text:
\begin{document}
\author[T.~Sepp\"al\"ainen]{Timo Sepp\"al\"ainen}
\address{Timo Sepp\"al\"ainen\\ University of Wisconsin-Madison\\
Mathematics Department\\ Van Vleck Hall\\ 480 Lincoln Dr.\\
Madison WI 53706-1388\\ USA.}
\email{seppalai@math.wisc.edu}
\urladdr{http://www.math.wisc.edu/~seppalai}
\thanks{T.~Sepp\"al\"ainen was partially supported by
National Science Foundation grant DMS-0701091 and by the
Wisconsin Alumni Research Foundation.}
\author[B.~Valk\'o]{Benedek Valk\'o}
\address{Benedek Valk\'o\\ University of Wisconsin-Madison\\
Mathematics Department\\ Van Vleck Hall\\ 480 Lincoln Dr.\\
Madison WI 53706-1388\\ USA.}
\email{valko@math.wisc.edu}
\urladdr{http://www.math.wisc.edu/~valko}
\thanks{B.~Valk\'o  was partially supported by
National Science Foundation grant DMS-0905820}

\keywords{Scaling exponent, directed polymer, random environment, Brownian queue,
superdiffusivity, Burke's theorem, partition function, geometric Brownian motion, Dufresne's
identity}
\subjclass[2000]{60K35, 60K37, 82B41, 82D60}
\date{\today}
\begin{abstract}
We study the scaling exponents of a  1+1-dimensional directed polymer in a Brownian  random
environment      introduced by O'Connell and  Yor.   For a version of the model with
  boundary conditions that are stationary in a space-time  sense we identify the
exact values of the exponents.  For the version without the boundary conditions we
get the conjectured upper bounds on the exponents.
\end{abstract}
\maketitle

\section{Introduction}

We study the scaling exponents of a directed polymer model in 1+1
  dimensions (one space dimension plus  time dimension)
   whose random environment is constructed from
Brownian motions.   For a positive integer $n$ and an inverse
temperature  parameter $\beta>0$,
the partition function is defined by
\be
\fren_n(\beta)=
\int\limits_{0<s_{1}<\dotsm<s_{n-1}<n}  \exp\bigl[ \beta\bigl(B_1(0,s_1) +B_{2}(s_1,s_{2})
+\dotsm + B_n(s_{n-1},n)\bigr)\bigr] \,ds_{1,n-1}
\label{zetadef1}\ee
where $\{B_j\}$ are independent one-dimensional standard Brownian motions,
$B(s,t)=B(t)-B(s)$, and
$ds_{1,n-1}$ $=$ $ds_1ds_2 \dotsm ds_{n-1}$.
This model was   introduced
 by O'Connell and  Yor   \cite{oconn-yor-01} in connection with a related polymer model
 that they named  the
{\sl generalized Brownian queue}.  Subsequently the exact limiting free energy density
${p(\beta)}$ was computed by
Moriarty and O'Connell \cite{mori-oconn-07}. To state their result, recall the
  gamma function   $\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}\,dx$ and the digamma

function   $\digamf=\Gamma'/\Gamma$.

\begin{theorem} [\cite{mori-oconn-07}]    For $\beta>0$ this almost sure limit holds:
\be \lim_{n\to\infty} n^{-1}\log \fren_n(\beta)=  {p(\beta)}=
\inf_{t>0}\left\{t \beta^2-\digamf(t)  \right\}-2\log\beta.
\nn \ee
 \label{zeta1thm}
  \end{theorem}

Moriarty and O'Connell extracted this result with the help of large deviation asymptotics  from
{the generalized Brownian queuing system} whose limit is readily computable.
In this related model  the polymer path is allowed to begin in the
infinite past and a decaying exponential factor is included inside the integral
to make  the partition function converge.  We also work with the Brownian queueing system
  to obtain  an estimate on the fluctuations:

\begin{theorem} There exist finite, positive  $\beta$-dependent constants
$b_0, n_0, C$ such that for  $b\ge b_0$ and $n\ge n_0$
\[  \bP\bigl( \,\abs{\log \fren_n(\beta) -n {p(\beta)}}\ge bn^{1/3} \bigr)
\le Cb^{-3/2}.  \]
\label{zeta2thm}\end{theorem}

The conjectured behavior for directed polymers  is that the
  order of magnitude of the  fluctuations of $\log\fren_n(\beta)$
is  $n^\chi$ and in 1+1 dimensions this exponent takes the value
 $\chi=1/3$  at all inverse temperatures $\beta>0$.
Theorem \ref{zeta2thm} gives the expected  upper
bound on the exponent:  $\chi\le 1/3$.  In Theorem \ref{thm:freepath2} below
we give the corresponding upper bound on the fluctuations
of the polymer path.

\medskip

 The generalized Brownian queueing system amounts
to putting boundary conditions on the polymer  \eqref{zetadef1}  that are stationary in a natural
two-dimensional manner.   This stationarity comes from a
  Burke-type  property
discovered by O'Connell and  Yor   \cite{oconn-yor-01}, see Lemma \ref{lemburke} below.
 For the  model with boundary conditions
we identify the exact scaling exponents.  Our
analysis of these models adapts  the steps of the recent work  \cite{sepp-poly} where a
discrete lattice model with analogous properties was discovered and its
scaling exponents studied.   The roots of the proofs in  \cite{sepp-poly}  can be traced back
to the seminal paper \cite{cato-groe-06}.  In the context of maximal increasing paths on planar
Poisson points, \cite{cato-groe-06}   was the first to chart a path to
the scaling exponents  of a two-dimensional growth  model  without asymptotic
analysis of Fredholm determinants.

 Suboptimal but still highly nontrivial
bounds on scaling exponents in 1+1 dimensional polymer models have been obtained
for Brownian polymers in Poisson environments
 \cite{come-yosh-05,wuth-98aihp,wuth-98aop},
 for  Gaussian random walk    in a Gaussian environment
   \cite{meja-04,petermann},  and  for the related model of  first passage percolation
\cite{lice-newm-piza,newm-piza}.

The overall situation in 1+1 dimensional polymers is now similar to that
for two-dimensional directed last-passage percolation models,
which are of course closely related as zero-temperature directed  polymers.  In both
areas  there is a Brownian model and some particular discrete models that are amenable
to explicit computations. Models with general distributions remain  beyond the reach of current techniques.

Currently results are farther along for last-passage models:
 in addition to exponents, explicit Tracy-Widom  limit distributions are known.
  Key results   appear in the papers
 \cite{baik-deif-joha-99, bala-cato-sepp, bary-01, cato-groe-06,
ferr-spoh-06,grav-trac-wido-01, joha, joha-ptrf-00}.
The connection between last passage models and random matrix theory has been one
of the major inspirations of the subject.  The recent article \cite{oconn-toda} finds
a connection between the Brownian polymer  \eqref{zetadef1}  and the quantum Toda lattice and
proposes this as the possible polymer analogue of the random matrix connection
of last-passage percolation.

There is one  directed polymer model which is essentially solved: the continuum random directed polymer in 1+1 dimension which describes a Brownian path in a white noise environment.
 The free energy is defined as
\be \nonumber
\log E \left[:\!\exp\!:  \left\{-\int_0^T  \dot{\mathcal W}(t,b(t)) dt    \right\}     \right]
\ee
where $\dot{\mathcal W}$ is a space-time white noise,  $b$ is a Brownian motion (or bridge) and $:\!\!\exp\!\!:$ is the Wick-ordered exponential. In \cite{bala-quas-sepp} the exact scaling exponent is determined in the case of a stationary boundary condition using a connection to the KPZ equation and the
weakly asymmetric simple exclusion process.
In \cite{amir-corw-quas} the same connection was used to compute the exact distribution of the free energy in the point-to-point setting, and the Tracy-Widom distribution is derived in the appropriate scaling limit. (See also \cite{sasa-spoh}).

\medskip

{\sl Some frequently used notation.}
We write $f(s,t)=f(t)-f(s)$ for increments, without assuming that $s\le t$.
$\bZ_+=\{0,1,2,\dotsc\}$, $\bN=\{1,2,3,\dotsc\}$ and $\bR_+=[0,\infty)$.
  $\overline{X}=X-\E X$ denotes a centered random variable.
In general a superscript $\om$ is added to a symbol whenever its dependence
on a particular realization of the environment needs to be made explicit.

\section{Models and results}

We begin with precise definitions of two polymer models, the one already encountered
and another one with a boundary.
$B$ and $\{B_k: k\in\bN\}$
 denote  independent  standard  Brownian motions indexed by $\bR$.
 They form the random environment $\om$ under probability measure $\bP$.

By Brownian scaling $\{B(ct)\}\eqd\{c^{1/2}B(t)\}$
the parameter $\beta$ can be removed  from the exponent in \eqref{zetadef1}
and replaced by a parameter that controls the upper limit of integration.
This is   convenient for us,  so instead of  $\fren_n(\beta)$
 we work with the family
\be Z_{j,k}(s,t) =
\int\limits_{s<s_{j}<\dotsm<s_{k-1}<t}  \exp\bigl[ B_j(s,s_j) +B_{j+1}(s_j,s_{j+1})
+\dotsm + B_k(s_{k-1},t)\bigr] \,ds_{j,k-1}
\label{Zdef1.1}\ee
where $1\le j\le k\in \bN$ and $s<t\in\bR$.  Occasionally we may also write
$Z_{(j,k),(s,t)} =Z_{j,k}(s,t)$.  The distributional identity is
\be \fren_n(\beta)\eqd\beta^{-2(n-1)}Z_{1,n}(0,{n}\beta^2).\label{brscal}\ee
 This is the last appearance of the partition function $Z_n(\beta)$ defined by \eqref{zetadef1}
 in the paper.   Similar notation will be used below for other partition functions.

The partition function is the normalizing constant for the quenched polymer
distribution $Q_{(j,k),(s,t)}$.  This is a probability measure on
nondecreasing cadlag paths
  $x:[s,t]\to\{j,j+1,\dotsc,k\}$ that go from $x(s)=j$ to $x(t)=k$.  We represent these
paths   in terms of the jump times
 $s<\sigma_j<\sigma_{j+1}<\dotsm<\sigma_{k-1}\le t $ {where}
$x(\sigma_i-) = i< i+1=x(\sigma_i)$.
The   measure $Q_{(j,k),(s,t)}$ is defined   by
\be\begin{aligned}
&E^{Q_ {(j,k),(s,t)}}f(\sigma_j,\dotsc,\sigma_{k-1})
=\frac1{Z_{{(j,k),(s,t)}}}
\int\limits_{s<s_{j}<\dotsm<s_{k-1}<t}  f(s_j,\dotsc, s_{k-1})
 \\[7pt]
 &\qquad\qquad\qquad  \times  \exp\bigl[ B_j(s,s_j) +B_{j+1}(s_j,s_{j+1})
+\dotsm + B_k(s_{k-1},t) \bigr] \,ds_{j,k-1}.
\end{aligned}\label{Qdefjkst}\ee
This measure is called quenched because the environment of Brownian motions
is fixed.  Integrating away  the environment gives the
annealed expectation  $E_{(j,k),(s,t)}(\cdot)=\bE E^{Q_{(j,k),(s,t)}}(\cdot)$.

In addition to the digamma function we also need its derivative, the trigamma function $\trigamf=\digamf'$.
$\digamf$ is concave and increasing and $\trigamf$ is positive, convex
and strictly decreasing with $\trigamf(0+)=\infty$ and $\trigamf(\infty)=0$.

Theorem \ref{zeta1thm} is equivalent to the statement
\be
\lim_{n\to\infty}n^{-1}\log Z_{1,n}(0, n\tau)= \trigamf(\para)\para- \digamf(\para)
\quad \text{$\P$-a.s.}
\label{Zlim5}\ee
where   $\tau>0$ and   $\para$ is the unique value such that
$\trigamf(\para)=\tau$.
 Theorem \ref{zeta2thm} will be proved in Section \ref{sec:freeZ} in terms of
$Z_{1,n}(0, n\tau)$.  In conjunction with Theorem \ref{zeta2thm} goes an upper
bound on the fluctuations of the path, also proved in Section \ref{sec:freeZ}.

\begin{theorem}\label{thm:freepath2}
Let $\tau>0$ and    $0<\gamma<1$. Then for all large enough $n$
 and $b$
\be
P_{(1,n),(0,n\tau)}\bigl(|\sigma_{\lfloor n \gamma\rfloor}-n\gamma \tau |>b n^{2/3}   \bigr)\le
C(\tau) b^{-3}.
\ee
\end{theorem}

Theorem \ref{thm:freepath2} says that the path   stays close to the diagonal of the rectangle
$[1,n]\times[0,n\tau]$, and typical  fluctuations away from the diagonal have order of
magnitude at most $n^{2/3}$.  This gives an upper bound $\zeta\le 2/3$ for the second
basic scaling exponent $\zeta$  which describes the fluctuations of the polymer path.

\medskip

These are the results for the polymer without a boundary, and we turn to discuss
the model with boundary.  For this model the upper bounds of Theorems
\ref{zeta2thm} and \ref{thm:freepath2} are combined with matching lower bounds,
so we have the precise values of the scaling exponents.    An additional
  parameter $\para>0$ is introduced in this model.

