[]F. Cinque []The Negative Reflection Principle and the Telegraph Process []
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# The Negative Reflection Principle and the Joint Distribution of the
Telegraph Process and its Maximum

Fabrizio Cinque1
Department of Statistical Sciences, Sapienza University of Rome, Italy
<EMAIL_ADDRESS>
(May 22, 2021)

###### Abstract

In this paper we study the joint distributions of the telegraph process and
its maximum conditioned on the number of changes of direction and the initial
velocity. We prove that in the case of positive starting velocity, a form of
the reflection principle holds. We call it the negative reflection principle
and we generalize it to the paths of the random motions moving with constant
finite velocity. In the last section we obtain the conditional distribution of
the first passage time of the telegraph process by using the known results on
the maximum of the motion. Finally, we derive the distribution of the first
returning time to the origin.

Keywords:  Telegraph Process; First Passage Time; Reflection Principle;
Induction Principle

2020 MSC:  Primary 60G51; 60K99

##  1 Introduction

The one-dimensional symmetric telegraph process describes the position of a
particle starting from the origin $x=0$ at time $t=0$ and moving forwards and
backwards on the real line. It moves alternatively with two finite constant
velocity $+c$ and $-c$, with $c>0$, and the initial speed is uniformly chosen
between the two possible alternatives. The changes of direction are paced by a
homogeneous Poisson process $\\{N(t)\\}_{t\geq 0}$ of rate $\lambda>0$,
meaning that the displacements of the particle are exponentially distributed
with average length $c/\lambda$. By denoting the initial velocity of the
motion with $V(0)\sim Unif\\{-c,+c\\}$, we can define the symmetric telegraph
process $\\{\mathcal{T}(t)\\}_{t\geq 0}$ as follows

$\mathcal{T}(t)\coloneqq
V(0)\int_{0}^{t}(-1)^{N(s)}\mathop{}\\!\mathrm{d}s=V(0)\sum_{i=0}^{N(t)-1}\Bigl{(}T_{i+1}-T_{i}\Bigr{)}(-1)^{i}+V(0)(-1)^{N(t)}\Bigl{(}t-T_{N(t)}\Bigr{)}$
(1.1)

where $T_{i}$ is the $i$-th arrival time of the Poisson process for
$i=1,\dots,N(t)$ and $T_{0}=0\ a.s.$

The telegraph process is the prototype of the finite speed random motions and
it has been formally presented in [12]. Several authors have studied the
symmetric telegraph process (1.1), see [14], [19], as well as its
generalizations. The most common one is the asymmetric motion with velocities
$c_{1}$ and $-c_{2},\ 0<c_{1}\not=c_{2}>0$, and two possible rates of
reversals $\lambda_{1}\not=\lambda_{2}>0$, whose probability law was first
obtained in [2] and has been further investigate in many papers, for instance
[7]. We also recall the motion with Erlang distributed displacements, see [6],
and the motion describing a particle that uniformly chooses its velocity in
the continuous set $[-c,c]$, see [4].
Some researchers also undertook the study of the multidimensional extensions
of the telegraph process. Among the most remarkable papers we recall [22],
where the author present a planar motion with orthogonal directions, [16],
regarding the planar random motion with infinite possible directions, and [21]
regarding random motions with finite speed in spaces of dimension $n\geq 3$.

The most relevant feature of the random motions with finite velocity is that
they preserve the natural property of moving with finite speed along the same
direction, therefore they are suitable to describe real motions and they
emerge in several fields. For instance, in finance they can model the stock
prices or the volatility of financial markets, see [15], [23], while in
ecology, they model the displacements of wild animals on the soil, see [13].
Hence, finite speed random processes represent a realistic alternative to the
widely used diffusion processes.

In this paper we will focus on the symmetric telegraph process (1.1) which has
the following conditional distributions. Let $x\in(-ct,ct)$ and $v=\pm c$

$P\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\ V(0)=v,\
N(t)=2k\\}=\frac{(2k)!}{k!(k-1)!}\frac{(c^{2}t^{2}-x^{2})^{k-1}(ct+sign(v)x)}{(2ct)^{2k}}\mathop{}\\!\mathrm{d}x,$
(1.2)

for $k\in\mathbb{N}$, and

$P\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\ V(0)=v,\
N(t)=2k+1\\}=P\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\ N(t)=2k+1\\}$
$=P\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\
N(t)=2k+2\\}=\frac{(2k+1)!}{k!^{2}}\frac{(c^{2}t^{2}-x^{2})^{k}}{(2ct)^{2k+1}}\mathop{}\\!\mathrm{d}x,$
(1.3)

for $k\in\mathbb{N}_{0}$, see [3] and [5] for the proof of (1.2) and (1.3).

We also need to recall the fascinating result of [3] concerning the
conditional distribution of the maximum of the telegraph process, for
$n\in\mathbb{N},\ \beta\in(0,ct)$,

$P\Big{\\{}\max_{0\leq s\leq t}\mathcal{T}(s)\in\mathop{}\\!\mathrm{d}\beta\
\Big{|}\ V(0)=c,\
N(t)=n\Big{\\}}=2P\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\ |\
N(t)=n\\}.$ (1.4)

The main result of this paper has been inspired by (1.4) and it concerns the
existence of a one-to-one correspondence between the two following sets of
trajectories, for $\beta\in[0,ct)$, $x\in(2\beta-ct,\beta]$ and natural $n$,

$\big{\\{}\omega\in\Omega\ :s\mapsto X(\omega,s)\ s.t.\ V(\omega,0)=+c,\
N(\omega,t)=n,\ \max_{0\leq s\leq t}X(\omega,s)>\beta,\
X(\omega,t)=x\big{\\}},$ (1.5) $\big{\\{}\omega\in\Omega\ :s\mapsto
X(\omega,s)\ s.t.\ V(\omega,0)=-c,\ N(\omega,t)=n,\
X(\omega,t)=2\beta-x\big{\\}},$ (1.6)

where $\\{X(t)\\}_{t\geq 0}$ is a finite constant velocity random motion
(meaning the it moves with velocities $\pm c$) whose changes of direction are
governed by a point process $\\{N(t)\\}_{t\geq 0}$. We call this property
negative reflection principle. This resembles the reflection principle, but
the trajectories have opposite starting velocities. This means that the
trajectories in the first set concerns a motion starting with the positive
speed $c$, while the samples in the other set start with the negative velocity
$-c$. In Section 2 we prove the existence of a bijective relationship between
sets (1.5) and (1.6) and we give a graphical method to negatively reflect the
samples.

In the third section we study the connection between the telegraph process at
time $t>0$ and its maximum in the time interval $[0,t]$, $\displaystyle
M(t)\coloneqq\max_{0\leq s\leq t}\mathcal{T}(s)$, conditionally on the initial
velocity and the number of changes of direction occurred up to time $t$.
Throughout the work, we use the following notation for the conditional
probability measure, for integer $n\geq 0$ and real $t>0$

$P_{n}^{\pm}\\{\ \cdot\ \\}\coloneqq P\\{\ \cdot\ |\ V(0)=\pm c,\ N(t)=n\\}.$
(1.7)

We provide a specific proof of the relationship, for $\beta\in(0,ct)$ and
$x\in(2\beta-ct,\beta]$,

$P_{n}^{+}\\{M(t)>\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{n}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}$ (1.8)

with $n\in\mathbb{N}$. We define this formula the negative reflection
principle for the telegraph process. Again, the important difference with the
classical reflection principle is the inversion of the initial velocity.

From the results regarding the positive initial velocity, we obtain the
conditional distributions of the motion starting with negative initial
velocity. In this case we must consider the possibility that the particle does
not overcome the level $x=0$ in the time interval $[0,t]$. The conditional
probability mass of this event has a cyclic behavior and it has been proved in
[3]. It reads, for $k\in\mathbb{N}$

$P\big{\\{}\max_{0\leq s\leq
t}\mathcal{T}(s)=0\,|\,V(0)=-c,\,N(t)=2k-1\big{\\}}=\binom{2k}{k}\frac{1}{2^{2k}}=P\big{\\{}\max_{0\leq
s\leq t}\mathcal{T}(s)=0\,|\,V(0)=-c,\,N(t)=2k\big{\\}}.$

We recall the incredible fact that the previous probability mass is
independent of both the current time $t$ and the velocity $c$. By joining the
position of the particle at time $t$, $\mathcal{T}(t)$, the probabilities lose
this characteristic as well as their cyclic behavior, but we can establish
some other interesting relationships. Finally, we exhibit the joint
distribution of the telegraph process and its maximum conditioned on a
negative starting velocity only.

We point out that the interested reader can easily obtain some other important
conditional and unconditional distributions by means of the presented results.
For instance, we can immediately derive the conditional distribution of the
maximum of the telegraph bridge as well as the conditional distribution of the
motion in presence of an absorbing barrier; we recall [20] and the recent
works [8] and [9] concerning some further kinds of barriers.

In the last section we focus on the hitting times of the symmetric telegraph
process. We describe a new method to study the distribution of the first
passage time across a level $\beta>0,\ F_{\beta}$. The results concerning this
random time appeared for the first time in [10] and [11]. The authors used the
Darlin-Siegert’s relationship to prove the distributions of the first passage
time conditioned on the starting velocity only. The first passage time of the
telegraph process has been investigated in the papers [17], [18], [24] and
[25], also for some generalization of the telegraph process. Here, we provide
some results on the distribution of $F_{\beta}$ conditioned on the number of
reversals in a time interval $[0,t],\ t>0$, and the starting speed of the
motion. We achieve these results by means of the conditional distributions of
the maximum of the telegraph process presented in [3]. Finally, we study the
conditional and unconditional distribution of the returning time to the
origin.

