KPZ Equation Limit of Stochastic Higher Spin Six Vertex Model

Abstract

We consider the stochastic higher spin six vertex (SHS6V) model introduced by Corwin and Petrov (Commun. Math. Phys., 343(2), 651–700 2016) with general integer spin parameters I, J. Starting from near stationary initial condition, we prove that the SHS6V model converges to the Kardar-Parisi-Zhang (KPZ) equation under weakly asymmetric scaling. This generalizes the result in Corwin et al. (2018, Theorem 1.1) from I = J = 1 to general I, J.

Appendices

Appendix A: Stationary Distribution of the SHS6V Model

In this section, we provide a one parameter family of stationary distribution for the unfused SHS6V model. It is worth to remark that in the recent work of [30], a translation-invariant Gibbs measure was obtained (using the idea from [2]) for the space-time inhomogeneous SHS6V model on the full lattice, see Proposition 4.5 of [30]. However, It is not obvious that the dynamic of SHS6V model under this Gibbs measure coincides with the one of the bi-infinite SHS6V model specified in Lemma 2.1. This being the case, we choose to proceed without relying on the result from [30].

We start with a well-known combinatoric lemma.

Lemma A.1 (q-binomial formula)

Set ν = q^−Ias usual, the following identity holds for all$q \in \mathbb {C}$,

$$ \sum\limits_{n=0}^{I} \frac{(\nu; q)_{n}}{(q; q)_{n}} z^{n} = \frac{(\nu z; q)_{\infty}}{(z; q)_{\infty}}. $$

Proof

According to q-binomial theorem [1],

$$ \sum\limits_{n=0}^{\infty} \frac{(\nu; q)_{n}}{(q; q)_{n}} z^{n} = \frac{(\nu z; q)_{\infty}}{(z; q)_{\infty}}. $$

When ν = q^−I, (ν, q)_n = 0 for n > I. Therefore,

$$ \sum\limits_{n=0}^{I} \frac{(\nu; q)_{n}}{(q; q)_{n}} z^{n} = \sum\limits_{n=0}^{\infty} \frac{(\nu; q)_{n}}{(q; q)_{n}} z^{n}= \frac{(\nu z; q)_{\infty}}{(z; q)_{\infty}}. $$

□

Lemma A.2

Fix q > 1, ν = q^−Iand ρ ∈ (0, I), define a probability measure π_ρon$\{0, 1, \dots , I\}$:

$$ \pi_\rho (i) = \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \frac{(\nu, q)_{i}}{(q, q)_{i}} \chi^{i}, \quad i \in \{0, 1, \dots, I\}, $$

(A.1)

where χis the unique negative real number satisfying

$$ \sum\limits_{i=1}^{I} \frac{\chi}{\chi - q^{i}} = \rho. $$

(A.2)

Furthermore, we have

$$ \mathbb{E}\big[\pi_\rho \big] = \rho, \qquad \text{Var}\big[\pi_\rho \big] = \rho - \sum\limits_{i=1}^{I} \frac{\chi^{2}}{(q^{i} - \chi)^{2}}. $$

Proof

We first show that π_ρ is indeed a probability measure. It is clear that $\pi _{\rho } (i) \geqslant 0$ for all $i \in \{0, 1, \dots , I\}$. By Lemma A.1,

$$ \sum\limits_{i=0}^{I} \pi_\rho (i) = \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \sum\limits_{i=0}^{I} \frac{(\nu, q)_{i}}{(q, q)_{i}} \chi^{i} = \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \frac{(\nu \chi, q)_{\infty}}{(\chi, q)_{\infty}} = 1. $$

Next, we compute the expectation and the variance of π_ρ. Using again Lemma A.1, the moment generating function is given by

$$ \begin{array}{@{}rcl@{}} {\Lambda}(z) &=& \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \sum\limits_{i=0}^{I} \frac{(\nu, q)_{i}}{(q, q)_{i}} \chi^{i} z^{i} = \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \frac{(\nu \chi z, q)_{\infty}}{(\chi z, q)_{\infty}}\\ &=& \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \prod\limits_{i=1}^{I} (1 - \nu q^{i-1} \chi z). \end{array} $$

(A.3)