The partition function is
\be\begin{aligned} Z_{n}^\para(t) &= Z^\para_{n,t} =
\int_{-\infty<s_{0}<s_1<\dotsm<s_{n-1}<t}  \exp\bigl[ -B(s_0)+\para s_0 \\[7pt]
 &\qquad  +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}.
\end{aligned}\label{Zdef2}\ee
 The quenched polymer measure $Q^\para_{n,t}$
lives on    nondecreasing cadlag paths
  $x:(-\infty,t]\to\{0,1,\dotsc,n\}$ that go from $x(-\infty)=0$ to $x(t)=n$.  Again
we represent these     in terms of  jump times
 $-\infty<\sigma_0<\sigma_1<\dotsm<\sigma_{n-1}\le t $ {where}
$x(\sigma_i-) = i< i+1=x(\sigma_i)$.
The   measure is defined   by
\be\begin{aligned}
&E^{Q^\para_{n,t}}f(\sigma_0,\sigma_1,\dotsc,\sigma_{n-1})
=\frac1{Z_{n}^\para(t) }
\int_{-\infty<s_{0}<\dotsm<s_{n-1}<t}  f(s_0, s_1,\dotsc, s_{n-1})
 \\[7pt]
 &\qquad  \times  \exp\bigl[ -B(s_0)+\para s_0 +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}.
\end{aligned}\label{Qdef2}\ee
Annealed probability and expectation are   denoted by
$P^\para_{n,t}(\cdot)=\bE Q^\para_{n,t}(\cdot)$  and
 $E^\para_{n,t}(\cdot)=\bE E^{Q^\para_{n,t}}(\cdot)$, and simply
 $E(\cdot)$ when the parameters are understood.

 \medskip

{\sl Notational remark.}   To simplify notation we shall not consistently
carry the superscript $\para$ in the notation for the objects of the polymer model
with boundary.  The notational distinction between the two models is that
the model without boundary has two space-time parameters represented by $((j,k),(s,t))$   in
definitions \eqref{Zdef1.1} and \eqref{Qdefjkst}, while the model with boundary
has only a single space-time parameter, namely $(n,t)$ in definitions
\eqref{Zdef2} and \eqref{Qdef2}.  When dependence on the environment $\om$
needs to be displayed explicitly, $\om$ is added as a superscript,  as   for example in
$Q^{\para,\om}_{n,t}$.

\medskip

The boundary conditions render the model stationary in a sense made precise
in Theorem \ref{burkethm}.  As a consequence some explicit computations can be
performed: from Theorems \ref{burkethm} and \ref{varthm} we obtain
\be
\E (\log Z_n^{\para}(t))=-n \digamf(\para)+\para t
\quad\text{and}\quad
 \Vvv(\log Z_n^\para(t)) = n\trigamf(\para)-t+2 E^\para_{n,t}(\sigma_0^+).
\label{meanvar}\ee

As we take the size of the polymer to
infinity,  the interesting exponents appear when the  endpoint follows approximately a   {\sl characteristic direction}
specified by the parameter $\para$.  The precise hypothesis is
  \be \text{$t>0$ and $n\in\bN$ satisfy $\abs{t-n \trigamf(\para)}\le A n^{2/3} $
  for a  constant $0\le A<\infty$. }   \label{assA}\ee
The purpose of this assumption is to kill the first two terms of the variance formula in
\eqref{meanvar}.
The first theorem says that the exponent  $\chi=1/3$
describes  the order of fluctuations of $\log Z_n^{\para}(t)$.

\begin{theorem}\label{varbdthm}  Assume \eqref{assA}.
 Then there exist  constants $0<C_1 <C_2 <\infty$
 that depend on $(\para, A)$  such that for all
 $n\ge 1$
\be
C_1 n^{2/3} \le \Vvv (\log Z_n^{\para}(t))\le  C_2 n^{2/3}.
\ee
\end{theorem}

The second theorem says that under the annealed distribution
  the path stays close to the diagonal of the rectangle $[0,n]\times[0,t]$, and
  $n^{2/3}$ is the correct order of typical fluctuations.

\begin{theorem}\label{path-thm1}
Assume \eqref{assA}   and   let $0\le \gamma<1$.   Then for
large enough $n$ and $b$
 \be \label{path11}
P^\para_{n,t}(\abs{\sigma_{\lfloor \gamma n \rfloor }-\gamma t}>b n^{2/3})\le C b^{-3}.
\ee
 For any  $\varepsilon>0$ there exists $\delta>0$ such that
\be\label{path21}
\varlimsup_{n\to \infty}P^\para_{n,t}(|\sigma_{\lfloor \gamma n \rfloor }-\gamma t|\le \delta n^{2/3})\le \varepsilon.
\ee

\end{theorem}

Variable  $\sigma_0$ is a special case of Theorem \ref{path-thm1}, but in fact
controlling this case turns out to be the key to {\sl both} Theorems  \ref{varbdthm} and
 \ref{path-thm1}.  Let us also mention that  our arguments give the
 moment bound $E^\para_{n,t}(\abs{\sigma_0}^p)\le Cn^{2p/3}$ for $1\le p<3$.

\section{Properties of the model with boundary}

Following O'Connell and Yor \cite{oconn-yor-01}
introduce the following processes to render the role of boundary conditions
clearer in \eqref{Zdef2}:
   $\uu_0(t)=B(t)$, and then define inductively for
$k\in\bN$
\be\begin{aligned}
  r_k(t)&=\log\int_{-\infty}^t e^{\uu_{k-1}(s,t)-\para(t-s)+B_k(s,t)}\,ds
\end{aligned}\label{defrk}\ee
and
\bea\begin{aligned}
\uu_k(t)&= \uu_{k-1}(t)+r_k(0)-r_k(t),\\
X_k(t)&=B_{k}(t)+r_k(0)-r_k(t).
\end{aligned}\label{defYk}\eea
From the definitions {and a simple induction argument} follow the equations
\be\begin{aligned}
\sum_{k=1}^n r_k(t)
&= \log \int_{-\infty<s_{0}<s_1\dotsm<s_{n-1}<t}  \exp\bigl[ B(s_0,t)-\para(t- s_0) \\[7pt]
 &\qquad \qquad +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}\\[4pt]
&= B(t)-\para t +\log Z_{n}^\para(t)
\end{aligned} \label{rB1}\ee
and
\be\begin{aligned}
Z_{n}^\para(t) =  \int_0^t   \exp[- B(s)+\para s] Z_{1,n}(s,t) \,ds

\;+ \; \sum_{j=1}^n  \Bigl( \;\prod_{k=1}^j e^{r_k(0)} \Bigr) Z_{j,n}(0,t).
\end{aligned}\label{ZB2}\ee
If we define  also
\be  Z_{0}^\para(t)= \exp[ -B(t)+\para t]  \label{Z0}\ee
then the above can  be expressed in the form
\be\begin{aligned}
Z_{n}^\para(t) =  \int_0^t  Z_{0}^\para(s) Z_{1,n}(s,t) \, ds

\; + \; \sum_{j=1}^n  Z_{j}^\para(0) Z_{j,n}(0,t).
\end{aligned}\label{ZB3}\ee
Combining \eqref{rB1} with \eqref{defYk}  gives the space and time increments
\be
\log Z_{n}^\para(t)- \log Z_{n-1}^\para(t) = r_n(t)  \qquad
 \text{for $n\in\bN,\, t\in\bR_+$}  \label{Zincr1}\ee
and
\be
\log Z_{n}^\para(t)- \log Z_{n}^\para(s) = \para(t-s)-\uu_n(s,t)  \qquad
 \text{for $n\in\bZ_+,\, s<t\in\bR_+$} .  \label{Zincr2}\ee

\begin{remark}\label{remark1}
The density of $(\sigma_0,\dots, \sigma_{n-1})$ under $Q_{n,t}$ can also be written as
\be
\frac1{\widehat Z_n^\para(t)}
\exp\bigl[
\widehat B(s_0,t)+\widehat B_1(s_0,s_1)+\widehat B_2(s_1,s_2)+\dots+\widehat B_n(s_{n-1},t)
\bigr]\ind\{s_0<\dotsm<s_{n-1}<t\}
\label{altQ}\ee
where $\widehat B(u)=B(u)-\para u/2$ (and similarly for $\widehat B_k$)
and $\widehat Z_n^\para(t)=Z_n^\para(t) \exp(B(t)-\para t)$. From this representation
and the stationarity of Brownian increments it is clear that
\be

E^{Q_{n,t}}f(\sigma_0,\sigma_1,\dotsc,\sigma_{n-1}) \eqd
E^{Q_{n,0}}f(t+\sigma_0,t+\sigma_1,\dotsc,t+\sigma_{n-1}).
\label{altQ2} \ee

\end{remark}

\subsection{Burke property}

The following lemma summarizes the Burke property of O'Connell and Yor  \cite[Thm.~5]{oconn-yor-01}.

\begin{lemma}[\cite{oconn-yor-01}]\label{lemburke} Let $B$ and $C$ be independent standard Brownian motions indexed by $\bR$ and $\para>0$ a fixed constant. For $t\in\bR$ set
\begin{align*}
&r(t)=\log\int_{-\infty}^t \exp\left(B(s,t)+C(s,t)-\para(t-s)\right)ds,&\\
&f(t)=B(t)+r(0)-r(t), \quad g(t)=C(t)+r(0)-r(t).&
\end{align*}
Then $f$ and $g$ are independent standard Brownian motions indexed by $\bR$ and  for each $t\ge0$  the processes $\{(f(s),g(s)):s\le t\}$ and $\{r(s):s\ge t\}$ are independent. Moreover the following identity holds almost surely:
\be\label{r_ident}
r(t)=\log\int^{\infty}_t \exp\left(f(t,s)+g(t,s)+\para(t-s)\right)ds.
\ee
\end{lemma}
Although   identity (\ref{r_ident}) is not stated explicitly in Theorem 5 of \cite{oconn-yor-01} it is
contained in the proof.
In the next theorem we generalize the Burke property in a form suitable for our use.

\begin{theorem}\label{burkethm}
Let $n\in\bN$ and  $0\le s_n\le s_{n-1}\le\dots\le s_1<\infty$.
Then over the index $j$
 the following random variables and processes are all mutually independent:
 \begin{align*}
& r_j(s_j) \ \text{ and } \    \{X_j(s): s\le s_j\} \ \text{ for } \    1\le j\le n, \quad  \{Y_n(s): s\le s_n\}, \quad \\
& \qquad\qquad\qquad\qquad\qquad   \text{and} \quad
\{Y_j(s_{j+1},s): s_{j+1}\le s\le s_j\} \ \text{ for } \    1\le j\le n-1.
 \end{align*}
Furthermore,  the $X_j$ and $Y_j$  processes are standard Brownian motions,  and $r_j(s_j)\overset{d}=-\log \eta$ with $\eta\sim Gamma(\para,1)$.
\end{theorem}
\begin{proof}
We use induction on $n$.   For $n=1$ the statement follows from Lemma \ref{lemburke} together with Dufresne's identity \cite[Cor.~4]{dufr-osaka01} which gives the distribution of $r_1(t)$.

We now assume that the statement is true for $n-1$ and prove it for $n$. From definitions \eqref{defrk}--\eqref{defYk},
$\{r_{n}(s), Y_{n}(s), X_n(s): s\le s_n\}$ is a function of
 $\{Y_{n-1}(s): s\le s_{n}\}$ and an
independent Brownian motion $B_{n}$. This means that $\{r_{n}(s), Y_{n}(s), X_n(s): s\le s_n\}$ is independent of  the processes $\{X_j(s): s\le s_j\}$, $\{Y_j(s_{j+1},s): s_{j+1}\le  s\le s_j\}$
for  $1\le j\le n-1$, and random variables $\{r_j(s_j): 1\le j \le n-1\}$.
 Besides the induction hypothesis we also used that $\{Y_{n-1}(s_{n},s), s\ge s_{n}\}$ is independent of  $\{Y_{n-1}(s): s\le s_{n}\}$. That $\{ Y_{n}(s), X_n(s): s\le s_n\}$ and $ r_{n}(s_n)$ have the right joint distribution again follows from Lemma \ref{lemburke} and Dufresne's identity.
\end{proof}
Applying the shift $\shift_uf(t)=f(u+t)-f(u)$ that preserves
the distributions of the  Brownian motions
shows that  $r_k(t)=r_k(0)\circ\shift_t$ is a stationary process. {We also note that $\bE (r_k(t))=-\digamf(\para)$ and $\Vvv (r_k(t))=\trigamf(\para)$.}
\subsection{Reversal}

Fix $T\in\bR$ and  $n\in \bN$.

