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Mathematics 2018, 6(6), 91; doi:10.3390/math6060091

Article
The Randomized First-Hitting Problem of Continuously Time-Changed Brownian Motion
Dipartimento di Matematica, Università Tor Vergata, 00133 Rome, Italy
Received: 4 April 2018 / Accepted: 25 May 2018 / Published: 28 May 2018

## Abstract

:
Let $X ( t )$ be a continuously time-changed Brownian motion starting from a random position $η , S ( t )$ a given continuous, increasing boundary, with $S ( 0 ) ≥ 0 ,$$P ( η ≥ S ( 0 ) ) = 1 ,$ and F an assigned distribution function. We study the inverse first-passage time problem for $X ( t ) ,$ which consists in finding the distribution of $η$ such that the first-passage time of $X ( t )$ below $S ( t )$ has distribution $F ,$ generalizing the results, valid in the case when $S ( t )$ is a straight line. Some explicit examples are reported.
Keywords:
first-passage time; inverse first-passage problem; diffusion

## 1. Introduction

This brief note is a continuation of [1,2]. Let $σ ( t )$ be a regular enough non random function, and let $X ( t ) = η + ∫ 0 t σ ( s ) d B s ,$ where $B t$ is standard Brownian motion (BM) and the initial position $η$ is a random variable, independent of $B t .$ Suppose that the quadratic variation $ρ ( t ) = ∫ 0 t σ 2 ( s ) d s$ is increasing and $ρ ( + ∞ ) = ∞ ,$ then there exists a standard BM $B ˜$ such that $X ( t ) = η + B ˜ ( ρ ( t ) ) ,$ namely $X ( t )$ is a continuously time-changed BM (see e.g., [3]). For a continuous, increasing boundary $S ( t ) ,$ such that $P ( η ≥ S ( 0 ) ) = 1 ,$ let
$τ = τ S = inf { t > 0 : X ( t ) ≤ S ( t ) }$
be the first-passage time (FPT) of $X ( t )$ below $S .$ We assume that $τ$ is finite with probability one and that it possesses a density $f ( t ) = d F ( t ) d t ,$ where $F ( t ) = P ( τ ≤ t ) .$ Actually, the FPT of continuously time-changed BM is a well studied problem for constant or linear boundary and a non-random initial value (see e.g., [4,5,6]).
Assuming that $S ( t )$ is increasing, and $F ( t )$ is a continuous distribution function, we study the following inverse first-passage-time (IFPT) problem:
given a distribution F, find the density g of η (if it exists) for which it results $P ( τ ≤ t ) = F ( t )$.
The function g is called a solution to the IFPT problem. This problem, also known as the generalized Shiryaev problem, was studied in [1,2,7,8], essentially in the case when $X ( t )$ is BM and $S ( t )$ is a straight line; note that the question of the existence of the solution is not a trivial matter (see e.g., [2,7]). In this paper, by using the properties of the exponential martingale, we extend the results to more general boundaries $S .$
The IFPT problem has interesting applications in mathematical finance , in particular in credit risk modeling, where the FPT represents a default event of an obligor (see [7]) and in diffusion models for neural activity ([9]).
Notice, however, that another type of inverse first-passage problem can be considered: it consists in determining the boundary shape S, when the FPT distribution F and the starting point $η$ are assigned (see e.g., [10,11,12,13]).
The paper is organized as follows: Section 2 contains the main results, in Section 3 some explicit examples are reported; Section 4 is devoted to conclusions and final remarks.

