Key words

1 Introduction

Motivated by the need of describing the price of a risky asset by means of a process with bounded variations, which seems quite realistic in true markets, [4] introduced the geometric telegrapher’s process expressed via an exponential transformation of the telegrapher’s process. Paper [8] proposed a similar financial market model that is free of arbitrage under suitable conditions, and is based on a continuous time random motion with alternating constant velocities and jumps occurring when the velocities are switching. Other contributions on stochastic processes characterized by alternating finite velocities are given in [2,7,9] and [10]. Moreover, the problem of estimating the parameters of the geometric telegrapher’s process has been faced in [1].

In this contribution we study a modified version of the geometric telegrapher’s process under the assumption that the random times between consecutive slope changes are exponentially distributed with linearly increasing parameters. This is suggested in the recent paper [3], where a damped telegrapher’s process is studied. In this framework the trajectories of damped processes are continuous curves composed by stochastically smaller and smaller paths. Some examples of damped diffusion processes can be found in the literature of financial modeling, such as [6].

The damped geometric telegrapher’s process is introduced in Section 2, where we obtain its probability law and study the asymptotic behavior. The moment generating function-approach is then used to evaluate the m-th moment of the process.

We remark that our contribution can be seen as an initial attempt to modify the geometric telegrapher’s process. Specific problems of mathematical finance, such as the problem of existence of arbitrage opportunities, will be the object of future investigations.

2 The stochastic model and probability laws

Let us assume that the price of risky assets is described by the following stochastic process, named damped geometric telegrapher’s process:

(1)

where s 0 >0, a ∈ ℝ, c>0, and where N t is an alternating counting process characterized by independent random times U k , D k , k ≥ 1. Hence,

where T 2k =U (k)+D (k) and T 2k+1=T 2k +U k+1 for k=0, 1,…, with U (0)=D (0)=0 and

(2)

We assume that {U k } and {D k } are mutually independent sequences of independent random variables characterized by exponential distribution with parameters

(3)

respectively. We remark that process S t has bounded variations and its sample-paths are constituted by connected lines having exponential behavior, characterized alternately by growth rates a+c and a−c, where a is the growth rate of risky assets’ price in the absence of randomness, and c is the intensity of the random factor of alternating type. Assumption (3) implies that the reversal rates λ k and μ k linearly increase with the number of reversals, so that the sample paths of S t are subject to an increasing number of slope changes when t increases, this giving a damped behavior. An example is shown in Fig. 1.

Fig. 1
figure 1

A simulated sample path of S t

Full size image

Denoting by F (k)(u) the distribution function of the k-fold convolution of random variables U j (see (2)), hereafter we show a suitable method to disclose it.

Proposition 1. For k=1, 2, … we have

(4)

Proof. We proceed by induction on k. For k=1, the result is obvious. Let us now assume (4) holding for all m=1, … , k−1. Hence, due to independence,

this giving (4).

Making use of a similar reasoning, for k=1, 2, … we also have

(5)

Note that (4) and (5) identify with the distribution functions of the maximum of k independent and exponentially distributed random variables with parameters λ and μ, respectively. Moreover, denoting by \( \tilde U_j (\tilde D_j ),j \geqslant 1 \), independent and exponentially distributed random variables with parameters λ (μ), recalling (2) and (3) we remark that

In order to obtain the distribution function of process X t , let us now introduce the compound process

Hereafter we obtain the distribution function of Y t .

Proposition 2. For any fixed t>0 and y ∈ [0, +∞), we have

(6)

Proof. For t>0 the distribution function of Y t can be expressed as

where, due to (4),

Hence, recalling (5), we obtain

so that (6) immediately follows. _

Notice that P(Y t =0)=e−λt.

Let us now define the stochastic process identifying the total time spent by S t going upward:

so that

(7)

Proposition 3. For all 0<τ<t, the distribution function of W t is:

(8)

Moreover,

Proof. Note that, for a fixed value t 0 >0,

(9)

Moreover, if W t0=τ, τ ≤ t 0, and Y τ =t 0−τ (Y τ >t 0−τ), then the motion is going upward (downward) at time t 0. Finally, since Y t is an increasing process, due to (9), the survival function P(W t >τ) is equal to H(t−τ, τ) for 0<τ≤t. Hence, (8) immediately follows from (6).

