On an Inverse First-Passage Problem for Jump-Diffusion Processes

Abstract

Computing the exact mathematical expression for a quantity defined in terms of a first-passage time random variable for a jump-diffusion process is in general very difficult. In this paper, we consider the following inverse problem: can we find a certain distribution for the size of the jumps that leads to a simple solution of the integro-differential equation satisfied by the quantity of interest, subject to the appropriate boundary conditions? Such distributions are found, in particular, for the mean of the first-passage time for important jump-diffusion processes.

1. Introduction

We consider the one-dimensional jump-diffusion process $\left\{X\right(t),t\ge 0\}$ starting at $X\left(0\right)=x$ and defined by

$$X\left(t\right)=x+{\int }_0^{t}f\left[X\left(s\right)\right]\mathrm{d}s+{\int }_0^t{\left\{v\left[X\right(s\left)\right]\right\}}^{1/2}\mathrm{d}B\left(s\right)+\sum _{k=1}^{N\left(t\right)}{Y}_k,$$

(1)

where $\left\{B\right(t),t\ge 0\}$ is a standard Brownian motion, $\left\{N\right(t),t\ge 0\}$ is a Poisson process with rate $\lambda$, and $\{Y_1,Y_2,\dots \}$ is an infinite set of independent random variables that are distributed as the random variable Y having a probability density function $f_Y\left(y\right)$, and are independent of the Poisson process. Moreover, $\left\{B\right(t),t\ge 0\}$ and $\left\{N\right(t),t\ge 0\}$ are independent stochastic processes. The continuous part $\{X_c\left(t\right),t\ge 0\}$ of the jump-diffusion process is a diffusion process with infinitesimal mean $f\left(x\right)$ and infinitesimal variance $v\left(x\right)$.

We define the first-passage time

$$\tau \left(x\right)=inf\{t>0:X(t)\in D\mid X(0)=x\notin D\},$$

(2)

where $D\subset \mathbb{R}$. It can be shown (see [1]) that the moment-generating function of $\tau \left(x\right)$, namely

$$M(x;\alpha ):=E\left[e^{-\alpha \tau \left(x\right)}\right],$$

(3)

where $\alpha >0$, satisfies the integro-differential equation (IDE)

$$\begin{array}{ccc}\hfill \alpha M(x;\alpha )& =& {\displaystyle \frac{1}{2}}v\left(x\right)M^{″}(x;\alpha )+f\left(x\right)M^{\prime }(x;\alpha )\hfill \\ & & +\lambda \left\{{\int }_{-\infty }^{\infty }M(x+y;\alpha )f_Y\left(y\right)\mathrm{d}y-M(x;\alpha )\right\}\hfill \end{array}$$

(4)

for $x\notin D$, subject to the boundary condition

$$M(x;\alpha )=1\mathrm{if}x\in D.$$

(5)

We assume that

• The functions $f\left(x\right)$ and $v\left(x\right)$ are continuous on $\mathbb{R}$, and $v\left(x\right)\ge v_0>0$.
• The density function $f_Y\left(y\right)$ is bounded and measurable. Moreover it is such that

  $${\int }_{-\infty }^{\infty }{f}_Y\left(y\right)\mathrm{d}y=1\mathrm{and}{\int }_{-\infty }^{\infty }\left|y\right|f_Y\left(y\right)\mathrm{d}y<\infty .$$

  (6)
• The solution of the boundary–value problem (4), (5) satisfies $0\le M(x;\alpha )\le 1$ for $x\notin D$, which follows from the definition of $M(x;\alpha )$.
• The set D is such that the first–passage time $\tau \left(x\right)$ is almost surely finite for all $x\notin D$, and the boundary condition $M(x;\alpha )=1$ for $x\in D$ is well posed.

Under these assumptions, it can be shown that the integro–differential operator in (4) is the generator of a Feller jump–diffusion, and that (4), (5) admit a unique bounded classical solution (see [1,2]).

Remark 1.

If $\lambda =0$, Equation (4) reduces to the well-known Kolmogorov backward equation for the function $M(x;\alpha )$ in the case of a one-dimensional diffusion process. Let

$$u(t,x):=E\left[F\left(X\right(T\left)\right)\mid X\left(t\right)=x\right],$$

(7)

where $F(·)$ is a real function, and $T>t$. We assume that $u(t,x)$ is differentiable with respect to t and twice differentiable with respect to x. By the Markov property,

$$u(t,x)=E\left[u\right(t+\Delta t,X(t+\Delta t))\mid X(t)=x].$$

(8)

