X(\varphi(X)\varphi(Y))=\varphi(X)\varphi(Y)-\mathbb{E}[\varphi(X)\varphi(Y)\mid y]=\varphi(Y)L_X\varphi(X)=\varphi(Y)\cdot R_X, \tag*{(44)} L X ( φ ( X ) φ ( Y )) = φ ( X ) φ ( Y ) − E [ φ ( X ) φ ( Y ) ∣ y ] = φ ( Y ) L X φ ( X ) = φ ( Y ) ⋅ R X , (44) Similarly, we have L Y ( φ ( X ) φ ( Y ) ) = φ ( X ) ⋅ R Y L_Y(\varphi(X)\varphi(Y))=\varphi(X)\cdot R_Y L Y ( φ ( X ) φ ( Y )) = φ ( X ) ⋅ R Y
By linearity for f = α φ ( X ) + β φ ( Y ) + γ φ ( X ) φ ( Y ) f=\alpha\varphi(X)+\beta\varphi(Y)+\gamma\varphi(X)\varphi(Y) f = α φ ( X ) + β φ ( Y ) + γ φ ( X ) φ ( Y ) L X f ( x , y ) = ( α + γ φ ( Y ) ) ⋅ R X L_Xf(x,y)=(\alpha+\gamma\varphi(Y))\cdot R_X L X f ( x , y ) = ( α + γ φ ( Y )) ⋅ R X L Y f ( x , y ) = ( β + γ φ ( X ) ) ⋅ R Y L_Yf(x,y)=(\beta+\gamma\varphi(X))\cdot R_Y L Y f ( x , y ) = ( β + γ φ ( X )) ⋅ R Y
( L X f ) ( L Y f ) = R X R Y ( α + γ φ ( Y ) ) ( β + γ φ ( X ) ) = R X R Y ( α β + α γ φ ( X ) + β γ φ ( Y ) + γ 2 φ ( X ) φ ( Y ) ) . (45) (L_Xf)(L_Yf)=R_XR_Y(\alpha+\gamma\varphi(Y))(\beta+\gamma\varphi(X))=R_XR_Y(\alpha\beta+\alpha\gamma\varphi(X)+\beta\gamma\varphi(Y)+\gamma^2\varphi(X)\varphi(Y)). \tag*{(45)} ( L X f ) ( L Y f ) = R X R Y ( α + γ φ ( Y )) ( β + γ φ ( X )) = R X R Y ( α β + α γ φ ( X ) + β γ φ ( Y ) + γ 2 φ ( X ) φ ( Y )) . (45) A necessary step to prove Theorem 3.1 is to calculate the Taylor expansion of several polynomials of M , φ ( ⋅ ) , R X , R Y M,\varphi(\cdot),R_X,R_Y M , φ ( ⋅ ) , R X , R Y
Lemma 4.2 (Taylor expansion lemma). Recall J ‾ = tanh ( J ) \overline{J}=\tanh(J) J = tanh ( J )
(i) E [ φ ( X ) 2 ] = 1 − 2 J ‾ X ‾ 0 Y ‾ 0 + O ( J ‾ 2 ) \mathbb{E}[\varphi(X)^2]=1-2\overline{J}\overline{X}_0\overline{Y}_0+O(\overline{J}^2) E [ φ ( X ) 2 ] = 1 − 2 J X 0 Y 0 + O ( J 2 ) E [ φ ( Y ) 2 ] = 1 − 2 J ‾ X ‾ 0 Y ‾ 0 ± O ( J ‾ 2 ) \mathbb{E}[\varphi(Y)^2]=1-2\overline{J}\overline{X}_0\overline{Y}_0\pm O(\overline{J}^2) E [ φ ( Y ) 2 ] = 1 − 2 J X 0 Y 0 ± O ( J 2 )
(ii) E [ R X 2 ] = 1 ± O ( ∣ J ‾ ∣ ) \mathbb{E}[R_X^2]=1\pm O(\lvert\overline{J}\rvert) E [ R X 2 ] = 1 ± O (∣ J ∣) E [ R Y 2 ] = 1 ± O ( ∣ J ‾ ∣ ) \mathbb{E}[R_Y^2]=1\pm O(\lvert\overline{J}\rvert) E [ R Y 2 ] = 1 ± O (∣ J ∣)
(iii) E [ R X R Y ] = − σ X 0 σ Y 0 J ‾ + 2 X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 J ‾ 2 ± O ( ∣ J ‾ ∣ 3 ) \mathbb{E}[R_XR_Y]=-\sigma_{X_0}\sigma_{Y_0}\overline{J}+2\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0}\overline{J}^2\pm O(\lvert\overline{J}\rvert^3) E [ R X R Y ] = − σ X 0 σ Y 0 J + 2 X 0 Y 0 σ X 0 σ Y 0 J 2 ± O (∣ J ∣ 3 )
(iv) E [ R X R Y φ ( X ) ] = ± O ( J ‾ 2 ) \mathbb{E}[R_XR_Y\varphi(X)]=\pm O(\overline{J}^2) E [ R X R Y φ ( X )] = ± O ( J 2 ) E [ R X R Y φ ( Y ) ] = ± O ( J ‾ 2 ) \mathbb{E}[R_XR_Y\varphi(Y)]=\pm O(\overline{J}^2) E [ R X R Y φ ( Y )] = ± O ( J 2 )
(v) E [ R X R Y φ ( X ) φ ( Y ) ] = 1 ± O ( J ‾ 2 ) \mathbb{E}[R_XR_Y\varphi(X)\varphi(Y)]=1\pm O(\overline{J}^2) E [ R X R Y φ ( X ) φ ( Y )] = 1 ± O ( J 2 )
(vi) E [ M R X R Y ] = σ X 0 σ Y 0 − ( s + 1 ) 2 X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 J ‾ ± O ( J ‾ 2 ) \mathbb{E}[MR_XR_Y]=\sigma_{X_0}\sigma_{Y_0}-(s+1)^2\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0}\overline{J}\pm O(\overline{J}^2) E [ M R X R Y ] = σ X 0 σ Y 0 − ( s + 1 ) 2 X 0 Y 0 σ X 0 σ Y 0 J ± O ( J 2 )
(vii) E [ M R X R Y φ ( X ) ] = ( s − 1 ) X ‾ 0 σ Y 0 ± O ( ∣ J ‾ ∣ ) \mathbb{E}[MR_XR_Y\varphi(X)]=(s-1)\overline{X}_0\sigma_{Y_0}\pm O(\lvert\overline{J}\rvert) E [ M R X R Y φ ( X )] = ( s − 1 ) X 0 σ Y 0 ± O (∣ J ∣) E [ M R X R Y φ ( Y ) ] = ( s − 1 ) Y ‾ 0 σ X 0 ± O ( ∣ J ‾ ∣ ) \mathbb{E}[MR_XR_Y\varphi(Y)]=(s-1)\overline{Y}_0\sigma_{X_0}\pm O(\lvert\overline{J}\rvert) E [ M R X R Y φ ( Y )] = ( s − 1 ) Y 0 σ X 0 ± O (∣ J ∣)
(viii) E [ M R X R Y φ ( X ) φ ( Y ) ] = ( s − 1 ) 2 X ‾ 0 Y ‾ 0 + O ( ∣ J ‾ ∣ ) \mathbb{E}[MR_XR_Y\varphi(X)\varphi(Y)]=(s-1)^2\overline{X}_0\overline{Y}_0+O(\lvert\overline{J}\rvert) E [ M R X R Y φ ( X ) φ ( Y )] = ( s − 1 ) 2 X 0 Y 0 + O (∣ J ∣)
All the E \mathbb{E} E μ J , a , b \mu_{J,a,b} μ J , a , b
Equipped with these, we are ready to prove Lemma 4.1.