Define the following new processes:
\bea
\label{defY*} Y_j^*(s)&=&Y_{n-j}(T)-Y_{n-j}(T-s), \qquad 0\le j\le n,\\
\label{defB*}B_j^*(s)&=&X_{n+1-j}(T)-X_{n+1-j}(T-s),\qquad 1\le j\le n,\\
\label{defr*}r_j^*(s)&=&r_{n+1-j}(T-s),\qquad 1\le j\le n,\\
\label{defX*}X_j^*(s)&=&B_{n+1-j}(T)-B_{n+1-j}(T-s),\qquad 1\le j\le n.
\eea
Define the {\sl dual   environment}  by  $\om^*=(Y^*_0, B^*_j: 1\le j\le n)$.
\begin{theorem}\label{revthm}  Fix $n\in\bN$.
We have the following equality in distribution for the  processes on $\bR$:
\be \begin{aligned}
&\left\{Y_{k}: 0\le k\le n; \, X_j, B_{j}, r_j: \, 1\le j\le n \right\}  \eqd
\left\{Y_{k}^*: 0\le k\le n; \, X_j^*, B_j^*, r_j^*: \, 1\le j\le n \right\}.
\end{aligned} \notag\ee
In particular, the dual environment $\om^*=(Y^*_0, B^*_j: 1\le j\le n)$
has the same distribution as the original environment $\om=(B, B_j: 1\le j\le n)$.
\end{theorem}
\begin{proof}
{By (\ref{r_ident}) from Lemma \ref{lemburke}}

\be \label{rXY}
r_k(t)=\log \int_{t}^{\infty} \exp(Y_k(t,s)+\para(t-s)+X_k(t,s)) ds.
\ee
Definitions (\ref{defY*})--(\ref{defX*}) together with (\ref{defYk}) and
\eqref{rXY}  give the identities
\be \begin{aligned}
r_k^*(t)&=\log\int_{-\infty}^t \exp(Y^*_{k-1}(s,t)-\para(t-s)+B_k^*(s,t))\,ds,\\
Y^*_k(t)&= Y^*_{k-1}(t)+r^*_k(0)-r^*_k(t),\\[3pt]
X^*_k(t)&=B_k^*(t)+r^*_k(0)-r^*_k(t).
\end{aligned}  \label{eqn*}\ee
  Theorem \ref{burkethm}
  tells us that  $\{Y_n, X_j, 1\le j\le n\}$  are independent Brownian motions on $(-\infty, t]$
for any $t>0$, hence over all of $\bR$.   Consequently by \eqref{defY*}--\eqref{defB*} processes
   $\{Y_0^*, B_j^*, 1\le j\le n\}$  have the same distribution as $\{Y_0, B_j, 1\le j\le n\}$.
The theorem follows because  \eqref{eqn*} defines the same recursions as  \eqref{defrk}--\eqref{defYk}.
\end{proof}

We define the dual quenched measure $Q^*_n$   on non-decreasing cadlag paths $x:\bR_+\to \{0,1,\dots,n\}$ with $x(0)=0$ and $x(\infty)=n$. These paths can be represented by    jump times
$0<\sigma^*_1<\dotsm<\sigma^*_{n}$ defined by $x(\sigma^*_j-)=j-1<j=x(\sigma^*_j)$.  The dual measure   $Q^*_n$  is defined by
 \be\label{Qstar}\begin{aligned}
&E^{Q^*_{n}}\bigl[f(\sigma^*_1,\sigma^*_2,\dotsc,\sigma^*_{n})\bigr]  =\frac1{Z_n^*}
\int_{0<s_1<\dotsm<s_n<\infty}  f(s_1, s_2,\dotsc, s_n)   \\[4pt]
&\qquad \times\exp\left[ X_1(0,s_1)+X_2(s_1,s_2)+\dotsm+ X_n(s_{n-1},s_{n})+Y_n(0,s_{n})-\para s_{n}\right]
\,d{s_{1,n}}.
\end{aligned}\ee
Because of the shifting described in \eqref{altQ2} we only need
the measure $Q^*_{n}$ instead of a family indexed also by $t$.

 The dual measure is naturally connected  with the representation of  $Q_{n,t}$ given
 in Remark \ref{remark1}.   The    next lemma is proved by a straightforward  change of
 variables in the integrals.  {Recall the definition of $\widehat Z_n^\para(t)$ from (\ref{altQ}).}
\begin{lemma}\label{LEMrev}  Fix $t\in\bR$ and $n\in\bN$  and for each
$\om$  define the dual environment $\om^*$
with $T=t$.
Then  $
Z_n^{*,\om} =  \widehat Z_{n}^{\para, \om^*}(t),
$
and
\[  E^{Q^{*,\om}_{n}}\bigl[f(\sigma^*_1,\sigma^*_2,\dotsc,\sigma^*_{n})  \bigr]
=    E^{Q_{n,t}^{\om^*}}\bigl[ f(t-\sigma_{n-1},t-\sigma_{n-2},\dotsc,t-\sigma_{0})\bigr].    \]
Consequently, by Theorem \ref{revthm},
\[
 E^{Q^{*}_{n}}\bigl[f(\sigma^*_1,\sigma^*_2,\dotsc,\sigma^*_{n}) \bigr]
   \eqd   E^{Q_{n,t}}\bigl[f(t-\sigma_{n-1},t-\sigma_{n-2},\dotsc,t-\sigma_{0})\bigr].
\]
 \end{lemma}

\subsection{Variance identity}

\begin{theorem} For $\para>0$ and  $(t,n)\in(0,\infty)\times\bN$
\be\begin{aligned}
 \Vvv[\log Z_n^\para(t)] &= n\trigamf(\para)-t+2 E^\para_{n,t}(\sigma_0^+)\\
&= -n\trigamf(\para)+t+2 E^\para_{n,t}(\sigma_0^-).
\end{aligned}\label{var1}\ee
\label{varthm}\end{theorem}
\begin{remark}
Adding the two equations gives $ \Vvv[\log Z_n^\para(t)] =E^\para_{n,t}|\sigma_0|$ while subtracting them yields $E^\para_{n,t} (\sigma_0)=t-n\trigamf(\para)$.
\end{remark}
\begin{proof}[Proof of Theorem \ref{varthm}]  While we keep $\para$ fixed we simplify the notation
to $Z_n(t)=Z_n^\para(t)$.

The proof begins the same way as the proof of
Theorem 3.7 in \cite{sepp-poly}. (The idea was
originally learned from the proof of Theorem 2.1 in \cite{cato-groe-06}.)
Abbreviate the increments with reference to compass directions:
\begin{align*}
\north &=\log Z_{n}(t)-\log Z_{n}(0),
\quad \south =\log Z_{0}(t)=\para t-B(t),\\
\quad \east &=\log Z_{n}(t)-\log Z_{0}(t),
\quad \west =\log Z_{n}(0)=\sum_{j=1}^n r_j(0).
\end{align*}
The south and west increments are given by the boundary conditions
as indicated above, while the east and {north} increments are computed
from \eqref{ZB2}.
Then
\be\begin{aligned}
\Vvv\bigl[\log Z_{n}(t)\bigr]&=\Vvv(\west +\north )=\Vvv(\west )+\Vvv(\north )+2\Cvv(\west ,\north )\\
&=\Vvv(\west )+\Vvv(\north )+2\Cvv(\south +\east -\north ,\north )\\
 &= \Vvv(\west )-\Vvv(\north )+2\Cvv(\south ,\north )\\
&=n\trigamf(\para)-t+2\Cvv(\south ,\north ).\\
\end{aligned}\label{aux3}\ee
{The third line comes from the independence of $\east $ and $\north $ while the  last equality came
 from (\ref{Zincr2}) and Theorem \ref{burkethm}.} Substituting $\north$ instead of
$\west$ in the covariance in the second line of (\ref{aux3}) leads to the formula
\be
\Vvv\bigl[\log Z_{n}(t)\bigr] =  -n\trigamf(\para)+t
+2\Cvv(\east,\west ).
\label{aux3.1}\ee
Equations \eqref{aux3} and \eqref{aux3.1} give the two lines of
\eqref{var1} once we evaluate the covariances.
We begin with \eqref{aux3}.

We need to vary separately the parameters of the boundary conditions
on the $x$ and $y$ axes.  So we rename the parameter on the
$y$-axis as $\paraa$, and rewrite the partition function with
boundaries
as follows:
\be\begin{aligned} Z_{n}(t) &=
\int_{0<s_{0}<s_1\dotsm<s_{n-1}<t}  \exp\bigl[ -B(s_0)+\para s_0 \\[7pt]
 &\qquad\qquad   +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}\\
&
+
\int_{-\infty<s_{0}<s_1\dotsm<s_{n-1}<t} \ind\{s_0<0\}
 \exp\bigl[ -B(s_0)+\paraa s_0 \\[7pt]
 &\qquad \qquad +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}.
\end{aligned}\label{Z5}\ee

Next we argue that
\be \text{$\bE(\north\vert\south=x)$ does not
depend on $\para$.}  \label{aux3.8}\ee

Indeed, if we condition $\{h(s)=\para s-B(s), 0\le s\le t\}$ on the event $h(t)=x$ then this (as a process) has the same distribution as $\para s-\tilde B_s$ where $\tilde B_s$ is a BM conditioned to hit $y=\para t-x$ at $t$. It is known that the latter has the same distribution as $B_s-sB_t/t  +{sy}/{t}$ which means that the conditional distribution of $\{h(s), 0\le s\le t\}$ is the same as
\[
\para s-\left(B_s- \frac{sB_t}{t}+\frac{s(\para t-x)}{t}\right)=\frac{s}{t} B_t-B_s+\frac{sx}{t}
\]
which does not depend on $\para$.

The density of $\south$ is
$f_\para(x)=(2\pi t)^{-1/2} \exp(-\frac1{2t}(x-\para t)^2)$.
Utilizing \eqref{aux3.8},
\be\begin{aligned}
\frac{\partial}{\partial\para} \E(\north )
&=\frac{\partial}{\partial\para} \int_\bR \E(\north \,\vert\, \south =x) f_\para(x)\,dx
=\int_\bR \E(\north \,\vert\, \south =x)
\frac{\partial   f_\para(x)}{\partial\para}\,dx\\
&=\int_\bR \E(\north \,\vert\, \south =x) ( x-\para t)f_\para(x)\,dx\\
&=\E(\north \south )- \E(\north )\E(\south) =
\Cvv(\north ,\south ).
\end{aligned}\label{aux4}\ee
On the other hand, utilizing \eqref{Z5},
\be\begin{aligned}
&\frac{\partial}{\partial\para} \E(\north )
= \E\Bigl[\; \frac{\partial}{\partial\para} \log Z_n(t)\Bigr] \\
&=\E\Bigl[\; \frac1{ Z_n(t)} \int_{0<s_{0}<s_1\dotsm<s_{n-1}<t}
s_0\, \exp\bigl\{ -B(s_0)+\para s_0 \\[7pt]
 &\qquad\qquad   +  B_1(s_0,s_1)
+B_{2}(s_1,s_{2}) +\dotsm + B_n(s_{n-1},t)\bigr\} \,ds_{0,n-1}\Bigr]\\
&=E(\sigma_0^+).
\end{aligned}\label{aux5}\ee
Combining \eqref{aux4}, \eqref{aux4} and \eqref{aux5} gives
the first line of  \eqref{var1}.

Proof of the second line of \eqref{var1} proceeds analogously.
$\bE(\east\,\vert\,\west=x)$ does not
depend on $\paraa$, and so the analogue of computation
\eqref{aux4} gives $\partial_\paraa \E(\east) = -\Cvv(\east, \west)$.
Then from  \eqref{Z5},  $\partial_\paraa \E(\east)
=\E(\partial_\paraa \log Z_n(t))
=E(\sigma_0\ind\{\sigma_0<0\})$.
\end{proof}

\subsection{Comparison lemma}

We find it useful here  to augment the family $Z_{j,k}(s,t)$ defined
for $j\ge 1$  by \eqref{Zdef1.1}
by introducing   $Z_{0,k}(t)=Z_{0,k}(0,t)$. These will be defined not exactly consistently
with \eqref{Zdef1.1}, but in a manner that gives us inequalities between ratios of
partition functions.
For $k\in \bN$ and $t\in\bR_+$ define
\be  Z_{0,0}(t)=e^{-B(t)} \label{Zdef1.01} \ee
and
 \be\begin{aligned} Z_{0,k}(t) &=
\int\limits_{0<s_{0}<\dotsm<s_{k-1}<t}  \exp\bigl[ -B(s_0) +B_1(s_0,s_1)
\\[4pt]   &\qquad \qquad \qquad \qquad
+B_{2}(s_1,s_{2})
+\dotsm + B_k(s_{k-1},t)\bigr] \,ds_{0,k-1}.
\end{aligned}\label{Zdef1.02}\ee

For $n\in\bN$ and events $D$ on the paths
 we write $Z_{n,t}^\para(D)=Z_{n,t}^\para Q_{n,t}^\para(D)$ for the unnormalized
quenched measure.  It is also convenient  to set, for $A\subseteq\bR$,
\be Z_{0,t}^\para(\sigma_0\in A)= \ind_{A\cap\bR_+}(t) \exp[-B(t)+\para t].  \label{0conv}\ee

\begin{lemma}  Let $\para>0$.  For $0< s<t$ and $n\in\bZ_+$
\be
\frac{Z_{n+1,t}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)}
\le \frac{Z_{0,n+1}(t)}{Z_{0, n}(t)}\le \frac{Z_{n+1,t}^\para(\sigma_0<0)}{Z_{n,t}^\para(\sigma_0<0)}
\label{comp1}\ee
and
\be
\frac{Z_{n,t}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)}
\ge \frac{Z_{0,n}(t)}{Z_{0, n}(s)}\ge \frac{Z_{n,t}^\para(\sigma_0<0)}{Z_{n,s}^\para(\sigma_0<0)}.
\label{comp2}\ee
The second inequality of \eqref{comp2} makes sense only for $n\ge 1$.
\end{lemma}
\begin{proof}
We check the  cases that initialize the inductive proofs.   For \eqref{comp1}
\begin{align*}
\frac{Z_{1,t}^\para(\sigma_0>0)}{Z_{0,t}^\para(\sigma_0>0)}
&=\frac{\int_0^t e^{-B(s)+\para s+B_1(s,t)}\,ds}{e^{-B(t)+\para t}}
=\int_0^t e^{B(s,t)+\para (s-t)+B_1(s,t)}\,ds\\
&\le  \int_0^t e^{B(s,t)+B_1(s,t)}\,ds  = \frac{Z_{0,1}(t)}{Z_{0, 0}(t)}
<\infty = \frac{Z_{1,t}^\para(\sigma_0<0)}{Z_{0,t}^\para(\sigma_0<0)} .
\end{align*}
For the first part of  \eqref{comp2}
\begin{align*}
\frac{Z_{0,t}^\para(\sigma_0>0)}{Z_{0,s}^\para(\sigma_0>0)}=e^{-B(s,t)+\para (t-s)}
>e^{-B(s,t)}=  \frac{Z_{0,0}(t)}{Z_{0, 0}(s)}
\end{align*}
and for the second part (now for $n=1$)
\begin{align*}
 \frac{Z_{0,1}(t)}{Z_{0, 1}(s)} = e^{B_1(s,t)}
 +\frac{\int_s^t e^{-B(u)+ B_1(u,t)}\,du}{\int_0^s e^{-B(u)+ B_1(u,s)}\,du}
 >e^{B_1(s,t)}  =  \frac{Z_{1,t}^\para(\sigma_0<0)}{Z_{1,s}^\para(\sigma_0<0)}.
\end{align*}

Next the induction steps.  We make use of the decomposition (for $0\le s<t$)
\[
 {Z_{n,t}^\para(\sigma_0\in A)}
=
 {Z_{n,s}^\para(\sigma_0\in A)} e^{B_n(s,t)}
+ \int_s^t   {Z_{n-1,u}^\para(\sigma_0\in A)}  e^{B_n(u,t)} \,du
\]
valid for $n\in\bN$ and for both $A=(-\infty, 0)$ and $A=(0,\infty)$.  For $n=1$
convention \eqref{0conv} is used on the right-hand side.

