text:
\section{Introduction}\label{;intro}

Let $B=\{ B_{t}\} _{t\ge 0}$ be a one-dimensional standard Brownian motion.
For every $\mu \in \R $, we denote by
$\db{\mu }=\{ \db{\mu }_{t}:=B_{t}+\mu t\} _{t\ge 0}$ the Brownian motion
with constant drift $\mu $ and set
\begin{align*}
 \da{\mu }_{t}:=\int _{0}^{t}e^{2\db{\mu }_{s}}ds, \quad t\ge 0;
\end{align*}
when $\mu =0$, we simply write $A_{t}$ for $\da{\mu }_{t}$.
This additive functional, together with the geometric Brownian motion
$e^{\db{\mu }_{t}},\,t\ge 0$, plays an important role in a number of areas
such as option pricing in mathematical finance,
diffusion processes in random environments, probabilistic study of
Laplacians on hyperbolic spaces, and so on; see the detailed surveys
\cite{mySI,mySII} by Matsumoto--Yor and references therein.

In \cite{yor92}, Yor proved that for every $t>0$, the joint law
of $B_{t}$ and $A_{t}$ is given by
\begin{align}\label{;jl1}
 \pr \!\left(
 B_{t}\in dx,\,A_{t}\in dv
 \right)
 =\frac{1}{v}\exp \left\{
 -\frac{1}{2v}\left( 1+e^{2x}\right)
 \right\} \Theta (e^{x}/v,t)\,dxdv,\quad x\in \R ,\,v>0,
\end{align}
or equivalently,
\begin{align}\label{;jl2}
 \pr \left( \eb{t}\in du,\,A_{t}\in dv\right)
 =\frac{1}{uv}\exp \left( -\frac{1+u^{2}}{2v}\right)
 \Theta (u/v,t)\,dudv, \quad u,v>0,
\end{align}
where for every $r>0$, the function $\Theta (r,t),\,t>0$, is the
(unnormalized) density of the so-called Hartman--Watson distribution
(\cite{hw}) which is characterized by the Laplace transform
\begin{align}\label{;LTt}
 \int _{0}^{\infty }dt\,
 \exp \left( -\frac{\la ^{2}}{2}t\right)
 \Theta (r,t)
 =I_{|\la |}(r),\quad \la \in \R .
\end{align}
Here for every index $\nu \in \R $,
the function $I_{\nu }$ is the modified Bessel function
of the first kind of order $\nu $; see \cite[Section~5.7]{leb}
for definition. By the Cameron--Martin relation, we see from
\eqref{;jl2} that for every $\mu \in \R $ and $t>0$, the law of
$\da{\mu }_{t}$ is expressed as
\begin{align}
 \frac{\pr (\da{\mu }_{t}\in dv)}{dv}
 &=\frac{1}{v}\exp \left(
 -\frac{1}{2v}-\frac{\mu ^{2}}{2}t
 \right) \int _{0}^{\infty }\frac{du}{u}\,u^{\mu }\exp \left(
 -\frac{u^{2}}{2v}\right) \Theta (u/v,t)\notag \\
 &=v^{\mu -1}\exp \left(
 -\frac{1}{2v}-\frac{\mu ^{2}}{2}t
 \right) \int _{0}^{\infty }dr\,r^{\mu -1}\exp \left(
 -\frac{v}{2}r^{2}
 \right) \Theta (r,t),\quad v>0, \label{;Amlaw0}
\end{align}
where for the second line, we changed the variables with $u=vr,\,r>0$.

\begin{rem}\label{;aderi}
 Two derivations of the formula \eqref{;jl1} different from the original one
 by \cite{yor92}, are found in the present paper; see \pref{;pjd} and
 \rref{;rfinal}, in the latter of which detailed derivation of
 \eqref{;Amlaw0} without using \eqref{;jl2} is provided as well. Those
 two derivations also differ from the one in \cite[Subsection~4.1]{mySI}
 based on the Sturm--Liouville theory.
\end{rem}

It is also proven by Yor \cite{yor80} that the function $\Theta $ admits the
following integral representation: for every $r>0$ and $t>0$,
\begin{align}\label{;irepr0}
 &\Theta (r,t)=\frac{r}{\sqrt{2\pi ^3t}}
 \int _{0}^\infty dy\,\exp \left(
 \frac{\pi ^2-y^2}{2t}
 \right) \exp \left(
 -r\cosh y
 \right) \sinh y\sin \left( \frac{\pi y}{t}\right) ,
\end{align}
which we may rephrase conveniently as
\begin{align}\label{;irepr0d}
  \Theta (r,t)=
  \frac{r}{2\pi }\exp \left(
  \frac{\pi ^{2}}{2t}
  \right)
  \ex \!\left[
  \exp \left( -r\cosh B_{t}\right) \sinh B_{t}
  \sin \left( \frac{\pi }{t}B_{t}\right)
  \right]
\end{align}
by the fact that the function $\R \ni y\mapsto \sinh y\sin (\pi y/t)$ is
symmetric. We also refer to \cite[p.~82]{bs} for \eqref{;irepr0},
as well as \cite[p.~175, Formula~1.10.8]{bs} for the joint law
\eqref{;jl1} above.

The formula \eqref{;irepr0} has continuously attracted
researchers' attention particularly from the view point of
evaluation of Asian options in the Black--Scholes framework; see, e.g.,
\cite{cs,bm} and references therein.  Yor obtained
\eqref{;irepr0} by inverting the Laplace transform \eqref{;LTt}, reasoning
of which is also reproduced in \cite[Appendix~A]{mySI}.
In \cite[Subsection~A.3]{har}, we explain
\eqref{;irepr0} via the relation
\begin{align}\label{;LTr}
 \int _{0}^{\infty }\frac{dr}{r}\,e^{-r\cosh x}\Theta (r,t)
 =\frac{1}{\sqrt{2\pi t}}\exp \left(
  -\frac{x^{2}}{2t}
  \right) ,\quad t>0,\, x\in \R ,
\end{align}
which is found, e.g., in \cite[Proposition~4.5\,(i)]{myPI} and,
as was observed in \cite[Proposition~4.2]{mySI}, may be obtained
by integrating both sides of \eqref{;jl1} with respect to $v$.
In this paper, we continue our discussion of \cite[Subsection~A.3]{har}
and, based on Yor's formula \eqref{;irepr0} (or \eqref{;irepr0d}),
aim at providing the following alternative representations of $\Theta $:

\begin{thm}\label{;mt}
For every $r>0$ and $t>0$, it holds that
\begin{align}
  \Theta (r,t)&=\frac{r}{\pi }\exp \left( \frac{\pi ^{2}}{8t}\right)
  \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] \label{;irepr1}\\
  &=\frac{r}{\pi }\exp \left( \frac{\pi ^{2}}{8t}\right)
  \ex \!\left[
  \cosh B_{t}\sin (r\sinh B_{t})\sin \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] \label{;irepr2}\\
  &=\frac{r}{2\pi }\exp \left( \frac{\pi ^{2}}{8t}\right)
  \ex \!\left[
  \cosh B_{t}\cos \left( r\sinh B_{t}-
  \frac{\pi }{2t}B_{t}
  \right)
  \right] . \label{;irepr3}
 \end{align}
 More generally, we have for every $r>0$ and $t>0$,
 \begin{align}\label{;irepr4}
  \Theta (r,t)&=\frac{r}{\pi }\exp \left( \frac{\pi ^{2}}{8t}\right)
  \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t}-\nu )\cos \left(
  \frac{\pi }{2t}B_{t}-\nu
  \right)
  \right] ,
 \end{align}
 where $\nu \in \R $ is arbitrary. The representations
 \eqref{;irepr1}, \eqref{;irepr2} and \eqref{;irepr3} may be seen as
 the case $\nu =0,\pi /2,\pi /4$, respectively.
\end{thm}

The third representation \eqref{;irepr3} follows by summing \eqref{;irepr1}
and \eqref{;irepr2}. If we multiply \eqref{;irepr1} and
\eqref{;irepr2} by $\cos ^{2}\nu $ and $\sin ^{2}\nu $, respectively,
then taking their sum leads to the fourth representation \eqref{;irepr4};
for details, see the proof of \tref{;mt} given in \sref{;prfmt}
whose reasoning will also reveal that $\nu $ may be replaced by any
complex number.

\begin{rem}\label{;rdiff}
\thetag{1} Taking the difference of \eqref{;irepr1} and \eqref{;irepr2}
leads to the following fact of interest:
\begin{align}\label{;diff0}
 \ex \!\left[
  \cosh B_{t}\cos \left( r\sinh B_{t}+
  \frac{\pi }{2t}B_{t}
  \right)
  \right] =0 \quad \text{for any $r>0$ and $t>0$,}
\end{align}
which we return to in \rref{;rcontr}. The representation \eqref{;irepr3}
should be considered jointly with \eqref{;diff0}.

\noindent
\thetag{2} Differentiating the representation \eqref{;irepr4} with
respect to $\nu $ yields
\begin{align*}
  \ex \!\left[
  \cosh B_{t}\sin \left(
  r\sinh B_{t}+\frac{\pi }{2t}B_{t}-2\nu
  \right)
  \right] =0,\quad r>0,\,t>0,
\end{align*}
for any $\nu \in \R $, which agrees with \eqref{;diff0} since
for any $r>0$ and $t>0$,
\begin{align*}
 \ex \!\left[
  \cosh B_{t}\sin \left( r\sinh B_{t}+
  \frac{\pi }{2t}B_{t}
  \right)
  \right] =0
\end{align*}
due to the fact that the function
$
 \sin \left( r\sinh x+
 \pi x/(2t)
 \right) ,\,x\in \R
$, is an odd function.

\noindent
\thetag{3} In view of \eqref{;irepr0}, we see from \eqref{;irepr1} that
\begin{align*}
 \lim _{r\to \infty }
 \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] =0,
\end{align*}
which may also be explained by the Riemann--Lebesgue lemma.
The same remark is true for \eqref{;irepr2}.
\end{rem}

\tref{;mt} has several applications. One of its immediate consequences
is that for every fixed $t>0$, the derivative of $\Theta (r,t)$ of any order
at $r=0+$ vanishes due to repeated occurrence of the multiple factor
$\sin (r\sinh B_{t})$ in integrands when taking derivatives
of \eqref{;irepr1} and \eqref{;irepr2} with respect to $r$.

\begin{prop}\label{;pderi} Fix $t>0$. It holds that
\begin{align}\label{;qpderi1}
 \lim _{r\to 0+}\frac{\partial ^{n}}{\partial r^{n}}\Theta (r,t)=0, \quad
 n=0,1,2,\ldots .
\end{align}
In particular, as $r\to 0+$,
\begin{align}\label{;qpderi2}
 \Theta (r,t)=o(r^{\kappa }) \quad \text{for any $\kappa >0$.}
\end{align}
\end{prop}
For the proof, see \ssref{;prfpderi}; the fact \eqref{;qpderi1}
can also be deduced readily from \eqref{;irepr3} and \eqref{;diff0},
for which we refer the reader to \rref{;rcontr} in the same subsection.
As a consequence of \eqref{;qpderi2},
we see that the integral in \eqref{;Amlaw0} does converge even when
$\mu \le 0$. We recall that \eqref{;qpderi1} is also observed in
\cite[Subsection~2.1]{my2003} from Yor's formula
\eqref{;irepr0} combined with a remark by Stieltjes in 1894, stating
that for any integer $n$,
\begin{align}\label{;stieltjes}
 \int _{\R }dx\,\exp \left(
 -\frac{x^{2}}{2t}
 \right) e^{nx}\sin \left(
 \frac{\pi x}{t}
 \right) =0
\end{align}
(see \cite[Equation~(6.3)]{my2003}). Our \tref{;mt} allows us to obtain
\eqref{;qpderi1} more directly, without relying on Stieltjes' remark.

