The Randomized First-Hitting Problem of Continuously Time-Changed Brownian Motion

Abstract

Let $X\left(t\right)$ be a continuously time-changed Brownian motion starting from a random position $\eta ,S\left(t\right)$ 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.

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

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

(1)

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]).

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]).

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 $\eta$ 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 $S\left(t\right)$ a continuous, increasing boundary with $S\left(0\right)\ge 0,\sigma \left(t\right)$ 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:

$$\widehat{g}\left(\theta \right)={\int }_0^{+\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)}dF\left(t\right).$$

(2)

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

$$E\left[Z_0\right]=E\left[e^{-\theta X\left(0\right)}\right]=E\left[e^{-\theta \eta }\right]=\underset{t\to \infty }{lim}E\left[e^{-\theta X(\tau \wedge t)-\frac{1}{2}{\theta }^2\rho (\tau \wedge t)}\right]$$

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

(3)

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

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

(4)

that is Equation (2). ☐

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, $X\left(t\right)=B_t(\sigma \left(t\right)=1);$ in fact, one obtains

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

(5)

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 $\tau .$

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 $a+bt(a=S\left(0\right)\ge 0,b={\tilde{S}}^{\prime }\left(0\right)\ge 0)$ 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

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

(6)

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

$$\widehat{g}\left(\theta \right)={\int }_0^{S(+\infty )}{e}^{-\theta x-\frac{{\theta }^2}{2}\rho \left(S^{-1}\left(x\right)\right)}dH\left(x\right).$$

(7)

In the special case when $S\left(t\right)=a+bt(a,b>0)$ and $\rho \left(t\right)=t,$ Equation (7) becomes (cf. the result in [8] for $a=0):$

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

(8)

where $h\left(x\right)=H^{\prime }\left(x\right)$ and $\widehat{h}$ denotes the Laplace transform of $h.$

In analogous way, the SE problem can be solved if the support of H is $(-\infty ,0];$ now, the FPT is understood as ${\tau }^{-}=inf\{t>0:\eta +B\left(\rho \left(t\right)\right)>-S\left(t\right)\}(\eta <0),$ 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 $X^{-}<0:$

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

(9)

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.

$${\eta }^{\pm }=\left\{\begin{array}{cc}{\eta }^{+}\hfill & \mathrm{with}\mathrm{probability}{p}^{+}\hfill \\ {\eta }^{-}\hfill & \mathrm{with}\mathrm{probability}{p}^{-}\hfill \end{array}\right.$$

(10)

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. $X\left(\tau \right).$

If $\widehat{g}$ is analytic in a neighbor of $\theta =0,$ then the moments of order n of $\eta ,E\left({\eta }^n\right),$ 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

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

(11)

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

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

(12)

and so

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

(13)

Thus, we obtain the compatibility conditions

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

(14)

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 $S\left(t\right)=a+bt(a,b\ge 0)$ 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

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

(15)

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

(16)

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

(17)

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 $\alpha ,\beta \ge 0,$ from Equation (4) we get

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

(18)

Thus, setting $\tilde{\tau }=\rho \left(\tau \right),$ we obtain (see Equation (5)):

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

(19)

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 $\tilde{f}.$

Under the assumption that $S\left(t\right)=\alpha +\beta \rho \left(t\right),$ with $\alpha ,\beta \ge 0,$ 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 $\alpha ,\beta \ge 0;$ for $\gamma ,\lambda >0,$ suppose that the FPT density $f=F^{\prime }$ is given by

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

(20)

(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:

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

(21)

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 ${\lambda }_i(i=1,2),$ where ${\lambda }_1=\beta -\sqrt{{\beta }^2-2\lambda }$ and ${\lambda }_2=\beta +\sqrt{{\beta }^2-2\lambda }.$

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 ${\lambda }_i,i=1,2.$

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 $\alpha ,\beta \ge 0;$ for $\beta >0,$ suppose that the Laplace transform of $\tilde{f}$ has the form:

$$\widehat{\tilde{f}}\left(\theta \right)=\sum _{k=1}^N\frac{A_k}{{(\theta +B_k)}^{c_k}},$$

(22)

for some $c_k>0,A_k,B_k>0,k=1,\cdots ,N.$ Then, there exists a value ${\beta }^{*}>0$ such that the solution to the IFPT problem exists, provided that $\beta \ge b^{*}.$

If $\beta =0$ and the Laplace transform of $\tilde{f}$ has the form:

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

(23)

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 ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2$ with $0\le {\alpha }_1\le {\alpha }_2$ and ${\beta }_2\ge {\beta }_1\ge 0,$ such that, for every $t\ge 0:$

$${\alpha }_1+{\beta }_1\rho \left(t\right)\le S\left(t\right)\le {\alpha }_2+{\beta }_2\rho \left(t\right),$$

(24)

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 $S_{{\alpha }_1,{\beta }_1}\left(t\right)={\alpha }_1+{\beta }_1\rho \left(t\right).$

Then, by using Proposition (3.13) of [16] (see also [1]), we obtain the following:

Proposition 3.

Let $S\left(t\right)$ a continuous, increasing boundary satisfying Equation (24) and suppose that the FPT τ of $X\left(t\right)=\eta +B\left(\rho \right(t\left)\right)(\eta >S(0\left)\right)$ 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:

$$e^{-{\alpha }_2(\theta +2({\beta }_2-{\beta }_1))}\left[\widehat{\tilde{f}}\left(\frac{\theta (\theta +2{\beta }_2)}{2}\right)-{({\alpha }_2-S\left(0\right))}^{\frac{p-1}{p}}{\left({\int }_{S\left(0\right)}^{{\alpha }_2}{g}^p\left(x\right)dx\right)}^{1/p}\right]\le \widehat{g}\left(\theta \right)$$

$$\le e^{-{\alpha }_1\theta }\widehat{\tilde{f}}\left(\frac{\theta (\theta +2{\beta }_1)}{2}\right);$$

(25)

(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

$$\widehat{\overline{g}}\left(\theta \right)=E\left[e^{-\theta S\left(\tau \right)-\frac{{\theta }^2}{2}\rho \left(\tau \right)-\mu \tau }\right]+\mu {\int }_0^{+\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)-\mu t}\left({\int }_t^{+\infty }f\left(s\right)ds\right)dt.$$

(26)

Proof. 