It is clear that

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\big[\pi_\rho \big] &= {\Lambda}^{\prime}(1),\\ \text{Var}\big[\pi_\rho \big] &= {\Lambda}^{\prime\prime}(1) + {\Lambda}^{\prime}(1) - {\Lambda}^{\prime}(1)^{2}. \end{array} $$

Via (A.3), one has

$$ \begin{array}{@{}rcl@{}} {\Lambda}^{\prime}(z) &=& \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \bigg(\prod\limits_{i=1}^{I} (1 - \nu q^{i-1} \chi z)\bigg) \bigg(\sum\limits_{i=1}^{I} \frac{-\nu q^{i-1} \chi}{1 - \nu q^{i-1} \chi z}\bigg),\\ {\Lambda}^{\prime\prime}(z) &=&\frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \bigg(\prod\limits_{i=1}^{I} (1 - \nu q^{i-1} \chi z)\bigg)\\ &&\times \bigg[\bigg(\sum\limits_{i=1}^{I} \frac{-\nu q^{i-1} \chi}{1 - \nu q^{i-1} \chi z}\bigg)^{2} - \sum\limits_{i=1}^{I} \frac{(\nu q^{i-1} \chi)^{2}}{(1 - \nu q^{i-1} \chi z)^{2}}\bigg]. \end{array} $$

Note that

$$ \frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \prod\limits_{i=1}^{I} (1 - \nu q^{i-1} \chi) = 1,$$

combining this with (A.2) yields

$$ {\Lambda}^{\prime}(1) = \rho, \qquad {\Lambda}^{\prime\prime}(1) = \rho^{2} - \sum\limits_{i=1}^{I} \frac{\chi^{2}}{(q^{i} - \chi)^{2}}, $$

which concludes the lemma. □

Theorem A.3

For ρ ∈ (0, I), the product measure$\bigotimes \pi _{\rho }$is stationary for the unfused SHS6V model$\vec {\eta }(t)$ (Definition 2.3).

Proof

It suffices to show that if $\vec {\eta }(t) \sim \bigotimes \pi _{\rho }$, then $\vec {\eta }(t+1) \sim \bigotimes \pi _{\rho }$.

Recall that K(t, y) = N(t, y) − N(t + 1, y) records the number of particles (either zero or one) that move across location y at time t. We first show that $K(t, y) \sim \text {Ber }(\frac {\alpha (t) \chi }{\alpha (t) \chi +1})$ (recall that α(t) = αq^{mod (t)}). To this end, referring to (??),

$$ K(t, y) =\sum\limits_{y^{\prime} = -\infty}^{y} \prod\limits_{z = y^{\prime}+1}^{y} \bigg(B^{\prime}(t, z, \eta_{z} (t)) - B(t, z, \eta_{z} (t))\bigg) B(t, y^{\prime}, \eta_{y^{\prime}}(t)). $$

(A.4)

Recalling from (??), $B(t, z, \eta ) \sim \text {Ber }\big (\frac {\alpha (t)(1 - q^{\eta })}{1 +\alpha (t)}\big ), B^{\prime }(t, z, \eta ) \sim \text {Ber }\big (\frac {\alpha (t) + \nu q^{\eta }}{1 +\alpha (t)}\big )$. Since the random variables $B, B^{\prime }$ are all independent,

$$ \begin{array}{@{}rcl@{}} &&\mathbb{E}\bigg[\prod\limits_{z = y^{\prime}+1}^{y} \bigg(B^{\prime}(t, z, \eta_{z} (t)) - B(t, z, \eta_{z} (t))\bigg) B(t, y^{\prime}, \eta_{y^{\prime}}(t)) \bigg| \mathcal{F}(t)\bigg]\\ &=& \frac{\alpha(t) (1-q^{\eta_{y^{\prime}}(t)})}{1 + \alpha(t)} \prod\limits_{z = y^{\prime}+1}^{y} \frac{(\alpha(t) + \nu) q^{\eta_{z} (t)}}{1 + \alpha(t)}. \end{array} $$