\begin{rem}\label{;rstieltjes}
 Since the left-hand side of \eqref{;stieltjes} is written as
 \begin{align*}
  \sqrt{2\pi t}\,\ex \!\left[
  e^{nB_{t}}\sin \left(
  \frac{\pi }{t}B_{t}
  \right)
  \right] ,
 \end{align*}
 the fact \eqref{;stieltjes} is verified by the Cameron--Martin relation:
 \begin{align*}
  e^{-n^{2}t/2}\,\ex \!\left[
  e^{nB_{t}}\sin \left(
  \frac{\pi }{t}B_{t}
  \right)
  \right] &=
  \ex \!\left[
  \sin \left\{
  \frac{\pi }{t}(B_{t}+nt)
  \right\}
  \right] \\
  &=(-1)^{n}\ex \!\left[
  \sin \left(
  \frac{\pi }{t}B_{t}
  \right)
  \right] ,
 \end{align*}
 which is zero by symmetry of Brownian motion.
\end{rem}

When inserting the representation \eqref{;irepr0} into
\eqref{;Amlaw0}, a double integral emerges in the description of
the law of $\da{\mu }_{t}$. The second application of \tref{;mt} is
that, when $\mu $ is a nonnegative integer, we easily reduce
that apparently complicated double integral to a single integral
using Fubini's theorem, thanks to the well-known formulae
(see \cite[Equations~(4.11.2) and (4.11.3)]{leb}) for the Hermite
polynomials
\begin{align*}
 H_{2n}(x)&=\frac{(-1)^{n}2^{2n+1}e^{x^{2}}}{\sqrt{\pi }}
 \int _{0}^{\infty }ds\,s^{2n}e^{-s^{2}}\cos (2xs), \quad x\in \R , \\
 H_{2n+1}(x)&=\frac{(-1)^{n}2^{2n+2}e^{x^{2}}}{\sqrt{\pi }}
 \int _{0}^{\infty }ds\,s^{2n+1}e^{-s^{2}}\sin (2xs), \quad x\in \R ,
\end{align*}
where $n$ is any nonnegative integer.
Dealing with other values of $\mu $ as well, we put the above-mentioned
reduction in \pref{;preduc} below, which recovers in part
Theorem~4.2 of \cite{duf} by Dufresne.
For every $\mu \in \R $, we denote by $H_{\mu }$ the Hermite
function of degree $\mu $ and recall its integral representation
when $\mu >-1$:
\begin{align}\label{;herf}
 H_{\mu }(x)=\frac{2^{\mu +1}e^{x^{2}}}{\sqrt{\pi }}
 \int _{0}^{\infty }ds\,s^{\mu }e^{-s^{2}}
 \cos \left(
 2xs-\frac{\pi \mu }{2}
 \right) ,\quad x\in \R
\end{align}
(see Section~10.2 and Equation~\thetag{10.5.5} in \cite{leb} for
the definition of the Hermite functions and the integral representation \eqref{;herf}, respectively).

\begin{prop}\label{;preduc}
Let $\mu >-1$. For every $t>0$, the law of $\da{\mu }_{t}$ admits the density
function expressed by
\begin{align}
  &\frac{\pr (\da{\mu }_{t}\in dv)}{dv} \notag \\
  &=\frac{C_{\mu }(t)}{\sqrt{v^{3-\mu }}}
  \ex \!\left[
  \exp \left(
  -\frac{\cosh ^{2}B_{t}}{2v}
  \right)
  H_{\mu }\!\left(
  \frac{\sinh B_{t}}{\sqrt{2v }}
  \right) \cosh B_{t}\cos \left\{
  \frac{\pi }{2}\left( \frac{B_{t}}{t}-\mu \right)
  \right\}
  \right] \label{;Amlaw}
\end{align}
for $v>0$, where
$
C_{\mu }(t)=\bigl( 1/\sqrt{2^{\mu +1}\pi }\bigr)
e^{\pi ^{2}/(8t)-\mu ^{2}t/2}
$.
\end{prop}

Recall that
\begin{align*}
 H_{0}(x)=1, \quad H_{1}(x)=2x, \quad x\in \R
\end{align*}
(see, e.g., \cite[p.~60]{leb}). The expression \eqref{;Amlaw}
in the case $\mu =0$ was obtained by several
authors, for which we refer the reader to \cite[p.~223]{duf} as well as
the beginning of \cite[Subsection~2.2]{my2003}; an alternative derivation
of \eqref{;Amlaw} in the case $\mu =0$ and $\mu =1$
without relying on \tref{;mt}, will be found in \sref{;sfd}.
If we consider the law of $1/(2\da{\mu }_{t})$,
then from \eqref{;Amlaw}, we partly recover Formula~\thetag{4.9} in
\cite[Theorem~4.2]{duf} due to Dufresne, who also shows that the formula is
valid for $\mu \le -1$ as well, by developing a recurrence relation
that connects the law of $1/(2\da{\nu }_{t})$ with that of
$1/(2\da{\mu }_{t})$ when $\nu <\mu $. We do not pursue it here with
generality, however, if we repeat integration by
parts as necessary appealing to the fact \eqref{;qpderi1},
then \tref{;mt} enables us to reduce the computation of the
case $\mu \le -1$ to a situation where the formula \eqref{;herf}
applies or the function $H_{-1}$ emerges; see \rref{;ribp}.

We explain in some detail the plan of the paper.
In \sref{;prfmt}, we prove \tref{;mt}; we do this by
preparing \lref{;lequiv} which shows equivalence of certain
three relations for expectations relative to Brownian motion.
Since in the proof of \tref{;mt}, one implication between
two of the three relations is used, we give its proof in
\sref{;prfmt} and the rest of the proof of the lemma is
provided in the appendix. \psref{;pderi} and \ref{;preduc} are
proven in \sref{;prfprops}. We prove in \sref{;sfd} a family of integral
identities that not only embraces relations in \tref{;mt}, but
can also be used to derive the formula \eqref{;jl1} with the representation
\eqref{;irepr1} of $\Theta $ inserted in; see \pref{;punif} and
discussions that follow the proof of the proposition.
Revealing its connection with \eqref{;Amlaw0},
we also explore an explicit integral representation for the
law of time integral of reciprocal squared Bessel process of any
dimension, and show that it is simplified into a single
integral when the dimension is an even integer not less than
$-2$. Finally in the appendix, we complete the proof of \lref{;lequiv}.

\section{Proof of \tref{;mt}}\label{;prfmt}
From now on, we fix $t>0$. This section is devoted to the proof of \tref{;mt}.
Let two real-valued functions $\f $ and $\g $ on $\R $ be continuous
for simplicity and suppose that they are even functions and satisfy
\begin{align}\label{;intcond}
 \ex \!\left[
 |\f (B_{t})|
 \right] <\infty  && \text{and} &&
 \ex \!\left[
 |\g (B_{t})|\cosh B_{t}
 \right] <\infty .
\end{align}

\begin{lem}\label{;lequiv}
Under the condition \eqref{;intcond}, the following three relations
are equivalent:
\begin{align}
  \thetag{i}\ &\frac{2}{\pi }\ex \!\left[
  \frac{\f (B_{t})\cosh B_{t}}{\cosh (2B_{t})+\cosh (2x)}
  \right]
  =\frac{1}{\sqrt{2\pi t}}\exp \left(
  -\frac{x^{2}}{2t}
  \right) \g (x) \quad \text{for any }x\in \R ; \label{;rel3}\\
  \thetag{ii}\ &\ex \!\left[
  e^{-r\cosh B_{t}}\f (B_{t})
  \right]
  =\ex \!\left[
  \g (B_{t})\cosh B_{t}\cos (r\sinh B_{t})
  \right] \quad \text{for any }r\ge 0; \label{;rel2}\\
  \thetag{iii}\ &\ex \!\left[
  \frac{\f (B_{t})}{\cosh B_{t}+\cosh x}
  \right]
  =\ex \!\left[
  \frac{\g (B_{t})}{\cosh (x+B_{t})}
  \right] \quad \text{for any }x\in \R .  \label{;rel1}
\end{align}
\end{lem}

Since, in the proof of \tref{;mt}, we only use the implication
from \thetag{i} to \thetag{ii}, we give a proof of it
below and postpone proofs of other implications to
the appendix.

\begin{proof}[Proof of \thetag{i} $\Rightarrow $ \thetag{ii} in \lref{;lequiv}]
For every $r\ge 0$, we integrate both sides of \eqref{;rel3} multiplied by
$\cosh x\cos (r\sinh x)$ with respect to $x\in \R$.
Then the right-hand side turns into that of \eqref{;rel2} in view of the
latter condition in \eqref{;intcond}. On the other hand, provided that we are
allowed to use Fubini's theorem, the left-hand side turns into
 \begin{align*}
 \frac{2}{\pi }\ex \!\left[
 F(B_{t})\cosh B_{t}\int _{\R }dx\,
 \frac{\cosh x\cos (r\sinh x)}{\cosh (2B_{t})+\cosh (2x)}
 \right] =\ex \!\left[
 e^{-r\cosh B_{t}}\f (B_{t})
 \right]
\end{align*}
as claimed. Here for the equality, we used the fact that
for any $b\in \R $ and $r\ge 0$,
\begin{align}\label{;fact1}
 \int _{\R }dx\,
 \frac{\cosh x\cos (r\sinh x)}{\cosh (2b)+\cosh (2x)}
 =\frac{\pi }{2\cosh b}e^{-r\cosh b},
\end{align}
which follows readily by noting
$
 \cosh (2b)+\cosh (2x)
 =2\left( \cosh ^{2}b+\sinh ^{2}x\right)
$,
and changing the variables
with $\sinh x=y\cosh b,\,y\in \R $. Usage of Fubini's theorem
mentioned above is justified by taking $r=0$ in \eqref{;fact1};
indeed,
\begin{align}
 \frac{2}{\pi }\int _{\R }dx\,
 \cosh x\,\ex \!\left[
 \frac{|F(B_{t})|\cosh B_{t}}{\cosh (2B_{t})+\cosh (2x)}
 \right] =\ex \!\left[
 |F(B_{t})|
 \right] , \label{;fubi1}
\end{align}
which is assumed to be finite in \eqref{;intcond}. The
proof is complete.
\end{proof}

