#  BOUNDARY CROSSING PROBLEMS AND FUNCTIONAL TRANSFORMATIONS FOR ORNSTEIN-
UHLENBECK PROCESSES

Aria<EMAIL_ADDRESS>Larbi<EMAIL_ADDRESS>Massimiliano<EMAIL_ADDRESS>
(1 Department of Statistics, University of Warwick, Coventry, CV4 7AL, United
Kingdom.)

###### Abstract

We are interested in the law of the first passage time of driftless Ornstein-
Uhlenbeck processes to time varying thresholds. We show that this problem is
connected to the law of the first passage time of the process to some two-
parameter family of functional transformations which, for specific values of
the parameters, appears in a realisation of a standard Ornstein-Uhlenbeck
bridge. We provide three different proofs of this connection. The first proof
is based on a similar result to the case of the Brownian motion, the second
uses a generalisation of the so-called Gauss-Markov processes and the third
relies on the Lie group symmetry method applied to the Fokker-Planck equation
of the Ornstein-Uhlenbeck process. We investigate the properties of this
transformation and study the algebraic and analytical properties of an
involution operator which is used in constructing it. We also show that this
transformation maps the space of solutions of Sturm-Liouville equations into
the space of solutions of the associated nonlinear ordinary differential
equations. Lastly, we discuss the interpretation of such transformations
through the method of images and give new examples of curves with explicit
first passage time densities.

Keywords: First passage times; Lie algebras; Sturm-Liouville equations;
Orstein-Uhlenbeck; Fokker Planck equation; Ornstein-Uhlenbck bridge; Brownian
motion.
2020 Mathematics Subject Classification: Primary 35K05, 60J50, 60J60.

##  1 Introduction

Let $U:=(U_{t})_{t\geq 0}$ be a one-dimensional driftless Ornstein-Uhlenbeck
(OU for short) process defined on a filtered probability space
$(\Omega,(\mathcal{F})_{t\geq 0},\mathcal{F},\mathbb{P})$ as the unique
solution to the following stochastic differential equation (SDE for short)

$dU_{t}=-kU_{t}dt+dB_{t},\;\;\;U_{0}=x\in\mathbb{R},$ (1)

where $(B_{t})_{t\geq 0}$ is a standard Brownian motion (BM) starting at 0 and
$k\in\mathbb{R}$ is a constant. The OU process is a Gauss-Markov process with
transition density function given by

$p_{t}(x,y):=\frac{\partial}{\partial y}\mathbb{P}(U_{t}\leq
y|U_{0}=x)=\frac{e^{kt}}{\sqrt{r(t)}}\phi\left(\frac{ye^{kt}-x}{\sqrt{r(t)}}\right),$
(2)

where $\phi(x)=e^{-\frac{x^{2}}{2}}/\sqrt{2\pi}$ is the probability density
function of the standard normal distribution and $r(t)=(e^{2kt}-1)/2k$ with
its inverse given by $s(t)=\ln{(2kt+1)}/2k$. It is well known by the Dambis,
Dubins-Schwartz theorem (see, e.g., Theorem V.1.6 in [25]), that the OU
process can be written in terms of a time changed BM $(W_{t})_{t\geq 0}$ as

$U_{t}=e^{-kt}\big{(}x+W_{r(t)}\big{)},\;\;\;t\geq 0.$ (3)

We are interested in the distribution of the first passage time (FPT) or the
random stopping time

$\hat{T}^{f}=\inf\\{t:U_{t}=f(t)\\},$

where $x=0$, the time-varying threshold
$f\in\mathcal{C}([0,\infty),\mathbb{R})$ satisfies $f(0)\neq 0$ and the usual
convention $\inf\\{\emptyset\\}=\infty$ applies. $\mathcal{C}(I,K)$, for some
subintervals $I$ and $K\subseteq\mathbb{R}$, denotes the space of continuous
functions from $I$ into $K$. Such a boundary crossing problem for the case
when the process is a BM, i.e. $k=0$, has been of extensive study for over a
century. We refer to [2] for some recent review on this topic. For the OU
process with drift, well known explicit results only exist for constant [1] or
hyperbolic type boundaries [9]. Such a hitting time problem is of great
interest, as the OU process has been used in many applications to model
objects such as interest rates in finance or the evolution of the neuronal
membrane voltages in neuroscience, see e.g. [1] and citations therein. In this
paper, we focus on the OU without drift, as the results for the OU process
with drift can be directly obtained from our results after some
transformations.
The goal of this paper is to derive, through different methods, an explicit
analytical expression between the distribution of the FPT of the OU process to
a continuously differentiable boundary $f$ and some two-parameter family of
curves, $\hat{S}^{\alpha,\beta}f$ (equation (13)), further developing the
results for the BM obtained in [3]. Note that the continuously differentiable
assumption of $f$ is linked to Starssen’s [29] result for the BM and is
crucial in ensuring that the distribution of $\hat{T}^{f}$ has a continuous
density with respect to the Lebesgue measure on $(0,\infty)$. After
introducing some notations and recalling some results for the FPT of the
Brownian motion, in Section 2 we recall the OU bridge and its properties,
which will be essential in understanding the properties of the two-parameter
transformation $\hat{S}^{\alpha,\beta}$ studied in Section 3, together with
the properties of other transformations of interest.

The main result of this paper, Theorem 4.1, is presented in Section 4, and it
is proved in three alternative ways. In the first proof, we directly use the
relationship between the FPT of an OU and that of a BM and use Theorem 1.2 in
[3] to obtain Theorem 4.1. The second proof consists of using a generalisation
of the Gauss-Markov processes which were introduced in Section 3.2 of [3] and
finding an analogue version of that proof in our case. We develop a third
different approach using the Lie algebra to find the symmetries of the OU
Fokker-Planck equation

$\frac{\partial h}{\partial t}(x,t)=\frac{1}{2}\frac{\partial^{2}h}{\partial
x^{2}}(x,t)+kx\frac{\partial h}{\partial x}(x,t)+kh(x,t),$ (4)

and then use these symmetries (Lemma 4.7) to construct a certain function,
$\hat{h}^{\alpha,\beta}(x,t)$ (equation (23)), derive our transformation
$\hat{S}^{\alpha,\beta}f$ and relate the FPT distribution of $\hat{T}^{f}$ to
$\hat{T}^{\hat{S}^{\alpha,\beta}f}$. Within the same section, we also discuss
the asymptotics of the distribution of $\hat{T}^{\hat{S}^{\alpha,\beta}f}$. We
provide the analogue of the Kolmogorov–Erdös–Petrovski transience test [11] in
the OU case and show its connection to the asymptotic behaviour of the FPT.
Lastly, in Section 5, we use the method of images to obtain new classes of
boundaries such that explicit FPT distributions can be obtained and use our
$\hat{S}^{\alpha,\beta}f$ transformation (13) to produce new examples. A
limitation of this method is that it only works for boundaries with certain
properties given in Lemma 5.3.

As we were finalising the paper, we discovered that the
$\hat{S}^{\alpha,\beta}$ transformation, a variant of the boundary crossing
identity (15) in Theorem 4.1 and the Lie symmetries (24) had previously
appeared in [20] (our (13) can be obtained by setting $A=k^{2},B=0$ in
equation (39) therein), using the Lie approach. When comparing our results, we
found misprints in one of their Lie symmetries and boundary crossing identity,
as discussed in Section 4.3.

###  1.1 Notations and previous results

Here, we introduce the notations that will be used throughout the paper. Some
of the discussed operators and notations can also be found in Alili and Patie
[3], but we recall them here for convenience.
The nonlinear operator $\tau$ is defined on the space of functions whose
reciprocals are square integrable in some (possibly infinite) interval of
$\mathbb{R}^{+}$ by $\tau f(t)=\int_{0}^{t}[f(s)]^{-2}ds$ and we use it to
define

$\hat{A}(a,b)=\left\\{\pm f\in\mathcal{C}([0,a],\mathbb{R}^{+}):\;\tau
f\left(\operatorname{sgn}(k)s\big{(}\operatorname{sgn}(k)a\big{)}\right)=\frac{b}{1-\mathbbm{1}_{\\{k<0\\}}2kb}\right\\},$

where $a,b\in\mathbb{R}^{+}$ and $\operatorname{sgn}(k)$ is the sign of k. We
also denote by $\hat{A}_{\infty}$ the set of continuous functions which are of
constant sign on some nonempty interval $[0,l],\;l>0,$ yielding the following
decomposition:
$\hat{A}_{\infty}=\bigcup\limits_{a>0}\bigcup\limits_{b>0}\hat{A}(a,b)$. Now,
for $\alpha\in\mathbb{R}^{*},\beta\in\mathbb{R}$, where
$\mathbb{R}^{*}=\mathbb{R}\setminus\\{0\\}$, we introduce the following family
of nonlinear operators acting on $\hat{A}_{\infty}$

$\Pi^{\alpha,\beta}f(t)=f(t)(\alpha+\beta\tau f(t)).$ (5)

As explained in the beginning of Section 2 of [3] and [25](Appendix 8), these
operators are closely related to the Sturm-Liouville equation

$\phi^{\prime\prime}=\mu\phi,$ (6)

where $\mu$ denotes a positive Radon measure on $\mathbb{R}^{+}$ and
$\phi^{\prime\prime}$ is the second derivative in the sense of distributions.
In fact, if $\phi$ solves (6), then the the vectorial space
$\\{\Pi^{\alpha,\beta}\phi;\alpha,\beta\in\mathbb{R}\\}$ is the set of
solutions to the same equation. Moreover, all positive solutions are convex
and described by the set $\\{\Pi^{\alpha,\beta}\varphi;\alpha>0,\beta\geq
0\\}$, where $\varphi$ is the unique, positive, decreasing solution such that
$\varphi(0)=1$. Next, we let $\rho$ to be the inversion operator acting on the
space of continuous monotone functions i.e., $\rho f\circ f(t)=t$, where
$\circ$ denotes the composition of functions. In the paper by Alili and Patie
[3], the problem of the first hitting time of a BM to a certain transformation
of the boundary $f$ is discussed. The latter transformation with its
decomposition is the following

$S^{\alpha,\beta}f(t)=\Sigma\circ\Pi^{\alpha,-\beta}\circ\Sigma
f=\left(\frac{1+\alpha\beta
t}{\alpha}\right)f\left(\frac{\alpha^{2}t}{1+\alpha\beta t}\right),$ (7)

where $f\in\mathcal{C}([0,\infty),\mathbb{R})$ satisfies $f(0)\neq 0$ and
$\alpha\in\mathbb{R}^{*},\beta\in\mathbb{R}$. The involution
$\Sigma:A_{\infty}\to A_{\infty}$, i.e., $\Sigma\circ\Sigma=id,$ is specificed
by

$\Sigma f(t)=\frac{1}{f(\rho\circ\tau f(t))},$

where

$A_{\infty}=\bigcup\limits_{a>0}\bigcup\limits_{b>0}A(a,b)\;\;\text{and}\;\;A(a,b)=\left\\{\pm
f\in\mathcal{C}([0,a],\mathbb{R}^{+}):\;\tau f\left(a\right)=b\right\\}.$

