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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 $X\left(t\right)$ be a continuously time-changed Brownian motion starting from a random position $\eta ,\phantom{\rule{4pt}{0ex}}S\left(t\right)$ a given continuous, increasing boundary, with $S\left(0\right)\ge 0,$$P(\eta \ge S(0\left)\right)=1,$ and F an assigned distribution function. We study the inverse first-passage time problem for $X\left(t\right),$ which consists in finding the distribution of $\eta $ such that the first-passage time of $X\left(t\right)$ below $S\left(t\right)$ has distribution $F,$ generalizing the results, valid in the case when $S\left(t\right)$ 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 $\sigma \left(t\right)$ be a regular enough non random function, and let $X\left(t\right)=\eta +{\int}_{0}^{t}\sigma \left(s\right)d{B}_{s},$ where ${B}_{t}$ is standard Brownian motion (BM) and the initial position $\eta $ is a random variable, independent of ${B}_{t}.$ Suppose that the quadratic variation $\rho \left(t\right)={\int}_{0}^{t}{\sigma}^{2}\left(s\right)ds$ is increasing and $\rho (+\infty )=\infty ,$ then there exists a standard BM $\tilde{B}$ such that $X\left(t\right)=\eta +\tilde{B}\left(\rho \left(t\right)\right),$ namely $X\left(t\right)$ is a continuously time-changed BM (see e.g., [3]). For a continuous, increasing boundary $S\left(t\right),$ such that $P(\eta \ge S(0\left)\right)=1,$ let
be the first-passage time (FPT) of $X\left(t\right)$ below $S.$ We assume that $\tau $ is finite with probability one and that it possesses a density $f\left(t\right)=\frac{dF\left(t\right)}{dt},$ where $F\left(t\right)=P(\tau \le t).$ 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]).

$$\tau ={\tau}_{S}=inf\{t>0:X\left(t\right)\le S\left(t\right)\}$$

Assuming that $S\left(t\right)$ is increasing, and $F\left(t\right)$ 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 $P(\tau \le t)=F\left(t\right)$.

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 $X\left(t\right)$ is BM and $S\left(t\right)$ 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]).

## 2. Main Results

The following holds:

**Theorem**

**1.**

Let be $S\left(t\right)$ a continuous, increasing boundary with $S\left(0\right)\ge 0,\phantom{\rule{4pt}{0ex}}\sigma \left(t\right)$ a bounded, non random continuous function of $t>0,$ and let $X\left(t\right)=\eta +{\int}_{0}^{t}\sigma \left(s\right)d{B}_{s}$ be the integral process starting from the random position $\eta \ge S\left(0\right);$ we assume that $\rho \left(t\right)={\int}_{0}^{t}{\sigma}^{2}\left(s\right)ds$ is increasing and satisfies $\rho (+\infty )=+\infty .$ Let F be the probability distribution of the FPT ${\tau}_{S}$ of X below the boundary S (${\tau}_{S}$ is a.s. finite by virtue of Remark 3). We suppose that the r.v. η admits a density $g\left(x\right);$ for $\theta >0,$ we denote by $\widehat{g}\left(\theta \right)=E\left({e}^{-\theta \eta}\right)$ the Laplace transform of $g.$

Then, if there exists a solution to the IFPT problem for $X,$ the following relation holds:

$$\widehat{g}\left(\theta \right)={\int}_{0}^{+\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)}dF\left(t\right).$$

**Proof.**

The process $X\left(t\right)$ is a martingale, we denote by ${\mathcal{F}}_{t}$ its natural filtration. Thanking to the hypothesis, by using the Dambis, Dubins–Schwarz theorem (see e.g., [3]), it follows that the process $\tilde{B}\left(t\right)=X\left({\rho}^{-1}\left(t\right)\right)$ is a Brownian motion with respect to the filtration ${\mathcal{F}}_{{\rho}^{-1}\left(t\right)};$ so the process $X\left(t\right)$ can be written as $X\left(t\right)=\eta +\tilde{B}\left(\rho \left(t\right)\right)$ and the FPT $\tau $ can be written as $\tau =inf\{t>0:\eta +\tilde{B}\left(\rho \left(t\right)\right)\le S\left(t\right)\}.$
For $\theta >0,$ let us consider the process ${Z}_{t}={e}^{-\theta X\left(t\right)-\frac{1}{2}{\theta}^{2}\rho \left(t\right)};$ as easily seen, ${Z}_{t}$ is a positive martingale; indeed, it can be represented as ${Z}_{t}={e}^{-\theta X\left(0\right)}-\theta {\int}_{0}^{t}{Z}_{s}\sigma \left(s\right)d{B}_{s}$ (see e.g., Theorem 5.2 of [14]). We observe that, for $t\le \tau $ the martingale ${Z}_{t}$ is bounded, because $X\left(t\right)$ is non negative and therefore $0<{Z}_{t}\le {e}^{-\theta X\left(t\right)}\le 1.$ Then, by using the fact that, for any finite stopping time $\tau $ one has $E\left[{Z}_{0}\right]=E\left[{Z}_{\tau \wedge t}\right]$ (see e.g., Formula (7.7) in [14]), and the dominated convergence theorem, we obtain that

$$E\left[{Z}_{0}\right]=E\left[{e}^{-\theta X\left(0\right)}\right]=E\left[{e}^{-\theta \eta}\right]=\underset{t\to \infty}{lim}E\left[{e}^{-\theta X(\tau \wedge t)-\frac{1}{2}{\theta}^{2}\rho (\tau \wedge t)}\right]$$

$$=E\left[\underset{t\to \infty}{lim}{e}^{-\theta X(\tau \wedge t)-\frac{1}{2}{\theta}^{2}\rho (\tau \wedge t)}\right]=E\left[{e}^{-\theta S\left(\tau \right)-\frac{1}{2}{\theta}^{2}\rho \left(\tau \right)}\right].$$

Thus, if $\widehat{g}\left(\theta \right)=E\left({e}^{-\theta \eta}\right)$ is the Laplace transform of the density of the initial position $\eta ,$ we finally get
that is Equation (2). ☐

$$\widehat{g}\left(\theta \right)=E\left[{e}^{-\theta S\left(\tau \right)-\frac{{\theta}^{2}}{2}\rho \left(\tau \right)}\right],$$

**Remark**

**1.**

If one takes in place of $X\left(t\right)$ a process of the form $\tilde{X}\left(t\right)=\eta S\left(t\right)+S\left(t\right)B\left(\rho \left(t\right)\right),$ with $\eta \ge 1,$ that is, a special case of continuous Gauss-Markov process ([15]) with mean $\eta S\left(t\right),$ then $\tilde{X}\left(t\right)/S\left(t\right)$ is still a continuously time-changed BM, and so the IFPT problem for $\tilde{X}\left(t\right)$ and $S\left(t\right)$ 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 $\tau .$ 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 $\widehat{g}$ 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 $\eta .$ In line of principle, the right-hand member of Equation (4) can be expressed in terms of the Laplace transform of $f\left(t\right)={F}^{\prime}\left(t\right),$ though it is not always possible to do this explicitly. A simple case is when $S\left(t\right)=a+bt,$ with $a,b\ge 0,$ and $\rho \left(t\right)=t,$ that is, $X\left(t\right)={B}_{t}\phantom{\rule{4pt}{0ex}}(\sigma \left(t\right)=1);$ in fact, one obtains
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 $\tau .$

$$\widehat{g}\left(\theta \right)=E\left[{e}^{-\theta (a+b\tau )-\frac{{\theta}^{2}}{2}\tau}\right]={e}^{-\theta a}E\left[{e}^{-\theta (b+\frac{\theta}{2})\tau}\right]={e}^{-\theta a}\widehat{f}\left(\frac{\theta (\theta +2b)}{2}\right),$$

**Remark**

**3.**

Let $S\left(t\right)$ be increasing and $S\left(0\right)\ge 0,$ then τ is a.s. finite; in fact $\tilde{\tau}=\rho \left(\tau \right)=inf\{t>0:\eta +{\tilde{B}}_{t}\le \tilde{S}\left(t\right)\}\le {\tilde{\tau}}_{1},$ where $\tilde{S}\left(t\right)=S\left({\rho}^{-1}\left(t\right)\right)$ is increasing and ${\tilde{\tau}}_{1}$ is the first hitting time to $S\left(0\right)$ of BM $\tilde{B}$ starting at $\eta ;$ since ${\tilde{\tau}}_{1}$ is a.s. finite, also $\tilde{\tau}$ is so. Next, from the finiteness of $\tilde{\tau}$ it follows that $\tau ={\rho}^{-1}\left(\tilde{\tau}\right)$ is finite, too. Moreover, if one seeks that $E\left(\tau \right)<\infty ,$ a sufficient condition for this is that $\rho \left(t\right)$ and $\tilde{S}\left(t\right)$ are both convex functions; indeed, $\tilde{\tau}\le {\tilde{\tau}}_{2},$ where ${\tilde{\tau}}_{2}$ is the FPT of BM $\tilde{B}$ starting from η below the straight line $a+bt\phantom{\rule{4pt}{0ex}}(a=S\left(0\right)\ge 0,\phantom{\rule{4pt}{0ex}}b={\tilde{S}}^{\prime}\left(0\right)\ge 0)$ which is tangent to the graph of $\tilde{S}\left(t\right)$ at $t=0.$ Thus, since $E\left({\tilde{\tau}}_{2}\right)<\infty ,$ it follows that $E\left(\tilde{\tau}\right)$ is finite, too; finally, being ${\rho}^{-1}$ concave, Jensen’s inequality for concave functions implies that $E\left(\tau \right)=E\left({\rho}^{-1}\left(\tilde{\tau}\right)\right)\le {\rho}^{-1}\left(E\left(\tilde{\tau}\right)\right)$ and therefore $E\left(\tau \right)<\infty .$

**Remark**

**4.**

Theorem 1 allows to solve also the so called Skorokhod embedding (SE) problem:

Given a distribution $H,$ find an integrable stopping time ${\tau}^{*},$ such that the distribution of $X\left({\tau}^{*}\right)$ is $H,$ namely $P(X\left({\tau}^{*}\right)\le x)=H\left(x\right).$

