# Existence in the inverse Shiryaev problem111The research of Yoann Potiron is
supported by Japanese Society for the Promotion of Science Grants-in-Aid for
Early-Career Scientists ”Econometric methods for high frequency data”
(20K13470), Grants-in-Aid for Young Scientists B ”Forecasting and model
selection in time-varying parameter models” (17K13718), Fukuzawa fund and an
internal grant ”Econometric methods for high frequency data” from Keio
University. The author would like to thank Steven Lalley, Per Mykland, Simon
Clinet, Masaaki Fukasawa, Goran Peskir, Cristina Zucca, Laura Sacerdote, Mario
Abundo, David Saunders, Martin Kolb, Erik Ekstrom,Svante Janson, and the
participants of the Research Webinar Series at Decision Sciences Areaa at
Indian Institute of Management Bangalore, the 14th International Conference of
the ERCIM WG on Computational and Methodological Statistics at King’s College
London and the 5th International Conference on Econometrics and Statistics
(EcoSta 2022) at Ryukoku University in Kyoto for helpful discussions and
advice.

Yoann Potiron222Faculty of Business and Commerce, Keio University. 2-15-45
Mita, Minato-ku, Tokyo, 108-8345, Japan. Phone: +81-3-5418-6571. E-mail:
<EMAIL_ADDRESS>website: http://www.fbc.keio.ac.jp/~potiron

(This version: )

###### Abstract

In this paper, we consider the inverse first-passage Shiryaev problem, i.e.
for $(W_{t})_{t\geq 0}$ a standard Brownian motion and any upper boundary
_continuous_ function $g:\mathbb{R}^{+}\rightarrow\mathbb{R}$ satisfying
$g(0)\geq 0$, we define

$\mathrm{T}_{g}^{W}:=\inf\\{t\in\mathbb{R}^{+}\text{ s.t. }W_{t}\geq g(t)\\},$

and $f_{g}^{W}(t)$ its related density. For any target density function of the
form $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ satisfying some smooth
assumptions and any arbitrarily big horizon time $T>0$, we show the existence
of a related _continuous_ boundary $g_{f,T}:[0,T]\rightarrow\mathbb{R}^{+}$,
with $g_{f,T}(0)\geq 0$, which satisfies

$f_{g_{f,T}}^{W}(t)=f(t)\text{ for }0\leq t\leq T.$

The novelty in this paper is that we consider a more direct proof of existence
based on Arzelà-Ascoli theorem, where $g_{f,T}$ is exhibited as a limit
boundary of a subsequence of a piecewise linear boundary. As this is not a
completely direct proof, we do not obtain any explicit formula for $g_{f,T}$
as a function of $f$. The results are also proved in the symmetrical two-
dimensional boundary case.

Keywords: Arzelà-Ascoli theorem, Brownian motion, inverse first-passage
problem, stopping time, time-varying boundary

##  1 Introduction

In the theory of stochastic processes, first-passage time related problems
have been studied extensively. When specified to a standard Brownian motion
$(W_{t})_{t\geq 0}$, the related first passage to an upper boundary
_continuous_ function $g:\mathbb{R}^{+}\rightarrow\mathbb{R}$ satisfying
$g(0)\geq 0$ is defined as

$\displaystyle\mathrm{T}_{g}^{W}:=\inf\\{t\in\mathbb{R}^{+}\text{ s.t.
}W_{t}\geq g(t)\\},$ (1.1)

with $f_{g}^{W}(t)$ its related density. There are two classes of problems
related to these stopping times. The first-passage problem investigates
$f_{g}^{W}(t)$ when $g$ is known. The _inverse_ first-passage problem aims at
finding a boundary $g_{f}:\mathbb{R}^{+}\rightarrow\mathbb{R}$, with
$g_{f}(0)\geq 0$, which satisfies

$\displaystyle f_{g_{f}}^{W}(t)=f(t)\text{ for }t\geq 0,$ (1.2)

where $f$ is a _target_ density function of the form
$f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$. Almost 50 years ago, A.
Shiryaev333during a Banach center meeting in 1976 asked if one can find such
_continuous_ $g_{f}$ when $f(t)=\lambda\exp(-\lambda t)$ for $\lambda>0$ is
exponentially distributed. Obviously, this question can be classified as an
_inverse_ problem.

First-passage problems are often described as being hard, and explicit
solutions of the density $f_{g}^{W}(t)$ are only exhibited for a few cases
$g$. These include and are not limited to: the linear case (see, e.g., [Doob,
1949] (Formula (4.2), p. 397) or [Malmquist, 1954] (p. 526)); the Daniels
boundary (see [Daniels, 1969]) when

$g(t)=\frac{\alpha}{2}-\frac{t}{\alpha}\log\left(\frac{\beta}{2}+\sqrt{\frac{\beta^{2}}{4}+\gamma\exp\left(\frac{-\alpha^{2}}{t}\right)}\right),$

with $\alpha>0$, $\beta\geq 0$ and $\gamma>\beta^{2}/4$; a formula in a
general boundary setup but which depends on asymptotic conditional
expectations (see [Durbin, 1971]); the one-sided square root boundary (see
[Novikov, 1971]); the quadratic boundary (see [Salminen, 1988]); the
piecewise-linear boundary (see [Wang and Pötzelberger, 1997]) ; the piecewise-
specific boundary ([Novikov et al., 1999]) where "specific" could mean any of
the aforementioned cases.

As for the Shiryaev inverse problem itself, there are two early papers
([Dudley and Gutmann, 1977] and [Anulova, 1981]) which provide the existence
of stopping times for any target density function $f(t)$, but the obtained
boundaries are _lower semi-continuous_. Later, key results were found in
[Peskir, 2002b], where the problem was reformulated as a nonlinear Volterra
integral equation of the second kind. Highly related to that, [Peskir, 2002a]
exhibits three boundaries in the neighborhood of 0, which respectively satisfy

$\lim_{t\rightarrow 0}f_{g}(t)=0\text{ , }\lim_{t\rightarrow
0}f_{g}(t)=\infty,\text{ , }\lim_{t\rightarrow 0}f_{g}(t)=c\in(0,\infty).$

Subsequently, [Zucca and Sacerdote, 2009] investigate two methods for
approximating the boundary. The first method is based on a piecewise-linear
scheme, while the second makes use of the Volterra equation form. Assuming
that the boundary exists, is unique, concave or convex, they in particular
prove that the error vanishes asymptotically in the first case. Finally, [Gür
and Pötzelberger, 2020] extends the latter method to provide a boundary
estimator based on a sample of hitting times. See also [Abundo, 2006] for
extensions to the general diffusion process case.

There is also a literature who builds a bridge between the inverse problem and
the free boundary problem in survival analysis. The idea goes back to
[Avellaneda and Zhu, 2001]. In an impressive series of papers, [Cheng et al.,
2006] and [Chen et al., 2011] respectively show the existence and uniqueness
of the viscosity solution in the related free boundary problem and that the
obtained resultant satisfies the inverse first-passage problem (1.1)-(1.2). In
a more recent paper, [Ekström and Janson, 2016] prove rigorously the
connection between the inverse problem and a related optimal stopping problem.
See also [Fukasawa and Obloj, 2020] for possible application of the method.

With the powerful use of variational inequalities and/or optimal stopping
theory, the aforementioned papers prove existence and unicity of the Shiryaev
problem when boundaries are _lower semi-continuous_. As for the case when the
extracted boundaries are _continuous_ , the results are more mitigated,
although conditions on $f$ are exhibited in [Chen et al., 2011] (Proposition
6) and also in [Ekström and Janson, 2016] (Theorem 8.2). In this paper, we
propose to investigate a more direct proving technology, which in particular
circumvents the use of variational inequalities and/or optimal stopping
theory. More specifically, the proof is based on a discretization of the
problem to approximate the boundary with a piecewise-linear boundary whose
definition is almost equal to that in [Zucca and Sacerdote, 2009] (Section 3).
The merit of our paper is that based on this approximation, we _show_ the
existence, whereas the cited paper _assumes_ such existence.

Key steps of the proof of existence are the definition, for any arbitrarily
big horizon time $T>0$, of a sequence of continuous piecewise-linear functions
$g_{f,T}^{(n)}:\mathbb{R}^{+}\rightarrow\mathbb{R}$ with $g_{f,T}^{(n)}(0)\geq
0$ which satisfies for any $n\in\mathbb{N}-\\{0\\}$, any
$l\in\mathbb{N}-\\{0\\}$ with $l\geq n$ and any $m=0,\cdots,2^{n}-1$

$\displaystyle\mathbb{P}\left(T_{g_{f,T}^{(l)}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\right)=\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds.$
(1.3)

Then, using a fundamental result of analysis, i.e. the Arzelà-Ascoli theorem,
we are able to extract a continuous boundary
$g_{f,T}:\mathbb{R}^{+}\rightarrow\mathbb{R}$, with $g_{f,T}(0)\geq 0$, as the
limit of a subsequence of $g_{f,T}^{(n)}$. Finally, we can show that (1.3) is
preserved at the limit, i.e. for any $p\in\mathbb{N}-\\{0\\}$ and
$k=0,\ldots,2^{p}-1$ we obtain

$\displaystyle\mathbb{P}\big{(}T_{g_{f,T}}^{W}\in[kT/2^{p},(k+1)T/2^{p}]\big{)}=\int_{kT/2^{p}}^{(k+1)T/2^{p}}f(s)ds.$
(1.4)

By Borel arguments, (1.4) implies that

$f_{g_{f,T}}^{W}(t)=f(t)\text{ for any }0\leq t\leq T.$

The assumptions required on $f(t)$ are rather smooth, and in particular, the
exponential distribution $f(t)=\lambda\exp(-\lambda t)$ for $\lambda>0$
satisfies the assumptions of this paper. The results are also derived in the
two-dimensional symmetrical boundary case, which to the best of our knowledge
is new.

As it stands, we do not obtain any explicit formula for the boundary limit as
a function of $f$, nor unicity of the solution. Rather, we obtain $g_{f,T}$
which a-priori depends on $T$. To obtain an explicit formula, one possibility
consists in investigating the highly nonlinear Volterra problem as the
question of uniqueness for this kind of equations has been affirmatively
resolved in [Peskir, 2005].

In terms of applications, first-passage problems can be useful in statistics
(see, e.g., [Sen, 1981] or [Siegmund, 1986]), in biology (see, e.g.,
[Ricciardi et al., 1999]), in mathematical finance such as in the default
barrier reproducing a given default distribution problem (see, e.g., [Hull and
White, 2001]), or in financial econometrics when modeling endogeneity in
sampling times (see [Fukasawa, 2010], [Robert and Rosenbaum, 2012] or [Potiron
and Mykland, 2017]) with high frequency data.

##  2 Notation

Prior to the exposition of the theory, we give some common definition which
will be used throughout the manuscript. For $A\subset\mathbb{R}^{+}$ and
$B\subset\mathbb{R}$ s.t. $0\in A$, we define:

$\displaystyle\mathcal{C}_{0}(A,B)$ $\displaystyle:=$
$\displaystyle\\{h:A\rightarrow B\text{ s.t. }h\text{ is continuous}\\},$
$\displaystyle\mathcal{C}_{0}^{+}(A,B)$ $\displaystyle:=$
$\displaystyle\\{h:A\rightarrow B\text{ s.t. }h\text{ is continuous and
}h(0)\geq 0\\},$ $\displaystyle\mathcal{C}_{0}^{-}(A,B)$ $\displaystyle:=$
$\displaystyle\\{h:A\rightarrow B\text{ s.t. }h\text{ is continuous and
}h(0)\leq 0\\},$ $\displaystyle\overline{\mathcal{C}_{0}^{-}(A,B)}$
$\displaystyle:=$
$\displaystyle\mathcal{C}_{0}^{-}(A,B)\cup\\{h:A\rightarrow-\infty\\},$

as respectively the functional space of continuous functions, the subspace
with non-negative starting values, the subspace with non-positive starting
values and the union of the subspace with non-positive starting values
together with function with only one value equal to $-\infty$. Finally, for
any vector function
$h=(h^{(1)},h^{(2)})\in\overline{\mathcal{C}_{0}^{-}(A,B)}\times\mathcal{C}_{0}^{+}(A,B)$,
we define the related infinite norm as:

$\displaystyle\left|\left|h\right|\right|$ $\displaystyle=$ $\displaystyle
2\sup_{t\in A}\left|h^{(2)}(t)\right|\text{ if }h^{(1)}\neq-\infty$
$\displaystyle=$ $\displaystyle\sup_{t\in A}\left|h^{(1)}(t)\right|+\sup_{t\in
A}\left|h^{(2)}(t)\right|\text{ else}$

##  3 Objective

We introduce the (complete) stochastic basis
$\mathcal{B}=(\Omega,\mathbb{P},\mathcal{F},\mathbf{F})$, where $\mathcal{F}$
is a $\sigma$-field and $\mathbf{F}=(\mathcal{F}_{t})_{t\in\mathbb{R}^{+}}$ is
a filtration. For any $\mathcal{F}_{t}$-standard Brownian motion $W_{t}$ and
any given pair of boundary continuous functions

$g=(g^{(1)},g^{(2)})\in\overline{\mathcal{C}_{0}^{-}(\mathbb{R}^{+},\mathbb{R})}\times\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R}),$

we associate the boundary crossing variable

$\displaystyle\mathrm{T}_{g}^{W}:=\inf\\{t\in\mathbb{R}^{+}\text{ s.t.
}W_{t}\leq g^{(1)}(t)\text{ or }W_{t}\geq g^{(2)}(t)\\}.$ (3.1)

In (3.1), the process $W_{t}$ is the _hitting_ process, and the function $g$
plays the role of the (two-dimensional: upper and lower) boundary.
Accordingly, $\mathrm{T}_{g}^{W}$ corresponds to the variable obtained when
the process first hits the upper boundary function or the lower boundary
function. If one or both functional components of $g$ are constant, we can
"abuse" notation and just refer to one or both constant components as real
constant. We also define the corresponding cumulative function as

$\displaystyle P_{g}^{W}(t):=\mathbb{P}(\mathrm{T}^{W}_{g}\leq t)$ (3.2)

and its related density as

$\displaystyle f_{g}^{W}(t):=\frac{dP_{g}^{W}(t)}{dt}.$ (3.3)

We will consider those two cases in the following of this paper:

  1. 1.

(upper boundary case) $g=(-\infty,g^{(2)})$ with
$g^{(2)}\in\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R})$

  2. 2.

(symmetrical boundary case) $g=(-g^{(2)},g^{(2)})$ with
$g^{(2)}\in\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R})$

Our objective in this paper is to show that for any target density function of
the form

$f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+},$

with related cumulative function defined as

$\displaystyle F:\mathbb{R}^{+}$ $\displaystyle\rightarrow$
$\displaystyle[0,1]$ (3.4) $\displaystyle F:t$ $\displaystyle\mapsto$
$\displaystyle\int_{0}^{t}f(s)ds,$ (3.5)

there exists (for each case) a boundary

$g_{f}=(g_{f}^{(1)},g_{f}^{(2)})\in\overline{\mathcal{C}_{0}^{-}(\mathbb{R}^{+},\mathbb{R})}\times\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R}),$

such that $P_{g_{f}}^{W}(t)=F(t)\text{ for }t\geq 0$, or equivalently

$\displaystyle f_{g_{f}}^{W}(t)=f(t)\text{ for }t\geq 0.$ (3.6)

##  4 Cases with explicit formula

Although the value of $f_{g}^{W}(t)$ is not known for a general function $g$,
it is known in a few cases. Of particular interest for this paper is when $g$
is linear or piecewise-linear. Add symmetric boundary case.

Before turning to the examples, we introduce a definition. By $p_{g}(t,x\mid
s,y)$, we denote the transition probability density function of the Brownian
motion $W_{t}$ at time $t$ constrained by the absorbing boundary $g_{t}$ over
$[s,t]$ given that $W_{s}=y$, i.e.,

$\displaystyle p_{g}^{W}(t,x\mid s,y)=\frac{\partial}{\partial
x}\mathbb{P}(W_{t}<x,\mathrm{T}_{g}^{W}>t\mid W_{s}=y)$ (4.1)

with $x<g(t)$, $t>s\geq 0$ and $y<g(s)$ given and fixed.

######  Example 4.1.

(Linear upper boundary) We assume that

$g(t)=Ct+D,$

where $C\in\mathbb{R}$ and $D\in\mathbb{R}^{+}-\\{0\\}$. According to [Doob,
1949] (p. 397) or [Malmquist, 1954] (p. 526), we have that the density
function is equal to

$\displaystyle f_{g}^{W}(t\mid t_{0},x_{0})$ $\displaystyle=$
$\displaystyle\frac{D-x_{0}}{\sqrt{2\pi(t-t_{0})^{3}}}\exp(-(D+C(t-t_{0})-x_{0})^{2}/(2(t-t_{0})).$
(4.2)

The transition probability density is equal to

$\displaystyle p_{g}^{W}(t_{1},x_{1}\mid t_{0},x_{0})$ $\displaystyle=$
$\displaystyle(1-\exp((\frac{-2(g(t_{1})-x_{1})(g(t_{0})-x_{0})}{t_{1}-t_{0}})))\frac{\exp(-\frac{(x_{1}-x_{0})^{2}}{2(t_{1}-t_{0})})}{\sqrt{2\pi(t_{1}-t_{0})}}$
(4.3)

following [Doob, 1949] (Equation (4.2) and Section 5) or [Durbin, 1971] (Lemma
1).

######  Example 4.2.

(Piecewise-linear upper boundary) We assume that

$g(t)=C_{i}t+D_{i},\text{ for }t\in[t_{i-1},t_{i}]\text{ and }i=1,\ldots,n$

with $t_{i}=iH$, where $H\in\mathbb{R}^{+}-\\{0\\}$ and
$C_{i},D_{i}\in\mathbb{R}$. If we choose
$C_{i+1}+D_{i+1}t_{i}=C_{i}+D_{i}t_{i}$, we get that the boundary is
continuous. In view of Definition (4.1), we can reexpress the constrained
transition density function as

$\displaystyle p_{g}^{W}(t_{1},x_{1},\ldots,t_{n},x_{n}\mid t_{0},x_{0})$
$\displaystyle=$ $\displaystyle\prod_{i=1}^{n}p_{g}^{W}(t_{i},x_{i}\mid
t_{i-1},x_{i-1})$ (4.4) $\displaystyle=$
$\displaystyle\prod_{i=1}^{n}(1-\exp((\frac{-2(g(t_{i})-x_{i})(g(t_{i-1})-x_{i-1})}{t_{i}-t_{i-1}})))\frac{\exp(-\frac{(x_{i}-x_{i-1})^{2}}{2(t_{i}-t_{i-1})})}{\sqrt{2\pi(t_{i}-t_{i-1})}}$
(4.5)

with $(x_{1},x_{2},\ldots,x_{n})\in\mathbb{R}^{n}$ and $x_{i}\leq g(t_{i})$
for $i=1,\ldots,n$ and $x_{0}<g(t_{0})$ where $t_{0}<t_{1}<t_{2}<\ldots<t_{n}$
are given and fixed. We can deduce that

$\displaystyle\mathbb{P}(W_{t_{1}}\in C_{1},\ldots,W_{t_{n}}\in
C_{n},\mathrm{T}_{g}^{W}>t_{n}\mid
W_{t_{0}}=x_{0})=\int_{C_{1}}\ldots\int_{C_{n}}$ $\displaystyle
p_{g}^{W}(t_{1},x_{1},\ldots,t_{n},x_{n}\mid t_{0},x_{0})dx_{1}\ldots dx_{n}$

for any Borel set $C_{i}\subset(-\infty,g(t_{i})]$ with $i=1,\ldots,n$. Note
that Equality (4.5) is already available in [Wang and Pötzelberger, 1997] and
[Zucca and Sacerdote, 2009] (see Section 2.1.3, pp. 1323-1324), and follows by
induction from Equation (4.3) in Example 4.1.

######  Example 4.3.

(Linear symmetric boundary) We assume that

$\displaystyle g(t)$ $\displaystyle=$
$\displaystyle(\gamma_{2}+\delta_{2}t,\gamma_{1}+\delta_{1}t),$

where $\gamma_{1}>0$, $\gamma_{2}<0$, $\delta_{1}\geq\delta_{2}$ and not
$\delta_{1}=\delta_{2}=0$. In view of [Anderson, 1960] (Theorem 5.1, p. 191),
we have that the density is equal to:

$\displaystyle f_{g}^{W}(0)$ $\displaystyle=$ $\displaystyle 0,$ (4.7)
$\displaystyle f_{g}^{W}(t)$ $\displaystyle=$
$\displaystyle\frac{1}{t^{3/2}}\Bigg{[}\phi(\frac{\delta_{1}t+\gamma_{1}}{\sqrt{t}})\sum_{r=0}^{\infty}\Big{\\{}((2r+1)\gamma_{1}-2r\gamma_{2})e^{-(2r/t)(r\gamma_{1}-(r+1)\gamma_{2})(\delta_{1}t+\gamma_{1}-(\delta_{2}t+\gamma_{2}))}$
$\displaystyle-(2(r+1)\gamma_{1}-2r\gamma_{2})e^{-(2(r+1)/t)((r+1)\gamma_{1}-r\gamma_{2})(\delta_{1}t+\gamma_{1}-(\delta_{2}t+\gamma_{2}))}\Big{\\}}$
$\displaystyle+\phi(\frac{\delta_{2}t+\gamma_{2}}{\sqrt{t}})\sum_{r=0}^{\infty}\Big{\\{}(2r\gamma_{1}-(2r+1)\gamma_{2})e^{-(2(r+1)/t)((r+1)\gamma_{1}-r\gamma_{2})(\delta_{1}t+\gamma_{1}-(\delta_{2}t+\gamma_{2}))}$
$\displaystyle-(2(r+1)\gamma_{1}-2r\gamma_{2})e^{-(2r/t)(r\gamma_{1}-(r+1)\gamma_{2})(\delta_{1}t+\gamma_{1}-(\delta_{2}t+\gamma_{2}))}\Big{\\}}\Bigg{]}\text{
for any }t>0,$

where $\phi$ is defined as the standard Gaussian density function. Now we
assume that the boundaries are symmetric, i.e. that $g=(-g^{(2)},g^{(2)})$
with