It is interesting to observe that the telegraph process is related to a wider
class of processes, the stochastic fluid models. We reference [1] for some
general results for the first passage time, the returning time to the origin
and other random times for these models.

##  2 The negative reflection principle

We begin by considering the positively initially oriented telegraph process
and we present the intuition for the negative reflection principle of the
telegraph process, which is the starting point of the whole study.

By considering the fundamental relationship (1.4), we can write

$\displaystyle
P_{n}^{+}\\{\mathcal{T}(t)>\beta\\}+P_{n}^{-}\\{\mathcal{T}(t)>\beta\\}$
$\displaystyle=2P\\{\mathcal{T}(t)>\beta\ |\
N(t)=n\\}=P_{n}^{+}\\{M(t)>\beta\\}$ $\displaystyle=P_{n}^{+}\\{M(t)>\beta,\
\mathcal{T}(t)>\beta\\}+P_{n}^{+}\\{M(t)>\beta,\mathcal{T}(t)\leq\beta\\},$

for $n\in\mathbb{N},\ \beta\in[0,ct)$, thus

$P_{n}^{+}\\{M(t)>\beta,\
\mathcal{T}(t)\leq\beta\\}=P_{n}^{-}\\{\mathcal{T}(t)>\beta\\}.$ (2.1)

By conditioning on $N(t)=n$, each path of the telegraph process is uniquely
determined by the Poisson times $T_{1},...,T_{n}$. Their joint random variable
is uniformly distributed on the simplex since $N(t)$ is a homogeneous Poisson
process. This means that each trajectory of the process has the same
probability (density). Thus relation (2.1) suggests that for each trajectory
of the set on the right-hand side there exits a trajectory in the set on the
left-hand side. With this at hand, it is reasonable to study if there exists a
one-to-one correspondence between the following sets of trajectories of the
telegraph process,, for $n\in\mathbb{N}$ and $\beta\in[0,ct)$

$\\{\omega\in\Omega\ :\ V(\omega,0)=+c,\ N(\omega,t)=n,\max_{0\leq s\leq
t}\mathcal{T}(\omega,s)>\beta,\ \mathcal{T}(\omega,t)\leq\beta\\},$ (2.2)
$\\{\omega\in\Omega\ :\ V(\omega,0)=-c,\ N(\omega,t)=n,\
\mathcal{T}(\omega,t)>\beta\\},$ (2.3)

with $\bigl{(}\Omega,\mathcal{F},P\bigr{)}$ probability space.

In Theorem 3.1 we generalize result (2.1), by proving that for
$\beta\in[0,ct)$ and $x\in(2\beta-ct,\beta]$

$P_{n}^{+}\\{M(t)>\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{n}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}$ (2.4)

with natural $n\in\mathbb{N}$. We refer to this formula as the negative
reflection principle for the telegraph process. The adjective negative is due
to the inversion of the starting speed. We point out that the classical
reflection principle, regarding Wiener process, does not take into account the
initial speed of the motion, in fact Brownian motion has both infinite
velocity and infinite rate of changes of direction.

Here, provide a proof of the negative reflection principle for a wide class of
finite speed random motions, where the symmetric telegraph process is just a
particular case.

######  Theorem 2.1 (Negative reflection principle).

Let $\bigl{(}\Omega,\mathcal{F},P\bigr{)}$ be a probability space. Let
$\\{X(t)\\}_{t\geq 0}$ be a finite velocity random motion such that

$X(t)=V_{0}\int_{0}^{t}(-1)^{N(s)}\mathop{}\\!\mathrm{d}s$ (2.5)

where $V_{0}\in\\{\pm c\\}\ a.s.,\ c>0$, and $\\{N(t)\\}_{t\geq 0}$ is an
independent point process whose waiting times have a common support
$(0,+\infty)$.
Let $\beta\in[0,ct)$ and $x\in(2\beta-ct,\beta]$. There exists a bijective
relationship between the two following sets of trajectories of the process
$X(t)$

$\displaystyle\mathcal{P}_{n}^{+}=\\{\omega\in\Omega\ :s\mapsto X(\omega,s)\
s.t.\ V_{0}(\omega)=+c,\ N(\omega,t)=n,\ \max_{0\leq s\leq
t}X(\omega,s)>\beta,\ X(\omega,t)=x\\}$ (2.6)

and

$\mathcal{P}_{n}^{-}=\\{\omega\in\Omega\ :s\mapsto X(\omega,s)\ s.t.\
V_{0}(\omega)=-c,\ N(\omega,t)=n,\ X(\omega,t)=2\beta-x\\}$ (2.7)

for $n\in\mathbb{N}$.

The reader can notice that this bijective relationship holds even if we remove
the condition on the number of changes of direction $N(t)=n$ from both the
sets in (2.6) and (2.7).

The process $X(t)$ is a non-jumping stochastic motion that moves with two
alternating velocities $\pm c$ and starts at $x=0$ at time $t=0$ with an
initial speed randomly chosen between the two possible ones. Its changes of
direction are governed by the point process $N(t)$.
Let $\omega^{+}\in\mathcal{P}_{n}^{+}$ and
$f:\mathcal{P}_{n}^{+}\longrightarrow\mathcal{P}_{n}^{-}$ being the bijective
application between sets (2.6) and (2.7). We call
$\omega^{-}=f(\omega^{+})\in\mathcal{P}_{n}^{-}$ the negatively reflected
sample of $\omega^{+}$ and vice versa.

###### Proof.

First of all we point out that we require $\beta\in[0,ct),\ x\in(2\beta-
ct,\beta]$ otherwise the sets are both negligible.

$\beta$$t_{1}$A$t_{2}$B$t$$x$C$\boldsymbol{s}$$\boldsymbol{X(s)}$O Figure 1:
Sample path $\omega^{+}\in\mathcal{P}_{5}^{+}$, with $n=5$.

$\beta$O’ $\ \ \ \
t_{2}-t_{1}$$t_{2}$B$t$$2\beta-x$C’$\boldsymbol{s}$$\boldsymbol{X(s)}$A’
Figure 2: $\omega^{-}\in\mathcal{P}_{5}^{-}$, the negatively reflected sample
of $\omega^{+}$.

Let $\omega^{+}\in\mathcal{P}_{n}^{+}$, then $\exists\ a.s.\
0<F_{\beta}^{(1)}(\omega^{+})=t_{1}<F_{\beta}^{(2)}(\omega^{+})=t_{2}<t$,
respectively the first and the second passage time across level $\beta$ of the
sample $\\{x\,:\,x=X(\omega^{+},s),\,s\in[0,t]\\}$. The negatively reflected
trajectory of $\omega^{+}$ is graphically obtained as follows (consider
Figures 2 and 2):

  * 1.

reflect the path from time $t_{1}$ to time $t_{2}$ across level $\beta$ (see
the polygonal curve from $A$ to $B$ in Figure 2);

  * 2.

translate the reflected poly-line in point 1 to the origin of the axes, so it
now starts at $(0,0)$ and ends at the point $(t_{2}-t_{1},0)$ (see the sample
from $A^{\prime}$ to $O^{\prime}$ in Figure 2);

  * 3.

consider the original path in the time interval $[0,t_{1}]$ and translate it
horizontally by the vector $(t_{2}-t_{1},0)$ so it starts at the point
$(t_{2}-t_{1},0)$, that is where the poly-line in steps 1-2 ends, and it ends
at $(t_{2},\beta)$ (see the poly-lines from $O^{\prime}$ to $B$ in Figure 2);

  * 4.

reflect the (original) path in the time interval $[t_{2},t]$ across level
$\beta$ (see the broken line from $B$ to $C$ in Figure 2 and its counterpart
from $B$ to $C^{\prime}$ in Figure 2).

This procedure produces one and only one negatively reflected sample of
$\omega^{+}\in\mathcal{P}_{n}^{+}$. By applying the method in the reverse way,
we can build one and only one trajectory in $\mathcal{P}_{n}^{+}$ starting
from an element $\omega^{-}\in\mathcal{P}_{n}^{-}$. Notice that in
$\omega^{-}$, $t_{2}-t_{1}$ represents the first crossing time through level 0
and $t_{2}$ represents the first crossing time through $\beta$. Figures 2 and
2 describe respectively a sample path in $\mathcal{P}_{n}^{+}$ and its
negatively reflected counterpart in $\mathcal{P}_{n}^{-}$, in the case $n=5$.