In fact, let be $S\left(t\right)$ increasing, with $S\left(0\right)=0;$ first suppose that the support of H is $[0,+\infty );$ then, from Equation (4) it follows that
and this solves the SE problem with ${\tau}^{*}=\tau ;$ it suffices to take the random initial point $X\left(0\right)=\eta >0$ in such a way that its Laplace transform $\widehat{g}$ satisfies

$$\widehat{g}\left(\theta \right)=E\left[{e}^{-\theta X\left(\tau \right)-\frac{{\theta}^{2}}{2}\rho \left({S}^{-1}\left(X\left(\tau \right)\right)\right)}\right],$$

$$\widehat{g}\left(\theta \right)={\int}_{0}^{S(+\infty )}{e}^{-\theta x-\frac{{\theta}^{2}}{2}\rho \left({S}^{-1}\left(x\right)\right)}dH\left(x\right).$$

In the special case when $S\left(t\right)=a+bt\phantom{\rule{4pt}{0ex}}(a,b>0)$ and $\rho \left(t\right)=t,$ Equation (7) becomes (cf. the result in [8] for $a=0):$
where $h\left(x\right)={H}^{\prime}\left(x\right)$ and $\widehat{h}$ denotes the Laplace transform of $h.$

$$\widehat{g}\left(\theta \right)={e}^{\frac{a{\theta}^{2}}{2b}}\widehat{h}\left(\frac{\theta (\theta +2b)}{2b}\right),$$

In analogous way, the SE problem can be solved if the support of H is $(-\infty ,0];$ now, the FPT is understood as ${\tau}^{-}=inf\{t>0:\eta +B\left(\rho \left(t\right)\right)>-S\left(t\right)\}\phantom{\rule{4pt}{0ex}}(\eta <0),$ that is, the first hitting time to the boundary ${S}^{-}\left(t\right)=-S\left(t\right)$ 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 $S\left(t\right)$ is a straight line).

The r.v. $X\left(\tau \right)$ can be represented as a mixture of the r.v. ${X}^{+}>0$ and ${X}^{-}<0:$

$$X\left(\tau \right)=\left\{\begin{array}{cc}{X}^{+}\hfill & \phantom{\rule{4pt}{0ex}}\mathrm{with}\phantom{\rule{4pt}{0ex}}\mathrm{probability}\phantom{\rule{4pt}{0ex}}{p}^{+}=P(X\left(\tau \right)\ge 0)\hfill \\ {X}^{-}\hfill & \phantom{\rule{4pt}{0ex}}\mathrm{with}\phantom{\rule{4pt}{0ex}}\mathrm{probability}\phantom{\rule{4pt}{0ex}}{p}^{-}=1-{p}^{+}.\hfill \end{array}\right.$$

Suppose that the SE problem for the r.v. ${X}^{+}$ and ${X}^{-}$ can be solved by ${S}^{+}\left(t\right)=S\left(t\right)$ and ${\eta}^{+}=\eta >0,$ and ${S}^{-}\left(t\right)=-S\left(t\right)$ and ${\eta}^{-}=-\eta <0,$ respectively. Then, we get that the r.v.
and the boundary ${S}^{\pm}\left(t\right)={S}^{+}\left(t\right)\cup {S}^{-}\left(t\right)$ solve the SE problem for the r.v. $X\left(\tau \right).$

$${\eta}^{\pm}=\left\{\begin{array}{cc}{\eta}^{+}\hfill & \phantom{\rule{4pt}{0ex}}\mathrm{with}\phantom{\rule{4pt}{0ex}}\mathrm{probability}\phantom{\rule{4pt}{0ex}}{p}^{+}\hfill \\ {\eta}^{-}\hfill & \phantom{\rule{4pt}{0ex}}\mathrm{with}\phantom{\rule{4pt}{0ex}}\mathrm{probability}\phantom{\rule{4pt}{0ex}}{p}^{-}\hfill \end{array}\right.$$

If $\widehat{g}$ is analytic in a neighbor of $\theta =0,$ then the moments of order n of $\eta ,\phantom{\rule{4pt}{0ex}}E\left({\eta}^{n}\right),$ exist finite, and they are given by $E\left({\eta}^{n}\right)={(-1)}^{n}\frac{{d}^{n}}{d{\theta}^{n}}\widehat{g}{|}_{\theta =0}.$ By taking the first derivative in Equation (4) and calculating it at $\theta =0,$ we obtain

$$E\left(\eta \right)=-{\widehat{g}}^{\prime}\left(0\right)=E\left(S\left(\tau \right)\right).$$

By calculating the second derivative of $\widehat{g}$ at $\theta =0,$ we get
and so

$$E\left({\eta}^{2}\right)={\widehat{g}}^{\prime \prime}\left(0\right)=E({S}^{2}\left(\tau \right)-\rho \left(\tau \right))),$$

$$Var\left(\eta \right)=E\left({\eta}^{2}\right)-{E}^{2}\left(\eta \right)=Var\left(S\left(\tau \right)\right)-E\left(\rho \left(\tau \right)\right).$$

Thus, we obtain the compatibility conditions

$$\left\{\begin{array}{c}E\left(\eta \right)=E\left(S\right(\tau \left)\right)\hfill \\ Var\left(S\right(\tau \left)\right)\ge E\left(\rho \right(\tau \left)\right).\hfill \end{array}\right.$$

