Stationary Distributions of the Atlas Model

TL;DR

This paper analyzes the Atlas model with Brownian particles, establishing explicit invariant distributions and gap distributions, and provides bounds on the fluctuation of the lowest-ranked particle.

Contribution

It derives explicit invariant and gap distributions for the Atlas model and proves fluctuation bounds for the lowest-ranked particle.

Findings

Invariant distribution $ u_a$ explicitly characterized

Product-of-exponential gap distribution $ ext{pi}_a$ confirmed

Bound on the fluctuation of the lowest particle established

Abstract

In this article we study the Atlas model, which constitutes of Brownian particles on $ \mathbb{R} $, independent except that the Atlas (i.e., lowest ranked) particle $ X_{(1)}(t) $ receive drift $ \gamma dt $, $ \gamma\in\mathbb{R} $. For any fixed shape parameter $ a>2\gamma_- $, we show that, up to a shift $ \frac{a}{2}t $, the entire particle system has an invariant distribution $ \nu_a $, written in terms an explicit Radon-Nikodym derivative with respect to the Poisson point process of density $ a e^{a\xi} d\xi $. We further show that $ \nu_a $ indeed has the product-of-exponential gap distribution $ \pi_a $ derived in Sarantsev and Tsai (2016). As a simple application, we establish a bound on the fluctuation of the Atlas particle $ X_{(1)}(t) $ uniformly in $ t $, with the gaps initiated from $ \pi_a $ and $ X_{(1)}(0)=0 $.

Equations

$$d{X}_i(t)=\gamma 1\{p_{X(t)}(i)=1\}dt+d{W}_i(t),i\in \mathbb{N}.$$

$$\displaystyle\mathcal{U}=\Big{\{}x=(x_{i})_{i=1}^{\infty}\,:\,\lim_{i\to\infty}x_{i}=\infty,\text{ and }\sum_{i=1}^{\infty}e^{-ax_{i}^{2}}<\infty,\forall a>0\Big{\}},$$

$$\displaystyle\mathcal{V}=\Big{\{}\{x_{i}\}_{i=1}^{\infty}\,:\,(x_{i})_{i=1}^{\infty}\in\mathcal{U}\Big{\}}$$

$${\nu }_a(\text{scalebox}1.8\cdot ):=\frac{1}{\Gamma (\frac{2\gamma }{a}+1)}{\mathbf{E}}_{{\mu }_a}(e^{2\gamma X_{(1)}}\text{scalebox}1.8\cdot )$$

$$Z:=(X_{(i+1)}-X_{(i)}{)}_{i=1}^{\infty }\sim {\pi }_a=\underset{i=1}{\overset{\infty }{⨂}}\mathrm{Exp}(2\gamma +ia).$$

$$d{X}_i(t)=\sum _{j=1}^m{\gamma }_{j}1\{p_{X(t)}(i)=j\}dt+d{W}_i(t),i\in \mathbb{N},$$

$$\mathbf{P}(|X_{(1)}(t)+\frac{at}{2}|\ge \xi )\le c{e}^{-\frac{1}{2}(2\gamma +a)\xi },\forall t,\xi \in {\mathbb{R}}_{+},$$

$${\mathbf{E}}_{{\mu }_a}(e^{2\gamma X_{(1)}}1\{X_{(1)}\le \xi \})={\int }_{-\infty }^{\xi }{e}^{2\gamma \zeta }\frac{d}{d\zeta }(1-e^{-e^{a\zeta }})d\zeta .$$

$$\frac{{\mathbf{E}}_{{\mu }_a}(e^{2\gamma X_{(1)}}{\prod }_{i=1}^{m-1}1\{Z_i\ge {\xi }_i\})}{{\mathbf{E}}_{{\mu }_a}(e^{2\gamma X_{(1)}})}=\prod _{i=1}^{m-1}{e}^{-(2\gamma +ia){\xi }_i}=:\eta .$$

$$\displaystyle(Y_{(k)})_{k=1}^{n}\stackrel{{\scriptstyle\text{d}}}{{=}}\Big{(}\sum_{i=k}^{n}G_{i}\Big{)}_{k=1}^{n},\quad\text{ where }\big{(}G_{i}\big{)}_{i=1}^{n}\sim\prod_{i=1}^{n}\operatorname{Exp}(ia).$$

$$\displaystyle\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}$$

$$\displaystyle=\mathbf{E}\Big{(}e^{2\gamma(N(\xi)\xi-\sum_{i=m}^{N(\xi)}G_{i})}\prod_{i=1}^{m-1}e^{-2\gamma G_{i}}\mathbf{1}{\{G_{i}\geq\xi_{i}\}}\Big{|}N(\xi)\geq m\Big{)}$$

$$\displaystyle=\mathbf{E}\Big{(}e^{2\gamma(N(\xi)\xi-\sum_{i=m}^{N(\xi)}G_{i}})\Big{|}N(\xi)\geq m\Big{)}\prod_{i=1}^{m-1}\frac{iae^{-(ai+2\gamma)\xi_{i}}}{ai+2\gamma}.$$