## 2. Main Results

The following holds:
Theorem 1.
Let be $S ( t )$ a continuous, increasing boundary with $S ( 0 ) ≥ 0 , σ ( t )$ a bounded, non random continuous function of $t > 0 ,$ and let $X ( t ) = η + ∫ 0 t σ ( s ) d B s$ be the integral process starting from the random position $η ≥ S ( 0 ) ;$ we assume that $ρ ( t ) = ∫ 0 t σ 2 ( s ) d s$ is increasing and satisfies $ρ ( + ∞ ) = + ∞ .$ Let F be the probability distribution of the FPT $τ S$ of X below the boundary S ($τ S$ is a.s. finite by virtue of Remark 3). We suppose that the r.v. η admits a density $g ( x ) ;$ for $θ > 0 ,$ we denote by $g ^ ( θ ) = E ( e − θ η )$ the Laplace transform of $g .$
Then, if there exists a solution to the IFPT problem for $X ,$ the following relation holds:
$g ^ ( θ ) = ∫ 0 + ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) d F ( t ) .$
Proof.
The process $X ( t )$ is a martingale, we denote by $F t$ its natural filtration. Thanking to the hypothesis, by using the Dambis, Dubins–Schwarz theorem (see e.g., [3]), it follows that the process $B ˜ ( t ) = X ( ρ − 1 ( t ) )$ is a Brownian motion with respect to the filtration $F ρ − 1 ( t ) ;$ so the process $X ( t )$ can be written as $X ( t ) = η + B ˜ ( ρ ( t ) )$ and the FPT $τ$ can be written as $τ = inf { t > 0 : η + B ˜ ( ρ ( t ) ) ≤ S ( t ) } .$ For $θ > 0 ,$ let us consider the process $Z t = e − θ X ( t ) − 1 2 θ 2 ρ ( t ) ;$ as easily seen, $Z t$ is a positive martingale; indeed, it can be represented as $Z t = e − θ X ( 0 ) − θ ∫ 0 t Z s σ ( s ) d B s$ (see e.g., Theorem 5.2 of [14]). We observe that, for $t ≤ τ$ the martingale $Z t$ is bounded, because $X ( t )$ is non negative and therefore $0 < Z t ≤ e − θ X ( t ) ≤ 1 .$ Then, by using the fact that, for any finite stopping time $τ$ one has $E [ Z 0 ] = E [ Z τ ∧ t ]$ (see e.g., Formula (7.7) in [14]), and the dominated convergence theorem, we obtain that
$E [ Z 0 ] = E [ e − θ X ( 0 ) ] = E [ e − θ η ] = lim t → ∞ E [ e − θ X ( τ ∧ t ) − 1 2 θ 2 ρ ( τ ∧ t ) ]$
$= E [ lim t → ∞ e − θ X ( τ ∧ t ) − 1 2 θ 2 ρ ( τ ∧ t ) ] = E [ e − θ S ( τ ) − 1 2 θ 2 ρ ( τ ) ] .$
Thus, if $g ^ ( θ ) = E ( e − θ η )$ is the Laplace transform of the density of the initial position $η ,$ we finally get
$g ^ ( θ ) = E e − θ S ( τ ) − θ 2 2 ρ ( τ ) ,$
that is Equation (2). ☐
Remark 1.
If one takes in place of $X ( t )$ a process of the form $X ˜ ( t ) = η S ( t ) + S ( t ) B ( ρ ( t ) ) ,$ with $η ≥ 1 ,$ that is, a special case of continuous Gauss-Markov process ([15]) with mean $η S ( t ) ,$ then $X ˜ ( t ) / S ( t )$ is still a continuously time-changed BM, and so the IFPT problem for $X ˜ ( t )$ and $S ( t )$ is reduced to that of continuously time-changed BM and a constant barrier, for which results are available (see e.g., [4,5,6]).
Remark 2.
By using Laplace transform inversion (when it is possible), Equation (4) allows to find the solution g to the IFPT problem for $X ,$ the continuous increasing boundary $S ,$ and the distribution F of the FPT $τ .$ Indeed, some care has to be used to exclude that the found distribution of η has atoms together with a density. However, as already noted in [2,7], the function $g ^$ may not be the Laplace transform of some probability density function, so in that case the IFPT problem has no solution; really, it may admit more than one solution, since the right-hand member of Equation (4) essentially furnishes the moments of η of any order $n ,$ but this is not always sufficient to uniquely determine the density g of $η .$ In line of principle, the right-hand member of Equation (4) can be expressed in terms of the Laplace transform of $f ( t ) = F ′ ( t ) ,$ though it is not always possible to do this explicitly. A simple case is when $S ( t ) = a + b t ,$ with $a , b ≥ 0 ,$ and $ρ ( t ) = t ,$ that is, $X ( t ) = B t ( σ ( t ) = 1 ) ;$ in fact, one obtains
$g ^ ( θ ) = E e − θ ( a + b τ ) − θ 2 2 τ = e − θ a E e − θ ( b + θ 2 ) τ = e − θ a f ^ θ ( θ + 2 b ) 2 ,$
which coincides with Equation (2.2) of [2], and it provides a relation between the Laplace transform of the density of the initial position η and the Laplace transform of the density of the FPT $τ .$
Remark 3.