If $Var\left(S\right(\tau \left)\right)<E\left(\rho \right(\tau \left)\right),$ a solution to the IFPT problem does not exist. In the special case when $S\left(t\right)=a+bt\phantom{\rule{4pt}{0ex}}(a,\phantom{\rule{4pt}{0ex}}b\ge 0)$ and $\rho \left(t\right)=t,$ Equation (11) becomes $E\left(\eta \right)=a+bE\left(\tau \right)$ and Equation (13) becomes $Var\left(\eta \right)={b}^{2}Var\left(\tau \right)-E\left(\tau \right),$ while Equation (14) coincides with Equation (2.3) of [2]. By writing the Taylor’s expansions at $\theta =0$ of both members of Equation (4), and equaling the terms with the same order in $\theta ,$ one gets the successive derivatives of $\widehat{g}\left(\theta \right)$ at $\theta =0;$ thus, one can write any moment of $\eta $ in terms of the expectation of a function of $\tau ;$ for instance, it is easy to see that

$$E\left({\eta}^{3}\right)=E\left[{\left(S\left(\tau \right)\right)}^{3}\right]-3E\left[S\left(\tau \right)\rho \left(\tau \right)\right],$$

$$E\left({\eta}^{4}\right)=E[\left(S{\left(\tau \right)}^{4}\right]-6E[\left(S{\left(\tau \right)}^{2}\rho \left(\tau \right)\right]+3E[\left(\rho {\left(\tau \right)}^{2}\right],$$

$$E\left({\eta}^{5}\right)=E[15S\left(\tau \right){\rho}^{2}\left(\tau \right)-240{S}^{3}\left(\tau \right)\rho \left(\tau \right)+{S}^{5}\left(\tau \right)].$$

#### 2.1. The Special Case $S\left(t\right)=\alpha +\beta \rho \left(t\right)$

If $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\phantom{\rule{4pt}{0ex}}\beta \ge 0,$ from Equation (4) we get

$$\widehat{g}\left(\theta \right)=E\left[{e}^{-\theta (\alpha +\beta \rho \left(\tau \right))-\frac{{\theta}^{2}}{2}\rho \left(\tau \right)}\right]={e}^{-\theta \alpha}E\left[{e}^{-\theta \rho \left(\tau \right)(\beta +\theta /2)}\right].$$

Thus, setting $\tilde{\tau}=\rho \left(\tau \right),$ we obtain (see Equation (5)):
having denoted by $\tilde{f}$ the density of $\tilde{\tau}.$ In this way, we reduce the IFPT problem of $X\left(t\right)=\eta +B\left(\rho \right(t\left)\right)$ below the boundary $S\left(t\right)=\alpha +\beta \rho \left(t\right)$ to that of BM below the linear boundary $\alpha +\beta t.$ For instance, taking $\rho \left(t\right)={t}^{3}/3,$ the solution to the IFPT problem of $X\left(t\right)$ through the cubic boundary $S\left(t\right)=\alpha +\frac{\beta}{3}{t}^{3},$ and the FPT density $f,$ is nothing but the solution to the IFPT problem of BM through the linear boundary $\alpha +\beta t,$ and the FPT density $\tilde{f}.$

$$\widehat{g}\left(\theta \right)={e}^{-\theta \alpha}E\left[{e}^{-\theta (\beta +\theta /2)\tilde{\tau}}\right]={e}^{-\theta \alpha}\widehat{\tilde{f}}\left(\theta (\beta +\theta /2)\right),$$

Under the assumption that $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\phantom{\rule{4pt}{0ex}}\beta \ge 0,$ 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 $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\phantom{\rule{4pt}{0ex}}\beta \ge 0;$ for $\gamma ,\phantom{\rule{4pt}{0ex}}\lambda >0,$ suppose that the FPT density $f={F}^{\prime}$ is given by
(namely the density $\tilde{f}$ of $\tilde{\tau}$ is the Gamma density with parameters $(\gamma ,\lambda )).$ Then, the IFPT problem has solution, provided that $\beta \ge \sqrt{2\lambda},$ and the Laplace transform of the density g of the initial position η is given by:
which is the Laplace transform of the sum of two independent random variables, ${Z}_{1}$ and ${Z}_{2},$ such that ${Z}_{i}-\alpha /2$ has distribution Gamma of parameters γ and ${\lambda}_{i}\phantom{\rule{4pt}{0ex}}(i=1,2),$ where ${\lambda}_{1}=\beta -\sqrt{{\beta}^{2}-2\lambda}$ and ${\lambda}_{2}=\beta +\sqrt{{\beta}^{2}-2\lambda}.$

$$f\left(t\right)=\left\{\begin{array}{cc}\frac{{\lambda}^{\gamma}}{\Gamma \left(\gamma \right)}\rho {\left(t\right)}^{\gamma -1}{e}^{-\lambda \rho \left(t\right)}{\rho}^{\prime}\left(t\right)\hfill & if\phantom{\rule{4pt}{0ex}}t>0\hfill \\ 0\hfill & otherwise\hfill \end{array}\right.$$

$$\widehat{g}\left(\theta \right)=\left[{e}^{-\alpha \theta /2}\frac{{(\beta -\sqrt{{\beta}^{2}-2\lambda})}^{\gamma}}{{(\theta +\beta -\sqrt{{\beta}^{2}-2\lambda})}^{\gamma}}\right]\xb7\left[{e}^{-\alpha \theta /2}\frac{{(\beta +\sqrt{{\beta}^{2}-2\lambda})}^{\gamma}}{{(\theta +\beta +\sqrt{{\beta}^{2}-2\lambda})}^{\gamma}}\right],$$

**Remark**

**5.**

If f is given by Equation (20), that is $\tilde{f}$ is the Gamma density, the compatibility condition in Equation (14) becomes $\beta \ge \sqrt{\lambda},$ which is satisfied under the assumption $\beta \ge \sqrt{2\lambda}$ required by Proposition 1. In the special case when $\gamma =1,$ then η has the same distribution as $\alpha +{Z}_{1}+{Z}_{2},$ where ${Z}_{i}$ are independent and exponential with parameter ${\lambda}_{i},\phantom{\rule{4pt}{0ex}}i=1,2.$

The following result also follows from Proposition 2.5 of [2].