Due to (7) and Proposition 3, the probability law of X t can be easily obtained.

Proposition 4. Let τ * =τ ∗(x, t)=(x+ct)/(2c). For all t>0 and x<ct we have

Moreover, P(X t <ct)=1−e−λt and P(Xt≤ct)=1.

In the following proposition we finally obtain the distribution function of S t .

Proposition 5. For all t>0 and x<s 0 e(a+c)t , we have

where \( A_\lambda (t) = \exp \left\{ { - \frac{\lambda }{2}\left( {1 - \frac{a}{c}} \right)t} \right\}\;and\;A_\mu (t) = \exp \left\{ { - \frac{\mu }{2}\left( {1 - \frac{a}{c}} \right)t} \right\} \) . Moreover,

Proof. It immediately follows from (1) and recalling Proposition 4.

By straightforward use of Proposition 5, hereafter we come to the probability law of S t , which is characterized by a discrete component on s 0 e(a+c)t having probability e−λt , and by an absolutely continuous component on (s 0 e(a−c)t , s 0 e(a+c)t).

Proposition 6. The absolutely continuous component of the probability law of S t for t>0 and x ∈ (s 0 e(a−c)t , s 0 e(a+c)t) is given by:

Some plots of density p(x, t) are shown in Fig. 2 for various choices of λ and t. Let us now analyze the behavior of p(x, t) in the limit as t tends to+∞.

Fig. 2
figure 2

Plot of p(x, t) for s 0=1, a=0.1, c=1, μ=2, and λ=2, 3, 4, 5, from bottom to top near the origin, with t=1 (left-hand-side) and t=3 (right-hand-side)

Full size image

Corollary 1. If λ(c−a)=μ(c+a) then

where β=λ/(c+a); whereas, if λ(c−a)≠μ(c+a) then

Hence, under condition λ(c−a)=μ(c+a), process S t has a stationary density which is of log-logistic type with shape parameter β and scale parameter s 0. We remark that a similar result also holds under the suitable scaling conditions given hereafter.

Corollary 2. Let α t =s 0 exp{at}. If λ=μ→+∞, c→+∞, with λ/c→θ, then

Let us now analyse the behavior of p(x, t) when x approaches the endpoints of its support, i.e. the interval [s 1 , s 2] := [s 0 e(a−c)t , s 0 e(a+c)t].

Corollary 3. For any fixed t>0, we have

Hereafter we express them-th moment of S t in terms of the Gauss hypergeometric function 2 F 1.

Proposition 7. Let m be a positive integer. Then, for t>0,

(10)

Proof. Due to Proposition 4, by setting y=(ct+x)/2c we have

(11)

After some calculations (11) gives

where I=(0, 1−e−λt). Hence, recalling the equation (3.194.1) of [5], and noting that \( E[S_t^m ] = s_0^m e^{mat} M_{X_t } (m) \), the right-hand-side of (10) immediately follows.

Figures 3 and 4 show some plots of mean and variance of S t , respectively, evaluated by using (10). The right-hand-sides of both figures show cases when condition λ(c−a)=μ(c+a) is fulfilled.

Fig. 3
figure 3

Plot of E(S t ) for (λ,μ)=(1.5, 0.9), (1.75, 1.05), (2, 1.2) (left-hand-side) and for (λ, μ)=(3, 1.8), (3.5, 2.1), (4, 2.4) (right-hand-side) from top to bottom, with s 0=1, a=0.5, c=2

Full size image
Fig. 4
figure 4

Plot of Var(S t ) for (λ, μ)=(1, 0.6), (1.5, 0.9), (2, 1.2) (left-hand-side) and for (λ, μ)=(6, 3.6), (7, 4.2), (8, 4.8) (right-hand-side) from top to bottom, with s 0=1, a=0.5, c=2.

Full size image

Remark 1. If λ=μ, then the moment (10) can be expressed as: