Limit of the extremal process of i.i.d. Gaussians see from the tip I'm trying to calculate the weak limit of $\mathcal{E}_N(x)=\sum_{k=1}^{2^N}\delta_{x-Z_k}$ , with $Z_k=X_k-\max_{k\leq 2^N}X_k$, $\{X_k\}$ being $2^N$ copies of i.i.d. Gaussians with mean zero and variance $N$(so it can be seen as discretization of Brownian motion).
I'm considering the Laplace transform, since it works well when $\max_{k\leq 2^N}X_k$ is replaced by the typical extrema $m_N=\sqrt{2\log 2}N-\frac{1}{2\sqrt{2\log 2}}\log N$ in the definition of $Z_k$. Which is to say, I want to calculate the limit of the following for every test function $\phi\in C_0^+(\mathbb{R})$: $$\mathbb{E}\exp\left(-\int\phi(x)\mathcal{E}_N(x)dx\right)=\mathbb{E}\exp\left(-\sum_{k=1}^{2^N}\phi(X_k-\max X_k)\right).$$
The main difficulty is that these $\{X_k-\max X_k\}$ are now not independent, while the previous ones $\{X_k-m_N\}$ are independent. I don't know how to deal with this, even if I use some simpler special functions eg. $\phi(x)=1(x\geq-A)$ to simplify the transform to $$\mathbb{E}\exp\left(-\#\{X_k:X_k\geq\max X_k-A\}\right).$$
How can I deal with the dependency here?
 A: $\newcommand\R{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}$For $\phi=c\,1_{[-A,\infty)}$ with $c\ge0$ and $A\ge0$, the expectation in question converges to
\begin{equation*}
    \frac1{1+(e^c-1)e^{A\sqrt{\ln4}}}. \tag{1}\label{1}
\end{equation*}