Let $S ( t )$ be increasing and $S ( 0 ) ≥ 0 ,$ then τ is a.s. finite; in fact $τ ˜ = ρ ( τ ) = inf { t > 0 : η + B ˜ t ≤ S ˜ ( t ) } ≤ τ ˜ 1 ,$ where $S ˜ ( t ) = S ( ρ − 1 ( t ) )$ is increasing and $τ ˜ 1$ is the first hitting time to $S ( 0 )$ of BM $B ˜$ starting at $η ;$ since $τ ˜ 1$ is a.s. finite, also $τ ˜$ is so. Next, from the finiteness of $τ ˜$ it follows that $τ = ρ − 1 ( τ ˜ )$ is finite, too. Moreover, if one seeks that $E ( τ ) < ∞ ,$ a sufficient condition for this is that $ρ ( t )$ and $S ˜ ( t )$ are both convex functions; indeed, $τ ˜ ≤ τ ˜ 2 ,$ where $τ ˜ 2$ is the FPT of BM $B ˜$ starting from η below the straight line $a + b t ( a = S ( 0 ) ≥ 0 , b = S ˜ ′ ( 0 ) ≥ 0 )$ which is tangent to the graph of $S ˜ ( t )$ at $t = 0 .$ Thus, since $E ( τ ˜ 2 ) < ∞ ,$ it follows that $E ( τ ˜ )$ is finite, too; finally, being $ρ − 1$ concave, Jensen’s inequality for concave functions implies that $E ( τ ) = E ( ρ − 1 ( τ ˜ ) ) ≤ ρ − 1 ( E ( τ ˜ ) )$ and therefore $E ( τ ) < ∞ .$
Remark 4.
Theorem 1 allows to solve also the so called Skorokhod embedding (SE) problem:
Given a distribution $H ,$ find an integrable stopping time $τ * ,$ such that the distribution of $X ( τ * )$ is $H ,$ namely $P ( X ( τ * ) ≤ x ) = H ( x ) .$
In fact, let be $S ( t )$ increasing, with $S ( 0 ) = 0 ;$ first suppose that the support of H is $[ 0 , + ∞ ) ;$ then, from Equation (4) it follows that
$g ^ ( θ ) = E [ e − θ X ( τ ) − θ 2 2 ρ ( S − 1 ( X ( τ ) ) ) ] ,$
and this solves the SE problem with $τ * = τ ;$ it suffices to take the random initial point $X ( 0 ) = η > 0$ in such a way that its Laplace transform $g ^$ satisfies
$g ^ ( θ ) = ∫ 0 S ( + ∞ ) e − θ x − θ 2 2 ρ ( S − 1 ( x ) ) d H ( x ) .$
In the special case when $S ( t ) = a + b t ( a , b > 0 )$ and $ρ ( t ) = t ,$ Equation (7) becomes (cf. the result in [8] for $a = 0 ) :$
$g ^ ( θ ) = e a θ 2 2 b h ^ θ ( θ + 2 b ) 2 b ,$
where $h ( x ) = H ′ ( x )$ and $h ^$ denotes the Laplace transform of $h .$
In analogous way, the SE problem can be solved if the support of H is $( − ∞ , 0 ] ;$ now, the FPT is understood as $τ − = inf { t > 0 : η + B ( ρ ( t ) ) > − S ( t ) } ( η < 0 ) ,$ that is, the first hitting time to the boundary $S − ( t ) = − S ( t )$ from below.
Therefore, the solution to the general SE problem, namely without restrictions on the support of the distribution $H ,$ can be obtained as follows (see [8], for the case when $S ( t )$ is a straight line).
The r.v. $X ( τ )$ can be represented as a mixture of the r.v. $X + > 0$ and $X − < 0 :$
$X ( τ ) = X + with probability p + = P ( X ( τ ) ≥ 0 ) X − with probability p − = 1 − p + .$
Suppose that the SE problem for the r.v. $X +$ and $X −$ can be solved by $S + ( t ) = S ( t )$ and $η + = η > 0 ,$ and $S − ( t ) = − S ( t )$ and $η − = − η < 0 ,$ respectively. Then, we get that the r.v.
$η ± = η + with probability p + η − with probability p −$
and the boundary $S ± ( t ) = S + ( t ) ∪ S − ( t )$ solve the SE problem for the r.v. $X ( τ ) .$
If $g ^$ is analytic in a neighbor of $θ = 0 ,$ then the moments of order n of $η , E ( η n ) ,$ exist finite, and they are given by $E ( η n ) = ( − 1 ) n d n d θ n g ^ | θ = 0 .$ By taking the first derivative in Equation (4) and calculating it at $θ = 0 ,$ we obtain
$E ( η ) = − g ^ ′ ( 0 ) = E ( S ( τ ) ) .$
By calculating the second derivative of $g ^$ at $θ = 0 ,$ we get
$E ( η 2 ) = g ^ ′ ′ ( 0 ) = E ( S 2 ( τ ) − ρ ( τ ) ) ) ,$
and so
$V a r ( η ) = E ( η 2 ) − E 2 ( η ) = V a r ( S ( τ ) ) − E ( ρ ( τ ) ) .$
Thus, we obtain the compatibility conditions
$E ( η ) = E ( S ( τ ) ) V a r ( S ( τ ) ) ≥ E ( ρ ( τ ) ) .$
If $V a r ( S ( τ ) ) < E ( ρ ( τ ) ) ,$ a solution to the IFPT problem does not exist. In the special case when $S ( t ) = a + b t ( a , b ≥ 0 )$ and $ρ ( t ) = t ,$ Equation (11) becomes $E ( η ) = a + b E ( τ )$ and Equation (13) becomes $V a r ( η ) = b 2 V a r ( τ ) − E ( τ ) ,$ while Equation (14) coincides with Equation (2.3) of [2]. By writing the Taylor’s expansions at $θ = 0$ of both members of Equation (4), and equaling the terms with the same order in $θ ,$ one gets the successive derivatives of $g ^ ( θ )$ at $θ = 0 ;$ thus, one can write any moment of $η$ in terms of the expectation of a function of $τ ;$ for instance, it is easy to see that
$E ( η 3 ) = E [ ( S ( τ ) ) 3 ] − 3 E [ S ( τ ) ρ ( τ ) ] ,$
$E ( η 4 ) = E [ ( S ( τ ) 4 ] − 6 E [ ( S ( τ ) 2 ρ ( τ ) ] + 3 E [ ( ρ ( τ ) 2 ] ,$
$E ( η 5 ) = E [ 15 S ( τ ) ρ 2 ( τ ) − 240 S 3 ( τ ) ρ ( τ ) + S 5 ( τ ) ] .$