$$\displaystyle\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}\prod_{i=1}^{m-1}\mathbf{1}{\{Z_{i}\geq\xi_{i}\}}\Big{|}N(\xi)\geq m\Big{)}=\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}\Big{|}N(\xi)\geq m\Big{)}\eta.$$

$$\displaystyle\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}\prod_{i=1}^{m-1}\mathbf{1}{\{Z_{i}\geq\xi_{i}\}}\Big{)}$$

$$\displaystyle=\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}\mathbf{1}{\{N(\xi)\geq m\}}\Big{)}\eta+\mathbf{E}_{\mu_{a}}\Big{(}e^{2\gamma X_{(1)}}\prod_{i=1}^{m-1}\mathbf{1}{\{Z_{i}\geq\xi_{i}\}})\mathbf{1}{\{N(\xi)<m\}}\Big{)}.$$

$${\mathbf{E}}_{{\mu }_a}(e^{2q\gamma X_{(1)}})={\mathbf{E}}_{{\mu }_a}(|e^{2\gamma X_{(1)}}{|}^q)=\Gamma (\frac{2q\gamma }{a}+1)<\infty .$$

$$\displaystyle\mathbf{E}_{\mu_{a}}\big{(}e^{2\gamma\overline{X}_{(1)}(0)}\phi(\overline{X}_{(1)}(t),\ldots,\overline{X}_{(m)}(t))\big{)}=\mathbf{E}_{\mu_{a}}\big{(}e^{2\gamma\overline{X}_{(1)}(0)}\phi(\overline{X}_{(1)}(0),\ldots,\overline{X}_{(m)}(0))\big{)}.$$

$${\varphi }^{\text{s}}:{\mathbb{R}}^n\to \mathbb{R},{\varphi }^{\text{s}}(x):=\varphi (x_{(1)},\dots ,x_{(m)}).$$

$$\displaystyle\overline{X}^{n}_{i}(t)=x_{i}+\int_{0}^{t}\big{(}\gamma\mathbf{1}{\{p_{\overline{X}^{n}(s)}(i)=1\}}+\tfrac{a}{2}\big{)}ds+W_{i}(t),\quad i=1,\ldots,n,$$

$$\frac{1}{2}{\partial }_i{V}^{\epsilon }(x)=\gamma 1\{p_{{\overline{X}}^n(s)}(i)=1\},\text{ on }{\Omega }_{\epsilon }:=\{x\in {\mathbb{R}}^n:|x_i-x_j|>\epsilon ,\forall i<j\}.$$

$$\displaystyle\overline{X}^{n,\varepsilon}_{i}(t)=x_{i}+\int_{0}^{t}\big{(}\tfrac{1}{2}\partial_{i}V^{\varepsilon}(\overline{X}^{n,\varepsilon}(s))+\tfrac{a}{2}\big{)}ds+W_{i}(t),\quad i=1,\ldots,n.$$

$$\displaystyle\partial_{t}u=\sum_{i=1}^{n}\big{(}\tfrac{1}{2}\partial_{ii}+\tfrac{a}{2}\partial_{i}+\tfrac{1}{2}\partial_{i}V^{\varepsilon}\big{)}u,\quad u(0,x)=\phi^{\text{s}}(x).$$

$$\displaystyle\sup_{s\leq t,x\in\mathbb{R}^{n}}\big{\{}|u(t,x)|e^{\xi(|x_{1}|+\ldots+|x_{n}|)}\big{\}}<\infty,\quad\forall\xi,t<\infty.$$

$$\displaystyle\partial_{t}(\partial_{i}u)=\sum_{j=1}^{n}\big{(}\big{(}\tfrac{1}{2}\partial_{jj}+\tfrac{a}{2}\partial_{j}+\partial_{j}V^{\varepsilon}\big{)}$$

$${\partial }_{i}u(0,x)={\partial }_i{\varphi }^{\text{s}}(x)\in C_c^{\infty }({\mathbb{R}}^n),$$

$$\displaystyle\sup_{s\leq t,x\in\mathbb{R}^{n},i=1,\ldots,n}\big{\{}|\partial_{i}u(t,x)|e^{\xi(|x_{1}|+\ldots+|x_{n}|)}\big{\}}<\infty,\quad\forall\xi,t<\infty.$$

$$\displaystyle\sup_{s\leq t,x\in\mathbb{R}^{n},|\beta|\leq k}\big{\{}|\partial_{\beta}u(t,x)|e^{\xi(|x_{1}|+\ldots+|x_{n}|)}\big{\}}<\infty,\quad\forall\xi,t,k<\infty.$$

$$v(t):={\int }_{{\mathbb{R}}^n}u(t,x)e^{V^{\epsilon }(x)}\prod _{i=1}^n{e}^{a{x}_i}d{x}_i.$$