We now show analytically that the procedure above is bijective.
Let integer $n\geq 1$, every trajectory $\omega\in\Omega$ of the process
$X(\omega,t)$ can be described by the position of the motion at the $n$
arrival times, $0<T_{1}(\omega)=t_{1}<...<T_{n}(\omega)=t_{n}<t$ of the
process $N(t)$. Let $\omega^{+}\in\mathcal{P}_{n}^{+}$ and let
$F_{\beta}^{(1)}(\omega^{+})=t_{\beta}^{(1)},F_{\beta}^{(2)}(\omega^{+})=t_{\beta}^{(2)}$,
respectively the first and the second passage time across level $\beta$ in the
sample $\omega^{+}$. Let us assume $t_{h-1}<t_{\beta}^{(1)}<t_{h}$ and
$t_{l-1}<t_{\beta}^{(2)}<t_{l}$, with $1\leq h<l\leq n$, meaning that the
sample $\omega^{+}$ crosses $\beta$ for the first time along the $h$-th
displacement and it crosses $\beta$ for the second time along the $l$-th
displacement. We can uniquely describe $\omega^{+}$ by means of a
$(n+1)$-dimensional vector
$v^{+}=\phi(\omega^{+})=\Bigl{(}X(\omega^{+},t_{i})=x_{i}:i=1,...,n+1\Bigr{)}$,
where $t_{n+1}=t$ (we do not strictly need the $(n+1)$-th coordinate since it
can be expressed as an affine combination of the other elements, but we
maintain it for the sake of clarity). By applying the constructive method
above, we obtain the negatively reflected sample
$\omega^{-}\in\mathcal{P}_{n}^{-}$ and we can uniquely describe it by means of
$v^{-}=\phi(\omega^{-})$. Vectors $v^{+},v^{-}$ are respectively:

$v^{+}=\begin{pmatrix}\begin{rcases}\begin{array}[]{l}0<x_{1}\\\ \ \ \ \ \ \
\cdot\\\ \ \ \ x_{h-1}\end{array}\end{rcases}<\beta\\\
\begin{rcases}\begin{array}[]{l}\ \ \ \ x_{h}\\\ \ \ \ \ \ \ \cdot\\\ \ \ \
x_{l-1}\end{array}\end{rcases}>\beta\\\ x_{l}<\beta\\\ \cdot\\\ \cdot\\\
x_{n+1}=x<\beta\end{pmatrix},\ \ \ \ \ \ \
v^{-}=\begin{pmatrix}\begin{rcases}\begin{array}[]{l}\ \ \ y_{1}=\beta-
x_{h}\\\ \ \ \ \ \ \ \ \ \ \ \ \cdot\\\ y_{i}=\beta-x_{h-1+i}\\\ \ \ \ \ \ \ \
\ \ \ \ \cdot\\\ y_{l-h}=\beta-x_{l-1}\end{array}\end{rcases}<0\\\
\begin{rcases}\begin{array}[]{l}0<y_{l-h+1}=x_{1}\\\ \ \ \ \ \ \ \ \ \ \ \
\cdot\\\ \ \ \ \ \ y_{l-h+j}=x_{j}\\\ \ \ \ \ \ \ \ \ \ \ \ \cdot\\\ \ \ \ \ \
y_{l-1}=x_{h-1}\end{array}\end{rcases}<\beta\\\ y_{l}=2\beta-x_{l}>\beta\\\
\cdot\\\ \cdot\\\ y_{n+1}=2\beta-x>\beta\end{pmatrix}$ (2.8)

where the elements $y_{1},\dots,y_{l-h}$ are consequences of points 1-2 in the
building procedure (which work on $x_{h},\dots,x_{l-1}$), elements
$y_{l-h+1},\dots,y_{l-1}$ are consequences of point 3 (which work on
$x_{1},\dots,x_{h-1}$) and $y_{l},\dots,y_{n+1}$ are consequences of point 4
(which work on $x_{l},\dots,x_{n+1}$). Hence, we have the following bijective
affine relationship between $v^{+}$ and $v^{-}$

$v^{-}=f(v^{+})=\begin{pmatrix}0\ &\ -I_{l-h}\ &\ 0\\\ I_{h-1}\ &\ 0\ &\ 0\\\
0\ &\ 0\ &\ -I_{n-l+2}\end{pmatrix}v^{+}+\begin{pmatrix}\beta_{l-h}\\\
0_{h-1}\\\ 2\cdot\beta_{n-l+2}\end{pmatrix}$ (2.9)

where $I_{k}$ is the $k\times k$ identity matrix, $\beta_{k}$ and $0_{k}$ are
$k$-dimensional vectors of all $\beta$s and $0$s respectively.
It is important to notice that the sample $\omega^{+}\in\mathcal{P}_{n}^{+}$
was characterized by the pair $(h,l)$ which identifies the displacements of
the path where the motion crosses level $\beta$ for the first and the second
time respectively. By observing the vector representation $v^{-}$ in (2.8) of
the negatively reflected path of $\omega^{+}$,
$\omega^{-}\in\mathcal{P}_{n}^{-}$, we have that this trajectory crosses level
$0$ for the first time along its $(l-h+1)$-th displacement and crosses level
$\beta$ for the first time along its $l$-th displacement, then we can
characterize it by means of the pair $(l-h+1,l)$. Now we notice that any
sample $\omega^{-}\in\mathcal{P}_{n}^{-}$ with pair $(l-h+1,l),\ 1\leq h<l\leq
n$, can be uniquely expressed in a vector form $v^{-}=\phi(\omega^{-})$ as in
(2.8) and we can obtain its negatively reflected path
$\omega^{+}\in\mathcal{P}_{n}^{+}$, with pair $(h,l)$, by means of $f^{-1}$.
Hence, we have proved the existence of a bijective relationship between the
two following sets

$\mathcal{P}_{n}^{+}\cap\\{\omega\in\Omega\ :\
T_{h-1}(\omega)<F_{\beta}^{(1)}(\omega)<T_{h}(\omega)\ ,\
T_{l-1}(\omega)<F_{\beta}^{(2)}(\omega)<T_{l}(\omega)\\}$ (2.10)

and

$\mathcal{P}_{n}^{-}\cap\\{\omega\in\Omega\ :\
T_{l-h}(\omega)<F_{0}^{(1)}(\omega)<T_{l-h+1}(\omega)\ ,\
T_{l-1}(\omega)<F_{\beta}^{(1)}(\omega)<T_{l}(\omega)\\}$ (2.11)

where $F_{0}^{(1)}$ is the first returning time to $0$. Finally, we simply
observe that the function $g(h,l)=(l-h+1,l)$ is an automorphism. Thus, any
negatively reflected trajectory of a sample $\omega\in\mathcal{P}_{n}^{+}$,
with pair $(h,l)$, is different from the negatively reflected path of a sample
$\omega^{\prime}\in\mathcal{P}_{n}^{+}$ with pair
$(h^{\prime},l^{\prime})\not=(h,l)$. This concludes the proof of the theorem.
∎

We point out that in building a transformation between trajectories of a
finite velocity stochastic motion, it is important not to create new
constraints on the paths. For instance, to force a change of direction at a
specific level is a serious mistake.

It is important to keep in mind that the negative reflection principle
presented in Theorem 2.1 does not consider probabilities, it regards
trajectories only. Result (2.4), that we prove in Theorem 3.1, regarding the
probabilities, is due to the uniform distribution of the arrival times in the
case of the homogeneous Poisson process which implies that the trajectories
have all the same probability (density).

The negative reflection principle holds for both continuous and discrete
processes (it is sufficient to suitably adapt the point process $N(t)\,$ and
consider $X(t)$ at the desired times only). For instance, the reader can prove
that it holds in the case of simple random walks. In particular, for a simple
symmetric random walk we can obtain an equality in terms of the probabilities
since the sample paths of a symmetric random walk have all the same
probability.

######  Example 2.1 (Simple symmetric random walk).

Let $S_{0}=0\ a.s.$ and $S_{n}=S_{0}+\sum_{i=1}^{n}X_{i}$, with $X_{i}$
independent uniformly distributed random variables in $\\{-1,1\\}$. Let $n\geq
2$, for $\beta\in\\{0,1,...,n-2\\}$ and $x\in\\{2(\beta+1)-n,...,\beta\\}$ we
have that

$\displaystyle P\\{\max_{0\leq k\leq n}S_{k}>\beta,S_{n}=x\ |\ S_{1}=1\\}$
$\displaystyle=P\\{\max_{0\leq k\leq n-1}S_{k}>\beta-1,S_{n-1}=x-1\\}$
$\displaystyle=P\\{S_{n-1}=2\beta-x+1\\}=P\\{S_{n}=2\beta-x\ |\ S_{1}=-1\\},$

where in the second equality we used the known result on the maximum of a
simple symmetric random walk, due to the classical reflection principle. The
sets in the first and the fourth member respectively correspond to the sets
(2.6) and (2.7), with no condition on the point process $N(t)$, in the case of
random walks. $\diamond$

##  3 Joint distribution of the symmetric telegraph process and its maximum

In this section we present the conditional and unconditional joint
distributions of the position of the telegraph particle at time $t>0$ and its
maximum in the time interval $[0,t]$. The results we present here, easily
yield other conditional distributions of the symmetric telegraph process.

###  3.1 Motion starting with positive velocity

We now give a direct proof of the negative reflection principle for the
telegraph process.

######  Theorem 3.1 (Negative reflection principle for the telegraph process).

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$n\in\mathbb{N}$ and $x\in(-ct,ct)$

$P_{n}^{+}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\begin{cases}\begin{array}[]{l
l}0&if\ \beta<\max\\{0,x\\},\\\
P_{n}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{n}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}&if\
\max\\{0,x\\}\leq\beta<\frac{ct+x}{2},\\\
P_{n}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}&if\
\beta\geq\frac{ct+x}{2}.\end{array}\end{cases}$ (3.1)

###### Proof.

When $N(t)=n=1$ distribution (3.1) reduces to

$P_{1}^{+}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\begin{cases}\begin{array}[]{l
l}0&\ if\ \beta<\frac{ct+x}{2},\\\
P_{1}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\displaystyle\frac{\mathop{}\\!\mathrm{d}x}{2ct}&\
if\ \beta\geq\frac{ct+x}{2},\end{array}\end{cases}$ (3.2)

since the random variable $\mathcal{T}(t)$ is uniformly distributed in
$(-ct,ct)$ if one Poisson event occurs in the time interval $[0,t]$, see (1.3)
with $k=0$. To prove (3.2) we observe that the process reaches level $\beta$
at time $\frac{\beta}{c}$, then it changes direction and it keeps moving with
speed $-c$ until time $t$ where it will be located in
$\beta-c(t-\frac{\beta}{c})=2\beta-ct=x$. Thus,
$P_{1}^{+}\Big{\\{}M(t)=\frac{ct+\mathcal{T}(t)}{2},\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\Big{\\}}=P_{1}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$.