For $t>0,$ one has:

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

(27)

Taking the derivative, one obtains the FPT density of $\overline{\tau }:$

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

(28)

where f is the density of $\tau .$ Then, by the same arguments used in the proof of Theorem 1, we obtain

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

$$={\int }_0^{\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)}\overline{f}\left(t\right)dt$$

$$={\int }_0^{\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)}\left[e^{-\mu t}f\left(t\right)+\mu e^{-\mu t}{\int }_t^{\infty }f\left(s\right)ds\right]dt$$

$$={\int }_0^{\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)-\mu t}f\left(t\right)dt+\mu {\int }_0^{\infty }{e}^{-\theta S\left(t\right)-\frac{{\theta }^2}{2}\rho \left(t\right)-\mu t}\left({\int }_t^{\infty }f\left(s\right)ds\right)dt$$

that is Equation (26). ☐

Remark 7.

(i) 

For $\mu =0,$ namely when no jump occurs, Equation (26) becomes Equation (4).

(ii) 

If τ is exponentially distributed with parameter $\lambda ,$ then Equation (26) provides:

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

(29)

(iii) 

In the special case when $S\left(t\right)=\alpha +\beta \rho \left(t\right)(\alpha ,\beta \ge 0),$ 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

$$\widehat{\overline{g}}\left(\theta \right)=e^{-\alpha \theta }\left[{\left(1-\frac{2\mu }{\theta (\theta +2\beta )}\right)}^{-1}\widehat{\overline{f}}\left(\frac{\theta (\theta +2\beta )}{2}-\mu \right)-\frac{2\mu }{\theta (\theta +2\beta )-2\mu }\right],$$

where, for simplicity of notation we have denoted again with $\widehat{\overline{f}}$ the Laplace transform of $\overline{\tilde{\tau }};$ of course, if $\rho \left(t\right)=t,$ then $\widehat{\overline{f}}$ is the Laplace transform of $\overline{\tau }.$ Notice that, if $\mu =0$ the last equation is nothing but Equation (5) with $\alpha ,\beta$ in place of $a,b.$

3. Some Examples

Example 1.

If $S\left(t\right)=a+bt,$ with $a,b\ge 0,$ and $X\left(t\right)=B_t(\rho \left(t\right)=1),$ 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 $\alpha ,\beta \ge 0,$ 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:

$$g\left(x\right)=\left\{\begin{array}{c}\frac{{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}{e}^{-{\lambda }_1(x-\alpha )}-e^{-{\lambda }_2(x-\alpha )},\mathrm{if}b>\sqrt{2\lambda }\hfill \\ 2\lambda (x-\alpha )e^{-\sqrt{2\lambda }(x-a)},\mathrm{if}b=\sqrt{2\lambda }.\hfill \end{array}\right.(x\ge \alpha )$$

(30)

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 $S\left(t\right)=t^2,\rho \left(t\right)=t$ 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:

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

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

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

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

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 $X\left(t\right)$ 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

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

(31)

where, for $\theta >0,\widehat{g}\left(\theta \right)$ 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

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

(32)

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:

$$dU\left(t\right)=-\frac{{\rho }^{\prime }\left(t\right)w^{″}\left(U\left(t\right)\right)}{2{\left(w^{\prime }\left(U\left(t\right)\right)\right)}^3}dt+\frac{\sqrt{{\rho }^{\prime }\left(t\right)}}{w^{\prime }\left(U\left(t\right)\right)}d{B}_t,$$

(33)

where $w^{\prime }\left(x\right)$ and $w^{″}\left(x\right)$ denote first and second derivative of $w\left(x\right).$

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

$$dU\left(t\right)=\mu \left(U\left(t\right)\right)dt+\sigma \left(U\left(t\right)\right)d{B}_t,U\left(0\right)=\eta ,$$

(34)

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 $Lw\left(x\right)=0,w\left(0\right)=0,w^{\prime }\left(0\right)=1,$ 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 ${\int }_0^t\frac{2\mu \left(z\right)}{{\sigma }^2\left(z\right)}dz$ converges, the scale function is explicitly given by

$$w\left(x\right)={\int }_0^{x}exp\left(-{\int }_0^t\frac{2\mu \left(z\right)}{{\sigma }^2\left(z\right)}dz\right)dt.$$

(35)

If $\zeta \left(t\right):=w\left(U\right(t\left)\right),$ by It$\widehat{\mathrm{o}}$’s formula one obtains

$$\zeta \left(t\right)=w\left(\eta \right)+{\int }_0^t{w}^{\prime }\left(w^{-1}\left(\zeta \left(s\right)\right)\right)\sigma \left(w^{-1}\left(\zeta \left(s\right)\right)\right)d{B}_s,$$

(36)

that is, the process $\zeta \left(t\right)$ is a local martingale, whose quadratic variation is

$$\rho \left(t\right)\doteq {\langle \zeta \rangle }_t={\int }_0^t{\left[w^{\prime }\left(U\left(s\right)\right)\sigma \left(U\left(s\right)\right)\right]}^{2}ds,t\ge 0.$$

(37)

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.