#### 2.1. The Special Case $S ( t ) = α + β ρ ( t )$

If $S ( t ) = α + β ρ ( t ) ,$ with $α , β ≥ 0 ,$ from Equation (4) we get
$g ^ ( θ ) = E [ e − θ ( α + β ρ ( τ ) ) − θ 2 2 ρ ( τ ) ] = e − θ α E [ e − θ ρ ( τ ) ( β + θ / 2 ) ] .$
Thus, setting $τ ˜ = ρ ( τ ) ,$ we obtain (see Equation (5)):
$g ^ ( θ ) = e − θ α E [ e − θ ( β + θ / 2 ) τ ˜ ] = e − θ α f ˜ ^ ( θ ( β + θ / 2 ) ) ,$
having denoted by $f ˜$ the density of $τ ˜ .$ In this way, we reduce the IFPT problem of $X ( t ) = η + B ( ρ ( t ) )$ below the boundary $S ( t ) = α + β ρ ( t )$ to that of BM below the linear boundary $α + β t .$ For instance, taking $ρ ( t ) = t 3 / 3 ,$ the solution to the IFPT problem of $X ( t )$ through the cubic boundary $S ( t ) = α + β 3 t 3 ,$ and the FPT density $f ,$ is nothing but the solution to the IFPT problem of BM through the linear boundary $α + β t ,$ and the FPT density $f ˜ .$
Under the assumption that $S ( t ) = α + β ρ ( t ) ,$ with $α , β ≥ 0 ,$ a number of explicit results can be obtained, by using the analogous ones which are valid for BM and a linear boundary (see [2]). As for the question of the existence of solutions to the IFPT problem, we have:
Proposition 1.
Let be $S ( t ) = α + β ρ ( t ) ,$ with $α , β ≥ 0 ;$ for $γ , λ > 0 ,$ suppose that the FPT density $f = F ′$ is given by
$f ( t ) = λ γ Γ ( γ ) ρ ( t ) γ − 1 e − λ ρ ( t ) ρ ′ ( t ) i f t > 0 0 o t h e r w i s e$
(namely the density $f ˜$ of $τ ˜$ is the Gamma density with parameters $( γ , λ ) ) .$ Then, the IFPT problem has solution, provided that $β ≥ 2 λ ,$ and the Laplace transform of the density g of the initial position η is given by:
$g ^ ( θ ) = e − α θ / 2 ( β − β 2 − 2 λ ) γ ( θ + β − β 2 − 2 λ ) γ · e − α θ / 2 ( β + β 2 − 2 λ ) γ ( θ + β + β 2 − 2 λ ) γ ,$
which is the Laplace transform of the sum of two independent random variables, $Z 1$ and $Z 2 ,$ such that $Z i − α / 2$ has distribution Gamma of parameters γ and $λ i ( i = 1 , 2 ) ,$ where $λ 1 = β − β 2 − 2 λ$ and $λ 2 = β + β 2 − 2 λ .$
Remark 5.
If f is given by Equation (20), that is $f ˜$ is the Gamma density, the compatibility condition in Equation (14) becomes $β ≥ λ ,$ which is satisfied under the assumption $β ≥ 2 λ$ required by Proposition 1. In the special case when $γ = 1 ,$ then η has the same distribution as $α + Z 1 + Z 2 ,$ where $Z i$ are independent and exponential with parameter $λ i , i = 1 , 2 .$
The following result also follows from Proposition 2.5 of [2].
Proposition 2.
Let be $S ( t ) = α + β ρ ( t ) ,$ with $α , β ≥ 0 ;$ for $β > 0 ,$ suppose that the Laplace transform of $f ˜$ has the form:
$f ˜ ^ ( θ ) = ∑ k = 1 N A k ( θ + B k ) c k ,$
for some $c k > 0 , A k , B k > 0 , k = 1 , ⋯ , N .$ Then, there exists a value $β * > 0$ such that the solution to the IFPT problem exists, provided that $β ≥ b * .$
If $β = 0$ and the Laplace transform of $f ˜$ has the form:
$f ˜ ^ ( θ ) = ∑ k = 1 N A k ( 2 θ + B k ) c k ,$
then, the solution to the IFPT problem exists.