$${\int }_{{\mathbb{R}}^n}\frac{1}{2}({\partial }_{ii}u(t,x))e^{V^{\epsilon }(x)}\prod _{j=1}^n{e}^{a{x}_j}d{x}_j=-{\int }_{{\mathbb{R}}^n}({\partial }_{i}u(t,x))(\frac{1}{2}{\partial }_i{V}^{\epsilon }(x)+\frac{a}{2})e^{V^{\epsilon }(x)}\prod _{j=1}^n{e}^{a{x}_j}d{x}_j,$$

$$\displaystyle\int_{\mathbb{R}^{n}}\mathbf{E}_{x}\big{(}\phi^{\text{s}}(\overline{X}^{n,\varepsilon}(t))\big{)}e^{V^{\varepsilon}(x)}\prod_{i=1}^{n}e^{ax_{i}}dx_{i}=\int_{\mathbb{R}^{n}}\phi^{\text{s}}(x)e^{V^{\varepsilon}(x)}\prod_{i=1}^{n}e^{ax_{i}}dx_{i}.$$

$$\displaystyle\mathbf{E}_{x}\big{(}\phi^{\text{s}}(\overline{X}^{n}(t))\big{)}$$

Full text

1. Introduction

In this article we study the (infinite) Atlas model. Such a model consists of a semi-infinite collection of particles $X_{i}(t)$, $i=1,2,\ldots$, performing independent Brownian motions on $\mathbb{R}$, except that the Atlas (i.e., lowest ranked) particle receives a drift of strength $\gamma\in\mathbb{R}$. To rigorously define the model, we recall that $x=(x_{i})_{i=1}^{\infty}\in\mathbb{R}^{\mathbb{N}}$ is rankable if there exists a ranking permutation $p:\mathbb{N}\to\mathbb{N}$ such that $x_{p(i)}\leq x_{p(j)}$, for all $i<j\in\mathbb{N}$. To ensure that such a ranking permutation is unique, we resolve ties in lexicographic order. That is, if $x_{p(i)}=x_{p(j)}$ for $i<j$, then $p(i)<p(j)$. We then let p_{x}({\raisebox{-2.15277pt}{\scalebox{1.8}{\cdot}}}):\mathbb{N}\to\mathbb{N} denote the unique ranking permutation for a given, rankable $x$. Fix independent standard Brownian motions $W_{1},W_{2},\ldots$. For suitable initial conditions, the infinite Atlas model $X(t)=(X_{i}(t))_{i=1}^{\infty}$ is given by the unique weak solution of the following system of Stochastic Differential Equations

[TABLE]

To state the well-posedness results of (1.1), consider the following configuration space

[TABLE]

and note that $\lim_{i\to\infty}x_{i}=\infty$ necessarily implies that $x$ is rankable. It is shown in [Sar17a, Theorem 3.2], for any fixed $\gamma\in\mathbb{R}$ and any given $x\in\mathcal{U}$, the system (1.1) admits a unique weak solution $X(t)$ starting from the initial condition $x$, such that $\mathbf{P}(X(t)\in\mathcal{U},\ \forall t\geq 0)=1$. See also [Shk11, IKS13].

The interest of the Atlas model originates from the study of diffusions with rank-based drifts [Fer02, FK09]. In particular, the Atlas model was first introduced, in finite dimensions, as a simple special case of rank-based diffusions [Fer02]. Due to their intriguing properties, rank-based diffusions have been intensively studied in various generality. See [BFK05, BFI*+*11, IKS13, Sar17b] and the references therein. The infinite-dimensional system (1.1) considered here was introduced by Pal and Pitman [PP08]. Parts of the motivation was to understand the effect of a drift exerted on a large (but finite) collection of Brownian particles [Ald02, TT15]. In particular, it was shown in [PP08] that, for $\gamma>0$, the system (1.1) admits a stationary gap distribution of i.i.d. $\operatorname{Exp}(2\gamma)$, which indicates that the drift $\gamma dt$ is balanced by the push-back of a crowd of particles of density $2\gamma$. To state the previous result more precisely, given a rankable $x=(x_{i})_{i=1}^{\infty}$, we let $(x_{(1)}\leq x_{(2)}\leq\ldots)$ denote the corresponding ranked points, i.e., $x_{(i)}=x_{(p_{x})^{-1}(i)}$, and consider the corresponding gaps $z_{i}:=x_{(i+1)}-x_{(i)}$. It was shown in [PP08] that $\pi:=\bigotimes_{i=1}^{\infty}\operatorname{Exp}(2\gamma)$ is a stationary distribution of the gap process $Z(t):=(X_{(i+1)}(t)-X_{(i)}(t))_{i=1}^{\infty}$ of the Atlas model (1.1).

It addition to the i.i.d. $\operatorname{Exp}(2\gamma)$ distribution, it was recently shown in [ST17] that the Atlas model has a different type of stationary gap distributions. That is, for each $a>2\gamma_{-}$, $\pi_{a}:=\bigotimes_{i=1}^{\infty}\operatorname{Exp}(2\gamma+ia)$ is also a stationary gap distribution of the Atlas model. Unlike $\pi$, the distribution $\pi_{a}$ has exponentially growing particle density away from the Atlas particle. In this article, we go one step further and show that, in fact, up to a deterministic shift $\frac{at}{2}$ of each particle, the entire particle system $\{X_{i}(t)+\frac{at}{2}\}_{i=1}^{\infty}$ has a stationary distribution. This extends the result of [ST17] on stationary gap distributions. In the following we use $\{x_{i}\}_{i=1}^{\infty}\subset\mathbb{R}$ to denote a configuration of indistinguishable particles, in contrast with $(x_{i})_{i=1}^{\infty}$, which denotes labeled (named) particles. Let

[TABLE]

denote the corresponding configuration space, and let $\mu_{a}$ denote the Poisson point process on $\mathbb{R}$ with density $ae^{a\xi}d\xi$. It is standard to show (e.g., using techniques from [Pan13, Section 2.2]) that $\mu_{a}$ is supported on $\mathcal{V}$. Let $\Gamma(\alpha):=\int_{0}^{\infty}\xi^{-1-\alpha}e^{-\xi}d\xi$ denote the Gamma function, and let $\operatorname{Gamma}(\alpha,\beta)\sim\frac{1}{\Gamma(\alpha)}\beta^{\alpha}\xi^{-1-\alpha}e^{-\beta\xi}\mathbf{1}_{\{\xi>0\}}d\xi$ denote the Gamma distribution. The following is the main result.

Theorem 1.1**.**

2. (a)

For any fixed $\gamma\in\mathbb{R}$ and $a>2\gamma_{-}$, $\mathbf{E}_{\mu_{a}}(e^{2\gamma X_{(1)}})=\Gamma(\frac{2\gamma}{a}+1)\in(0,\infty),$ so that

[TABLE]

defines a probability distribution supported on $\mathcal{V}$. Furthermore, under $\nu_{a}$, we have that $e^{aX_{(1)}}\sim\operatorname{Gamma}(\frac{2\gamma}{a}+1,1)$, and that

[TABLE] 3. (b)

The distribution $\nu_{a}$ is a stationary distribution of $\{X_{i}(t)+\frac{at}{2}\}_{i=1}^{\infty}$, where $(X_{i}(t))_{i=1}^{\infty}$ evolves under (1.1).

Remark 1.2**.**

Under $\nu_{a}$, the Atlas particle $X_{(1)}$ and the gap process $Z=(Z_{i})_{i=1}^{\infty}$ are not independent.

For the special case $\gamma=0$, the Atlas model (1.1) reduces to independent Brownian motions. In this case, it is well known that the Poisson point process $\mu_{a}$ is quasi-stationary [Lig78], and the shift $-\frac{a}{2}t$ can be easily calculated from the motion of independent Brownian particles. Here we show that, with a drift $\gamma dt$ exerted on the Atlas particle $X_{(1)}(t)$, a stationary distribution is obtained by taking $V(x):=2\gamma x_{(1)}$ to be the potential. Indeed, under such a choice of $V(x)$, we have that $\gamma\mathbf{1}{\{p_{x}(i)=1\}}=\tfrac{1}{2}\partial_{x_{i}}e^{V(x)}$. This explains why we should expect the stationary distribution $\nu_{a}$ as in (1.3). The proof of Theorem 1.1 amounts to justifying the aforementioned heuristic in the setting of infinite dimensional diffusions with discontinuous drift coefficients. We achieve this through finite-dimensional, smooth approximations, and using the explicit expressions of semigroups from Girsanov’s theorem to take limits.

Due to their simplicity, product-of-exponential stationary gap distributions have been intensively searched within competing Brownian particle systems, in both finite and infinite dimensions. See [Sar17a] and the references therein. To date, derivations of product-of-exponential stationary gap distributions have been relying on the theory of Semimartingale Reflecting Brownian Motions (SRBM), e.g., [Wil95]. On the other hand, given the expression (1.3) of $\nu_{a}$, the gap distribution (1.4) follows straightforwardly from Rényi’s representation [Rén53]. Theorem 1.1 hence provides an alternative derivation of the product-of-exponential distribution $\pi_{a}$ without going through SRBM.

Our methods should generalize to the case of competing Brownian particle systems with finitely many non-zero drift coefficients, i.e.,

[TABLE]

yielding the stationary distribution \nu_{a}({\raisebox{-2.15277pt}{\scalebox{1.8}{\cdot}}}):=\frac{1}{J}\mathbf{E}_{\mu_{a}}(e^{2\sum_{j=1}^{m}\gamma_{j}X_{(j)}}{\raisebox{-2.15277pt}{\scalebox{1.8}{\cdot}}}), for some normalizing constant $J<\infty$. Here we consider only the Atlas model for simplicity of notations.