We begin with  the first inequalities in  \eqref{comp1}--\eqref{comp2}.
 Assume the first inequality of both  \eqref{comp1} and \eqref{comp2} holds for $n-1$.
We verify first \eqref{comp2} for $n$.
\begin{align*}
\frac{Z_{n,t}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)}
&=  e^{B_n(s,t)}
+ \int_s^t  \frac{Z_{n-1,u}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)} \,e^{B_n(u,t)} \,du\\
&=  e^{B_n(s,t)}
+ \int_s^t  \frac{Z_{n-1,u}^\para(\sigma_0>0)}{Z_{n-1,s}^\para(\sigma_0>0)}
\cdot  \frac{Z_{n-1,s}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)} \,e^{B_n(u,t)} \,du\\
&\ge   e^{B_n(s,t)}
+ \int_s^t  \frac{Z_{0,n-1}(u)}{Z_{0,n-1}(s)}
\cdot  \frac{Z_{0,n-1}(s)}{Z_{0,n}(s)} \,e^{B_n(u,t)} \,du\\
&= \frac{Z_{0,n}(t)}{Z_{0, n}(s)}
\end{align*}
and the inequality came from the induction assumption.

Deriving the first inequality of  \eqref{comp1} for $n$ requires an extra step.  First decompose and use
 \eqref{comp2} for $n$ that we just proved:
\begin{align*}
&\frac{Z_{n+1,t}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)}
= \frac{Z_{n+1,s}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)}\,e^{B_n(s,t)}
+ \int_s^t  \frac{Z_{n,u}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)} \,e^{B_n(u,t)} \,du\\[7pt]
&\qquad\qquad= \frac{Z_{n+1,s}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)}
\cdot \frac{Z_{n,s}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)}\,e^{B_n(s,t)} \,
+ \int_s^t  \frac{Z_{n,u}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)} \,e^{B_n(u,t)} \,du\\
&\qquad\qquad\le  \frac{Z_{n+1,s}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)}
\cdot \frac{Z_{0,n}(s)}{Z_{0,n}(t)}\,e^{B_n(s,t)} \,
+ \int_s^t  \frac{Z_{0,n}(u)}{Z_{0,n}(t)}  \,e^{B_n(u,t)} \,du\\
&\qquad\qquad\equiv \frac{Z_{n+1,s}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)} \,
a(s,t) + \int_s^t a(u,t)\,du.
\end{align*}
The last equality defines the continuous function $a(s,t)$.  The same steps applied
to the middle member of  \eqref{comp1} gives an equality
\be
\frac{Z_{0,n+1}(t)}{Z_{0, n}(t)}  = \frac{Z_{0,n+1}(s)}{Z_{0, n}(s)}
\,a(s,t) + \int_s^t a(u,t)\,du \notag\ee
from which
\be
\frac{Z_{n+1,t}^\para(\sigma_0>0)}{Z_{n,t}^\para(\sigma_0>0)}  -  \frac{Z_{0,n+1}(t)}{Z_{0, n}(t)}
\; \le \; a(s,t) \biggl( \frac{Z_{n+1,s}^\para(\sigma_0>0)}{Z_{n,s}^\para(\sigma_0>0)}
- \frac{Z_{0,n+1}(s)}{Z_{0, n}(s)}\biggr).   \label{compaux8}\ee
Each of the three terms on the right-hand side above vanishes as we take $s\searrow 0$.  We have proved that the  first inequalities in  \eqref{comp1}--\eqref{comp2}  hold for $n$.

To derive the second inequalities in  \eqref{comp1}--\eqref{comp2}   for $n$
the induction proofs work the same way: once $(\sigma_0>0)$ has been replaced
by $(\sigma_0<0)$   all inequalities in the above calculations have to be reversed.
In \eqref{compaux8}  on the right the ratio  $Z_{n+1,s}^\para(\sigma_0<0)/Z_{n,s}^\para(\sigma_0<0)$
does not vanish as $s\searrow 0$ but it stays bounded a.s.~and again $a(0+,t)=0$ takes
the entire right-hand side to $0$.
\end{proof}

\section{Upper bound for the variance }
The parameter $\para\in(0,\infty)$ can be considered  fixed throughout.  The constants
such as
$C(\para)$ that depend on $\para$
and  appear in all our statements are locally bounded functions of $\para$.  {Sometimes we will suppress the dependence on $\para$ and the constants in the intermediate estimates may change from line to line.}
Recall also the key assumption \eqref{assA} on $t>0$ and $n\in\bN$, namely that

$\abs{t-n \trigamf(\para)}\le A n^{2/3} $.
We develop the upper bounds so that the effect of the constant  $A$ is explicitly present.

\begin{theorem}\label{UBthm}  Assume \eqref{assA}.
 Then there exists a constant $C(\para)<\infty$ such that
\be
\Vvv (\log Z_n^{\para}(t))\le (9A+C(\para)) n^{2/3}.
\ee
\end{theorem}
The value 9 in the statement has no particular meaning, except as a constant
independent of $\para$.
 As a first step we give a bound on the tails of $\sigma_0^\pm$.

\begin{lemma}\label{lem_sigtail} Assume \eqref{assA} and fix $K>0$. Then
we can fix
$C=C(\para)<\infty$ large enough  and $s=s(K,\para)>0$ small enough   such that,
 for $3An^{2/3}\le u\le Kn $,
we have the bound
\be \P(Q_{n,t}(\sigma_0^\pm\ge u)\ge  e^{-s u^2/n})\le \frac{C n^2}{u^4}
E_{n,t} (\sigma_0^\pm) + \frac{C n^2}{u^3}. \label{sigtail_a}
\ee
 \end{lemma}
\begin{proof}We first consider the proof for $\sigma^+$.
Set $\lambda=\para+\frac{bu}{n}$ where $b>0$ will be small.
Superscripts $\para$ and $\lambda$ indicate which parameter is used.
 We have
\bea
Q^\para_{n,t}(\sigma_0^+\ge u)&=&\frac1{Z_{n}^\para(t) } \nn
\int_{-\infty<s_{0}<\dotsm<s_{n-1}<t}  \m1(s_0\ge u)
 \\
 &&\qquad  \times  \exp\bigl[ -B(s_0)+\para s_0 +  B_1(s_0,s_1)
+\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}.\nn\\
&\le&\frac1{Z_{n}^\para(t) } \nn
\int_{-\infty<s_{0}<\dotsm<s_{n-1}<t}  \m1(s_0\ge u) e^{(\para-\lambda)u}
 \\
 &&\qquad  \times  \exp\bigl[ -B(s_0)+\lambda s_0 +  B_1(s_0,s_1)
+\dotsm + B_n(s_{n-1},t)\bigr] \,ds_{0,n-1}.\nn\\
&\le& \frac{Z_n^{\lambda}(t)}{Z_{n}^\para(t) }e^{(\para-\lambda) u}.\nn
\eea
Thus
\bea \nn
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su^2/n})&\le&
\P\bigl(\,\frac{Z_n^{\lambda}(t)}{Z_{n}^\para(t) }e^{(\para-\lambda) u}\ge e^{-su^2/n}\bigr)\\
&=&\P(\log( Z_n^{\lambda}(t))-\log( Z_n^{\para}(t))\ge (\lambda-\para)u-su^2/n)\nn
\eea
From \eqref{rB1} and $e^{-r_k(t)}\sim$ Gamma($\para$,1) (Theorem \ref{burkethm})
\be\label{ElogZ}
\E \log( Z_n^{\para}(t))=-n \digamf(\para)+\para t
\ee
so upon centering
\bea\nn
&&\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su^2/n})\le \\
\nn
&&\hskip15mm\P\bigl\{\overline{\log( Z_n^{\lambda}(t))}-\overline{\log( Z_n^{\para}(t))}\ge
n(\digamf(\lambda)-\digamf(\para))-{t}(\lambda-\para)+(\lambda-\para)u-su^2/n).
\eea

Given $K$ we can restrict $b=b(K,\para)$ small enough so that
$\abs{\lambda-\para}\le bK\le \para/2$ and then a
  Taylor expansion gives
\[
|\digamf(\lambda)-\digamf(\para)-(\lambda-\para) \trigamf(\para)|\le C(\para) (\lambda-\para)^2.
\]
Together with $|t-n\trigamf(\para)|\le A n^{2/3}$ this leads to
\begin{align*}
&n(\digamf(\lambda)-\digamf(\para)-\frac{t}{n}(\lambda-\para))+(\lambda-\para)u-su^2/n\\
&\qquad \ge
-C(\para) b^2 u^2/n-A b u n^{-1/3}+  b u^2/n-s u^2/n\\
&\qquad\ge \frac{bu^2}{3n} \bigl(1-3C(\para)b\bigr)  +  \frac{bu}{3n^{1/3}}(un^{-2/3}-3A)
 + \frac{bu^2}{3n}( 1-3s/b)  \\
&\qquad \ge C(\para) u^2/n.
\end{align*}
The last line, with  $C(\para)>0$,  follows  by enforcing $u\ge 3An^{2/3}$,
  choosing $s=b/3$ and then taking $b=b(\para)$ small enough. Put  this back above and apply Chebyshev's inequality:
\bea \nn
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su^2/n})&\le&\P(\overline{\log( Z_n^{\lambda}(t))}-\overline{\log( Z_n^{\para}(t))}\ge C u^2/n)\\ \nn
 &\le& C \frac{n^2}{u^4}
\Vvv\bigl[\log Z_n^{\lambda}(t)-\log Z_n^{\para}(t)\bigr]\\ \label{anotherUP}
&\le& 2C \frac{n^2}{u^4}\bigl(\Vvv[\log( Z_n^{\lambda}(t))]+\Vvv[\log( Z_n^{\para}(t))] \bigr).
\eea
{Using Lemma \ref{4.1lem} below together with  the definition of  $\lambda$:}
\begin{align*}
\Vvv(\log( Z_n^{\lambda}(t))&\le \Vvv(\log( Z_n^{\para}(t)) + n\abs{\trigamf(\lambda)-\trigamf(\para)}\\
&\le  \Vvv(\log( Z_n^{\para}(t)) +  C(\para) bu  \le   \Vvv(\log( Z_n^{\para}(t)) +  C(\para) u.
\end{align*}
We can drop $b$ from the upper bound above because the earlier requirements force
$b$ to be small. So we can safely assume  form the beginning that $b\le 1$.

Continuing from (\ref{anotherUP}),
\bea\nn
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su^2/n})

&\le& C \frac{n^2}{u^4}\Vvv(\log( Z_n^{\para}(t)) + C\frac{n^2}{u^3}.
\eea
Using Theorem \ref{varthm} and once more $u\ge 3An^{2/3}$  we get
\bea\nn
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su^2/n})
&\le& C \frac{n^2}{u^4}E_{n,t} (\sigma_0^+) +CA n^{8/3} u^{-4} + C\frac{n^2}{u^3}\\ \nn
&\le&  C \frac{n^2}{u^4}E_{n,t} (\sigma_0^+) +C \frac{n^2}{u^3}
\eea
which is exactly what we wanted to prove.