Using Taylor expansion in both variables t and x, and the properties of $\left\{B\right(t),t\ge 0\}$, we can write that

$$\begin{array}{ccc}\hfill E\left[u\right(t+\Delta t,X(t+\Delta t)\mid X\left(t\right)=x]& =& u(t,x)+\Delta t{u}_t(t,x)+\Delta tf\left(x\right)u_x(t,x)\hfill \\ & & +{\displaystyle \frac{1}{2}}\Delta tv\left(x\right)u_{xx}(t,x)+o(\Delta t).\hfill \end{array}$$

(9)

Now, in a short interval of length $\Delta t$, the number of jumps is equal to

$$\left\{\begin{array}{cc}1& \mathrm{with}\mathrm{probability}\lambda \Delta t+o(\Delta t),\hfill \\ 0& \mathrm{with}\mathrm{probability}1-\lambda \Delta t+o(\Delta t),\hfill \end{array}\right.$$

(10)

so that the probability of at least two jumps is $o(\Delta t)$. Hence,

$$E\left[u\right(t+\Delta t,X(t+\Delta t)\left)\right]=E\left[E\right[u(t+\Delta t,x+Y)\left]\right]\lambda \Delta t+E\left[u\right(t+\Delta t,x\left)\right](1-\lambda \Delta t)+o(\Delta t).$$

(11)

We have

$$E\left[u(t+\Delta t,x+Y)\right]={\int }_{-\infty }^{\infty }u(t+\Delta t,x+y)f_Y\left(y\right)\mathrm{d}y.$$

(12)

Making use of the above results, we find that

$$0=u_t(t,x)+f\left(x\right)u_x(t,x)+{\displaystyle \frac{1}{2}}v\left(x\right)u_{xx}(t,x)+\lambda \left\{{\int }_{-\infty }^{\infty }u(t,x+y)f_Y\left(y\right)\mathrm{d}y-u(t,x)\right\}.$$

(13)

Finally, the probability density function $f_{\tau \left(x\right)}\left(t\right)$ of the first-passage time $\tau \left(x\right)$ satisfies the same partial integro-differential equation as $u(t,x)$. Equation (4) is then obtained by taking the Laplace transform of the above equation with $f_{\tau \left(x\right)}\left(t\right)$ instead of $u(t,x)$.

Next, the mean $m\left(x\right)$ of the random variable $\tau \left(x\right)$ satisfies (if it exists) the IDE

$${\displaystyle \frac{1}{2}}v\left(x\right)m^{″}\left(x\right)+f\left(x\right)m^{\prime }\left(x\right)+\lambda \left\{{\int }_{-\infty }^{\infty }m(x+y)f_Y\left(y\right)\mathrm{d}y-m\left(x\right)\right\}=-1.$$

(14)

The boundary condition is $m\left(x\right)=0$ if $x\in D$.

Suppose that the complement of the set D is the interval $(a,b)$. Sometimes, we are also interested in the first-passage-place probability

$$p\left(x\right):=P\left[X\right(\tau \left(x\right))\le a].$$

(15)

We assume that $P\left[\tau \right(x)<\infty ]=1$. The function $p\left(x\right)$ can be obtained by solving the IDE

$${\displaystyle \frac{1}{2}}v\left(x\right)p^{″}\left(x\right)+f\left(x\right)p^{\prime }\left(x\right)+\lambda \left\{{\int }_{-\infty }^{\infty }p(x+y)f_Y\left(y\right)\mathrm{d}y-p\left(x\right)\right\}=0.$$

(16)

This time, the boundary conditions are $p\left(a\right)=1$ and $p\left(b\right)=0$. If we are looking for $p_b\left(x\right):=P[X\left(\tau \left(x\right)\right)\ge b]$ instead, then we must have $p_b\left(a\right)=0$ and $p_b\left(b\right)=1$. Notice that, since we assumed that $P\left[\tau \right(x)<\infty ]=1$, we simply have $p_b\left(x\right)=1-p\left(x\right)$.

First-passage times have applications in numerous fields. Inverse first-passage-time (IFPT) problems have been studied by various authors mainly for pure diffusion processes, but also for jump-diffusion processes. A classical inverse first-passage-time problem consists in trying to find a barrier b such that for a Brownian motion $\left\{B\right(t),t\ge 0\}$ and a positive random variable $\xi$ we can write that $P\left[B\right(s)>b(s),0\le s\le t]=P[\xi >t]\forall t\ge 0$; see [3]. A related problem for killed Brownian motion was studied by Ettinger et al. [4].

A seminal paper on the inverse first-passage problem is by Capocelli and Ricciardi [5], published in 1972, in which the authors posed the problem, providing constructive criteria and examples. In Zucca and Sacerdote [6], the IFPT problem was treated for a Brownian motion. The authors derived explicit and semi-explicit boundary behavior for given first-passage distributions. Cheng et al. [7] studied the existence and uniqueness of solutions to the inverse boundary crossing problem for diffusions processes. Civallero and Zucca [8] considered the IFPT problem in two dimensions. They solved numerically for boundaries corresponding to target distributions. Klump [9] provided a novel Monte Carlo algorithm that in the limit recovers the boundary solving the IFPT problem.