Proof of Lemma 4.1. We make a case for every entry of A A A J ˉ = J + O ( J 3 ) \bar{J}=J+O(J^3) J ˉ = J + O ( J 3 )
A_{11}, A_{22}. Plugging in f = φ ( X ) f=\varphi(X) f = φ ( X ) [ 1 0 0 ] [1\ \ 0\ \ 0] [ 1 0 0 ] φ ( X ) \varphi(X) φ ( X )
A 11 = ( λ J ) 2 2 ⋅ E [ R X 2 ] + d J 2 2 ⋅ E [ X ˉ ( Y ) 2 R X 2 ] . (46) A_{11} = \frac{(\lambda J)^2}{2} \cdot\mathbb{E}[R_X^2] + \frac{dJ^2}{2} \cdot\mathbb{E}[\bar{X}(Y)^2R_X^2]. \tag*{(46)} A 11 = 2 ( λ J ) 2 ⋅ E [ R X 2 ] + 2 d J 2 ⋅ E [ X ˉ ( Y ) 2 R X 2 ] . (46) where we used (38) so only the last two terms remain. We bound the first term using Lemma 4.2(i). For the second term, we combine Lemmas 2.4 and 4.2(i) to get
E [ X ˉ ( Y ) 2 R X 2 ] = E [ X ˉ ( Y ) 2 E [ R X 2 ∣ Y ] ] = ( X ˉ 0 2 + O ( ∣ J ˉ ∣ ) ) ( 1 ± O ( ∣ J ˉ ∣ ) ) = X ˉ 0 2 ± O ( ∣ J ∣ ) . \mathbb{E}[\bar{X}(Y)^2R_X^2] = \mathbb{E}[\bar{X}(Y)^2\mathbb{E}[R_X^2\mid Y]] = (\bar{X}_0^2+O(|\bar{J}|))(1\pm O(|\bar{J}|)) = \bar{X}_0^2\pm O(|J|). E [ X ˉ ( Y ) 2 R X 2 ] = E [ X ˉ ( Y ) 2 E [ R X 2 ∣ Y ]] = ( X ˉ 0 2 + O ( ∣ J ˉ ∣ )) ( 1 ± O ( ∣ J ˉ ∣ )) = X ˉ 0 2 ± O ( ∣ J ∣ ) . All in all, we get
A 11 = J 2 2 ( λ 2 + d X ˉ 0 2 ) ± O ( ∣ J ∣ 3 ) . A_{11} = \frac{J^2}{2}\left(\lambda^2+d\bar{X}_0^2\right)\pm O(|J|^3). A 11 = 2 J 2 ( λ 2 + d X ˉ 0 2 ) ± O ( ∣ J ∣ 3 ) . Similarly, we have A 22 = J 2 2 ( λ 2 + d Y ˉ 0 2 ) ± O ( ∣ J ∣ 3 ) A_{22}=\frac{J^2}{2}\left(\lambda^2+d\bar{Y}_0^2\right)\pm O(|J|^3) A 22 = 2 J 2 ( λ 2 + d Y ˉ 0 2 ) ± O ( ∣ J ∣ 3 )
A_{33}. Plugging in f = φ ( X ) φ ( Y ) f=\varphi(X)\varphi(Y) f = φ ( X ) φ ( Y ) [ 0 0 1 ] [0\ \ 0\ \ 1] [ 0 0 1 ]
A 33 = E [ R X R Y φ ( X ) φ ( Y ) ] + J E [ M R X R Y φ ( X ) φ ( Y ) ] + ( λ J ) 2 2 E [ R X 2 φ ( Y ) 2 + R Y 2 φ ( X ) 2 ] + d J 2 2 E [ X ˉ ( Y ) 2 R X 2 φ ( Y ) 2 + Y ˉ ( X ) 2 R Y 2 φ ( X ) 2 ] = = Lemma 4.2(iv) 1 ± O ( ∣ J ∣ ) . (47) \begin{aligned}
A_{33}={}&\mathbb{E}[R_XR_Y\varphi(X)\varphi(Y)]+J\mathbb{E}[MR_XR_Y\varphi(X)\varphi(Y)]\\
&+\frac{(\lambda J)^2}{2}\mathbb{E}[R_X^2\varphi(Y)^2+R_Y^2\varphi(X)^2]\\
&+\frac{dJ^2}{2}\mathbb{E}[\bar{X}(Y)^2R_X^2\varphi(Y)^2+\bar{Y}(X)^2R_Y^2\varphi(X)^2]
=\mathop{=}\limits_{\text{Lemma 4.2(iv)}}1\pm O(|J|). \tag*{(47)}
\end{aligned} A 33 = E [ R X R Y φ ( X ) φ ( Y )] + J E [ M R X R Y φ ( X ) φ ( Y )] + 2 ( λ J ) 2 E [ R X 2 φ ( Y ) 2 + R Y 2 φ ( X ) 2 ] + 2 d J 2 E [ X ˉ ( Y ) 2 R X 2 φ ( Y ) 2 + Y ˉ ( X ) 2 R Y 2 φ ( X ) 2 ] = Lemma 4.2(iv) = 1 ± O ( ∣ J ∣ ) . (47) To see the that third and fourth term are bounded by O ( J 2 ) O(J^2) O ( J 2 )
E μ J , a , b [ R X 2 φ ( Y ) 2 ] ≤ ( † ) E μ J , a , b [ φ ( X ) 2 φ ( Y ) 2 ] \mathbb{E}_{\mu_{J,a,b}}[R_X^2\varphi(Y)^2]\mathrel{\mathop{\leq}\limits^{(\dagger)}}\mathbb{E}_{\mu_{J,a,b}}[\varphi(X)^2\varphi(Y)^2] E μ J , a , b [ R X 2 φ ( Y ) 2 ] ≤ ( † ) E μ J , a , b [ φ ( X ) 2 φ ( Y ) 2 ] ≤ Lemma 2.5 ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + O ( J ˉ 2 ) ) E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) 2 ] \mathrel{\mathop{\leq}\limits^{\text{Lemma 2.5}}}
\left(1+\bar{J}(xy-\bar{X}_0\bar{Y}_0)+O(\bar{J}^2)\right)\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2\varphi(Y)^2] ≤ Lemma 2.5 ( 1 + J ˉ ( x y − X ˉ 0 Y ˉ 0 ) + O ( J ˉ 2 ) ) E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) 2 ] ≤ independence ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] ≤ O ( 1 ) , \mathrel{\mathop{\leq}\limits^{\text{independence}}}
\left(1+2\bar{J}+O(\bar{J}^2)\right)\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)^2]\leq O(1), ≤ independence ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] ≤ O ( 1 ) , where ( † ) (\dagger) ( † )
E μ J , a , b [ R X 2 ∣ Y ] = Var μ J , a , b ( φ ( X ) ∣ Y ) ≤ E μ J , a , b [ φ ( X ) 2 ∣ Y ] . (48) \mathbb{E}_{\mu_{J,a,b}}[R_X^2\mid Y]=\operatorname{Var}_{\mu_{J,a,b}}(\varphi(X)\mid Y)\leq\mathbb{E}_{\mu_{J,a,b}}[\varphi(X)^2\mid Y]. \tag*{(48)} E μ J , a , b [ R X 2 ∣ Y ] = Var μ J , a , b ( φ ( X ) ∣ Y ) ≤ E μ J , a , b [ φ ( X ) 2 ∣ Y ] . (48) and so the tower property implies that
E μ J , a , b [ φ ( Y ) 2 R X 2 ] = E μ J , a , b [ φ ( Y ) 2 E μ J , a , b [ R X 2 ∣ Y ] ] ≤ ( 42 ) E μ J , a , b [ φ ( Y ) 2 E μ J , a , b [ φ ( X ) 2 ∣ Y ] ] = E μ J , a , b [ φ ( Y ) 2 φ ( X ) 2 ] . (49) \mathbb{E}_{\mu_{J,a,b}}[\varphi(Y)^2R_X^2]=\mathbb{E}_{\mu_{J,a,b}}[\varphi(Y)^2\mathbb{E}_{\mu_{J,a,b}}[R_X^2\mid Y]]\mathrel{\mathop{\leq}\limits^{(42)}}\mathbb{E}_{\mu_{J,a,b}}[\varphi(Y)^2\mathbb{E}_{\mu_{J,a,b}}[\varphi(X)^2\mid Y]]=\mathbb{E}_{\mu_{J,a,b}}[\varphi(Y)^2\varphi(X)^2]. \tag*{(49)} E μ J , a , b [ φ ( Y ) 2 R X 2 ] = E μ J , a , b [ φ ( Y ) 2 E μ J , a , b [ R X 2 ∣ Y ]] ≤ ( 42 ) E μ J , a , b [ φ ( Y ) 2 E μ J , a , b [ φ ( X ) 2 ∣ Y ]] = E μ J , a , b [ φ ( Y ) 2 φ ( X ) 2 ] . (49) The above shows the third term is bounded. To show the boundedness of the fourth term we combine the above with X ˉ ( y ) 2 ≤ 1 \bar{X}(y)^2\leq1 X ˉ ( y ) 2 ≤ 1 y ∈ { ± 1 } y\in\{\pm1\} y ∈ { ± 1 }
A_{12}. Plugging in f = α φ ( X ) + β φ ( Y ) f=\alpha\varphi(X)+\beta\varphi(Y) f = α φ ( X ) + β φ ( Y ) A 12 A_{12} A 12 α β \alpha\beta α β
A 12 = 1 2 E [ R X R Y ] + J ˉ 2 E [ M R X R Y ] = − 1 2 σ X 0 σ Y 0 J ˉ + X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ 2 + J ˉ 2 ( σ X 0 σ Y 0 − ( s + 1 ) 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ ) + O ( ∣ J ˉ ∣ 3 ) (50) \begin{aligned}
A_{12} = \frac{1}{2}\mathbb{E}[R_XR_Y] + \frac{\bar{J}}{2}\mathbb{E}[MR_XR_Y] \\
&= -\frac{1}{2}\sigma_{X_0}\sigma_{Y_0}\bar{J}+\bar{X}_0\bar{Y}_0\sigma_{X_0}\sigma_{Y_0}\bar{J}^2+\frac{\bar{J}}{2}\left(\sigma_{X_0}\sigma_{Y_0}-(s+1)^2\bar{X}_0\bar{Y}_0\sigma_{X_0}\sigma_{Y_0}\bar{J}\right)+O(|\bar{J}|^3) \tag*{(50)}
\end{aligned} A 12 = 2 1 E [ R X R Y ] + 2 J ˉ E [ M R X R Y ] = − 2 1 σ X 0 σ Y 0 J ˉ + X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ 2 + 2 J ˉ ( σ X 0 σ Y 0 − ( s + 1 ) 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ ) + O ( ∣ J ˉ ∣ 3 ) (50) Lemma 4.2(ii),(iv)
= − J 2 2 ( s 2 + 2 s − 1 ) X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 ± O ( ∣ J ∣ 3 ) . = -\frac{J^2}{2}(s^2+2s-1)\bar{X}_0\bar{Y}_0\sigma_{X_0}\sigma_{Y_0}\pm O(|J|^3). = − 2 J 2 ( s 2 + 2 s − 1 ) X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 ± O ( ∣ J ∣ 3 ) . A_{13}, A_{23}. Plugging in f = α φ ( X ) + γ φ ( X ) φ ( Y ) f=\alpha\varphi(X)+\gamma\varphi(X)\varphi(Y) f = α φ ( X ) + γ φ ( X ) φ ( Y ) A 13 A_{13} A 13 α γ \alpha\gamma α γ
A 13 = 1 2 E [ R X R Y φ ( X ) ] + J 2 E [ M R X R Y φ ( X ) ] + ( λ J ) 2 E [ φ ( Y ) R Y 2 ] + d J 2 E [ X ˉ ( y ) 2 R X 2 φ ( Y ) ] ≤ J 2 ( s − 1 ) X ˉ 0 σ Y 0 + J 2 ( λ 2 ⋅ E [ φ ( Y ) R X 2 ] + d E [ X ˉ ( y ) 2 R X 2 φ ( Y ) ] ) ≤ J ( s − 1 ) X ˉ 0 σ Y 0 ± O ( J 2 ) . \begin{aligned}
A_{13}=\frac{1}{2}\mathbb{E}[R_XR_Y\varphi(X)]+\frac{J}{2}\mathbb{E}[MR_XR_Y\varphi(X)]+(\lambda J)^2\mathbb{E}[\varphi(Y)R_Y^2]+dJ^2\mathbb{E}[\bar{X}(y)^2R_X^2\varphi(Y)] \\
&\leq\frac{J}{2}(s-1)\bar{X}_0\sigma_{Y_0}+J^2\left(\lambda^2\cdot\mathbb{E}[\varphi(Y)R_X^2]+d\mathbb{E}[\bar{X}(y)^2R_X^2\varphi(Y)]\right)\leq J(s-1)\bar{X}_0\sigma_{Y_0}\pm O(J^2).