A natural question, following the discovery a stationary gap distribution, is the longtime behavior of the Atlas particle $X_{(1)}(t)$ under such a gap distribution. For the i.i.d. $\operatorname{Exp}(2)$ gap distribution $\pi$, this question was raised in [PP08] and answered in [DT17]. It was shown in [DT17] that $X_{(1)}(t)$ fluctuates at order $t^{\frac{1}{4}}$ around its starting location, and scales to a $\frac{1}{4}$-fractional Brownian motion, as $t\to\infty$. As a simple application of Theorem 1.1, under the stationary gap distribution $\pi_{a}$ and $X_{(1)}(0)=0$, we establish an exponential tail bound, uniformly in $t$, of the fluctuation Atlas particle around its expected location $-\frac{at}{2}$. This shows that the fluctuation of $X_{(1)}(t)$ stays bounded under $\pi_{a}$, in sharp contrast with the $t^{\frac{1}{4}}$ fluctuation obtained in [DT17].

Corollary 1.3**.**

Fix $\gamma\in\mathbb{R}$ and $a>2\gamma_{-}$. Starting the Atlas model (1.1) from the initial distribution $X_{(i)}(0)=0$, $(X_{(i+1)}(0)-X_{(i)}(0))_{i=1}^{\infty}\sim\pi_{a}$, we have that

[TABLE]

for some constant $c=c(a,\gamma)<\infty$ depending only on $a,\gamma$.

Acknowledgements

I thank Amir Dembo, Ioannis Karatzas, Andrey Sarantsev and Ofer Zeitouni for enlightening discussions. I am grateful to Andrey Sarantsev for many useful comments on the presentation of this article. This work was partially supported by a Junior Fellow award from the Simons Foundation to Li-Cheng Tsai.

2. Proof

2.1. Theorem 1.1(a)

Fix $\gamma\in\mathbb{R}$, $a>2\gamma_{-}$, and let $\{X_{i}\}_{i=1}^{\infty}$ denote a sample from the Poisson point process $\mu_{a}$. Let $N(\xi):=\#\{X_{i}\in(-\infty,\xi]\}$ denote the number of particles in $(-\infty,\xi]$, whereby $N(\xi)\sim\operatorname{Pois}(e^{a\xi})$. Indeed, $\mathbf{P}_{\mu_{a}}(X_{(1)}>\xi)=\mathbf{P}(N(\xi)=0)=e^{-e^{a\xi}}$. From this we calculate

[TABLE]

Performing the change of variable $\zeta^{\prime}:=e^{a\zeta}$, we see that

$\mathbf{E}_{\mu_{a}}(e^{2\gamma X_{(1)}}\mathbf{1}{\{e^{aX_{(1)}}\leq e^{\xi}\}})=\int_{0}^{e^{\xi}}{\zeta^{\prime}}^{\frac{2\gamma}{a}}e^{-\zeta^{\prime}}d\zeta^{\prime}.$ From this it follows that $\mathbf{E}_{\mu_{a}}(e^{2\gamma X_{(1)}})=\Gamma(\frac{2\gamma}{a}+1)$ and that $e^{aX_{(1)}}\sim\operatorname{Gamma}(\frac{2\gamma}{a}+1,1)$ under $\nu_{a}$.

Turning to showing (1.4), we let $\{X_{i}\}_{i=1}^{\infty}$ be sampled from $\mu_{a}$ and let $(Z_{k})_{k=1}^{\infty}=(X_{(i+1)}-X_{(i)})_{i=1}^{\infty}$ denote the gap process. Fix arbitrary $m<\infty$. Our goal is to show that

[TABLE]

For any given threshold $\xi\in\mathbb{R}$, we let $\mu_{a,\xi}$ denote the restriction of the Poisson point process $\mu_{a}$ onto $(-\infty,\xi]$. For the restricted process, we have $\mu_{a,\xi}\sim\{\xi-Y_{1},\ldots,\xi-Y_{N(\zeta)}\}$, where $Y_{1},Y_{2},\ldots$ are i.i.d. $\operatorname{Exp}(a)$ variables, independent of $N(\xi)$. Let $Y_{(1)}<Y_{(2)}<\ldots<Y_{(n)}$ denote the ranking of $(Y_{1},\ldots,Y_{n})$. We then have that, conditionally on $N(\xi)\geq m$, $(X_{(1)},\ldots,X_{(m)})=(\xi-Y_{N(\xi)},\ldots,\xi-Y_{N(\xi)-m+1})$. Further, by Rényi’s representation [Rén53],

[TABLE]

Using this we calculate

[TABLE]

Further use $\frac{ia}{ai+2\gamma}=\mathbf{E}e^{-2\gamma G_{i}}$ to write $\prod_{i=1}^{m-1}\frac{iae^{-(ai+2\gamma)\xi_{i}}}{ai+2\gamma}=\mathbf{E}(e^{-2\gamma(G_{1}+\ldots+G_{m-1})})\eta$, We then obtain