Therefore, by tower property

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\big[K(t, y)\big] &=& \sum\limits_{y^{\prime} = -\infty}^{y} \mathbb{E} \bigg[\prod\limits_{z = y^{\prime}+1}^{y} \frac{\alpha(t) (1-q^{\eta_{y^{\prime}}(t)})}{1 + \alpha(t)} \prod\limits_{z = y^{\prime}+1}^{y} \frac{(\alpha(t) + \nu) q^{\eta_{z} (t)}}{1 + \alpha(t)}\bigg], \\ &=& \sum\limits_{y^{\prime} = -\infty}^{y} \frac{\alpha(t)}{1 + \alpha(t)} \bigg(\frac{\alpha(t) + \nu}{1 + \alpha(t)}\bigg)^{y- y^{\prime}} \big(\mathbb{E}\big[q^{\eta_{y}(t)}\big]\big)^{y- y^{\prime}} (1 - \mathbb{E}\big[q^{\eta_{y}(t)}\big]).\\ \end{array} $$

(A.5)

As $\eta _{y} (t) \sim \pi _{\rho }$, we obtain using Lemma A.1

$$ \mathbb{E}\big[q^{\eta_{y} (t)}\big] =\frac{(\chi, q)_{\infty}}{(\chi \nu, q)_{\infty}} \sum\limits_{i=0}^{\infty} \frac{(\nu, q)_{i}}{(q, q)_{i}} (\chi q)^{i} = \frac{(\chi \nu q; q)_{\infty}}{(\chi q; q)_{\infty}} \frac{(\chi; q)_{\infty}}{(\chi \nu; q)_{\infty}} = \frac{1 - \chi}{1 - \chi \nu}. $$

Inserting the value of $\mathbb {E}\big [q^{\eta _{y} (t)}\big ]$ into the RHS of (A.5) yields that

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\big[K(t, y)\big] &=& \sum\limits_{y^{\prime} = -\infty}^{y} \frac{\alpha(t)}{1 + \alpha(t)} \bigg(\frac{(\alpha(t) + \nu)(1-\chi)}{(1 + \alpha(t))(1-\chi\nu)}\bigg)^{y- y^{\prime}} \bigg(1 - \frac{1 - \chi}{1 - \chi \nu}\bigg)\\ &=& \frac{\alpha(t) \chi}{\alpha(t) \chi +1}. \end{array} $$

Since K(t, y) ∈{0, 1}, we conclude that

$$ K(t, y) \sim \text{Ber }(\frac{\alpha(t) \chi}{\alpha(t) \chi +1}). $$

(A.6)

The next step is to show that the marginal of $\vec {\eta }(t+1)$ is distributed as π_ρ for each coordinate. Referring to (A.4), it is straightforward that the following recursion holds

$$ \begin{array}{@{}rcl@{}} K(t, y) = &B(t, y, \eta_{y} (t)) + \Big(B^{\prime}(t, y, \eta_{y} (t)) - B(t, y, \eta_{y} (t))\Big) K(t, y-1). \end{array} $$

(A.7)

Therefore,

$$ \begin{array}{@{}rcl@{}} \eta_{y} (t) - \eta_{y} (t+1) &=& N(t, y) - N(t, y-1) + N(t+1, y-1) - N(t+1, y) , \\ &=& K(t, y) - K(t, y-1), \\ &=& K(t, y-1) \Big(B^{\prime}(t, y, \eta_{y}(t)) - B(t, y, \eta_{y}(t)) - 1\Big)\\ &&+ B(t, y, \eta_{y} (t)). \end{array} $$

For the second equality above, we used K(t, y) = N(t, y) − N(t + 1, y). Therefore,

$$ \eta_{y} (t+1) = \left\{\begin{array}{llll} \eta_{y} (t) - B(t, y, \eta_{y} (t)), \qquad &K(t, y-1) = 0,\\ \eta_{y} (t) + 1 - B^{\prime}(t, y, \eta_{y} (t)), \qquad &K(t, y-1) = 1. \end{array}\right. $$

(A.8)