$g^{(2)}(t)=Ct+D,$

where $C\in\mathbb{R}^{+}$ and $D\in\mathbb{R}^{+}-\\{0\\}$. From Equation
(4.7)-(4.3), we can deduce that

$\displaystyle f_{g}^{W}(0)$ $\displaystyle=$ $\displaystyle 0,$ (4.9)
$\displaystyle f_{g}^{W}(t)$ $\displaystyle=$
$\displaystyle\frac{2}{t^{3/2}}\phi(\frac{Ct+D}{\sqrt{t}})\sum_{r=0}^{\infty}\Big{\\{}(4r+1)De^{-(8r(r+1)D)(Ct+D)/t)}$
$\displaystyle-(4r+2)De^{-(4(r+1)(2r+1)D(Ct+D)/t)}\Big{\\}}\text{ for any
}t>0.$

From Equation (4.9)-(4.3), we can now infer that the conditional density
function is equal to

$\displaystyle f_{g}^{W}(t_{0}\mid t_{0},x_{0})$ $\displaystyle=$
$\displaystyle 0,$ (4.11) $\displaystyle f_{g}^{W}(t\mid t_{0},x_{0})$
$\displaystyle=$
$\displaystyle\frac{2}{(t-t_{0})^{3/2}}\phi(\frac{Ct+D-x_{0}}{\sqrt{t-t_{0}}})\sum_{r=0}^{\infty}\Big{\\{}(4r+1)(D-x_{0})e^{-(8r(r+1)(D-x_{0}))(Ct+D-x_{0})/t)}$
$\displaystyle-(4r+2)(D-x_{0})e^{-(4(r+1)(2r+1)(D-x_{0})(C(t-t_{0})+D-x_{0})/(t-t_{0}))}\Big{\\}}\text{
for any }t>t_{0}.$

We can also obtain the formula for transition probability density in a similar
way to that of Equation (4.3).

######  Example 4.4.

(Piecewise-linear symmetric boundary) Following the same line of reasoning as
that in Example 4.2, we can establish similar formulae.

##  5 Nonlinear Equation

Ideally, we would like to find a closed-form expression where we would be able
to express $g_{f}(t)$ in terms of $f(t)$. For this particular purpose, we
introduce some notation, namely the set of functions

$\displaystyle
lin:(\overline{\mathcal{C}_{0}^{-}(\mathbb{R}^{+},\mathbb{R})}\times\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R}))\times\mathbb{R}^{+}\times\left(\\{(-z,z)\in\mathbb{R}^{2}\text{
s.t. }z\in\mathbb{R}\\}\cup(-\infty\times\mathbb{R})\right)$
$\displaystyle\rightarrow$
$\displaystyle\overline{\mathcal{C}_{0}^{-}(\mathbb{R}^{+},\mathbb{R})}\times\mathcal{C}_{0}^{+}(\mathbb{R}^{+},\mathbb{R})$
$\displaystyle(g,t,a)$ $\displaystyle\mapsto$ $\displaystyle g^{lin}_{t,a},$
(5.1)

which is formally defined as

$\displaystyle g^{lin}_{t,a}(u)$ $\displaystyle=$ $\displaystyle g(u)\text{
when }0\leq u\leq t,$ (5.2) $\displaystyle g^{lin}_{t,a}(u)$ $\displaystyle=$
$\displaystyle g(t)+a(u-t)\text{ when }u\geq t.$ (5.3)

The function $g^{lin}_{t,a}$ corresponds exactly to $g$ up to $t$ and to a
linear one dimensional boundary (in the upper boundary case) or symmetrical
two dimensional boundary (in the symmetrical boundary case) starting at the
same value $g(t)$ and of coefficient $a$. Therefore, $g$ needs only to be
defined on $[0,t]$ for $g^{lin}_{t,a}$ to exist. We also introduce the related
probability that the Brownian motion hits such function between $t$ and $s$:

$\displaystyle p_{g}^{W}:\\{(t,s)\in(\mathbb{R}^{+})^{2}\text{ s.t. }t\leq
s\\}\times\left(\\{(-z,z)\in\mathbb{R}^{2}\text{ s.t.
}z\in\mathbb{R}\\}\cup(-\infty\times\mathbb{R})\right)$
$\displaystyle\rightarrow$ $\displaystyle[0,1]$ (5.4) $\displaystyle
p_{g}^{W}(t,s,a)$ $\displaystyle=$
$\displaystyle\mathbb{P}\left(\mathrm{T}_{g^{lin}_{t,a}}^{W}\in[t,s]\right).$

As the definition (5.4) is based on the definition of $g^{lin}_{t,a}$, $g$
needs only to be defined on $[0,t]$ for $p_{g}^{W}(t,s,a)$ to exist. Now we
can express $g_{f}$ as solution of the following equation

$\displaystyle
f_{g_{f}}^{W}(t)=\frac{dp_{g_{f}}}{ds}(t,t,g_{f}^{\prime}(t))\text{ for }t\geq
0,$ (5.5)

which in view of (3.6) can be re-expressed as

$\displaystyle f(t)=\frac{dp_{g}}{ds}(t,t,g^{\prime}(t))\text{ for }t\geq 0.$
(5.6)

The above differential equation (5.6) would be key to our analysis had we
found any explicit way to solve the equation. This is not surprising that the
direct route does not yield a closed-form expression, as the literature was
not able to track down such expression despite many attempts over the years
(such as, e.g., the Volterra integral equation of the second kind developed in
[Peskir, 2002b] and [Peskir, 2002a]).

##  6 Discretization of the problem

As we have not found any closed-form expression, we discretize the problem and
introduce a piecewise-linear approximation of the boundary on $[0,T]$, where
$T>0$ is the horizon time. In particular, note that this approximation is not
based on the differential equation (5.6) derived in Section 5. On the
contrary, we consider a piecewise-linear approximation whose definition is
almost the same as that of Section 3 in [Zucca and Sacerdote, 2009]. There is
a difference in objective though, as the cited paper _assumes_ the existence
of $g_{f}$, whereas we will _show_ the existence of $g_{f}$ in Section 7.

For any $n\in\mathbb{N}-\\{0\\}$, we define the subspace of piecewise linear
functions as

$\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})_{n}^{PL}:=\Big{\\{}g\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})\text{
s.t. }g$ $\text{ is piecewise-linear with related junction points
}\\{mT/2^{n}\\}_{m=0}^{n}\Big{\\}}.$

For any $n\in\mathbb{N}-\\{0\\}$, we can then define the sequence of boundary
approximation
$g_{f,T}^{(n)}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})_{n}^{PL}$
recursively on $m$ as:

$\displaystyle g_{f,T}^{(n)}(u)$ $\displaystyle=$
$\displaystyle\alpha_{0,n}\text{ for }u\in[0,T/2^{n}],$ (6.1) $\displaystyle
g_{f,T}^{(n)}(u)$ $\displaystyle=$ $\displaystyle
g_{f,T}^{(n)}(mT/2^{n})+\alpha_{m,n}(u-mT/2^{n})\text{ for
}u\in[mT/2^{n},(m+1)T/2^{n}],m=1,\cdots,2^{n}-1$ (6.2)

and where

$\displaystyle\alpha_{0,n}$ $\displaystyle:=$
$\displaystyle(-\infty,\alpha_{0,n}^{(2)})\text{ in upper boundary case}$
$\displaystyle\alpha_{0,n}$ $\displaystyle:=$
$\displaystyle(-\alpha_{0,n}^{(2)},\alpha_{0,n}^{(2)})\text{ in symmetrical
boundary case}$

with $\alpha_{0,n}^{(2)}\in\mathbb{R}^{+}-\\{0\\}$ and

$\displaystyle\alpha_{m,n}$ $\displaystyle:=$
$\displaystyle(-\infty,\alpha_{m,n}^{(2)})\text{ in upper boundary case}$
$\displaystyle\alpha_{m,n}$ $\displaystyle:=$
$\displaystyle(-\alpha_{m,n}^{(2)},\alpha_{m,n}^{(2)})\text{ in symmetrical
boundary case}$

with $\alpha_{m,n}^{(2)}\in\mathbb{R}$ for $m=1,\cdots,2^{n}-1$ satisfy

$\displaystyle\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}$
$\displaystyle=$ $\displaystyle\int_{0}^{T/2^{n}}f(s)ds,$ (6.3) $\displaystyle
p_{g_{f,T}^{(n)}}(mT/2^{n},(m+1)T/2^{n},\alpha_{m,n})$ $\displaystyle=$
$\displaystyle\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds\text{ for
}m=1,\cdots,2^{n}-1.$ (6.4)

In view of the recursiveness of Definitions (6.1)-(6.4), we note that
$g_{f,T}^{(n)}$ is used as an argument in the function $p_{g}$ on the left
hand side of Equation (6.4), although it is yet only defined on
$[0,mT/2^{n}]$. This is actually in line with the definition of $p_{g}$ which
states that $g$ needs only to be defined on $[0,t]$ for $p_{g}^{W}(t,s,a)$ to
exist. The driving idea behind (6.3)-(6.4) is to equate the probability of
hitting the boundary in the discrete problem to the target probability on each
block.

##  7 Proof of existence

In this section, we show the existence of a continuous boundary function
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
satisfying

$\displaystyle f_{g_{f,T}}^{W}=f_{t}\text{ for }0\leq t\leq T$ (7.1)

for any target density $f$ and any horizon time $T>0$. The proof goes in two
steps. First, we show that there exists a subsequence of $g_{f,T}^{(n)}$ which
converges uniformly to some $g_{f,T}$ using a fundamental result of analysis,
i.e. Arzelà-Ascoli theorem. Second, we show that the obtained limit $g_{f,T}$
satisfies Equation (7.1).

More specifically, the key result for the proof of existence is the obtention
of the limit boundary $g_{f,T}$ via Arzelà-Ascoli theorem. We claim that the
assumption of uniformly boundedness and uniformly equicontinuity in Arzelà-
Ascoli theorem are satisfied if we can show that $\alpha_{m,n}^{(2)}$ are
bounded uniformly, i.e. there exists a constant $K\in\mathbb{R}^{+}-\\{0\\}$
such that

$\displaystyle\sup_{\underset{m=0,\cdots,2^{n}-1}{n\in\mathbb{N}}}\left|\alpha_{m,n}^{(2)}\right|\leq
K.$ (7.2)

Then, the proof of obtention of the limit boundary $g_{f,T}$ goes as follows:

(i) We show the claim, i.e. we show that conditions of Arzelà-Ascoli theorem
are satisfied whenever (7.2) holds. This is done in Proposition 7.9.

(ii) We show that Equation (7.2) is satisfied. This is carried in Proposition
7.11.

(iii) We can then conclude the existence of limit boundary in Corollary 7.12.

We start this proof by giving a more explicit form to Equations (6.3)-(6.4).
Recalling Definition (4.1), this first lemma gives for any
$n\in\mathbb{N}-\\{0\\}$ and any $m=0,\cdots,2^{n}-1$ the transition
probability density function of the Brownian motion $W_{t}$ at time
$\frac{(m+1)T}{2^{n}}$ constrained by the absorbing boundary $g_{f,T}^{(n)}$
over $[\frac{mT}{2^{n}},\frac{(m+1)T}{2^{n}}]$ given that
$W_{\frac{mT}{2^{n}}}=x_{m}$.

######  Lemma 7.1.

For any $n\in\mathbb{N}-\\{0\\}$ and any $m=0,\cdots,2^{n}-1$ we have

$\displaystyle
p_{g_{f,T}^{(n)}}^{W}(\frac{(m+1)T}{2^{n}},x_{m+1}\mid\frac{mT}{2^{n}},x_{m})$
$\displaystyle=$
$\displaystyle(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{(m+1)T}{2^{n}})-x_{m+1})(g_{f,T}^{(n)}(\frac{mT}{2^{n}})-x_{m})}{\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{m+1}-x_{m})^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}.$

###### Proof.

This is a direct application of Equation (4.3) in Example 4.1. ∎

The next lemma re-expresses Equation (6.3). Note that the only unknown value
in Equation (7.4) is $\alpha_{0,n}$.

######  Lemma 7.2.

For any $n\in\mathbb{N}-\\{0\\}$, Equation (6.3) can be reexpressed as

$\displaystyle
1-\int_{-\infty}^{\alpha_{0,n}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}-\int_{0}^{T/2^{n}}f(s)ds=0.$
(7.4)

###### Proof.

We have that

$\displaystyle\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}(T_{\alpha_{0,n}}^{W}>T/2^{n}\big{)}^{C})$
(7.5) $\displaystyle=$ $\displaystyle
1-\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}>T/2^{n}\big{)}$ $\displaystyle=$
$\displaystyle
1-\mathbb{P}\big{(}W_{T/2^{n}}\in(-\infty,\alpha_{0,n}],T_{\alpha_{0,n}}^{W}>T/2^{n}\big{)}$
$\displaystyle=$ $\displaystyle
1-\int_{-\infty}^{\alpha_{0,n}}p_{\alpha_{0,n}}^{W}(T/2^{n},x_{1}\mid
0,0)dx_{1}$ $\displaystyle=$ $\displaystyle
1-\int_{-\infty}^{\alpha_{0,n}}(1-\exp(\frac{-2(\alpha_{0,n}(\frac{T}{2^{n}})-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}$

where we used the fact that $T_{\alpha_{0,n}}^{W}\geq 0$ in the first
equality, the fact that
$T_{\alpha_{0,n}}^{W}\subset\\{W_{T/2^{n}}\in(-\infty,\alpha_{0,n}]\\}$ in the
third equality, Equation (4.2) from Example 4.2 in the fourth equality and
Equation (7.1) from Lemma 7.1 in the fifth equality. Then, as an application
of Equation (6.3) together with Equation (7.5), we can deduce that

$\displaystyle
1-\int_{-\infty}^{\alpha_{0,n}}(1-\exp(\frac{-2(\alpha_{0,n}(\frac{T}{2^{n}})-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}=\int_{0}^{T/2^{n}}f(s)ds.$
(7.6)

Rearranging the terms, this can be reexpressed as

$\displaystyle
1-\int_{-\infty}^{\alpha_{0,n}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}-\int_{0}^{T/2^{n}}f(s)ds=0.$

∎

In the next lemma, we reexpress Equation (6.4) in the case $m=1$. Note that
since $\alpha_{0,n}$ is known as solution of Equation (7.4), then
$\alpha_{1,n}$ is the only unknown variable of Equation (7.7).

######  Lemma 7.3.

For any $n\in\mathbb{N}-\\{0\\}$ and $m=1$, Equation (6.4) can be reexpressed
as

$\displaystyle
1-\int_{0}^{2T/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{(2T}{2^{n}})-x_{2})(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})}{\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})g_{f,T}^{(n)}(\frac{T}{2^{n}})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}dx_{2}$
$\displaystyle=$ $\displaystyle 0$ (7.7)

###### Proof.

We have that

$\displaystyle p_{g_{f,T}^{(n)}}(T/2^{n},2T/2^{n},\alpha_{1,n})$
$\displaystyle=$
$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}\in[T/2^{n},2T/2^{n}]\right)$
(7.8) $\displaystyle=$
$\displaystyle\mathbb{P}\left((\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}<T/2^{n},\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n})^{C}\right)$
$\displaystyle=$ $\displaystyle
1-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}<T/2^{n},\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\right)$
$\displaystyle=$ $\displaystyle
1-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}<T/2^{n}\right)-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\right)$

where we used Def. (5.1)-(5.4) in the first equality, the fact that
$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}<T/2^{n}\\}$
and
$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\\}$
are disjoint events in the fourth equality. Then, on the one hand, we have
that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}<T/2^{n}\right)$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}$
(7.9) $\displaystyle=$ $\displaystyle\int_{0}^{T/2^{n}}f(s)ds,$

where the first equality is due to the fact that
$(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}(t)=\alpha_{0,n}$ for any $0\leq
t\leq T/2^{n}$, the second equality corresponds to Equation (6.3). On the
other hand, we have that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\right)$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}W_{T/2^{n}}\in(-\infty,\alpha_{0,n}],W_{2T/2^{n}}\in(-\infty,g_{f,T}^{(n)}(2T/2^{n})],$
$\displaystyle\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\big{)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}p_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}(T/2^{n},x_{1},2T/2^{n},x_{2}\mid
0,0)dx_{1}dx_{2}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}p_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}(2T/2^{n},x_{2}\mid
T/2^{n},x_{1})$ $\displaystyle\times
p_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}(T/2^{n},x_{1}\mid
0,0)dx_{1}dx_{2}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{(2T}{2^{n}})-x_{2})(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})}{\frac{T}{2^{n}}}))$
$\displaystyle\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})g_{f,T}^{(n)}(\frac{T}{2^{n}})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}dx_{2},$

where we obtain the first equality since
$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}}^{W}>2T/2^{n}\\}\subset\\{W_{T/2^{n}}\in(-\infty,\alpha_{0,n}],W_{2T/2^{n}}\in(-\infty,g_{f,T}^{(n)}(2T/2^{n})]\\}$,
the second equality corresponds to Equation (4.2), the third equality is due
to Equation (4.4) and we have used Equation (7.1) from Lemma 7.1 in the fourth
equality. Then, as an application of Equation (7.8), Equation (7.9), Equation
(7) along with Equation (6.3), we obtain that

$\displaystyle
1-\int_{0}^{2T/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{(2T}{2^{n}})-x_{2})(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})}{\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})g_{f,T}^{(n)}(\frac{T}{2^{n}})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}dx_{2}$
$\displaystyle=$ $\displaystyle 0$

∎

In the lemma that follows, we reexpress Equation (6.4) in the case $m>1$. Note
that $\alpha_{m,n}$ is the only unknown variable of the general equation
(7.11), as $(\alpha_{k,n})_{1\leq k\leq m-1}$ are known recursively.

######  Lemma 7.4.

For any $n\in\mathbb{N}-\\{0\\}$ and $m>1$, Equation (6.4) can be reexpressed
as

$\displaystyle
1-\int_{0}^{(m+1)T/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}\ldots\int_{-\infty}^{g_{f,T}^{(n)}((m+1)T/2^{n})}\prod_{k=0}^{m}(1-$
$\displaystyle\exp(\frac{-2(g_{f,T}^{(n)}(\frac{((k+1)T}{2^{n}})-x_{k+1})(g_{f,T}^{(n)}(\frac{kT}{2^{n}})-x_{k})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{(x_{k+1}-x_{k})^{2}}{\frac{T}{2^{n-1}}})}{\sqrt{\pi\frac{T}{2^{n-1}}}}dx_{1}dx_{2}\ldots
dx_{m+1}$ $\displaystyle=$ $\displaystyle 0$ (7.11)

###### Proof.

We compute that

$\displaystyle p_{g_{f,T}^{(n)}}(mT/2^{n},(m+1)T/2^{n},\alpha_{m,n})$
$\displaystyle=$
$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\right)$
$\displaystyle=$
$\displaystyle\mathbb{P}\left((\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<mT/2^{n},\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n})^{C}\right)$
$\displaystyle=$ $\displaystyle
1-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<mT/2^{n},\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\right)$
$\displaystyle=$ $\displaystyle
1-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<mT/2^{n}\right)$
$\displaystyle-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\right)$

where we used Def. (5.1)-(5.4) in the first equality, the fact that
$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<mT/2^{n}\\}$
and
$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\\}$
are disjoint events in the fourth equality. Then, on the one hand, we have
that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<mT/2^{n}\right)$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}\cup_{k=0}^{m-1}\\{kT/2^{n}\leq\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<(k+1)T/2^{n}\\}\big{)}$
(7.13) $\displaystyle=$
$\displaystyle\sum_{k=0}^{m-1}\mathbb{P}\big{(}kT/2^{n}\leq\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<(k+1)T/2^{n}\big{)}$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}+\sum_{k=1}^{m-1}p_{(g_{f,T}^{(n)})^{lin}_{kT/2^{n},\alpha_{k,n}}}(kT/2^{n},(k+1)T/2^{n},\alpha_{k,n})$
$\displaystyle=$ $\displaystyle\int_{0}^{mT/2^{n}}f(s)ds$

where we used the fact that for $k=0,\ldots,m-1$ we have
$\\{kT/2^{n}\leq\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}<(k+1)T/2^{n}\\}$
are disjoint events in the second equality, the third equality is due to the
fact that $(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{1,n}}(t)=\alpha_{0,n}$ for
any $0\leq t\leq T/2^{n}$ and for $k=1,\ldots,m-1$ we have
$(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}=(g_{f,T}^{(n)})^{lin}_{kT/2^{n},\alpha_{k,n}}$
are equal on $[0,(k+1)T/2^{n}]$, the fourth equality is a direct consequence
of Equation (6.3). On the other hand, we can calculate that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\right)$
$\displaystyle=$
$\displaystyle\mathbb{P}\big{(}W_{T/2^{n}}\in(-\infty,\alpha_{0,n}],W_{2T/2^{n}}\in(-\infty,g_{f,T}^{(n)}(2T/2^{n})],\ldots,$
(7.14) $\displaystyle
W_{(m+1)T/2^{n}}\in(-\infty,g_{f,T}^{(n)}((m+1)T/2^{n})],$
$\displaystyle\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\big{)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}\ldots\int_{-\infty}^{g_{f,T}^{(n)}((m+1)T/2^{n})}p_{(g_{f,T}^{(n)})^{lin}_{T/2^{n},\alpha_{m,n}}}^{W}(T/2^{n},$
$\displaystyle x_{1},2T/2^{n},x_{2},\ldots,(m+1)T/2^{n},x_{m+1}\mid
0,0)dx_{1}dx_{2}\ldots dx_{m+1}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}\ldots\int_{-\infty}^{g_{f,T}^{(n)}((m+1)T/2^{n})}\prod_{k=0}^{m}p_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}$
$\displaystyle((k+1)T/2^{n},x_{k+1}\mid k/2^{n},x_{k})dx_{1}dx_{2}\ldots
dx_{m+1}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}\ldots\int_{-\infty}^{g_{f,T}^{(n)}((m+1)T/2^{n})}\prod_{k=0}^{m}(1-$
$\displaystyle\exp(\frac{-2(g_{f,T}^{(n)}(\frac{((k+1)T}{2^{n}})-x_{k+1})(g_{f,T}^{(n)}(\frac{kT}{2^{n}})-x_{k})}{\frac{T}{2^{n}}}))$
$\displaystyle\frac{\exp(-\frac{(x_{k+1}-x_{k})^{2}}{\frac{T}{2^{n-1}}})}{\sqrt{\pi\frac{T}{2^{n-1}}}}dx_{1}dx_{2}\ldots
dx_{m+1},$

where we obtain the first equality since

$\\{\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},\alpha_{m,n}}}^{W}>(m+1)T/2^{n}\\}\subset$
$\\{W_{T/2^{n}}\in(-\infty,\alpha_{0,n}],W_{2T/2^{n}}\in(-\infty,g_{f,T}^{(n)}(2T/2^{n})],\ldots,W_{(m+1)T/2^{n}}\in(-\infty,g_{f,T}^{(n)}((m+1)T/2^{n})\\},$

the second equality corresponds to Equation (4.2), the third equality is due
to Equation (4.4) and we have used Equation (7.1) from Lemma 7.1 in the fourth
equality. Then an application of Equation (7), together with Equation (7.13),
Equation (7.14) and Equation (6.4) yields

$\displaystyle
1-\int_{0}^{(m+1)T/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}\ldots\int_{-\infty}^{g_{f,T}^{(n)}((m+1)T/2^{n})}\prod_{k=0}^{m}(1-$
$\displaystyle\exp(\frac{-2(g_{f,T}^{(n)}(\frac{((k+1)T}{2^{n}})-x_{k+1})(g_{f,T}^{(n)}(\frac{kT}{2^{n}})-x_{k})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{(x_{k+1}-x_{k})^{2}}{\frac{T}{2^{n-1}}})}{\sqrt{\pi\frac{T}{2^{n-1}}}}dx_{1}dx_{2}\ldots
dx_{m+1}$ $\displaystyle=$ $\displaystyle 0$

∎

We continue this proof by stating this lemma, which will be useful in showing
the existence and unicity of $\alpha_{0,n}^{(2)}$ in Proposition 7.7. This
basically states that the probability of hitting a constant boundary on
$[0,T/2^{n}]$ is strictly decreasing from $1$ to $0$ as respectively the
constant grows.

######  Lemma 7.5.

For any $n\in\mathbb{N}-\\{0\\}$, we define the functions

$\displaystyle z_{1}^{(n)}:\mathbb{R}^{+}$ $\displaystyle\rightarrow$
$\displaystyle[0,1]$ (7.15) $\displaystyle x$ $\displaystyle\mapsto$
$\displaystyle\mathbb{P}\big{(}T_{(-\infty,x)}^{W}\in[0,T/2^{n}]\big{)}$

and

$\displaystyle z_{2}^{(n)}:\mathbb{R}^{+}$ $\displaystyle\rightarrow$
$\displaystyle[0,1]$ (7.16) $\displaystyle x$ $\displaystyle\mapsto$
$\displaystyle\mathbb{P}\big{(}T_{(-x,x)}^{W}\in[0,T/2^{n}]\big{)},$

where $z_{i}^{(n)}$ for $i=1,2$ correspond respectively to the upper boundary
case and the symmetrical boundary case. Then, we have that for any $i=1,2$ and
any $n\in\mathbb{N}-\\{0\\}$:

$\displaystyle z_{i}^{(n)}(x)\overset{x\rightarrow 0}{\rightarrow}1,$ (7.17)
$\displaystyle z_{i}^{(n)}(x)\overset{x\rightarrow+\infty}{\rightarrow}0,$
(7.18)

and that $z_{i}^{(n)}$ is continuous and strictly decreasing in $x$.

###### Proof.

We have for any $x\in\mathbb{R}^{+}$ that

$\displaystyle z_{1}^{(n)}(x)$ $\displaystyle=$
$\displaystyle\mathbb{P}\big{(}T_{(-\infty,x)}^{W}\in[0,T/2^{n}]\big{)}$
$\displaystyle=$ $\displaystyle\int_{0}^{T/2^{n}}\frac{x}{\sqrt{2\pi
t^{3}}}\exp(-x^{2}/2t)dt,$

where we used Formula (4.2) from Example 4.1 in the second equality. It is
then direct to see that for any $n\in\mathbb{N}-\\{0\\}$, we have
$z_{1}^{(n)}(x)\overset{x\rightarrow 0}{\rightarrow}1$,
$z_{1}^{(n)}(x)\overset{x\rightarrow+\infty}{\rightarrow}0$, and that
$z_{1}^{(n)}$ is continuous and strictly decreasing in $x$. The proof in the
case $i=2$ (i.e. corresponding to the symmetrical boundary case) is very
similar applying the formula in the symmetrical case from [Anderson, 1960]. ∎

The next lemma is the counterpart of Lemma 7.5 when considering
$\alpha_{m,n}^{(2)}$ for any $n\in\mathbb{N}-\\{0\\}$ and any
$m=1,\cdots,2^{n}-1$.

######  Lemma 7.6.

For any $n\in\mathbb{N}-\\{0\\}$ and any $m=1,\cdots,2^{n}-1$, we consider the
functions

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}:\mathbb{R}$
$\displaystyle\rightarrow$ $\displaystyle(0,\int_{mT/2^{n}}^{+\infty}f(s)ds)$
(7.19) $\displaystyle z$ $\displaystyle\mapsto$ $\displaystyle
p_{g_{f,T}^{(n)}}(mT/2^{n},(m+1)T/2^{n},(-\infty,z)).$

and

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},2}^{(g_{f,T}^{(n)})}:\mathbb{R}$
$\displaystyle\rightarrow$ $\displaystyle(0,\int_{mT/2^{n}}^{+\infty}f(s)ds)$
(7.20) $\displaystyle z$ $\displaystyle\mapsto$ $\displaystyle
p_{g_{f,T}^{(n)}}(mT/2^{n},(m+1)T/2^{n},(-z,z)),$

where $R^{(g_{f,T}^{(n)})}_{mT/2^{n},(m+1)T/2^{n},i}$ for $i=1,2$ correspond
respectively to the upper boundary case and the symmetrical boundary case.
Then, we obtain that for $i=1,2$:

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle\overset{z\rightarrow-\infty}{\rightarrow}$
$\displaystyle\int_{mT/2^{n}}^{+\infty}f(s)ds,$ (7.21) $\displaystyle
R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle\overset{z\rightarrow+\infty}{\rightarrow}$ $\displaystyle 0,$
(7.22)

and that $R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}$ is continuous and
strictly decreasing. In particular, this implies that
$R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}$ is invertible, i.e. that there
exists

$\displaystyle(R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})})^{-1}:(0,\int_{mT/2^{n}}^{+\infty}f(s)ds)$
$\displaystyle\rightarrow$ $\displaystyle\mathbb{R}$ (7.23)

such that
$(R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})})^{-1}(R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(z))=z$
for any $z\in\mathbb{R}$ and

$R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}((R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})})^{-1}(w))=w$

for any $w\in(0,\int_{mT/2^{n}}^{+\infty}f(s)ds)$.

###### Proof.

For any $n\in\mathbb{N}-\\{0\\}$, any $m=1,\cdots,2^{n}-1$ and any
$z\in\mathbb{R}$, we can compute

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle=$ $\displaystyle p_{g_{f,T}^{(n)}}(mT/2^{n},(m+1)T/2^{n},z)$
$\displaystyle=$ $\displaystyle
1-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}}^{W}<mT/2^{n}\right)$
$\displaystyle-\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}}^{W}>(m+1)T/2^{n}\right),$

where the first equality corresponds to Definition (7.19) and the second
equality corresponds to Equation (7). On the one hand, we have by Equation
(7.13) that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}}^{W}<mT/2^{n}\right)$
$\displaystyle=$ $\displaystyle\int_{0}^{mT/2^{n}}f(s)ds.$ (7.25)

On the other hand, we obtain by Equation (7.14) that

$\displaystyle\mathbb{P}\left(\mathrm{T}_{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}}^{W}>(m+1)T/2^{n}\right)$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(2T/2^{n})}\ldots\int_{-\infty}^{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}((m+1)T/2^{n})}\prod_{k=0}^{m}(1-$
(7.26)
$\displaystyle\exp(\frac{-2((g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(\frac{((k+1)T}{2^{n}})-x_{k+1})((g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(\frac{kT}{2^{n}})-x_{k})}{\frac{T}{2^{n}}}))$
$\displaystyle\frac{\exp(-\frac{(x_{k+1}-x_{k})^{2}}{\frac{T}{2^{n-1}}})}{\sqrt{\pi\frac{T}{2^{n-1}}}}dx_{1}dx_{2}\ldots
dx_{m+1}.$

Then, by an application of Equation (7), Equation (7.25) and Equation (7.26),
we can deduce that

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle=$ $\displaystyle
1-\int_{0}^{mT/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(2T/2^{n})}\ldots\int_{-\infty}^{(g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}((m+1)T/2^{n})}$
(7.27)
$\displaystyle\prod_{k=0}^{m}(1-\exp(\frac{-2((g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(\frac{((k+1)T}{2^{n}})-x_{k+1})((g_{f,T}^{(n)})^{lin}_{mT/2^{n},z}(\frac{kT}{2^{n}})-x_{k})}{\frac{T}{2^{n}}}))$
$\displaystyle\frac{\exp(-\frac{(x_{k+1}-x_{k})^{2}}{\frac{T}{2^{n-1}}})}{\sqrt{\pi\frac{T}{2^{n-1}}}}dx_{1}dx_{2}\ldots
dx_{m+1}.$

In view of the expression of
$R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}(z)$ in Equation (7.27), we can
deduce that

$\displaystyle R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle\overset{z\rightarrow-\infty}{\rightarrow}$
$\displaystyle\int_{mT/2^{n}}^{+\infty}f(s)ds,$ $\displaystyle
R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(z)$
$\displaystyle\overset{z\rightarrow+\infty}{\rightarrow}$ $\displaystyle 0,$

and that $R_{mT/2^{n},(m+1)T/2^{n},1}^{(g_{f,T}^{(n)})}$ is continuous and
strictly decreasing. Finally, we can deduce the existence of the inverse by an
application of the intermediate value theorem. The case $i=2$ follows in a
similar way.

∎

Our next result establishes existence and uniqueness of the sequence
$\alpha_{0,n}^{(2)}\in\mathbb{R}^{+}-\\{0\\}$ (which is equivalent to the
existence and uniqueness of the sequence $\alpha_{0,n}$) and that of
$\alpha_{m,n}^{(2)}\in\mathbb{R}$ (which is equivalent to that of the sequence
$\alpha_{m,n}$) for any $m=1,\cdots,2^{n}-1$ and for any
$n\in\mathbb{N}-\\{0\\}$. We make the positivity assumption on $f$ for that
result:

[A

] We assume that $f(t)>0$ for any $t\geq 0$.