\end{aligned} A 13 = 2 1 E [ R X R Y φ ( X )] + 2 J E [ M R X R Y φ ( X )] + ( λ J ) 2 E [ φ ( Y ) R Y 2 ] + d J 2 E [ X ˉ ( y ) 2 R X 2 φ ( Y )] ≤ 2 J ( s − 1 ) X ˉ 0 σ Y 0 + J 2 ( λ 2 ⋅ E [ φ ( Y ) R X 2 ] + d E [ X ˉ ( y ) 2 R X 2 φ ( Y )] ) ≤ J ( s − 1 ) X ˉ 0 σ Y 0 ± O ( J 2 ) . Claim 4.2(iii), (vi)
Here the final inequality holds because similarly to (43), we can show that
E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] = E μ J , a , b [ ∣ φ ( Y ) ∣ E μ J , a , b [ R X 2 ∣ Y ] ] ≤ E μ J , a , b [ ∣ φ ( Y ) ∣ E μ J , a , b [ φ ( X ) 2 ∣ Y ] ] = E μ J , a , b [ ∣ φ ( Y ) ∣ φ ( X ) 2 ] (51) \mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|R_X^2]=\mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|\mathbb{E}_{\mu_{J,a,b}}[R_X^2\mid Y]]\leq\mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|\mathbb{E}_{\mu_{J,a,b}}[\varphi(X)^2\mid Y]]=\mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|\varphi(X)^2] \tag*{(51)} E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] = E μ J , a , b [ ∣ φ ( Y ) ∣ E μ J , a , b [ R X 2 ∣ Y ]] ≤ E μ J , a , b [ ∣ φ ( Y ) ∣ E μ J , a , b [ φ ( X ) 2 ∣ Y ]] = E μ J , a , b [ ∣ φ ( Y ) ∣ φ ( X ) 2 ] (51) E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] ≤ E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] ≤ ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) E μ 0 , a , b [ ∣ φ ( Y ) ∣ φ ( X ) 2 ] (52) \mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|R_X^2]\leq\mathbb{E}_{\mu_{J,a,b}}[|\varphi(Y)|R_X^2]\leq(1+2\bar{J}+O(\bar{J}^2))\mathbb{E}_{\mu_{0,a,b}}[|\varphi(Y)|\varphi(X)^2] \tag*{(52)} E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] ≤ E μ J , a , b [ ∣ φ ( Y ) ∣ R X 2 ] ≤ ( 1 + 2 J ˉ + O ( J ˉ 2 )) E μ 0 , a , b [ ∣ φ ( Y ) ∣ φ ( X ) 2 ] (52) Lemma 2.5
≤ ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) E μ 0 , a , b [ ∣ φ ( Y ) ∣ ] E μ 0 , a , b [ φ ( X ) 2 ] ≤ ( 1 + 2 J ˉ + O ( J ˉ 2 ) ) ( E μ 0 , a , b [ φ ( Y ) 2 ] ) 1 / 2 E μ 0 , a , b [ φ ( X ) 2 ] = O ( 1 ) . \begin{aligned}
\leq(1+2\bar{J}+O(\bar{J}^2))\mathbb{E}_{\mu_{0,a,b}}[|\varphi(Y)|]\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2] \\
&\leq(1+2\bar{J}+O(\bar{J}^2))\left(\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)^2]\right)^{1/2}\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]=O(1).
\end{aligned} ≤ ( 1 + 2 J ˉ + O ( J ˉ 2 )) E μ 0 , a , b [ ∣ φ ( Y ) ∣ ] E μ 0 , a , b [ φ ( X ) 2 ] ≤ ( 1 + 2 J ˉ + O ( J ˉ 2 )) ( E μ 0 , a , b [ φ ( Y ) 2 ] ) 1/2 E μ 0 , a , b [ φ ( X ) 2 ] = O ( 1 ) . Cauchy-Schwarz
This shows that λ 2 E [ φ ( Y ) R X 2 ] \lambda^2\mathbb{E}[\varphi(Y)R_X^2] λ 2 E [ φ ( Y ) R X 2 ] d E [ X ˉ ( y ) 2 R X 2 φ ( Y ) ] d\mathbb{E}[\bar{X}(y)^2R_X^2\varphi(Y)] d E [ X ˉ ( y ) 2 R X 2 φ ( Y )] X ˉ ( y ) 2 ≤ 1 \bar{X}(y)^2\leq1 X ˉ ( y ) 2 ≤ 1 y ∈ { ± 1 } y\in\{\pm1\} y ∈ { ± 1 }
Similarly, we have A 23 = J 2 ( s − 1 ) Y ˉ 0 σ X 0 ± O ( J 2 ) A_{23}=\frac{J}{2}(s-1)\bar{Y}_0\sigma_{X_0}\pm O(J^2) A 23 = 2 J ( s − 1 ) Y ˉ 0 σ X 0 ± O ( J 2 ) A 0 A_0 A 0 E E E O ( ∣ J ∣ ) O(|J|) O ( ∣ J ∣ ) ∥ E ∥ o p ≤ O ( ∣ J ∣ ) \|E\|_{\mathrm{op}}\leq O(|J|) ∥ E ∥ op ≤ O ( ∣ J ∣ )
Proof of Lemma 4.2
In this section, we prove Lemma 4.2. Before proving the statement we record a few equations which will be useful:
E μ 0 , a , b [ X φ ( X ) ] = E μ 0 , a , b [ X 2 ] − X ˉ 0 E μ 0 , a , b [ X ] σ X 0 = 1 − X ˉ 0 2 σ X 0 = σ X 0 (53) \mathbb{E}_{\mu_{0,a,b}}[X\varphi(X)]=\frac{\mathbb{E}_{\mu_{0,a,b}}[X^2]-\bar{X}_0\mathbb{E}_{\mu_{0,a,b}}[X]}{\sigma_{X_0}}=\frac{1-\bar{X}_0^2}{\sigma_{X_0}}=\sigma_{X_0} \tag*{(53)} E μ 0 , a , b [ X φ ( X )] = σ X 0 E μ 0 , a , b [ X 2 ] − X ˉ 0 E μ 0 , a , b [ X ] = σ X 0 1 − X ˉ 0 2 = σ X 0 (53) and we note that ( X − X ‾ 0 ) 2 = ( 1 − X ‾ 0 2 ) − 2 X X ‾ 0 + 2 X ‾ 0 2 = σ X 0 2 − 2 X ‾ 0 ( X − X ‾ 0 ) (X-\overline{X}_0)^2=(1-\overline{X}_0^2)-2X\overline{X}_0+2\overline{X}_0^2=\sigma_{X_0}^2-2\overline{X}_0(X-\overline{X}_0) ( X − X 0 ) 2 = ( 1 − X 0 2 ) − 2 X X 0 + 2 X 0 2 = σ X 0 2 − 2 X 0 ( X − X 0 )
E μ 0 , a , b [ X φ ( X ) 2 ] = E μ 0 , a , b [ X ⋅ σ X 0 2 − 2 X ‾ 0 ( X − X ‾ 0 ) σ X 0 2 ] = E μ 0 , a , b [ X ⋅ ( 1 − 2 X ‾ 0 φ ( X ) σ X 0 ) ] = X ‾ 0 − 2 X ‾ 0 σ X 0 ⋅ E μ 0 , a , b [ X φ ( X ) ] = X ‾ 0 − 2 X ‾ 0 σ X 0 ⋅ σ X 0 = − X ‾ 0 . (54) \begin{aligned}
\mathbb{E}_{\mu_0,a,b}[X\varphi(X)^2]
&=\mathbb{E}_{\mu_0,a,b}\left[X\cdot\frac{\sigma_{X_0}^2-2\overline{X}_0(X-\overline{X}_0)}{\sigma_{X_0}^2}\right]\\
&=\mathbb{E}_{\mu_0,a,b}\left[X\cdot\left(1-\frac{2\overline{X}_0\varphi(X)}{\sigma_{X_0}}\right)\right]\\
&=\overline{X}_0-\frac{2\overline{X}_0}{\sigma_{X_0}}\cdot\mathbb{E}_{\mu_0,a,b}[X\varphi(X)]\\
&=\overline{X}_0-\frac{2\overline{X}_0}{\sigma_{X_0}}\cdot\sigma_{X_0}=-\overline{X}_0. \tag*{(54)}
\end{aligned} E μ 0 , a , b [ X φ ( X ) 2 ] = E μ 0 , a , b [ X ⋅ σ X 0 2 σ X 0 2 − 2 X 0 ( X − X 0 ) ] = E μ 0 , a , b [ X ⋅ ( 1 − σ X 0 2 X 0 φ ( X ) ) ] = X 0 − σ X 0 2 X 0 ⋅ E μ 0 , a , b [ X φ ( X )] = X 0 − σ X 0 2 X 0 ⋅ σ X 0 = − X 0 . (54) Throughout we assume that ∣ J ‾ ∣ ≤ 1 2 |\overline{J}|\leq\frac{1}{2} ∣ J ∣ ≤ 2 1 ∣ X ‾ 0 ∣ , ∣ Y ‾ 0 ∣ ≤ 1 |\overline{X}_0|,|\overline{Y}_0|\leq1 ∣ X 0 ∣ , ∣ Y 0 ∣ ≤ 1
R X ( y ) = X − X ‾ ( y ) σ X 0 = ( X − X ‾ 0 ) ( 1 − J ‾ X y ) σ X 0 ( 1 + J ‾ X ‾ 0 y ) = φ ( X ) ⋅ 1 − J ‾ X y 1 + J ‾ X ‾ 0 y ≤ O ( φ ( X ) ) and R Y ( x ) ≤ O ( φ ( Y ) ) . (55) \begin{aligned}
R_X(y)
&=\frac{X-\overline{X}(y)}{\sigma_{X_0}}
=\frac{(X-\overline{X}_0)(1-\overline{J}Xy)}{\sigma_{X_0}(1+\overline{J}\overline{X}_0y)}\\
&=\varphi(X)\cdot\frac{1-\overline{J}Xy}{1+\overline{J}\overline{X}_0y}
\leq O(\varphi(X))
\quad\text{and}\quad
R_Y(x)\leq O(\varphi(Y)). \tag*{(55)}
\end{aligned} R X ( y ) = σ X 0 X − X ( y ) = σ X 0 ( 1 + J X 0 y ) ( X − X 0 ) ( 1 − J X y ) = φ ( X ) ⋅ 1 + J X 0 y 1 − J X y ≤ O ( φ ( X )) and R Y ( x ) ≤ O ( φ ( Y )) . (55) Furthermore,
∣ R ˙ X ( y ) ∣ = ∣ φ ( X ) ⋅ d d J ‾ 1 − J ‾ X y 1 + J ‾ X ‾ 0 y ∣ = ∣ φ ( X ) ⋅ y ( X + X ‾ 0 ) ( 1 + J ‾ X ‾ 0 y ) 2 ∣ ≤ O ( φ ( X ) ) and ∣ R ˙ Y ( x ) ∣ ≤ O ( φ ( Y ) ) . (56) \left|\dot{R}_X(y)\right|
=\left|\varphi(X)\cdot\frac{d}{d\overline{J}}\frac{1-\overline{J}Xy}{1+\overline{J}\overline{X}_0y}\right|
=\left|\varphi(X)\cdot\frac{y(X+\overline{X}_0)}{(1+\overline{J}\overline{X}_0y)^2}\right|
\leq O(\varphi(X))
\quad\text{and}\quad
|\dot{R}_Y(x)|\leq O(\varphi(Y)). \tag*{(56)} R ˙ X ( y ) = φ ( X ) ⋅ d J d 1 + J X 0 y 1 − J X y = φ ( X ) ⋅ ( 1 + J X 0 y ) 2 y ( X + X 0 ) ≤ O ( φ ( X )) and ∣ R ˙ Y ( x ) ∣ ≤ O ( φ ( Y )) . (56) We also have
∣ R ¨ X ( y ) ∣ = ∣ φ ( X ) ⋅ d d J ‾ y ( X + X ‾ 0 ) ( 1 + J ‾ X ‾ 0 y ) 2 ∣ = ∣ φ ( X ) ⋅ 2 X ‾ 0 ( X + X ‾ 0 ) ( 1 + J ‾ X ‾ 0 y ) 3 ∣ ≤ O ( φ ( X ) ) and ∣ R ¨ Y ( x ) ∣ ≤ O ( φ ( Y ) ) . (57) \left|\ddot{R}_X(y)\right|
=\left|\varphi(X)\cdot\frac{d}{d\overline{J}}\frac{y(X+\overline{X}_0)}{(1+\overline{J}\overline{X}_0y)^2}\right|
=\left|\varphi(X)\cdot\frac{2\overline{X}_0(X+\overline{X}_0)}{(1+\overline{J}\overline{X}_0y)^3}\right|
\leq O(\varphi(X))
\quad\text{and}\quad
|\ddot{R}_Y(x)|\leq O(\varphi(Y)). \tag*{(57)} R ¨ X ( y ) = φ ( X ) ⋅ d J d ( 1 + J X 0 y ) 2 y ( X + X 0 ) = φ ( X ) ⋅ ( 1 + J X 0 y ) 3 2 X 0 ( X + X 0 ) ≤ O ( φ ( X )) and ∣ R ¨ Y ( x ) ∣ ≤ O ( φ ( Y )) . (57) By Lemma 2.4, we also have
R X ( y ) = ( 39 ) φ ( X ) − J ‾ σ X 0 y ± O ( J ‾ 2 ) and R Y ( x ) = ( 39 ) φ ( Y ) − J ‾ σ Y 0 x ± O ( J ‾ 2 ) . (58) R_X(y)\underset{(39)}{=}\varphi(X)-\overline{J}\sigma_{X_0}y\pm O(\overline{J}^2)
\quad\text{and}\quad
R_Y(x)\underset{(39)}{=}\varphi(Y)-\overline{J}\sigma_{Y_0}x\pm O(\overline{J}^2). \tag*{(58)} R X ( y ) ( 39 ) = φ ( X ) − J σ X 0 y ± O ( J 2 ) and R Y ( x ) ( 39 ) = φ ( Y ) − J σ Y 0 x ± O ( J 2 ) . (58) Next, we write,
M ( J ‾ , x , y ) = Theorem 3.1 ( x + s X ‾ ) ( y + s Y ‾ ) = Lemma 2.4 s ≤ 1 ( x + s ( X ‾ 0 + J ‾ σ X 0 2 y ) ) ( y + s ( Y ‾ 0 + J ‾ σ Y 0 2 x ) ) ± O ( J ‾ 2 ) = ( x + s X ‾ 0 ) ( y + s Y ‾ 0 ) + J ‾ ( ( x + s X ‾ 0 ) s σ Y 0 2 x + ( y + s Y ‾ 0 ) s σ X 0 2 y ) ± O ( J ‾ 2 ) . (59) \begin{aligned}
M(\overline{J},x,y)
&\underset{\text{Theorem 3.1}}{=}(x+s\overline{X})(y+s\overline{Y})\\
&\underset{\substack{\text{Lemma 2.4}\\s\leq1}}{=}
\left(x+s\left(\overline{X}_0+\overline{J}\sigma_{X_0}^2y\right)\right)
\left(y+s\left(\overline{Y}_0+\overline{J}\sigma_{Y_0}^2x\right)\right)
\pm O(\overline{J}^2)\\
&=(x+s\overline{X}_0)(y+s\overline{Y}_0)
+\overline{J}\left((x+s\overline{X}_0)s\sigma_{Y_0}^2x
+(y+s\overline{Y}_0)s\sigma_{X_0}^2y\right)
\pm O(\overline{J}^2). \tag*{(59)}
\end{aligned} M ( J , x , y ) Theorem 3.1 = ( x + s X ) ( y + s Y ) Lemma 2.4 s ≤ 1 = ( x + s ( X 0 + J σ X 0 2 y ) ) ( y + s ( Y 0 + J σ Y 0 2 x ) ) ± O ( J 2 ) = ( x + s X 0 ) ( y + s Y 0 ) + J ( ( x + s X 0 ) s σ Y 0 2 x + ( y + s Y 0 ) s σ X 0 2 y ) ± O ( J 2 ) . (59) and we note that
∣ M ˙ ( J ‾ , x , y ) ∣ ≤ s ∣ X ‾ ˙ ( y ) ∣ ⋅ ∣ y + s Y ‾ ( x ) ∣ + s ∣ Y ‾ ˙ ( x ) ∣ ⋅ ∣ x + s X ‾ ( y ) ∣ ≤ x , y , s , X ‾ , Y ‾ ≤ 1 Lemma 2.3 O ( y ( 1 − X ‾ 0 2 ) ( 1 + X ‾ 0 J ‾ y ) 2 ) + O ( x ( 1 − Y ‾ 0 2 ) ( 1 + Y ‾ 0 J ‾ x ) 2 ) ≤ x , y , X ‾ 0 , Y ‾ 0 ≤ 1 ∣ J ‾ ∣ ≤ 1 2 O ( 1 ) . (60) \begin{aligned}
|\dot{M}(\overline{J},x,y)|
&\leq s|\dot{\overline{X}}(y)|\cdot|y+s\overline{Y}(x)|
+s|\dot{\overline{Y}}(x)|\cdot|x+s\overline{X}(y)|\\
&\underset{\substack{x,y,s,\overline{X},\overline{Y}\leq1\\\text{Lemma 2.3}}}{\leq}
O\left(\frac{y(1-\overline{X}_0^2)}{(1+\overline{X}_0\overline{J}y)^2}\right)
+O\left(\frac{x(1-\overline{Y}_0^2)}{(1+\overline{Y}_0\overline{J}x)^2}\right)
\underset{\substack{x,y,\overline{X}_0,\overline{Y}_0\leq1\\|\overline{J}|\leq\frac12}}{\leq}O(1). \tag*{(60)}
\end{aligned} ∣ M ˙ ( J , x , y ) ∣ ≤ s ∣ X ˙ ( y ) ∣ ⋅ ∣ y + s Y ( x ) ∣ + s ∣ Y ˙ ( x ) ∣ ⋅ ∣ x + s X ( y ) ∣ x , y , s , X , Y ≤ 1 Lemma 2.3 ≤ O ( ( 1 + X 0 J y ) 2 y ( 1 − X 0 2 ) ) + O ( ( 1 + Y 0 J x ) 2 x ( 1 − Y 0 2 ) ) x , y , X 0 , Y 0 ≤ 1 ∣ J ∣ ≤ 2 1 ≤ O ( 1 ) . (60) Finally, we have
∣ M ¨ ( J ‾ , x , y ) ∣ ≤ s ∣ X ‾ ¨ ( y ) ∣ ⋅ ∣ y + s Y ‾ ( x ) ∣ + s ∣ Y ‾ ¨ ( x ) ∣ ⋅ ∣ x + s X ‾ ( y ) ∣ + 2 s 2 ∣ X ‾ ˙ ( y ) ∣ ∣ Y ‾ ˙ ( x ) ∣ ≤ x , y , s , X ‾ , Y ‾ ≤ 1 Lemma 2.3 O ( X ‾ 0 ( 1 − X ‾ 0 2 ) ( 1 + X ‾ 0 J ‾ y ) 3 ) + O ( Y ‾ 0 ( 1 − Y ‾ 0 2 ) ( 1 + Y ‾ 0 J ‾ x ) 3 ) + O ( 1 ) ≤ x , y , X ‾ 0 , Y ‾ 0 ≤ 1 ∣ J ‾ ∣ ≤ 1 2 O ( 1 ) . (61) \begin{aligned}
|\ddot{M}(\overline{J},x,y)|
&\leq s|\ddot{\overline{X}}(y)|\cdot|y+s\overline{Y}(x)|
+s|\ddot{\overline{Y}}(x)|\cdot|x+s\overline{X}(y)|
+2s^2|\dot{\overline{X}}(y)||\dot{\overline{Y}}(x)|\\
&\underset{\substack{x,y,s,\overline{X},\overline{Y}\leq1\\\text{Lemma 2.3}}}{\leq}
O\left(\frac{\overline{X}_0(1-\overline{X}_0^2)}{(1+\overline{X}_0\overline{J}y)^3}\right)
+O\left(\frac{\overline{Y}_0(1-\overline{Y}_0^2)}{(1+\overline{Y}_0\overline{J}x)^3}\right)
+O(1)
\underset{\substack{x,y,\overline{X}_0,\overline{Y}_0\leq1\\|\overline{J}|\leq\frac12}}{\leq}O(1). \tag*{(61)}
\end{aligned} ∣ M ¨ ( J , x , y ) ∣ ≤ s ∣ X ¨ ( y ) ∣ ⋅ ∣ y + s Y ( x ) ∣ + s ∣ Y ¨ ( x ) ∣ ⋅ ∣ x + s X ( y ) ∣ + 2 s 2 ∣ X ˙ ( y ) ∣∣ Y ˙ ( x ) ∣ x , y , s , X , Y ≤ 1 Lemma 2.3 ≤ O ( ( 1 + X 0 J y ) 3 X 0 ( 1 − X 0 2 ) ) + O ( ( 1 + Y 0 J x ) 3 Y 0 ( 1 − Y 0 2 ) ) + O ( 1 ) x , y , X 0 , Y 0 ≤ 1 ∣ J ∣ ≤ 2 1 ≤ O ( 1 ) . (61) Lemma 5.1. Let k ≥ 1 k \ge1 k ≥ 1 F : R × { ± 1 } 2 → R F:\mathbb{R} \times\{\pm1\}^{2} \to\mathbb{R} F : R × { ± 1 } 2 → R
F ( J ˉ , x , y ) = R X ( J ˉ , y ) n R X R Y ( J ˉ , x ) n R Y φ ( X ) n X φ ( Y ) n Y M ( J ˉ , x , y ) n M F(\bar{J},x,y)=R_X(\bar{J},y)^{n_{R_X}}R_Y(\bar{J},x)^{n_{R_Y}}\varphi(X)^{n_X}\varphi(Y)^{n_Y}M(\bar{J},x,y)^{n_M} F ( J ˉ , x , y ) = R X ( J ˉ , y ) n R X R Y ( J ˉ , x ) n R Y φ ( X ) n X φ ( Y ) n Y M ( J ˉ , x , y ) n M such that n R X + n X ≤ 2 n_{R_X}+n_X \le2 n R X + n X ≤ 2 n R Y + n Y ≤ 2 n_{R_Y}+n_Y \le2 n R Y + n Y ≤ 2 n M ≤ 1 n_M \le1 n M ≤ 1 C C C
sup ∣ J ˉ ∣ ≤ 1 / 2 E μ 0 , a , b ∣ F ∣ , sup ∣ J ˉ ∣ ≤ 1 / 2 E μ 0 , a , b ∣ ∂ F ∂ J ˉ ∣ , sup ∣ J ˉ ∣ ≤ 1 / 2 E μ 0 , a , b ∣ ∂ 2 F ∂ J ˉ 2 ∣ ≤ C . \sup_{|\bar{J}|\le1/2}\mathbb{E}_{\mu_{0,a,b}}|F|,\quad\sup_{|\bar{J}|\le1/2}\mathbb{E}_{\mu_{0,a,b}}\left|\frac{\partial F}{\partial\bar{J}}\right|,\quad\sup_{|\bar{J}|\le1/2}\mathbb{E}_{\mu_{0,a,b}}\left|\frac{\partial^2F}{\partial\bar{J}^2}\right|\le C. ∣ J ˉ ∣ ≤ 1/2 sup E μ 0 , a , b ∣ F ∣ , ∣ J ˉ ∣ ≤ 1/2 sup E μ 0 , a , b ∂ J ˉ ∂ F , ∣ J ˉ ∣ ≤ 1/2 sup E μ 0 , a , b ∂ J ˉ 2 ∂ 2 F ≤ C . Proof. By (47), (48), (49) and ∣ J ˉ ∣ ≤ 1 2 |\bar{J}|\le\frac{1}{2} ∣ J ˉ ∣ ≤ 2 1