Due to (A.4), we see that $K(t, y-1) \in \sigma \Big (B(t, z, \eta ), B^{\prime }(t, z, \eta ), \eta _{z}(t): z \leqslant y-1, \eta \in \{0, 1, \dots , I\} \Big )$. Note that we have assumed $\vec {\eta }(t) \sim \bigotimes \pi _{\rho }$, which implies the independence between η_y(t) and η_z(t) for z≠y. Therefore, η_y(t) and K(t, y − 1) are independent. Using (A.8) we get

$$ \begin{array}{@{}rcl@{}} \mathbb{P}\big(\eta_{y} (t+1) = i\big) &=& \mathbb{P}\big(K(t, y-1) = 0\big) \mathbb{P}\big(\eta_{y} (t) - B(t, y, \eta_{y} (t)) = i\big)\\ &&+\mathbb{P}\big(K(t, y-1) = 1\big) \mathbb{P}\big(\eta_{y} (t) - B^{\prime}(t, y, \eta_{y} (t)) = i-1\big). \end{array} $$

By $K(t, y-1) \sim \text {Ber }(\frac {\alpha (t) \chi }{\alpha (t) \chi +1})$ and $\eta _{y} (t) \sim \pi _{\rho }$, one readily has

$$ \begin{array}{@{}rcl@{}} &&\mathbb{P}\big(\eta_{y} (t+1) = i\big)\\ &=& \frac{1}{1 + \alpha(t) \chi} \bigg[\pi_\rho (i) \frac{1 + \alpha(t) q^{i}}{1 + \alpha(t)} + \pi_\rho (i+1) \frac{\alpha(t) (1 - q^{i+1})}{1 +\alpha(t)}\bigg]\\ &&+ \frac{\alpha(t) \chi}{1 + \alpha(t) \chi} \bigg[\pi_\rho (i) \frac{\alpha(t) + \nu q^{i}}{1 +\alpha(t)} + \pi_\rho (i-1) \frac{1 - \nu q^{i-1}}{1 + \alpha(t)}\bigg]\\ &=& \pi_\rho (i). \end{array} $$

To conclude Theorem A.3, it suffices to show the independence among η_y(t + 1) for different value of y. It is enough to show that

$$ \eta_{y} (t+1) \text{ is independent with } \{\eta_{y+1} (t+1), \eta_{y+2}(t+1), \dots\} \text{ for all } y \in \mathbb{Z}. $$

(A.9)

We need the following lemma.

Lemma A.4

For all$y \in \mathbb {Z}$, η_y(t + 1) is independent with K(t, y).

Let us first see how this lemma leads to (A.9). We have via (A.4),

$$K(t, y) \in \sigma\Big(B(t, z, \eta), B^{\prime}(t, z, \eta), \eta_{z} (t): z \leqslant y, \eta \in \{0, 1, \dots, I\}\Big).$$

Combining this with (A.8),

$$\eta_{y}(t+1) \in \sigma\Big(B(t, z, \eta), B^{\prime}(t, z, \eta), \eta_{z} (t): z \leqslant y, \eta \in \{0, 1, \dots, I\}\Big).$$

Since η_i(t) are all independent for different i, one has

$$ \begin{array}{@{}rcl@{}} &&\Big(B(t, z, \eta), B^{\prime}(t, z, \eta), \eta_{z} (t): z \leqslant y, \eta \in \{0, 1, \dots, I\}\Big) \text{ is independent with }\\ &&(\eta_{y+1}(t), \eta_{y+2}(t), \dots). \end{array} $$

We achieve

$$ \big(K(t, y), \eta_{y} (t+1)\big) \text{ is independent with } \big(\eta_{y+1}(t), \eta_{y+2} (t), {\dots} \big). $$

Using Lemma A.4, we conclude

$$ \eta_{y}(t+1) \text{ is independent with } \big(K(t, y), \eta_{y+1}(t), \eta_{y+2}(t), {\dots} \big). $$

Therefore,

$$ \begin{array}{@{}rcl@{}} &&\eta_{y}(t+1) \text{ is independent with } \sigma\Big(K(t, y), \eta_{z}(t), B(t, z, \eta), B^{\prime}(t, z, \eta): z \!\geqslant\! y+1,\\ &&\eta \in \{0, 1, \dots, I\}\Big). \end{array} $$

(A.10)

On the other hand, by (A.7) and (A.8), we conclude for all $y \in \mathbb {Z}$

$$ \begin{array}{@{}rcl@{}} \big(\eta_{y+1}(t+1), \eta_{y+2}(t+1), {\dots} \big) &\in& \sigma\Big(K(t, y), B(t, z, \eta), B^{\prime}(t, z, \eta),\eta_{z}(t): z \geqslant y\\ &&\quad +1, \eta \in \{0, 1, \dots, I\}\Big). \end{array} $$