#### 2.2. Approximate Solution to the IFPT Problem for Non Linear Boundaries

Now, we suppose that there exist $α 1 , α 2 , β 1 , β 2$ with $0 ≤ α 1 ≤ α 2$ and $β 2 ≥ β 1 ≥ 0 ,$ such that, for every $t ≥ 0 :$
$α 1 + β 1 ρ ( t ) ≤ S ( t ) ≤ α 2 + β 2 ρ ( t ) ,$
namely $S ( t )$ is enveloped from above and below by the functions $S α 2 , β 2 ( t ) = α 2 + β 2 ρ ( t )$ and $S α 1 , β 1 ( t ) = α 1 + β 1 ρ ( t ) .$
Then, by using Proposition (3.13) of [16] (see also [1]), we obtain the following:
Proposition 3.
Let $S ( t )$ a continuous, increasing boundary satisfying Equation (24) and suppose that the FPT τ of $X ( t ) = η + B ( ρ ( t ) ) ( η > S ( 0 ) )$ below the boundary $S ( t )$ has an assigned probability density f and that there exists a density g with support $( S ( 0 ) , + ∞ ) ,$ which is solution to the IFPT problem for $X ( t )$ and the boundary $S ( t ) ;$ as before, denote by $f ˜ ( t )$ the density of $ρ ( τ )$ and by $f ˜ ^ ( θ )$ its Laplace transform, for $θ > 0 .$ Then:
(i)
If $α 2 > α 1$ and the function $g ∈ L p ( S ( 0 ) , α 2 )$ for some $p > 1 ,$ its Laplace transform $g ^ ( θ )$ must satisfy:
$e − α 2 ( θ + 2 ( β 2 − β 1 ) ) f ˜ ^ θ ( θ + 2 β 2 ) 2 − ( α 2 − S ( 0 ) ) p − 1 p ∫ S ( 0 ) α 2 g p ( x ) d x 1 / p ≤ g ^ ( θ )$
$≤ e − α 1 θ f ˜ ^ θ ( θ + 2 β 1 ) 2 ;$
(ii)
If $α 1 = α 2 = S ( 0 ) ,$ then Equation (25) holds without any further assumption on g (and the term $( α 2 − S ( 0 ) ) p − 1 p ∫ S ( 0 ) α 2 g p ( x ) d x 1 / p$ vanishes).
Remark 6.
The smaller $α 2 − α 1$ and $β 2 − β 1 ,$ the better the approximation to the Laplace transform of $g .$ Notice that, if g is bounded, then the term $( α 2 − S ( 0 ) ) p − 1 p ∫ S ( 0 ) α 2 g p ( x ) d x 1 / p$ can be replaced with $( α 2 − S ( 0 ) ) | | g | | ∞ .$

#### 2.3. The IFPT Problem for $X ¯ ( t ) = η + B ( ρ ( t ) ) +$ Large Jumps

As an application of the previous results, we consider now the piecewise-continuous process $X ¯ ( t )$, obtained by superimposing to $X ( t )$ a jump process, namely we set $X ¯ ( t ) = η + B ( ρ ( t ) )$ for $t < T ,$ where T is an exponential distributed time with parameter $μ > 0 ;$ we suppose that, for $t = T$ the process $X ¯ ( t )$ makes a downward jump and it crosses the continuous increasing boundary $S ,$ irrespective of its state before the occurrence of the jump. This kind of behavior is observed e.g. in the presence of a so called catastrophes (see e.g., [17]). For $η ≥ S ( 0 ) ,$ we denote by $τ ¯ S = inf { t > 0 : X ¯ ( t ) ≤ S ( t ) }$ the FPT of $X ¯ ( t )$ below the boundary $S ( t ) .$ The following holds:
Proposition 4.
If there exists a solution $g ¯$ to the IFPT problem of $X ¯ ( t )$ below $S ( t )$ with $X ¯ ( 0 ) = η ≥ S ( 0 ) ,$ then its Laplace transform is given by
$g ¯ ^ ( θ ) = E e − θ S ( τ ) − θ 2 2 ρ ( τ ) − μ τ + μ ∫ 0 + ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) − μ t ∫ t + ∞ f ( s ) d s d t .$
Proof.
For $t > 0 ,$ one has:
$P ( τ ¯ S ≤ t ) = P ( τ ¯ S ≤ t | t < T ) P ( t < T ) + 1 · P ( t ≥ T ) = P ( τ S ≤ t ) e − μ t + ( 1 − e − μ t ) .$
Taking the derivative, one obtains the FPT density of $τ ¯ :$
$f ¯ ( t ) = e − μ t f ( t ) + μ e − μ t ∫ t + ∞ f ( s ) d s ,$
where f is the density of $τ .$ Then, by the same arguments used in the proof of Theorem 1, we obtain
$g ¯ ^ ( θ ) = E e − θ S ( τ ¯ ) − θ 2 2 ρ ( τ ¯ )$
$= ∫ 0 ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) f ¯ ( t ) d t$
$= ∫ 0 ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) e − μ t f ( t ) + μ e − μ t ∫ t ∞ f ( s ) d s d t$
$= ∫ 0 ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) − μ t f ( t ) d t + μ ∫ 0 ∞ e − θ S ( t ) − θ 2 2 ρ ( t ) − μ t ∫ t ∞ f ( s ) d s d t$
that is Equation (26). ☐
Remark 7.
(i)
For $μ = 0 ,$ namely when no jump occurs, Equation (26) becomes Equation (4).
(ii)
If τ is exponentially distributed with parameter $λ ,$ then Equation (26) provides:
$g ¯ ^ ( θ ) = λ + μ λ E e − θ S ( τ ) − θ 2 2 ρ ( τ ) − μ τ .$
(iii)
In the special case when $S ( t ) = α + β ρ ( t ) ( α , β ≥ 0 ) ,$ we can reduce to the FPT $τ ˜ ¯$ of BM + large jumps below the linear boundary $α + β t ;$ then, it is possible to write $g ¯ ^$ in terms of the Laplace transform of $τ ˜ ¯ .$ Really, by using Proposition 3.10 of [16] one gets
$g ¯ ^ ( θ ) = e − α θ 1 − 2 μ θ ( θ + 2 β ) − 1 f ¯ ^ θ ( θ + 2 β ) 2 − μ − 2 μ θ ( θ + 2 β ) − 2 μ ,$
where, for simplicity of notation we have denoted again with $f ¯ ^$ the Laplace transform of $τ ˜ ¯ ;$ of course, if $ρ ( t ) = t ,$ then $f ¯ ^$ is the Laplace transform of $τ ¯ .$ Notice that, if $μ = 0$ the last equation is nothing but Equation (5) with $α , β$ in place of $a , b .$