**Proposition**

**2.**

Let be $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\phantom{\rule{4pt}{0ex}}\beta \ge 0;$ for $\beta >0,$ suppose that the Laplace transform of $\tilde{f}$ has the form:
for some ${c}_{k}>0,\phantom{\rule{4pt}{0ex}}{A}_{k},{B}_{k}>0,\phantom{\rule{4pt}{0ex}}k=1,\cdots ,N.$ Then, there exists a value ${\beta}^{*}>0$ such that the solution to the IFPT problem exists, provided that $\beta \ge {b}^{*}.$

$$\widehat{\tilde{f}}\left(\theta \right)=\sum _{k=1}^{N}\frac{{A}_{k}}{{(\theta +{B}_{k})}^{{c}_{k}}},$$

If $\beta =0$ and the Laplace transform of $\tilde{f}$ has the form:
then, the solution to the IFPT problem exists.

$$\widehat{\tilde{f}}\left(\theta \right)=\sum _{k=1}^{N}\frac{{A}_{k}}{{(\sqrt{2\theta}+{B}_{k})}^{{c}_{k}}},$$

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

Now, we suppose that there exist ${\alpha}_{1},{\alpha}_{2},{\beta}_{1},{\beta}_{2}$ with $0\le {\alpha}_{1}\le {\alpha}_{2}$ and ${\beta}_{2}\ge {\beta}_{1}\ge 0,$ such that, for every $t\ge 0:$
namely $S\left(t\right)$ is enveloped from above and below by the functions ${S}_{{\alpha}_{2},{\beta}_{2}}\left(t\right)={\alpha}_{2}+{\beta}_{2}\rho \left(t\right)$ and ${S}_{{\alpha}_{1},{\beta}_{1}}\left(t\right)={\alpha}_{1}+{\beta}_{1}\rho \left(t\right).$

$${\alpha}_{1}+{\beta}_{1}\rho \left(t\right)\le S\left(t\right)\le {\alpha}_{2}+{\beta}_{2}\rho \left(t\right),$$

**Proposition**

**3.**

Let $S\left(t\right)$ a continuous, increasing boundary satisfying Equation (24) and suppose that the FPT τ of $X\left(t\right)=\eta +B\left(\rho \right(t\left)\right)\phantom{\rule{4pt}{0ex}}(\eta >S(0\left)\right)$ below the boundary $S\left(t\right)$ has an assigned probability density f and that there exists a density g with support $\left(S\right(0),+\infty ),$ which is solution to the IFPT problem for $X\left(t\right)$ and the boundary $S\left(t\right);$ as before, denote by $\tilde{f}\left(t\right)$ the density of $\rho \left(\tau \right)$ and by $\widehat{\tilde{f}}\left(\theta \right)$ its Laplace transform, for $\theta >0.$ Then:

- (i)
- If ${\alpha}_{2}>{\alpha}_{1}$ and the function $g\in {L}^{p}(S\left(0\right),{\alpha}_{2})$ for some $p>1,$ its Laplace transform $\widehat{g}\left(\theta \right)$ must satisfy:$${e}^{-{\alpha}_{2}(\theta +2({\beta}_{2}-{\beta}_{1}))}\left[\widehat{\tilde{f}}\left(\frac{\theta (\theta +2{\beta}_{2})}{2}\right)-{({\alpha}_{2}-S\left(0\right))}^{\frac{p-1}{p}}{\left({\int}_{S\left(0\right)}^{{\alpha}_{2}}{g}^{p}\left(x\right)dx\right)}^{1/p}\right]\le \widehat{g}\left(\theta \right)$$$$\le {e}^{-{\alpha}_{1}\theta}\widehat{\tilde{f}}\left(\frac{\theta (\theta +2{\beta}_{1})}{2}\right);$$
- (ii)
- If ${\alpha}_{1}={\alpha}_{2}=S\left(0\right),$ then Equation (25) holds without any further assumption on g (and the term ${({\alpha}_{2}-S\left(0\right))}^{\frac{p-1}{p}}{\left({\int}_{S\left(0\right)}^{{\alpha}_{2}}{g}^{p}\left(x\right)dx\right)}^{1/p}$ vanishes).