In [10,11], Abundo considered the following problem: suppose that $\left\{X\right(t),t\ge 0\}$ is a one-dimensional diffusion process whose initial value $X\left(0\right)$ is a random variable, and $S\left(t\right)$ is a given boundary. Assuming that $P\left[X\right(0)>S(0\left)\right]=1$, can we find a probability distribution for $X\left(0\right)$ such that the first-passage-time distribution of $X\left(t\right)$ below $S\left(t\right)$ has a certain distribution function $F\left(x\right)$? He obtained results that generalized the ones in Jackson et al. [12] who treated the case when $\left\{X\right(t),t\ge 0\}$ is a Wiener process and $S\left(t\right)$ is a straight line across the origin. See also [13,14] and the references therein. Abundo [15] has also treated a related inverse first-passage-place problem for jump-diffusion processes, and in [16] he extended the IFPT problem to diffusions with stochastic resetting.

The author has published a number of papers on inverse first-passage problems for diffusion processes; see [17,18]. In [19], an inverse first-passage-place problem was studied.

The jump size Y is sometimes assumed to be exponentially or uniformly distributed. This is a simplifying assumption. In practical applications, the distribution of Y is unknown. Moreover, even if we accept this type of distribution as an acceptable approximation for a particular application, calculating the exact mathematical expression for the functions $M(x;\alpha )$, $m\left(x\right)$ and/or $p\left(x\right)$ is very difficult, and the expressions in question are generally very complicated. In [20], Kou and Wang were able to obtain exact and explicit results for quantities associated to first-passage times of jump-diffusion processes when the jump size has an asymmetric double exponential distribution. Kou mentioned in a related paper [21] that this assumption, made in the context of option pricing, aimed to strike a balance between reality and tractability.

The random variable Y can actually be of any type. If it is of discrete type, or of mixed type, the density function $f_Y$ will involve one or several Dirac delta functions. When Y is a discrete random variable, instead of its density function, we can use its probability mass function $p_Y$. The IDE (4) will become a difference-differential equation.

Suppose that $Y\in \{y_1,y_2,\dots \}$. Then Equation (4) becomes

$$\begin{array}{ccc}\hfill \alpha M(x;\alpha )& =& {\displaystyle \frac{1}{2}}v\left(x\right)M^{″}(x;\alpha )+f\left(x\right)M^{\prime }(x;\alpha )\hfill \\ & & +\lambda \left\{\sum _{j=1}^{\infty }M(x+y_j;\alpha )p_Y\left(y_j\right)-M(x;\alpha )\right\}.\hfill \end{array}$$

(17)

Solving this type of equation is also a difficult task. We must take into account the fact that, for at least some of the values of Y, the quantity $x+y_j$ may belong to the stopping set D, so that $M(x+y_j;\alpha )=1$.

If Y is a constant equal to $\eta$ (a degenerate random variable), the IDE simplifies to

$$\alpha M(x;\alpha )={\displaystyle \frac{1}{2}}v\left(x\right)M^{″}(x;\alpha )+f\left(x\right)M^{\prime }(x;\alpha )+\lambda \left\{M(x+\eta ;\alpha )-M(x;\alpha )\right\}.$$

(18)

Again, $x+\eta$ may or may not belong to D. If $x+\eta \in D$ for any x, so that $M(x+\eta ;\alpha )=1$, then the above equation is a non-homogeneous second-order linear ordinary differential equation (ODE).

Finally, in the case of a mixed type random variable, the equation satisfied by the function $M(x;\alpha )$ will involve an integral plus at least one term of the form $M(x+y_1)p_Y\left(y_1\right)$, thus rendering the problem still more difficult to tackle. Similarly for the functions $m\left(x\right)$ and $p\left(x\right)$.

In this paper, we consider the following inverse problem for jump-diffusion processes: can we find a probability density function $f_Y\left(y\right)$ for which the function $m\left(x\right)$, for instance, is of a given simple form?

In the next section, we will try to find simple solutions to Equation (4). Next, in Section 3 and Section 4, the same problem will be studied for the functions $m\left(x\right)$ and $p\left(x\right)$, respectively. In each case, particular density functions $f_Y\left(y\right)$ leading to simple solutions will be derived. We will conclude this paper with a few remarks in Section 5.