######  Proposition 7.7.

Under Assumption [A], for any $n\in\mathbb{N}-\\{0\\}$, Equation (6.3) defines
a unique $\alpha_{0,n}^{(2)}\in\mathbb{R}^{+}-\\{0\\}$ and Equation (6.4)
defines a unique $\alpha_{m,n}^{(2)}\in\mathbb{R}$ for any
$m=0,\cdots,2^{n}-1$. Moreover, for any $n\in\mathbb{N}-\\{0\\}$ and any
$m=0,\cdots,2^{n}-1$, the approximated boundary satisfies

$\displaystyle\mathbb{P}\left(T_{g_{f,T}^{(n)}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\right)=\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds.$
(7.28)

###### Proof.

For any $n\in\mathbb{N}-\\{0\\}$, we prove Proposition 7.7, i.e. the statement
and Equality (7.28), by recurrence on $m=0,\cdots,2^{n}-1$. We start with the
$m=0$ case. By Definition (6.3), it is clear that

$\displaystyle\mathbb{P}\left(T_{g_{f,T}^{(n)}}^{W}\in[0,T/2^{n}]\right)=\int_{0}^{T/2^{n}}f(s)ds.$
(7.29)

We have thus shown Equality (7.28). Now, it remains to show the existence and
uniqueness of $\alpha_{0,n}^{(2)}\in\mathbb{R}^{+}-\\{0\\}$. To do that, we
note that in view of Assumption [A] we deduce that

$\displaystyle 0<\int_{0}^{T/2^{n}}f(s)ds<1.$ (7.30)

Moreover, rewriting the left hand-side of Equality (7.29) as a function of
$z_{i}^{(n)}$ for $i=1,2$ (following the definitions (7.15) and (7.16)) yields
that $\alpha_{0,n}^{(2)}$ solves the equation

$\displaystyle z_{i}^{(n)}(\alpha_{0,n}^{(2)})=\int_{0}^{T/2^{n}}f(s)ds.$
(7.31)

From Inequality (7.30) and Equality (7.31), we can easily deduce that

$\displaystyle 0<z_{i}^{(n)}(\alpha_{0,n}^{(2)})<1.$ (7.32)

Finally, an application of the Intermediate Value Theorem together with Lemma
7.5 and Equality (7.32) provides the existence and uniqueness of
$\alpha_{0,n}^{(2)}\in\mathbb{R}^{+}-\\{0\\}$.

We consider now the case when $m=1,\cdots,2^{n}-1$. By hypothesis of
recurrence, we deduce that

$\displaystyle\mathbb{P}\big{(}T_{g_{f,T}^{(n)}}^{W}\in[0,mT/2^{n}]\big{)}$
$\displaystyle=$ $\displaystyle\int_{0}^{mT/2^{n}}f(s)ds.$ (7.33)

Then, in view of Equation (7.33) together with Definition (6.1)-(6.4), we
infer that

$\displaystyle\mathbb{P}\big{(}T_{g_{f,T}^{(n)}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\big{)}$
$\displaystyle=$ $\displaystyle\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds,$ (7.34)

so that we have proved Equality (7.28). In what follows, we prove the
existence and uniqueness of $\alpha_{m,n}^{(2)}\in\mathbb{R}$. We note that
applying Assumption [A] we get

$\displaystyle
0<\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds<\int_{mT/2^{n}}^{+\infty}f(s)ds.$
(7.35)

In addition, reexpressing the left hand-side of Equation (7.34) as a function
of $R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}$ for $i=1,2$ (following the
definitions (7.19) and (7.20)) rends that $\alpha_{m,n}^{(2)}$ solves the
equation

$\displaystyle
R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(\alpha_{m,n}^{(2)})=\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds.$
(7.36)

From Equation (7.35) and Equation (7.36), we can infer that

$\displaystyle
0<R_{mT/2^{n},(m+1)T/2^{n},i}^{(g_{f,T}^{(n)})}(\alpha_{m,n}^{(2)})<\int_{mT/2^{n}}^{+\infty}f(s)ds.$
(7.37)

To conclude, an application of the Intermediate Value Theorem along with Lemma
7.6 and Equation (7.37) provides the existence and uniqueness of
$\alpha_{m,n}^{(2)}\in\mathbb{R}$. ∎

Key to our proof of existence is the next corollary, as it extends Equality
(7.28) to the case where the approximated boundary is at level $l\geq n$,
everything else being frozen in the equality. The proof is a direct
consequence of the fact that the grid is nested. This is the reason why we
chose an approximation length of $T/2^{n}$ rather than, say, $T/n$.

######  Corollary 7.8.

Under Assumption [A], for any $n\in\mathbb{N}-\\{0\\}$, any
$l\in\mathbb{N}-\\{0\\}$ with $l\geq n$ and any $m=0,\cdots,2^{n}-1$, the
approximated boundary satisfies

$\displaystyle\mathbb{P}\left(T_{g_{f,T}^{(l)}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\right)=\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds.$
(7.38)

###### Proof.

For any $n\in\mathbb{N}-\\{0\\}$, any $l\in\mathbb{N}-\\{0\\}$ with $l\geq n$
and any $m=0,\cdots,2^{n}-1$, we have

$\displaystyle\mathbb{P}\left(T_{g_{f,T}^{(l)}}^{W}\in[mT/2^{n},(m+1)T/2^{n}]\right)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{2^{l-n}-1}\mathbb{P}\left(T_{g_{f,T}^{(l)}}^{W}\in[(m2^{l-n}+i)T/2^{l},(m2^{l-n}+i+1)T/2^{l}]\right)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{2^{l-n}-1}\int_{(m2^{l-n}+i)T/2^{l}}^{(m2^{l-n}+i+1)T/2^{l}}f(s)ds$
$\displaystyle=$ $\displaystyle\int_{mT/2^{n}}^{(m+1)T/2^{n}}f(s)ds,$

where we used the fact that the probability of an union of disjoint events is
equal to the sum of probability of each individual event in the first
equality, and Equality (7.28) from Proposition 7.7 in the second equality. ∎

To prove the proposition, we use a fundamental result of analysis, namely
Arzelà-Ascoli theorem. In the notation of our paper, the theorem states:
_(Arzelà-Ascoli theorem) The sequence
$g_{f,T}^{(n)}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
of continuous functions defined on the interval $[0,T]$ is uniformly bounded
if there is a constant number $M>0$ such that_

$\displaystyle\left|\left|g_{f,T}^{(n)}(t)\right|\right|\leq M$ (7.39)

_for every function $g_{f,T}^{(n)}$ belonging to the sequence, and every
$t\in[0,T]$. (Here, $M$ must be independent of $n$ and $t$, but not of $T$.)
The sequence is said to be uniformly equicontinuous if, for every
$\varepsilon>0$, there exists a $\delta>0$ such that_

$\displaystyle\left|\left|g_{f,T}^{(n)}(t)-g_{f,T}^{(n)}(s)\right|\right|\leq\varepsilon$
(7.40)

_whenever $|t-s|<\delta$ for all functions $g_{f,T}^{(n)}$ in the sequence.
(Here, $\delta$ may depend on $\varepsilon$ and $T$, but not $t$, $s$ or $n$.)
If the sequence
$g_{f,T}^{(n)}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
is uniformly bounded and uniformly continuous, then there exists a subsequence
which converges uniformly to some
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$._

In the following proposition, we show that the assumption of uniformly
boundedness and uniformly equicontinuity are satisfied if we can show (7.2),
namely that $\alpha_{m,n}^{(2)}$ are bounded uniformly, i.e. there exists a
constant $K\in\mathbb{R}^{+}-\\{0\\}$ such that

$\displaystyle\sup_{\underset{m=0,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\alpha_{m,n}^{(2)}\right|\leq K.$

######  Proposition 7.9.

Under Assumption [A], the conditions (7.39) and (7.40) of Arzelà-Ascoli
theorem are satisfied whenever (7.2) holds.

###### Proof.

We start with Inequality (7.39). By recurrence, we can rewrite (6.1)-(6.2) as

$\displaystyle g_{f,T}^{(n)}(u)$ $\displaystyle=$
$\displaystyle\alpha_{0,n}\text{ for }u\in[0,T/2^{n}],$ (7.41) $\displaystyle
g_{f,T}^{(n)}(u)$ $\displaystyle=$
$\displaystyle\alpha_{0,n}+(T/2^{n})\sum_{i=1}^{m-1}\alpha_{i,n}+\alpha_{m,n}(u-mT/2^{n})\text{
for }u\in[mT/2^{n},(m+1)T/2^{n}],$ $\displaystyle m=1,\cdots,2^{n}-1.$

We obtain that for $m=0,\cdots,2^{n}-1$ and $u\in[mT/2^{n},(m+1)T/2^{n}]$:

$\displaystyle\left|\left|g_{f,T}^{(n)}(u)\right|\right|$ $\displaystyle\leq$
$\displaystyle\left|\left|\alpha_{0,n}\right|\right|+(T/2^{n})\sum_{i=1}^{m-1}\left|\left|\alpha_{i,n}\right|\right|+\left|\left|\alpha_{m,n}\right|\right|(u-mT/2^{n})$
$\displaystyle\leq$
$\displaystyle\left|\left|\alpha_{0,n}\right|\right|+(T/2^{n})\sum_{i=1}^{m}\left|\left|\alpha_{i,n}\right|\right|$
$\displaystyle\leq$
$\displaystyle\left|\left|\alpha_{0,n}\right|\right|+(T/2^{n})\sum_{i=1}^{2^{n}-1}\left|\left|\alpha_{i,n}\right|\right|$
$\displaystyle\leq$
$\displaystyle\left|\left|\alpha_{0,n}\right|\right|+(2^{n}-1)T/2^{n}\sup_{\underset{i=1,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\left|\alpha_{i,n}\right|\right|$ $\displaystyle\leq$
$\displaystyle(1+(2^{n}-1)T/2^{n})\sup_{\underset{i=0,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\left|\alpha_{i,n}\right|\right|$ $\displaystyle=$ $\displaystyle
2(1+(2^{n}-1)T/2^{n})\sup_{\underset{i=0,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\alpha_{i,n}^{(2)}\right|$ $\displaystyle\leq$ $\displaystyle
2(1+(2^{n}-1)T/2^{n})K,$

where we used triangular inequality in the first inequality, the definition of
$\alpha_{i,n}$ in the equality and (7.2) in the last inequality. We have thus
shown that (7.2) $\implies$ (7.39).

