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?



Limit of the extremal process of i.i.d. Gaussians see from the tip
$\newcommand\R{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}$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.