**Remark**

**6.**

The smaller ${\alpha}_{2}-{\alpha}_{1}$ and ${\beta}_{2}-{\beta}_{1},$ the better the approximation to the Laplace transform of $g.$ Notice that, if g is bounded, then the term ${({\alpha}_{2}-S\left(0\right))}^{\frac{p-1}{p}}{\left({\int}_{S\left(0\right)}^{{\alpha}_{2}}{g}^{p}\left(x\right)dx\right)}^{1/p}$ can be replaced with $({\alpha}_{2}-S\left(0\right)){\left|\right|g\left|\right|}_{\infty}.$

#### 2.3. The IFPT Problem for $\overline{X}\left(t\right)=\eta +B\left(\rho \left(t\right)\right)+$ Large Jumps

As an application of the previous results, we consider now the piecewise-continuous process $\overline{X}\left(t\right)$, obtained by superimposing to $X\left(t\right)$ a jump process, namely we set $\overline{X}\left(t\right)=\eta +B\left(\rho \left(t\right)\right)$ for $t<T,$ where T is an exponential distributed time with parameter $\mu >0;$ we suppose that, for $t=T$ the process $\overline{X}\left(t\right)$ 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 $\eta \ge S\left(0\right),$ we denote by ${\overline{\tau}}_{S}=inf\{t>0:\overline{X}\left(t\right)\le S\left(t\right)\}$ the FPT of $\overline{X}\left(t\right)$ below the boundary $S\left(t\right).$ The following holds:

**Proposition**

**4.**

If there exists a solution $\overline{g}$ to the IFPT problem of $\overline{X}\left(t\right)$ below $S\left(t\right)$ with $\overline{X}\left(0\right)=\eta \ge S\left(0\right),$ then its Laplace transform is given by

$$\widehat{\overline{g}}\left(\theta \right)=E\left[{e}^{-\theta S\left(\tau \right)-\frac{{\theta}^{2}}{2}\rho \left(\tau \right)-\mu \tau}\right]+\mu {\int}_{0}^{+\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)-\mu t}\left({\int}_{t}^{+\infty}f\left(s\right)ds\right)dt.$$

**Proof.**

For $t>0,$ one has:

$$P({\overline{\tau}}_{S}\le t)=P({\overline{\tau}}_{S}\le t|t<T)P(t<T)+1\xb7P(t\ge T)=P({\tau}_{S}\le t){e}^{-\mu t}+(1-{e}^{-\mu t}).$$

Taking the derivative, one obtains the FPT density of $\overline{\tau}:$
where f is the density of $\tau .$ Then, by the same arguments used in the proof of Theorem 1, we obtain
that is Equation (26). ☐

$$\overline{f}\left(t\right)={e}^{-\mu t}f\left(t\right)+\mu {e}^{-\mu t}{\int}_{t}^{+\infty}f\left(s\right)ds,$$

$$\widehat{\overline{g}}\left(\theta \right)=E\left[{e}^{-\theta S\left(\overline{\tau}\right)-\frac{{\theta}^{2}}{2}\rho \left(\overline{\tau}\right)}\right]$$

$$={\int}_{0}^{\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)}\overline{f}\left(t\right)dt$$

$$={\int}_{0}^{\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)}\left[{e}^{-\mu t}f\left(t\right)+\mu {e}^{-\mu t}{\int}_{t}^{\infty}f\left(s\right)ds\right]dt$$

$$={\int}_{0}^{\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)-\mu t}f\left(t\right)dt+\mu {\int}_{0}^{\infty}{e}^{-\theta S\left(t\right)-\frac{{\theta}^{2}}{2}\rho \left(t\right)-\mu t}\left({\int}_{t}^{\infty}f\left(s\right)ds\right)dt$$

**Remark**

**7.**

- (i)
- (ii)
- If τ is exponentially distributed with parameter $\lambda ,$ then Equation (26) provides:$$\widehat{\overline{g}}\left(\theta \right)=\frac{\lambda +\mu}{\lambda}E\left[{e}^{-\theta S\left(\tau \right)-\frac{{\theta}^{2}}{2}\rho \left(\tau \right)-\mu \tau}\right].$$
- (iii)
- In the special case when $S\left(t\right)=\alpha +\beta \rho \left(t\right)\phantom{\rule{4pt}{0ex}}(\alpha ,\beta \ge 0),$ we can reduce to the FPT $\overline{\tilde{\tau}}$ of BM + large jumps below the linear boundary $\alpha +\beta t;$ then, it is possible to write $\widehat{\overline{g}}$ in terms of the Laplace transform of $\overline{\tilde{\tau}}.$ Really, by using Proposition 3.10 of [16] one gets$$\widehat{\overline{g}}\left(\theta \right)={e}^{-\alpha \theta}\left[{\left(1-\frac{2\mu}{\theta (\theta +2\beta )}\right)}^{-1}\phantom{\rule{4pt}{0ex}}\widehat{\overline{f}}\left(\frac{\theta (\theta +2\beta )}{2}-\mu \right)-\frac{2\mu}{\theta (\theta +2\beta )-2\mu}\right],$$

## 3. Some Examples

**Example**

**1.**

If $S\left(t\right)=a+bt,$ with $a,\phantom{\rule{4pt}{0ex}}b\ge 0,$ and $X\left(t\right)={B}_{t}\phantom{\rule{4pt}{0ex}}(\rho \left(t\right)=1),$ examples of solution to the IFPT problem, for $X\left(t\right)$ and various FPT densities $f,$ can be found in [2].