Let $T^{f}=\inf\\{t:B_{t}=f(t)\\}$ be the FPT of the standard BM
$(B_{t})_{t\geq 0}$. For $t<-1/\alpha\beta$ if $\alpha\beta<0$ or $t<\infty$
otherwise, Theorem 1.2 in [3] connects the FPT distribution of $B_{t}$ to
$S^{\alpha,\beta}f$ to that to $f$ as

$\mathbb{P}(T^{S^{\alpha,\beta}f}\in dt)=\alpha^{3}(1+\alpha\beta
t)^{-\frac{5}{2}}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
t)}(S^{\alpha,\beta}f(t))^{2}}S^{\alpha,\beta}(\mathbb{P}(T^{f}\in dt)).$ (8)

This relationship is crucial and will be used later in the direct proof of
Theorem 4.1. We now introduce a new nonlinear operator $\hat{\Sigma}$ defined
on $\hat{A}_{\infty}$ by

$\hat{\Sigma}=\Lambda^{-1}\circ\Sigma\circ\Lambda,$ (9)

where $\Lambda f(t)=e^{ks(t)}f(s(t))=\sqrt{2kt+1}f(s(t))$, where $\Lambda$ is
defined on $[0,\infty)$ for $k>0$ and on $[0,-1/2k)$ for $k<0$, with its
inverse given by $\Lambda^{-1}f(t)=e^{-kt}f(r(t))$. Note that, $\hat{\Sigma}$
can also be represented in the following way

$\hat{\Sigma}f(t)=\frac{e^{-k\rho\circ\tau f(r(t))-kt}}{f(\rho\circ\tau
f(r(t)))}.$ (10)

We then define the family
$(\hat{S}^{\alpha,\beta})_{\alpha\in\mathbb{R}^{*},\beta\in\mathbb{R}}$ of
operators, $\hat{S}^{\alpha,\beta}:\hat{A}_{\infty}\to\hat{A}_{\infty}$, as
follows

$\hat{S}^{\alpha,\beta}=\hat{\Sigma}\circ\Pi^{\alpha,-\beta}\circ\hat{\Sigma}.$
(11)

Finally, for all $a>0$, we set

$a_{\alpha,\beta}=\begin{cases}\frac{a}{\alpha^{2}-\alpha\beta
a}&\text{if}\;\;\;\alpha^{2}-\alpha\beta a>0,\;k\geq 0;\\\
\frac{a}{\alpha^{2}-(\alpha\beta+2k\alpha^{2})a}&\text{if}\;\;\;0<\frac{2ka}{(\alpha\beta+2k\alpha^{2})a-\alpha^{2}}<1,\;k<0;\\\
+\infty&\text{otherwise},\end{cases}$

and for $f\in A_{\infty}$, we write

$a_{\alpha,\beta}^{f}=\begin{cases}a&\text{if}\;\;\alpha^{2}-\alpha\beta
a>0,\;k\geq 0;\\\
\frac{a}{1-2ka}&\text{if}\;\;0<\frac{2ka}{(\alpha\beta+2k\alpha^{2})a-\alpha^{2}}<1,\;k<0;\\\
\rho\circ\tau
f\left(s\left(\frac{\alpha}{\beta}\right)\right)&\text{otherwise}.\end{cases}$

Notice that, as $a\to\infty$, we have $a_{\alpha,\beta}\to
r(\zeta_{\alpha,\beta})$ where

$\zeta_{\alpha,\beta}=\begin{cases}s\left(-\frac{1}{\alpha\beta}\right)&\text{if}\;\alpha\beta<0,\;k\geq
0;\\\
s\left(-\frac{1}{\alpha\beta+2k\alpha^{2}}\right)&\text{if}\;0<\frac{2k}{\alpha\beta+2k\alpha^{2}}<1,\;k<0;\\\
+\infty&\text{otherwise}.\end{cases}$

For convenience, all key transformations that are used throughout this paper
are listed in Table 1 in Appendix A.

##  2 Ornstein-Uhlenbeck bridges

We start by defining the OU bridge and then recall its different
representations, using the detailed analysis of both Wiener and OU bridges
provided in [5].

######  Definition 2.1.

An OU bridge process $U^{\text{br}}=\\{U^{\text{br}}_{t}:t\in[0,T]\\}$ from a
to b of length T is characterised by the following properties:

  1. (i)

$U^{\text{br}}_{0}=a$ and $U^{\text{br}}_{T}=b$ (each with probability 1).

  2. (ii)

$U^{\text{br}}$ is a Gaussian process.

  3. (iii)

$\mathbb{E}[U^{\text{br}}_{t}]=a\frac{\sinh{(k(T-t))}}{\sinh{(kt)}}+b\frac{\sinh{(kt)}}{\sinh{(kT)}}$.

  4. (iv)

$\text{Cov}(U^{\text{br}}_{s},U^{\text{br}}_{t})=\frac{\sinh{(ks)}\sinh{(k(T-t))}}{k\sinh{(kT)}},\;\;\;0\leq
s\leq t<T$.

  5. (v)

The process $U^{\text{br}}_{t}$ is almost surely continuous on $[0,T)$.

In what follows, we present three different representations of OU bridge
processes. First, consider the following linear SDE

$dU^{\text{br}}_{t}=\left(-k\coth{(k(T-t))}U^{\text{br}}_{t}+k\frac{b}{\sinh{(k(T-t))}}\right)dt+dB_{t},\;\;\;0\leq
t<T,$ (12)

with initial condition $U^{\text{br}}_{0}=a$. This SDE has unique strong
solution given by

$U_{t}^{ir}:=U_{t}^{\text{br}}=\begin{cases}a\frac{\sinh{(k(T-t))}}{\sinh{(kT)}}+b\frac{\sinh{(kt)}}{\sinh{(kT)}}+\int_{0}^{t}\frac{\sinh{(k(T-t))}}{\sinh{(k(T-s))}}dB_{s}\;\;\;\;\text{if}\;\;0\leq
t<T,\\\
b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{if}\;\;t=T.\end{cases}$

This is referred to as the integral representation (ir) of the OU bridge. The
anticipative (av) and the space-time (st) transform versions can also be found
by transforming the different representations of the Wiener bridge

$U_{t}^{av}=a\frac{\sinh{(k(T-t))}}{\sinh{(kT)}}+b\frac{\sinh{(kt)}}{\sinh{(kT)}}+\left(U_{t}-\frac{\sinh{(kt)}}{\sinh{(kT)}}U_{T}\right),\;\;\;\;0\leq
t<T,$
$U_{t}^{st}=a\frac{\sinh{(k(T-t))}}{\sinh{(kT)}}+b\frac{\sinh{(kt)}}{\sinh{(kT)}}+e^{-kt}\left(\frac{r(T)-r(t)}{r(T)}\right)W_{\left(\frac{r(T)r(t)}{r(T)-r(t)}\right)},\;\;\;\;0\leq
t<T.$

The space-time transformation is of our main interest and we will later show
its connection to our transformation $\hat{S}^{\alpha,\beta}$. One thing to
notice is that the space-time and anticipative versions are only weak
solutions to the SDE (12). If this was not the case, then by the definition of
strong solutions we would get
$\mathbb{P}(U_{t}^{av}=U_{t}^{ir}\;\;\forall\;t\in[0,T))=1$, which is not
true. This tells us that the sample path behaviours differ for each version.
Section 3.2 of [5] provides a detailed analysis of such behaviours. Of course,
$U^{br}$ can be thought of as $(U_{t},t\leq T)$ conditioned on $U_{T}=b$. For
conditioned processes and Markovian bridges, see [12], [25].

##  3 Properties of transformations

We now state a result that gives an insight into our $\hat{S}^{\alpha,\beta}$
transformation.

######  Theorem 3.1.

  1. (1)

Let $(\alpha,\beta)$ and
$(\alpha^{\prime},\beta^{\prime})\in\mathbb{R}^{*}\times\mathbb{R}$. The
mapping $\hat{S}^{\alpha,\beta}:\hat{A}_{\infty}\to\hat{A}_{\infty}$ defined
by (11) admits the following representation

$\hat{S}^{\alpha,\beta}f(t)=\left(\frac{1+\alpha\beta
r(t)}{\alpha}\right)\sqrt{2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1}\;e^{-kt}\;f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right),$ (13)

with the property
$\hat{S}^{\alpha,\beta}\circ\hat{S}^{\alpha^{\prime},\beta^{\prime}}=\hat{S}^{\alpha\alpha^{\prime},\alpha\beta^{\prime}+\frac{\beta}{\alpha^{\prime}}}$.
In particular, $\hat{S}^{1,\beta}$ is a semigroup. Furthermore, if
$f\in\hat{A}(a,b)$, then
$\hat{S}^{\alpha,\beta}f\in\hat{A}(a_{\alpha,\beta},b_{\alpha,\beta}^{f})$.

  2. (2)

Let $\mu$ be a positive Radon measure on $\mathbb{R}^{+}$. Then, there exists
a unique positive and differentiable function $\mathfrak{f}$ with
$\mathfrak{f}(0)=1$ and
$\mathfrak{f}^{\prime}+k\mathfrak{f}+\frac{k}{(2k\tau\mathfrak{f}+1)\mathfrak{f}}>0$,
which satisfies the following nonlinear differential equation

$f^{3}f^{\prime\prime}-k^{2}f^{4}=\frac{-\mu(s(\tau f))+k^{2}}{(2k(\tau
f)+1)^{2}}.$ (14)

Moreover,
$\\{\hat{S}^{\alpha,\beta}\mathfrak{f};\alpha>0,\beta\in\mathbb{R}\\}$ spans
the set of positive solutions of (14).

Before proving Theorem 3.1, we start by discussing some of the properties of
the operators $\hat{\Sigma}$ and $\Pi^{\alpha,\beta}$.

######  Proposition 3.2.

  1. For any $a,b>0,\lambda\in\mathbb{R}$ and $f\in\hat{A}_{\infty}$, we have the following assertions.

  2. (1)

$\hat{\Sigma}$ is an involution operator, that is
$\hat{\Sigma}\circ\hat{\Sigma}=I_{d}$ on $\hat{A}_{\infty}$.

  3. (2)

$\tau f(t)=r(\rho\circ\tau\circ\hat{\Sigma}f(r(t)))$.

  4. (3)

$\hat{\Sigma}(\hat{A}(a,b))=\hat{A}(b,a)$.

  5. (4)

$\Pi^{\alpha,\beta}\Lambda=\Lambda\Pi^{\alpha,\beta}$.

###### Proof.

  1. (1)

Use the decomposition (9) of $\hat{\Sigma}$ and the fact that $\Sigma$ is an
involution.

  2. (2)

By using the functional representation of $\hat{\Sigma}$ (10),

$\tau\circ\hat{\Sigma}f(t)=\int_{0}^{t}f^{2}(\rho\circ\tau
f(r(y)))e^{2ky+2k\rho\circ\tau f(r(y))}dy=\int_{0}^{\rho\circ\tau
f(r(t))}e^{2kz}dz=r(\rho\circ\tau f(r(t))),$

which follows by a change of variable. Now use the fact that $\hat{\Sigma}$ is
an involution.

  3. (3)

First, consider $k<0$ and take $f\in\hat{A}(a,b)$. Observe that $\rho\circ\tau
f(b/1-2kb)=-s\big{(}-a\big{)}$. Now, by part (2),
$\tau\circ\hat{\Sigma}f\left(-s\left(-b\right)\right)=r(\rho\circ\tau
f(b/1-2kb))=a/1-2ka$ and so $\hat{\Sigma}f\in\hat{A}(b,a)$. Similarly, the
result holds for $k>0$.

  4. (4)

$\tau\Lambda f(t)=\int_{0}^{t}\left[e^{2ks(y)}f^{2}(s(y))\right]^{-1}dy=\tau
f(s(t))$, which follows by a change of variable. The identity can easily be
derived now.

∎

###### proof of Theorem 3.1.

  1. (1)

By using identity (4) from Proposition 3.2 and the decomposition of
$S^{\alpha,\beta}$ given in (7) respectively, we can write

$\hat{S}^{\alpha,\beta}=\hat{\Sigma}\circ\Pi^{\alpha,-\beta}\circ\hat{\Sigma}=\Lambda^{-1}\circ(\Sigma\circ\Pi^{\alpha,-\beta}\circ\Sigma)\circ\Lambda=\Lambda^{-1}\circ
S^{\alpha,\beta}\circ\Lambda.$

Now, if we use the functional definitions of the operators, we get the desired
result, as illustrated in Figure 1 in Appendix A. By construction, the
$S^{\alpha,\beta}$ transformation plays a key role in obtaining our
$\hat{S}^{\alpha,\beta}$ transformation. This procedure can be interpreted as
we first take the boundary $f$ to its corresponding first hitting time
boundary for the standard BM using the $\Lambda$ transformation, then use
$S^{\alpha,\beta}$ on this curve and then revert it back to the OU problem.
Next, the property of $\hat{S}$ stated in the theorem follows by the similar
semigroup property of the operator $S$ (Proposition 2.3 of [3]). Lastly, using
the identity $\tau\hat{S}^{\alpha,\beta}f(t)=\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right)\right)$ which
follows readily by a change of variables, we get that if $f\in\hat{A}(a,b)$,
then
$\hat{S}^{\alpha,\beta}f\in\hat{A}(a_{\alpha,\beta},b_{\alpha,\beta}^{f})$.

  2. (2)

As discussed in Section 1.1, if $\phi$ solves the Sturm-Liouville equation
(6), then the the vectorial space
$\\{\Pi^{\alpha,\beta}\phi;\alpha,\beta\in\mathbb{R}\\}$ is the set of
solutions to the same equation. Moreover, all positive solutions are convex
and described by the set $\\{\Pi^{\alpha,\beta}\varphi;\alpha>0,\beta\geq
0\\}$, where $\varphi$ is the unique, positive, decreasing solution such that
$\varphi(0)=1$. Now, we need to show that the image of (14) by $\hat{\Sigma}$
is (6) and vice-versa. We start by investigating the operator $\hat{\Sigma}$.
Let $\phi(t)=\hat{\Sigma}f(t)$. Then, we obtain $\phi(s(\tau
f(t)))f(t)=e^{-ks(\tau f(t))-kt}$. Differentiating both side once, we get

$\phi^{\prime}(s(\tau f(t)))=-e^{-ks(\tau f(t))-kt}(2k\tau
f(t)+1)\left(kf(t)+f^{\prime}(t)+\frac{k}{(2k\tau f(t)+1)f(t)}\right).$

Differentiating again, we obtain

$\phi^{\prime\prime}(s(\tau f(t)))=-(2k\tau f(t)+1)^{2}e^{-ks(\tau
f(t))-kt}\left(f^{2}(t)f^{\prime\prime}(t)-k^{2}f^{3}(t)-\frac{k^{2}}{(2k\tau
f(t)+1)^{2}f(t)}\right).$

So if $\phi$ satisfies (6), we get (14) and vice versa. Now, by letting
$\mathfrak{f}=\hat{\Sigma}\varphi$, we see that it satisfies the required
properties.

∎

######  Remark 3.3.

Notice that the law of the process $(\hat{S}^{1,-\beta}U_{t},0\leq
t<\zeta_{1,-\beta})$ for a given $\beta$ corresponds to the law of a bridge
associated to the OU. Indeed, set $\beta=-1/r(T)$ for some $T>0$. Then, for
any $0\leq t<T$, by using (13), we see that $(\hat{S}^{1,-1/r(T)}U_{t},t\leq
T)$ is a representation of the OU bridge of length T from 0 to 0.

##  4 Main result and proofs

We are now ready to state the main result of this paper which relates the
distributions of the family of stopping times
$(\hat{T}^{\hat{S}^{\alpha,\beta}f})_{\alpha\in\mathbb{R}^{*},\beta\in\mathbb{R}}$
to that of $\hat{T}^{f}$. Recall that $\mathcal{C}^{1}(I,K)$, for some
subintervals $I$ and $K\subseteq\mathbb{R}$, denotes the space of continuously
differentiable functions from $I$ into $K$.

######  Theorem 4.1.

Let $f\in C^{1}([0,\infty),\mathbb{R})$ be such that $f(0)\neq 0$. Let
$\alpha,\beta\in\mathbb{R},\alpha\neq 0$. Then, for $t<\zeta_{\alpha,\beta}$,
we have the following relationship

$\displaystyle\mathbb{P}\Big{(}\hat{T}^{\hat{S}^{\alpha,\beta}f}\in
dt\Big{)}=$ (15) $\displaystyle e^{3kt}\alpha^{3}(1+\alpha\beta
r(t))^{-5/2}\bigg{(}2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\bigg{)}^{-3/2}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f)^{2}e^{2kt}}\hat{S}^{\alpha,\beta}\Big{(}\mathbb{P}(\hat{T}^{f}\in
dt)\Big{)}.$

We provide three different proofs of Theorem 4.1. The first method is a direct
approach that uses results from [3]. The second approach is more interesting,
as it uses a generalisation of the so-called Gauss-Markov processes and gives
an insight into some of the results given in Theorem 3.1. The third proof
relies on the Lie group techniques applied to the Fokker-Planck equation of
the OU. Before proving the theorem, we see an example of its applications.

######  Example 4.2.

Theorem 3.1 in [1] gives an expression for the FPT density of an OU process
starting at $0$ hitting a constant threshold $f(t)=a>0$. Writing
$\hat{p}^{a}(t)dt=\mathbb{P}\Big{(}\hat{T}^{a}\in dt\Big{)}$, the latter
expression is

$\hat{p}^{a}(t)=-ke^{-ka^{2}/2}\sum_{j=1}^{\infty}\frac{D_{\nu_{j,-a\sqrt{t}}}(0)}{D^{\prime}_{\nu_{j,-a\sqrt{t}}}(-a\sqrt{2k})}\exp(-k\nu_{j,-a\sqrt{2k}}t),$

where $D_{\nu}(.)$ is the parabolic cylinder function with index
$\nu\in\mathbb{R}$ [16], $\nu_{j,b}$ the ordered sequence of positive zeros of
$D_{\nu}(b)$,
$D_{\nu_{j,-a\sqrt{t}}}(0)=2^{\frac{\nu_{j,-a\sqrt{t}}}{2}}\frac{\Gamma(1/2)}{\Gamma\left((1-\nu_{j,-a\sqrt{t}})/2\right)}$
and $D^{\prime}_{\nu_{j,b}}(b)=\frac{\partial
D_{\nu}(b)}{\partial\nu}|_{\nu=\nu_{j,b}}$. Now, we use the
$\hat{S}^{\alpha,\beta}$ transformation with
$f(t)=a,\;\alpha,\beta\in\mathbb{R},\alpha\neq 0$ to get the following family
of curves

$\hat{S}^{\alpha,\beta}f(t)=\frac{ae^{-kt}}{2k\alpha}\left\\{(\alpha\beta
e^{2kt}-\alpha\beta+2k)\left[(2k\alpha^{2}+\alpha\beta)e^{2kt}-2k\alpha^{2}-\alpha\beta+2k\right]\right\\}^{1/2}.$

By Theorem 4.1,

$\displaystyle\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)$
$\displaystyle=e^{2kt}\alpha^{2}\frac{(1+\alpha\beta
r(t))^{-3/2}}{\big{(}2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\big{)}}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f(t))^{2}\;e^{2kt}}\;\hat{p}^{a}\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)$ $\displaystyle=\frac{-ke^{2kt}\alpha^{2}(1+\alpha\beta
r(t))^{-5/2}}{\big{(}2\alpha^{2}r(t)+\alpha\beta
r(t)+1\big{)}}\exp\left(-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f(t))^{2}\;e^{2kt}-ka^{2}/2\right)$
$\displaystyle\hskip
113.81102pt\times\sum_{j=1}^{\infty}\frac{D_{\nu_{j,-a\sqrt{t}}}(0)}{D^{\prime}_{\nu_{j,-a\sqrt{t}}}(-a\sqrt{2k})}\exp\left(-k\nu_{j,-a\sqrt{2k}}\;s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right).$

Such curves for $\beta=1,\;k=1/2$ and different values of $\alpha$ are plotted
in figure 2 in Appendix A.
In the special case of $\alpha\beta=2k,$ we get the family of curves
$\hat{S}^{\alpha,\beta}f(t)=\frac{a}{\alpha}\sqrt{(\alpha^{2}+1)e^{2kt}-\alpha^{2}}$,
with FPT density given by

$\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)=\frac{-k\alpha^{2}e^{-ka^{2}\left(\frac{\alpha^{2}+1}{\alpha^{2}}\right)e^{2kt}+\frac{ka^{2}}{2}+kt}}{(\alpha^{2}+1)e^{2kt}-\alpha^{2}}\sum_{j=1}^{\infty}\frac{D_{\nu_{j,-a\sqrt{t}}}(0)}{D^{\prime}_{\nu_{j,-a\sqrt{t}}}(-a\sqrt{2k})}e^{-k\nu_{j,-a\sqrt{2k}}\;s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)}.$

Before providing the proofs of Theorem 4.1, we mention that, similarly as in
the BM case, it is sufficient to only consider the case where $\alpha>0$. This
is because, by the symmetry of the BM and the OU process starting at 0 without
drift, hitting negative or positive valued thresholds for the first time have
the same probability. Hence for any
$(\alpha,\beta)\in\mathbb{R}^{*}\times\mathbb{R}$, we have
$\hat{T}^{\hat{S}^{\alpha,\beta}f}\stackrel{{\scriptstyle\text{d}}}{{=}}\hat{T}^{\hat{S}^{|\alpha|,\operatorname{sgn}(\alpha)\beta}f},$
where $\stackrel{{\scriptstyle\text{d}}}{{=}}$ denotes equality in
distribution.

###  4.1 First proof of Theorem 4.1 via a direct approach

Theorem 1.2 in [3] connects the FPT distribution of $S^{\alpha,\beta}f$ to
that of $f$. We can connect the FPTs of OU and BM in the following manner,

$\hat{T}^{f}=\inf\\{t:U_{t}=f(t)\\}=\inf\\{s(t):W_{t}=e^{ks(t)}f(s(t))\\}=s(T^{\Lambda
f}).$ (16)

Now, we proceed with the first proof.

###### First proof of Theorem 4.1.

Note that $\hat{T}^{\hat{S}^{\alpha,\beta}f}=s(T^{S^{\alpha,\beta}\Lambda
f}).$ Writing $\hat{p}^{f}(t)dt=\mathbb{P}\left(\hat{T}^{f}\in dt\right)$ and
$p^{f}(t)dt=\mathbb{P}\left(T^{f}\in dt\right)$, we get

$\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)=e^{2kt}\;p^{S^{\alpha,\beta}\Lambda
f}(r(t)).$ (17)

Now, using (7) and (8), we can rewrite the right hand side as follows:

$\displaystyle p^{S^{\alpha,\beta}\Lambda f}(r(t))$
$\displaystyle=\alpha^{3}(1+\alpha\beta
r(t))^{-5/2}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(S^{\alpha,\beta}\Lambda f(r(t)))^{2}}S^{\alpha,\beta}\Big{(}p^{\Lambda
f}(r(t))\Big{)}$ (18) $\displaystyle=\alpha^{2}(1+\alpha\beta
r(t))^{-3/2}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(S^{\alpha,\beta}\Lambda f(r(t)))^{2}}p^{\Lambda
f}\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right).$

By using (16), we get

$p^{\Lambda f}\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)=\frac{1}{2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1}\;\hat{p}^{f}\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right).$ (19)

Notice that $S^{\alpha,\beta}\Lambda
f(r(t))=e^{kt}\hat{S}^{\alpha,\beta}f(t)$. Now, plugging (19) into (18) and
then into (17), we get

$\displaystyle\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)$
$\displaystyle=e^{2kt}\alpha^{2}\frac{(1+\alpha\beta
r(t))^{-3/2}}{\big{(}2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\big{)}}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f(t))^{2}\;e^{2kt}}\;\hat{p}^{f}\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)$ $\displaystyle=e^{3kt}\alpha^{3}(1+\alpha\beta
r(t))^{-5/2}\bigg{(}2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\bigg{)}^{-3/2}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f(t))^{2}\;e^{2kt}}\;\hat{S}^{\alpha,\beta}\bigg{(}\hat{p}^{f}(t)\bigg{)},$

where we used (13) to obtain the last equation. ∎

###  4.2 Second proof of Theorem 4.1 via generalised Gauss-Markov processes

We take $\phi\in\hat{A}(a,b)\cap AC([0,b))$, where $AC([0,b))$ is the space of
absolutely continuous functions on $[0,b)$. We introduce the process
$X:=(X_{t})_{0\leq t<b}$ to be the generalised Gauss-Markov process of OU type
with parameters $(\phi,k)$, which is defined as the unique strong solution to
the following SDE

$dX_{t}=\left(\frac{\phi^{\prime}(t)}{\phi(t)}+k\right)X_{t}dt+e^{kt}dB_{t},\;\;\;\;\;0\leq
t<b,$

with $X_{0}=x\in\mathbb{R}$, i.e.,

$X_{t}=\phi(t)e^{kt}\left(x+\int_{0}^{t}\frac{1}{\phi(s)}dB_{s}\right),\;\;\;\;\;0\leq
t<b.$

We also denote by
$\mathbb{P}^{(\phi,k)}=\left(\mathbb{P}^{(\phi,k)}_{x}\right)_{x\in\mathbb{R}}$
the family of probability measures corresponding to the process X. We assume
throughout that $\phi(0)=1$. Notice that when $\phi(t)=e^{-kt}$, we get
$X_{t}=W_{r(t)}$, where $(W_{t},t\geq 0)$ is another BM. The motivation for
introducing the generalised Gauss-Markov process is the connection between its
boundary crossing problem and that of the OU.

######  Lemma 4.3.

For any $y\in\mathbb{R}$, set $T_{y}=\inf\\{0<t<b:\;\phi(t)e^{kt}\
\int_{0}^{t}\frac{1}{\phi(s)}dB_{s}=y\\}$. Then, for any $f\in\hat{A}(a,b)$,
setting $\phi=\hat{\Sigma}f$, the identity

$\hat{T}^{f}=s(\tau\phi(T_{1}))$

holds almost surely.

###### Proof.

Using (3), we have a.s.

$\displaystyle T_{1}$
$\displaystyle=\inf\\{t>0:\;\phi(t)e^{kt}W_{\tau\phi(t)}=1\\}=\inf\left\\{t:\;U_{s(\tau\phi(t))}=\frac{e^{-kt-
ks(\tau\phi(t))}}{\phi(t)}\right\\}=\rho\circ\tau\phi(r(\hat{T}^{f}))$

and the result follows. ∎

Next, we introduce the following notation

$H_{t}^{k}(x)=\sqrt{\frac{\alpha\phi(t)}{\Pi^{\alpha,\beta}\phi(t)}}\exp\left(\frac{\beta}{2}\frac{x^{2}e^{-2kt}}{\phi(t)\Pi^{\alpha,\beta}\phi(t)}\right).$

Our aim now is to show that the parametric families of distributions
$\left(\mathbb{P}^{(\Pi^{\alpha,\beta}\phi,k)}\right)_{(\alpha,\beta)\in\mathbb{R}^{*}\times\mathbb{R}}$
of generalised Gauss-Markov processes are related by some space-time harmonic
transformations.

######  Proposition 4.4.

For $(\alpha,\beta)\in\mathbb{R}^{*}\times\mathbb{R}$ and $\phi$ as above, the
process $(H_{t}^{k}(X_{t}))_{0\leq t<a_{\alpha,-\beta}^{\phi}}$ is a
$\mathbb{P}^{(\phi,k)}$-martingale. Furthermore, the absolute-continuity
relationship

$d\mathbb{P}^{(\Pi^{\alpha,\beta}\phi,k)}_{x|\mathcal{F}_{t}}=\frac{H_{t}^{k}(X_{t})}{H_{0}^{k}(x)}d\mathbb{P}^{(\phi,k)}_{x|\mathcal{F}_{t}}$

holds for all $x\in\mathbb{R}$ and $t<a_{\alpha,-\beta}^{\phi}$. Consequently,
for any reals x and y, we have

$\mathbb{P}^{(\Pi^{\alpha,\beta}\phi,k)}_{x}(T_{y}\in
dt)=\frac{H_{t}^{k}(y)}{H_{0}^{k}(x)}\mathbb{P}_{x}^{(\phi,k)}(T_{y}\in
dt),\;\;\;t<a_{\alpha,-\beta}^{\phi}.$

###### Proof.

The proof is similar to that of Lemma 3.2 in [3]. As that proof has a few
typos, we will give the full proof here. From the Itô formula, we have

$\frac{\beta}{2}\frac{X_{t}^{2}e^{-2kt}}{\phi(t)\Pi^{\alpha,\beta}\phi(t)}=\beta\int_{0}^{t}\frac{X_{s}e^{-ks}}{\phi(s)\Pi^{\alpha,\beta}\phi(s)}dB_{s}-\frac{\beta^{2}}{2}\int_{0}^{t}\left(\frac{X_{s}e^{-ks}}{\phi(s)\Pi^{\alpha,\beta}\phi(s)}\right)^{2}ds+\frac{1}{2}\ln\left(\frac{\alpha+\beta\tau\phi(t)}{\alpha}\right).$

Now, as $\mathbb{E}[e^{-\frac{\lambda}{2}B_{t}^{2}}]=(1+\lambda
t)^{-1/2},\;\lambda>-1/t$, we deduce that, for all
$t<a_{\alpha,-\beta}^{\phi}$, we have

$\mathbb{E}[H_{t}^{k}(X_{t})]=\mathbb{E}\left[\sqrt{\frac{\alpha}{\alpha+\beta\tau\phi}}e^{-\frac{\beta}{2}\frac{W_{\tau\phi}^{2}}{\alpha+\beta\tau\phi}}\right]=1$

Hence, it is a true martingale. Next, notice that

$d\left<\beta\int_{0}^{.}\frac{X_{s}e^{-ks}}{\phi(s)\Pi^{\alpha,\beta}\phi(s)}dB_{s},X_{.}\right>_{t}=\beta\frac{X_{t}}{\phi(t)\Pi^{\alpha,\beta}\phi(t)}dt,$

where $\left<.,.\right>$ is the quadratic variation process. Also,

$\frac{(\Pi^{\alpha,\beta}\phi(t))^{\prime}}{\Pi^{\alpha,\beta}\phi(t)}=\frac{\phi^{\prime}(t)}{\phi(t)}+\frac{\beta}{\phi(t)\Pi^{\alpha,\beta}\phi(t)},$

and hence the absolute continuity relationship follows by an application of
Girsanov’s theorem. Now, Doob’s optional stopping theorem implies that

$\displaystyle\mathbb{P}^{(\Pi^{\alpha,\beta}\phi,k)}_{x}(T_{y}\leq t)$
$\displaystyle=\mathbb{E}_{x}^{(\phi,k)}\left[\mathbbm{1}_{\\{T_{y}\leq
t\\}}\frac{H_{t}^{k}(X_{t})}{H_{0}^{k}(x)}\right]=\mathbb{E}_{x}^{(\phi,k)}\left[\mathbbm{1}_{\\{T_{y}\leq
t\\}}\mathbb{E}_{x}^{(\phi,k)}\left[\frac{H_{t}^{k}(X_{t})}{H_{0}^{k}(x)}\bigg{|}\mathcal{F}_{t\wedge
T_{y}}\right]\right]$
$\displaystyle=\mathbb{E}_{x}^{(\phi,k)}\left[\mathbbm{1}_{\\{T_{y}\leq
t\\}}\frac{H_{T_{y}}^{k}(y)}{H_{0}^{k}(x)}\right],$

where $\mathbbm{1}_{A}$ is the indicator function of the set $A$. The result
follows by differentiation. ∎

Now, we are ready to give the second proof of Theorem 4.1.