To proof for $\sigma^-$ starts with  $\lambda=\para-\frac{bu}{n}$ with $b>0$ small.
Using  \be\nn \ind(s_0\le -u) e^{\para s_0}\le e^{-(\para-\lambda) u} \ind(s_0\le -u) e^{\lambda s_0}\ee
we get
\be\nn
Q_{n,t}^{\para}(\sigma_0^{-}\ge u)\le \frac{Z_{n}^{\lambda}(t)}{Z_n^{\para}(t)} e^{-(\para-\lambda)u}
\ee
and the rest of the proof goes the same way.
\end{proof}

\begin{lemma}\label{4.1lem}
For  $\para,  \lambda>0$,
\[
\bigl\lvert \Vvv(\log( Z_n^{\lambda}(t)) - \Vvv(\log( Z_n^{\para}(t)) \bigr\rvert
\le  n\abs{\trigamf(\lambda)-\trigamf(\para)}.
\]
\end{lemma}
\begin{proof}
Assume $\lambda>\para$.  From Theorem \ref{varthm}
\begin{align*}
\Vvv(\log( Z_n^{\lambda}(t))-\Vvv(\log( Z_n^{\para}(t))&=n(\trigamf(\lambda)-\trigamf(\para))
+2 \bigl[ E^\lambda_{n,t} (\sigma_0^+) -  E^\para_{n,t} (\sigma_0^+)\bigr]\\
&=-n(\trigamf(\lambda)-\trigamf(\para))+2 \bigl[ E^\lambda_{n,t} (\sigma_0^-) -
 E^\para_{n,t} (\sigma_0^-)\bigr].
\end{align*}
From the definition (\ref{Qdef2}) of  the quenched expectation
\begin{align*}
\frac\partial{\partial\para}E^{Q^{\para}} (\sigma_0^{\pm})= \Cov^{Q^\para}(\sigma_0,\sigma_0^{\pm}).
\end{align*}
For any random variable
\begin{align*}
\Cov(X,X^{\pm})&=E ((X^+-X^-)X^\pm)-E (X^+-X^-)E X^\pm \\
&=\pm\Var (X^\pm) \pm E X^+ E X^-.
\end{align*}
Consequently
\[
 E^\lambda_{n,t} (\sigma_0^-) -  E^\para_{n,t} (\sigma_0^-)\le 0 \le
E^\lambda_{n,t} (\sigma_0^+) -  E^\para_{n,t} (\sigma_0^+).
\]
The claim follows.
\end{proof}
Next the tail bound for $\sigma_0^\pm$ for larger deviations.

\begin{lemma}   Assume \eqref{assA}.  Let $\delta>0$.
 Then there exist  $c=c(\para, \delta)<\infty$ and
  $s=s(\para, \delta)>0$ such that for
  $u\ge \max\{\delta n, 3An^{2/3}\} $
\be
\P(Q_{n,t}(\sigma_0^\pm\ge u)\ge e^{-su})\le 2 e^{-c u}.
\ee
\label{ubldlemma}\end{lemma}
\begin{proof}
We do the case $\sigma_0^+$.  The argument for $\sigma_0^-$
is analogous. Set $\lambda=\para+\nu$ where $\nu>0$ is small.
Then
\bea
\nn
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su})&\le& \P(\frac{Z_n^{\lambda}(t)}{Z_{n}^\para(t) }e^{(\para-\lambda) u}\ge e^{-su})\\
&=&\P(\log( Z_n^{\lambda}(t))-\log( Z_n^{\para}(t))\ge (\lambda-\para)u-su).\nn
\eea
After centering:
\bea \nn
&&\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su})\le\\
&&\hskip15mm
\P\bigl(\overline{\log( Z_n^{\lambda}(t))}-\overline{\log( Z_n^{\para}(t))}\ge
n(\digamf(\lambda)-\digamf(\para))-{t}(\lambda-\para)+(\lambda-\para)u-su\bigr).\label{ldp1}
\eea
From a Taylor expansion, $u\ge  \max\{\delta n, 4An^{2/3}\} $, and then fixing $\nu$   small
enough and $s=\nu/2$,
\begin{align*}
&n(\digamf(\lambda)-\digamf(\para))- {t}(\lambda-\para)+(\lambda-\para)u-su \\
&\qquad \ge
-C(\para) \nu^2 n -A \nu n^{2/3} +\nu u-s u\\
&\qquad\ge C(\delta, \para) u.  \nn
\end{align*}
From (\ref{Zincr1}) and (\ref{Z0})
\be \label{diff}
\overline{\log( Z_n^{\lambda}(t))}-\overline{\log( Z_n^{\para}(t))}=\sum_{j=1}^n \overline{r_j^{\lambda}(t)}-\sum_{j=1}^n \overline{r_j^{\para}(t)}
\ee
where, for a fixed $t$,
 $e^{-r^\para_j(t)}$  are i.i.d.~$\textup{Gamma}(\para,1)$ variables, and similarly $e^{-r^\lambda_j(t)}$with parameter $\lambda$.
 Thus for  certain sums $S'_n, S''_n$ of i.i.d.\ mean zero variables with an exponential moment
\[
\P(Q^\para_{n,t}(\sigma_0^+\ge u)\ge e^{-su})\le  \P( S'_n-S''_n\ge Cu) \le {\P(S'_n\ge Cu/2)+\P(S''_n<-Cu/2)\le} e^{-cu}
 \]
where the last inequality follows from $u\ge\delta n$ and   standard large deviation theory.
\end{proof}

Now we combine the two deviation estimates into a  moment bound.

\begin{lemma}   Assume \eqref{assA}.
 Then there exists  $C(\para)<\infty$ such that
\be  E_{n,t}(\sigma_0^\pm) \le (4A+C(\para))n^{2/3}.
\label{sigmamom}\ee
\end{lemma}
\begin{proof}  We do the computation for $\sigma_0^+$. It is identical for $\sigma_0^-$.
  Let $r\ge 1$, to be chosen at the end,
and $B=r\vee (3A)$.  Begin with
\begin{align}
 E_{n,t}(\sigma_0^+) \le Bn^{2/3} + \int_{Bn^{2/3}}^{n\vee (Bn^{2/3})} P(\sigma_0^+\ge u)\,du
 + \int_{n\vee (Bn^{2/3})}^\infty P(\sigma_0^+\ge u)\,du.
\label{int3}\end{align}
The last integral in \eqref{int3} is bounded by a constant that depends on $\para$,
 uniformly over $r>0$, $A>0$ and  $n\ge 1$,
as can be seen by an application of  Lemma \ref{ubldlemma} with $\delta=1$.
The middle integral is bounded as follows, utilizing Lemma \ref{lem_sigtail} with $K=1$.
\begin{align*}
  &\int_{Bn^{2/3}}^{n\vee (Bn^{2/3})} P(\sigma_0^+\ge u)\,du \\
 &\le   \int_{Bn^{2/3}}^{n\vee (Bn^{2/3})} \left\{ e^{-s u^2/n}+\P(Q_{n,t}(\sigma_0^+\ge u) \ge e^{-su^2/n})\right\}\,du
\\
   &\le \int_{Bn^{2/3}}^\infty   \biggl(   \frac{C(\para) n^2}{u^4}
E_{n,t} (\sigma_0^+) + \frac{C(\para) n^2}{u^3}  \biggr) \,du   + C(\para) \\[3pt]
&\le  \frac{C(\para)}{B^3} E_{n,t} (\sigma_0^+) + \frac{C(\para) }{B^2}n^{2/3}  + C(\para).   \end{align*}
Combine the estimates, noting that $B\ge 1$ and $n\ge 1$,  to
\[    E_{n,t}(\sigma_0^+) \le (B+C(\para)) n^{2/3} + \frac{C(\para)}{r^3} E_{n,t} (\sigma_0^+). \]
Choose $r=(4C(\para))^{1/3}$.  Rearranging gives the conclusion.
\end{proof}

Theorem \ref{UBthm} is now proved by \eqref{sigmamom}, the variance identity \eqref{var1}
and assumption \eqref{assA}.   For future reference let us also state tail bounds
on $\sigma_0^\pm$ that we obtain by combining \eqref{sigmamom} with   Lemmas
 \ref{lem_sigtail} and \ref{ubldlemma}.

 \begin{proposition}
Under assumption \eqref{assA} we have these tail bounds, for finite positive  constants
$C$, $c$ and  $s$ that depend on $\para$, and for $b>0$:
\be  \label{sigmatail1}
\P(Q_{n,t}({\sigma_0}^\pm \ge b n^{2/3})\ge  e^{-s b^2 n^{1/3}})\le C b^{-3}
\quad\text{for $3A\le b\le n^{1/3}$,}
\ee
\be  \label{sigmatail2}
\P(Q_{n,t}({\sigma_0}^\pm \ge b n^{2/3})\ge  e^{-s b n^{2/3}})\le 2e^{-cbn^{2/3}}
\quad\text{for $b\ge n^{1/3}\vee(3A)$}
\ee
and
\be
\label{sigmatail}
P_{n,t}(\sigma_0^\pm \ge b n^{2/3})\le C b^{-3}\quad\text{for $b\ge 3A$}.
\ee
\label{sigmaprop}\end{proposition}
From \eqref{sigmatail} we get the moment bound
\be   E_{n,t}(\,\abs{\sigma_0}^p\,) \le (3^pA^p+C(\para))n^{2p/3}\quad
\text{for $1\le p<3$.}
\label{sigmapmom}\ee

\section{Lower bound on the variance}
\begin{theorem}\label{LBthm}
Assume that   \eqref{assA} holds. Then there exists a constant
$C_1=C_1(A,\para)$ such that
\be
\Vvv (\log Z_n^{\para}(t))\ge C_1 n^{2/3}.
\ee
\end{theorem}

The   estimate that gives the theorem is in the next proposition.

\begin{proposition} Assume that  \eqref{assA} holds with a constant
$A\in\bR_+$.
  Then there exist
finite positive $\para$-dependent
constants $C(\para), c(\para), D(\para) $ so that,  if  $0<\delta\le 1$ and $K\ge 1$
satisfy \[  D(\para)(A+1)\delta^{1/2}\le K\le c(\para)(A+1)^{-4}\delta^{-1/2}, \]
then
\be \nn   \varlimsup_{n\to \infty} \P\bigl(Q^\para_{n,t}(0< \sigma_0\le \delta n^{2/3})>e^{-K n^{1/3} \sqrt{\delta}}\,\bigr)\le C(\para)(e^{-K^2/16}+ K^{3/4} \delta ^{3/8}).
\ee
\label{LBprop1}\end{proposition}

\begin{remark}
As a corollary we   get the following more general statement. Fix $x\in\bR$ and assume
that
 \be  D(\para)(A+\abs{x}+1)\delta^{1/2}\le K\le c(\para)(A+\abs{x}+1)^{-4}\delta^{-1/2}.
 \label{LBass9}\ee
Then
\be  \varlimsup_{n\to \infty} \P(Q^\para_{n,t}(x n^{2/3}< \sigma_0\le (x+\delta) n^{2/3})>e^{-K n^{1/3} \sqrt{\delta}})\le C(e^{-K^2/16}+ K^{3/4} \delta ^{3/8}).
\ee
This follows because by the translation invariance   \eqref{altQ2}
\be\nn
Q^\para_{n,t}(x n^{2/3}< \sigma_0\le (\delta+x) n^{2/3})\eqd Q^\para_{n,t-x n^{2/3}}(0< \sigma_0\le \delta n^{2/3})
\ee
and  $\abs{t-xn^{2/3}-n \trigamf(\para)}\le (A+\abs{x}) n^{2/3}$.
In particular, with $x=-\delta$, we get the  matching estimate for $\sigma_0^-$, and we can
combine the estimates for $\sigma^\pm_0$:  under assumptions \eqref{assA} and \eqref{LBass9} with $x=-\delta$,
\be    \varlimsup_{n\to \infty} \P\bigl(Q^\para_{n,t}(\abs{\sigma_0}\le \delta n^{2/3})>2e^{-K n^{1/3} \sqrt{\delta}}\,\bigr)\le C(\para)(e^{-K^2/16}+ K^{3/4} \delta ^{3/8}).
\label{LBprop1bound3}\ee
\end{remark}

Before proving
  Proposition \ref{LBprop1}  let us observe how   Theorem \ref{LBthm} is proved.
Estimate \eqref{LBprop1bound3} gives the annealed limit
\[
 \lim_{\delta \searrow 0} \varlimsup_{n\to \infty} P^\para_{n,t}(\abs{\sigma_0}\le \delta n^{2/3})=0.
\]
Then by the  variance identity  \eqref{var1}
\[
\Vvv(\log Z_n^{\para}(t))
=E_{n,t}^\para\bigl(\abs{\sigma_0}\bigr) \ge \delta n^{2/3} P^\para_{n,t}(\abs{\sigma_0}\ge \delta n^{2/3}) \]
and  Theorem \ref{LBthm} follows.

\begin{proof}[Proof of Proposition \ref{LBprop1}]
Set $u=\delta n^{2/3}, \upsilon(\delta)=K \sqrt{\delta}$ and begin by writing
\begin{align}
&\P(Q^\para_{n,t}(0< \sigma_0\le \delta n^{2/3})>e^{-n^{1/3} \upsilon(\delta)})=\P\left(\frac{Q^\para( \sigma_0>u\textup{ or } \sigma_0<0)}{Q^\para( 0< \sigma_0 \le u)}<e^{n^{1/3} \upsilon(\delta)}-1\right)\\
&\qquad\qquad
=\P\left(\frac{Z^\para_{n,t}( \sigma_0>u \textup{ or } \sigma_0<0)}{Z^\para_{n,t}( 0< \sigma_0 \le u)}<e^{n^{1/3} \upsilon(\delta)}-1\right)\nn \\
&\qquad\qquad
=\P\left(\frac{Z^\para_{n,t}( \sigma_0>u\textup{ or } \sigma_0<0)}{Z^\para_{n,t}( 0< \sigma_0 \le u)}\cdot\frac{Z_{1,n}(0,t)}{Z_{1,n}(0,t)}<e^{n^{1/3} \upsilon(\delta)}-1\right)\nn .
\end{align}
Split the last probability to get
\begin{align}
&\P(Q^\para_{n,t}(0< \sigma_0\le\delta n^{2/3})>e^{-n^{1/3} \upsilon(\delta)})
\ \le \ \P\left(\frac{Z^\para_{n,t}( \sigma_0>u\textup{ or } \sigma_0<0)}{Z_{1,n}(0,t)}<e^{2 n^{1/3} \upsilon(\delta)}\right)\nn\\[4pt]
&\qquad\qquad\qquad \qquad\qquad
 +\P\left(\frac{Z^\para_{n,t}( 0< \sigma_0\le u)}{Z_{1,n}(0,t)}> e^{n^{1/3}  \upsilon(\delta)} \frac1{1-e^{-n^{1/3} \upsilon(\delta)}}\right)\nn\\[4pt]
&\qquad\qquad
 \le \ \P\left(\frac{Z^\para_{n,t}( \sigma_0>u)}{Z_{1,n}(0,t)}<e^{2 n^{1/3} \upsilon(\delta)}\right)
\label{LB1}\\[4pt]
&\qquad\qquad \qquad \qquad\qquad
+ \ \P\left(\frac{Z^\para_{n,t}( 0< \sigma_0\le u)}{Z_{1,n}(0,t)}> e^{n^{1/3}  \upsilon(\delta)}\right). \label{LB2}
\end{align}
We   bound    probabilities in (\ref{LB1}) and (\ref{LB2})  separately.