2. The Moment-Generating Function of the First-Passage Time

Let $\{B_0\left(t\right),t\ge 0\}$ be a one-dimensional standard Brownian motion, and define

$${\tau }_0\left(x\right)=inf\{t>0:B_0\left(t\right)=0\mid B_0\left(0\right)=x>0\}.$$

(19)

It is well known (and can be deduced at once from Equation (4)) that the moment-generating function $M_0(x;{\alpha }_0)$ of ${\tau }_0\left(x\right)$ satisfies the Kolmogorov backward equation

$${\displaystyle \frac{1}{2}}{M}_0^{″}(x;{\alpha }_0)={\alpha }_0{M}_0(x;{\alpha }_0)$$

(20)

for $x>0$, subject to the boundary condition $M_0(0;{\alpha }_0)=1$. The solution to Equation (20) that satisfies the boundary condition, and is such that $M_0(x;{\alpha }_0)<1$$\forall x>0$, is

$$M_0(x;{\alpha }_0)=e^{-\sqrt{2{\alpha }_0}x}.$$

(21)

First, we will try to determine whether a jump-diffusion process whose continuous part $\{X_c\left(t\right),t\ge 0\}$ is Wiener process can have a moment-generating function $M(x;\alpha )$ of the same form as $M_0(x;{\alpha }_0)$ for a certain non-negative jump size distribution, when the set D is the interval $(-\infty ,0]$ and $X\left(0\right)=x>0$.

We take $f\left(x\right)\equiv \mu \in \mathbb{R}$ and $v\left(x\right)\equiv {\sigma }^2$ (with $\sigma >0$), and we assume that $M(x;\beta )=e^{-\beta x}$, where $\beta >0$. Notice that $M(0;\beta )=1$ and that $M(x;\beta )<1$ for any $x>0$, as required.

With this choice, Equation (4) becomes, after simplification,

$$\lambda {\int }_0^{\infty }{e}^{-\beta y}{f}_Y\left(y\right)\mathrm{d}y=\beta +\lambda +\mu \beta -{\displaystyle \frac{1}{2}}{\sigma }^2{\beta }^2.$$

(22)

This equation can be rewritten as follows:

$$M_Y\left(\beta \right)=1+{\displaystyle \frac{\mu +1}{\lambda }}\beta -{\displaystyle \frac{1}{2\lambda }}{\sigma }^2{\beta }^2,$$

(23)

where $M_Y\left(\beta \right)$ is the moment-generating function of the random variable Y. Since

$$\underset{\beta \to \infty }{lim}{M}_Y\left(\beta \right)=0,$$

(24)

we must conclude that there is no legitimate random variable for which Equation (22) is satisfied for any $\beta$. We can however find a valid density function $f_Y\left(y\right)$ such that Equation (23) is satisfied for a given value of the parameter $\beta$, provided that the right-hand member of the equation is in the interval $(0,1)$.

For example, suppose that $\lambda =\sigma =1$ and $\mu =-1$. Then, if $\beta =1$, we must find a density function $f_Y\left(y\right)$ for which

$$E\left[e^{-Y}\right]={\displaystyle \frac{1}{2}}.$$

(25)

Let us consider the density function

$$f_Y\left(y\right)={\displaystyle \frac{1}{e^r-1}}{e}^y\mathrm{for}0\le y\le r.$$

(26)

Then, Equation (25) is satisfied if we choose r such that

$${\displaystyle \frac{r}{e^r-1}}={\displaystyle \frac{1}{2}}.$$

(27)

We find that $r\approx$ 1.2564. Thus, with this density function, we have

$$E\left[e^{-\tau \left(x\right)}\right]=e^{-x}.$$

(28)

There are actually many important random variables Y such that Equation (25) is satisfied. For example, Y can have an exponential distribution with parameter 1.

For any fixed value of $\beta$, Y can also be a constant $\eta$. We must choose the constant $\eta$ that satisfies the equation

$$E\left[e^{-\beta Y}\right]=e^{-\beta \eta }=1+{\displaystyle \frac{\mu +1}{\lambda }}\beta -{\displaystyle \frac{1}{2\lambda }}{\sigma }^2{\beta }^2.$$

(29)

That is,

$$\eta =-{\displaystyle \frac{1}{\beta }}ln\left(1+{\displaystyle \frac{\mu +1}{\lambda }}\beta -{\displaystyle \frac{1}{2\lambda }}{\sigma }^2{\beta }^2\right).$$

(30)

We can state the following proposition.

Proposition 1.

Assume that

$$h\left(\beta \right):=1+{\displaystyle \frac{\mu +1}{\lambda }}\beta -{\displaystyle \frac{1}{2\lambda }}{\sigma }^2{\beta }^2\in (0,1).$$

(31)

Then, we have that

$$E\left[e^{-\beta \tau \left(x\right)}\right]=e^{-\beta x}$$

(32)

for any random variable Y such that

$$E\left[e^{-\beta Y}\right]=h\left(\beta \right),$$

(33)

where β is a fixed constant.