The first and third cases of (3.1) are trivial for $n\geq 2$. We focus on the
second case, which can be written as follows, for $\beta\in[0,ct)$ and
$x\in(2\beta-ct,\beta]$,

$P_{n}^{+}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{n}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{n}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}.$ (3.3)

For $n=2$, denoting with $T_{j}$ the $j$-th Poisson arrival time, we consider
that

$P^{+}_{2}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\int_{0}^{\frac{\beta}{c}}P\Big{\\{}T_{1}\in\mathop{}\\!\mathrm{d}t_{1},\
T_{2}\in\frac{ct-\mathop{}\\!\mathrm{d}x}{2c}+T_{1}\ |\
N(t)=2\Big{\\}}=\frac{\beta}{c^{2}t^{2}}\mathop{}\\!\mathrm{d}x$

which coincides with result (3.3) when $n=2$.

We now prove (3.3) for all natural $n>2$ by means of an induction argument.
The case above, $n=2$, represents the induction base for $n$ even. Let natural
$n\geq 3$, at time $T_{2}=t_{2}<t$:

\- Case 1. The motion has enough time to cross level $\beta$ and reach $x$ at
time $t$, so if $\\\ c(t-t_{2})\geq(\beta-2ct_{1}+ct_{2})+(\beta-x)\iff
t_{2}\leq\frac{ct-2\beta+x}{2c}+t_{1}$.

\- Case 2. The motion has not enough time to cross level $\beta$, but it has
time to reach $x$ at time $t$, then if $t_{2}>\frac{ct-2\beta+x}{2c}+t_{1}$
and $c(t-t_{2})\geq|x-2ct_{1}+ct_{2}|$.
Thus we can write the following recurrence relationship, for natural $n\geq 3$

$\displaystyle
P_{n}^{+}\\{M(t)\leq\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=$ (3.4)
$\displaystyle=\int_{0}^{\frac{\beta}{c}}\int_{t_{1}}^{\frac{ct-2\beta+x}{2c}+t_{1}}P_{n-2}^{+}\\{M(t-t_{2})\leq\beta-2ct_{1}+ct_{2},\mathcal{T}(t-t_{2})\in\mathop{}\\!\mathrm{d}x-2ct_{1}+ct_{2}\\}$
$\displaystyle\ \ \ \times
P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1},T_{2}\in\mathop{}\\!\mathrm{d}t_{2}|N(t)=n\\}$
$\displaystyle\ \ \
+\int_{0}^{\frac{\beta}{c}}\int_{\frac{ct-2\beta+x}{2c}+t_{1}}^{\frac{ct-x}{2c}+t_{1}}P_{n-2}^{+}\\{\mathcal{T}(t-t_{2})\in\mathop{}\\!\mathrm{d}x-2ct_{1}+ct_{2}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1},\
T_{2}\in\mathop{}\\!\mathrm{d}t_{2}\ |\ N(t)=n\\}$

Let $k\in\mathbb{N}$, we assume that distribution (3.3) holds for $n=2k$
(induction hypothesis). Hence, for $n=2k+2$, formula (3.4) reads

$\displaystyle P_{2k+2}^{+}$
$\displaystyle\\{M(t)\leq\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\int_{0}^{\frac{\beta}{c}}\int_{t_{1}}^{\frac{ct-x}{2c}+t_{1}}P_{2k}^{+}\\{\mathcal{T}(t-t_{2})\in\mathop{}\\!\mathrm{d}x-2ct_{1}+ct_{2}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1},\
T_{2}\in\mathop{}\\!\mathrm{d}t_{2}\ |\ N(t)=2k+2\\}$ $\displaystyle\ \ \
-\int_{0}^{\frac{\beta}{c}}\int_{t_{1}}^{\frac{ct-2\beta+x}{2c}+t_{1}}P_{2k}^{-}\\{\mathcal{T}(t-t_{2})\in
2\beta-\mathop{}\\!\mathrm{d}x-2ct_{1}+ct_{2}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1},\
T_{2}\in\mathop{}\\!\mathrm{d}t_{2}\ |\ N(t)=2k+2\\}$
$\displaystyle=\mathop{}\\!\mathrm{d}x\int_{0}^{\frac{\beta}{c}}\mathop{}\\!\mathrm{d}t_{1}\int_{t_{1}}^{\frac{ct-x}{2c}+t_{1}}\frac{(ct-x+2ct_{1}-2ct_{2})^{k-1}(ct+x-2ct_{1})^{k}}{\bigl{[}2c(t-t_{2})\bigr{]}^{2k}}\frac{(2k+2)!}{k!(k-1)!}\frac{(t-t_{2})^{2k+2}}{t^{2k+2}}\mathop{}\\!\mathrm{d}t_{2}$
$\displaystyle\ \ \
-\mathop{}\\!\mathrm{d}x\frac{(2k+2)!}{k!(k-1)!}\int_{0}^{\frac{\beta}{c}}\mathop{}\\!\mathrm{d}t_{1}\int_{t_{1}}^{\frac{ct-2\beta+x}{2c}+t_{1}}\frac{(ct-2\beta+x+2ct_{1}-2ct_{2})^{k}(ct+2\beta-x-2ct_{1})^{k-1}}{(2c)^{2k}t^{2k+2}}\mathop{}\\!\mathrm{d}t_{2}$
$\displaystyle=\frac{(2k+2)!}{(k+1)!\,k!}\frac{\mathop{}\\!\mathrm{d}x}{(2ct)^{2k+2}}\Bigl{[}(c^{2}t^{2}-x^{2})^{k}(ct+x)-\bigl{[}c^{2}t^{2}-(2\beta-x)^{2}\bigr{]}^{k}\bigl{[}ct-(2\beta-x)\bigr{]}\Bigr{]}$
$\displaystyle=P_{2k+2}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k+2}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}$

which concludes the proof of the Theorem for $n$ even.

In the case of $n$ odd the proof works in the same way and therefore it is
omitted. We note only that in formula (3.4) for $n=3$, by considering (3.2),
the first term is equal to $0$ and it is trivial to see that (3.3) holds when
$n=3$. This is the induction base in the case of $n$ odd. ∎

Thanks to the negative reflection principle for the telegraph process (3.3),
we easily achieve the joint density of the telegraph process and its maximum
as we show in the next corollaries.

######  Corollary 3.1.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$\beta\in(0,ct),\ x\in(2\beta-ct,\beta)$

$P_{2k+1}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\frac{(2k+1)!}{k!(k-1)!}\frac{(2\beta-x)\bigl{(}c^{2}t^{2}-(2\beta-x)^{2}\bigr{)}^{k-1}}{2^{2k-1}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x,$
(3.5)

for $k\in\mathbb{N}$. For $|x|<ct$

$P_{1}^{+}\\{M(t)=\frac{ct+\mathcal{T}(t)}{2},\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{1}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\},$
(3.6)

or alternatively, for $\beta\in(0,ct)$

$P_{1}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)=2M(t)-ct\\}=P_{1}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}.$
(3.7)

###### Proof.

For natural $n\geq 2,\ \beta\in(0,ct)$ and $x\in(2\beta-ct,\beta)$

$P_{n}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}/\mathop{}\\!\mathrm{d}\beta=\frac{\partial}{\partial\beta}P_{n}^{+}\\{M(t)\leq\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=-\frac{\partial}{\partial\beta}P_{n}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}$ (3.8)

where in the last equation we used the negative reflection principle for the
telegraph process (3.3). Formula (3.5) follows from (3.8) and (1.3) by
performing simple calculations.

Result (3.6) was shown at the beginning of the proof of Theorem 3.1. ∎

######  Remark 3.1.

Let integer $k\geq 1,\ \beta\in(0,ct)$ and $2\beta-ct<x\leq\beta$

$\displaystyle P_{2k+1}^{+}$
$\displaystyle\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<x\\}=\mathop{}\\!\mathrm{d}\beta\int_{2\beta-
ct}^{x}\frac{(2k+1)!}{k!(k-1)!}\frac{(2\beta-y)\bigl{(}c^{2}t^{2}-(2\beta-y)^{2}\bigr{)}^{k-1}}{2^{2k-1}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}y$
$\displaystyle=\frac{(2k+1)!}{k!^{2}}\frac{\bigl{(}c^{2}t^{2}-(2\beta-x)^{2}\bigr{)}^{k}}{2^{2k}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}\beta=P_{2k+1}^{+}\\{\mathcal{T}(t)\in
2\mathop{}\\!\mathrm{d}\beta-x\\}=\frac{P_{2k+1}^{+}\\{M(t)\in
2\mathop{}\\!\mathrm{d}\beta-x\\}}{2}$

which for $x=\beta$ coincides with distribution (1.4). $\diamond$

We now give the details about the case when $N(t)=2k$ for natural $k$. In this
case we must consider that the motion moves with positive speed at time $t$
and it may occur that it reaches its maximum at the end of the time interval.
Therefore it can happen that the position at time $t$ coincides with the
maximum in $[0,t]$.

######  Corollary 3.2.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N}$

$P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\frac{(2k)!}{k!(k-1)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k-1}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-2}}{2^{2k-1}(ct)^{2k}}\bigl{[}ct+(2k-1)(2\beta-x)\bigr{]}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x,$
(3.9)

for $\beta\in(0,ct),\ x\in(2\beta-ct,\beta)$, and

$\displaystyle
P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
(3.10)
$\displaystyle=\frac{(2k)!}{k!(k-1)!}\frac{2\beta(c^{2}t^{2}-\beta^{2})^{k-1}}{(2ct)^{2k}}\mathop{}\\!\mathrm{d}\beta,$
(3.11)

for $\beta\in(0,ct)$.