In what follows, we denote by $\C $ the complex plane and write
$i=\sqrt{-1}$. A pair of functions $F$ and $G$ fulfilling the relation
\eqref{;rel3} may be obtained by the residue theorem applied to a
meromorphic function $f$ of the form
\begin{align*}
 f(z)=\frac{J(z)}{\cosh (2z)+\cosh (2x)}\exp \left(
 -\frac{z^{2}}{2t}
 \right) ,\quad z\in \C ,
\end{align*}
where $x\in \R $ and $J(z),\,z\in \C$, is an odd entire function which will be
taken to be either $\sinh (2z)$ or $\sinh z$ below. When $x\neq 0$,
the poles $w$ of $f$ each of whose imaginary part $\mathrm{Im}\,w$
lies between $0$ and $\pi $, are two points $\pm x+ (\pi /2)i$.
By taking a rectangular contour circling these poles and having its two
sides on the two lines $\mathrm{Im}\,z=0$ and $\mathrm{Im}\,z=\pi $,
residue calculus yields, at least heuristically,
\begin{align*}
 &\frac{1}{2\pi i}
 \int _{\R }\frac{d\xi }{\cosh (2\xi )+\cosh (2x)}
 \exp \left(
 -\frac{\xi ^{2}}{2t}+\frac{\pi ^{2}}{2t}
 \right) J(\xi +\pi i)\exp \left(
 -\frac{\pi \xi }{t}i
 \right) \\
 &=\frac{1}{2\sinh (2x)}\exp \left(
 -\frac{x^{2}}{2t}+\frac{\pi ^{2}}{8t}
 \right) \left\{
 J(x+\pi i/2)\exp \left(
 -\frac{\pi x}{2t}i
 \right) +J(x-\pi i/2)\exp \left(
 \frac{\pi x}{2t}i
 \right)
 \right\}
\end{align*}
for $x\neq 0$. When $J(z)=\sinh (2z)$ and $\sinh z$, the above
computation is justified, yielding the following lemma:

\begin{lem}\label{;liden}
 It holds that for any $x\in \R $,
 \begin{align}
  \frac{1}{\pi }\ex \!\left[
  \frac{
  \sinh B_{t}\cosh B_{t}\sin \left( \pi B_{t}/t\right)
  }
  {
  \cosh (2B_{t})+\cosh (2x)
  }
  \right]
  &=\frac{1}{\sqrt{2\pi t}}
  \exp \left(
  -\frac{x^{2}}{2t}-\frac{3\pi ^{2}}{8t}
  \right) \cos \left(
  \frac{\pi x}{2t}
  \right) , \label{;iden1}\\
  \frac{1}{\pi }\ex \!\left[
  \frac{
  \sinh B_{t}\sin \left( \pi B_{t}/t\right)
  }
  {
  \cosh (2B_{t})+\cosh (2x)
  }
  \right]
  &=\frac{1}{\sqrt{2\pi t}}
  \exp \left(
  -\frac{x^{2}}{2t}-\frac{3\pi ^{2}}{8t}
  \right) S(x), \label{;iden2}
 \end{align}
 where in the latter identity, the function $S(x)\equiv S(x,t),\,x\in \R $,
 is given by
 \begin{align*}
  S(x)=
  \begin{cases}
   \sin (\frac{\pi }{2t}x)/\sinh x & \text{for $x\neq 0$}, \\
   \pi /(2t) & \text{for $x=0$}.
  \end{cases}
 \end{align*}
\end{lem}

Validity of \eqref{;iden1} and \eqref{;iden2} at $x=0$ follows by passing
to the limit as $x\to 0$. We remark that these two identities are found
in Lemmas~3.1 and 3.2 of \cite{my2003} by Matsumoto--Yor, in which
paper those two lemmas are used to show that, in the case
$\mu =0$ and $1$, the expression \eqref{;Amlaw0}
with Yor's formula \eqref{;irepr0} inserted in coincides with
\eqref{;Amlaw}.
What is revealed in the present paper is that coincidence for any $\mu >-1$
is also reduced to the above two identities \eqref{;iden1} and
\eqref{;iden2} via \tref{;mt}.

We are in a position to prove the theorem.

\begin{proof}[Proof of \tref{;mt}]
 First we prove \eqref{;irepr1}. The identity \eqref{;iden1} tells us that
 we may take
 \begin{align*}
  \f (x)=\frac{1}{2}\sinh x\sin \left(
  \frac{\pi x}{t}
  \right)  && \text{and} &&
  \g (x)=\exp \left( -\frac{3\pi ^{2}}{8t}\right)
  \cos \left(
  \frac{\pi x}{2t}
  \right)
 \end{align*}
 in the relation \eqref{;rel3}. It is clear that these functions
 fulfill the integrability condition \eqref{;intcond}. Therefore
 by \lref{;lequiv}, we have for every $r\ge 0$,
 \begin{align*}
  \frac{1}{2}\ex \!\left[
  e^{-r\cosh B_{t}}\sinh B_{t}\sin \left(
  \frac{\pi }{t}B_{t}
  \right)
  \right]
  =\exp \left( -\frac{3\pi ^{2}}{8t}\right) \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] .
 \end{align*}
 Now the representation \eqref{;irepr1} follows from this and Yor's
 formula \eqref{;irepr0d}.

 We proceed to the proof of \eqref{;irepr2}. The second identity
 \eqref{;iden2} in \lref{;liden} shows that we may take
 in \eqref{;rel3}
 \begin{align*}
  \f (x)=\frac{1}{2}\frac{\sinh x}{\cosh x}\sin \left(
  \frac{\pi x}{t}
  \right) && \text{and} &&
  \g (x)=\exp \left( -\frac{3\pi ^{2}}{8t}\right)
  S(x),
 \end{align*}
 which pair also fulfills \eqref{;intcond}. Therefore by
 \lref{;lequiv}, we have for every $r\ge 0$,
 \begin{align*}
  \frac{1}{2}\ex \!\left[
  e^{-r\cosh B_{t}}\frac{\sinh B_{t}}{\cosh B_{t}}\sin \left(
  \frac{\pi }{t}B_{t}
  \right)
  \right]
  =\exp \left( -\frac{3\pi ^{2}}{8t}\right) \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})S(B_{t})
  \right] .
 \end{align*}
 Thanks to the fact that $\cosh B_{t}$, and hence $|\sinh B_{t}|$ as well,
 is integrable, by differentiating both sides with respect to $r$ and by
 appealing to the formula \eqref{;irepr0d} again, we arrive at \eqref{;irepr2}.

 The representation \eqref{;irepr3} is a consequence of summation of
 \eqref{;irepr1} and \eqref{;irepr2}. To prove \eqref{;irepr4},
 fix $\nu \in \R $. Using the addition theorem, we develop
 \begin{align*}
  &\cos (r\sinh B_{t}-\nu )\cos \left(
  \frac{\pi }{2t}B_{t}-\nu
  \right) \\
  &=\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \cos ^{2}\nu
  +\sin (r\sinh B_{t})\sin \left(
  \frac{\pi }{2t}B_{t}
  \right) \sin ^{2}\nu \\
  &\quad +R(B_{t})\sin \nu \cos \nu
 \end{align*}
 with $R(x),\,x\in \R $, an odd function such that
 $
 \ex \left[
 |R(B_{t})|\cosh B_{t}
 \right] <\infty
 $.
 Hence the right-hand side
 of the claimed identity \eqref{;irepr4} is equal to
 \begin{align*}
  &\frac{r}{\pi }\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) \Bigl\{
  \ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] \cos ^{2}\nu \\
  &\qquad \qquad \qquad +
  \ex \!\left[
  \cosh B_{t}\sin (r\sinh B_{t})\sin \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] \sin ^{2}\nu
  \Bigr\} \\
  &=\Theta (r,t)\!\left( \cos ^{2}\nu +\sin ^{2}\nu \right)
 \end{align*}
 by \eqref{;irepr1} and \eqref{;irepr2}, which shows \eqref{;irepr4}
 and completes the proof of the theorem.
\end{proof}

\section{Proof of \psref{;pderi} and \ref{;preduc}}\label{;prfprops}

In the sequel, we set the function $g(r)\equiv g(r,t),\,r>0$, by
\begin{align*}
 g(r):=\ex \!\left[
  \cosh B_{t}\cos (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \right] .
\end{align*}
As seen in the previous section, it holds that for any $r>0$,
\begin{align}
 &g(r)=\ex \!\left[
 \cosh B_{t}\sin (r\sinh B_{t})\sin \left(
 \frac{\pi }{2t}B_{t}
 \right) \right] \notag
\intertext{and}
 &\qquad \ \ \quad \Theta (r,t)=\frac{r}{\pi }\exp \left(
 \frac{\pi ^{2}}{8t}
 \right) g(r) . \label{;Theta}
\end{align}

\subsection{Proof of \pref{;pderi}}\label{;prfpderi}

In this subsection, we prove \pref{;pderi}.

\begin{proof}[Proof of \pref{;pderi}]
 By the relation \eqref{;Theta}, it suffices to show that
 \begin{align}\label{;deri0d}
  \lim _{r\to 0+}g^{(n)}(r)=0,\quad n=0,1,2,\ldots .
 \end{align}
 By observing the fact that
 \begin{align}\label{;finite}
  \ex \!\left[
  \cosh B_{t}\left| \sinh B_{t}\right| ^{n}
  \right] <\infty
 \end{align}
 for any nonnegative integer $n$, successive differentiation
 of the above two representations of $g$ yields
 \begin{align*}
  g^{(2n)}(r)&=(-1)^{n}
  \ex \!\left[
  \cosh B_{t}\sinh ^{2n}\!B_{t}\sin (r\sinh B_{t})\sin \left(
  \frac{\pi }{2t}B_{t}
  \right) \right] , \\
  g^{(2n+1)}(r)&=(-1)^{n+1}
  \ex \!\left[
  \cosh B_{t}\sinh ^{2n+1}\!B_{t}\sin (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \right]
 \end{align*}
 for every nonnegative integer $n$, from which \eqref{;deri0d}
 follows readily by the dominated convergence theorem.
\end{proof}

\begin{rem}\label{;rcontr}
 Another direct way of convincing ourselves of \eqref{;deri0d} is to
 consider the mapping
 \begin{align*}
  \R \ni r\mapsto
  \ex \!\left[
  \cosh B_{t}\cos \left(
  \frac{\pi }{2t}B_{t}-r\sinh B_{t}
  \right)
  \right] ,
 \end{align*}
 which determines a $C^{\infty }$-function
 thanks to \eqref{;finite}, agrees with $2g(r)$ for $r>0$,
 and is identically zero on $(-\infty ,0)$ in view of \eqref{;diff0},
 and hence would yield a contradiction if \eqref{;deri0d} were not
 the case.
\end{rem}

\subsection{Proof of \pref{;preduc}}\label{;prfpreduc}

In this subsection, we prove \pref{;preduc}.

\begin{proof}[Proof of \pref{;preduc}]
 We insert the representation \eqref{;irepr4} into \eqref{;Amlaw0}
 putting $\nu =\pi \mu /2$. Then for $\mu >-1$,
 Fubini's theorem entails that \eqref{;Amlaw0} is rewritten as
 \begin{align}\label{;fubini}
  \frac{1}{\pi }\exp \left(
  \frac{\pi ^{2}}{8t}-\frac{\mu ^{2}}{2}t
  \right)
  v^{\mu -1}\exp \left(
  -\frac{1}{2v}
  \right) \ex \!\left[
  h(B_{t})\cosh B_{t}\cos \left\{
  \frac{\pi }{2}\left(
  \frac{B_{t}}{t}-\mu
  \right)
  \right\}
  \right] ,
 \end{align}
 where we set the function $h(x),\,x\in \R $, by
 \begin{align*}
  h(x)=\int _{0}^{\infty }dr\,r^{\mu }\exp \left(
  -\frac{v}{2}r^{2}
  \right) \cos \left(
  r\sinh x-\frac{\pi \mu }{2}
  \right) ,
 \end{align*}
 which is equal, by changing the variables with
 $r=\sqrt{(2/v)}s$, to
 \begin{align*}
  &\left(
  \frac{2}{v}
  \right) ^{(\mu +1)/2}
  \int _{0}^{\infty }ds\,s^{\mu }e^{-s^{2}}
  \cos \left(
  2\frac{\sinh x}{\sqrt{2v}}s-\frac{\pi \mu }{2}
  \right) \\
  &=\sqrt{\frac{\pi }{(2v)^{\mu +1}}}
  \exp \left(
  -\frac{\sinh ^{2}x}{2v}
  \right)
  H_{\mu }\!\left(
  \frac{\sinh x}{\sqrt{2v}}
  \right)
 \end{align*}
 by the formula \eqref{;herf} for $H_{\mu }$ with $\mu >-1$.
 Inserting the last expression of $h$ into \eqref{;fubini} and rearranging terms
 lead to \eqref{;Amlaw} and conclude the proof.
\end{proof}