(A.11)

Combining (A.10) and (A.11), we find that for all $y \in \mathbb {Z}$

$$ \eta_{y} (t+1) \text{ is independent with } \big(\eta_{y+1}(t+1), \eta_{y+2}(t+1), {\dots} \big), $$

which concludes (A.9). □

Proof of Lemma A.4

As K(t, y) ∈{0, 1}, it suffices to show that for all $j \in \{0, 1, \dots , I\}$, one has

$$ \mathbb{P}\big(\eta_{y}(t+1) = j, K(t, y) = 1\big) = \mathbb{P}\big(\eta_{y}(t+1) = j\big) \mathbb{P}\big(K(t, y) = 1 \big). $$

Due to (A.7),

$$ K(t, y) = \left\{\begin{array}{llll} B(t, y, \eta_{y} (t)), \qquad &K(t, y-1) = 0,\\ B^{\prime}(t, y, \eta_{y} (t)), \qquad &K(t, y-1) = 1. \end{array}\right. $$

Together with (A.8), we obtain that if K(t, y − 1) = 0,

$$ \big(\eta_{y} (t+1), K(t, y)\big) = (j, 1) \text{ is equivalent to } \big(\eta_{y} (t), B(t, y, \eta_{y} (t))\big) = (j+1, 1). $$

If K(t, y − 1) = 1,

$$ \big(\eta_{y} (t+1), K(t, y)\big) = (j, 1) \text{ is equivalent to } \big(\eta_{y} (t), B(t, y, \eta_{y} (t))\big) = (j, 1). $$

The discussion above yields (using the independence between η_y(t) and K(t, y − 1))

$$ \begin{array}{@{}rcl@{}} &&\mathbb{P}\big(\eta_{y} (t+1) = j, K(t, y) = 1\big), \\ &=& \mathbb{P}\big(K(t, y-1) = 0\big) \mathbb{P}\big(\eta_{y} (t) = j+1, B(t, y, \eta_{y} (t)) = 1\big) + \mathbb{P}\big(K(t, y-1) = 1\big) \\ &&\times \mathbb{P}\big(\eta_{y} (t) = j, B^{\prime}(t, y, \eta_{y} (t)) = 1\big),\\ &=& \frac{1}{1 + \alpha(t) \chi} \frac{\alpha(t) (1- q^{j+1})}{1 + \alpha(t)} \pi_\rho (j+1) + \frac{\alpha(t) \chi}{1 + \alpha(t) \chi} \frac{\alpha(t) + \nu q^{j}}{1 + \alpha(t)} \pi_\rho (j), \\ &=& \frac{\alpha(t) \chi \pi_\rho(j) }{\alpha(t) \chi +1} = \mathbb{P}\big(\eta_{y+1} (t+1) = j\big) \mathbb{P}\big(K(t, y) = 1\big), \end{array} $$

which concludes Lemma A.4. □

Remark A.5

Since $\vec {g}(t) = \vec {\eta }(Jt)$, it is clear that for all ρ ∈ (0, I), $\bigotimes \pi _{\rho }$ is also stationary for the fused SHS6V model $\vec {g}(t)$.

Appendix B: KPZ Scaling Theory

The KPZ scaling theory has been developed in a landmark contribution by [31]. The scaling theory is a physics approach which makes prediction for the non-universal coefficients of the KPZ equation. In this appendix, we show how the coefficients of the KPZ (??) arise from the microscopic observables of the fused SHS6V model using the KPZ scaling theory.