###### Proof.

Thesis (3.9) follows from (3.8) and (1.2)

$\displaystyle P_{2k}^{+}$
$\displaystyle\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}/\mathop{}\\!\mathrm{d}\beta=-\frac{\partial}{\partial\beta}P_{2k}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\frac{(2k)!}{k!(k-1)!}\Biggl{[}\frac{k\bigl{[}c^{2}t^{2}-(2\beta-x)^{2}\bigr{]}^{k-1}-(k-1)\bigl{[}ct-(2\beta-x)\bigr{]}^{k}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-2}}{2^{2k-1}(ct)^{2k}}\Biggr{]}\mathop{}\\!\mathrm{d}x$
(3.12)

which coincides with (3.9).

$\displaystyle
P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{+}\\{M(t)>\beta,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\},$

where in the last equation we used the negative reflection principle for the
telegraph process (2.4). Alternatively we can simply use (3.3) with $x=\beta$.
∎

We point out that by comparing probabilities (3.10) and (4.18) we obtain that

$\displaystyle
P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=\frac{\mathop{}\\!\mathrm{d}\beta}{c\,\mathop{}\\!\mathrm{d}t}P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\
|\ V(0)=c,\ N(t)=2k\\}$ (3.13)
$\displaystyle=\frac{\beta}{ct}P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\ |\
V(0)=c,\ N(t)=2k\\}.$

By keeping in mind the result of [3] concerning the expected value of the
maximum of the telegraph process, that is, for integer $k\geq 1$,
$\mathbb{E}[M(t)\ |\ V(0)=c,\ N(t)=2k]=\binom{2k}{k}ct/2^{2k}$, we easily
obtain the conditional probability that the maximum in $[0,t]$ is equal to the
position of the motion at time $t$

$\displaystyle P\\{M(t)=\mathcal{T}(t)\ |\ V(0)=c,\ N(t)=2k\\}$
$\displaystyle=\int_{0}^{ct}P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
(3.14) $\displaystyle=\binom{2k}{k}\frac{1}{2^{2k}}=P\\{M(t)=0\ |\ V(0)=-c,\
N(t)=2k\\}.$

Formula (3.14) shows that the probability of the particle, moving with
positive initial velocity, of being at time $t$ at the maximum level it
reached in the interval $[0,t]$ is equal to the probability that the motion,
starting with negative speed, never crosses the level $0$ before time $t$. We
point out that probability (3.14) is independent of both the current time $t$
and the velocity $c$.

######  Corollary 3.3.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N},\ \beta\in(0,ct),\ x\in(2\beta-ct,\beta]$

$\displaystyle P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<x\\}$ $\displaystyle=P_{2k}^{-}\\{\mathcal{T}(t)\in
2\mathop{}\\!\mathrm{d}\beta-x\\}=\frac{P_{2k}^{-}\\{M(t)\in
2\mathop{}\\!\mathrm{d}\beta-x\\}}{2}$ (3.15)
$\displaystyle=\frac{2(2k)!}{k!(k-1)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}}{(2ct)^{2k}}\mathop{}\\!\mathrm{d}\beta.$

###### Proof.

By means of formula (3.12), we have

$\displaystyle P_{2k}^{+}$
$\displaystyle\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\ \mathcal{T}(t)<x\\}$
$\displaystyle=\frac{2(2k)!\
\mathop{}\\!\mathrm{d}\beta}{k!(k-1)!(2ct)^{2k}}\int_{2\beta-
ct}^{x}\biggl{(}k\bigl{[}c^{2}t^{2}-(2\beta-y)^{2}\bigr{]}^{k-1}-(k-1)\bigl{[}ct-(2\beta-y)\bigr{]}^{k}\bigl{[}ct+(2\beta-y)\bigr{]}^{k-2}\biggr{)}\mathop{}\\!\mathrm{d}y$
$\displaystyle=\frac{2(2k)!\
\mathop{}\\!\mathrm{d}\beta}{k!(k-1)!(2ct)^{2k}}\Biggl{[}k\int_{2\beta-
ct}^{x}\bigl{[}c^{2}t^{2}-(2\beta-y)^{2}\bigr{]}^{k-1}\mathop{}\\!\mathrm{d}y+\bigl{[}ct-(2\beta-x)\bigr{]}^{k}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}$
$\displaystyle\ \ \ -k\int_{2\beta-
ct}^{x}\bigl{[}ct-(2\beta-y)\bigr{]}^{k-1}\bigl{[}ct+(2\beta-y)\bigr{]}^{k-1}\mathop{}\\!\mathrm{d}y\Biggr{]},$

which coincides with (3.15). Note that in the last equality we integrated by
parts the second integral. ∎

By means of (3.15) we can easily check the following

$\displaystyle P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\leq\beta\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<\beta\\}+P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=2P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}+P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\},$

where we used (1.4) in the last equation.

######  Remark 3.2.

In light of (3.15), we can write

$\displaystyle
P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<\beta\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}-P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\},$
(3.16)

that is another form of (3.10).
Furthermore, we have that, for $\beta\in(0,ct)$ and $0<x<\beta$

$\displaystyle\int_{x}^{\beta}P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}w,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\},$ (3.17)

where in the last equation we used formulas (3.3) and (3.10). $\diamond$

It is important to establish these relationships in order to show the strong
connection between all the distributions. Furthermore, we are often able to
avoid the use of the explicit form of the joint distributions in the
calculations.

Now we present the joint distribution of the motion and its maximum
conditioned on a positive starting velocity.

######  Theorem 3.2.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. For
$\beta\in(0,ct)$

$P\Big{\\{}M(t)\in\mathop{}\\!\mathrm{d}\beta,\ \mathcal{T}(t)=2M(t)-ct\
\Big{|}\ V(0)=c\Big{\\}}=\frac{\lambda\,e^{-\lambda
t}}{c}\mathop{}\\!\mathrm{d}\beta,$ (3.18)

or alternatively, for $x\in(-ct,ct)$

$P\Big{\\{}M(t)=\frac{\mathcal{T}(t)+ct}{2},\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ \Big{|}\
V(0)=c\Big{\\}}=\frac{\lambda\,e^{-\lambda t}}{2c}\mathop{}\\!\mathrm{d}x.$
(3.19)

For $\beta\in(0,ct),\ x\in(2\beta-ct,\beta)$

$\displaystyle P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\ \mathcal{T}$
$\displaystyle(t)\in\mathop{}\\!\mathrm{d}x\ |\ V(0)=c\\}$ (3.20)
$\displaystyle=\frac{\lambda\,e^{-\lambda
t}}{c\sqrt{ct+(2\beta-x)}}\Biggl{[}\frac{\lambda(2\beta-x)}{c\sqrt{ct+(2\beta-x)}}\,I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}$
$\displaystyle\ \ \
+\biggl{(}\frac{\lambda(2\beta-x)}{c\sqrt{ct-(2\beta-x)}}+\frac{\sqrt{ct-(2\beta-x)}}{ct+(2\beta-x)}\biggr{)}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x$

and

$P\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\ |\
V(0)=c\\}=\frac{\lambda\beta\,e^{-\lambda
t}}{c\sqrt{c^{2}t^{2}-\beta^{2}}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}\beta.$
(3.21)

Finally, there is a singular component

$P\\{M(t)=\mathcal{T}(t)=ct\ |\ V(0)=c\\}=e^{-\lambda t}.$ (3.22)

As usual,
$I_{r}(x)=\sum_{j=0}^{\infty}\bigl{(}\frac{x}{2}\bigr{)}^{2j+r}\frac{1}{j!\Gamma(j+1+r)}$
is the modified Bessel function of first type with imaginary argument and
order $r\in\mathbb{R}$.

###### Proof.

Formulas (3.18) and (3.19) can be respectively derived from (3.6) and (3.7)
where only one change of direction occurs in the time interval $[0,t]$.

To prove result (3.20) we consider the following joint distributions. For
$\beta\in(0,ct)$ and $x\in(2\beta-ct,\beta)$, by means of (3.5) we obtain that

$P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\,N(t)\
\text{odd}\,|\,V(0)=c\\}=\frac{\lambda^{2}\,(2\beta-x)\,e^{-\lambda
t}}{c^{2}\,\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x$
(3.23)

and by means of (3.12) we obtain that

$\displaystyle P$ $\displaystyle\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\ N(t)\ \text{odd}\ |\
V(0)=c\\}/(\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x)$
$\displaystyle=\frac{\lambda^{2}\,e^{-\lambda
t}}{2c^{2}}\Biggl{[}I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}-\frac{ct-(2\beta-x)}{ct+(2\beta-x)}\,I_{2}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\Biggr{]}$
$\displaystyle=\frac{\lambda\,e^{-\lambda
t}}{c\bigl{[}ct+(2\beta-x)\bigr{]}}\,\Biggl{[}\frac{\lambda\,(2\beta-x)}{c}\,I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}+\sqrt{\frac{ct-(2\beta-x)}{ct+(2\beta-x)}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\Biggr{]},$
(3.24)

where in the last equality we used the recurrence relationship for the Bessel
functions, with $n$ integer

$I_{n+1}(x)=I_{n-1}(x)-\frac{2n}{x}I_{n}(x).$ (3.25)

By summing up (3.23) and (3.24) we obtain the claimed result (3.20).

To prove (3.21) it is sufficient to consider (3.11) and perform some simple
calculation.