**Example**

**2.**

Let be $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\phantom{\rule{4pt}{0ex}}\beta \ge 0,$ and suppose that τ has density $f\left(t\right)=\lambda {e}^{-\rho \left(t\right)}{\rho}^{\prime}\left(t\right){\mathbf{1}}_{(0,+\infty )}\left(t\right)$ (that is, the density $\tilde{f}$ of $\tilde{\tau}=\rho \left(\tau \right)$ is exponential with parameter $\lambda ).$ By using Proposition 1 we get that $\eta =\alpha +{Z}_{1}+{Z}_{2},$ where ${Z}_{i}$ are independent random variable, such that ${Z}_{i}-\alpha /2$ has exponential distribution with parameter ${\lambda}_{i}$ ($i=1,2),$ where ${\lambda}_{1}=\beta -\sqrt{{\beta}^{2}-2\lambda}$ and ${\lambda}_{2}=\beta +\sqrt{{\beta}^{2}-2\lambda}.$ Then, the solution g to the IFPT problem for $X\left(t\right)=\eta +B\left(\rho \right(t\left)\right),$ the boundary S and the exponential FPT distribution, is:

$$g\left(x\right)=\left\{\begin{array}{c}\frac{{\lambda}_{1}{\lambda}_{2}}{{\lambda}_{2}-{\lambda}_{1}}{e}^{-{\lambda}_{1}(x-\alpha )}-{e}^{-{\lambda}_{2}(x-\alpha )},\phantom{\rule{4pt}{0ex}}\mathrm{if}\phantom{\rule{4pt}{0ex}}b>\sqrt{2\lambda}\hfill \\ 2\lambda (x-\alpha ){e}^{-\sqrt{2\lambda}(x-a)},\phantom{\rule{4pt}{0ex}}\mathrm{if}\phantom{\rule{4pt}{0ex}}b=\sqrt{2\lambda}.\hfill \end{array}\right.\phantom{\rule{4pt}{0ex}}(x\ge \alpha )$$

In general, for a given continuous increasing boundary $S\left(t\right)$ and an assigned distribution of $\tau ,$ it is difficult to calculate explicitly the expectation on the right-hand member of Equation (4) to get the Laplace transform of $\eta .$ Thus, a heuristic solution to the IFPT problem can be achieved by using Equation (4) to calculate the moments of $\eta $ (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 $\eta $ of any order, this would not determinate the distribution of $\eta .$ However, this procedure is useful to study the properties of the distribution of $\eta ,$ provided that the solution to the IFPT problem exists.

**Example**

**3.**

Let be $S\left(t\right)={t}^{2},\phantom{\rule{4pt}{0ex}}\rho \left(t\right)=t$ and suppose that τ is exponentially distributed with parameter $\lambda ;$ we search for a solution $\eta >0$ to the IFPT problem by using the method of moments, described above. The compatibility condition in Equation (14) requires that ${\lambda}^{3}<20$ (for instance, one can take $\lambda =1).$ From Equations (11), (12) and (15)–(17), and calculating the moments of τ up to the eighth order, we obtain:

$$E\left(\eta \right)=E\left({\tau}^{2}\right)=\frac{2}{{\lambda}^{2}};\phantom{\rule{4pt}{0ex}}E\left({\eta}^{2}\right)=E\left({\tau}^{4}\right)-E\left(\tau \right)=\frac{24-{\lambda}^{3}}{{\lambda}^{4}};\phantom{\rule{4pt}{0ex}}{\sigma}^{2}\left(\eta \right)=Var\left(\eta \right)=\frac{20-{\lambda}^{3}}{{\lambda}^{4}};$$

$$E\left({\eta}^{3}\right)=E\left({\tau}^{6}\right)-3E\left({\tau}^{3}\right)=\frac{720-18{\lambda}^{3}}{{\lambda}^{6}};\phantom{\rule{4pt}{0ex}}E\left({\eta}^{4}\right)=E\left({\tau}^{8}\right)-6E\left({\tau}^{3}\right)+3E\left({\tau}^{2}\right)=\frac{8!-36{\lambda}^{5}+6{\lambda}^{6}}{{\lambda}^{8}}.$$

Notice that, under the condition ${\lambda}^{3}<20$ 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
meaning that the candidate η has an asymmetric distribution with a tail toward the left.

$$\frac{E{(\eta -E\left(\eta \right))}^{3}}{\sigma {\left(\eta \right)}^{3}}=-12\frac{24-{\lambda}^{3}}{{(20-{\lambda}^{3})}^{3/2}}<0,$$