\medskip
\noindent \textbf{The term (\ref{LB1}).}
\be \nn
\frac{Z^\para_{n,t}( \sigma_0>u)}{Z_{1,n}(0,t)}=\int_u^t \exp(-B(s)+\para s) \frac{Z_{1,n}(s,t)}{Z_{1,n}(0,t)} ds.
\ee
Construct a new environment $\tilde\om$  with
\be\nn
\tilde B(s)=-(B_n(t)-B_n(t-s)), \quad \tilde B_i(s)=B_{n-i}(t)-B_{n-i}(t-s), \quad 1\le i \le n-1,
\ee
and take a new parameter $\lambda=\para+a(\delta) n^{-1/3}$ where $a(\delta)=K^{-1/4} \delta^{-1/8}$.
Quantities  that use environment $\tilde\om$ are marked with a tilde.
From the definitions one checks that
\be
Z_{1,n}(s,t)=\tilde Z_{0,n-1}(0,t-s) \quad\text{ for any $t>0$ and  $s\in(-\infty,t)$.}
\label{ZZtil}\ee
Use  (\ref{comp2}) for the new system to get
\bea \nn
\frac{Z_{1,n}(s,t)}{Z_{1,n}(0,t)}&=&\frac{\tilde Z_{0,n-1}(0,t-s)}{\tilde Z_{0,n-1}(0,t)}\ge \frac{\tZ_{n-1,t-s}^{\lambda}(\sigma_0>0)}{\tZ_{n-1,t}^{\lambda}(\sigma_0>0)}=\frac{\tilde Q^\lambda_{n-1,t-s}(\sigma_0>0) \tZ_{n-1,t-s}^{\lambda}}{\tilde Q^\lambda_{n-1,t}(\sigma_0>0)\tZ_{n-1,t}^{\lambda}}\\
\nn &\ge&\tilde Q^\lambda_{n-1,t-s}(\sigma_0>0)\frac{ \tZ_{n-1,t-s}^{\lambda}}{\tZ_{n-1,t}^{\lambda}}=
\tilde Q^\lambda_{n-1,t-s}(\sigma_0>0) \exp(\tilde Y_{n-1}(t-s,t)-\lambda s).
\eea
Thus, denoting   the probability in \eqref{LB1}  by $p_1$,
\begin{align*}
p_1&=\P\left(\frac{Z^\para_{n,t}( \sigma_0>u)}{Z_{1,n}(0,t)}<e^{2 n^{1/3} \upsilon(\delta)}\right)\\
&\le
\P\left(
\int_u^t  e^{-B(s)+\tilde Y_{n-1}(t-s,t)+(\para-\lambda) s}

\,\tilde Q^\lambda_{n-1,t-s}(\sigma_0>0) \,ds<e^{2 n^{1/3} \upsilon(\delta)}
\right) \\
&=  \P\left(
\int_u^t  e^{-B(s)+\tilde Y^*_{n-1}(t-s,t)+(\para-\lambda) s}

\,\tilde Q^{\lambda,*}_{n-1}(t-s-\sigma^*_{n-1}>0) \,ds<e^{2 n^{1/3} \upsilon(\delta)}
\right).
\end{align*}
On the last line above we applied the $*$ transformation to the $\tilde \omega$ system and Lemma \ref{LEMrev}
to replace the measure $\tilde Q^{\lambda, \tilde \om^*}$ with the dual measure $\tilde Q^{\lambda,*}$.

 Set
\[
\bar u=n\trigamf(\para)-(n-1)\trigamf(\lambda).
\]
Given any $\delta$ and $K$,  there exists a constant $c_0(\para)>0$ such that
   $\bar u/2\ge c_0(\para)a(\delta)n^{2/3}$  for large enough $n$.
Furthermore, from the hypothesis on    $\delta$ and $K$ it follows that
  $\bar u/2\ge u$ and $\bar u/2\ge 3An^{2/3}$.
Restrict the integration inside the probability and decompose  again:
\begin{align}
p_1&\le \P\left(
\int_u^{\bar u/2}   e^{-B(s)+\tilde Y^*_{n-1}(t-s,t)+(\para-\lambda) s}

\,\tilde Q^{\lambda,*}_{n-1}(t-\sigma^*_{n-1}>\bar u/2) \,ds<e^{2 n^{1/3} \upsilon(\delta)}
\right)\nn\\
&\label{eq:lower0}\le \P\left(\tilde Q^{\lambda,*}_{n-1}(t-\sigma^*_{n-1}>\bar u/2)\le 1/2\right)\\
&\label{eq:lower1} \qquad +\;
\P\left(
\int_u^{\bar u/2} \exp(-B(s)+\tilde Y^*_{n-1}(t-s,t)+(\para-\lambda) s) ds<2e^{2 n^{1/3} \upsilon(\delta)}
\right).
\end{align}
For probability (\ref{eq:lower0}) switch to complements, apply Lemma \ref{LEMrev} again,
and then  the upper bound (\ref{sigmatail1}):
\be\begin{aligned}
 &\P\left(\tilde Q^{\lambda,*}_{n-1}(t-\sigma^*_{n-1}>\bar u/2)\le  1/2\right)
 = \P\left(\tilde Q^{\lambda,*}_{n-1}(t-\bar u-\sigma^*_{n-1}\le -\bar u/2)> 1/2\right)\\
&\qquad   =\P\left( \tilde Q^\lambda_{n-1,t-\bar u}(\sigma_0 \le -\bar u/2) >1/2\right)
 \le C(\para) a(\delta)^{-3}=C(\para)K^{3/4}\delta^{3/8}. \end{aligned}\label{eq:lower2}\ee
On the last line above assumption \eqref{assA} continues to be valid with the
same constant $A$ because
$
\abs{t-\bar u -(n-1)\trigamf(\lambda)}=\abs{t-n\trigamf(\para)}\le An^{2/3}.
$
Property $\bar u/2\ge 3An^{2/3}$ is the assumption needed for (\ref{sigmatail1}).

Next we  estimate probability (\ref{eq:lower1}).
Process $s\mapsto \tilde Y^*_{n-1}(t-s,t)$  is
  a standard Brownian motion, and   independent of $B$ because
$\tilde Y^*$ was constructed from the new environment $\tilde\om$.   Define another
standard Brownian motion
\be\nonumber B^\dagger (s)= 2^{-1/2}n^{-1/3}  \bigl(-B(n^{2/3}s)+\tilde Y^*_{n-1}(t-n^{2/3}s,t)\bigr).\ee
Then  probability \eqref{eq:lower1} equals
\begin{align}
&\P\left(
\int_u^{\bar u/2} \exp( \sqrt{2}n^{1/3}B^\dagger(n^{-2/3}s)+(\para-\lambda) s) \,ds<2e^{2 n^{1/3} \upsilon(\delta)}
\right)\nn\\
&\qquad \le
\P\left(n^{2/3}
\int_\delta^{c_0(\para)a(\delta)} \exp(\sqrt{2} n^{1/3} B^{\dagger}(s)-a(\delta) n^{1/3} s) \,ds<2e^{2 n^{1/3} \upsilon(\delta)}
\right)\nn\\
&\qquad =
\P\left(n^{-1/3} \log
\int_\delta^{c_0(\para)a(\delta)}  e^{\sqrt{2} n^{1/3} B^{\dagger}(s)-a(\delta) n^{1/3} s}

\,ds\le2 \upsilon(\delta) +n^{-1/3} \log(2n^{-2/3})
\right)  \nn\label{eq:lower2}
\end{align}
 As $n\to\infty$ the probability on the last line above converges to
\[
\P\left(\sup_{\delta\le s\le c_0(\para)a(\delta)}(
\sqrt{2} B^{\dagger}(s)-a(\delta)  s) \le2 \upsilon(\delta)
\right).
\]
 Introduce one more   Brownian motion
$B(s)= B^{\dagger}(\delta+s)-B^{\dagger}(\delta)$.
Abbreviate temporarily  $\tau=a(\delta)^2 (c_0(\para)a(\delta)-\delta)/\sqrt 2 >1$
where the inequality is a consequence of the assumption on $\delta$ and $K$.
 Then the
probability above is
\begin{align*}
&\le \P\bigl(\sqrt{2} B^{\dagger}(\delta)<- \upsilon(\delta)\bigr)
+\P\left(\sup_{0\le s\le c_0(\para)a(\delta)-\delta}(
\sqrt{2} B(s)-a(\delta)  s) \le 3 \upsilon(\delta)+\delta a(\delta)
\right)\nn\\ \nn
&\le  e^{- \frac14\upsilon(\delta)^2 \delta^{-1}}
+\P\left(\sup_{0\le s\le  \tau}(
 B(s)-  s) \le 2^{-1/2}a(\delta)(3 \upsilon(\delta)+\delta a(\delta) )
\right)\\
&\le  e^{- \frac14\upsilon(\delta)^2 \delta^{-1}}
+   \P\left(\sup_{0\le s\le 1}(
 B(s)-  s) \le 2^{-1/2}a(\delta)(3 \upsilon(\delta)+\delta a(\delta))
\right)\\[3pt]
&\le e^{- \frac14\upsilon(\delta)^2 \delta^{-1}}
+  C a(\delta)(3\upsilon(\delta)+\delta a(\delta) ).
\end{align*}
The last inequality comes because  $\sup_{0\le s\le 1} ( B(s)-  s)$ is a.s.~positive with a bounded density function. Including the estimate from \eqref{eq:lower2}  we get
\be \label{eq:LB1}
\varlimsup_{n\to \infty} \, (\ref{LB1})\le  C(\para) (e^{- K^2/4  }+ K^{3/4}\delta^{3/8}
 +  K^{-1/2}\delta^{3/4}  ) \le   C(\para)(e^{- K^2/4  }+ K^{3/4}\delta^{3/8}) .
\ee
{In the last inequality we used $\delta\le 1\le K$.}

\medskip
\noindent \textbf{The term (\ref{LB2}).}

For  probability  (\ref{LB2}) we separate the argument into a lemma because the
same estimate will be needed again, though with different parameters.

\begin{lemma} Assume \eqref{assA} with the constant $A$ and let $\aaa,\bbb, \kappa>0$.
Then there exist finite, positive
constants $C(\para)$, $C_1(\para)$ and  $n_0(a,b,\kappa,\para)$ such that, if $b/a\ge C_1(\para)(A+1)$, then
for $n\ge n_0(a,b, \kappa, \para)$
\[ \P\left(\frac{Z^\para_{n,t}( 0< \sigma^\pm_0\le \aaa n^{2/3})}{Z_{1,n}(0,t)}\ge \kappa e^{n^{1/3} \bbb}\right)
\le
C(\para) \bigl( a^3b^{-3} + \exp(-b^2a^{-1}/16) \bigr).
  \]
\label{LBlemma3}\end{lemma}

Before proving the lemma let us use it to conclude the proof of Proposition \ref{LBprop1}.
In Lemma \ref{LBlemma3}  take $\kappa=1$, $\aaa=\delta$ and $\bbb=K\sqrt\delta$.
 Then for large enough $n$,
\be
 \text{probability (\ref{LB2}) } \le C(\para) (\delta^{3/2}K^{-3}+ e^{-K^2/16}).
\label{eq:LB2}\ee

Combine (\ref{eq:LB1}) and  (\ref{eq:LB2})  with $\delta\le 1\le K$, and we have
\bea\nn
 \varlimsup_{n\to \infty} \P(Q^\para_{n,t}(0< \sigma_0\le \delta n^{2/3})>e^{-n^{1/3} \upsilon(\delta)})
  \le   C(\para) ( e^{-K^2/16}+  K^{3/4} \delta ^{3/8}   )
\nn
\eea
and  the proposition is proved.
 \end{proof}

\begin{proof}[Proof of Lemma \ref{LBlemma3}]
  We do the case of $\sigma^+_0$ in full detail.  Abbreviate $u=\aaa n^{2/3}$.
Introduce the new environment $\tilde \omega$ as before, and a new parameter
  $\lambda=\para-r  n^{-1/3}$ with $r =b/(4a)$.
 We must restrict $n$ large enough so that for example $rn^{-1/3}<\para/2$ so that
 $\lambda$ is a legitimate parameter.

Begin with \eqref{ZZtil}   and then apply   comparison  \eqref{comp2}:
\begin{align*}
\frac{Z_{1,n}(s,t)}{Z_{1,n}(0,t)}& =\frac{\tilde Z_{0,n-1}(0,t-s)}{\tilde Z_{0,n-1}(0,t)} \le \frac{\tZ_{n-1,t-s}^{\lambda}(\sigma_0<0)}{\tZ_{n-1,t}^{\lambda}(\sigma_0<0)}=
\frac{\tZ_{n-1,t-s}^{\lambda}}{\tZ_{n-1,t}^{\lambda}}\cdot \frac{\tilde Q^{\lambda}_{n-1,t-s}(\sigma_0<0)}{\tilde Q^{\lambda}_{n-1,t}(\sigma_0<0)}\\
&\le  \exp\bigl(\tilde Y_{n-1}(t-s,t)-\lambda s\bigr)\cdot \frac{1}{\tilde Q^{\lambda}_{n-1,t}(\sigma_0<0)}.
\end{align*}
Substitute the above bound in the probability that is to be bounded:
\bea \nn&&
\P\left(\frac{Z^\para_{n,t}( 0< \sigma_0\le u)}{Z_{1,n}(0,t)}\ge\kappa e^{n^{1/3}\bbb }\right)\\
&&\hskip10mm \nn =\P\left(\int_0^u \exp(-B(s)+\para s) \frac{Z_{1,n}(s,t)}{Z_{1,n}(0,t)}\, ds
\ge \kappa  e^{n^{1/3}  b}  \right)\\
&&\hskip10mm \nn \le
\P\left(\int_0^u \frac{ \exp(-B(s)+\tilde Y_{n-1}(t-s,t)+(\para-\lambda) s)  }{\tilde Q^{\lambda}_{n-1,t}(\sigma_0<0)}\,ds\ge\kappa e^{n^{1/3}  \bbb}\right)\\[3pt]
&&\hskip10mm
\le\P\left(\tilde Q^{\lambda}_{n-1,t}(\sigma_0<0)\le 1/2\right)\label{eq:lower3.9} \\
&&\hskip20mm
+ \ \P\left(\int_0^u \exp(-B(s)+\tilde Y_{n-1}(t-s,t)+(\para-\lambda) s) \, ds\ge \frac\kappa{2}  e^{n^{1/3}  \bbb}\right).\label{eq:lower4}
\eea

To treat probability (\ref{eq:lower3.9}) set
$\bar u=  (n-1)\trigamf(\lambda)-n\trigamf(\para)$.
$\trigamf$ is positive, convex and strictly decreasing, so one can check that
 $\bar u \ge\tfrac14 \abs{\trigamf'(\para)} r n^{2/3}$ for  all $n\ge 1$ provided $C_1(\para)$ in the hypothesis
is large enough.   Use the shift
invariance property of $Q$ described in Remark \ref{remark1} and the upper bound (\ref{sigmatail1}):
\be\begin{aligned}
&\P(\tilde Q^{\lambda}_{n-1,t}(\sigma_0<0)<1/2)=\P(\tilde Q^{\lambda}_{n-1,t}(\sigma_0>0)\ge1/2)\\
&\qquad =

\P(\tilde Q^{\lambda}_{n-1,t+\bar u}(\sigma_0>\bar u)\ge1/2) \le C(\para) r^{-3}
\le C(\para)(a/b)^3.
 \end{aligned}\label{auxlemma8}\ee
The choice of $\bar u$ makes \eqref{assA} valid again with the same $A$, and
a large enough $C_1(\para)$  guarantees that $\bar u\ge 3An^{2/3}$ so that
(\ref{sigmatail1}) can be applied.

For probability (\ref{eq:lower4}), after rescaling the integral and introducing a new Brownian motion,
\begin{align*}
&\text{(\ref{eq:lower4})}  \le \P\left(n^{2/3}\int_0^\aaa \exp(n^{1/3}(\sqrt{2}B^{\dagger}(s)+r  s))  ds>\frac{\kappa}2  e^{n^{1/3}  \bbb}\right)\\
&  =\P\left(n^{-1/3}\log\int_0^\aaa \exp(n^{1/3}(\sqrt{2}B^{\dagger}(s)+r  s))  ds> \bbb +n^{-1/3} \log( \kappa n^{-2/3}/2)\right)\\
&\le
\P\left(\sup_{0\le s\le \aaa}(
\sqrt{2} B^{\dagger}(s)+r   s) \ge \tfrac34\bbb
\right).
\end{align*}
In the last inequality we took $n$ large enough so that $n^{-1/3} \log( \kappa n^{-2/3}/2)< b/4$.
Via $\sup_{0\le s\le \aaa}
 B^{\dagger}(s)\eqd {\aaa}^{1/2} \abs{B^{\dagger}(1)}$  bound the last probability by
\[
\P\left(\sup_{0\le s\le \aaa}
\sqrt{2} B^{\dagger}(s)\ge \tfrac34b-r  \aaa
\right)\le C\exp\left( -\,\frac1{4\aaa}{( \,\tfrac34b-r  \aaa)^2}\right)
=C\exp\left(-\,\frac{b^2}{16\aaa}\right).
\]
Combining estimate \eqref{auxlemma8} with above gives the conclusion for $\sigma^+_0$.

The case of $\sigma^-_0$ goes similarly, with small alterations.
Now $\lambda=\para+rn^{-1/3}$.  Utilizing
\eqref{ZZtil} and comparison \eqref{comp2}
the ratio is  developed as follows:
\begin{align*}
&\frac{Z^\para_{n,t}( -u\le  \sigma_0<0)}{Z_{1,n}(0,t)}
=
 \int_{-u}^0 \exp(-B(s)+\para s) \frac{Z_{1,n}(s,t)}{Z_{1,n}(0,t)}\, ds\\
 &\qquad \le  \int_{-u}^0 \frac{ \exp(-B(s)-\tilde Y_{n-1}(t,t-s)-(\para-\lambda) s)  }{\tilde Q^{\lambda}_{n-1,t}(\sigma_0>0)}\,ds.
\end{align*}
The rest follows along the same lines as above. With this we consider
Lemma \ref{LBlemma3}   proved.
\end{proof}

\section{Fluctuations of the path under boundary conditions}

\begin{theorem}
Assume \eqref{assA} holds,  let $0<\gamma<1$ and assume $b\ge 3(A+1)$.  Then for
$n\ge (1-\gamma)^{-1}$
\be \label{path1}
P(\abs{\sigma_{\lfloor \gamma n \rfloor }-\gamma t}>b n^{2/3})\le C(\para) b^{-3}.
\ee

Also, for any $0<\gamma<1, \varepsilon>0$ there exists $\delta>0$ with
\be\label{path2}
\varlimsup_{n\to \infty}P(|\sigma_{\lfloor \gamma n \rfloor }-\gamma t|\le \delta n^{2/3})\le \varepsilon.
\ee

\end{theorem}

\begin{proof}
For the first statement it is enough to prove that
\be \label{eq:path1} Q_{n,t}(\sigma_k-v>u)\eqd Q_{n-k,t -v }(\sigma_0>u).\ee
Indeed, from this identity we get
\be
P_{n,t}(|\sigma_{\lfloor\gamma n\rfloor}-\gamma t |>b n^{2/3})=P_{n-\lfloor\gamma n\rfloor, (1-\gamma)t} (|\sigma_0|>b n^{2/3})\le C(\para) b^{-3}
\ee
where the last inequality comes from  applying (\ref{sigmatail}).  This is legitimate
because
\[\abs{(1-\gamma)t-(n-\lfloor\gamma n\rfloor)\trigamf(\para)}\le (A+1)n^{2/3}.\]
Condition $n\ge (1-\gamma)^{-1}$ ensures that $n-\lfloor\gamma n\rfloor\ge 1$.

By Lemma \ref{LEMrev}, to prove (\ref{eq:path1})   it is enough to show that the  distribution of $(\sigma_1^*,\dots, \sigma_{n-1}^*)$ is the same under $Q^*_n$ and $Q^*_{n-1}$.  For this
 we check that integrating out $\sigma_{n}^*$ from the density function of $Q^*_n$ results
in the density of $Q^*_{n-1}$.  In the next calculation use \eqref{rXY} and \eqref{defYk}.
\begin{align*}
&\int_{s_{n-1}}^{\infty}\frac1{Z_n^*} \exp\left[ X_1(0,s_1)+X_2(s_1,s_2)+\dotsm+ X_n(s_{n-1},s_{n})+Y_n(0,s_{n})-\para s_{n}\right] ds_{n}\\
&  =\frac1{Z_n^*} \exp\left[ X_1(0,s_1) +\dotsm+ X_{n-1} (s_{n-2},s_{n-1})+Y_n(0,s_{n-1})-\para s_{n-1}\right] \\ \nn
& \qquad \times
\int_{s_{n-1}}^\infty  \exp\left[X_{n}(s_{n-1},s_n)+Y_n(s_{n-1}, s_n)-\para (s_n-s_{n-1}) \right]ds_{n}\\
& =
\frac1{Z_n^*} \exp\left[ X_1(0,s_1) +\dotsm+ X_{n-1} (s_{n-2},s_{n-1})+Y_n(0,s_{n-1})-\para s_{n-1}\right] e^{r_n(s_{n-1})}\nn
\\
& =\frac{e^{r_n(0)}}{Z_n^*} \exp\left[ X_1(0,s_1) +\dotsm+ X_{n-1} (s_{n-2},s_{n-1})+Y_{n-1}(0,s_{n-1})-\para s_{n-1}\right] \nn
\end{align*}
which is exactly the density of $Q^*_{n-1}$ and also shows that $Z_n^*=Z_{n-1}^* e^{r_{n}(0)}$.

 To prove   (\ref{path2}), use  (\ref{eq:path1}) to write
\be\nn
Q_{n,t}(|\sigma_{\lfloor \gamma n\rfloor} -\gamma t|\le \delta n^{2/3})\eqd Q_{n-\lfloor \gamma n\rfloor,(1-\gamma)t}(|\sigma_{0} |\le \delta n^{2/3})
\ee
and apply Proposition \ref{LBprop1}.
 \end{proof}

\section{Upper bounds   without boundary conditions}
\label{sec:freeZ}

\begin{theorem}\label{thm:freeZ}  Let $\tau>0$ and pick $\para$ so that
$\trigamf(\para)=\tau$.
Then for $n\ge n_0(\tau)$ and $b\ge b_0(\tau)$  we have
\be \label{free0}
\P(\abs{\log Z_{1,n}(0,n\tau)-n(\trigamf(\para)\para- \digamf(\para)) }
\ge b n^{1/3})\le C(\tau) b^{-3/2}.
\ee
 \end{theorem}
\begin{proof}
The choice of $\para$ gives $\bE(\log  Z_ {n,n\tau}^{\para})=n(\trigamf(\para)\para- \digamf(\para))$, so by
 Theorem \ref{UBthm} we only need to prove the bound
\be
\P(|\log Z_{1,n}(0,n\tau)-\log Z_ {n,n\tau}^{\para}|\ge b n^{1/3})
\le C b^{-3/2}.
\ee
Abbreviate $t=n\tau$.
By (\ref{ZB2})
\be\label{bound}
Z_{n,t}^\para\ge e^{r_1(0)} Z_{1,n}(0,t).
\ee
This gives
\bea
\P\left( \log Z_ {n,t}^{\para} - \log Z_{1,n}(0,t) \le -b n^{1/3} \right)
\le \P\left(e^{r_1(0)}\le e^{-b n^{1/3}}  \right) \le C e^{-b n^{1/3}}, \nn
\eea
the last inequality follows from  $e^{r_1(0)}\sim \textup{Gamma}(\para,1)^{-1}$ which has bounded density   near 0.

To get the opposite bound set $u=\sqrt{b} n^{2/3}$  and write
\be\begin{aligned}
&\P\left(\frac{Z_{n,t}^\para}{Z_{1,n}(0,t)}\ge e^{b n^{1/3}}\right)=\P\left(\frac{Z_{n,t}^\para(|\sigma_0|\le u)}{Z_{1,n}(0,t)\, Q_{n,t}^\para(|\sigma_0|\le u)}\ge e^{b n^{1/3}}   \right)\\
& \qquad \qquad \le \P\left( \frac{Z_{n,t}^\para(|\sigma_0|\le u)}{Z_{1,n}(0,t) }\ge \frac12 e^{b n^{1/3}}   \right)+
\P\left(Q_{n,t}^\para(|\sigma_0|\le u)\le 1/2\right)\\
&\qquad \qquad \le  C(\para)b^{-3/2}.
\end{aligned} \label{free3}\ee
To get the last inequality, apply Lemma \ref{LBlemma3} with $\aaa=\sqrt b$
to the first probability,
and the upper bound  (\ref{sigmatail1}) to the second probability,
and take both $n$ and $b$ large
enough.
\end{proof}

Theorem \ref{zeta2thm}  is a restatement of Theorem \ref{thm:freeZ}. The next theorem proves Theorem \ref{thm:freepath2}.

\begin{theorem}\label{thm:freepath}
Assume \eqref{assA} holds,

 and  $0<\gamma<1$. Then for large enough $n$
 and $b$    we have
\be
P_{(1,n),(0,t)}\left(|\sigma_{\lfloor n \gamma\rfloor}-\gamma t|>b n^{2/3}   \right)\le C(\para) b^{-3}.
\ee
\end{theorem}
\begin{proof}
Let $\ell =\lfloor n \gamma\rfloor, t'=\gamma t$ and $u=b n^{2/3}$. By the definitions and   (\ref{bound})
\begin{align*}
&Q_{(1,n),(0,t)}\left(| \sigma_{\ell }-t'|>u  \right)=
\frac1{Z_{1,n}(0,t)}   \int\limits_{\abs{s-t'}>u}  {Z_{1,\ell }(0,s) Z_{\ell +1,n}(s,t)} \,ds\\
 &\quad \le \frac{e^{-r_1(0)}}{Z_{1,n}(0,t)}   \int\limits_{\abs{s-t'}>u}  {Z^\para_{\ell ,s}\, Z_{\ell +1,n}(s,t)} \,ds  \; = \;
 \frac{e^{-r_1(0)}Z_{n,t}^\para}{Z_{1,n}(0,t)} \,Q_{n,t}\left(| \sigma_{\ell }-t'|>u  \right) .
\end{align*}

Consider  $h\in(b^{-3},1)$.
\bea \nn
\P\left(Q_{(1,n),(0,t)}\left(| \sigma_{\ell }-t'|>u  \right)>h  \right)&\le& \P(e^{r_1(0)}\le b^{-3})+\P\left[
\frac{Z_{n,t}^\para}{Z_{1,n}(0,t)} \ge e^{r n^{1/3}}
\right]\\
\nn&&+\P\left[
Q_{n,t}\left(| \sigma_{\ell }-t'|>u  \right) >e^{-r n^{1/3}} h b^{-3}
\right]
\eea
where we set $r=s b^2/(3(1-\gamma))$ with $s$ from Proposition \ref{sigmaprop}.
The first term is bounded by $C b^{-3}$ as $e^{r_1(0)}$ has bounded density near zero. The second term is bounded by $C r^{-3/2}\le C b^{-3}$ by (\ref{free3}). Finally, \eqref{eq:path1} and Lemma \ref{lem_sigtail} give, for large enough $n$ and $b$ and uniformly for $h\in(b^{-3},1)$,
\begin{align*}
&\P\left[   Q_{n,t}\left(| \sigma_{\ell }-t'|>u  \right) >e^{-r n^{1/3}} h b^{-3}
\right]\le \P\left[
Q_{n,t}\left(| \sigma_{\ell }-t'|>u  \right) >e^{-2r n^{1/3}}
\right]\\
 &\qquad\qquad =\P\left[Q_{n-\ell ,t-t'}(|\sigma_0|>u)>e^{-s u^2/(n-\ell) }\right]
\le   C b^{-3}.
\end{align*}
Collecting the estimates
\[
\P\left[Q_{(1,n),(0,t)}\left(| \sigma_{\ell }-t'|>u  \right)>h  \right]\le C  b^{-3}
\]
and from this
\begin{align*}
P_{(1,n),(0,t)}\left(|\sigma_{\lfloor n \gamma\rfloor}-\gamma t|>b n^{2/3}   \right)&\le b^{-3}+\int_{b^{-3}}^1 \P\left[Q_{(1,n),(0,t)}\left(| \sigma_{\ell }-t'|>u  \right)>h  \right] dh\le C b^{-3}.
\end{align*}
This completes the proof of the theorem.
\end{proof}

\bibliography{growthrefs}

\bibliographystyle{abbrv}

\end{document}


TITLE: Prove that $z_1^2 + z_2^2 + z_3^2 = 0$
QUESTION [0 upvotes]: Let $z_1,\,z_2$ and $z_3$ with the following properties: $z_1+z_2+z_3=0$ and $|z_1|=|z_2|=|z_3|=1$.
Prove that $z_1^2 + z_2^2 + z_3^2 = 0$.