Finally, the probability mass (3.22) coincides with the probability that no
switches occur in $[0,t]$. ∎

We point out that, as well as we showed in the conditional case (3.13),
probabilities (3.21) and (4.19) are related by

$P\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\ |\
V(0)=c\\}=\frac{\mathop{}\\!\mathrm{d}\beta}{c\,\mathop{}\\!\mathrm{d}t}P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\
|\ V(0)=c\\}.$ (3.26)

###  3.2 Motion starting with negative velocity

We now study the case when the motion starts with a negative initial velocity.
It is important to recall that, when $V(0)=-c$, the particle may not cross the
level $x=0$ in the time interval $[0,t]$ with a positive probability,
displayed in formula (1).

We begin by recalling the result of Theorem $3.1$ of [3], for $\beta\in(0,ct)$
and natural $k$, we have

$P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}=\frac{2(2k)!}{k!(k-1)!}\frac{(ct-\beta)^{k}(ct+\beta)^{k-1}}{(2ct)^{2k}}\mathop{}\\!\mathrm{d}\beta=2\,P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}.$
(3.27)

This distribution already shows that a classical reflection principle holds.
The next results are confirming this incredible behavior.

######  Theorem 3.3.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N}$, for $x\in(-ct,0]$

$\displaystyle P_{2k}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)=\mathcal{T}(t)\in-\mathop{}\\!\mathrm{d}x\\}$
(3.28)
$\displaystyle=P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
(3.29)

and, for $\beta\in(0,ct),\ x\in(2\beta-ct,\beta)$

$P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}.$ (3.30)

###### Proof.

For $k=1$, the reader can prove the theorem by means of (3.6) and by
proceeding as shown below.

We begin by showing formula (3.28). Let natural $k\geq 2$ and $x\in(-ct,0]$.
At time $T_{1}=t_{1}$ the particle is at level $-ct_{1}$ and:

\- Case 1. It has time to overcome level $0$ and to be in $x$ at time $t$,
i.e. if $c(t-t_{1})>ct_{1}+|x|=ct_{1}-x\iff
t_{1}<\frac{ct+x}{2c}\in\bigl{(}0,\frac{t}{2}\bigr{)}$.

\- Case 2. It has no time to overcome level $0$, but it must be in $x$ at time
$t$, i.e. if $c(t-t_{1})>ct_{1}+x\iff
t_{1}<\frac{ct-x}{2c}\in\bigl{(}\frac{t}{2},t\bigr{)}$.

In light of these two cases, we proceed as follows

$\displaystyle P_{2k}^{-}\\{M(t)=0,\ $
$\displaystyle\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\int_{0}^{\frac{ct+x}{2c}}P_{2k-1}^{+}\\{M(t-t_{1})\leq
ct_{1},\
\mathcal{T}(t-t_{1})\in\mathop{}\\!\mathrm{d}x+ct_{1}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\
|\ N(t)=2k\\}$ $\displaystyle\ \ \
+\int_{\frac{ct+x}{2c}}^{\frac{ct-x}{2c}}P_{2k-1}^{+}\\{\mathcal{T}(t-t_{1})\in\mathop{}\\!\mathrm{d}x+ct_{1}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\
|\ N(t)=2k\\}$
$\displaystyle=\int_{0}^{\frac{ct-x}{2c}}P_{2k-1}^{+}\\{\mathcal{T}(t-t_{1})\in\mathop{}\\!\mathrm{d}x+ct_{1}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\
|\ N(t)=2k\\}$ $\displaystyle\ \ \
-\int_{0}^{\frac{ct+x}{2c}}P_{2k-1}^{-}\\{\mathcal{T}(t-t_{1})\in
ct_{1}-\mathop{}\\!\mathrm{d}x\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\ |\
N(t)=2k\\}$
$\displaystyle=P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\},$

where in the second step we applied the negative reflection principle for the
telegraph process (3.3).

We now prove the joint density (3.30). Let natural $k\geq 2,\ \beta\in(0,ct)$
and $x\in(2\beta-ct,\beta)$. At time $T_{1}=t_{1}$ the motion is at position
$-ct_{1}$ and it starts moving with speed $+c$. It must have time to reach
$\beta$ and to be in $x$ at time $t$, then we need
$c(t-t_{1})>ct_{1}+\beta+|\beta-x|=ct_{1}+2\beta-x\iff
t_{1}<\frac{ct-2\beta+x}{2c}\in(0,t/2)$. Now, we can write

$\displaystyle P_{2k}^{-}$
$\displaystyle\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\int_{0}^{\frac{ct-2\beta+x}{2c}}P_{2k-1}^{+}\\{M(t-t_{1})\in\mathop{}\\!\mathrm{d}\beta+ct_{1},\,\mathcal{T}(t-t_{1})\in\mathop{}\\!\mathrm{d}x+ct_{1}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\,|\,N(t)=2k\\}$

and by keeping in mind (3.5) we obtain that it coincides with the joint
density (3.9) regarding the positively initially oriented motion with an even
number of reversals. ∎

######  Remark 3.3.

We want to underline the following relationships. From (3.29), we have, for
natural $k$ and $x\in(-ct,0]$

$P_{2k}^{-}\\{M(t)=0,\ \mathcal{T}(t)\leq
x\\}=\int_{-ct}^{x}P_{2k}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}y\\}=P_{2k+1}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}\frac{2ct}{(2k+1)\mathop{}\\!\mathrm{d}x}.$
(3.31)

By keeping in mind formulas (3.28) and (3.14) we easily check the following

$\int_{-ct}^{0}P_{2k}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}y\\}=\binom{2k}{k}\frac{1}{2^{2k}}=P\\{M(t)=0\
|\ V(0)=-c,\ N(t)=2k\\},$ (3.32)

that is the probability mass that the motion never crosses the level $x=0$
when it starts with a negative initial velocity and changes its direction $2k$
times in the time interval $[0,t]$. $\diamond$

By taking into account the properties of the symmetric telegraph process, it
is easy to show that, for $v=\pm c,\ \beta\in(0,ct)$ and natural $n$,

$P\\{\max_{0\leq s\leq t}\mathcal{T}(s)\in\mathop{}\\!\mathrm{d}\beta\ |\
V(0)=v,\ N(t)=n\\}=P\\{\min_{0\leq s\leq
t}\mathcal{T}(s)\in-\mathop{}\\!\mathrm{d}\beta\ |\ V(0)=-v,\ N(t)=n\\}.$
(3.33)

By considering (3.33) and (3.28) we obtain that

$P\\{\max_{0\leq s\leq
t}\mathcal{T}(s)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\,|\,V(0)=c,\,N(t)=2k\\}=P\\{\min_{0\leq
s\leq
t}\mathcal{T}(s)=0,\,\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\,|\,V(0)=c,\,N(t)=2k\\},$

with $\beta\in(0,ct)$ and $k\in\mathbb{N}$. This formula describes an
important relationship between the paths of the telegraph process. An
analogous equation can be derived for the case of a negative initial velocity
(we must switch the maximum and the minimum and consider $\beta\in(-ct,0)$ ).

######  Corollary 3.4.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N}$, for $\beta\in(0,ct),\\\ x\in(2\beta-ct,\beta]$

$P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<x\\}=P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<x\\},$ (3.34)

and, for $x\in(-ct,ct)$

$\displaystyle P_{2k}^{-}\\{M(t)$ $\displaystyle\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\begin{cases}\begin{array}[]{l
l}P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\},&\ if\
\max\\{0,x\\}\leq\beta<\frac{ct+x}{2},\\\
P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\},&\ if\
\beta\geq\frac{ct+x}{2}.\end{array}\end{cases}$ (3.35)

The reader can notice the incredible fact that (3.35) and (3.27) show that a
classical reflection principle holds in the case of a motion that starts
moving with a negative velocity and changes direction an even number of times.

###### Proof.

By means of (3.30), we have $P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<x\\}=\int_{2\beta-
ct}^{x}P_{2k}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}y\\}$ that proves equation $(\ref{met-
pripx})$.

For the second case of result (3.35) there is nothing to prove. We focus on
the first one, where $\max\\{0,x\\}\leq\beta<\frac{ct+x}{2}$. We split it into
two cases, $x<0$ and $x\geq 0$.
Let $x<0$, by means of (3.30) and (3.29), we obtain that

$\displaystyle P_{2k}^{-}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\int_{0}^{\beta}P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}w,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}+P_{2k}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=P_{2k}^{+}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}+\Bigl{(}P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-P_{2k}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}\Bigr{)}$
$\displaystyle=-P_{2k}^{-}\\{\mathcal{T}(t)\in
2\beta-\mathop{}\\!\mathrm{d}x\\}+P_{2k}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\},$

where in the last equation we used the negative reflection principle for the
telegraph process (3.3).
For $x\geq 0$, we have $P_{2k}^{-}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=\int_{x}^{\beta}P_{2k}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}w,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$, and the claimed result follows
by considering (3.30) and (3.17). ∎

We now consider the case when $N(t)=2k+1$ for natural $k\geq 0$. In this case,
the motion moves with positive speed at time $t$ and therefore it can reach
the maximum level also at the end of the time interval, like in the case of an
initially positively oriented motion that changes direction an even number of
times.