Next, if $\{X_c\left(t\right),t\ge 0\}$ is a geometric Brownian motion with infinitesimal mean x and infinitesimal variance $2x^2$, and if we define

$${\tau }_1\left(x\right)=inf\{t>0:X_c\left(t\right)=1\mid X_c\left(0\right)=x\in (0,1)\},$$

(34)

we find that

$$E\left[e^{-\alpha {\tau }_1\left(x\right)}\right]=x^{\sqrt{\alpha }}.$$

(35)

Remember that the origin is a natural boundary for a geometric Brownian motion since it can be expressed as the exponential of a Wiener process.

We will try to find jump-diffusion processes, with $f\left(x\right)=\mu x$ and $v\left(x\right)={\sigma }^2{x}^2$, having the moment-generating function $M(x;\gamma )=x^{\gamma }$, where $\gamma >0$, when

$$\tau \left(x\right)=inf\{t>0:X(t)\ge 1\mid X(0)=x\in (0,1\left)\right\}.$$

(36)

We assume that the jumps are non-negative, so that the process cannot attain or cross the origin. The IDE (4) reduces to

$${\displaystyle \frac{1}{2}}{\sigma }^2\gamma (\gamma -1)x^{\gamma }+\mu \gamma x^{\gamma }+\lambda \left\{{\int }_0^{\infty }{(x+y)}^{\gamma }{f}_Y\left(y\right)\mathrm{d}y-x^{\gamma }\right\}=\gamma x^{\gamma }.$$

(37)

That is,

$$\left({\displaystyle \frac{1}{2}}{\sigma }^2\gamma (\gamma -1)+\mu \gamma -\lambda -\gamma \right)x^{\gamma }+\lambda E\left[{(Y+x)}^{\gamma }\right]=0.$$

(38)

Setting $\gamma$ equal to 1, we obtain the following proposition.

Proposition 2.

If $f\left(x\right)=\mu x$, where $\mu <1$, and $v\left(x\right)={\sigma }^2{x}^2$, then we have $E\left[e^{-\tau \left(x\right)}\right]=x$ for any non-negative random variable Y for which

$$E\left[Y\right]={\displaystyle \frac{1-\mu }{\lambda }}x.$$

(39)

Remark 2.

(i) The jump size can indeed be state-dependent. (ii) For any $\gamma \in \{1,2,\dots \}$, we can obtain conditions on the moments of Y that give $M(x;\gamma )=x^{\gamma }$ by using Newton’s binomial theorem.

To conclude this section, we mention that there is a case when the function $M(x;\alpha )$ will not be of the same form as the function $M_0(x;{\alpha }_0)$ defined in Equation (21), but close. Suppose that

$$f_Y\left(y\right)=\delta (y+x),$$

(40)

where $\delta (·)$ is the Dirac delta function. The jump is negative, but there is no overshoot at the origin. Indeed, with this density function, the first jump of the Poisson process will immediately bring the jump-diffusion process from its current value to zero.

In the case of a standard Brownian motion with jumps, the IDE (4) reduces to

$$\alpha M(x;\alpha )={\displaystyle \frac{1}{2}}{M}^{″}(x;\alpha )+\lambda \left\{1-M(x;\alpha )\right\}.$$

(41)

That is,

$$(\alpha +\lambda )M(x;\alpha )={\displaystyle \frac{1}{2}}{M}^{″}(x;\alpha )+\lambda .$$

(42)

It is a simple matter to show that the function $M(x;\alpha )$ that satisfies the conditions $M(0;\alpha )=1$ and $M(x;\alpha )<1$ for any $x>0$ is

$$M(x;\alpha )={\displaystyle \frac{\alpha }{\alpha +\lambda }}{e}^{-\sqrt{2\alpha +2\lambda }x}+{\displaystyle \frac{\lambda }{\alpha +\lambda }}\mathrm{for} x\ge 0.$$

(43)

This result can be generalized to any diffusion process with jumps. The function $M(x;\alpha )$ will satisfy an ODE of the same form as that satisfied by $M_0(x;{\alpha }_0)$, but non-homogeneous:

$${\displaystyle \frac{1}{2}}v\left(x\right)M^{″}(x;\alpha )+f\left(x\right)M^{\prime }(x;\alpha )+\lambda =(\alpha +\lambda )M(x;\alpha ).$$

(44)

A particular solution to the above equation is $M_p(x;\alpha )=\lambda /(\alpha +\lambda )$. It follows that

$$M(x;\alpha )=M_0(x;\alpha +\lambda )+{\displaystyle \frac{\lambda }{\alpha +\lambda }}\mathrm{for} x\ge 0.$$