REPLY [3 votes]: Hint:
$$
\require{cancel}
\begin{align}
z_1^2 + z_2^2 + z_3^2 & = \bcancel{(z_1 + z_2 + z_3)^2} - 2(z_1z_2+z_2z_3+z_3z_1) \\[3px]
 & = -2z_1z_2z_3\left(\frac{1}{z_1}+\frac{1}{z_3}+\frac{1}{z_3}\right) \\[3px]
 & = -2z_1z_2z_3(\bar z_1+\bar z_2 + \bar z_3) \\[3px]
 & = -2z_1z_2z_3(\overline{z_1+z_2+z_3})
\end{align}
$$


TITLE: Why am I getting two different answers for this diff. equation? Completing the square vs quad form.
QUESTION [1 upvotes]: With this differential equation, after seperating and integrating, and using the initial condition to solve for C, and then substituting that value of C into the general solution, I must solve for Y to get the particular solution.  Why am I getting two different solutions, where in method 1 I complete the square, and method 2, use the quadratic formula?
The ODE is:

$\frac{dy}{dx} = \frac{2x}{1+2y}$
The initial condition is:

$y(2) = 0$
Separating and integrating, I get the general soln.:

$y + y^2 = x^2 + c$
Plugging in the initial condition:

$0 + 0^2 = (2)^2 + c$
$c = -4$
Now, plugging in (-4) into the general solution:

$y + y^2 = x^2 - 4$
Now, method 1, completing the square:

$y^2 + y = x^2 - 4$
$(y + \frac{1}{2})^2 = x^2 - 4 + \frac{1}{4}$
$y = -\frac{1}{2} ^+_- (4x^2 - 15)^{\frac{1}{2}}$
Now, method 2, quad. formula:

$y = \frac{-1 ^+_- (1^2 - 4(1)(-x^2+4))^{\frac{1}{2}}}{2}$
$y = -\frac{1}{2} ^+_- \frac{1}{2}(4x^2-15)^{\frac{1}{2}}$
So, what did I do wrong with completing the square?
The quad. formula is the same answer from the back of the
book.  There is a 1/2 in front of the square root, whereas with
completing the square, the 1/2 is not there.   What did I do wrong?

REPLY [2 votes]: $$\left( y+ \frac{1}{2}\right)^2= x^2-\frac{15}{4} \Rightarrow y+ \frac{1}{2}= \pm \sqrt{\left( x^2-\frac{15}{4}\right) }\Rightarrow y=-\frac{1}{2} \pm \sqrt{\left( x^2-\frac{15}{4}\right)} \\=-\frac{1}{2} \pm \sqrt{ \frac{1}{4}\left( 4x^2-15\right)}= -\frac{1}{2} \pm \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$$
EDIT: But since only $y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$ satisfies the initial condition $y(2)=0$, we reject $y(x)= -\frac{1}{2} - \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$ and so our only solution is:
$$y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$$


TITLE: How would I mathematically prove that 1+1=2?
QUESTION [0 upvotes]: Everyone knows that 1+1=2, but how would one mathematically prove that this equation is true? Or can you?

REPLY [0 votes]: You can! Math is axiomatic, so the exact proof would depend on which axioms you choose (some might choose $1 + 1 = 2$ as an axiom itself, instead of a theorem).
Whitehead and Russell's Principia Mathematica was an attempt to prove all of math from first principles. They famously got to showing that $1 + 1 = 2$ on page 379.


TITLE: What's the optimal parameter for vector B given x and y
QUESTION [0 upvotes]: I need to brush up my Maths, for sure, but, meanwhile I hope you could help me with this.
This is a Linear Regression problem.
Question: given X and Y compute the optimale B (from what I understand, B is the constant and the slope )
I've this 2 variables x and y  :
X=(-0.78, -1.51, 0.74, -0.62)
Y=(-34.5, -30.79, 19.31, -19.44)
The result choices are :
52, 13, 26, 32
I've check the formula y=mx+b, and of course f(x)=b0+b1x1+b2x2+...+bnxn but I'm a little bit lost.
So, if you could help me with this, that would be great.
Thank you

REPLY [0 votes]: The optional parameter $t$ is obtained by minimizing
$$(t\vec{X} - \vec{Y})^2$$
which is also a linear regression problem. The optimal solution is
$$t=\frac {\vec{X}\cdot\vec{Y}}{X^2}=\frac{99.9}{3.82}=26.1$$


TITLE: A deck of 52 cards is revealed 1 at a time till an ace appears. Let $X$ be the number of cards revealed. Find the PMF and Expectation
QUESTION [1 upvotes]: A deck of 52 cards is revealed 1 at a time till an ace appears. Let $X$ be the number of cards revealed until the ace. Find the PMF and Expectation(Can leave as a summation).
Let me know if any of my thought process is wrong
Since it's asking for a PMF it's going to be probabilities clumped up at different Xs(aka not a continuous function). The probability of drawing an Ace on the first draw is 4/52 then assuming you didn't draw it on the first draw it becomes 4/51 and this keeps going so I assume the PMF=P(X=k)=$\prod_{i=1}^k$(48-i/(52-i)) for $1\le k<48$. 0 otherwise.
With the PMF we know the expectation is going to be E[X]=P(X)*X. So that gives us $\sum^{47}_{k=1}k\prod_{i=1}^k(48-i/(52-i))$ because in worst can we'll have to draw 47 cards before we get an Ace and that's the case that the last 4 cards are aces.

REPLY [1 votes]: Now that the answer is in a comment, I'll write it as an answer with some extra stuff.
So we want to find $P(X=k)$, for $1\le k \le 49$. Let's take $k=4$ as an example. We first need a non-ace card, which has probability $\frac{48}{52}$. Then another one, $\frac{47}{51}$, and another, $\frac{46}{50}$. Then we need an ace, $\frac{4}{49}$. So $P(X=4)=\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{4}{49}$. This generalises to:
$$
P(X=k)
= \frac{4}{53-k}\cdot\prod_{j=0}^{k-2}\frac{48-j}{52-j}
$$
(Note that if $k=1$, the product is the empty product, which is $1$).
You seem to have no problem writing up the expectation, so I won't go into that (it turns out to be $53/5=10.6$ by the way). Note however that the possible values for $k$ are $\{1,\ldots,49\}$, not $\{1,\ldots,47\}$.
Now for the extra bit. We can rewrite the probability as:
$$
P(X=k) = \frac{4}{52}\cdot\frac{52-k}{51}\cdot\frac{51-k}{50}\cdot\frac{50-k}{49}
$$
This form actually gives another interpretation. We think about the positions of the aces in the deck. The $k$'th card needs to be an ace, which has probabailty $\frac{4}{52}$. The other three aces have to be placed after position $k$, which gives the rest of the fractions (e.g. there are 51 spots left for the second ace, $52-k$ of these being after position $k$).


TITLE: Eigenvalues of $I + v v^T$
QUESTION [1 upvotes]: I'm trying to calculate the eigenvalues of an $N\times N$ matrix $M$ whose matrix elements are $m_{ij} = \delta_{ij} + c$, where c is a real constant. Anybody knows if there is a closed formula for this?
Notice that the matrix can also be written as $M=I + v v^T$, where the vector $v$ has elements $v_i = \sqrt{c}$.
I have search through the stack.exchange questions with the words "diagonalization" and "low-rank" in the title, but they seem to concern more general cases than the simple one I'm interested in.
Thank you!

REPLY [3 votes]: The $N\times N$ matrix $M=I+cvv^T$ (where $v$ is a vector and $c$ a scalar) has eigenvalue $1$ with multipilicity $N-1$ and the eigenvalue $\lambda=1+c\|v\|^2$ with multipicity $1$.  The corresponding eigenvector is $v$, which upon multiplication by $M$ yields $$Mv = v + (cvv^T)v=  v + cv(v^Tv)=v+cv\|v\|^2 = (1+c\|v\|^2)v.$$  In the special case where $v$ is the vector of all $1$ recovers the formula for your $m_{ij}=\delta_{ij}+c$ case.

REPLY [2 votes]: Let $\mathbf 1$ be the vector with each element being $1$.
$$(I+c\mathbf 1\mathbf 1^T)\mathbf 1=(1+cN)\mathbf 1$$
For each one $x$ of the $N-1$ indepedent vectors orthogonal to $\mathbf 1$
$$(I+c\mathbf 1\mathbf 1^T)x = x$$
So the eigenvalues of $I+c\mathbf 1\mathbf 1^T$ are $1+cN$ with multiplicity $1$, and $1$ with multiplicity $N-1$.


TITLE: Hydrogen in an electromagnetic field bound states for full quantum electrodynamics
QUESTION [2 upvotes]: I am looking for a reference or derivation of the non-relativistic bound states of hydrogen in an electromagnetic field that include Zeeman effect, Stark shift, and Lamb shift.  I am looking for a full QED treatment.  I have seen Scully and Zubairy Quantum Optics but they only provide a rough approximation.
I am finding that those who are doing QED are mostly interested in high-energy particle physics, whereas I am interested in atomic physics and the terms relevant to corrections of the spectral line energies (resonances in QFT parlance).
I have used the Dirac equation in the low energy limit to arrive at terms for relativistic correction, spin-orbit interaction, and Darwin term, what I would like to be able to do is write out a derivation that includes the electron self-interaction term; that is a derivation with a more explicit treatment than Scully of the quantum vacuum for the Lamb shift.

REPLY [1 votes]: Let us state first the idea behind QED. QED stands for quantum electrodynamics and it has two main components that are merged. One the one hand special relativity (SR) on the other quantum mechanics. The first one concerns very high speeds or high energies, the second one pertains very small things. If you have a Hydrogen atom just sitting there there is no need to include (SR) effects. Same as, if your system is to big, there is no need to use quantum mechanics.
With that cleared out, you should understand the different origins of the effects you are speaking of. The Lamb shift has to do with the vacuum polarization (or electron self-energy), or if you want the "dressing" of the photon propagator. As can be seen in most books of quantum field theory (QFT), it is related to the Uehling term correction to the Coulomb potential. We could call this an intrinsic effect.
The Zeeman and Stark effects are different in origin. They appear through external fields. This means they will not pop out of QED without modifying anything else. You should be asking then how to model external fields within QED and you will have to look for effective field theories (EFTs).
In the context of EFTs, one essentially finds an effective field theory by some method/approximation. There are several ways to speak about effective theories, but I believe you might be interested in expanding your theory around external backgrounds.
I hope there are enough keywords to point you in some useful direction.


TITLE: Consider the following two player game. A pile of coins is places on a table
QUESTION [0 upvotes]: Consider the following two player game. A pile of coins is places on a table. There are two
players, Alice and Bob, who alternate turns removing one, two or three coins from the pile.
Alice goes first. The player who takes the last coin wins. Use induction to prove that Alice
has a winning strategy (i.e. she can play so as to guarantee a win) if the number of coins in
the pile is not a multiple of 4, and Bob has a winning strategy if the number of coins in the
pile is a multiple of 4.
Interesting question. I can prove it in my brain but i'm not quite sure how to prove it using induction. thanks

REPLY [2 votes]: The standard strategy for solving problems like this starts with thinking that you're in the middle of playing this game with $0$ coins in the pile and your turn, and see who wins (it has to be your opponent, because on their previous turn, they removed the last coin and won).
The problem already tells you that modulo 4 is important, so next we look at what happens if there instead are $1, 2, 3$ or $4$ coins and your turn. Can the you win any of them? Can your opponent guarantee a win in any of them?
Next, for $5, 6, 7$ or $8$, can you force the opponent into a position that the previous paragraph deemed as "the current player loses"? If not, can your opponent do that to you no matter what you try?
A pattern should start to emerge. What does the winning strategy seem to be? With the intuition you've built from working through the above, it shouldn't be too hard to prove by doing the above paragraph of $5, 6, 7, 8$ in general using induction. (The induction parameter isn't really "How many coins there are on the table", but rather "How many groups of four coins there are", and then the induction step itself is split into four cases depending on how many coins are left modulo 4.)