[TABLE]

Taking into account the case $N(\xi)<m$, we write

[TABLE]

Since $a>-2\gamma$, fixing $q>1$ with $|q-1|$ small enough, we have

[TABLE]

That is, $e^{2q\gamma X_{(1)}}$ has bounded $q$-th moment with $q>1$, so in particular $\{e^{2\gamma X_{(1)}}\mathbf{1}{\{N(\xi)\geq m\}}\}_{\xi>0}$ is uniformly integrable. For fixed $m<\infty$, $\mathbf{1}{\{N(\xi)<m\}}\to_{\text{P}}0$, as $\xi\to\infty$. Using this to take the limit $\xi\to\infty$ in (2.2), we thus obtain $\mathbf{E}_{\mu_{a}}(e^{2\gamma X_{1}}\prod_{i=1}^{m-1}\mathbf{1}{\{Z_{i}\geq\xi_{i}\}})=\mathbf{E}_{\mu_{a}}(e^{2\gamma X_{1}})\eta.$ This concludes (2.1).

2.2. Theorem 1.1(b)

Samples from $\mu_{a}$ have, almost surely, no repeated points, i.e., $X_{(1)}<X_{(2)}<X_{(3)}<\ldots$. Fix arbitrary $m<\infty$ and $\phi\in C^{\infty}_{c}(\mathcal{W})$, where $\mathcal{W}:=\{(x_{1}<x_{2}<\ldots<x_{m})\}$ denote the Weyl chamber. Let $\overline{X}_{i}(t):=X_{i}(t)+\frac{a}{2}t$, and $\overline{X}_{(i)}(t):=X_{(i)}(t)+\frac{a}{2}t$ denote the compensated particle locations. It then suffices to show that

[TABLE]

As will be convenient for notations, for $n\geq m$, we consider the symmetric extension $\phi^{\text{s}}$ of $\phi$, defined for $n\geq m$ as

[TABLE]

We have slightly abused notations by using the same symbol $\phi^{\text{s}}$ to denote the function for all $n\in\mathbb{N}_{\geq m}\cup\{\infty\}$. Note that, by definition, the function $\phi$ vanishes near the boundary $\{(x_{1}\leq\ldots x_{i}=x_{i+1}\ldots\leq x_{m}):i=1,\ldots,n-1\}$ of $\mathcal{W}$, so, for $n<\infty$, $\phi^{\text{s}}\in C_{\text{c}}^{\infty}(\mathbb{R}^{n})$.

The strategy of proving (2.4) is to approximate the infinite system $\overline{X}(t)$ by finite systems. Fixing $m\leq n<\infty$, we consider the following $n$-dimensional analog of $\overline{X}(t)$:

[TABLE]

where the ranking permutation p_{x}({\raisebox{-2.15277pt}{\scalebox{1.8}{\cdot}}}):\{1,\ldots,n\}\to\{1,\ldots,n\} is defined similarly to the case of infinite particles. As the discontinuity of $x\mapsto\mathbf{1}{\{p_{x}(i)=1\}}$ imposes unwanted complication in the subsequence analysis, we consider further the mollified system as follows. Fix a mollifier $r\in C^{\infty}(\mathbb{R}^{n})$, i.e., $r\geq 0$, $r|_{\|x\|\geq 1}=0$ and $\int_{\mathbb{R}^{n}}r(y)dy=1$. Let $V(x):=2\gamma x_{(1)}=2\gamma\min(x_{1},\ldots,x_{n})$. For $\varepsilon\in(0,1)$, we define the mollified potential as $V^{\varepsilon}(x):=\int_{\mathbb{R}^{n}}V(y)r(\varepsilon^{-1}(x-y))\varepsilon^{-n}dy$. Under these notations, we have that

[TABLE]

We then consider the following mollified system

[TABLE]

With $\partial_{i}V^{\varepsilon}$ being smooth and bounded, the well-posedness of (2.8) follows from standard theory, e.g., [SV07]. Furthermore, letting $u(t,x):=\mathbf{E}_{x}(\phi^{\text{s}}(\overline{X}^{n,\varepsilon}(t)))$, we have that $u\in C^{\infty}(\mathbb{R}_{+}\times\mathbb{R}^{n})$, and that $u$ solves the following PDE:

[TABLE]

With $\partial_{i}V^{\varepsilon}$ being bounded and $\phi^{\text{s}}$ being compactly supported, applying the Feynman-Kac formula to the solution $u$ of (2.9), we see that $u$ decays exponentially as $|x|\to\infty$, i.e.,

[TABLE]

Such an exponential estimate (2.10) progresses to higher order derivatives of $u$. More precisely, with $\partial_{i}V^{\varepsilon}\in C^{\infty}(\mathbb{R}^{n})$ and $u\in C^{\infty}(\mathbb{R}_{+}\times\mathbb{R}^{n})$, taking derivative $\partial_{i}$ in (2.9), we see that $\partial_{i}u$ solves the following equation:

[TABLE]

A similarly procedure applied to the solution $\partial_{i}u$ of (2.11) yields

[TABLE]

Iterating this argument to higher order derivatives, we obtain

[TABLE]

The PDE (2.9) has stationary distribution $e^{V^{\varepsilon}(x)}\prod_{i=1}^{n}e^{ax_{i}}dx_{i}$ (not a probability distribution, since the total mass is infinite). More precisely, integrate $u(t,x)$ against the aforementioned distribution to get

[TABLE]

Taking time derivative using (2.9) and (2.12), followed by integrations by parts

[TABLE]

we obtain that $\frac{d~{}}{dt}v(t)=0$. Consequently,

[TABLE]

The next step is to take the limit $\varepsilon\to 0$ in (2.13), for fixed $n$. This amounts to establishing the convergence of the term \mathbf{E}_{x}\big{(}\phi^{\text{s}}(\overline{X}^{n,\varepsilon}(t))\big{)}. To this end, we use Girsanov’s theorem to write

[TABLE]

where $H(t):=(W_{i}(t)+\frac{at}{2}+x_{i})_{i=1}^{n}$ consists of independent, drifted Brownian motions starting from $x=(x_{i})_{i=1}^{n}$, and the terms $F(t)$ and $F^{\varepsilon}(t)$ are stochastic exponentials given by

[TABLE]

Taking the difference of (2.14)–(2.15), followed by using the Cauchy–Schwarz inequality, we obtain

[TABLE]

For the two terms in (2.18), we next show that: i) the first term is bounded; and ii) the second term vanishes as $\varepsilon\to 0$. Hereafter, we use $c(a_{1},a_{2},\ldots)$ to denote a finite, deterministic constant, that may change from line to line, but depends only on the designated variables $a_{1},a_{2},\ldots$.

i) Recall that $\phi^{\text{s}}$ is defined in terms of $\phi$ through (2.5). We fix $\lambda<\infty$, independently of $n$, such that $\operatorname{supp}(\phi^{\text{s}})\subset[-\lambda,\lambda]^{n}$. Under these notations,

[TABLE]

With $F(t)$ defined in (2.16), and $\langle M\rangle(t)=\gamma^{2}t$, it follows that

[TABLE]

Let $\Phi(x):=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^{2}}{2}}dy$ denote the Gaussian distribution function. With $H_{i}(t)=x_{i}+\frac{a}{2}t+W_{i}(t)$, we have

[TABLE]

Inserting these bounds into (2.19), we obtain

[TABLE]

ii) Expand the expression $\mathbf{E}_{x}(1-\tfrac{F^{\varepsilon}(t)}{F(t)})^{2}$ into

[TABLE]

From (2.16)–(2.17), we have

[TABLE]

Set $b^{\varepsilon}_{i}(s):=\frac{1}{2}\partial_{i}V^{\varepsilon}(H(s))$ to simplify notations. As $V(x)$ is Lipschitz with Lipschitz seminorm $2|\gamma|$, (i.e., $|V(x)-V(y)|\leq 2\gamma|x-y|$, $\forall x,y\in\mathbb{R}^{n}$), we have $|b^{\varepsilon}_{i}(s)|\leq|\gamma|$. Consequently,

[TABLE]

To estimate the expression (2.23), we use (2.7) and $|b^{\varepsilon}_{i}(s)|\leq|\gamma|$ to write

[TABLE]

Let $L_{i,j}(s,\xi)$ denote the localtime process of $H_{i}(s)-H_{j}(s)=W_{i}(s)-W_{j}(s)+(x_{j}-x_{i})$ at a given level $\xi$. We further bound the r.h.s. of (2.25) as

[TABLE]

Consequently, $|\langle M-M^{\varepsilon}\rangle(t)|\to_{\text{P}}0$. Since, by (2.24), $\langle M\rangle(t)$ and $\langle M^{\varepsilon}\rangle(t)$ are bounded (for fixed $t$), it also follows that $\mathbf{E}_{x}|\langle M-M^{\varepsilon}\rangle(t)|\to 0$ and hence $M(t)-M^{\varepsilon}(t)\to_{\text{P}}0$. Referring back to the expression (2.23), we see that $\frac{F^{\varepsilon}(t)}{F(t)}\to_{\text{P}}1$. Using again the fact that $\langle M\rangle(t)$ and $\langle M^{\varepsilon}\rangle(t)$ are bounded, (which implies the uniform integrability of $(\frac{F^{\varepsilon}(t)}{F(t)})^{k}$, $k=1,2$), we obtain $\mathbf{E}_{x}(\frac{F^{\varepsilon}(t)}{F}),\mathbf{E}_{x}(\frac{F^{\varepsilon}(t)}{F})^{2}\to 1$. Inserting these into (2.22) yields

[TABLE]