We continue with (7.40) case. We consider any arbitrarily small
$\varepsilon>0$. Accordingly, we set

$\displaystyle\delta=\frac{\varepsilon}{2K}.$ (7.43)

For any $t\in[0,T]$, we define the corresponding $m_{t}$ such that

$t\in[m_{t}T/2^{n},(m_{t}+1)T/2^{n}].$

From Equality (7), we can deduce that

$\displaystyle g_{f,T}^{(n)}(t)$ $\displaystyle=$
$\displaystyle\alpha_{0,n}+(T/2^{n})\sum_{i=1}^{m_{t}-1}\alpha_{i,n}+\alpha_{m_{t},n}(t-m_{t}T/2^{n}).$

Thus, for any $0\leq s\leq t$ such that

$\displaystyle|t-s|<\delta,$ (7.44)

we obtain

$\displaystyle g_{f,T}^{(n)}(t)-g_{f,T}^{(n)}(s)$ $\displaystyle=$
$\displaystyle\alpha_{0,n}+(T/2^{n})\sum_{i=1}^{m_{t}-1}\alpha_{i,n}+\alpha_{m_{t},n}(t-m_{t}T/2^{n})-$
$\displaystyle(\alpha_{0,n}+(T/2^{n})\sum_{i=1}^{m_{s}-1}\alpha_{i,n}+\alpha_{m_{s},n}(s-m_{s}T/2^{n}))$
$\displaystyle=$
$\displaystyle(T/2^{n})\sum_{i=m_{s}}^{m_{t}-1}\alpha_{i,n}+\alpha_{m_{t},n}(t-m_{t}T/2^{n})-\alpha_{m_{s},n}(s-m_{s}T/2^{n})$

Taking the norm, we obtain that

$\displaystyle\left|\left|g_{f,T}^{(n)}(t)-g_{f,T}^{(n)}(s)\right|\right|$
$\displaystyle=$
$\displaystyle\left|\left|(T/2^{n})\sum_{i=m_{s}}^{m_{t}-1}\alpha_{i,n}+\alpha_{m_{t},n}(t-m_{t}T/2^{n})-\alpha_{m_{s},n}(s-m_{s}T/2^{n})\right|\right|$
$\displaystyle\leq$
$\displaystyle\left|t-s\right|\sup_{\underset{i=1,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\left|\alpha_{i,n}\right|\right|$ $\displaystyle\leq$
$\displaystyle 2K\left|t-s\right|,$ $\displaystyle\leq$
$\displaystyle\varepsilon,$

where we used the definition of $\alpha_{i,n}$, of the norm and (7.2) in the
second inequality, the definition (7.43) and Inequality (7.44) in the last
inequality. We have thus shown that (7.2) $\implies$ (7.40). ∎

The following lemma establishes almost surely convergence of
$T_{h_{T}^{(n)}}^{W}$ towards $T_{h_{T}}^{W}$ when $h_{T}^{(n)}$ converges
uniformly to $h_{T}$. It appeared to be missing in the literature, so we added
that result. For our proof of existence, we only require the convergence in
distribution.

######  Lemma 7.10.

For any sequence
$h_{T}^{(n)}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
which converges uniformly to some
$h_{T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$,
we have that $T_{h_{T}^{(n)}}^{W}$ converges almost surely to $T_{h_{T}}^{W}$.
As a by-product, we deduce that $T_{h_{T}^{(n)}}^{W}$ converges in
distribution to $T_{h_{T}}^{W}$.

###### Proof.

To prove that $T_{h_{T}^{(n)}}^{W}$ converges almost surely to
$T_{h_{T}}^{W}$, it is sufficient to show that for any arbitrarily small
$\epsilon>0$, there exists $N_{\epsilon}\in\mathbb{N}$ such that for any
$n\in\mathbb{N},n\geq N_{\epsilon}$, we have:

$\displaystyle\left|T_{h_{T}^{(n)}}^{W}-T_{h_{T}}^{W}\right|\leq\epsilon.$
(7.45)

As $h_{T}^{(n)}$ converges uniformly to $h_{T}$, we have that for any
$\epsilon_{h}>0$, there exists $N_{\epsilon_{h}}\in\mathbb{N}$ such that for
any $n\in\mathbb{N},n\geq N_{\epsilon_{h}}$, we have

$\displaystyle\left|\left|h_{T}^{(n)}-h_{T}\right|\right|\leq\epsilon_{h}.$
(7.46)

We set

$\displaystyle\epsilon_{h}=\min\left(\frac{1}{2}\inf_{0\leq t\leq
T_{h_{T}}^{W}-\epsilon,i=1,2}\left|h_{T}^{(i)}(t)-W_{t}\right|,\sup_{T_{h_{T}}^{W}\leq
t\leq T_{h_{T}}^{W}+\epsilon}\left|W_{t}-h_{T}^{(i_{h})}(t)\right|\right),$
(7.47)

where $i_{h}=1$ if the lower boundary was hitten first and $i_{h}=2$ if the
upper boundary was hitten first. First, note that both terms of (7.47) are
strictly positive, since $T_{h_{T}}^{W}$ is defined as the first time when
$W_{t}$ hits $h_{T}$. Second, note that the first term in (7.47) prevents
(almost surely) $W_{t}$ from touching $h_{T}^{(n)}$ before
$T_{h_{T}}^{W}-\epsilon$, whereas the second term in (7.47) ensures (almost
surely) that $W_{t}$ first hits $g_{T}^{(n)}$ during
$[T_{h_{T}}^{W}-\epsilon,T_{h_{T}}^{W}+\epsilon]$. In other words, we have
shown that
$T_{h_{T}^{(n)}}^{W}\in[T_{h_{T}}^{W}-\epsilon,T_{h_{T}}^{W}+\epsilon]$
whenever (7.46) holds with $\epsilon_{h}$ set as in (7.47). This means that we
have shown (7.45) with $N_{\epsilon}:=N_{\epsilon_{h}}$. ∎

In the following proposition, we prove that there exists a subsequence of
$g_{f,T}^{(n)}$ which converges uniformly to some
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
via Arzelà-Ascoli theorem. We make some stronger assumption on $f$ for that
proof. Note that this assumption is natural, and comparable (although slightly
more restrictive) to Assumptions (30) or (31) required for Theorem 8.2 in
[Ekström and Janson, 2016]. In that paper, the authors do require boundedness
away from $0$, but do not need boundedness of $f(t)$.

[B

] We assume that there exist $f_{T}^{-}\in\mathbb{R}^{+}-\\{0\\}$ and
$f_{T}^{+}\in\mathbb{R}^{+}-\\{0\\}$ such that for any $t\in[0,T]$,

$\displaystyle f_{T}^{-}\leq f(t)\leq f_{T}^{+}.$ (7.48)

Note that Assumption [B] obviously implies Assumption [A].

######  Proposition 7.11.

Under Assumption [B], we have that Equation (7.2) is satisfied, namely that
$\alpha_{m,n}^{(2)}$ are bounded uniformly, i.e. there exists a constant
$K\in\mathbb{R}^{+}-\\{0\\}$ such that

$\displaystyle\sup_{\underset{m=0,\cdots,2^{n}-1}{n\in\mathbb{N}-\\{0\\}}}\left|\alpha_{m,n}^{(2)}\right|\leq
K.$ (7.49)

###### Proof.

As the statement of the proposition is "asymptotic", without loss of
generality we can drop the first terms of the sequence, and start from the
term

$n_{s}=\lceil\log_{2}(T)-\log_{2}(\eta)\rceil,$

where $\lceil.\rceil$ is the ceiling function. Then, for any
$n\in\mathbb{N}-\\{0\\},n\geq n_{s}$, we have that

$\displaystyle\frac{T}{2^{n}}\leq\eta.$ (7.50)

To prove (7.49), it is enough then to show that

$\displaystyle\sup_{\underset{m=0,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\alpha_{m,n}^{(2)}\right|\leq K.$ (7.51)

Actually, we will break up the proof in two cases (i) $m=0$ case, and (ii)
$m=1,\cdots,2^{n}-1$ case. To show Equation (7.51), we will then show that
there exist two constants $K_{0}\in\mathbb{R}^{+}-\\{0\\}$ and
$K_{1}\in\mathbb{R}^{+}-\\{0\\}$ such that

$\displaystyle\sup_{n\in\mathbb{N},n\geq
n_{s}}\left|\alpha_{0,n}^{(2)}\right|$ $\displaystyle\leq$ $\displaystyle
K_{0},$ (7.52)
$\displaystyle\sup_{\underset{m=1,\cdots,2^{n}-1}{n\in\mathbb{N},n\geq
n_{s}}}\left|\alpha_{m,n}^{(2)}\right|\leq K_{1}.$ (7.53)

Indeed, taking $K=\sup(K_{0},K_{1})$, we then obtain Equation (7.51).

We start with the case (i) $\alpha_{0,n}$. We want to show Equation (7.52). By
the boundedness away from 0 of Equation (7.48) in Assumption [B], there exist
a small $\eta>0$,

$\displaystyle g_{1}$ $\displaystyle:=$
$\displaystyle(-\infty,g_{1}^{(2)})\text{ in upper boundary case}$
$\displaystyle g_{1}$ $\displaystyle:=$
$\displaystyle(-g_{1}^{(2)},g_{1}^{(2)})\text{ in symmetrical boundary case}$

with $g_{1}^{(2)}\in\mathcal{C}_{0}^{+}([0,\eta],\mathbb{R}^{+}-\\{0\\})$ s.t.
for $0\leq t\leq\eta$:

$\displaystyle f_{g_{1}}^{W}(t)\leq f(t).$ (7.54)

If we write

$\displaystyle\max g_{1}$ $\displaystyle:=$ $\displaystyle(-\infty,\max_{0\leq
t\leq\eta}g_{1}^{(2)}(t))\text{ in upper boundary case}$ $\displaystyle\max
g_{1}$ $\displaystyle:=$ $\displaystyle(-\max_{0\leq
t\leq\eta}g_{1}^{(2)}(t),\max_{0\leq t\leq\eta}g_{1}^{(2)}(t))\text{ in
symmetrical boundary case},$

it is clear by definition that the boundaries of $g_{1}$ are within that of
$\max g_{1}$, i.e. the upper boundary of $g_{1}$ is smaller than that of $\max
g_{1}$ and the lower boundary of $g_{1}$ is greater than that of $\max g_{1}$
(if we are in the symmetrical boundary case). This implies that $g_{1}$ will
be hitten first by the Brownian motion, i.e. conditionally on the hit
happening during $[0,\eta]$ that

$T_{g_{1}}^{W}\leq T_{\max g_{1}}^{W}\text{ a.s.}$

In particular, this implies that

$\mathbb{P}(T_{\max g_{1}}^{W}\leq T/2^{n})\leq\mathbb{P}(T_{g_{1}}^{W}\leq
T/2^{n}).$

By Definition (3.2)-(3.3), this can be expressed as

$\displaystyle\mathbb{P}(T_{\max g_{1}}^{W}\leq
T/2^{n})\leq\int_{0}^{T/2^{n}}f_{g_{1}}^{W}(s)ds.$ (7.55)

Combining Equation (7.54) along with Equation (7.55), we can deduce that

$\displaystyle\mathbb{P}(T_{\max g_{1}}^{W}\leq
T/2^{n})\leq\int_{0}^{T/2^{n}}f(s)ds.$ (7.56)

By Definition (6.3), we have $\alpha_{0,n}$ defined such that

$\displaystyle\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}=\int_{0}^{T/2^{n}}f(s)ds,$
(7.57)

From Equation (7.56) and Equation (7.57), we can deduce that

$\displaystyle\mathbb{P}\big{(}T_{\max
g_{1}}^{W}\in[0,T/2^{n}]\big{)}\leq\mathbb{P}\big{(}T_{\alpha_{0,n}}^{W}\in[0,T/2^{n}]\big{)}.$
(7.58)

This can be rewritten as

$\displaystyle z_{i}^{(n)}\big{(}\max_{0\leq
t\leq\eta}g_{1}^{(2)}(t)\big{)}\leq
z_{i}^{(n)}\big{(}\alpha_{0,n}^{(2)}\big{)},$ (7.59)

where $z_{i}^{(n)}$ for $i=1,2$ correspond respectively to the upper boundary
case and the symmetrical boundary case. By an application of Lemma 7.5, we
then obtain that

$\displaystyle\alpha_{0,n}^{(2)}\leq\max_{0\leq t\leq\eta}g_{1}^{(2)}(t).$
(7.60)

This means that we have shown Equation (7.52).