F ≤ O ( 1 ) ∣ φ ( X ) ∣ n X + n R X ∣ φ ( Y ) ∣ n Y + n R Y ∣ M ∣ n M ≤ O ( 1 ) ∣ φ ( X ) ∣ n X + n R X ∣ φ ( Y ) ∣ n Y + n R Y . (62) F\le O(1)|\varphi(X)|^{n_X+n_{R_X}}|\varphi(Y)|^{n_Y+n_{R_Y}}|M|^{n_M}\le O(1)|\varphi(X)|^{n_X+n_{R_X}}|\varphi(Y)|^{n_Y+n_{R_Y}}. \tag*{(62)} F ≤ O ( 1 ) ∣ φ ( X ) ∣ n X + n R X ∣ φ ( Y ) ∣ n Y + n R Y ∣ M ∣ n M ≤ O ( 1 ) ∣ φ ( X ) ∣ n X + n R X ∣ φ ( Y ) ∣ n Y + n R Y . (62) and
∣ F ˙ ∣ , ∣ F ¨ ∣ ≤ ( 51 ) , ( 52 ) , ( 53 ) O ( 1 ) ( 1 + ∣ φ ( X ) ∣ n X + n R X ) ( 1 + ∣ φ ( Y ) ∣ n Y + n R Y ) . |\dot{F}|,|\ddot{F}|\underset{(51),(52),(53)}{\le}O(1)(1+|\varphi(X)|^{n_X+n_{R_X}})(1+|\varphi(Y)|^{n_Y+n_{R_Y}}). ∣ F ˙ ∣ , ∣ F ¨ ∣ ( 51 ) , ( 52 ) , ( 53 ) ≤ O ( 1 ) ( 1 + ∣ φ ( X ) ∣ n X + n R X ) ( 1 + ∣ φ ( Y ) ∣ n Y + n R Y ) . By assumption, n X + n R X , n Y + n R Y ∈ { 0 , 1 , 2 } n_X+n_{R_X},n_Y+n_{R_Y}\in\{0,1,2\} n X + n R X , n Y + n R Y ∈ { 0 , 1 , 2 } E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ∣ φ ( Y ) ∣ m Y ] \mathbb{E}_{\mu_{0,a,b}}[|\varphi(X)|^{m_X}|\varphi(Y)|^{m_Y}] E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ∣ φ ( Y ) ∣ m Y ] m X ∈ { 0 , 1 , 2 } m_X\in\{0,1,2\} m X ∈ { 0 , 1 , 2 } m Y ∈ { 0 , 1 , 2 } m_Y\in\{0,1,2\} m Y ∈ { 0 , 1 , 2 }
E μ 0 , a , b ∣ φ ( X ) ∣ ≤ Cauchy–Schwarz ( E [ φ ( X ) 2 ] ) 1 / 2 = 1 \mathbb{E}_{\mu_{0,a,b}}|\varphi(X)|\underset{\text{Cauchy--Schwarz}}{\le}\left(\mathbb{E}[\varphi(X)^2]\right)^{1/2}=1 E μ 0 , a , b ∣ φ ( X ) ∣ Cauchy–Schwarz ≤ ( E [ φ ( X ) 2 ] ) 1/2 = 1 and E μ 0 , a , b ∣ φ ( Y ) ∣ ≤ 1 \mathbb{E}_{\mu_{0,a,b}}|\varphi(Y)|\le1 E μ 0 , a , b ∣ φ ( Y ) ∣ ≤ 1
Consequently, it follows that
E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ∣ φ ( Y ) ∣ m Y ] ≤ independence E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ] E μ 0 , a , b [ ∣ φ ( Y ) ∣ m Y ] ≤ ( 54 ) 1. \mathbb{E}_{\mu_{0,a,b}}[|\varphi(X)|^{m_X}|\varphi(Y)|^{m_Y}]
\underset{\text{independence}}{\le}
\mathbb{E}_{\mu_{0,a,b}}[|\varphi(X)|^{m_X}]
\mathbb{E}_{\mu_{0,a,b}}[|\varphi(Y)|^{m_Y}]
\underset{(54)}{\le}1. E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ∣ φ ( Y ) ∣ m Y ] independence ≤ E μ 0 , a , b [ ∣ φ ( X ) ∣ m X ] E μ 0 , a , b [ ∣ φ ( Y ) ∣ m Y ] ( 54 ) ≤ 1. Lemma 5.2 (Taylor expansion lemma). Recall J ˉ = tanh ( J ) \bar{J}=\tanh(J) J ˉ = tanh ( J )
(i) E [ φ ( X ) 2 ] = 1 − 2 J ˉ X ˉ 0 Y ˉ 0 + O ( J ˉ 2 ) \mathbb{E}[\varphi(X)^2]=1-2\bar{J}\bar{X}_0\bar{Y}_0+O(\bar{J}^2) E [ φ ( X ) 2 ] = 1 − 2 J ˉ X ˉ 0 Y ˉ 0 + O ( J ˉ 2 ) E [ φ ( Y ) 2 ] = 1 − 2 J ˉ X ˉ 0 Y ˉ 0 + O ( J ˉ 2 ) \mathbb{E}[\varphi(Y)^2]=1-2\bar{J}\bar{X}_0\bar{Y}_0+O(\bar{J}^2) E [ φ ( Y ) 2 ] = 1 − 2 J ˉ X ˉ 0 Y ˉ 0 + O ( J ˉ 2 )
(ii) E [ R X 2 ] = 1 ± O ( ∣ J ˉ ∣ ) \mathbb{E}[R_X^2]=1\pm O(|\bar{J}|) E [ R X 2 ] = 1 ± O ( ∣ J ˉ ∣ ) E [ R Y 2 ] = 1 ± O ( ∣ J ˉ ∣ ) \mathbb{E}[R_Y^2]=1\pm O(|\bar{J}|) E [ R Y 2 ] = 1 ± O ( ∣ J ˉ ∣ )
(iii) E [ R X R Y ] = − σ X 0 σ Y 0 J ˉ + 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ 2 ± O ( ∣ J ˉ ∣ 3 ) \mathbb{E}[R_XR_Y]=-\sigma_{X_0}\sigma_{Y_0}\bar{J}+2\bar{X}_0\bar{Y}_0\sigma_{X_0}\sigma_{Y_0}\bar{J}^2\pm O(|\bar{J}|^3) E [ R X R Y ] = − σ X 0 σ Y 0 J ˉ + 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ 2 ± O ( ∣ J ˉ ∣ 3 )
(iv) E [ R X R Y φ ( X ) ] = ± O ( J ˉ 2 ) \mathbb{E}[R_XR_Y\varphi(X)]=\pm O(\bar{J}^2) E [ R X R Y φ ( X )] = ± O ( J ˉ 2 ) E [ R X R Y φ ( Y ) ] = ± O ( J ˉ 2 ) \mathbb{E}[R_XR_Y\varphi(Y)]=\pm O(\bar{J}^2) E [ R X R Y φ ( Y )] = ± O ( J ˉ 2 )
(v) E [ R X R Y φ ( X ) φ ( Y ) ] = 1 ± O ( J ˉ 2 ) \mathbb{E}[R_XR_Y\varphi(X)\varphi(Y)]=1\pm O(\bar{J}^2) E [ R X R Y φ ( X ) φ ( Y )] = 1 ± O ( J ˉ 2 )
(vi) E [ M R X R Y ] = σ X 0 σ Y 0 − ( s + 1 ) 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ ± O ( J ˉ 2 ) \mathbb{E}[MR_XR_Y]=\sigma_{X_0}\sigma_{Y_0}-(s+1)^2\bar{X}_0\bar{Y}_0\sigma_{X_0}\sigma_{Y_0}\bar{J}\pm O(\bar{J}^2) E [ M R X R Y ] = σ X 0 σ Y 0 − ( s + 1 ) 2 X ˉ 0 Y ˉ 0 σ X 0 σ Y 0 J ˉ ± O ( J ˉ 2 )
(vii) E [ M R X R Y φ ( X ) ] = ( s − 1 ) X ˉ 0 σ Y 0 ± O ( ∣ J ˉ ∣ ) \mathbb{E}[MR_XR_Y\varphi(X)]=(s-1)\bar{X}_0\sigma_{Y_0}\pm O(|\bar{J}|) E [ M R X R Y φ ( X )] = ( s − 1 ) X ˉ 0 σ Y 0 ± O ( ∣ J ˉ ∣ ) E [ M R X R Y φ ( Y ) ] = ( s − 1 ) Y ˉ 0 σ X 0 ± O ( ∣ J ˉ ∣ ) \mathbb{E}[MR_XR_Y\varphi(Y)]=(s-1)\bar{Y}_0\sigma_{X_0}\pm O(|\bar{J}|) E [ M R X R Y φ ( Y )] = ( s − 1 ) Y ˉ 0 σ X 0 ± O ( ∣ J ˉ ∣ )
(viii) E [ M R X R Y φ ( X ) φ ( Y ) ] = ( s − 1 ) 2 X ˉ 0 Y ˉ 0 ± O ( ∣ J ˉ ∣ ) \mathbb{E}[MR_XR_Y\varphi(X)\varphi(Y)]=(s-1)^2\bar{X}_0\bar{Y}_0\pm O(|\bar{J}|) E [ M R X R Y φ ( X ) φ ( Y )] = ( s − 1 ) 2 X ˉ 0 Y ˉ 0 ± O ( ∣ J ˉ ∣ )
All the E \mathbb{E} E μ J , a , b \mu_{J,a,b} μ J , a , b
Proof. (i). Let F = φ ( X ) 2 F=\varphi(X)^2 F = φ ( X ) 2 E [ F ( 0 ) ] ≤ E [ ∣ F ( 0 ) ∣ ] ≤ O ( 1 ) \mathbb{E}[F(0)]\leq\mathbb{E}[|F(0)|]\leq O(1) E [ F ( 0 )] ≤ E [ ∣ F ( 0 ) ∣ ] ≤ O ( 1 ) E [ ∣ F ˙ ( 0 ) ∣ ] ≤ O ( 1 ) \mathbb{E}[|\dot{F}(0)|]\leq O(1) E [ ∣ F ˙ ( 0 ) ∣ ] ≤ O ( 1 ) E [ ∣ F ¨ ( J ˉ ) ∣ ] ≤ O ( 1 ) \mathbb{E}[|\ddot{F}(\bar{J})|]\leq O(1) E [ ∣ F ¨ ( J ˉ ) ∣ ] ≤ O ( 1 ) F F F F F F d d J ˉ φ ( X ) 2 ∣ J ˉ = 0 = 0 \left.\frac{d}{d\bar{J}}\varphi(X)^2\right|_{\bar{J}=0}=0 d J ˉ d φ ( X ) 2 J ˉ = 0 = 0 φ ( X ) \varphi(X) φ ( X ) J ˉ \bar{J} J ˉ
E μ J ˉ , a , b [ φ ( X ) 2 ] = E μ 0 , a , b [ φ ( X ) 2 ] + J ˉ E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) φ ( X ) 2 ] ± O ( J ˉ 2 ) = 1 + J ˉ ( E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y ] − X ˉ 0 Y ˉ 0 ) ± O ( J ˉ 2 ) = 1 − 2 X ˉ 0 Y ˉ 0 J ˉ ± O ( J ˉ 2 ) . \begin{aligned}
\mathbb{E}_{\mu_{\bar{J},a,b}}\left[\varphi(X)^2\right]=\mathbb{E}_{\mu_{0,a,b}}\left[\varphi(X)^2\right]+\bar{J}\mathbb{E}_{\mu_{0,a,b}}\left[(XY-\bar{X}_0\bar{Y}_0)\varphi(X)^2\right]\pm O(\bar{J}^2) \\
&=1+\bar{J}\left(\mathbb{E}_{\mu_{0,a,b}}\left[X\varphi(X)^2\right]\mathbb{E}_{\mu_{0,a,b}}[Y]-\bar{X}_0\bar{Y}_0\right)\pm O(\bar{J}^2)=1-2\bar{X}_0\bar{Y}_0\bar{J}\pm O(\bar{J}^2).
\end{aligned} E μ J ˉ , a , b [ φ ( X ) 2 ] = E μ 0 , a , b [ φ ( X ) 2 ] + J ˉ E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) φ ( X ) 2 ] ± O ( J ˉ 2 ) = 1 + J ˉ ( E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y ] − X ˉ 0 Y ˉ 0 ) ± O ( J ˉ 2 ) = 1 − 2 X ˉ 0 Y ˉ 0 J ˉ ± O ( J ˉ 2 ) . The second assertion follows by exchanging the roles of X X X Y Y Y
(ii). Let F = R X 2 F=R_X^2 F = R X 2 F F F F F F
R X ( y ) 2 = φ ( X ) 2 − 2 J ˉ σ X 0 φ ( X ) y ± O ( J ˉ 2 ) , R_X(y)^2=\varphi(X)^2-2\bar{J}\sigma_{X_0}\varphi(X)y\pm O(\bar{J}^2), R X ( y ) 2 = φ ( X ) 2 − 2 J ˉ σ X 0 φ ( X ) y ± O ( J ˉ 2 ) , where there is an absolute constant in the O ( ⋅ ) O(\cdot) O ( ⋅ ) E μ 0 , a , b [ F ( 0 ) ] = E μ 0 , a , b [ φ ( X ) 2 ] = 1 \mathbb{E}_{\mu_{0,a,b}}[F(0)]=\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]=1 E μ 0 , a , b [ F ( 0 )] = E μ 0 , a , b [ φ ( X ) 2 ] = 1
E μ 0 , a , b [ F ˙ ( 0 ) ] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 ) ] ≤ triangle ineq E μ 0 , a , b [ ∣ F ˙ ( 0 ) ∣ ] + E μ 0 , a , b [ ∣ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 ) ∣ ] ≤ E μ 0 , a , b [ ∣ F ˙ ( 0 ) ∣ ] + 2 E μ 0 , a , b [ ∣ F ( 0 ) ∣ ] ≤ Lemma 5.1 O ( 1 ) . \begin{aligned}
\mathbb{E}_{\mu_{0,a,b}}[\dot{F}(0)]+\mathbb{E}_{\mu_{0,a,b}}\left[(XY-\bar{X}_0\bar{Y}_0)F(0)\right]\underset{\text{triangle ineq}}{\leq}\mathbb{E}_{\mu_{0,a,b}}[|\dot{F}(0)|]+\mathbb{E}_{\mu_{0,a,b}}\left[|(XY-\bar{X}_0\bar{Y}_0)F(0)|\right] \\
&\leq\mathbb{E}_{\mu_{0,a,b}}[|\dot{F}(0)|]+2\mathbb{E}_{\mu_{0,a,b}}[|F(0)|]\underset{\text{Lemma 5.1}}{\leq}O(1).
\end{aligned} E μ 0 , a , b [ F ˙ ( 0 )] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 ) ] triangle ineq ≤ E μ 0 , a , b [ ∣ F ˙ ( 0 ) ∣ ] + E μ 0 , a , b [ ∣ ( X Y − X ˉ 0 Y ˉ 0 ) F ( 0 ) ∣ ] ≤ E μ 0 , a , b [ ∣ F ˙ ( 0 ) ∣ ] + 2 E μ 0 , a , b [ ∣ F ( 0 ) ∣ ] Lemma 5.1 ≤ O ( 1 ) . Consequently, the linear term in J ˉ \bar{J} J ˉ R X 2 R_X^2 R X 2 O ( J ˉ ) O(\bar{J}) O ( J ˉ )
(iii). We have the following explicit expressions:
R X ( y ) = X − X ˉ ( y ) σ X 0 = Lemma 2.3 ( X − X ˉ 0 ) ( 1 − J ˉ X y ) σ X 0 ( 1 + J ˉ X ˉ 0 y ) and R Y ( x ) = Lemma 2.3 ( Y − Y ˉ 0 ) ( 1 − J ˉ Y x ) σ Y 0 ( 1 + J ˉ Y ˉ 0 x ) . R_X(y)=\frac{X-\bar{X}(y)}{\sigma_{X_0}}\underset{\text{Lemma 2.3}}{=}\frac{(X-\bar{X}_0)(1-\bar{J}Xy)}{\sigma_{X_0}(1+\bar{J}\bar{X}_0y)}
\quad\text{and}\quad
R_Y(x)\underset{\text{Lemma 2.3}}{=}\frac{(Y-\bar{Y}_0)(1-\bar{J}Yx)}{\sigma_{Y_0}(1+\bar{J}\bar{Y}_0x)}. R X ( y ) = σ X 0 X − X ˉ ( y ) Lemma 2.3 = σ X 0 ( 1 + J ˉ X ˉ 0 y ) ( X − X ˉ 0 ) ( 1 − J ˉ X y ) and R Y ( x ) Lemma 2.3 = σ Y 0 ( 1 + J ˉ Y ˉ 0 x ) ( Y − Y ˉ 0 ) ( 1 − J ˉ Y x ) . It follows that
μ ( X , Y ) R X ( Y ) R Y ( X ) = Lemma 2.3 σ X 0 σ Y 0 ( 1 − J ˉ 2 ) 4 ( 1 + J ˉ X ˉ 0 Y ˉ 0 ) ⋅ X Y ( 1 − J ˉ X Y ) ( 1 + J ˉ X ˉ 0 Y ) ( 1 + J ˉ Y ˉ 0 X ) . \mu(X,Y)R_X(Y)R_Y(X)\underset{\text{Lemma 2.3}}{=}\frac{\sigma_{X_0}\sigma_{Y_0}(1-\bar{J}^2)}{4(1+\bar{J}\bar{X}_0\bar{Y}_0)}\cdot\frac{XY(1-\bar{J}XY)}{(1+\bar{J}\bar{X}_0Y)(1+\bar{J}\bar{Y}_0X)}. μ ( X , Y ) R X ( Y ) R Y ( X ) Lemma 2.3 = 4 ( 1 + J ˉ X ˉ 0 Y ˉ 0 ) σ X 0 σ Y 0 ( 1 − J ˉ 2 ) ⋅ ( 1 + J ˉ X ˉ 0 Y ) ( 1 + J ˉ Y ˉ 0 X ) X Y ( 1 − J ˉ X Y ) . so that
E [ R X R Y ] = ( − σ X 0 σ Y 0 J ˉ ) ⋅ ( 1 − J ˉ 2 ) ( 1 − J ˉ 2 X ˉ 0 2 ) ( 1 − J ˉ 2 Y ˉ 0 2 ) ⋅ 1 − J ˉ X ˉ 0 Y ˉ 0 1 + J ˉ X ˉ 0 Y ˉ 0 = ( − σ X 0 σ Y 0 J ˉ ) ⋅ ( 1 ± O ( J ˉ 2 ) ) ⋅ ( 1 − 2 J ˉ X ˉ 0 Y ˉ 0 ± O ( ∣ J ˉ ∣ 2 ) ) . \begin{aligned}
\mathbb{E}[R_XR_Y]=(-\sigma_{X_0}\sigma_{Y_0}\bar{J})\cdot\frac{(1-\bar{J}^2)}{(1-\bar{J}^2\bar{X}_0^2)(1-\bar{J}^2\bar{Y}_0^2)}\cdot\frac{1-\bar{J}\bar{X}_0\bar{Y}_0}{1+\bar{J}\bar{X}_0\bar{Y}_0} \\
&=(-\sigma_{X_0}\sigma_{Y_0}\bar{J})\cdot(1\pm O(\bar{J}^2))\cdot(1-2\bar{J}\bar{X}_0\bar{Y}_0\pm O(|\bar{J}|^2)).
\end{aligned} E [ R X R Y ] = ( − σ X 0 σ Y 0 J ˉ ) ⋅ ( 1 − J ˉ 2 X ˉ 0 2 ) ( 1 − J ˉ 2 Y ˉ 0 2 ) ( 1 − J ˉ 2 ) ⋅ 1 + J ˉ X ˉ 0 Y ˉ 0 1 − J ˉ X ˉ 0 Y ˉ 0 = ( − σ X 0 σ Y 0 J ˉ ) ⋅ ( 1 ± O ( J ˉ 2 )) ⋅ ( 1 − 2 J ˉ X ˉ 0 Y ˉ 0 ± O ( ∣ J ˉ ∣ 2 )) . Expanding this gives the desired.