(45)

3. The Expected Value of the First-Passage Time

In this section, we consider the first-passage time

$$\tau \left(x\right)=inf\{t>0:X(t)\le 0\mid X(0)=x>0\}.$$

(46)

In the case of a Wiener process with drift $\mu <0$ and dispersion parameter $\sigma$, the expected value of $\tau \left(x\right)$ is equal to $-x/\mu$. We look for jump-diffusion processes for which $m\left(x\right):=E\left[\tau \right(x\left)\right]$ is of the form $c{x}^k$, where $c>0$ and $k\in \{1,2,\dots \}$. With this choice, the IDE (14) becomes

$${\displaystyle \frac{1}{2}}v\left(x\right)k(k-1)c{x}^{k-2}+f\left(x\right)kc{x}^{k-1}+\lambda \left\{{\int }_{-\infty }^{\infty }c{(x+y)}^k{f}_Y\left(y\right)\mathrm{d}y-c{x}^k\right\}=-1.$$

(47)

The function $m\left(x\right)=c{x}^k$ satisfies the conditions $m\left(0\right)=0$ and ${lim}_{x\to \infty }m\left(x\right)=\infty$.

We assume that the random variable Y is non-negative. If Y can take any negative value, then $m(x+y)=0$ for any $y\le -x$. Therefore, $m\left(x\right)$ cannot be of the form $c{x}^k$ for any y in Equation (47).

If $Y\in \mathbb{R}$, the integral in Equation (14) becomes

$${\int }_{-\infty }^{\infty }m(x+y)f_Y\left(y\right)\mathrm{d}y={\int }_{-x}^{\infty }m(x+y)f_Y\left(y\right)\mathrm{d}y.$$

(48)

If we are looking for the moment-generating function $M(x;\alpha )$, then the integral in Equation (4) must be divided into two parts:

$$\begin{array}{ccc}\hfill {\int }_{-\infty }^{\infty }M(x+y;\alpha )f_Y\left(y\right)\mathrm{d}y& =& {\int }_{-\infty }^{-x}1\times f_Y\left(y\right)\mathrm{d}y+{\int }_{-x}^{\infty }M(x+y;\alpha )f_Y\left(y\right)\mathrm{d}y\hfill \\ & =& P[Y\le -x]+{\int }_{-x}^{\infty }M(x+y;\alpha )f_Y\left(y\right)\mathrm{d}y.\hfill \end{array}$$

(49)

Finding a random variable Y for which Equation (4) is satisfied for a given form of the function $M(x;\alpha )$ then becomes even more challenging.

Proposition 3.

Let Y be a non-negative random variable having expected value

$$E\left[Y\right]=-{\displaystyle \frac{cf\left(x\right)+1}{\lambda c}}.$$

(50)

If the right-hand member of the above equation is non-negative, then the function $m\left(x\right)$ is given by $cx$. Moreover, if $E\left[Y\right]$ and $E\left[Y^2\right]$ are such that

$$2xE\left[Y\right]+E\left[Y^2\right]=-{\displaystyle \frac{cv\left(x\right)+2cxf\left(x\right)+1}{\lambda c}}\ge 0,$$

(51)

then $m\left(x\right)=c{x}^2$.

Proof. 

These results follow from Equation (47) with $k=1$ and $k=2$, respectively. □

Remark 3.

(i) In the case when $f\left(x\right)\equiv \mu <-1/c$, Equation (50) is easily satisfied for a wide variety of random variables. Moreover, notice that the mean of Y does not depend on the function $v\left(x\right)$. (ii) For $k\ge 3$, the conditions on the moments of Y can become difficult to satisfy.

The second moment of $\tau \left(x\right)$, $m_2\left(x\right):=E\left[{\tau }^2\left(x\right)\right]$, satisfies the IDE

$${\displaystyle \frac{1}{2}}v\left(x\right)m_2^{″}\left(x\right)+f\left(x\right)m_2^{\prime }\left(x\right)+\lambda \left\{{\int }_0^{\infty }{m}_2(x+y)f_Y\left(y\right)\mathrm{d}y-m_2\left(x\right)\right\}=-2m\left(x\right).$$

(52)

For a Wiener process without jumps and with infinitesimal parameters $\mu =-1$ and ${\sigma }^2=1$, we find that $m_2\left(x\right)=x(x+1)$. Let us try a solution of the form $m_2\left(x\right)=cx(x+1)$, where $c>0$, in the above equation with $v\left(x\right)\equiv {\sigma }^2$ and $f\left(x\right)\equiv \mu <0$. We obtain that

$$c{\sigma }^2+\mu c(2x+1)+\lambda c\left\{(2x+1)E\left[Y\right]+E\left[Y^2\right]\right\}={\displaystyle \frac{2x}{\mu }}.$$

(53)

Assume that $\lambda =1$ and $\mu =-1$. Then, if $E\left[Y\right]=1/2$, Equation (53) is satisfied if and only if

$$E\left[Y^2\right]={\displaystyle \frac{(c-2)x}{c}}-{\sigma }^2+{\displaystyle \frac{1}{2}}$$

(54)

is positive for any $x>0$. It follows that the condition ${\sigma }^2<1/2$ must be fulfilled.