We end this section with a remark on the case $\mu \le -1$.

\begin{rem}\label{;ribp}
 We take $\mu =-3/2$ and $-2$ as an illustration. Noting the relation
 \eqref{;Theta}, we use the function
 $g$ to rewrite the integral in \eqref{;Amlaw0} with respect to $r$ as
 \begin{align*}
  \frac{1}{\pi }\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) I(\mu ) && \text{with} &&
  I(\mu ):=\int _{0}^{\infty }dr\,r^{\mu }\exp \left(
  -\frac{v}{2}r^{2}
  \right) g(r).
 \end{align*}
 The proof of \pref{;preduc} shows that $I(\mu )$ may be expressed
 in terms of $H_{\mu }$ when $\mu >-1$.
 If we take $\mu =-3/2$, then integration by parts yields
 \begin{align}\label{;qribp1}
  I(-3/2)=-2vI(1/2)+2\int _{0}^{\infty }\frac{dr}{r^{1/2}}\,\exp \left(
  -\frac{v}{2}r^{2}
  \right) g'(r)
 \end{align}
 owing to \eqref{;qpderi1}.
 Recalling \eqref{;irepr4}, we have
 \begin{align*}
  g'(r)&=-\ex \!\left[
  \cosh B_{t}\sinh B_{t}\sin (r\sinh B_{t}-\nu )\cos \left(
  \frac{\pi }{2t}B_{t}-\nu
  \right)
  \right] \\
  &=-\ex \!\left[
  \cosh B_{t}\sinh B_{t}\cos \left( r\sinh B_{t}-\nu -\frac{\pi }{2}\right)
  \cos \left(
  \frac{\pi }{2t}B_{t}-\nu
  \right)
  \right]
 \end{align*}
 for every $r>0$ and $\nu \in \R $. Therefore choosing
 $\nu =-(3/4)\pi $ and appealing to Fubini's theorem and \eqref{;herf},
 we may express the second term on the right-hand side of \eqref{;qribp1}
 in terms of $H_{-1/2}$. In the case $\mu =-2$, we have
 in the same way as above,
 \begin{align}\label{;qribp2}
  I(-2)=-vI(0)+\int _{0}^{\infty }\frac{dr}{r}\,\exp \left(
  -\frac{v}{2}r^{2}
  \right) g'(r).
 \end{align}
 Noting that
 \begin{align*}
  g'(r)=-\ex \left[
  \cosh B_{t}\sinh B_{t}\sin (r\sinh B_{t})\cos \left(
  \frac{\pi }{2t}B_{t}
  \right)
  \right] ,\quad r>0,
 \end{align*}
 and that $|\sin (r\sinh B_{t})/r|\le |\sinh B_{t}|$ for any $r>0$,
 Fubini's theorem entails that the second term on the right-hand side
 of \eqref{;qribp2} is written as
 \begin{align*}
  -\ex \!\left[
  \cosh B_{t}\sinh B_{t}\cos \left( \frac{\pi }{2t}B_{t}
  \right) \int _{0}^{\infty }\frac{dr}{r}\,\exp \left(
  -\frac{v}{2}r^{2}
  \right) \sin (r\sinh B_{t})
  \right] .
 \end{align*}
 By the formulae
 \begin{align}
  \int _{0}^{\infty }\frac{ds}{s}\,e^{-s^{2}}\sin (2xs)
  &=\sqrt{\pi }\int _{0}^{x}dy\,e^{-y^{2}},\quad x\ge 0, \label{;erf}\\
  H_{-1}(x)&=e^{x^{2}}\int _{x}^{\infty }dy\,e^{-y^{2}},
  \quad x\in \R , \label{;herf-1}
 \end{align}
 the integrand in the last expectation may be expressed in terms of $H_{-1}$.
 As for the formula \eqref{;herf-1}, see \cite[Equation~(10.5.3)]{leb}. The formula
 \eqref{;erf} may easily be verified by writing
 \begin{align*}
  \frac{\sin (2xs)}{s}=2\int _{0}^{x}dy\,\cos (2sy), \quad s>0,
 \end{align*}
 and using Fubini's theorem.
\end{rem}

\section{Further discussions}\label{;sfd}
In this section, we develop further the discussion used in
deriving \lref{;liden} to obtain a family of integral identities
that includes the relations \eqref{;irepr3} and \eqref{;diff0}
as its special cases, which, in turn, yield \eqref{;irepr1} and
\eqref{;irepr2}; that family also enables us to give
an independent proof for the joint density \eqref{;jl1}.

We keep $t>0$ fixed. We prove

\begin{prop}\label{;punif}
 For every $r,\la \ge 0$ and $\ga \in \R $, it holds that
 \begin{equation}\label{;equnif}
  \begin{split}
   &\ex \!\left[
   e^{-r\cosh B_{t}}\cos \left(
   \frac{\pi }{2t}B_{t}+\la \sinh B_{t}-\ga B_{t}
   \right)
   \right] \\
   &=\exp \left(
   \frac{\pi \ga }{2}-\frac{\pi ^{2}}{8t}
   \right)
   \ex \!\left[
   e^{-\la \cosh B_{t}}
   \cos \left(
   r\sinh B_{t}+\ga B_{t}
   \right)
   \right] .
  \end{split}
 \end{equation}
\end{prop}

Before giving a proof of the above proposition,
we show in the remark below how to obtain
several relations in \sref{;intro} from \eqref{;equnif}.

\begin{rem}\label{;rrecover}
\thetag{1} If we differentiate both sides of \eqref{;equnif}
at $\la =0$, then we obtain
\begin{equation}\label{;equnifd}
 \begin{split}
   &\ex \!\left[
   e^{-r\cosh B_{t}}\sinh B_{t}\sin \left\{
   \left( \frac{\pi }{2t}-\ga \right) B_{t}
   \right\}
   \right] \\
   &=\exp \left(
   \frac{\pi \ga }{2}-\frac{\pi ^{2}}{8t}
   \right)
   \ex \!\left[
   \cosh B_{t}
   \cos \left(
   r\sinh B_{t}+\ga B_{t}
   \right)
   \right] .
 \end{split}
\end{equation}
Taking $\ga =-\pi /(2t)$ yields the relation \eqref{;irepr3}.
By taking $\ga =\pi /(2t)$, the relation \eqref{;diff0} also
follows, which, together with \eqref{;irepr3}, entails
\eqref{;irepr1} and \eqref{;irepr2}.

\noindent
\thetag{2} Moreover, if we take $\ga =0$ in \eqref{;equnifd}, then
in view of the implication from \thetag{ii} to \thetag{i}
in \lref{;lequiv}, we have
\begin{align}\label{;iden3}
 \frac{1}{\pi }\exp \left(
 \frac{\pi ^{2}}{8t}
 \right)
 \ex \!\left[
 \frac{
 \sinh (2B_{t})\sin \left( \frac{\pi }{2t}B_{t}\right)
 }
 {
 \cosh (2B_{t})+\cosh (2x)
 }
 \right]
 =\frac{1}{\sqrt{2\pi t}}\exp \left(
 -\frac{x^{2}}{2t}
 \right) \quad \text{for any $x\in \R $}
\end{align}
(see also \rref{;riden3} below),
which may be restated, by replacing $x$ and $t$ by $x/2$ and
$t/4$, respectively, and by using the scaling property of Brownian
motion, as
\begin{align*}
 \frac{1}{2\pi }\exp \left(
 \frac{\pi ^{2}}{2t}
 \right)
 \ex \!\left[
 \frac{
 \sinh B_{t}\sin \left( \frac{\pi }{t}B_{t}\right)
 }
 {
 \cosh B_{t}+\cosh x
 }
 \right]
 =\frac{1}{\sqrt{2\pi t}}\exp \left(
 -\frac{x^{2}}{2t}
 \right) .
\end{align*}
Note that the left-hand side is rewritten as
\begin{align*}
 \frac{1}{2\pi }\exp \left(
 \frac{\pi ^{2}}{2t}
 \right)
 \int _{0}^{\infty }dr\,e^{-r\cosh x}\,
 \ex \!\left[
 e^{-r\cosh B_{t}}\sinh B_{t}\sin \left(
 \frac{\pi }{t}B_{t}
 \right)
 \right]
\end{align*}
by Fubini's theorem. Therefore once the characterization
\eqref{;LTr} of $\Theta $, namely the Laplace transform
of $\Theta (r,t)/r$ in variable $r>0$, is at our disposal,
the integral representation \eqref{;irepr0} follows by
the injectivity of Laplace transform. We also note that in view of
\lref{;lequiv}, the relation \eqref{;iden3} is equivalent to
\begin{align*}
 \exp \left(
 \frac{\pi ^{2}}{8t}
 \right)
 \ex \!\left[
 \frac{
 \sinh B_{t}\sin \left( \frac{\pi }{2t}B_{t}\right)
 }
 {
 \cosh B_{t}+\cosh x
 }
 \right]
 =\ex \!\left[
 \frac{1}{\cosh (x+B_{t})}
 \right] \quad \text{for any }x\in \R ,
\end{align*}
which relation has been observed in \cite[Subsection~A.3]{har}.
\end{rem}

It would also be of interest to note that by taking
$\ga =\pi /(4t)$ in \eqref{;equnif}, there holds the
following symmetry with respect to variables
$r,\la \ge 0$:
\begin{align*}
 \ex \!\left[
 e^{-r\cosh B_{t}}\cos \left(
 \la \sinh B_{t}+\frac{\pi }{4t}B_{t}
 \right)
 \right]
 =\ex \!\left[
 e^{-\la \cosh B_{t}}\cos \left(
 r \sinh B_{t}+\frac{\pi }{4t}B_{t}
 \right)
 \right] .
\end{align*}