(iv). Let F = R X R Y φ ( X ) F=R_XR_Y\varphi(X) F = R X R Y φ ( X ) F F F
R X R Y φ ( X ) = φ ( X ) 2 φ ( Y ) − J ˉ ( φ ( X ) 2 σ Y 0 X + φ ( X ) φ ( Y ) σ X 0 Y ) ± O ( J ˉ 2 ) . R_XR_Y\varphi(X)=\varphi(X)^2\varphi(Y)-\bar{J}\left(\varphi(X)^2\sigma_{Y_0}X+\varphi(X)\varphi(Y)\sigma_{X_0}Y\right)\pm O(\bar{J}^2). R X R Y φ ( X ) = φ ( X ) 2 φ ( Y ) − J ˉ ( φ ( X ) 2 σ Y 0 X + φ ( X ) φ ( Y ) σ X 0 Y ) ± O ( J ˉ 2 ) . The expectation of the leading term is E μ 0 , a , b [ F ( 0 ) ] = E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) ] = 0 \mathbb{E}_{\mu_{0,a,b}}[F(0)]=\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)]=0 E μ 0 , a , b [ F ( 0 )] = E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y )] = 0 J ˉ \bar{J} J ˉ
− E μ 0 , a , b [ φ ( X ) 2 σ Y 0 X + φ ( X ) φ ( Y ) σ X 0 Y ] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) φ ( X ) 2 φ ( Y ) ] = − σ Y 0 E μ 0 , a , b [ φ ( X ) 2 X ] − σ X 0 E μ 0 , a , b [ φ ( X ) ] E μ 0 , a , b [ φ ( Y ) Y ] \begin{aligned}
-\mathbb{E}_{\mu_{0,a,b}}\left[\varphi(X)^2\sigma_{Y_0}X+\varphi(X)\varphi(Y)\sigma_{X_0}Y\right]+\mathbb{E}_{\mu_{0,a,b}}\left[(XY-\bar{X}_0\bar{Y}_0)\varphi(X)^2\varphi(Y)\right] \\
&=-\sigma_{Y_0}\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2X]-\sigma_{X_0}\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)Y]
\end{aligned} − E μ 0 , a , b [ φ ( X ) 2 σ Y 0 X + φ ( X ) φ ( Y ) σ X 0 Y ] + E μ 0 , a , b [ ( X Y − X ˉ 0 Y ˉ 0 ) φ ( X ) 2 φ ( Y ) ] = − σ Y 0 E μ 0 , a , b [ φ ( X ) 2 X ] − σ X 0 E μ 0 , a , b [ φ ( X )] E μ 0 , a , b [ φ ( Y ) Y ] + E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y φ ( Y ) ] − X ˉ 0 Y ˉ 0 E μ 0 , a , b [ φ ( X ) 2 ] E [ φ ( Y ) ] = σ Y 0 X ˉ 0 − X ˉ 0 σ Y 0 = 0. (63) \begin{aligned}
\qquad+\mathbb{E}_{\mu_{0,a,b}}[X\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[Y\varphi(Y)]-\bar{X}_0\bar{Y}_0\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]\mathbb{E}[\varphi(Y)] \\
&=\sigma_{Y_0}\bar{X}_0-\bar{X}_0\sigma_{Y_0}=0. \tag*{(63)}
\end{aligned} + E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y φ ( Y )] − X ˉ 0 Y ˉ 0 E μ 0 , a , b [ φ ( X ) 2 ] E [ φ ( Y )] = σ Y 0 X ˉ 0 − X ˉ 0 σ Y 0 = 0. (63) This gives E μ J , a , b [ R X R Y φ ( X ) ] = O ( J ‾ 2 ) \mathbb{E}_{\mu_{J,a,b}}[R_XR_Y\varphi(X)] = O(\overline{J}^{2}) E μ J , a , b [ R X R Y φ ( X )] = O ( J 2 ) X X X Y Y Y E μ J , a , b [ R X R Y φ ( Y ) ] = O ( J ‾ 2 ) \mathbb{E}_{\mu_{J,a,b}}[R_XR_Y\varphi(Y)] = O(\overline{J}^{2}) E μ J , a , b [ R X R Y φ ( Y )] = O ( J 2 )
(v). Let F = R X R Y φ ( X ) φ ( Y ) F=R_XR_Y\varphi(X)\varphi(Y) F = R X R Y φ ( X ) φ ( Y ) F F F
R X R Y φ ( X ) φ ( Y ) = φ ( X ) 2 φ ( Y ) 2 − J ‾ ( φ ( X ) 2 φ ( Y ) σ Y 0 X + φ ( X ) φ ( Y ) 2 σ X 0 Y ) ± O ( J ‾ 2 ) . (64) R_XR_Y\varphi(X)\varphi(Y)=\varphi(X)^2\varphi(Y)^2-\overline{J}\left(\varphi(X)^2\varphi(Y)\sigma_{Y_0}X+\varphi(X)\varphi(Y)^2\sigma_{X_0}Y\right)\pm O(\overline{J}^{2}). \tag*{(64)} R X R Y φ ( X ) φ ( Y ) = φ ( X ) 2 φ ( Y ) 2 − J ( φ ( X ) 2 φ ( Y ) σ Y 0 X + φ ( X ) φ ( Y ) 2 σ X 0 Y ) ± O ( J 2 ) . (64) The expected value of the leading term is E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) 2 ] = E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] = 1 \mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2\varphi(Y)^2]=\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)^2]=1 E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) 2 ] = E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] = 1 J ‾ \overline{J} J
− E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) σ Y 0 X + φ ( X ) φ ( Y ) 2 σ X 0 Y ] + E μ 0 , a , b [ ( X Y − X ‾ 0 Y ‾ 0 ) φ ( X ) 2 φ ( Y ) 2 ] = − σ Y 0 E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) ] − σ X 0 E μ 0 , a , b [ Y φ ( Y ) 2 ] E μ 0 , a , b [ φ ( X ) ] + E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y φ ( Y ) 2 ] − X ‾ 0 Y ‾ 0 E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] = 0. (65) \begin{aligned}
&-\mathbb{E}_{\mu_{0,a,b}}\left[\varphi(X)^2\varphi(Y)\sigma_{Y_0}X+\varphi(X)\varphi(Y)^2\sigma_{X_0}Y\right]+\mathbb{E}_{\mu_{0,a,b}}\left[(XY-\overline{X}_0\overline{Y}_0)\varphi(X)^2\varphi(Y)^2\right]\\
&=-\sigma_{Y_0}\mathbb{E}_{\mu_{0,a,b}}[X\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)]-\sigma_{X_0}\mathbb{E}_{\mu_{0,a,b}}[Y\varphi(Y)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)]\\
&\quad+\mathbb{E}_{\mu_{0,a,b}}[X\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[Y\varphi(Y)^2]-\overline{X}_0\overline{Y}_0\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)^2]\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)^2]=0. \tag*{(65)}
\end{aligned} − E μ 0 , a , b [ φ ( X ) 2 φ ( Y ) σ Y 0 X + φ ( X ) φ ( Y ) 2 σ X 0 Y ] + E μ 0 , a , b [ ( X Y − X 0 Y 0 ) φ ( X ) 2 φ ( Y ) 2 ] = − σ Y 0 E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y )] − σ X 0 E μ 0 , a , b [ Y φ ( Y ) 2 ] E μ 0 , a , b [ φ ( X )] + E μ 0 , a , b [ X φ ( X ) 2 ] E μ 0 , a , b [ Y φ ( Y ) 2 ] − X 0 Y 0 E μ 0 , a , b [ φ ( X ) 2 ] E μ 0 , a , b [ φ ( Y ) 2 ] = 0. (65) (vi). Let F = M R X R Y F=MR_XR_Y F = M R X R Y F F F
E μ 0 , a , b [ M ( 0 ) R X ( 0 ) R Y ( 0 ) ] = E μ 0 , a , b [ ( X + s X ‾ 0 ) φ ( X ) ] ⋅ E μ 0 , a , b [ ( Y + s Y ‾ 0 ) φ ( Y ) ] = ( E μ 0 , a , b [ X φ ( X ) ] + s X ‾ 0 E μ 0 , a , b [ φ ( X ) ] ) ( E μ 0 , a , b [ Y φ ( Y ) ] + s Y ‾ 0 E μ 0 , a , b [ φ ( Y ) ] ) = σ X 0 σ Y 0 . (66) \begin{aligned}
\mathbb{E}_{\mu_{0,a,b}}[M(0)R_X(0)R_Y(0)]
&=\mathbb{E}_{\mu_{0,a,b}}[(X+s\overline{X}_0)\varphi(X)]\cdot\mathbb{E}_{\mu_{0,a,b}}[(Y+s\overline{Y}_0)\varphi(Y)]\\
&=(\mathbb{E}_{\mu_{0,a,b}}[X\varphi(X)]+s\overline{X}_0\mathbb{E}_{\mu_{0,a,b}}[\varphi(X)])(\mathbb{E}_{\mu_{0,a,b}}[Y\varphi(Y)]+s\overline{Y}_0\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)])\\
&=\sigma_{X_0}\sigma_{Y_0}. \tag*{(66)}
\end{aligned} E μ 0 , a , b [ M ( 0 ) R X ( 0 ) R Y ( 0 )] = E μ 0 , a , b [( X + s X 0 ) φ ( X )] ⋅ E μ 0 , a , b [( Y + s Y 0 ) φ ( Y )] = ( E μ 0 , a , b [ X φ ( X )] + s X 0 E μ 0 , a , b [ φ ( X )]) ( E μ 0 , a , b [ Y φ ( Y )] + s Y 0 E μ 0 , a , b [ φ ( Y )]) = σ X 0 σ Y 0 . (66) The linear in J ‾ \overline{J} J M R X R Y MR_XR_Y M R X R Y F ˙ 1 ( 0 ) \dot{F}_1(0) F ˙ 1 ( 0 ) μ 0 , a , b \mu_{0,a,b} μ 0 , a , b