Indeed, let $n:=2^N\to\infty$. Let
\begin{equation*}
    h:=\phi=c\,1_{[-A,\infty)} \tag{2}\label{2}
\end{equation*}
for some real  with $c\ge0$ and $A\ge0$.
Let $V_i:=X_i-\max_{1\le j\le n}X_j$. Let $g\colon\R^n\to\R$ be
any nonnegative Borel-measurable function that is symmetric (with respect to any permutation of its arguments). Let $f$ denote the pdf of each $X_j$, so that
\begin{equation*}
    f(x)=\frac1s\,\vpi\Big(\frac xs\Big)
\end{equation*}
for all real $x$, where $\vpi$ is the standard normal pdf and
\begin{equation*}
    s:=\sqrt N=\sqrt{\log_2 n}=\sqrt{\frac{\ln n}{\ln2}}. \tag{3}\label{3}
\end{equation*}
Then
\begin{equation*}
\begin{aligned}
    &Eg(V_1,\dots,V_n) \\ 
    &=n\,Eg(X_1-X_1,X_2-X_1,\dots,X_n-X_1)\,1(X_1>\max_{2\le j\le n}X_j) \\ 
    &=n\,\int_{\R^n}g(0,x_2-x_1,\dots,x_n-x_1)\,1(x_1>\max_{2\le j\le n}x_j)
    \prod_{j=1}^n f(x_j)dx_j\, \\ 
    &=n\,\int_\R f(x_1)\,dx_1\,\int_{(-\infty,0)^n}g(0,v_2,\dots,v_n)\,
    \prod_{j=2}^n f(x_1+v_j)dv_j.  
\end{aligned}
\end{equation*}
Then the expectation in question is
\begin{equation*}
    L:=L_n:=ne^{-c}I, \tag{3.5}\label{3.5}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
    I:=I_n&:=\int_\R f(x_1)\,dx_1\,\Big(\int_{-\infty}^0 e^{-h(v)} f(x_1+v)\,dv\Big)^{n-1} \\ 
    &=\int_\R dz\,\vpi(z)\,H(z)^{n-1}, 
\end{aligned}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
    H(z)&:=\int_{-\infty}^z e^{-h(s(w-z))}\vpi(w)\,dw \\ 
    &=e^{-c}\Phi(z)+(1-e^{-c})\Phi(z-A/s), 
\end{aligned}
\end{equation*}
where $\Phi$ is the standard normal cdf (and $h$ and $s$ are as defined in \eqref{2} and \eqref{3}).
Let
\begin{equation*}
    z_\ep:=\sqrt{(2-\ep)\ln n},
\end{equation*}
where $\ep\in(0,2)$.
Note that
\begin{equation*}
    I=J_1+J_2+J_3, \tag{4}\label{4}
\end{equation*}
where
\begin{equation*}
    J_1:=\int_{-\infty}^{z_\ep} dz\,\vpi(z)\,H(z)^{n-1},\quad 
    J_2:=\int_{z_\ep}^{z_0} dz\,\vpi(z)\,H(z)^{n-1},\quad 
    J_3:=\int_{z_0}^\infty dz\,\vpi(z)\,H(z)^{n-1}.    
\end{equation*}
Noting that $0<H\le\Phi<1$ and letting
\begin{equation*}
    G:=1-\Phi,
\end{equation*}
we see that, for each $\ep\in(0,2)$,
\begin{equation*}
    nJ_1\le\int_{-\infty}^{z_\ep} dz\,\vpi(z)\,n\Phi(z)^{n-1}
    =\Phi(z_\ep)^n \\ 
    =(1-G(z_\ep))^n\le\exp\{-nG(z_\ep)\}=o(1/n), \tag{4.5}\label{4.5}
\end{equation*}
since
\begin{equation*}
    nG(z_\ep)=n\exp\Big\{-\frac{z_\ep^2}{2+o(1)}\Big\}
    =n\exp\Big\{-\frac{2-\ep}{2+o(1)}\,\ln n\Big\}=n^{\ep/(2+o(1))}. 
\end{equation*}
So, the conclusion
\begin{equation*}
    nJ_1=o(1/n) \tag{5}\label{5}
\end{equation*}
will hold if $\ep=\ep_n$, for some sequence $\ep_n\downarrow0$. In what follows, it is indeed assumed that $\ep=\ep_n\downarrow0$.
Next,
\begin{equation*}
    J_3\le\int_{z_0}^\infty dz\,\vpi(z)=G(z_0)<\frac{\vpi(z_0)}{z_0}=o(\vpi(z_0))=o(1/n). 
    \tag{6}\label{6}
\end{equation*}
Further,
\begin{equation*}
\begin{aligned}
    H'(z)&=e^{-c}\vpi(z)+(1-e^{-c})\vpi(z-A/s) \\ 
    &=\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}e^{-A^2/(2s^2)}] \\ 
    &\sim\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}], 
\end{aligned}
\end{equation*}
since $s\to\infty$.
Also, for $z\in[z_\ep,z_0]$ we have $z\sim z_0$ (since $\ep\downarrow0$) and hence
\begin{equation*}
    z/s\to z_0/s=\sqrt{\ln4}. 
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
    J_2&=\int_{z_\ep}^{z_0} dz\,\frac{\vpi(z)}{H'(z)}\,H'(z)H(z)^{n-1} \\ 
    &\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,\int_{z_\ep}^{z_0} dz\,H'(z)H(z)^{n-1}.  
\end{aligned}
\tag{7}\label{7}
\end{equation*}
Also,
\begin{equation*}
\begin{aligned}
n\int_{-\infty}^{z_\ep} dz\,H'(z)H(z)^{n-1}= H(z_\ep)^n\le\Phi(z_\ep)^n =o(1),
\end{aligned}
\tag{8}\label{8}
\end{equation*}
as shown in \eqref{4.5}.
Also,
\begin{equation*}
\begin{aligned}
\int_{z_0}^\infty dz\,H'(z)H(z)^{n-1}
&\le\int_{z_0}^\infty dz\,H'(z) \\ 
&=e^{-c}G(z_0)+(1-e^{-c})G(z_0-A/s) \\ 
&\le G(z_0-A/s) \\
&\le \frac{\vpi(z_0-A/s)}{z_0-A/s} \\ 
&=O\Big(\frac{\vpi(z_0)}{z_0-A/s}\Big)
=o(\vpi(z_0))=o(1/n). 
\end{aligned}
\tag{9}\label{9}
\end{equation*}
Collecting \eqref{7},  \eqref{8}, and  \eqref{9}, we get
\begin{equation*}
\begin{aligned}
    J_2&\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\, \\ 
    &\times\Big(\int_{-\infty}^\infty dz\,H'(z)H(z)^{n-1} \\ 
    &\quad-\int_{-\infty}^{z_\ep} dz\,H'(z)H(z)^{n-1}-\int_{z_0}^\infty dz\,H'(z)H(z)^{n-1}\Big) \\ 
    &=\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,(1/n-o(1/n)-o(1/n)).    
\end{aligned}
\tag{10}\label{10}
\end{equation*}
Finally, collecting \eqref{3.5},  \eqref{4}, \eqref{5}, \eqref{6}, and  \eqref{10}, we get the limit \eqref{1} for $L$.

It follows that $\sum_{k=1}^n h(X_k-\max_{1\le i\le n} X_i)$ converges in distribution to a geometrically distributed random variable $Y$ such that
\begin{equation}
    P(Y=j)=\tfrac1B\,(1-\tfrac1B)^{j-1}\,1(j\in\{1,2,\dots\}),
\end{equation}
where
\begin{equation}
    B:=e^{A\sqrt{\ln4}}>1. 
\end{equation}
This result could probably be obtained directly, without using the Laplace transform.
A: I just found another way of representing this in terms of Poisson Point Process.
To simplify the notation we assume now the number of i.i.d. Gaussians are $e^t$ and have distribution $N(0,t)$. Then the approach suggested by @Iosif Pinelis will give us that for $\phi(x)=c1_{[-A,0)}$, the Laplace functional is $$\frac{e^c}{1+(e^c-1)e^{\sqrt{2}A}}.$$
Note that now there is another $e^c$ term because we do not count the zero.
Meanwhile, it is easy to show that

*

*The point process $\mathcal{E}_t$ centered at some critical speed $m_t=\sqrt{2}t-\frac{\log t}{2\sqrt{2}}$ has limiting distribution as Poisson Point Process with intensity measure $\frac{1}{\sqrt{2\pi}}e^{-\sqrt{2}x}1(x<0)dx$,


*and the maximum $M_t=\max_{1\leq k\leq e^t}X_t^{(k)}$, when centered at $m_t$, converges in distribution to a shifted Gumbell distribution: $$\mathbb{P}(M_t-m_t\leq x)\to \exp(-\frac{1}{2\sqrt{\pi}}e^{-\sqrt{2}x}).$$
This suggests if we let $c=-\frac{1}{\sqrt{2}}\log(2\sqrt{\pi})$, $\sqrt{2}(M_t-m_t-c)$ converges in distribution to the standard Gumbell distribution, say $V$.
Therefore the extreme point process seen from the tip, i.e. $\bar{\mathcal{E}}_t=\sum_{k\leq e^t}\delta_{x=X_t^{(k)}-M_t}=T_{m_t-M_t}\mathcal{E}_t$, where $T_a$ is the right-translation by $a$.
Therefore $\bar{\mathcal{E}}_t$ has a limiting distribution as Poisson Point Process, with intensity measure:
\begin{align*}
    \frac{1}{\sqrt{2\pi}}e^{-\sqrt{2}(x+M_t-m_t)}1(x<0)dx &\sim \frac{1}{\sqrt{2\pi}}e^{-\sqrt{2}(x+\frac{V}{\sqrt{2}}+c)}1(x<0)dx\\
&= \frac{1}{\sqrt{2\pi}}e^{-\sqrt{2}x}e^{-V}2\sqrt{\pi}1(x<0)dx\\
&= \sqrt{2}Ze^{-\sqrt{2}x}1(x<0)dx,
\end{align*}
given the standard exponential distributed random variable $Z$, since $e^{-V}$ is $Exp(1)$.
In conclusion, given an $Exp(1)$ random variable $Z$, the limiting distribution of $\bar{\mathcal{E}}_t$ is the Poisson Point Process with intensity measure as above, together with an atom at zero(for the point measure $M_t-M_t=0$).
We can easily check this will give the same Laplace functional as above by @Iosif. Very interesting!