Now, combine (2.21), (2.26) with (2.18), and insert the result into the l.h.s. of (2.13). After taking the $\varepsilon\to 0$ limit with $n<\infty$ being fixed, we obtain

[TABLE]

Recall that $\mu_{a,\zeta}$ denote the restriction of the Poisson point process $\mu_{a}$ on $(-\infty,\zeta]$ and that $N(\zeta)$ denote the number of particles on $(-\infty,\zeta]$. As mentioned previously, $\mu_{a,\zeta}\sim\{\zeta-Y_{1},\ldots,\zeta-Y_{N(\zeta)}\}$, where $Y_{1},Y_{2},\ldots$ are i.i.d. $\operatorname{Exp}(a)$ variables, independent of $N(\zeta)$. Conditionally on $N(\zeta)=n$, the process $\{\zeta-Y_{1},\ldots,\zeta-Y_{N(\zeta)}\}$ have joint distribution $\prod_{i=1}^{n}ae^{a(x_{i}-\zeta)}dx_{i}\mathbf{1}_{\{x_{i}\leq\zeta\}}$. With this, multiplying both sides of (2.27) by $a^{n}e^{-an\zeta}$, and averaging over $\{N(\zeta)\geq m\}$, we obtain that

[TABLE]

where the terms $R_{n}(\zeta)$ and $S_{n}(\zeta)$ are given by

[TABLE]

Recall that $\operatorname{supp}(\phi^{\text{s}})\subset[-\lambda,\lambda]^{n}$. Hence

[TABLE]

As for $R_{n}(\zeta)$, inserting the bound (2.20) into (2.29) gives

[TABLE]

Indeed, $x_{(1)}\leq\zeta+\sum_{i=1}^{n}(x_{i}-\zeta)_{+}$, so, after a change of variable $x_{i}-\zeta\mapsto x_{i}$, we obtain

[TABLE]

To bound the last integral, we split the integration over $x_{i}$ into $\{x_{i}>0\}$ and $\{x_{i}\leq 0\}$ for each $x_{i}$, and thereby express the integral as

[TABLE]

where $\{i_{1},\ldots,i_{k}\}$ ranges over all distinct $k$-indices from $\{1,\ldots,n\}$. Further, for each integral over $\{x>0\}$ and over $\{x\leq 0\}$, we have that

[TABLE]

Consequently,

[TABLE]

Now, with $N(\zeta)\sim\operatorname{Pois}(e^{a\zeta})$, we have $\mathbf{E}(\binom{N(\zeta)}{k})=\frac{1}{k!}\mathbf{E}(N(\zeta)\cdots(N(\zeta)-k+1))=\frac{1}{k!}e^{ka\zeta}$. Given this identity, setting $n=N(\zeta)$ and taking expected value, we obtain

[TABLE]

which converges to zero as $\zeta\to\infty$.

Using (2.30)–(2.31) in (2.28), and taking the limit $\zeta\to\infty$, we arrive at

[TABLE]

It remains to show that, under the limit $\zeta\to\infty$, we can exchange the finite system $\overline{X}^{N(\zeta)}$ for the infinite system $\overline{X}$ within the expressions in (2.32). As $\zeta\to\infty$, we have that

[TABLE]

where $\overline{X}^{N(\zeta)}(0)\sim\mu_{a,\zeta}$ and $\overline{X}(0)\sim\mu_{a}$. Such a statement (2.33) can be proven by techniques from [Sar17a] and [ST17, Section 3(a)]. We omit repeating the standard arguments here. Combining (2.33) and (2.3), we obtain that

[TABLE]

Combining (2.34)–(2.35) with (2.32), we thus obtain (2.4), and hence complete the proof.

2.3. Corollary 1.3

Fixing $\gamma\in\mathbb{R}$ and $a>2\gamma_{-}$, we let $c=c(a,\gamma)<\infty$ denote a generic finite constant that depends only on these two variables. Let $Y(t)=(Y_{i}(t))_{i=1}^{\infty}$ be a solution to (1.1) starting from the distribution $\{Y_{i}(0)\}_{i=1}^{\infty}\sim\nu_{a}$, so that $\{Y_{i}(t)+\frac{at}{2}\}_{i=1}^{\infty}\sim\nu_{a}$, for all $t\in\mathbb{R}_{+}$. Since, by (1.4), the gap process $(Y_{(i+1)}(0)-Y_{(i)}(0))_{i=1}^{\infty}$ is distributed as $\pi_{a}$, setting $X_{i}(t)=Y_{i}(t)-Y_{(1)}(0)$, we have that $X(t)$ is a solution to (1.1) with the designated initial distribution as in Corollary 1.3. Under these notations, for any given $\xi\geq 0$,

[TABLE]

With $e^{aY_{(1)}}\sim\operatorname{Gamma}(\frac{2\gamma}{a},1)$, we have that

[TABLE]

Combining these bounds with (2.36) yields the desired result.