In order to prove \pref{;punif}, we consider a contour
integral of a meromorphic function $f(z),\,z\in \C $,
of the form
\begin{align*}
 f(z)=\frac{\sinh (2z)}{\cosh (2z)+\cosh (2x)}
 \exp \left\{
 -\frac{1}{2t}\left(
 z-\frac{\pi }{2}i
 \right) ^{2}
 \right\} \g \left(
 z-\frac{\pi }{2}i
 \right) ,
\end{align*}
where $x\in \R $ is fixed and $\g $ denotes an
even entire function. If we take the same contour
as used in \sref{;prfmt} supposing $x\neq 0$, then we have
for a suitable choice of $\g $,
\begin{equation}\label{;rc}
 \begin{split}
 &\frac{1}{2\pi i}
 \int _{\R }d\xi \,\frac{\sinh (2\xi )}{\cosh (2\xi )+\cosh (2x)}
 \exp \left(
 -\frac{\xi ^{2}}{2t}+\frac{\pi ^{2}}{8t}
 \right) \\
 &\qquad \times \left\{
 \exp \left(
 \frac{\pi \xi }{2t}i
 \right) \g \left(
 \xi -\frac{\pi }{2}i
 \right)
 -\exp \left(
 -\frac{\pi \xi }{2t}i
 \right) \g \left(
 \xi +\frac{\pi }{2}i
 \right)
 \right\} \\
 &=\exp \left(
 -\frac{x^{2}}{2t}
 \right) \g (x)
 \end{split}
\end{equation}
by residue calculus applied to $f$. As will be seen, we are
allowed to take $\g (z)=\exp (-\la \cosh z)\cos (\ga z)$ and
$\exp (-\la \cosh z)\sin (\ga z)\sinh z$ for $\la \ge 0$ and
$\ga \in \R $, obtaining the following lemma: set
\begin{align*}
 \Sigma (x,\ga )
 \equiv \Sigma _{\la }(x,\ga )
 :=e^{\pi \ga /2}
 \sin \left(
 \frac{\pi x}{2t}+\la \sinh x+\ga x
 \right) ,\quad x\in \R .
\end{align*}

\begin{lem}\label{;lrc}
 For any $\la \ge 0$ and $\ga \in \R $, the relation \eqref{;rel3}
 holds for the following pairs of functions $\f (x)$ and $\g (x)$,
 $x\in \R $:
 \begin{align*}
  \thetag{i}& \
  \begin{cases}
   \f (x)=\dfrac{1}{2}\exp \left(
   \dfrac{\pi ^{2}}{8t}
   \right) \sinh x\left\{
   \Sigma (x,\ga )+\Sigma (x,-\ga )
   \right\} ,\\
   \g (x)=\exp (-\la \cosh x)\cos (\ga x);
  \end{cases}\\
  \thetag{ii}& \
  \begin{cases}
   \f (x)=-\dfrac{1}{2}\exp \left(
   \dfrac{\pi ^{2}}{8t}
   \right) \sinh x\cosh x\left\{
   \Sigma (x,\ga )-\Sigma (x,-\ga )
   \right\} ,\\
   \g (x)=\exp (-\la \cosh x)\sin (\ga x)\sinh x.
  \end{cases}
 \end{align*}
\end{lem}

\begin{proof}
 In the case $\g (z)=\exp (-\la \cosh z)\cos (\ga z),\,z\in \C $,
 because of the fact that
 \begin{align*}
  \left|
  \exp \left\{
  -\la \cosh \left(
  L+i\eta -\frac{\pi }{2}i
  \right)
  \right\}
  \right| =\exp (-\la \cosh L\sin \eta )\le 1
\end{align*}
for any $L\in \R $ and $0\le \eta \le \pi $ when $\la \ge 0$,
it follows readily that
\begin{align*}
 \left|
 \int _{0}^{\pi }d\eta \,f(L+i\eta )
 \right| \xrightarrow[|L|\to \infty ]{}0,
\end{align*}
which justifies \eqref{;rc}, yielding the pair \thetag{i}. The same justification
is also true in the case
$\g (z)=\exp (-\la \cosh z)\sin (\ga z)\sinh z,\,z\in \C $,
and leads to  \thetag{ii}. As it is clear that $\ga $ may be any
complex number in the above argument, it is also possible to
obtain \thetag{ii} from \thetag{i} by using the following
relations:
\begin{align*}
 \frac{\Sigma (x,\ga -i)-\Sigma (x,\ga +i)}{2i}&=-\Sigma (x,\ga )\cosh x, \\
 \frac{\cos \left\{ (\ga -i)x\right\}
 -\cos \left\{ (\ga +i)x\right\}
 }{2i}&=\sin (\ga x)\sinh x
\end{align*}
for any $x,\ga \in \R $. We conclude the proof of the lemma.
\end{proof}

\begin{rem}\label{;riden3}
 The relation \eqref{;iden3} is nothing but the case
 $\la =\ga =0$ in \thetag{i}; in other words, it is obtained
 simply by taking $\g \equiv 1$ in \eqref{;rc}.
\end{rem}

Using the above lemma, we prove \pref{;punif}.

\begin{proof}[Proof of \pref{;punif}]
 By \lref{;lequiv} and \thetag{i} of \lref{;lrc}, we have for any
 $r,\la \ge 0$ and $\ga \in \R $,
 \begin{align*}
  &\frac{1}{2}\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) \ex\!\left[
  e^{-r\cosh B_{t}}
  \sinh B_{t}\left\{
  \Sigma _{\la }(B_{t},\ga )+\Sigma _{\la }(B_{t},-\ga )
  \right\}
  \right] \\
  &=\ex \!\left[
  e^{-\la \cosh B_{t}}\cos (\ga B_{t})
  \cosh B_{t}\cos (r\sinh B_{t})
  \right] ,
 \end{align*}
 which shows that the first derivatives with respect to
 $\la $ of the following two expressions agree:
 \begin{align}
  &\frac{1}{2}\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) \ex\!\left[
  e^{-r\cosh B_{t}}
  \left\{
  \Tilde{\Sigma }_{\la }(B_{t},\ga )+\Tilde{\Sigma }_{\la }(B_{t},-\ga )
  \right\}
  \right] , \label{;ftn1}\\
  &\ex \!\left[
  e^{-\la \cosh B_{t}}\cos (\ga B_{t})\cos (r\sinh B_{t})
  \right] , \label{;ftn2}
 \end{align}
where we set
\begin{align*}
 \Tilde{\Sigma }_{\la }(x,\ga )
 :=e^{\pi \ga /2}\cos \left(
 \frac{\pi x}{2t}+\la \sinh x+\ga x
 \right) ,\quad x,\ga \in \R .
\end{align*}
(Since $\ex \!\left[ |\sinh B_{t}|\right] $ and $\ex [\cosh B_{t}]$
are finite, we are allowed to interchange the order of differentiation
and expectation.) Moreover, by the Riemann--Lebesgue lemma,
the former expression \eqref{;ftn1} converges to $0$ as $\la \to \infty $,
and by the bounded convergence theorem, the latter expression
\eqref{;ftn2} does as well. Therefore two expressions
\eqref{;ftn1} and \eqref{;ftn2} agree. Similarly,
by \lref{;lequiv} and \thetag{ii} of \lref{;lrc}, we have
\begin{align*}
  &-\frac{1}{2}\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) \ex\!\left[
  e^{-r\cosh B_{t}}
  \sinh B_{t}\cosh B_{t}\left\{
  \Sigma _{\la }(B_{t},\ga )-\Sigma _{\la }(B_{t},-\ga )
  \right\}
  \right] \\
  &=\ex \!\left[
  e^{-\la \cosh B_{t}}\sin (\ga B_{t})
  \sinh B_{t}\cosh B_{t}\cos (r\sinh B_{t})
  \right] .
 \end{align*}
 If we consider the following two expressions
 \begin{align}
  &\frac{1}{2}\exp \left(
  \frac{\pi ^{2}}{8t}
  \right) \ex\!\left[
  e^{-r\cosh B_{t}}
  \left\{
  \Tilde{\Sigma }_{\la }(B_{t},\ga )-\Tilde{\Sigma }_{\la }(B_{t},-\ga )
  \right\}
  \right] , \label{;ftn3}\\
  &\ex \!\left[
  e^{-\la \cosh B_{t}}\sin (\ga B_{t})\sin (r\sinh B_{t})
  \right] , \label{;ftn4}
 \end{align}
 then by differentiating them with respect to $\la $ and $r$
 successively, and by using the same reasoning as above,
 the last identity entails that those two expressions
 also agree. Consequently, the difference of \eqref{;ftn1} and \eqref{;ftn3}
 coincides with that of \eqref{;ftn2} and \eqref{;ftn4}, which proves
 the proposition.
\end{proof}

We recall from \cite[Proposition~3.3]{har} that for any $\la ,r\ge 0$,
\begin{align}\label{;jlt}
 \ex\!\left[
 \exp \left(
 -\la \eb{t}-\frac{\la ^{2}+r^{2}}{2}A_{t}
 \right)
 \right]
 =\ex\!\left[
 e^{-\la \cosh B_{t}}\cos (r\sinh B_{t})
 \right] .
\end{align}
Taking $\ga =0$ in \eqref{;equnif}, we thus have the following
relation:
\begin{align}\label{;jltd}
 \ex\!\left[
 \exp \left(
 -\la \eb{t}-\frac{\la ^{2}+r^{2}}{2}A_{t}
 \right)
 \right]
 =\exp \left(
 \frac{\pi ^{2}}{8t}
 \right) \ex\!\left[
 e^{-r\cosh B_{t}}\cos \left( \frac{\pi }{2t}B_{t}+\la \sinh B_{t}\right)
 \right] .
\end{align}
To our knowledge, the latter relation has not been noticed before,
and we will show below that the expression of the joint density
\eqref{;jl1} with the integral representation
\eqref{;irepr1} of $\Theta $ inserted in, can also be derived from it.
Since, in the above-cited paper \cite{har}, the relation \eqref{;jlt} is
obtained independently of the formula \eqref{;jl1}, by using
so-called Bougerol's identity and a certain invariance formula for
Cauchy variable, reasoning developed
below provides an independent, alternative proof of \eqref{;jl1}.

Before we proceed, we explain how we can easily derive from
\eqref{;jltd} simple explicit expressions of laws of
$A_{t}$ and $A^{(1)}_{t}$. Taking $\la =0$ in \eqref{;jltd}, we have
\begin{align}\label{;A0lt}
 \ex \!\left[
 \exp \left(
 -\frac{r^{2}}{2}A_{t}
 \right)
 \right]
 =\exp \left(
 \frac{\pi ^{2}}{8t}
 \right) \ex \!\left[
 e^{-r\cosh B_{t}}\cos \left(
 \frac{\pi }{2t}B_{t}
 \right)
 \right] .
\end{align}
Moreover, if we differentiate both sides of \eqref{;jltd} at $\la =0$,
we also have
\begin{align}\label{;A1lt}
 e^{t/2}\ex \!\left[
 \exp \left\{
 -\frac{r^{2}}{2}A^{(1)}_{t}
 \right\}
 \right]
 =\exp \left(
 \frac{\pi ^{2}}{8t}
 \right) \ex \!\left[
 e^{-r\cosh B_{t}}\sinh B_{t}\sin \left(
 \frac{\pi }{2t}B_{t}
 \right)
 \right] ,
\end{align}
where on the left-hand side, we used the Cameron--Martin relation.
Inserting the rewriting
\begin{align}\label{;rewr}
  e^{-r\cosh B_{t}}
  =\int _{0}^{\infty }dv\,\frac{\cosh B_{t}}{\sqrt{2\pi v^{3}}}
  \exp \left(
  -\frac{\cosh ^{2}B_{t}}{2v}
  \right) \exp \left(
  -\frac{r^{2}}{2}v
  \right)
\end{align}
into the right-hand sides of \eqref{;A0lt} and \eqref{;A1lt}, and
using Fubini's theorem, we see that for $v>0$,
\begin{align}
 \frac{\pr (A_{t}\in dv)}{dv}
 &=\exp \left(
 \frac{\pi ^{2}}{8t}
 \right)
 \ex \!\left[
 \frac{\cosh B_{t}}{\sqrt{2\pi v^{3}}}
 \exp \left(
 -\frac{\cosh ^{2}B_{t}}{2v}
 \right) \cos \left( \frac{\pi }{2t}B_{t}\right)
 \right] , \label{;A0law}\\
 \frac{\pr (\da{1}_{t}\in dv)}{dv}
 &=\exp \left(
 \frac{\pi ^{2}}{8t}-\frac{t}{2}
 \right)
 \ex \!\left[
 \frac{\sinh (2B_{t})}{\sqrt{2^{3}\pi v^{3}}}
 \exp \left(
 -\frac{\cosh ^{2}B_{t}}{2v}
 \right) \sin \left( \frac{\pi }{2t}B_{t}\right)
 \right] , \label{;A1law}
\end{align}
thanks to the injectivity of Laplace transform. Notice that
these two expressions agree with \eqref{;Amlaw} when
$\mu =0$ and $\mu =1$, respectively.