We consider now the case (ii) $m=1,\cdots,2^{n}-1$. We aim to show Equation
(7.53). For tractability purposes, we keep the following derivation in the
case $m=1$. We recall that from Equation (7.7) in Lemma 7.3, $\alpha_{1,n}$
satisfies the following equation:

$\displaystyle
1-\int_{0}^{2T/2^{n}}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{g_{f,T}^{(n)}(2T/2^{n})}(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{(2T}{2^{n}})-x_{2})(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}$
$\displaystyle\times(1-\exp(\frac{-2(g_{f,T}^{(n)}(\frac{T}{2^{n}})-x_{1})g_{f,T}^{(n)}(\frac{T}{2^{n}})}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}dx_{2}=0.$

To obtain uniform boundedness (i.e. Equation (7.53) in the case $m=1$),
derivation in the above equation does not seem tractable. To circumvent that
difficulty, we consider an interpolation of left hand-side of Equation (7.7)
for $t\in(T/2^{n},2T/2^{n}]$:

$\displaystyle I_{n}(t)$ $\displaystyle=$ $\displaystyle
1-\int_{0}^{t}f(s)ds-\int_{-\infty}^{\alpha_{0,n}}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}(1-\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}dx_{2}.$

We now differentiate $I_{n}(t)$ with respect to $t$:

$\displaystyle I_{n}^{\prime}(t)$ $\displaystyle=$
$\displaystyle-f(t)-\int_{-\infty}^{\alpha_{0,n}}\frac{d}{dt}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}(1-\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{2}\Big{]}dx_{1}$
$\displaystyle:=$ $\displaystyle-f(t)+\widetilde{I}_{n}^{\prime}(t).$

Note that $\alpha_{1,n}$ is solution of $I_{n}^{\prime}(t)=0$ if and only if
$\alpha_{1,n}$ is solution of Equation (7.7). Then, to obtain uniform
boundedness (i.e. Equation (7.53) in the case $m=1$), we can work with the
equation $I_{n}^{\prime}(t)=0$. In view of Equation (7.48) in Assumption [B],
we have that $f_{T}^{-}\leq f(t)\leq f_{T}^{+}$, and thus we have that
$I_{n}^{\prime}(t)=0$ implies that
$f_{T}^{-}\leq\widetilde{I}_{n}^{\prime}(t)\leq f_{T}^{+}$ for any
$n\in\mathbb{N}$ with $n\geq n_{s}$, i.e. that $\widetilde{I}_{n}^{\prime}(t)$
is uniformly bounded away from 0 and bounded. Thus, if we can show that

$\displaystyle(\forall n\in\mathbb{N}\text{ , }n\geq n_{s}\text{
}f_{T}^{-}\leq\widetilde{I}_{n}^{\prime}(t)\leq
f_{T}^{+})\implies\sup_{n\in\mathbb{N},n\geq
n_{s}}\left|\alpha_{m,n}^{(2)}\right|\leq K_{1},$ (7.61)

then Equation (7.53) in the case $m=1$ is shown. In the remainder of this
proof, we thus show Equation (7.61). For that purpose, we then derive that

$\displaystyle\frac{d}{dt}\int_{-\infty}^{h(t)}f(t,s)ds=\int_{-\infty}^{h(t)}\frac{df}{dt}(t,s)ds+h^{\prime}(t)f(t,h(t)).$
(7.62)

Applying Formula (7.62), we can calculate that

$\displaystyle\widetilde{I}_{n}^{\prime}(t)$ $\displaystyle=$
$\displaystyle-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}(1-\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle-\alpha_{1,n}\int_{-\infty}^{\alpha_{0,n}}(1-\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})))(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}dx_{1}.$

Note that the last term is actually equal to $0$. We can then calculate
$\widetilde{I}_{n}^{\prime}(t)$ as

$\displaystyle\widetilde{I}_{n}^{\prime}(t)$ $\displaystyle=$
$\displaystyle-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}(1-\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}}))$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}(1-\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}}))\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle-f(t)-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}$
$\displaystyle\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle:=-A_{n}-B_{n}-C_{n}-D_{n}.$

First, we have that

$\displaystyle A_{n}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{(t-\frac{T}{2^{n}})}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}\dfrac{\left(-t+\frac{T}{2^{n}}-2\left(x_{2}-x_{1}\right)^{2}\right)\exp(\frac{\left(x_{2}-x_{1}\right)^{2}}{t-\frac{T}{2^{n}}})}{2\left(t-\frac{T}{2^{n}}\right)^{\frac{5}{2}}}dx_{2}\Big{]}dx_{1}$

Second, we can calculate that

$\displaystyle B_{n}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{(t-\frac{T}{2^{n}})}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}$
$\displaystyle\dfrac{\left(-t+\frac{T}{2^{n}}-2\left(x_{2}-x_{1}\right)^{2}\right)\exp(\frac{\left(x_{2}-x_{1}\right)^{2}}{t-\frac{T}{2^{n}}})}{2\left(t-\frac{T}{2^{n}}\right)^{\frac{5}{2}}}dx_{2}\Big{]}dx_{1}$

Third, we can deduce that

$\displaystyle C_{n}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})$
$\displaystyle\times\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}$
$\displaystyle\times\frac{d}{dt}\Big{[}\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{(t-\frac{T}{2^{n}})}}\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle-\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}\exp(-\frac{2\left(\alpha_{0,n}-x_{1}\right)\left(\alpha_{1,n}\left(t-T/2^{n}\right)-x_{2}+\alpha_{0,n}\right)-\left(x_{2}-x_{1}\right)^{2}}{t-T/2^{n}})$
$\displaystyle\times\dfrac{\left(t+2x_{1}^{2}+\left(4\alpha_{0,n}-8x_{2}\right)x_{1}+2x_{2}^{2}+4\alpha_{0,n}x_{2}-T/2^{n}-4\alpha_{0,n}^{2}\right)}{2\left(t-T/2^{n}\right)^{\frac{5}{2}}}dx_{2}\Big{]}dx_{1}$

Finally, calculations lead to

$\displaystyle D_{n}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\frac{d}{dt}\Big{[}\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{\pi(t-\frac{T}{2^{n}})}}$
$\displaystyle\times\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\sqrt{\pi\frac{T}{2^{n}}}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}$
$\displaystyle\frac{d}{dt}\Big{[}\exp(\frac{-2(\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})-x_{2})(\alpha_{0,n}-x_{1})}{t-\frac{T}{2^{n}}})\frac{\exp(-\frac{(x_{2}-x_{1})^{2}}{t-\frac{T}{2^{n}}})}{\sqrt{(t-\frac{T}{2^{n}})}}\Big{]}dx_{2}\Big{]}dx_{1}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\alpha_{0,n}}\Big{[}\int_{-\infty}^{\alpha_{0,n}+\alpha_{1,n}(t-T/2^{n})}\exp(\frac{-2(\alpha_{0,n}-x_{1})\alpha_{0,n}}{\frac{T}{2^{n}}})\frac{\exp(-\frac{x_{1}^{2}}{\frac{T}{2^{n}}})}{\pi\sqrt{\frac{T}{2^{n}}}}$
$\displaystyle\times\exp(-\frac{2\left(\alpha_{0,n}-x_{1}\right)\left(\alpha_{1,n}\left(t-T/2^{n}\right)-x_{2}+\alpha_{0,n}\right)-\left(x_{2}-x_{1}\right)^{2}}{t-T/2^{n}})$
$\displaystyle\times\dfrac{\left(t+2x_{1}^{2}+\left(4\alpha_{0,n}-8x_{2}\right)x_{1}+2x_{2}^{2}+4\alpha_{0,n}x_{2}-T/2^{n}-4\alpha_{0,n}^{2}\right)}{2\left(t-T/2^{n}\right)^{\frac{5}{2}}}dx_{2}\Big{]}dx_{1}.$

Then, further similar derivation can lead to prove Equation (7.61). We have
thus shown Equation (7.53) in the case $m=1$. The proof in the uniform case
$m=1,\ldots,2^{n}-1$ involves a similar series of calculation, which we do not
report for tractability purposes. ∎

######  Corollary 7.12.

Under Assumption [A] and Assumption [B], there exists a subsequence of
$g_{f,T}^{(n)}$ which converges uniformly to some
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$.

###### Proof.

This is an application of Arzelà-Ascoli theorem together with Proposition 7.9
and Proposition 7.11. ∎

We provide now the main result of our paper, which states the existence of the
continuous boundary. This proof is basically a combination of application to
the previous obtained results.

######  Theorem 7.13.

(Existence) Under Assumption [A] and Assumption [B], there exists a
corresponding continuous boundary function
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
such that the density of $\mathrm{T}_{g_{f,T}}^{W}$ is $f$ on $[0,T]$.

###### Proof.

To prove Theorem 7.13, we have to show that there exists
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
such that the density of $T_{g_{f,T}}^{W}$ is exactly $f$ on $[0,T]$. By Borel
arguments, it is sufficient to show that for any $p\in\mathbb{N}-\\{0\\}$ and
$k=0,\ldots,2^{p}-1$ we have

$\displaystyle\mathbb{P}\big{(}T_{g_{f,T}}^{W}\in[kT/2^{p},(k+1)T/2^{p}]\big{)}=\int_{kT/2^{p}}^{(k+1)T/2^{p}}f(s)ds.$
(7.63)

To prove (7.63), we will use
$g_{f,T}\in\overline{\mathcal{C}_{0}^{-}([0,T],\mathbb{R})}\times\mathcal{C}_{0}^{+}([0,T],\mathbb{R})$
from Proposition 7.11, which is obtained as the uniform limit of a subsequence
of $g_{f,T}^{(n)}$. We refer to such subsequence as $g_{f,T}^{(n_{q})}$. Then,
for any $p\in\mathbb{N}-\\{0\\}$ and $k=0,\ldots,2^{p}-1$, we obtain

$\displaystyle\mathbb{P}\big{(}T_{g_{f,T}}^{W}\in[kT/2^{p},(k+1)T/2^{p}]\big{)}$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\mathbb{P}\big{(}T_{g_{f,T}^{(n_{q})}}^{W}\in[kT/2^{p},(k+1)T/2^{p}]\big{)}$
$\displaystyle=$ $\displaystyle\int_{kT/2^{p}}^{(k+1)T/2^{p}}f(s)ds,$

where the first equality corresponds to the convergence in distribution of
$T_{g_{f}^{(n_{s})}}^{W}$ to $T_{g_{f}}^{W}$ as an application of Lemma 7.10
along with the fact that there is uniform convergence of $g_{f,T}^{(n_{q})}$
towards $g_{f,T}$ as an application of Proposition 7.11, we have used
Corollary 7.8 in the second equality. In other words, we have shown (7.63). ∎

##  8 Conclusion

In this paper, we have investigated the question of existence in the inverse
first-passage problem. Correspondingly, we have provided smooth assumptions on
the target density for such boundary to exist. We have provided a more direct
proof to the almost-50-year-old Shiryaev question. There is room for
improvement, as we have not provided any closed-formula for the solution,
although this may be very challenging to obtain.

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