Recall that Theorem 1.6 reads

$$ \begin{array}{@{}rcl@{}} &&\sqrt{\epsilon} \big(N^{\mathsf{f}}_{\epsilon} (\epsilon^{-2} t, \epsilon^{-1} x + \epsilon^{-2} \mu_{\epsilon} t ) - \rho (\epsilon^{-1} x + \epsilon^{-2} \mu_{\epsilon} t ) - t \log \lambda_{\epsilon} \big)\\ &&\Rightarrow \mathcal{H}(t, x) \text{ in } C([0, \infty), C(\mathbb{R})) \text{ as } \epsilon \downarrow 0. \end{array} $$

Here, $N^{\mathsf {f}}_{\epsilon } (t, x)$ is the fused height function and ${\mathscr{H}}(t, x)$ solves the KPZ equation

$$ \begin{array}{@{}rcl@{}} \partial_{t} \mathcal{H}(t, x) = \frac{\alpha_{1}}{2} {\partial_{x}^{2}} \mathcal{H}(t, x) - \frac{\alpha_{2}}{2} \big(\partial_{x} \mathcal{H}(t, x) \big)^{2} + \sqrt{\alpha_{3}} \xi(t, x), \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} \alpha_{1} &=& \alpha_{2} = JV_{*} = \frac{J\big((I +J)b - (I + J - 2)\big)}{I^{2} (1-b)},\\ \alpha_{3} &=& J D_{*} = \frac{\rho(I-\rho)}{I} \cdot \frac{J \big((I + J) b - (I + J -2)\big)}{I^{2} (1-b)}. \end{array} $$

The first step in the KPZ scaling theory is to derive the stationary distribution of the fused SHS6V model, which is exactly what we did in Appendix A (see Remark A.5). Under stationary distribution $\bigotimes \pi _{\rho }$, we proceed to define two natural quantities of the models:

• The average steady state currentj(ρ) is defined as

  $$ j(\rho) = \epsilon^{-\frac{1}{2}} \big(\big\langle N^{\mathsf{f}}(t, x) - N^{\mathsf{f}}(t, x+1) \big\rangle_{\rho} - \rho \mu_{\epsilon}\big), $$

  (B.1)

  where 〈⋅〉_ρ means that we are taking the expectation under stationary distribution $\bigotimes \pi _{\rho }$ and μ is given in (??). Note that under stationary distribution, the average steady state current j(ρ) depends neither on space or time. Let us explain the meaning of (B.1). Note that N^f(t, x) − N^f(t + 1, x) records the number of particles in the fused SHS6V model that move across location x at time t, we subtract ρμ_𝜖 here because we are in a reference frame that moves to right with speed ρμ_𝜖.

• The integrated covariance is defined as

  $$ \begin{array}{@{}rcl@{}} &A(\rho):= \lim\limits_{r \to \infty} \frac{1}{2 r} \bigg\langle N^{\mathsf{f}} (t, x+r) - N^{\mathsf{f}} (t, x - r) - \big\langle N^{\mathsf{f}} (t, x+r) - N^{\mathsf{f}} (t, x - r) \big\rangle_{\rho} \bigg\rangle_{\!\rho}. \end{array} $$

The KPZ scaling theory (equation (12) and (15) of [31]) predicts that

$$(i)\ \alpha_{2} = -\lim\limits_{\epsilon \downarrow 0} j^{\prime\prime}_{\epsilon} (\rho), \qquad (ii)\ \frac{\alpha_{3}}{\alpha_{1}} = \lim\limits_{\epsilon \downarrow 0} A_{\epsilon} (\rho), $$

A_𝜖(ρ) and j_𝜖(ρ) depend on 𝜖 under weakly asymmetry scaling (??).

Let us first verify (ii), note that under stationary distribution, N𝜖f(t, x + r) − N𝜖f(t, x − r) is the sum of 2r i.i.d. random variables with the same distribution π_ρ, hence A_𝜖(ρ) = Var[π_ρ]. By Lemma A.2, we know that

$$ \text{Var}\big[\pi_\rho \big] = \rho - \sum\limits_{i=1}^{I} \frac{\chi^{2}}{(q^{i} - \chi)^{2}}, $$

where χ is the unique negative solution of

$$ \sum\limits_{i=1}^{I} \frac{\chi}{\chi - q^{i}} = \rho. $$

(B.2)

Under weakly asymmetric scaling, one has $q = e^{\sqrt {\epsilon }}$, which yields $\lim _{\epsilon \downarrow 0} \chi _{\epsilon } = \frac {\rho }{\rho - I}$. Therefore,

$$ \lim\limits_{\epsilon \downarrow 0} A_{\epsilon} (\rho) = \lim\limits_{\epsilon \downarrow 0} \text{Var}\big[\pi_\rho \big] = \frac{\rho(I - \rho)}{I}. $$

This matches with the value of $\frac {\alpha _{3}}{\alpha _{1}}$.