######  Theorem 3.4.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N}_{0}$. For $x\in(-ct,0]$

$\displaystyle P_{2k+1}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\binom{2k+1}{k}\frac{(ct-x)^{k-1}(ct+x)^{k}}{(2ct)^{2k+1}}\bigl{[}ct-(2k+1)x\bigr{]}\mathop{}\\!\mathrm{d}x$
(3.36)
$\displaystyle=P_{2k+1}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-\frac{(2k+1)!}{(k+1)!(k-1)!}\frac{(ct-x)^{k-1}(ct+x)^{k+1}}{(2ct)^{2k+1}}\mathop{}\\!\mathrm{d}x.$
(3.37)

For $\beta\in(0,ct)$ and $x\in(2\beta-ct,\beta)$

$\displaystyle P_{2k+1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
$\displaystyle=\frac{(2k+1)!}{k!(k-1)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}}{2^{2k}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x$
(3.38) $\displaystyle\ \ \
-\frac{(2k+1)!}{(k+1)!(k-2)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k+1}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-2}}{2^{2k}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x.$

For $\beta\in(0,ct)$

$\displaystyle
P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=\binom{2k+1}{k}\frac{(ct-\beta)^{k}(ct+\beta)^{k-1}}{(2ct)^{2k+1}}\bigl{[}ct+(2k+1)\beta\bigr{]}\mathop{}\\!\mathrm{d}\beta$
(3.39)
$\displaystyle=P_{2k+1}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-\frac{(2k+1)!}{(k+1)!(k-1)!}\frac{(ct-x)^{k+1}(ct+x)^{k-1}}{(2ct)^{2k+1}}\mathop{}\\!\mathrm{d}x.$

###### Proof.

First of all we prove the theorem for the particular case of $k=0$, i.e.
$N(t)=1$. In this case, the joint density (3.38) is identically null. On the
other hand, we have that the maximum is $0$ if $\mathcal{T}_{1}^{-}(t)\leq 0$
and therefore equation (3.37) reduces to the first term. Finally, if the
position of the motion at time $t$ is greater than $0$, then the maximum
coincides with $\mathcal{T}(t)$ and equation (3.39) reads
$P_{1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}=P_{1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}=P_{1}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}=\mathop{}\\!\mathrm{d}\beta/(2ct)$.

For $k\in\mathbb{N}$ the proof of (3.36) works like the proof of (3.29) in
Theorem 3.3 and therefore it is omitted.

The joint density (3.38) follows by applying the same arguments we used in the
proof of density (3.30).

We conclude by showing formula (3.39). For $\beta\in(0,ct)$,

$P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}=\int_{0}^{\frac{ct-\beta}{2c}}P_{2k}^{+}\\{M(t-t_{1})=\mathcal{T}(t-t_{1})\in\mathop{}\\!\mathrm{d}\beta+ct_{1}\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\,|\,N(t)=2k+1\\},$

where the integration set is obtained by considering that the first
probability in the integral is $0$ if $c(t-t_{1})<\beta+ct_{1}$. The claimed
result follows by using (3.11). ∎

######  Remark 3.4.

Let integer $k\geq 0$ and $x\in(-ct,0]$, by means of (3.37) we obtain

$P_{2k+1}^{-}\\{M(t)=0,\ \mathcal{T}(t)\leq
x\\}=\frac{2ctP_{2k+2}^{+}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}}{(2k+2)\mathop{}\\!\mathrm{d}x}=\frac{(2k+1)!}{(k+1)!k!}\frac{(c^{2}t^{2}-x^{2})^{k}(ct+x)}{(2ct)^{2k+1}},$

which resembles the relationship (3.31) in the case of an even number of
changes of direction. Furthermore, we immediately verify that

$P_{2k+1}^{-}\\{M(t)=0,\ \mathcal{T}(t)\leq
0\\}=\binom{2k+1}{k}\frac{1}{2^{2k+1}}=P_{2k+1}^{-}\\{M(t)=0\\},$ (3.40)

that is the probability that an initially negatively oriented motion, when
$2k+1$ changes of direction occur before time $t$, does not cross the starting
level $x=0$ in the time interval $[0,t]$. Incredibly, this probability mass is
independent of both $t$ and $c$. $\diamond$

######  Remark 3.5.

We point out that taking into account formulas (3.36) and (3.39), for
$\beta\in(0,ct)$

$P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}=P_{2k+1}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in-\mathop{}\\!\mathrm{d}\beta\\}.$ (3.41)

Therefore, by considering equation (3.40), we obtain that

$P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\\}=\binom{2k+1}{k}\frac{1}{2^{2k+1}}=P_{2k+1}^{-}\\{M(t)=0\\}.$
(3.42)

This equality suggests an interesting connection between the samples of the
sets in the first and the third members, concerning the initially negatively
oriented motion when an odd number of reversals occur. $\diamond$

######  Corollary 3.5.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$k\in\mathbb{N}_{0}$, for $\beta\in(0,ct),\ x\in(2\beta-ct,\beta]$

$P_{2k+1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\ \mathcal{T}(t)<x\\}\
=\frac{(2k+1)!}{(k+1)!(k-1)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k+1}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}}{2^{2k}(ct)^{2k+1}}\mathop{}\\!\mathrm{d}\beta$
(3.43)

and, for $x\in(-ct,ct)$

$\displaystyle P$ ${}_{2k+1}^{-}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$ (3.44)
$\displaystyle=\bigg{\\{}\begin{array}[]{l
l}P_{2k+1}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}-\frac{(2k+1)!}{(k+1)!(k-1)!}\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k+1}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}}{(2ct)^{2k+1}}\mathop{}\\!\mathrm{d}x&\
if\ \max\\{0,x\\}\leq\beta<\frac{ct+x}{2}\\\
P_{2k+1}^{-}\\{\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}&\ if\
\beta\geq\frac{ct+x}{2}\\\ \end{array}$ (3.47)

###### Proof.

The proof of probability (3.43) follows by using formula (3.38) and by
proceeding as we showed in the proof of Corollary 3.3.

For the second case of result (3.44) there is nothing to prove. We focus on
the first one, where $\max\\{0,x\\}\leq\beta<\frac{ct+x}{2}$. We consider two
cases, $x<0$ and $x\geq 0$.
Let $x<0$, by using (3.38), some calculation yield

$\displaystyle\int_{0}^{\beta}P_{2k+1}^{-}\\{$ $\displaystyle
M(t)\in\mathop{}\\!\mathrm{d}w,\ \mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$
(3.48)
$\displaystyle=\frac{(2k+1)!}{(k+1)!(k-1)!}\Biggl{[}\frac{(ct-x)^{k-1}(ct+x)^{k+1}}{(2ct)^{2k+1}}-\frac{\bigl{[}ct-(2\beta-x)\bigr{]}^{k+1}\bigl{[}ct+(2\beta-x)\bigr{]}^{k-1}}{(2ct)^{2k+1}}\Biggr{]}\mathop{}\\!\mathrm{d}x.$

Now, we have

$P_{2k+1}^{-}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{2k+1}^{-}\\{M(t)=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}+\int_{0}^{\beta}P_{2k+1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}w,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}$

and the first equation in (3.44) follows by using (3.37) and (3.48).
For $x\geq 0$, we have

$P_{2k+1}^{-}\\{M(t)\leq\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}=P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}+\int_{x}^{\beta}P_{2k+1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}w,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\\}.$

The second term follows by means of (3.38). By also considering (3.39) we
obtain (3.44). ∎

By applying (3.39) and (3.43) it is easy to check that

$\displaystyle P_{2k+1}^{-}\\{M(t)$
$\displaystyle\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)<\beta\\}+P_{2k+1}^{-}\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
$\displaystyle=\binom{2k+1}{k}\frac{(ct-\beta)^{k}(ct+\beta)^{k-1}}{(2ct)^{2k+1}}\bigl{[}(2k+1)ct+\beta\bigr{]}\mathop{}\\!\mathrm{d}\beta=P_{2k+1}^{-}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}$

which coincides with the result of Theorem 4.2 of [3].

We conclude this section by showing the distribution of the telegraph process
and its maximum conditioned on a negative starting velocity.

######  Theorem 3.5.

Let $\\{\mathcal{T}(t)\\}_{t\geq 0}$ be a symmetric telegraph process. Let
$x\in(-ct,0]$

$\displaystyle P\\{M(t)$ $\displaystyle=0,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\ V(0)=-c\\}$ (3.49)
$\displaystyle=e^{-\lambda t}\Biggl{[}\frac{-\lambda
x}{c(ct-x)}\,I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-x^{2}}\Bigr{)}+\frac{1}{\sqrt{c^{2}t^{2}-x^{2}}}\biggl{(}\frac{ct+x}{ct-x}-\frac{\lambda
x}{c}\biggr{)}I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-x^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}x,$

and

$P\\{M(t)=0,\ \mathcal{T}(t)=-ct\ |\ V(0)=-c\\}=e^{-\lambda t}.$ (3.50)

For $\beta\in(0,ct),\ x\in(2\beta-ct,\beta)$

$\displaystyle P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta$ $\displaystyle,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x\ |\ V(0)=-c\\}$ (3.51)
$\displaystyle=\ \frac{\lambda^{2}\,e^{-\lambda t}}{2c^{2}}\,\Biggl{[}\
\sum_{j=0}^{1}\biggl{(}\sqrt{\frac{ct-(2\beta-x)}{ct+(2\beta-x)}}\biggr{)}^{j}\,I_{j}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}$
$\displaystyle\ \ \
-\sum_{j=2}^{3}\biggl{(}\sqrt{\frac{ct-(2\beta-x)}{ct+(2\beta-x)}}\biggr{)}^{j}\,I_{j}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x,$

and

$P\\{M(t)=\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}\beta\ |\
V(0)=-c\\}=\frac{e^{-\lambda
t}}{ct+\beta}\Biggl{[}\frac{\lambda\beta}{c}\,I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}+\sqrt{\frac{ct-\beta}{ct+\beta}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}\beta.$
(3.52)

###### Proof.

Let $x\in(-ct,0]$. By means of (3.29) we obtain that

$P\\{M(t)=0,\ \mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\ N(t)\ \text{even}\ |\
V(0)=-c\\}=\frac{-\lambda x\,e^{-\lambda
t}}{\sqrt{c^{2}t^{2}-x^{2}}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-x^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}x.$
(3.53)

By using (3.37) and the recurrence relationship for the Bessel function (3.25)
with $n=1$, we have

$\displaystyle P\\{M(t)=0,$ $\displaystyle\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\ N(t)\ \text{odd}\ |\ V(0)=-c\\}$
$\displaystyle=\frac{e^{-\lambda t}}{ct-x}\Biggl{[}\frac{-\lambda
x}{c}\,I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-x^{2}}\Bigr{)}+\sqrt{\frac{ct+x}{ct-x}}\,I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-x^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}x.$
(3.54)

By summing up (3.53) and (3.54) we obtain the claimed result (3.49).