###### Second proof of Theorem 4.1.

We assume that our boundary $f\in\mathcal{C}^{1}((0,\infty),\mathbb{R}^{+})$.
This assumption is crucial, as it implies that the curves $\Lambda f$ are also
continuously differentiable and, according to Strassen [29], that $T^{\Lambda
f}$ has a continuous density with respect to the Lebesgue measure on
$(0,\infty)$. By the relation between the two stopping times (16),
$\hat{T}^{f}$ also has a continuous density. Let $\phi=\hat{\Sigma}f$,
$\hat{f}^{\alpha,\beta}=\hat{S}^{\alpha,\beta}f$ and thus, by definition,
$\hat{f}^{\alpha,\beta}=\hat{\Sigma}\circ\Pi^{\alpha,-\beta}\phi$. Since
$\hat{\Sigma}$ is an involution, from Lemma 4.3 we get
$\hat{T}^{\hat{f}^{\alpha,\beta}}=s(\tau\circ\Pi^{\alpha,-\beta}\phi(T_{1}))$
a.s Using identity (2) from Proposition 3.2, we get

$r(\rho\circ\tau\circ\Pi^{\alpha,-\beta}\phi(r(t)))=r(\rho\circ\tau\circ\hat{\Sigma}\hat{f}^{\alpha,\beta}(r(t))=\tau\hat{f}^{\alpha,\beta}(t).$

Now, for $t<\zeta_{\alpha,\beta}$, writing
$\hat{p}^{\hat{f}^{\alpha,\beta}}(t)dt=\mathbb{P}\Big{(}\hat{T}^{\hat{f}^{\alpha,\beta}}\in
dt\Big{)}$ and
$\tilde{p}^{(\phi,k)}(t)dt=\mathbb{P}^{(\phi,k)}_{0}\left(T_{1}\in dt\right)$,
we get

$\displaystyle\hat{p}^{\hat{f}^{\alpha,\beta}}(t)$
$\displaystyle=\frac{1}{(2k(\tau\hat{f}^{\alpha,\beta})+1)(\hat{f}^{\alpha,\beta})^{2}}\tilde{p}^{(\Pi^{\alpha,-\beta}\phi,k)}\left(s(\tau\hat{f}^{\alpha,\beta}(t))\right)$
$\displaystyle=\frac{1}{(2k(\tau\hat{f}^{\alpha,\beta})+1)(\hat{f}^{\alpha,\beta})^{2}}\tilde{p}^{(\Pi^{\alpha,-\beta}\phi,k)}\left(s\left(\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right),$

where we used the identity $\tau\hat{f}^{\alpha,\beta}(t)=\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right)\right)$ which
follows readily by a change of variable. Now, using Proposition 4.4 with the
identities

$\displaystyle\left(\phi\left(s\circ\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right)^{-1}$
$\displaystyle=\left(\phi\left(\rho\circ\tau\phi\left(r\circ
s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right)^{-1}$
$\displaystyle=e^{k\rho\circ\tau\hat{\Sigma}f\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)+ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)}f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)$
$\displaystyle=e^{ks(\tau\hat{f}^{\alpha,\beta}(t))+ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)}f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right),$

and

$\displaystyle(\Pi^{\alpha,-\beta}\phi(s(\tau\hat{f}^{\alpha,\beta}(t))))^{-1}$
$\displaystyle=(\Pi^{\alpha,-\beta}\phi(\rho\circ\tau\circ\Pi^{\alpha,-\beta}\phi(r(t))))^{-1}=e^{k\rho\circ\tau\Pi^{\alpha,-\beta}\phi(r(t))+kt}\hat{f}^{\alpha,\beta}(t)$
$\displaystyle=e^{ks(\tau\hat{f}^{\alpha,\beta}(t))+kt}\hat{f}^{\alpha,\beta}(t),$

yields, for any $t<\zeta_{\alpha,\beta}$,

$\displaystyle\tilde{p}^{(\Pi^{\alpha,-\beta}\phi,k)}\left(s\left(\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right)=$ $\displaystyle\sqrt{1+\alpha\beta
r(t)}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{f}^{\alpha,\beta}(t))^{2}\;e^{2kt}}\tilde{p}^{(\phi,k)}\left(s\left(\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right).$

Hence,

$\hat{p}^{\hat{f}^{\alpha,\beta}}(t)=\frac{e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{f}^{\alpha,\beta}(t))^{2}\;e^{2kt}}\sqrt{1+\alpha\beta
r(t)}}{(2k(\tau\hat{f}^{\alpha,\beta})+1)(\hat{f}^{\alpha,\beta})^{2}}\tilde{p}^{(\phi,k)}\left(s\left(\tau
f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)\right)\right).$

Using Lemma 4.3 again, we finally get (15). ∎

###  4.3 Third proof of Theorem 4.1 via the Lie group symmetries

We now provide the last proof for Theorem 4.1 using the Lie symmetries
approach [23]. In general, this technique can be used to find solutions to
differential equations with new boundary conditions from known ones. For
example, the Lie point symmetries of the heat equation

$\frac{\partial h}{\partial t}=\frac{1}{2}\frac{\partial^{2}h}{\partial
x^{2}}$

can be found in Section 3.3 of [3] and an excellent book on this technique is
[23].

Before going through the proof, we first discuss the connections between the
two boundary value problems (20) corresponding to the heat equation and the OU
Fokker-Planck equation. As in the second proof via generalised Gauss-Markov
processes, we let $f$ to be a continuously differentiable boundary on
$\mathbb{R}^{+}$. Set

$D^{f}=\\{(x,t)\in\mathbb{R\times\mathbb{R}^{+}}:x\leq f(t)\\}.$

We introduce the following boundary value problems $\mathcal{H}(f)$ and
$\mathcal{\hat{H}}(f)$ for the BM and OU, respectively:

$\mathcal{H}(f)=\begin{cases}\frac{\partial}{\partial
t}h(x,t)=\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}h(x,t)\;\;\;\;\text{on}\;D^{f};\\\ h(f(t),t)=0\;\;\;\;\;\text{for
all}\;t>0;\\\
h(.,0)=\delta_{0}(.)\;\;\;\;\text{on}\;(-\infty,f(0)),\end{cases}\mathcal{\hat{H}}(f)=\begin{cases}\frac{\partial}{\partial
t}\hat{h}(x,t)=k\frac{\partial}{\partial
x}(x\hat{h}(x,t))+\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\hat{h}(x,t)\;\;\text{on}\;D^{f};\\\
\hat{h}(f(t),t)=0\;\;\;\;\;\text{for all}\;t>0;\\\
\hat{h}(.,0)=\delta_{0}(.)\;\;\;\;\text{on}\;(-\infty,f(0)).\end{cases}$ (20)

Note that the first equations in $\mathcal{H}(f)$ and $\mathcal{\hat{H}}(f)$
are the heat equation and the OU Fokker-Planck equation, respectively. By
Proposition 5.4.3.1 of [6], solutions to both problems admit the following
probabilistic representations:

$h(x,t)dx=\mathbb{P}(W_{t}\in
dx,\;t<T^{f})\;\;\;\text{and}\;\;\;\hat{h}(x,t)dx=\mathbb{P}(U_{t}\in
dx,\;t<\hat{T}^{f}).$ (21)

Using (3) and (16), we connect the two solutions directly in the following
way:

$\mathbb{P}(W_{t}<x,\;t<T^{\Lambda
f})=\mathbb{P}(e^{-ks(t)}W_{t}<e^{-ks(t)}x,\;s(t)<s(T^{\Lambda
f}))=\mathbb{P}(U_{s(t)}<e^{-ks(t)}x,\;s(t)<\hat{T}^{f})$

and so

$\displaystyle
h(x,t)=e^{-ks(t)}\hat{h}(e^{-ks(t)}x,s(t))\;\;\;\text{and}\;\;\;\hat{h}(x,t)$
$\displaystyle=e^{kt}h(e^{kt}x,r(t)),$ (22)

where $h$ now denotes a solution to $\mathcal{H}(\Lambda f)$. This shows that
the solutions to the Fokker-Plank equation of the OU on $D^{f}$ are directly
connected to the solutions to the heat equation on $D^{\Lambda f}$. As will
become clear in the third proof of Theorem 4.1 at the end of this section, the
aim is to find a solution to the Fokker-Plank equation of the OU such that it
vanishes on our desired transformed boundary $\hat{S}^{\alpha,\beta}f$. To do
this, we first present a proposition for the OU which resembles Proposition
3.5 in [3] for the Brownian motion.

######  Proposition 4.5.

Let $\hat{h}$ be the solution to the boundary value problem
$\mathcal{\hat{H}}(f)$ in (20). Then, for any $\alpha>0$,
$\beta\in\mathbb{R}$, the following mapping

$\hat{h}^{\alpha,\beta}(x,t)=\frac{\alpha e^{kt}}{\sqrt{1+\alpha\beta
r(t)}}e^{-\frac{\alpha\beta e^{2kt}x^{2}}{2(1+\alpha\beta
r(t))}-ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)}\hat{h}\Bigg{(}\frac{\alpha
e^{-ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)+kt}}{1+\alpha\beta
r(t)}x,s\bigg{(}\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\bigg{)}\Bigg{)}$
(23)

is the solution to the boundary value problem
$\mathcal{\hat{H}}(\hat{S}^{\alpha,\beta}f)$.

###### Proof.

Firstly, if $f$ is an infinitely continuously differentiable function, then so
is its transformation $\hat{S}^{\alpha,\beta}f$. From Proposition 3.5 in [3],
the following function

$h^{\alpha,\beta}(x,t)=\frac{\alpha}{\sqrt{1+\alpha\beta
t}}e^{-\frac{\alpha\beta x^{2}}{2(1+\alpha\beta t)}}h\left(\frac{\alpha
x}{1+\alpha\beta t},\frac{\alpha^{2}t}{1+\alpha\beta t}\right),$

is a solution to $\mathcal{H}(S^{\alpha,\beta}\Lambda f)$ because $h$ solves
$\mathcal{H}(\Lambda f)$. Now using relation (22), we define
$\tilde{h}^{\alpha,\beta}$ as
$\tilde{h}^{\alpha,\beta}=e^{kt}h^{\alpha,\beta}(e^{kt}x,r(t))$. Then, this
expression is a solution to the boundary value problem
$\mathcal{\hat{H}}(\hat{S}^{\alpha,\beta}f)$. Using relation (22) again, we
can write

$h\left(\frac{\alpha e^{kt}x}{1+\alpha\beta
r(t)},\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)=e^{-ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)}\hat{h}\bigg{(}\frac{\alpha
e^{-ks\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)+kt}}{1+\alpha\beta
r(t)}x,s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right)\bigg{)},$

and so we get that $\tilde{h}^{\alpha,\beta}$ and $\hat{h}^{\alpha,\beta}$
coincide, which implies that $\hat{h}^{\alpha,\beta}$ is indeed a solution to
$\mathcal{\hat{H}}(\hat{S}^{\alpha,\beta}f)$. Moreover, we have that
$\hat{h}^{\alpha,\beta}(x,t)=0\iff x=\hat{S}^{\alpha,\beta}f(t)$, because by
assumption $\hat{h}$ is a solution to $\mathcal{\hat{H}}(f)$, hence
$\hat{h}(f(.),.)=0$. Also, $\hat{h}^{\alpha,\beta}(x,0)=e^{-\frac{\alpha\beta
x^{2}}{2}}\hat{h}^{\alpha,0}(x,0)$. Now, let us investigate
$\hat{h}^{\alpha,0}$. Using relation (22) again,

$\displaystyle\hat{h}^{\alpha,0}(x,t)$ $\displaystyle=\alpha e^{kt-
ks(\alpha^{2}r(t))}\hat{h}(\alpha
e^{-ks(\alpha^{2}r(t))+kt}x,s(\alpha^{2}r(t)))$ $\displaystyle=\alpha e^{kt-
ks(e^{-2(-\ln(\alpha))}r(t))}\hat{h}(e^{-ks(e^{-2(-\ln(\alpha))}r(t))+kt-(-\ln(\alpha))}x,s(e^{-2(-\ln(\alpha))}r(t)))$
$\displaystyle=e^{kt}h(e^{-\epsilon}e^{kt}x,e^{-2\epsilon}r(t))=e^{kt}h^{(4)}_{\epsilon}(e^{kt}x,r(t)),$

where $\epsilon=-\ln(\alpha)$. Now, the latter $h^{(4)}_{\epsilon}$ is the
fourth symmetry of the heat equation listed in Section 3.3 of [3], meaning
that it satisfies the heat equation. By using relation (22), we see that
$e^{kt}h^{(4)}_{\epsilon}(e^{kt}x,r(t))$ indeed satisfies the Fokker-Plank
equation of the OU and so $\hat{h}^{\alpha,0}$ is a solution to
$\mathcal{\hat{H}}(\hat{S}^{\alpha,0}f)$ and thus in particular
$\hat{h}^{\alpha,0}(.,0)=\delta_{0}(.)$ on $(-\infty,\frac{f(0)}{\alpha})$.
Hence, we get

$\hat{h}^{\alpha,\beta}(x,0)=e^{-\frac{\alpha\beta
x^{2}}{2}}\hat{h}^{\alpha,0}(x,0)=\delta_{0}(x),$

which concludes the proof. ∎

####  4.3.1 Lie symmetries of OU Fokker-Planck equation

The interesting result given in this section is constructing the
$\hat{h}^{\alpha,\beta}$ function (23) directly from the symmetries of the OU
Fokker Planck equation (4). Using similar techniques discussed in Section 2.4
in [23], after some lengthy and tedious calculations, one finds that the Lie
algebra of infinitesimal symmetries of the OU Fokker-Planck equation is
spanned by six vector fields, where $x,t$ are the two independent variables
and $h$ is the dependent variable,

$\mathbf{v_{1}}=e^{-kt}\frac{\partial}{\partial
x},\;\;\;\mathbf{v_{2}}=\frac{\partial}{\partial
t},\;\;\;\mathbf{v_{3}}=h\frac{\partial}{\partial
h},\;\;\;\mathbf{v_{4}}=e^{kt}\frac{\partial}{\partial
x}-2kxhe^{kt}\frac{\partial}{\partial h}$
$\mathbf{v_{5}}=-kxe^{-2kt}\frac{\partial}{\partial
x}+e^{-2kt}\frac{\partial}{\partial t}+khe^{-2kt}\frac{\partial}{\partial
h},\;\;\;\mathbf{v_{6}}=kxe^{2kt}\frac{\partial}{\partial
x}+e^{2kt}\frac{\partial}{\partial
t}-2k^{2}x^{2}he^{2kt}\frac{\partial}{\partial h}$

and by the infinite-dimensional subalgebra
$\mathbf{v_{\alpha}}=u(x,t)\frac{\partial}{\partial h}$, where $u(x,t)$ is an
arbitrary solution of the OU Fokker-Planck equation. Now, by exponentiating
the basis (for more details on this procedure, see Section 2.1, page 89 of
[23]), we can produce the one-parameter group of transformations leaving
invariant $\mathcal{\hat{H}}(f)$. Doing this procedure for each of the vectors
fields, we obtain the one-parameter groups $G_{i}$ generated by the
$\mathbf{v_{i}}$. Since each group $G_{i}$ is a symmetry group, if
$h=\hat{h}(x,t)$ is a solution to (4), then the following are also solutions
to (4) (examples of this procedure are given in Section 2.2 [23]):

$\displaystyle\hat{h}^{(1)}_{\epsilon}(x,t)=$ $\displaystyle\hat{h}(x-\epsilon
e^{-kt},t)$ (24) $\displaystyle\hat{h}^{(2)}_{\epsilon}(x,t)=$
$\displaystyle\hat{h}(x,t-\epsilon)$
$\displaystyle\hat{h}^{(3)}_{\epsilon}(x,t)=$ $\displaystyle
e^{\epsilon}\hat{h}(x,t)$ $\displaystyle\hat{h}^{(4)}_{\epsilon}(x,t)=$
$\displaystyle e^{-2kx\epsilon e^{kt}+k\epsilon^{2}e^{2kt}}\hat{h}(x-\epsilon
e^{kt},t)$ $\displaystyle\hat{h}^{(5)}_{\epsilon}(x,t)=$
$\displaystyle\frac{e^{kt}}{\sqrt{e^{2kt}-2k\epsilon}}\hat{h}\bigg{(}\frac{xe^{kt}}{\sqrt{e^{2kt}-2k\epsilon}},\frac{\ln{(e^{2kt}-2k\epsilon)}}{2k}\bigg{)}$
$\displaystyle\hat{h}^{(6)}_{\epsilon}(x,t)=$ $\displaystyle
e^{\frac{-2k^{2}x^{2}\epsilon}{e^{-2kt}+2k\epsilon}}\hat{h}\bigg{(}\frac{xe^{-kt}}{\sqrt{e^{-2kt}+2k\epsilon}},\frac{\ln{(e^{-2kt}+2k\epsilon)}}{-2k}\bigg{)}.$

In Proposition 4.5, we proved that the function $\hat{h}^{\alpha,\beta}$ is a
solution to the boundary value problem
$\mathcal{\hat{H}}(\hat{S}^{\alpha,\beta}f)$ by directly using the relation
between the two boundary value problems in (20). We can also construct this
function directly using the symmetries above.

######  Remark 4.6.

Proposition 5 of [20] gives a general class of symmetries for SDEs with a
drift term specified in Proposition 4 therein. The expression therein contains
misprints in $P^{\epsilon}_{1,2}$ and $\mathcal{T}_{p_{1,2}}$. Using the
notation in the original article, the correct expressions are

$\displaystyle P^{\epsilon}_{1,2}$ $\displaystyle:$
$\displaystyle\frac{u(x,t)}{\mathcal{U}(\mathcal{X},\mathcal{T})}=\frac{\theta_{\mathcal{F}_{2}}(x)}{\theta_{\mathcal{F}_{2}}(\mathcal{X})}\frac{e^{\pm
2\sqrt{A}\left(\nu-\frac{1}{4}\pm\frac{1}{4}\right)t\mp\frac{\sqrt{A}}{2}(x+2B/A)^{2}}}{e^{\pm
2\sqrt{A}\left(\nu-\frac{1}{4}\pm\frac{1}{4}\right)\mathcal{T}\mp\frac{\sqrt{A}}{2}(\mathcal{X}+2B/A)^{2}}}$
$\displaystyle\mathcal{T}_{p_{1,2}}$ $\displaystyle=$
$\displaystyle\mp\ln(e^{\mp 2\sqrt{A}t}+\epsilon)/(2\sqrt{A}).$

If we set $A=k^{2},B=0,\nu=0$ and $\theta_{\mathcal{F}_{2}}(x)=e^{-kx^{2}/2}$,
we recover the above symmetries (24) for the OU process.

######  Lemma 4.7.

$\hat{h}^{\alpha,\beta}=\left(\hat{h}^{(2)}_{\left(\frac{-\ln(\alpha)}{k}\right)}\circ\hat{h}^{(5)}_{\left(\frac{\alpha^{2}-1}{2k}\right)}\right)\circ\left(\hat{h}^{(6)}_{\left(\frac{\beta}{2k(2k\alpha-\beta)}\right)}\circ\hat{h}^{(2)}_{\left(\frac{1}{k}\ln{\left(\frac{2k\alpha-\beta}{2k\alpha}\right)}\right)}\circ\hat{h}^{(5)}_{\left(\frac{\beta}{2k(2k\alpha-\beta)}\right)}\right).$

Although the calculation is tedious, it can easily be shown that the
expression given in Lemma 4.7 holds. One thing to notice is that
$h^{\alpha,\beta}(x,t)$ from Proposition 3.5 of [3] is a composition of three
symmetries of the heat equation and so it satisfies the heat equation. Now,
because the function $\hat{h}^{\alpha,\beta}$ is a composition of solutions to
the OU Fokker-Planck equation, then is itself a solution to the OU Fokker-
Planck equation. The Lemma 4.7 can then be used to shorten the first part of
the proof given in Proposition 4.5. Now, we give the third proof of Theorem
4.1.

###### Third proof of Theorem 4.1.

By proposition 4.5 and Lemma 4.7, we deduce that $\hat{h}^{\alpha,\beta}$ in
(23) is a solution to the boundary value problem
$\mathcal{\hat{H}}(\hat{S}^{\alpha,\beta}f)$. Now, for a general boundary $f$,
$\mathbb{P}(\hat{T}^{f}\leq t)=1-\int_{-\infty}^{f(t)}\hat{h}(x,t)dx$. Setting
$\hat{p}^{f}(t)dt=\mathbb{P}\Big{(}\hat{T}^{f}\in dt\Big{)}$, we get

$\hat{p}^{f}(t)=-\frac{d}{dt}\left(\int_{-\infty}^{f(t)}\hat{h}(x,t)dx\right)=-\hat{h}(f(t),t)f^{\prime}(t)-\int_{-\infty}^{f(t)}\frac{\partial}{\partial
t}\hat{h}(x,t)dx=-\int_{-\infty}^{f(t)}\frac{\partial}{\partial
t}\hat{h}(x,t)dx,$

because $\hat{h}$ vanishes on the boundary. Using the OU Fokker-Planck
equation, we get

$\hat{p}^{f}(t)=-\int_{-\infty}^{f(t)}-\frac{\partial}{\partial
x}[-k(x)\hat{h}(x,t)]+\frac{1}{2}\frac{\partial^{2}}{\partial
x^{2}}\hat{h}(x,t)\;dx=-\frac{1}{2}\frac{\partial}{\partial
x}\hat{h}(x,t)|_{x=f(t)}.$ (25)

Using the $\hat{h}^{\alpha,\beta}$ function with its corresponding boundary
$\hat{S}^{\alpha,\beta}f$ gives us expression (15) from Theorem 4.1.

∎

######  Remark 4.8.

When we take the limit as $k$ goes to 0 of the symmetries in (24), we only
recover the first three symmetries of the heat equation (Section 3.3 of [3]).
The rest of the symmetries of the heat equation can be recovered by using a
combination of the symmetries of the OU Fokker-Planck equation.

######  Remark 4.9.

This Lie approach led to a variant of our boundary crossing identity (15) in
[20] (their equation 40) up to some misprints therein. In particular, by
setting
$\beta^{\prime}=\beta/2k\alpha,\alpha^{\prime}=1/\alpha^{2}-\beta/2k\alpha-1,A=k^{2},B=0,\nu=0,x_{0}=0$
and $\theta_{\mathcal{F}_{2}}(x)=e^{-kx^{2}/2}$, we noted that the first term
in (40) should be
$\mathcal{T}_{-}^{-1}\mathcal{T}_{+}^{-1/2}\sqrt{\alpha^{\prime}+\beta^{\prime}+1}$
instead of $\mathcal{T}_{+}^{-1/2}/\sqrt{\alpha^{\prime}+\beta^{\prime}+1}$.