## 4. Conclusions and Final Remarks

We have dealt with the IFPT problem for a continuously time-changed Brownian motion $X\left(t\right)$ starting from a random position $\eta .$ For a given continuous, increasing boundary $S\left(t\right)$ with $\eta \ge S\left(0\right)\ge 0,$ and an assigned continuous distribution function $F,$ the IFPT problem consists in finding the distribution, or the density g of $\eta ,$ such that the first-passage time $\tau $ of $X\left(t\right)$ below $S\left(t\right)$ has distribution $F.$ In this note, we have provided some extensions of the results, already known in the case when $X\left(t\right)$ is BM and $S\left(t\right)$ is a straight line, and we have reported some explicit examples. Really, the process we considered has the form $X\left(t\right)=\eta +{\int}_{0}^{t}\sigma \left(s\right)d{B}_{s},$ where ${B}_{t}$ is standard Brownian motion, and $\sigma \left(t\right)$ is a non random continuous function of time $t\ge 0,$ such that the function $\rho \left(t\right)={\int}_{0}^{t}{\sigma}^{2}\left(s\right)ds$ is increasing and it satisfies the condition $\rho (+\infty )=+\infty .$ Thus, a standard BM $\widehat{B}$ exists such that $X\left(t\right)=\eta +\widehat{B}\left(\rho \left(t\right)\right).$ Our main result states that
$$\widehat{g}\left(\theta \right)=E\left[{e}^{-\theta S\left(\tau \right)-\frac{{\theta}^{2}}{2}\rho \left(\tau \right)}\right],$$
where, for $\theta >0,\phantom{\rule{4pt}{0ex}}\widehat{g}\left(\theta \right)$ 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 $X\left(t\right)$ considered, for instance to a process of the form
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 $w\left(U\left(t\right)\right)=w\left(\eta \right)+\widehat{B}\left(\rho \left(t\right)\right),$ the IFPT problem for the process $U,$ the boundary $S\left(t\right)$ and the FPT distribution $F,$ is reduced to the analogous IFPT problem for $X\left(t\right)={\eta}_{1}+\widehat{B}\left(\rho \left(t\right)\right),$ starting from ${\eta}_{1}=w\left(\eta \right),$ instead of $\eta ,$ the boundary ${S}_{1}\left(t\right)=w\left(S\left(t\right)\right)$ and the same FPT distribution $F.$ When $\sigma \left(t\right)=1,$ i.e. $\rho \left(t\right)=t,$ the process $U\left(t\right)$ 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 $U\left(t\right)$ given by Equation (32) is indeed a weak solution of the SDE:
where ${w}^{\prime}\left(x\right)$ and ${w}^{\u2033}\left(x\right)$ denote first and second derivative of $w\left(x\right).$

$$U\left(t\right)={w}^{-1}(\widehat{B}\left(\rho \left(t\right)\right)+w\left(\eta \right)),$$

$$dU\left(t\right)=-\frac{{\rho}^{\prime}\left(t\right){w}^{\u2033}\left(U\left(t\right)\right)}{2{\left({w}^{\prime}\left(U\left(t\right)\right)\right)}^{3}}dt+\frac{\sqrt{{\rho}^{\prime}\left(t\right)}}{{w}^{\prime}\left(U\left(t\right)\right)}d{B}_{t}\phantom{\rule{4pt}{0ex}},$$

Provided that the deterministic function $\rho \left(t\right)$ 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
where the drift $\left(\mu \right)$ and diffusion coefficients $\left(\sigma \right)$ satisfy the usual conditions (see e.g., [18]) for existence and uniqueness of the solution of Equation (34). In fact, let $w\left(x\right)$ be the scale function associated to the diffusion $U\left(t\right)$ driven by the SDE Equation (34), that is, the solution of $Lw\left(x\right)=0,\phantom{\rule{4pt}{0ex}}w\left(0\right)=0,\phantom{\rule{4pt}{0ex}}{w}^{\prime}\left(0\right)=1,$ where L is the infinitesimal generator of U given by $Lh=\frac{1}{2}{\sigma}^{2}\left(x\right)\frac{{d}^{2}h}{d{x}^{2}}+\mu \left(x\right)\frac{dh}{dx}.$ As easily seen, if the integral ${\int}_{0}^{t}\frac{2\mu \left(z\right)}{{\sigma}^{2}\left(z\right)}\phantom{\rule{4pt}{0ex}}dz$ converges, the scale function is explicitly given by

$$dU\left(t\right)=\mu \left(U\left(t\right)\right)dt+\sigma \left(U\left(t\right)\right)d{B}_{t},\phantom{\rule{4pt}{0ex}}U\left(0\right)=\eta ,$$

$$w\left(x\right)={\int}_{0}^{x}exp\left(-{\int}_{0}^{t}\frac{2\mu \left(z\right)}{{\sigma}^{2}\left(z\right)}\phantom{\rule{4pt}{0ex}}dz\right)dt.$$

If $\zeta \left(t\right):=w\left(U\right(t\left)\right),$ by It$\widehat{\mathrm{o}}$’s formula one obtains
that is, the process $\zeta \left(t\right)$ is a local martingale, whose quadratic variation is

$$\zeta \left(t\right)=w\left(\eta \right)+{\int}_{0}^{t}{w}^{\prime}\left({w}^{-1}\left(\zeta \left(s\right)\right)\right)\sigma \left({w}^{-1}\left(\zeta \left(s\right)\right)\right)d{B}_{s}\phantom{\rule{4pt}{0ex}},$$

$$\rho \left(t\right)\doteq {\langle \zeta \rangle}_{t}={\int}_{0}^{t}{\left[{w}^{\prime}\left(U\left(s\right)\right)\sigma \left(U\left(s\right)\right)\right]}^{2}ds,\phantom{\rule{4pt}{0ex}}t\ge 0.$$

The (random) function $\rho \left(t\right)$ is differentiable and $\rho \left(0\right)=0;$ if it is increasing to $\rho (+\infty )=+\infty ,$ by the Dambis, Dubins–Schwarz theorem (see e.g., [3]) one gets that there exists a standard BM $\widehat{B}$ such that $\zeta \left(t\right)=\widehat{B}\left(\rho \left(t\right)\right)+w\left(\eta \right).$ 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 $\rho $ is a deterministic function. In fact, if $\rho \left(t\right)$ given by Equation (37) is random, it results that $\rho \left(t\right)$ and the FPT $\tau $ 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 $\left(\rho \right(t),\tau )$ 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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