## 3. Some Examples

Example 1.
If $S ( t ) = a + b t ,$ with $a , b ≥ 0 ,$ and $X ( t ) = B t ( ρ ( t ) = 1 ) ,$ examples of solution to the IFPT problem, for $X ( t )$ and various FPT densities $f ,$ can be found in [2].
Example 2.
Let be $S ( t ) = α + β ρ ( t ) ,$ with $α , β ≥ 0 ,$ and suppose that τ has density $f ( t ) = λ e − ρ ( t ) ρ ′ ( t ) 1 ( 0 , + ∞ ) ( t )$ (that is, the density $f ˜$ of $τ ˜ = ρ ( τ )$ is exponential with parameter $λ ) .$ By using Proposition 1 we get that $η = α + Z 1 + Z 2 ,$ where $Z i$ are independent random variable, such that $Z i − α / 2$ has exponential distribution with parameter $λ i$ ($i = 1 , 2 ) ,$ where $λ 1 = β − β 2 − 2 λ$ and $λ 2 = β + β 2 − 2 λ .$ Then, the solution g to the IFPT problem for $X ( t ) = η + B ( ρ ( t ) ) ,$ the boundary S and the exponential FPT distribution, is:
$g ( x ) = λ 1 λ 2 λ 2 − λ 1 e − λ 1 ( x − α ) − e − λ 2 ( x − α ) , if b > 2 λ 2 λ ( x − α ) e − 2 λ ( x − a ) , if b = 2 λ . ( x ≥ α )$
In general, for a given continuous increasing boundary $S ( t )$ and an assigned distribution of $τ ,$ it is difficult to calculate explicitly the expectation on the right-hand member of Equation (4) to get the Laplace transform of $η .$ Thus, a heuristic solution to the IFPT problem can be achieved by using Equation (4) to calculate the moments of $η$ (those up to the fifth order are given by Equations (11), (12) and (15)–(17)). Of course, even if one was able to find the moments of $η$ of any order, this would not determinate the distribution of $η .$ However, this procedure is useful to study the properties of the distribution of $η ,$ provided that the solution to the IFPT problem exists.
Example 3.
Let be $S ( t ) = t 2 , ρ ( t ) = t$ and suppose that τ is exponentially distributed with parameter $λ ;$ we search for a solution $η > 0$ to the IFPT problem by using the method of moments, described above. The compatibility condition in Equation (14) requires that $λ 3 < 20$ (for instance, one can take $λ = 1 ) .$ From Equations (11), (12) and (15)–(17), and calculating the moments of τ up to the eighth order, we obtain:
$E ( η ) = E ( τ 2 ) = 2 λ 2 ; E ( η 2 ) = E ( τ 4 ) − E ( τ ) = 24 − λ 3 λ 4 ; σ 2 ( η ) = V a r ( η ) = 20 − λ 3 λ 4 ;$
$E ( η 3 ) = E ( τ 6 ) − 3 E ( τ 3 ) = 720 − 18 λ 3 λ 6 ; E ( η 4 ) = E ( τ 8 ) − 6 E ( τ 3 ) + 3 E ( τ 2 ) = 8 ! − 36 λ 5 + 6 λ 6 λ 8 .$
Notice that, under the condition $λ 3 < 20$ the first four moments of η are positive, as it must be. However, they do not match those of a Gamma distribution.
An information about the asymmetry is given by the skewness value
$E ( η − E ( η ) ) 3 σ ( η ) 3 = − 12 24 − λ 3 ( 20 − λ 3 ) 3 / 2 < 0 ,$
meaning that the candidate η has an asymmetric distribution with a tail toward the left.