###  4.4 Asymptotic behaviour of the FPT densities

Here, we discuss the asymptotic behaviour of the FPT distribution of
$\hat{T}^{\hat{S}^{\alpha,\beta}f}$. As before, we assume that our boundary
$f\in\mathcal{C}^{1}((0,\infty),\mathbb{R}^{+})$ and, as discussed before
Section 4.1, we only consider $\alpha>0$. We write $\mathbb{P}(\hat{T}^{f}\in
dt)=\hat{p}^{f}(t)dt$ and from Theorem 4.1, for $\beta>0$, observe that we get
the following asymptotic identity

$\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)\sim e^{2kt}\alpha^{2}\frac{(\alpha\beta
r(t))^{-1/2}}{\big{(}2k\alpha^{2}r(t)+\alpha\beta
r(t)\big{)}}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f(t))^{2}\;e^{2kt}}\;\hat{p}^{f}\left(s\left(\frac{\alpha}{\beta}\right)\right)\;\;\;\text{as}\;t\to\infty,$
(26)

where $h(t)\sim I(t)$ as $t\to\zeta$ means that $h(t)/I(t)\to 1$ as
$t\to\zeta$, for some $\zeta\in[0,\infty]$. Now, for $\beta<0$, we need a
different approach. Set $r=\mathbb{P}(\hat{T}^{f}<\infty)$. We say that $f$ is
transient if $r<1$. We now give a necessary and sufficient condition for when
a given curve is transient.

######  Theorem 4.10.

Let $f\in\mathcal{C}^{1}((0,\infty),\mathbb{R}^{+})$. If
$f(t)e^{kt}/\sqrt{r(t)}$ is increasing for sufficiently large t, then $f$ is
transient if and only if

$\int_{s(1)}^{\infty}\frac{e^{3kt}f(t)}{r(t)^{3/2}}e^{-e^{2kt}f(t)^{2}/2r(t)}dt<\infty.$

###### Proof.

By relation (16), we deduce that

$r=\mathbb{P}(\hat{T}^{f}<\infty)=\mathbb{P}(s(T^{\Lambda
f})<\infty)=\mathbb{P}(T^{\Lambda f}<\infty).$

Now, by the classical Kolmogorov-Eröds-Petrovski theorem [11], if
$t^{-1/2}\Lambda f(t)$ (i.e. $f(t)e^{kt}/\sqrt{r(t)}$) is increasing for
sufficiently large $t$, then $\Lambda f$ is a transient curve for the Brownian
motion if and only if

$\int_{1}^{\infty}t^{-3/2}\Lambda f(t)e^{-(\Lambda f)^{2}/2t}dt<\infty,$

and the result follows by a change of variable. ∎

For $\beta<0$, the lifetime of our boundaries is
$\zeta_{\alpha,\beta}=s(-1/(\alpha\beta))$, so
$s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right)\to\infty$ as
$t\to\zeta_{\alpha,\beta}$. In this case, whenever $f$ is transient and
satisfies suitable conditions, we can derive the asymptotic expression of the
FPT density of the OU by making use of the corresponding result for the BM by
Anderson and Pitt [4], as proved in the following Proposition.

######  Proposition 4.11.

Assume that $f$ is transient and satisfies the following conditions

  1. (i)

$\Lambda f$ is increasing, concave, twice differentiable on $(0,\infty)$ and
of regular variation at $\infty$ with index $\alpha\in[1/2,1)$,

  2. (ii)

$\Lambda f(t)/\sqrt{t}$ is increasing at $\infty$, and $\Lambda f(t)/t$ is
convex and decreases to 0 for sufficiently large t,

  3. (iii)

There exist positive constants $c<1$ and $c^{\prime}$ such that the
inequalities $tf^{\prime}(s(t))\leq(c(2kt+1)-kt)f(s(t))$ and
$|t^{2}/(2kt+1)^{2}(f^{\prime\prime}(s(t))-k^{2}f(s(t)))|\leq
c^{\prime}f(s(t))$ are met for sufficiently large t.

Then, for any $\beta<0$, we have

$\hat{p}^{\hat{S}^{\alpha,\beta}f}(t)\sim\sqrt{\frac{|\alpha\beta|^{3}}{2\pi}}\left(\frac{2k}{|\alpha\beta|}+1\right)(1-r)\tilde{f}(t)\;\;\;\text{as}\;\;t\to\zeta_{\alpha,\beta},$

where

$\tilde{f}(t)=\frac{\left(\frac{k\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\right)f\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)-\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)f^{\prime}\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)}{\sqrt{2k\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}+1}}.$

###### Proof.

By (17), we have $\hat{p}^{f}(t)=e^{2kt}p^{\Lambda f}(r(t))$. By Anderson and
Pitt [4] result for the BM, if conditions (i)-(iii) are satisfied, we can
write

$\hat{p}^{f}(t)\sim e^{2kt}(1-r)\frac{\Lambda f(r(t))-r(t)(\Lambda
f(r(t)))^{\prime}}{\sqrt{2\pi}r(t)^{3/2}}e^{-(\Lambda
f(r(t)))^{2}/2r(t)}\;\;\text{as}\;\;t\to\infty.$

The result follows by simplifying this, combining it with (15) and using the
assumption that $\Lambda f(t)/t\downarrow 0$ as $t\to\infty$. ∎

For $\beta>0$, regardless of whether $f$ is transient or not, if
$\hat{S}^{\alpha,\beta}f(t)e^{kt}/\sqrt{r(t)}$ is increasing for sufficiently
large $t$, $0<\beta\sqrt{2k(\alpha/\beta)+1}f(s(\alpha/\beta))<\infty$ and
$f(0)>0$, we have

$\displaystyle\int_{s(1)}^{\infty}\frac{e^{3kt}\hat{S}^{\alpha,\beta}f(t)}{r(t)^{3/2}}e^{-e^{2kt}(\hat{S}^{\alpha,\beta}f(t))^{2}/2r(t)}dt$
$\displaystyle=$
$\displaystyle\int_{1}^{\infty}\frac{1}{\sqrt{z}}g(s(z))e^{-g(s(z))^{2}z/2}dz$
$\displaystyle<$
$\displaystyle\int_{1}^{\infty}g(s(z))e^{-g(s(z))^{2}z/2}dz<\infty,$

where we performed a change of variable and noted that
$g(t)=e^{kt}\hat{S}^{\alpha,\beta}f(t)/r(t)\sim$
$\beta\sqrt{2k(\alpha/\beta)+1}f(s(\alpha/\beta))$ as $t\to\infty$. The
finiteness of the integral follows by the fact that $g$ eventually stabilises
and the assumptions ensure that $g$ never blows up or vanishes, so its minimum
and the maximum are always attained and by Theorem 4.10,
$\hat{S}^{\alpha,\beta}f$ is always transient under these assumptions.
However, we cannot use Proposition 4.11 directly, because
$\Lambda\hat{S}^{\alpha,\beta}f(t)/t\sim\beta\sqrt{2k(\alpha/\beta)+1}f(s(\alpha/\beta))>0$
as $t\to\infty$ and so it does not decrease to 0, violating assumption (ii).
However, the general formula (26) still holds.

######  Remark 4.12.

We noted a misprint in Section 4.3 of [2]. For the Brownian motion, when
$\beta>0$, the one parameter family of transformed functions $S^{\beta}f$ is
always transient whenever $t^{-1/2}S^{\beta}f$ is increasing for sufficiently
large t, $0<\beta f(\alpha/\beta)<\infty$ and $f(0)>0$.

##  5 Interpretation via the method of images

In his lecture notes [17], Lerche did a thorough investigation of the method
of images for the standard BM. Here, we apply the method of images to the OU
process and use it to produce new examples of curves with explicit FPT
densities. As in the BM case, we would like to construct a function $\hat{h}$
satisfying the OU Fokker-Planck equation. Then, by the uniqueness of such
solutions, as seen in Section 4.3, $\hat{h}$ would also satisfy (21) and be
the solution to the boundary value problem $\mathcal{\hat{H}}(f)$. We proceed
as follows. First, we assume to have a positive $\sigma-$finite measure $F$
with $\int_{0}^{\infty}\phi(\sqrt{\epsilon\theta})F(d\theta)<\infty$ for all
$\epsilon>0$. Then, for $a>0$, we define the $\hat{h}$ function by

$\displaystyle\hat{h}(x,t)$
$\displaystyle:=p_{t}(0,x)-\frac{1}{a}\int_{0}^{\infty}p_{t}(\theta,x)F(d\theta)=\frac{e^{kt}}{\sqrt{r(t)}}\phi\left(\frac{xe^{kt}}{\sqrt{r(t)}}\right)-\frac{1}{a}\int_{0}^{\infty}\frac{e^{kt}}{\sqrt{r(t)}}\phi\bigg{(}\frac{xe^{kt}-\theta}{\sqrt{r(t)}}\bigg{)}F(d\theta).$

We know that $\hat{h}$ vanishes on the boundary $f$, and so by simplifying, we
get that $\hat{h}(x,t)=0$ is equivalent to

$l\left(\frac{xe^{kt}}{r(t)},\frac{1}{r(t)}\right)=a,$ (27)

where $l(y,s)=\int_{0}^{\infty}e^{\theta y-\frac{1}{2}\theta^{2}s}F(d\theta)$.
The following theorem gives us a way of calculating the density of
$\hat{T}^{f}$ explicitly.

######  Theorem 5.1.

The FPT density of $\hat{T}^{f}$ is given by

$\mathbb{P}(\hat{T}^{f}\in
dt)=\frac{e^{2kt}}{2r(t)^{3/2}}\phi\bigg{(}\frac{f(t)e^{kt}}{\sqrt{r(t)}}\bigg{)}E(\theta|(f(t),t))dt,$

where

$E(\theta|(f(t),t))=\frac{\int_{0}^{\infty}\theta\phi\bigg{(}\frac{f(t)e^{kt}-\theta}{\sqrt{r(t)}}\bigg{)}F(d\theta)}{\int_{0}^{\infty}\phi\bigg{(}\frac{f(t)e^{kt}-\theta}{\sqrt{r(t)}}\bigg{)}F(d\theta)}.$

###### Proof.

We just use relation (25) to get the result. ∎

######  Remark 5.2.

If the support of the measure $F$ is on $\mathbb{R}$ and $F(\\{0\\})=0$, then
we define

$\hat{h}(x,t):=p_{t}(0,x)-\frac{1}{a}\int_{-\infty}^{\infty}p_{t}(\theta,x)F(d\theta).$

Then, there exist positive and negative valued functions $f_{+}$ and $f_{-}$,
with $f_{-}<f_{+}$ and the properties $\hat{h}(f_{+}(t),t)=0$ and
$\hat{h}(f_{-}(t),t)=0$ for all $t<t_{a}$ for a certain $t_{a}\leq\infty$.
Then, the stopping time can also be defined as the first time the process
leaves the region $(f_{-}(t),f_{+}(t))$, i.e.

$\hat{T}^{f_{\pm}}=\inf\\{0<t<t_{a}\;|\;U_{t}\notin(f_{-}(t),f_{+}(t))\\}.$
(28)

The following lemma gives us a characterisation of the boundaries obtained
through the method of images.