Next, suppose that the first-passage time is rather defined by

$$\tau \left(x\right)=inf\{t>0:X(t)\notin (0,1)\mid X(0)=x\in (0,1\left)\right\}.$$

(55)

We try to find random variables Y that yield $m\left(x\right)=cx(1-x)$. Substituting this function into Equation (14), we obtain the following proposition.

Proposition 4.

Suppose that the distribution of the random variable Y is x-dependent and that there can be no overshoot. If the moments $E\left[Y\right]$ and $E\left[Y^2\right]$ of the random variable Y satisfy the condition

$$(1-2x)E\left[Y\right]-E\left[Y^2\right]={\displaystyle \frac{cv\left(x\right)+c(2x-1)f\left(x\right)-1}{\lambda c}},$$

(56)

then $m\left(x\right)=cx(1-x)$.

Remark 4.

(i) The assumption that Y is x-dependent means that the jump size distribution depends on the current value of the jump-diffusion process. (ii) In general, in the case of jump-diffusion processes, the process can jump over a lower or an upper boundary. But if the process starts at $x\in [0,d]$ and if Y is uniformly distributed over the interval $[-x,0]$ (for example), then there can be no jump below the boundary at 0 or above the one at d.

Finally, if we take $f_Y\left(y\right)=\delta (y+x)$, the IDE (14) simplifies to

$${\displaystyle \frac{1}{2}}v\left(x\right)m^{″}\left(x\right)+f\left(x\right)m^{\prime }\left(x\right)-\lambda m\left(x\right)=-1.$$

(57)

Contrary to the case of the moment-generating function $M(x;\alpha )$, this ODE is not of the same form as the one satisfied by the corresponding function $m_0\left(x\right)$ when there are no jumps (that is, $\lambda =0$). If $\{X_c\left(t\right),t\ge 0\}$ is a standard Brownian motion and the complement of D is the interval $(0,1)$, we must solve

$${\displaystyle \frac{1}{2}}{m}^{″}\left(x\right)-\lambda m\left(x\right)=-1$$

(58)

subject to the boundary conditions $m\left(0\right)=m\left(1\right)=0$. The solution to this simple ODE that satisfies these conditions involves exponential functions. It is therefore very different from $m_0\left(x\right)=x(1-x)$.

4. A First-Passage-Place Probability

In this section, we consider the first-passage time $\tau \left(x\right)$ defined in Equation (55), and we define the function

$$p\left(x\right)=P\left[X\right(\tau \left(x\right))=1].$$

(59)

That is, we are interested in a first-passage-place probability. The function $p\left(x\right)$ must satisfy the boundary conditions $p\left(0\right)=0$ and $p\left(1\right)=1$, and be strictly increasing with x in $(0,1)$. The interval $(0,1)$ can obviously be generalized to any finite interval $(a,b)$. We assume that there can be no overshoot.

If $\{X_c\left(t\right),t\ge 0\}$ is a standard Brownian motion, the function $p\left(x\right)$ is equal to x. If we try this solution in the IDE (16), we obtain that

$$f\left(x\right)+\lambda E\left[Y\right]=0⟹E\left[Y\right]=-{\displaystyle \frac{f\left(x\right)}{\lambda }}.$$

(60)

Proposition 5.

As in Proposition 4, assume that the distribution of the random variable Y is x-dependent and that there can be no overshoot. For any jump-diffusion process with $f\left[X\right(s\left)\right]\equiv 0$ in Equation (1), the function $p\left(x\right)$ defined in Equation (59) is equal to x if the average jump size is equal to zero.

Remark 5.

Let us consider the x-dependent discrete random variable $Y=Y_x$ defined as follows:

$$Y_x=\left\{\begin{array}{cc}1-x,\hfill & \mathit{with}\mathit{probability}x,\hfill \\ -x,\hfill & \mathit{with}\mathit{probability}1-x\hfill \end{array}\right.$$

(61)

for any $x\in (0,1)$. With this function, there is no overshoot possible, and we indeed have $E\left[Y_x\right]=0$. Moreover, the jump-diffusion process will leave the interval $(0,1)$ at the moment of the first jump of the Poisson process $\left\{N\right(t),t\ge 0\}$.

Next, we assume that $p\left(x\right)=x^2$. Then, we find that Equation (16), under the same hypotheses as in Proposition 5, becomes

$$v\left(x\right)+2xf\left(x\right)+\lambda \left(2xE\left[Y\right]+E\left[Y^2\right]\right)=0.$$

(62)

With the random variable defined in Equation (61), this equation reduces to

$$v\left(x\right)+2xf\left(x\right)+\lambda x(1-x)=0,$$

(63)

which is possible. For example, if $\lambda =1$, then we could have $v\left(x\right)=x^2$ and $f\left(x\right)=-1/2$.

Now, let us try the function

$$p\left(x\right)={\displaystyle \frac{e^x-1}{e-1}}\mathrm{for}0\le x\le 1.$$

(64)

This function could give the probability $P\left[X\right(\tau \left(x\right))=1]$ if the jump size Y belongs to the interval $(-x,1-x)\forall x\in (0,1)$ and if

$$E\left[e^Y\right]={\displaystyle \frac{\lambda -f\left(x\right)-\frac{1}{2}v\left(x\right)}{\lambda }}$$

(65)

is positive.

When the jump size Y has the density function $\delta (y+x)$, Equation (16) simplifies to

$${\displaystyle \frac{1}{2}}v\left(x\right)p^{″}\left(x\right)+f\left(x\right)p^{\prime }\left(x\right)-\lambda p\left(x\right)=0.$$

(66)

When $\{X_c\left(t\right),t\ge 0\}$ is a standard Brownian motion, we find that

$$p\left(x\right)={\displaystyle \frac{{\mathit{e}}^{\sqrt{2\lambda }x}-{\mathit{e}}^{-\sqrt{2\lambda }x}}{{\mathit{e}}^{\sqrt{2\lambda }}-{\mathit{e}}^{-\sqrt{2\lambda }}}}\mathrm{for}0\le x\le 1.$$

(67)

As in the case of the function $m\left(x\right)$, we see that the function $p\left(x\right)$ is very different from the corresponding function for the pure diffusion process (namely, $p_0\left(x\right)=x$).

Finally, when there can be an overshoot, the problem sometimes becomes easy to solve. For example, suppose that the jump size Y is equal to $\pm 1$ with probability 1/2. Then, we can write that $p(x+1)=1$ and $p(x-1)=0$ for every $x\in (0,1)$. It follows that the IDE (16) is reduced to a simple non-homogeneous second-order linear ODE:

$${\displaystyle \frac{1}{2}}v\left(x\right)p^{″}\left(x\right)+f\left(x\right)p^{\prime }\left(x\right)+\lambda \left[1-p\left(x\right)\right]=0.$$

(68)

For a jump-diffusion process whose continuous part $\{X_c\left(t\right),t\ge 0\}$ is a standard Brownian motion, we easily find that

$$p\left(x\right)=1+{\displaystyle \frac{e^{\omega (x-1)}+e^{-\omega (x-1)}}{e^{\omega }-e^{-\omega }}}\mathrm{for}0\le x\le 1,$$

(69)

where $\omega :=\sqrt{2\lambda }$.

An important problem when there may be an overshoot of the boundary by the jump-diffusion process is to determine the distribution, or at least the mean, of this overshoot.

5. Discussion

The problem of computing the exact distribution of first-passage times or first-passage places for jump-diffusion processes is a difficult one. It entails solving an integro-differential equation, subject to appropriate boundary conditions. When the random jumps follow a discrete distribution, the IDE becomes a difference-differential equation. This type of equation is also hard to solve explicitly.

In this paper, instead of choosing a well-known distribution for the jump size, such as an exponential or a uniform distribution, and trying to solve the corresponding IDE, we considered the inverse problem: we chose a simple form for the solution of the IDE, and we tried to determine whether there exist distributions for the random jumps that indeed yield this simple form.

First, we treated the problem when the function that we are looking for is the moment-generating function of the first-passage time $\tau \left(x\right)$. Next, we sought simple expressions for $E\left[\tau \right(x\left)\right]$ as well as for $E\left[{\tau }^2\left(x\right)\right]$. Finally, we studied the problem of finding explicit solutions to a first-passage-place problem. We obtained several exact solutions for important jump-diffusion processes, in particular when the continuous part $\{X_c\left(t\right),t\ge 0\}$ is a Wiener process or a geometric Brownian motion.

The inverse first-passage problem that we considered is different from those treated by other authors for diffusion or jump-diffusion processes.

Future work could focus on obtaining explicit results for first-passage problems for two-dimensional jump-diffusion processes, where jumps may occur in both or only one of the two components. In such cases, when the jumps are continuous random variables, one would need to solve a partial integro-differential equation.