We proceed to verify (i). First, note that by N^f(t, x) = N(Jt, x),

$$ \begin{array}{@{}rcl@{}} N^{\mathsf{f}}(t, x) - N^{\mathsf{f}} (t+1, x) = N(Jt, x) - N((J+1)t, x) = \sum\limits_{s = Jt}^{(J+1)t - 1} K(s, x), \end{array} $$

where K(s, x) = N(s, x) − N(s + 1, x). We have shown in (A.6) that $K(s, x) \sim \text {Ber }(\frac {\alpha (s) \chi }{1 + \alpha (s) \chi })$, where α(s) = αq^{mod (s)}. Therefore,

$$\mathbb{E}\big[N^{\mathsf{f}}(t, x) - N^{\mathsf{f}} (t+1, x)\big] = \mathbb{E}\bigg[\sum\limits_{s = Jt}^{(J+1)t - 1} K (s, x)\bigg] = \sum\limits_{k=0}^{J-1} \frac{\alpha q^{k} \chi}{1 + \alpha q^{k} \chi},$$

which yields

$$ j(\rho) = \epsilon^{-\frac{1}{2}} \bigg(\sum\limits_{k=0}^{J-1} \frac{\alpha q^{k} \chi}{1 + \alpha q^{k} \chi} - \rho \mu\bigg). $$

We proceed to taylor expand j_𝜖(ρ) around 𝜖 = 0. Note that χ is implicitly defined through (B.2), we expand χ_𝜖 around 𝜖 = 0

$$ \chi_{\epsilon} = \frac{\rho}{\rho - I} + \frac{(I+1)\rho}{2(\rho - I)} \sqrt{\epsilon} + \mathcal{O}(\epsilon). $$

Note that α depends on 𝜖 through $\alpha _{\epsilon } = \frac {1 - b}{b - e^{\sqrt {\epsilon }}}$. Via straightforward calculation, one has

$$ \frac{\alpha q^{k} \chi}{1 + \alpha q^{k} \chi} = \frac{\alpha_\epsilon e^{k\sqrt{\epsilon}} \chi_{\epsilon}}{1 + \alpha_\epsilon e^{k \sqrt{\epsilon}} \chi_{\epsilon}} = \frac{\rho}{I} + \frac{(I\rho - \rho^{2})((2k + I + 1)b + 1 - I - 2k)}{2(b-1) I^{2}} \sqrt{\epsilon} + \mathcal{O}(\epsilon), $$

which implies

$$ \sum\limits_{k=0}^{J-1} \frac{\alpha q^{k} \chi}{1 + \alpha q^{k} \chi} = \frac{J\rho}{I} + \frac{J(I\rho - \rho^{2})\big((I + J) b - (I+J-2)\big)}{2(b-1) I^{2}} \sqrt{\epsilon} + \mathcal{O}(\epsilon). $$

Referring to the expression of μ in (??), one has the asymptotic expansion

$$ \begin{array}{@{}rcl@{}} \mu_{\epsilon} =\frac{J}{I} + \frac{J (I - 2\rho) (2 + (b-1)(I+J)) }{2(b -1) I^{2}} \sqrt{\epsilon} + \mathcal{O}(\epsilon). \end{array} $$

Consequently,

$$ \begin{array}{@{}rcl@{}} j_{\epsilon}(\rho) &= \epsilon^{-\frac{1}{2}} \bigg(\sum\limits_{k=0}^{J-1} \frac{\alpha q^{k} \chi}{1 + \alpha q^{k} \chi} - \rho \mu\bigg) = \frac{\rho^{2} J(b(I + J) - (I+J-2)}{2(b -1) I^{2}} + \mathcal{O}(\epsilon^{\frac{1}{2}}). \end{array} $$

We have

$$ \lim\limits_{\epsilon \downarrow 0} -j_{\epsilon}^{\prime\prime} (\rho) = \frac{J (b(I+J) - (I+J-2))}{(1-b) I^{2}}, $$

which coincides with the value of α₂.