\begin{rem}\label{;rboug}
 On the other hand, by taking $\la =0$ in the former relation
 \eqref{;jlt}, we have
 \begin{align*}
  \ex \!\left[
  \exp \left(
  -\frac{r^{2}}{2}A_{t}
  \right)
  \right] =\ex \!\left[
  \cos (r\sinh B_{t})
  \right]
 \end{align*}
 for any $r\ge 0$, which relation is explained by Bougerol's original
 identity
 \begin{align*}
  \beta (A_{t})\stackrel{(d)}{=}\sinh B_{t}.
 \end{align*}
 Here and below, $\{ \beta (s)\} _{s\ge 0}$ denotes a one-dimensional
 standard Brownian motion independent of $B$. Moreover, if we differentiate
 both sides of \eqref{;jlt} at $\la =0$, then by
 the Cameron--Martin relation, we have
 \begin{align*}
  \ex \!\left[
  \exp \left\{
  -\frac{r^{2}}{2}A^{(1)}_{t}
  \right\}
  \right]
  =\ex \!\left[
  \cos \left\{
  r\sinh (B_{t}+\ve t)
  \right\}
  \right]
 \end{align*}
 for any $r\ge 0$, where on the right-hand side, $\ve $ is
 a Rademacher (or symmetric Bernoulli) random variable
 taking values $\pm 1$ with probability $1/2$, independently of
 $B$. By rewriting the left-hand side of the last identity as
 \[
 \ex \!\left[
 \cos \left\{
 r\beta \bigl( A^{(1)}_{t}\bigr)
 \right\}
 \right] ,
 \
 the injectivity of Fourier transform entails the
 following variant of Bougerol's identity:
 \begin{align*}
  \beta \bigl( A^{(1)}_{t}\bigr) \stackrel{(d)}{=}\sinh (B_{t}+\ve t).
 \end{align*}
 For Bougerol's identity and its variants including the above one,
 see the survey \cite{vak} by Vakeroudis; different kinds of extensions
 of Bougerol's identity may be found in \cite{har}.
\end{rem}

In what follows, we fix a measurable set
$\La \subset (0,\infty )$ arbitrarily
and denote $c_{t}=e^{\pi ^{2}/(8t)}$. We start with

\begin{lem}\label{;lher}
It holds that for any nonnegative integer $n$,
\begin{equation}\label{;eqlher}
 \begin{split}
 &\ex \!\left[
 H_{n}\!\left(
 -\frac{\eb{t}}{\sqrt{2A_{t}}}
 \right) ;\,A_{t}\in \La
 \right] \\
 &=c_{t}\!

 \int _{\La }dv\,
 \ex \!\left[
 \frac{\cosh B_{t}}{\sqrt{2\pi v^{3}}}\exp \left(
 -\frac{\cosh ^{2}B_{t}}{2v}
 \right) \cos \left(
 \frac{\pi }{2t}B_{t}+\frac{\pi }{2}n
 \right)
 \!\left(
 \sqrt{
 \frac{2}{v}
 }\sinh B_{t}
 \right) ^{n}
 \right] .
 \end{split}
\end{equation}
\end{lem}

\begin{proof}
 By the relations that for every $\la \ge 0$,
 \begin{align*}
  \exp \left(
  -\la \eb{t}-\frac{\la ^{2}}{2}A_{t}
  \right)
  =\sum _{n=0}^{\infty }
  \frac{1}{n!}H_{n}\!\left(
  -\frac{\eb{t}}{\sqrt{2A_{t}}}
  \right) \!
  \left(
  \frac{A_{t}}{2}
  \right) ^{n/2}\!\!\la ^{n}
 \end{align*}
 (cf.\ \cite[Equation~(4.9.3)]{leb}) and
 \begin{align*}
  \cos \left(
 \frac{\pi }{2t}B_{t}+\la \sinh B_{t}
 \right) =\sum _{n=0}^{\infty }\frac{1}{n!}
 \cos \left(
 \frac{\pi }{2t}B_{t}+\frac{\pi }{2}n
 \right) \! (\la \sinh B_{t})^{n},
 \end{align*}
 and by the fact that for any nonnegative integers $n,m$,
 \begin{align*}
  \ex \!\left[
  e^{nB_{t}}(A_{t})^{m}
  \right] <\infty && \text{and} &&
  \ex \!\left[
  |\sinh B_{t}|^{n}
  \right] <\infty ,
 \end{align*}
 successive differentiation
 of both sides of \eqref{;jltd} at $\la =0$ yields
 \begin{align*}
  \ex \!\left[
  H_{n}\!\left(
  -\frac{\eb{t}}{\sqrt{2A_{t}}}
  \right) \left(
  \frac{A_{t}}{2}
  \right) ^{n/2}\!\exp \left(
  -\frac{r^{2}}{2}A_{t}
  \right)
  \right]
  =c_{t}\,

  \ex \!\left[
  e^{-r\cosh B_{t}}\cos \left(
 \frac{\pi }{2t}B_{t}+\frac{\pi }{2}n
 \right) \sinh ^{n}B_{t}
  \right]
 \end{align*}
 for any nonnegative integer $n$. Plugging
 the rewriting \eqref{;rewr} into the right-hand side
 and using Fubini's theorem, we may appeal to the injectivity of
 Laplace transform to obtain the relation
 \begin{equation}\label{;jdh}
  \begin{split}
  &\left(
  \frac{v}{2}
  \right) ^{n/2}\ex \!\left[
  H_{n}\!\left(
  -\frac{\eb{t}}{\sqrt{2v}}
  \right) ;\,A_{t}\in dv
  \right] \\
  &=c_{t}\,

  \ex \!\left[
  \frac{\cosh B_{t}}{\sqrt{2\pi v^{3}}}\exp \left(
  -\frac{\cosh ^{2}B_{t}}{2v}
  \right) \cos \left(
 \frac{\pi }{2t}B_{t}+\frac{\pi }{2}n
 \right) \sinh ^{n}B_{t}
  \right] dv
  \end{split}
 \end{equation}
 for $v>0$, from which the assertion of the lemma follows.
\end{proof}

\begin{lem}\label{;lrecip}
 For any $\theta <1/2$, it holds that
 \begin{align*}
  \ex \!\left[
  \exp \left(
  \theta \frac{e^{2B_{t}}}{A_{t}}
  \right)
  \right] <\infty .
 \end{align*}
\end{lem}

\begin{proof}
 From \eqref{;A0law} (which corresponds to the case $n=0$
 in \eqref{;jdh} as well), it follows that for $v>0$,
 \begin{align*}
  \frac{\pr \!\left(
  1/A_{t}\in dv
  \right) }{dv}
  &\le c_{t}\,

  \ex \!\left[
  \frac{\cosh B_{t}}{\sqrt{2\pi v^{}}}
  \exp \left(
  -\frac{\cosh ^{2}B_{t}}{2}v
  \right)
  \right] \\
  &\le \frac{C}{\sqrt{v}}\exp \left(
  -\frac{v}{2}
  \right) ,
 \end{align*}
 where $C$ is a positive constant depending only on $t$.
 Proof completes by noting the identity in law
 $e^{-2B_{t}}A_{t}\stackrel{(d)}{=}A_{t}$; in fact, it holds more generally
 that
 \begin{align}\label{;jidenlaw}
  \left(
  e^{-2B_{t}}A_{t},\,A_{t}
  \right)
  \stackrel{(d)}{=}\left(
  A_{t},\,e^{-2B_{t}}A_{t}
  \right)
 \end{align}
 thanks to the time reversal:
 $\{ B_{t}-B_{t-s}\} _{0\le s\le t}\stackrel{(d)}{=}\{ B_{s}\} _{0\le s\le t}$.
\end{proof}

\begin{lem}\label{;lltrecip}
 For any $\al \ge 0$, it holds that
 \begin{align}
  &\ex \!\left[
  \exp \left(
  -\al \frac{e^{2B_{t}}}{2A_{t}}
  \right) ;\,A_{t}\in \La
  \right] \notag \\
  &=c_{t}\!

  \int _{\La }\frac{dv}{\sqrt{2\pi v^{3}}}\,
  \exp \left(
  -\frac{1}{2v}
  \right) \ex \!\left[
  \cosh B_{t}\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \frac{1}{\sqrt{1+\al }}
  \exp \left\{
  -\frac{\sinh ^{2}B_{t}}{2(1+\al )v}
  \right\}
  \right] . \label{;eqljd}
 \end{align}
\end{lem}

\begin{proof}
 We use the orthogonal expansion (see, e.g., \cite[Equation~(4.16.4)]{leb})
 \begin{align*}
  e^{-\al x^{2}}
  =\sum _{n=0}^{\infty }
  \frac{(-1)^{n}\al ^{n}}{2^{2n}n!(1+\al )^{n+1/2}}H_{2n}(x),\quad x\in \R
 \end{align*}
 (which is valid for $\al >-1$), together with Fubini's theorem, to
 reach the conclusion from \eqref{;eqlher}. To this end, we recall the
 fact that for any nonnegative integer $n$,
 \begin{align*}
  |H_{2n}(x)|\le C'\sqrt{(2n)!}\,2^{n}e^{x^{2}/2},\quad x\in \R ,
 \end{align*}
 where $C'$ is a constant whose approximate value is
 $1.086435$, and
 \begin{align*}
  \frac{(2n)!}{(n!)^{2}}\le 2^{2n}.
 \end{align*}
 (For the former inequality, see, e.g., \cite[p.~1006]{gr}; the latter is
 easily proven by induction.) Combining these inequalities, we see that
 \begin{align*}
  \sum _{n=0}^{\infty }\frac{\al ^{n}}{2^{2n}n!(1+\al )^{n}}
  \ex \!\left[
  \left|
  H_{2n}\!\left(
  -\frac{\eb{t}}{\sqrt{2A_{t}}}
  \right)
  \right|
  \right]
  \le
  C'\sum _{n=0}^{\infty }\left(
  \frac{\al }{1+\al }
  \right) ^{n}\ex \!\left[
  \exp \left(
  \frac{e^{2B_{t}}}{4A_{t}}
  \right)
  \right] ,
 \end{align*}
 which is finite thanks to \lref{;lrecip}. On the other hand,
 \begin{align*}
  &\int _{\La }dv\,
  \ex \!\left[
  \frac{\cosh B_{t}}{\sqrt{v^{3}}}
  \exp \left(
  -\frac{\cosh ^{2}B_{t}}{2v}
  \right) \sum _{n=0}^{\infty }
  \frac{\al ^{n}}{2^{2n}n!(1+\al )^{n}}
  \left(
  \frac{2}{v}\sinh ^{2}B_{t}
  \right) ^{n}
  \right] \\
  &=\int _{\La }\frac{dv}{\sqrt{v^{3}}}\,
  \exp \left(
  -\frac{1}{2v}
  \right) \ex \!\left[
  \cosh B_{t}\exp \left\{
  -\frac{\sinh ^{2}B_{t}}{2(1+\al )v}
  \right\}
  \right] ,
 \end{align*}
 which is also finite. Therefore if we multiply by
 \begin{align*}
  \frac{(-1)^{n}\al ^{n}}{2^{2n}n!(1+\al )^{n+1/2}}
 \end{align*}
 both sides of \eqref{;eqlher} with $n$ therein replaced by $2n$,
 and sum up over all nonnegative integers $n$, then
 Fubini's theorem leads to the conclusion.
\end{proof}

Appealing to the injectivity of Laplace transform in the last lemma,
we arrive at the following proposition: recall $c_{t}=e^{\pi ^{2}/(8t)}$.

\begin{prop}\label{;pjd}
 The joint law of $e^{2B_{t}}/A_{t}$ and $A_{t}$ is expressed as
 \begin{align*}
  &\pr \!\left(
  \frac{e^{2B_{t}}}{A_{t}}\in du,\,A_{t}\in dv
  \right) \\
  &=\frac{1}{2\pi \sqrt{uv^{3}}}
  \exp \left(
  \frac{\pi ^{2}}{8t}-\frac{u}{2}-\frac{1}{2v}
  \right)
  \ex \!\left[
  \cosh B_{t}\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \cos \left(
  \sqrt{
  \frac{u}{v}
  }\sinh B_{t}
  \right)
  \right] dudv
 \end{align*}
 for $u,v>0$.
\end{prop}

\begin{proof}
 We plug the rewriting
 \begin{align*}
  \frac{1}{\sqrt{1+\al }}
  \exp \left\{
  -\frac{\sinh ^{2}B_{t}}{2(1+\al )v}
  \right\}
  =\int _{0}^{\infty }\frac{du}{\sqrt{2\pi u}}
  \exp \left(
  -\frac{1+\al }{2}u
  \right) \cos \left(
  \sqrt{
  \frac{u}{v}
  }\sinh B_{t}
  \right)
 \end{align*}
 into the right-hand side of \eqref{;eqljd} and use Fubini's theorem
 to see that for any $\al \ge 0$,
 \begin{align*}
  &\ex \!\left[
  \exp \left(
  -\frac{\al }{2}\cdot \frac{e^{2B_{t}}}{A_{t}}
  \right) ;\,A_{t}\in \La
  \right] \\
  &=\int _{0}^{\infty }du\,\exp \left(
  -\frac{\al }{2}u
  \right)
  \int _{\La }\frac{du}{2\pi \sqrt{uv^{3}}}
  \exp \left(
  \frac{\pi ^{2}}{8t}-\frac{u}{2}-\frac{1}{2v}
  \right) \\
  &\qquad \qquad \qquad  \qquad \qquad \quad \times
  \ex \!\left[
  \cosh B_{t}\cos \left(
  \frac{\pi }{2t}B_{t}
  \right) \cos \left(
  \sqrt{
  \frac{u}{v}
  }\sinh B_{t}
  \right)
  \right] .
 \end{align*}
 Since a measurable set $\La \subset (0,\infty )$
 is also arbitrary, we obtain the proposition.
\end{proof}

It then follows readily that the above proposition, together with
the relation \eqref{;irepr1}, entails \eqref{;jl2} and hence \eqref{;jl1}.
We also see from the proposition that for $u,v>0$,
\begin{align*}
 &\pr \!\left(
 e^{-2B_{t}}A_{t}\in du,\,A_{t}\in dv
 \right) \\
 &=\frac{1}{2\pi \sqrt{u^{3}v^{3}}}
 \exp \left\{
 \frac{\pi ^{2}}{8t}-\frac{1}{2}\left(
 \frac{1}{u}+\frac{1}{v}
 \right)
 \right\}
 \ex \!\left[
 \cosh B_{t}\cos \left(
 \frac{\pi }{2t}B_{t}
 \right) \cos \left(
 \frac{\sinh B_{t}}{\sqrt{uv}}
 \right)
 \right] dudv,
\end{align*}
which is indeed symmetric with respect to $u$ and
$v$, reflecting the fact \eqref{;jidenlaw}.

Because of an intimate connection with exponential functionals
of Brownian motion, we conclude this section with exploration of
the law of some additive functional of Bessel process.

For every $\nu \in \R $ and $a\ge 0$, we denote by
$\bigl( R=\{ R_{t}\} _{0\le t<\tau _{0}},\prb{\nu }{a}\bigr) $
a Bessel process of index $\nu $ (or dimension
$\delta =2(\nu +1)$)
starting from $a$: $\prb{\nu }{a}(R_{0}=a)=1$. Here $\tau _{0}$
is the first hitting time of $R$ to the origin; recall that
for $a>0$, $\prb{\nu }{a}(\tau _{0}=\infty )=1$ when $\nu \ge 0$,
whereas $\prb{\nu }{a}(\tau _{0}<\infty )=1$ when $\nu <0$.
Fix $\nu \in \R $ and $a,t>0$ below. We recall from
\cite[Equation~(2.b$'$)]{yor92} that for any nonnegative,
measurable functional $f(w(s),s\le t),\,w\in C([0,t];\R )$,
on the space of real-valued continuous functions $w$ over $[0,t]$,
\begin{align}\label{;rcm}
 \exb{\nu }{a}\!\left[
 f(R_{s},s\le t);\,t<\tau _{0}
 \right]
 =\exb{0}{a}\!\left[
 f(R_{s},s\le t)\left(
 \frac{R_{t}}{a}
 \right) ^{\nu }\!
 \exp \left(
 -\frac{\nu ^{2}}{2}\int _{0}^{t}
 \frac{ds}{R^{2}_{s}}
 \right)
 \right] .
\end{align}
Here $\exb{\nu }{a}$ stands for expectation
relative to $\prb{\nu }{a}$. From the above relation, we
have in particular, for every measurable set $\La \subset (0,\infty )$
and for every $\mu \in \R $,
\begin{equation}\label{;rcmd}
 \begin{split}
 &\exb{\nu }{a}\!\left[
 \exp \left(
 -\frac{\mu ^{2}}{2}\int _{0}^{t}
 \frac{ds}{R^{2}_{s}}
 \right) ;\,R_{t}\in \La ,t<\tau _{0}
 \right] \\
 &=\exb{0}{a}\!\left[
 \left(
 \frac{R_{t}}{a}
 \right) ^{\nu }\!
 \exp \left(
 -\frac{\rho ^{2}}{2}\int _{0}^{t}
 \frac{ds}{R^{2}_{s}}
 \right) ;\,R_{t}\in \La
 \right] \\
 &=\exb{\rho }{a}\!\left[
 \left( \frac{R_{t}}{a}\right) ^{\nu -\rho }\!;\,R_{t}\in \La
 \right] ,
 \end{split}
\end{equation}
where we denote $\rho =\sqrt{\mu ^{2}+\nu ^{2}}$.
Using the explicit form of the transition density of Bessel
process (see, e.g., Equation~\thetag{2.d} in the above-cited
reference), we see that the last expression is written as
\begin{align*}
 \int _{\La }dr\,\frac{r}{t}\left(
 \frac{r}{a}
 \right) ^{\nu }\!\exp \left(
 -\frac{a^{2}+r^{2}}{2t}
 \right) I_{\rho }\left(
 \frac{ar}{t}
 \right) .
\end{align*}
Recalling the characterization \eqref{;LTt} of $\Theta $, we use
Fubini's theorem to see that the above expression is further
rewritten as
\begin{align*}
 \int _{0}^{\infty }du\,\exp \left(
 -\frac{\mu ^{2}}{2}u
 \right)
 \frac{1}{ta^{\nu }}\exp \left(
 -\frac{a^{2}}{2t}-\frac{\nu ^{2}}{2}u
 \right)
 \int _{\La }dr\,r^{\nu +1}\exp \left(
 -\frac{r^{2}}{2t}
 \right) \Theta \left(
 \frac{ar}{t},u
 \right) .
\end{align*}
Therefore, by comparing the last expression with the left-hand side of
\eqref{;rcmd}, the injectivity of Laplace transform, as well as
arbitrariness of $\La $, entails
\begin{equation}\label{;rjlaw}
 \begin{split}
  &\prb{\nu }{a}\!\left( R_{t}\in dr,\,
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,t<\tau _{0}
  \right) \\
  &=\frac{1}{ta^{\nu }}r^{\nu +1}\exp \left(
 -\frac{a^{2}}{2t}-\frac{r^{2}}{2t}-\frac{\nu ^{2}}{2}u
 \right) \Theta \left(
 \frac{ar}{t},u
 \right) drdu
 \end{split}
\end{equation}
for $r,u>0$. For the above formula in the case $\nu \ge 0$,
we refer to \cite[p.~392, Formula~1.20.8]{bs}; note that the
formula in the case $\nu <0$ is related to that in the case
$\nu >0$ via the following absolute continuity relationship
deduced from \eqref{;rcm}:
\begin{align*}
 \exb{\nu }{a}\!\left[
 f(R_{s},s\le t);\,t<\tau _{0}
 \right]
 =\exb{|\nu |}{a}\!\left[ f(R_{s},s\le t)
 \left(
 \frac{R_{t}}{a}
 \right) ^{\nu -|\nu |}
 \right] ,
\end{align*}
where $\nu $ is arbitrary.
We also remark that the relation \eqref{;rcm} may be proven  (see, e.g.,
\cite[Section~2]{mySI}, \cite[Subsection~A.1]{har15})
by using the Cameron--Martin
formula combined with so-called Lamperti's relation asserting
that there exists a Bessel process
$R$ of index $\nu $ starting from $a$, such that
\begin{align}\label{;lam}
 ae^{B^{(\nu )}_{s}}=R_{a^{2}A^{(\nu )}_{s}},\quad s\ge 0.
\end{align}

By \eqref{;rjlaw}, it holds in particular that for $u>0$,
\begin{equation}\label{;rlaw}
 \begin{split}
  &\prb{\nu }{a}\!\left(
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,t<\tau _{0}
  \right) \\
  &=\frac{1}{ta^{\nu }}\exp \left(
 -\frac{a^{2}}{2t}-\frac{\nu ^{2}}{2}u
 \right)
 \int _{0}^{\infty }dr\,r^{\nu +1}\exp \left(
 -\frac{r^{2}}{2t}
 \right) \Theta \left(
 \frac{ar}{t},u
 \right) du.
 \end{split}
\end{equation}
In view of \eqref{;Amlaw0}, if we define
\begin{align*}
 D(\mu ,t,v):=\frac{\pr (A^{(\mu )}_{t}\in dv)}{dv},
 \quad \mu \in \R ,\ t,v>0,
\end{align*}
then \eqref{;rlaw} is expressed as
\begin{align}\label{;rlawd}
 e^{\delta u}D(\nu +2,u,t/a^{2})\,du.
\end{align}
Therefore, when $\nu $ is an integer not less than $-2$, or
equivalently when the dimension $\delta =2(\nu +1)$ is
an even integer not less than $-2$, the expression
\eqref{;rlaw} of the density function with respect to the
Lebesgue measure $du$, is simplified into a single integral;
for example, when $\nu =-2$,
\begin{align*}
 &\prb{-2}{a}\!\left(
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,t<\tau _{0}
  \right) \\
  &=\exp \left(
  \frac{\pi ^{2}}{8u}
  -2u
  \right) \ex \!\left[
  \frac{a^{3}\cosh B_{u}}{\sqrt{2\pi t^{3}}}
  \exp \left(
  -\frac{a^{2}\cosh ^{2}B_{u}}{2t}
  \right) \cos \left(
  \frac{\pi }{2u}B_{u}
  \right)
  \right] du,
\intertext{and when $\nu =-1$, }
 &\prb{-1}{a}\!\left(
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,t<\tau _{0}
  \right) \\
  &=\exp \left(
  \frac{\pi ^{2}}{8u}
  -\frac{u}{2}
  \right) \ex \!\left[
  \frac{a^{3}\sinh (2B_{u})}{\sqrt{2^{3}\pi t^{3}}}
  \exp \left(
  -\frac{a^{2}\cosh ^{2}B_{u}}{2t}
  \right) \sin \left(
  \frac{\pi }{2u}B_{u}
  \right)
  \right] du,
\end{align*}
by virtue of \eqref{;A0law} and \eqref{;A1law}.

\begin{rem}\label{;rfinal}
\thetag{1} It is possible to obtain the expression \eqref{;rlawd} directly
 from Lamperti's relation \eqref{;lam}. To see this, note that if we define
 $\anu _{t},\,0\le t<a^{2}\da{\nu }_{\infty }$, by the inverse process
 of $a^{2}\da{\nu }_{t},\,t\ge 0$, then the relation \eqref{;lam}
 entails in particular that under $\prb{\nu }{a}$,
 \begin{align}\label{;idenlawr}
  \left\{
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}
  \right\} _{0\le t<\tau _{0}}\stackrel{(d)}{=}
  \Bigl\{
  \anu _{t}
  \Bigr\} _{0\le t<a^{2}\da{\nu }_{\infty }},
 \end{align}
 where we set $\da{\nu }_{\infty }:=\lim \limits_{t\to \infty }\da{\nu }_{t}$.
 Using the above identity in law, we have for any $T,u>0$,
 \begin{equation}\label{;lastexp}
  \begin{split}
  \int _{0}^{T}dt\,\prb{\nu }{a}\! \left(
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}\le u,\,t<\tau _{0}
  \right)
  &=\exb{\nu }{a}\! \left[
  \int _{0}^{\min \{ T,\,\tau _{0}\} }\!dt\,
  \ind _{\left\{
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}\le u
  \right\} }
  \right] \\
  &=\ex \! \left[
  \int _{0}^{\min \{ T,\,a^{2}\da{\nu }_{\infty }\} }\!dt\,
  \ind _{\{ \anu _{t}\le u\} }
  \right] ,
  \end{split}
 \end{equation}
 where we used Fubini's theorem for the first equality. The integrand
 in the last expectation is equal, by changing the
 variables with $t=a^{2}\da{\nu }_{s},\,0\le s\le \anu _{T}$, to
 \begin{align*}
  a^{2}\int _{0}^{\min \{ \anu _{T},\,u\} }\!ds\,e^{2\db{\nu }_{s}}.
 \end{align*}
 Therefore, by Fubini's theorem, we see from \eqref{;lastexp} that
 for any $T,u>0$,
 \begin{align*}
  \int _{0}^{T}dt\,\prb{\nu }{a}\! \left(
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}\le u,\,t<\tau _{0}
  \right)
  &=a^{2}\int _{0}^{u}ds\,\ex \!\left[
  e^{2\db{\nu }_{s}};\,s\le \anu _{T}
  \right] \\
  &=a^{2}\int _{0}^{u}ds\,\ex \!\left[
  e^{2\db{\nu }_{s}};\,a^{2}\da{\nu }_{s}\le T
  \right] \\
  &=a^{2}\int _{0}^{u}ds\,e^{\delta s}\,\pr \!\left(
  \da{\nu +2}_{s}\le T/a^{2}
  \right) ,
 \end{align*}
 which entails the expression \eqref{;rlawd}. Here in the
 last displayed equations, the second line
 is due to the definition of $\anu _{\cdot }$, and
 we used the Cameron--Martin relation for the third.
 The above reasoning, in turn, reveals that once the
 formula \eqref{;rlaw} is established, the representation
 \eqref{;Amlaw0} can be obtained from it by using
 Lamperti's relation; as far as we know, such
 derivation of \eqref{;Amlaw0} has not been performed
 in the existing literature. Recall that as noted above,
 the formula \eqref{;rlaw} may be seen as a consequence
 of Lamperti's relation as well, combined with the
 Cameron--Martin formula and the explicit form of the
 transition density of Bessel process.

 In addition, it also holds that by \eqref{;idenlawr},
 \begin{align*}
  \prb{\nu }{a}\! \left(
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}\le u,\,t<\tau _{0}
  \right)
  &=\pr \!\left(
  \anu _{t}\le u,\,t<a^{2}\da{\nu }_{\infty }
  \right) \\
  &=\pr \!\left(
  \da{\nu }_{u}\ge t/a^{2}
  \right) ,
 \end{align*}
 which seems less useful but yields the following relation
 of interest as to the function $D$:
 \begin{align*}
  \int _{0}^{u}ds\,e^{\delta s}D(\nu +2,s,t)
  =\int _{t}^{\infty }dv\,D(\nu ,u,v)
  \quad \text{for any $u,t>0$ and $\nu \in \R $.}
 \end{align*}

 \noindent
 \thetag{2} More generally than \eqref{;idenlawr}, it also follows
 from \eqref{;lam} that under $\prb{\nu }{a}$,
 \begin{align*}
  \left\{ \left( R_{t},\,
  \int _{0}^{t}\frac{ds}{R_{s}^{2}}
  \right)
  \right\} _{0\le t<\tau _{0}}\stackrel{(d)}{=}
  \left\{ \left( a\exp \bigl( \db{\nu }_{\anu _{t}}\bigr) ,\,
  \anu _{t}\right)
  \right\} _{0\le t<a^{2}\da{\nu }_{\infty }}
 \end{align*}
 for every $\nu \in \R $ and $a>0$. Then by repeating the same reasoning
 as in \thetag{1}, one may show that for $r,t,u>0$,
 \begin{equation}\label{;rjlawb}
 \begin{split}
  &\prb{\nu }{a}\!\left( R_{t}\in dr,\,
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,t<\tau _{0}
  \right) dt\\
  &=a^{2}e^{\delta u}\,\pr \!\left(
  ae^{\db{\nu +2}_{u}}\in dr,\,a^{2}\da{\nu +2}_{u}\in dt
  \right) du.
 \end{split}
 \end{equation}
 By taking $\nu =-2$ and $a=1$ in \eqref{;rjlawb}, the expression \eqref{;jl2},
 hence \eqref{;jl1}, is recovered from \eqref{;rjlaw} as well.
 Although both of them are based on Lamperti's relation \eqref{;lam},
 the above derivation of \eqref{;jl1} is different from the original one
 by \cite{yor92}, in the sense that what is essentially considered there
 is the integral (more precisely, the Laplace transform) with respect to $u$
 of the joint distribution on the right-hand side of \eqref{;rjlawb},
 whereas as reasoning in \thetag{1} shows, we consider the integral
 with respect to $t$ of the joint distribution on the left-hand side of
 \eqref{;rjlawb}, which readily leads us to a more specific relation than ever
 noticed between the two joint distributions as exhibited in
 \eqref{;rjlawb}.

 Finally, it is also immediate to see that the relation
 \eqref{;rjlawb} further extends in such a way that  for $r,t,u,v>0$,
 \begin{equation*}
 \begin{split}
  &\prb{\nu }{a}\!\left( R_{t}\in dr,\,
  \int _{0}^{t}\frac{ds}{R^{2}_{s}}\in du,\,
  \int _{0}^{t}V(R_{s})\,ds\in dv,\,
  t<\tau _{0}
  \right) dt\\
  &=a^{2}e^{\delta u}\,\pr \!\left(
  ae^{\db{\nu +2}_{u}}\in dr,\,a^{2}\da{\nu +2}_{u}\in dt,\,
  a^{2}\!\int _{0}^{u}V\bigl( ae^{\db{\nu +2}_{s}}\bigr) e^{2\db{\nu +2}_{s}}\!ds\in dv
  \right) du ,
 \end{split}
 \end{equation*}
 where $V$ is any measurable function on $(0,\infty )$.
\end{rem}

redpajama_set_name: RedPajamaArXiv


to celebrate Valentine's Day or anytime with a snazzy pencil!
double knot. Then, glue two felt hearts (or another shape) together on the other end.
You could also stick two craft foam stickers together for the same effect.
or for your child to make as inexpensive Valentine Gifts for their classmates!


text:
Hull Pilgrims
After the local elections in May, I wrote about Unison's countdown to power in Hull. Clearly they thought 'it was them what won it' as the picture below proudly demonstrates. The campaign they fought on behalf of Labour in the city could not have been as effective if it wasn't for the money we paid their Pilgrims - those who are paid to work full-time for their union instead of the job we pay them for.
It's not just Unison who benefit of course. Unite's convenor in the city, Dave Mathieson, is also a Pilgrim. Last year he called for councillors to take a pay cut. He wasn't advocating a pay cut for himself though, and I challenged him to leave his job at the council and work full-time for his union, with them picking up the tab. Unsurprisingly, he didn't take me up on my suggestion.
Mr Mathieson is a very busy man. Not only is he on Unite's executive council, he is also their Chair of Local Government, as well as being a member of the Labour Party. On 30 June he spoke at a rally of striking public sector workers in Hull. He said he was disgusted that his union had not joined the strike, and went on to say:
"Unless the Government backs off, there will be more strikes in August, September and October and I am sure Unite will be there alongside you. This is not just a public sector fight, it's a fight for society."
Even though he wasn't on strike, we still had to pay his salary whilst he encouraged others to do so.
Thanks to a supporter who sent a Freedom of Information request to Hull City Council, I now know how much Mr Mathieson and his colleagues in other unions cost taxpayers in Hull, and this information was not easy to obtain. The council was not willing to let us know the full facts. He earns a salary of £31,754, and when you include other costs of £11,033.28, the total bill to taxpayers of Mr Mathieson's political campaigning is £42,787.28. The figures for Unison are £128,361 and the GMB £39,371. In total those three unions cost Hull taxpayers over £219K, and this is not including teachers who could easily double that figure.
Unions argue taxpayers are not funding them. They say we are merely giving union reps time off to represent their members' interests in industrial disputes. Facilities time. That's all.
Claiming you are now in power, and swanning around the country and making speeches at various rallies cannot be described as time spent looking after your members working for the council. It is blatant political campaigning, and because we pay for it, unions can divert more of their resources into organising rallies such as the TUC rally in London last March. They can spend more time and money organising strikes, and even allow Mr Mathieson time off to be part of a trade union delegation to Venezuela.
I will not only challenge Mr Mathieson again to leave his job at the council, I will challenge the other four full-time Pilgrims too. As I said a year ago, if they do, I will be one of the first to congratulate them.
In other news, a pig was seen flying over the Guildhall.

redpajama_set_name: RedPajamaCommonCrawl


It is with great pride that Androïde joins the new Lenovo Data Center Partner program, specifically recognizing its partners' expertise with data center solutions.
In recognition of the Lenovo Data Center Sales and Lenovo Data Center Technical Sales certifications held by several staff at Androïde and its significant business volume in terms of Lenovo Data Center solutions, Androïde is delighted to have been directly promoted to the Gold partner level.
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Soyez à l'affût des dernières avancées technologiques pouvant contribuer au succès de votre organisation!
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Tory and I get together to catch up on all of the recent developments with Scientology. Tory told me that she has now done over 500 videos and have over 2 million views. She told me that people are coming out of the group in record numbers. We both agree the group is failing, despite all of the efforts to do PR. Yeah!