E [ d d J ‾ M R X R Y ∣ J ‾ = 0 ] = E [ φ ( X ) φ ( Y ) ( ( X + s X ‾ 0 ) s σ Y 0 2 X + ( Y + s Y ‾ 0 ) s σ X 0 2 Y − σ X 0 Y φ ( Y ) ( X + s X ‾ 0 ) ( Y + s Y ‾ 0 ) − σ Y 0 X φ ( X ) ( X + s X ‾ 0 ) ( Y + s Y ‾ 0 ) ) ] = s σ Y 0 2 E μ 0 , a , b [ φ ( Y ) ] E [ X φ ( X ) ( X + s X ‾ 0 ) ] + s σ X 0 2 E [ φ ( X ) ] E μ 0 , a , b [ Y φ ( Y ) ( Y + s Y ‾ 0 ) ] − σ X 0 ( E [ φ ( Y ) ] s X ‾ 0 + s 2 X ‾ 0 Y ‾ 0 E [ Y φ ( Y ) ] + E [ φ ( Y ) ] E [ X ] + s Y ‾ 0 E [ Y φ ( Y ) ] E [ X ] ) − σ Y 0 ( E [ φ ( X ) ] s Y ‾ 0 + s 2 Y ‾ 0 X ‾ 0 E [ X φ ( X ) ] + E [ φ ( X ) ] E [ Y ] + s X ‾ 0 E [ X φ ( X ) ] E [ Y ] ) = − 2 ( s 2 + s ) σ X 0 σ Y 0 X ‾ 0 Y ‾ 0 . (67) \begin{aligned}
\mathbb{E}\left[\left.\frac{d}{d\overline{J}}MR_XR_Y\right|_{\overline{J}=0}\right]
&=\mathbb{E}\left[\varphi(X)\varphi(Y)\left((X+s\overline{X}_0)s\sigma_{Y_0}^{2}X+(Y+s\overline{Y}_0)s\sigma_{X_0}^{2}Y\right.\right.\\
&\qquad\left.\left.-\sigma_{X_0}Y\varphi(Y)(X+s\overline{X}_0)(Y+s\overline{Y}_0)-\sigma_{Y_0}X\varphi(X)(X+s\overline{X}_0)(Y+s\overline{Y}_0)\right)\right]\\
&=s\sigma_{Y_0}^{2}\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)]\mathbb{E}[X\varphi(X)(X+s\overline{X}_0)]\\
&\quad+s\sigma_{X_0}^{2}\mathbb{E}[\varphi(X)]\mathbb{E}_{\mu_{0,a,b}}[Y\varphi(Y)(Y+s\overline{Y}_0)]\\
&\quad-\sigma_{X_0}\left(\mathbb{E}[\varphi(Y)]s\overline{X}_0+s^2\overline{X}_0\overline{Y}_0\mathbb{E}[Y\varphi(Y)]+\mathbb{E}[\varphi(Y)]\mathbb{E}[X]+s\overline{Y}_0\mathbb{E}[Y\varphi(Y)]\mathbb{E}[X]\right)\\
&\quad-\sigma_{Y_0}\left(\mathbb{E}[\varphi(X)]s\overline{Y}_0+s^2\overline{Y}_0\overline{X}_0\mathbb{E}[X\varphi(X)]+\mathbb{E}[\varphi(X)]\mathbb{E}[Y]+s\overline{X}_0\mathbb{E}[X\varphi(X)]\mathbb{E}[Y]\right)\\
&=-2(s^2+s)\sigma_{X_0}\sigma_{Y_0}\overline{X}_0\overline{Y}_0. \tag*{(67)}
\end{aligned} E [ d J d M R X R Y J = 0 ] = E [ φ ( X ) φ ( Y ) ( ( X + s X 0 ) s σ Y 0 2 X + ( Y + s Y 0 ) s σ X 0 2 Y − σ X 0 Y φ ( Y ) ( X + s X 0 ) ( Y + s Y 0 ) − σ Y 0 X φ ( X ) ( X + s X 0 ) ( Y + s Y 0 ) ) ] = s σ Y 0 2 E μ 0 , a , b [ φ ( Y )] E [ X φ ( X ) ( X + s X 0 )] + s σ X 0 2 E [ φ ( X )] E μ 0 , a , b [ Y φ ( Y ) ( Y + s Y 0 )] − σ X 0 ( E [ φ ( Y )] s X 0 + s 2 X 0 Y 0 E [ Y φ ( Y )] + E [ φ ( Y )] E [ X ] + s Y 0 E [ Y φ ( Y )] E [ X ] ) − σ Y 0 ( E [ φ ( X )] s Y 0 + s 2 Y 0 X 0 E [ X φ ( X )] + E [ φ ( X )] E [ Y ] + s X 0 E [ X φ ( X )] E [ Y ] ) = − 2 ( s 2 + s ) σ X 0 σ Y 0 X 0 Y 0 . (67) Second, we compute the coefficient coming from ( X Y − X ‾ 0 Y ‾ 0 ) M ( 0 ) R X ( 0 ) R Y ( 0 ) (XY-\overline{X}_0\overline{Y}_0)M(0)R_X(0)R_Y(0) ( X Y − X 0 Y 0 ) M ( 0 ) R X ( 0 ) R Y ( 0 ) E μ 0 , a , b [ φ ( X ) ] = E μ 0 , a , b [ φ ( Y ) ] = 0 \mathbb{E}_{\mu_{0,a,b}}[\varphi(X)]=\mathbb{E}_{\mu_{0,a,b}}[\varphi(Y)]=0 E μ 0 , a , b [ φ ( X )] = E μ 0 , a , b [ φ ( Y )] = 0
E μ 0 , a , b [ ( X Y − X ‾ 0 Y ‾ 0 ) ( X + s X ‾ 0 ) φ ( X ) ( Y + s Y ‾ 0 ) φ ( Y ) ] = E μ 0 , a , b [ s 2 X Y X ‾ 0 Y ‾ 0 φ ( X ) φ ( Y ) ] − X ‾ 0 Y ‾ 0 σ X 0 σ Y 0 = ( s 2 − 1 ) σ X 0 σ Y 0 X ‾ 0 Y ‾ 0 . (68) \begin{aligned}
\mathbb{E}_{\mu_{0,a,b}}[(XY-\overline{X}_0\overline{Y}_0)(X+s\overline{X}_0)\varphi(X)(Y+s\overline{Y}_0)\varphi(Y)]
&=\mathbb{E}_{\mu_{0,a,b}}[s^2XY\overline{X}_0\overline{Y}_0\varphi(X)\varphi(Y)]-\overline{X}_0\overline{Y}_0\sigma_{X_0}\sigma_{Y_0}\\
&=(s^2-1)\sigma_{X_0}\sigma_{Y_0}\overline{X}_0\overline{Y}_0. \tag*{(68)}
\end{aligned} E μ 0 , a , b [( X Y − X 0 Y 0 ) ( X + s X 0 ) φ ( X ) ( Y + s Y 0 ) φ ( Y )] = E μ 0 , a , b [ s 2 X Y X 0 Y 0 φ ( X ) φ ( Y )] − X 0 Y 0 σ X 0 σ Y 0 = ( s 2 − 1 ) σ X 0 σ Y 0 X 0 Y 0 . (68) We add (56) and (57) to get the term linear in J ‾ \overline{J} J X X X Y Y Y
(vii). Let F = M R X R Y φ ( X ) F = M R_X R_Y \varphi(X) F = M R X R Y φ ( X ) F F F
E μ 0 , a , b [ F ( 0 ) ] = E μ 0 , a , b [ ( X + s X ‾ 0 ) φ ( X ) 2 ] ⋅ E μ 0 , a , b [ ( Y + s Y ‾ 0 ) φ ( Y ) ] = ( E μ 0 , a , b [ X φ ( X ) 2 ] + s X ‾ 0 E μ 0 , a , b [ φ ( X ) 2 ] ) ( E μ 0 , a , b [ Y φ ( Y ) ] + s Y ‾ 0 E μ 0 , a , b [ φ ( Y ) ] ) = ( s − 1 ) X ‾ 0 ⋅ σ Y ‾ 0 . (69) \begin{aligned}
\mathbb{E}_{\mu_0,a,b}[F(0)] &= \mathbb{E}_{\mu_0,a,b}\left[(X+s\overline{X}_0)\varphi(X)^2\right]\cdot\mathbb{E}_{\mu_0,a,b}\left[(Y+s\overline{Y}_0)\varphi(Y)\right] \\
&= \left(\mathbb{E}_{\mu_0,a,b}[X\varphi(X)^2]+s\overline{X}_0\mathbb{E}_{\mu_0,a,b}[\varphi(X)^2]\right)\left(\mathbb{E}_{\mu_0,a,b}[Y\varphi(Y)]+s\overline{Y}_0\mathbb{E}_{\mu_0,a,b}[\varphi(Y)]\right) \\
&= (s-1)\overline{X}_0\cdot\sigma\overline{Y}_0. \tag*{(69)}
\end{aligned} E μ 0 , a , b [ F ( 0 )] = E μ 0 , a , b [ ( X + s X 0 ) φ ( X ) 2 ] ⋅ E μ 0 , a , b [ ( Y + s Y 0 ) φ ( Y ) ] = ( E μ 0 , a , b [ X φ ( X ) 2 ] + s X 0 E μ 0 , a , b [ φ ( X ) 2 ] ) ( E μ 0 , a , b [ Y φ ( Y )] + s Y 0 E μ 0 , a , b [ φ ( Y )] ) = ( s − 1 ) X 0 ⋅ σ Y 0 . (69) As in (ii), using another application of Lemma 5.1, the linear term in J ‾ \overline{J} J M R X R Y φ ( X ) M R_X R_Y\varphi(X) M R X R Y φ ( X ) O ( J ‾ ) O(\overline{J}) O ( J ) X X X Y Y Y
(viii). Let F = M R X R Y φ ( X ) φ ( Y ) F = M R_X R_Y\varphi(X)\varphi(Y) F = M R X R Y φ ( X ) φ ( Y ) F F F
E μ 0 , a , b [ F ( 0 ) ] = E μ 0 , a , b [ ( X + s X ‾ 0 ) φ ( X ) 2 ] ⋅ E μ 0 , a , b [ ( Y + s Y ‾ 0 ) φ ( Y ) 2 ] = ( E μ 0 , a , b [ X φ ( X ) 2 ] + s X ‾ 0 E μ 0 , a , b [ φ ( X ) 2 ] ) ( E μ 0 , a , b [ Y φ ( Y ) 2 ] + s Y ‾ 0 E μ 0 , a , b [ φ ( Y ) 2 ] ) = ( s − 1 ) 2 X ‾ 0 Y ‾ 0 . (70) \begin{aligned}
\mathbb{E}_{\mu_0,a,b}[F(0)] &= \mathbb{E}_{\mu_0,a,b}\left[(X+s\overline{X}_0)\varphi(X)^2\right]\cdot\mathbb{E}_{\mu_0,a,b}\left[(Y+s\overline{Y}_0)\varphi(Y)^2\right] \\
&= \left(\mathbb{E}_{\mu_0,a,b}[X\varphi(X)^2]+s\overline{X}_0\mathbb{E}_{\mu_0,a,b}[\varphi(X)^2]\right)\left(\mathbb{E}_{\mu_0,a,b}[Y\varphi(Y)^2]+s\overline{Y}_0\mathbb{E}_{\mu_0,a,b}[\varphi(Y)^2]\right) \\
&= (s-1)^2\overline{X}_0\overline{Y}_0. \tag*{(70)}
\end{aligned} E μ 0 , a , b [ F ( 0 )] = E μ 0 , a , b [ ( X + s X 0 ) φ ( X ) 2 ] ⋅ E μ 0 , a , b [ ( Y + s Y 0 ) φ ( Y ) 2 ] = ( E μ 0 , a , b [ X φ ( X ) 2 ] + s X 0 E μ 0 , a , b [ φ ( X ) 2 ] ) ( E μ 0 , a , b [ Y φ ( Y ) 2 ] + s Y 0 E μ 0 , a , b [ φ ( Y ) 2 ] ) = ( s − 1 ) 2 X 0 Y 0 . (70) As in (ii), using another application of Lemma 5.1, the linear term in J ‾ \overline{J} J M R X R Y φ ( X ) φ ( Y ) M R_X R_Y\varphi(X)\varphi(Y) M R X R Y φ ( X ) φ ( Y ) O ( J ‾ ) O(\overline{J}) O ( J )