## 4. Conclusions and Final Remarks

We have dealt with the IFPT problem for a continuously time-changed Brownian motion $X ( t )$ starting from a random position $η .$ For a given continuous, increasing boundary $S ( t )$ with $η ≥ S ( 0 ) ≥ 0 ,$ and an assigned continuous distribution function $F ,$ the IFPT problem consists in finding the distribution, or the density g of $η ,$ such that the first-passage time $τ$ of $X ( t )$ below $S ( t )$ has distribution $F .$ In this note, we have provided some extensions of the results, already known in the case when $X ( t )$ is BM and $S ( t )$ is a straight line, and we have reported some explicit examples. Really, the process we considered has the form $X ( t ) = η + ∫ 0 t σ ( s ) d B s ,$ where $B t$ is standard Brownian motion, and $σ ( t )$ is a non random continuous function of time $t ≥ 0 ,$ such that the function $ρ ( t ) = ∫ 0 t σ 2 ( s ) d s$ is increasing and it satisfies the condition $ρ ( + ∞ ) = + ∞ .$ Thus, a standard BM $B ^$ exists such that $X ( t ) = η + B ^ ( ρ ( t ) ) .$ Our main result states that
$g ^ ( θ ) = E e − θ S ( τ ) − θ 2 2 ρ ( τ ) ,$
where, for $θ > 0 , g ^ ( θ )$ denotes the Laplace transform of the solution g to the IFPT problem.
Notice that the above result can be extended to diffusions which are more general than the process $X ( t )$ considered, for instance to a process of the form
$U ( t ) = w − 1 ( B ^ ( ρ ( t ) ) + w ( η ) ) ,$
where w is a regular enough, increasing function; such a process U is obtained from BM by a space transformation and a continuous time-change (see e.g., the discussion in [2]). Since $w ( U ( t ) ) = w ( η ) + B ^ ( ρ ( t ) ) ,$ the IFPT problem for the process $U ,$ the boundary $S ( t )$ and the FPT distribution $F ,$ is reduced to the analogous IFPT problem for $X ( t ) = η 1 + B ^ ( ρ ( t ) ) ,$ starting from $η 1 = w ( η ) ,$ instead of $η ,$ the boundary $S 1 ( t ) = w ( S ( t ) )$ and the same FPT distribution $F .$ When $σ ( t ) = 1 ,$ i.e. $ρ ( t ) = t ,$ the process $U ( t )$ is conjugated to BM, according to the definition given in [2]; two examples of diffusions conjugated to BM are the Feller process, and the Wright–Fisher like (or CIR) process, (see e.g., [2]). The process $U ( t )$ given by Equation (32) is indeed a weak solution of the SDE:
$d U ( t ) = − ρ ′ ( t ) w ″ ( U ( t ) ) 2 ( w ′ ( U ( t ) ) ) 3 d t + ρ ′ ( t ) w ′ ( U ( t ) ) d B t ,$
where $w ′ ( x )$ and $w ″ ( x )$ denote first and second derivative of $w ( x ) .$
Provided that the deterministic function $ρ ( t )$ is replaced with a random function, the representation in Equation (32) is valid also for a time homogeneous one-dimensional diffusion driven by the SDE
$d U ( t ) = μ ( U ( t ) ) d t + σ ( U ( t ) ) d B t , U ( 0 ) = η ,$
where the drift $( μ )$ and diffusion coefficients $( σ )$ satisfy the usual conditions (see e.g., [18]) for existence and uniqueness of the solution of Equation (34). In fact, let $w ( x )$ be the scale function associated to the diffusion $U ( t )$ driven by the SDE Equation (34), that is, the solution of $L w ( x ) = 0 , w ( 0 ) = 0 , w ′ ( 0 ) = 1 ,$ where L is the infinitesimal generator of U given by $L h = 1 2 σ 2 ( x ) d 2 h d x 2 + μ ( x ) d h d x .$ As easily seen, if the integral $∫ 0 t 2 μ ( z ) σ 2 ( z ) d z$ converges, the scale function is explicitly given by
$w ( x ) = ∫ 0 x exp − ∫ 0 t 2 μ ( z ) σ 2 ( z ) d z d t .$
If $ζ ( t ) : = w ( U ( t ) ) ,$ by It$o ^$’s formula one obtains
$ζ ( t ) = w ( η ) + ∫ 0 t w ′ ( w − 1 ( ζ ( s ) ) ) σ ( w − 1 ( ζ ( s ) ) ) d B s ,$
that is, the process $ζ ( t )$ is a local martingale, whose quadratic variation is
$ρ ( t ) ≐ 〈 ζ 〉 t = ∫ 0 t [ w ′ ( U ( s ) ) σ ( U ( s ) ) ] 2 d s , t ≥ 0 .$
The (random) function $ρ ( t )$ is differentiable and $ρ ( 0 ) = 0 ;$ if it is increasing to $ρ ( + ∞ ) = + ∞ ,$ by the Dambis, Dubins–Schwarz theorem (see e.g., [3]) one gets that there exists a standard BM $B ^$ such that $ζ ( t ) = B ^ ( ρ ( t ) ) + w ( η ) .$ Thus, since w is invertible, one obtains the representation in Equation (32).
Notice, however, that the IFPT problem for the process U given by Equation (32) cannot be addressed as in the case when $ρ$ is a deterministic function. In fact, if $ρ ( t )$ given by Equation (37) is random, it results that $ρ ( t )$ and the FPT $τ$ are dependent. Thus, in line of principle it would be possible to obtain information about the Laplace transform of $g ,$ only in the case when the joint distribution of $( ρ ( t ) , τ )$ was explicitly known.

## Funding

This research was funded by the MIUR Excellence Department Project awarded tothe Department of Mathematics, University of Rome Tor Vergata, CUP E83C18000100006.

## Acknowledgments

I would like to express particular thanks to the anonymous referees for their constructive comments and suggestions leading to improvements of the paper.

## Conflicts of Interest

The author declares no conflict of interest.

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