######  Lemma 5.3 (Characterisation of boundaries).

The boundaries $f$ obtained through the method of images have the following
properties:

  1. 1.

$f$ is infinitely often continuously differentiable;

  2. 2.

$\Lambda f(t)/t$ (i.e. $f(t)e^{kt}/r(t)$) is monotone decreasing;

  3. 3.

$f^{\prime\prime}(t)-k^{2}f(t)\leq 0$.

###### Proof.

In Lemma 1.1 of [17], it is shown that the curves $\eta(t)$ that satisfy the
equation $l(\frac{\eta(t)}{t},\frac{1}{t})=a$ are infinitely continuously
differentiable and concave, with $\eta(t)/t$ monotone decreasing. As in our
case we have $a=l\left(\frac{f(t)e^{kt}}{r(t)},\frac{1}{r(t)}\right)$, by
changing $t\to s(t)$, we get $a=l\left(\frac{\Lambda
f(t)}{t},\frac{1}{t}\right)$. So $\eta(t):=\Lambda f(t)$ must satisfy the
three properties given in Lemma 1.1 of [17], so (1) and (2) follow directly.
For (3),

$f^{\prime\prime}(t)=k^{2}e^{-kt}\eta(r(t))+e^{3kt}\eta^{\prime\prime}(r(t))=k^{2}f(t)+e^{3kt}\eta^{\prime\prime}(r(t))$

and the result follows as $\eta(t)$ is concave. ∎

######  Remark 5.4.

From Theorem 3.1, if $\mu(s(\tau f))>k^{2}$, then we get that the solutions to
nonlinear ordinary differential equations (14) satisfy
$f^{\prime\prime}-k^{2}f\leq 0$.

Let us now consider some examples of the boundaries that arise for specific
measures $F$ via Theorem 5.1.

######  Example 5.5.

Consider $F(d\theta)=\delta_{2z}$ for some $z>0$. Using $\hat{h}(f(t),t)=0$
and simplifying, we get

$f(t)=\frac{\ln{(a)}}{2zk}\sinh{(kt)}+ze^{-kt}.$

By Theorem 5.1,

$\mathbb{P}(\hat{T}^{f}\in
dt)=\frac{ze^{2kt}}{r(t)^{3/2}}\phi\left(\frac{\frac{\ln(a)}{2z}r(t)+z}{\sqrt{r(t)}}\right)dt.$

######  Remark 5.6.

In [9], the FPT of a mean-reverting OU with parameter $\mu$ hitting a
hyperbolic type boundary of the form $\mu+Ae^{kt}+Be^{-kt}$ is studied. This
reduces to the FPT of a standard OU hitting a curve of the form
$Ae^{kt}+Be^{-kt}$ for arbitrary constants A and B which is the curve given in
this example. Also, applying the $\hat{S}$ transformation to these curves
gives the same family of curves with different constant coefficients.

######  Example 5.7.

Consider $F(d\theta)=\frac{d\theta}{\sqrt{2\pi}}$ on $\mathbb{R}$. Then,

$f_{\pm}(t)=\pm
e^{-kt}\sqrt{r(t)\ln\left(\frac{a^{2}e^{2kt}}{r(t)}\right)},\;\;\;\;\;t\leq
t_{a}=\begin{cases}-s(-a^{2})\;\;\;\;\;\;\;\text{if}\;a^{2}<1/(2k),\\\
+\infty\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\textrm{otherwise}.\end{cases}$

This is a two sided boundary, with exit time $\hat{T}^{f_{\pm}}$ defined in
(28) and FPT density given by

$\mathbb{P}(\hat{T}^{f_{\pm}}\in
dt)=\frac{e^{2kt}}{2r(t)}\phi\left(\sqrt{\ln\left(\frac{a^{2}e^{2kt}}{r(t)}\right)}\right)\sqrt{\ln\left(\frac{a^{2}e^{2kt}}{r(t)}\right)}dt.$

Applying the $\hat{S}$ transformation (13) to this curve, we get

$\hat{S}^{\alpha,\beta}f_{\pm}(t)=\pm\sqrt{r(t)}\sqrt{1+\alpha\beta
r(t)}e^{-kt}\sqrt{\ln{\bigg{[}\frac{a^{2}(2k\alpha^{2}r(t)+\alpha\beta
r(t)+1)}{\alpha^{2}r(t)}\bigg{]}}}.$ (29)

The FPT density is given by

$\displaystyle\hat{p}^{\hat{S}^{\alpha,\beta}f_{\pm}}(t)$
$\displaystyle=e^{2kt}\alpha^{2}\frac{(1+\alpha\beta
r(t))^{-3/2}}{\big{(}2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1\big{)}}e^{-\frac{\alpha\beta}{2(1+\alpha\beta
r(t))}(\hat{S}^{\alpha,\beta}f_{\pm}(t))^{2}\;e^{2kt}}\;\hat{p}^{f_{\pm}}\left(s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\right)\right)$
$\displaystyle=\frac{e^{2kt}}{\sqrt{8\pi}}\sqrt{\frac{\ln{\bigg{[}\frac{a^{2}(2k\alpha^{2}r(t)+\alpha\beta
r(t)+1)}{\alpha^{2}r(t)}\bigg{]}}}{1+\alpha\beta
r(t)}}e^{-\frac{1}{2}\ln{\left[\frac{a^{2}(2k\alpha^{2}r(t)+\alpha\beta
r(t)+1)}{\alpha^{2}r(t)}\right]}(\alpha\beta r(t)+1)}.$

Now, for example, by setting $a=1,\alpha=2,\beta=k$, we get

$\hat{S}^{2,k}f_{\pm}(t)=\pm\sqrt{\frac{e^{2kt}-1}{2k}}\;\sqrt{\ln{\left[\frac{k}{4}\big{(}\coth{(kt)}+9\big{)}\right]}},$

with FPT density given by

$\hat{p}^{\hat{S}^{2,k}f_{\pm}}(t)=\frac{e^{kt}}{\sqrt{8\pi}}\sqrt{\ln{\left[\frac{k}{4}\big{(}\coth{(kt)}+9\big{)}\right]}}e^{-\frac{e^{2kt}}{2}\ln{\left[\frac{k}{4}\big{(}\coth{(kt)}+9\big{)}\right]}}.$

In Appendix A, Figures 3 and 4 illustrate such curves for different values of
$k,\alpha$ and $\beta$. When $\beta=0$, $\hat{S}^{\alpha,0}f_{\pm}$ converges
to a constant level as $t\to\infty$, see Figure 3. Moreover, whenever $\alpha$
and $\beta$ have opposite signs, we get a closed shaped graph defined on
$t\in[0,\zeta_{0.1,-0.5}]$, see Figure 4.

######  Remark 5.8.

Back to our $\hat{S}$ transformation, using (27) with the time-transformation
$t\rightarrow s\bigg{(}\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\bigg{)}$, we
get

$a=\int_{0}^{\infty}e^{\frac{\theta}{\alpha}[\frac{\hat{f}^{\alpha,\beta}(t)e^{kt}}{r(t)}]-\frac{1}{2}[\frac{(\theta/\alpha)^{2}}{r(t)}]}F^{\alpha,\beta}(d\theta),$

where
$F^{\alpha,\beta}(d\theta)=e^{-\frac{\beta}{\alpha}\frac{\theta^{2}}{2}}F(d\theta)$
is the measure corresponding to the curve $\hat{S}^{\alpha,\beta}f(t)$.

## Acknowledgements

We are grateful to Dmitry Muravey for a discussion on the Lie symmetries while
finalising the paper. AA was funded by EPSRC grant [EP/V520226/1] as part of
the Warwick CDT in Mathematics and Statistics. For the purpose of open access,
the authors have applied a Creative Commons Attribution (CC BY) licence to any
Author Accepted Manuscript version arising from this submission.

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##  Appendix A Figures and charts

Notations |  Transformations
---|---
$r(t)$ |  $(e^{2kt}-1)/2k$
$s(t)$ |  $\ln{(2kt+1)}/2k$
$\Lambda f$ |  $\sqrt{2kt+1}f(s(t))$
$\Lambda^{-1}f$ |  $e^{-kt}f(r(t))$
$\tau f$ |  $\int_{0}^{t}[f(s)]^{-2}ds$
$\Pi^{\alpha,\beta}f$ |  $f(t)(\alpha+\beta\tau f(t))$
$\Sigma f$ |  $\frac{1}{f(\rho\circ\tau f(t))}$
$\hat{\Sigma}f$ |  $\Lambda^{-1}\circ\Sigma\circ\Lambda f(t)=\frac{e^{-k\rho\circ\tau f(r(t))-kt}}{f(\rho\circ\tau f(r(t)))}$
$S^{\alpha,\beta}f$ |  $\Sigma\circ\Pi^{\alpha,-\beta}\circ\Sigma f(t)=\left(\frac{1+\alpha\beta t}{\alpha}\right)f\left(\frac{\alpha^{2}t}{1+\alpha\beta t}\right)$
$\hat{S}^{\alpha,\beta}f$ |  $\hat{\Sigma}\circ\Pi^{\alpha,-\beta}\circ\hat{\Sigma}f(t)=\left(\frac{1+\alpha\beta r(t)}{\alpha}\right)\sqrt{2k\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}+1}\;e^{-kt}\;f\bigg{(}s\left(\frac{\alpha^{2}r(t)}{1+\alpha\beta r(t)}\right)\bigg{)}$
Table 1: Key transformations introduced in the paper.
OU${f(t)}$${\bigg{(}\frac{1+\alpha\beta
r(t)}{\alpha}\bigg{)}\sqrt{2k\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}+1}\;e^{-kt}\;f\bigg{(}s\bigg{(}\frac{\alpha^{2}r(t)}{1+\alpha\beta
r(t)}\bigg{)}\bigg{)}}$BM${\sqrt{2kt+1}f(s(t))}$${\bigg{(}\frac{1+\alpha\beta
t}{\alpha}\bigg{)}\sqrt{2k\frac{\alpha^{2}t}{1+\alpha\beta
t}+1}\;f\bigg{(}s\bigg{(}\frac{\alpha^{2}t}{1+\alpha\beta
t}\bigg{)}\bigg{)}}$$\scriptstyle{\hat{S}^{\alpha,\beta}}$$\scriptstyle{\Lambda}$$\scriptstyle{S^{\alpha,\beta}}$$\scriptstyle{\Lambda^{-1}}$
Figure 1: Flow chart of $\hat{S}$. Figure 2: $\hat{S}^{\alpha,1}a$ for fixed
$\beta=1,\;k=1/2,\;a=1$ and
$\alpha=1\;(red),\;2\;(green),\;3\;(orange),\;5\;(purple),\;20\;(black)$.
Figure 3: Transformed curves (29) for $\beta=0,k=1,a=1$ and $\alpha=-1$
(orange), 0.5 (green), 0.1 (blue). Figure 4: Transformed curves (29) for fixed
$k=1,a=1$. Orange: $\beta=-0.5,\alpha=0.1$. Green: $\beta=0.1,\alpha=0.1$.
Blue: $\beta=1,\alpha=0.1$.