The probability mass (3.50) coincides with the probability that no Poisson
events occur in $[0,t]$.

To prove the joint density (3.51) we consider the following distributions. By
means of (3.30) and (3.12) we obtain that

$\displaystyle P\\{M$ $\displaystyle(t)\in\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\ N(t)\ \text{even}\ |\ V(0)=-c\\}$
(3.55) $\displaystyle=\frac{\lambda^{2}\,e^{-\lambda t}}{2c^{2}}\,\Biggl{[}\
I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}-\frac{ct-(2\beta-x)}{ct+(2\beta-x)}\,I_{2}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x.$

and by means of (3.38) we have that

$\displaystyle P\\{M(t)\in$ $\displaystyle\mathop{}\\!\mathrm{d}\beta,\
\mathcal{T}(t)\in\mathop{}\\!\mathrm{d}x,\ N(t)\ \text{odd}\ |\ V(0)=-c\\}$
(3.56) $\displaystyle=\frac{\lambda^{2}\,e^{-\lambda t}}{2c^{2}}\
\sum_{j=0}^{1}(-1)^{j}\biggl{(}\sqrt{\frac{ct-(2\beta-x)}{ct+(2\beta-x)}}\biggr{)}^{2j+1}\,I_{2j+1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-(2\beta-x)^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}\beta\mathop{}\\!\mathrm{d}x.$

By summing up (3.55) and (3.56) we obtain the distribution (3.51).

Finally, to prove (3.52) we just need to consider equation (3.41) and
probability (3.54). ∎

######  Remark 3.6.

From the distributions we displayed in this section, the interested reader can
obtain some interesting conditional and unconditional probabilities. For
instance it is easy to study the distribution of the maximum conditioned on
the position of the process at the end of the time interval, that is the
maximum of a telegraph bridge. It is also easy to compute the probability law
of the telegraph process in presence of an absorbing barrier and the
distribution of some kind of the telegraph meander. $\diamond$

##  4 On the random times of the symmetric telegraph process

In this section we focus on the conditional and unconditional distributions of
the first passage time and the returning time to the origin of the telegraph
process.

######  Remark 4.1.

We show the following interesting relationship between the time derivative of
the cumulative distribution functions of the maximum of the telegraph process
and its density, when we know the starting speed $V(0)$ and the number of
changes of direction $N(t)$. For $0<\frac{\beta}{c}<t,\ n\in\mathbb{N}_{0}$

$-\frac{\partial}{\partial
t}P^{\pm}_{n}\\{M(t)\leq\beta\\}=\frac{\beta}{t\,\mathop{}\\!\mathrm{d}\beta}P_{n}^{\pm}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}.$
(4.1)

In the case of $V(0)=+c$, from Corollary 5.1 of [3], we have that, for natural
$k\geq 0,\ \beta\in[0,ct]$

$P^{+}_{2k+1}\\{M(t)\leq\beta\\}=P^{+}_{2k+2}\\{M(t)\leq\beta\\}=\frac{\beta}{ct}\sum_{j=0}^{k}\binom{2j}{j}\frac{(c^{2}t^{2}-\beta^{2})^{j}}{(2ct)^{2j}}$
(4.2)

and we point out that (1.4) has the same cyclic behavior displayed by equation
(4.2).
Now, by suitably adapting (1.4) we have

$\displaystyle-$ $\displaystyle\frac{\partial}{\partial
t}P^{+}_{2k+1}\\{M(t)\leq\beta\\}=-\frac{\partial}{\partial
t}P^{+}_{2k+2}\\{M(t)\leq\beta\\}=-\frac{\partial}{\partial
t}\int_{0}^{\beta}\frac{2(2k+1)!}{k!^{2}}\frac{(c^{2}t^{2}-w^{2})^{k}}{(2ct)^{2k+1}}\mathop{}\\!\mathrm{d}w$
$\displaystyle=-\frac{2(2k+1)!}{k!^{2}\>2^{2k+1}}\,\frac{\partial}{\partial
t}\int^{\frac{\beta}{ct}}_{0}(1-y^{2})^{k}\mathop{}\\!\mathrm{d}y=\frac{2(2k+1)!}{k!^{2}\>2^{2k+1}}\,\frac{\beta}{ct^{2}}\Bigl{(}1-\frac{\beta^{2}}{c^{2}t^{2}}\Bigr{)}^{k}=\frac{\beta}{t\,\mathop{}\\!\mathrm{d}\beta}P_{2k+1}^{+}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}$
(4.3)

which concludes the proof when $V(0)=c$.

In the case of a negative starting speed, $V(0)=-c$, by means of (4.2) and
Corollary 5.2 of [3] we can write the following relationship for the
cumulative distribution function, for $\beta\in[0,ct],\ k\in\mathbb{N}$

$P_{2k}^{-}\\{M(t)\leq\beta\\}\ =\
P_{2k}^{+}\\{M(t)\leq\beta\\}+\binom{2k}{k}\frac{(c^{2}t^{2}-\beta^{2})^{k}}{(2ct)^{2k}}.$
(4.4)

Now, in light of result (4.3) and by means of (4.4), we can write

$\displaystyle-\frac{\partial}{\partial t}P^{-}_{2k}\\{M(t)\leq\beta\\}$
$\displaystyle=\frac{\beta}{t\,\mathop{}\\!\mathrm{d}\beta}P^{+}_{2k}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}-\binom{2k}{k}\frac{(2k)(c^{2}t^{2}-\beta^{2})^{k-1}}{(2ct)^{2k}\,t}\beta^{2}$
$\displaystyle=\frac{2(2k)!}{k!(k-1)!}\frac{(c^{2}t^{2}-\beta^{2})^{k-1}(ct-\beta)}{2^{2k}(ct)^{2k}}\frac{\beta}{t}=\frac{\beta}{t\,\mathop{}\\!\mathrm{d}\beta}P^{-}_{2k}\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\\}.$
(4.5)

For the odd case, from [3] we can write

$P_{2k+1}^{-}\\{M(t)\leq\beta\\}\ =\
\frac{2k+1}{2k+2}P_{2k}^{-}\\{M(t)\leq\beta\\}+\frac{1}{2k+2}P_{2k+1}^{+}\\{M(t)\leq\beta\\}$
(4.6)

and (4.1) follows by using results (4.3) and (4.5). $\diamond$

It is interesting to observe that relationship (4.1) is typical of self-
similar processes. However, the telegraph process is not self-similar in the
classical sense, but it enjoys a kind of quasi-similarity property with $H=1$.
In fact, the time-scaled process $\\{\mathcal{T}(at)\\}_{t\geq 0}$, with
$a>0$, can be explicitly written as

$\mathcal{T}(at)=V(0)\int_{0}^{at}(-1)^{N(s)}\mathop{}\\!\mathrm{d}s=aV(0)\int_{0}^{t}(-1)^{N(as)}\mathop{}\\!\mathrm{d}s,$

that is a scaled telegraph motion, but with rate $a\lambda>0$.

###  4.1 First passage time: motion starting with positive velocity

Let $F_{\beta}=\inf\\{s\geq 0\>:\>\mathcal{T}(s)\geq\beta\\}$ the first
passage time of the symmetric telegraph process across level $\beta$. Note
that for $\beta>0$ and natural $n$

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=c,\
N(t)=n\\}=P\\{F_{-\beta}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=-c,\ N(t)=n\\}$
(4.7)

and therefore we limit ourselves to study the first passage time across a
level $\beta>0$ for both the initial possible velocities.

It is clear that the telegraph process starting with positive velocity can hit
level $\beta>0$ at time $t\geq\frac{\beta}{c}$ and only if $N(t)$ is even. In
particular, if $N(t)=0$, the first passage through $\beta$ occurs at time
$\frac{\beta}{c}$ with probability one. In the following theorem we study the
distribution of the first passage time, by assuming the number of changes of
direction. Also in this case, $T_{\beta}\geq\beta/c\ a.s.$

######  Theorem 4.1.

Let $F_{\beta}$ be the first passage time of a symmetric telegraph process
across level $\beta>0$. Let $0<\frac{\beta}{c}<s\leq t$. For $k\in\mathbb{N}$

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=c,\
N(t)=2k\\}=\frac{(2k)!}{t^{2k}}\beta\sum_{j=1}^{k}\frac{(t-s)^{2k-2j}}{j!(j-1)!(2k-2j)!}\frac{(c^{2}s^{2}-\beta^{2})^{j-1}}{(2c)^{2j-1}}\mathop{}\\!\mathrm{d}s$
(4.8)

and

$P\Big{\\{}F_{\beta}=\frac{\beta}{c}\ \Big{|}\ V(0)=c,\
N(t)=2k\Big{\\}}=\Bigl{(}1-\frac{\beta}{ct}\Bigr{)}^{2k}.$

For $k\in\mathbb{N}_{0}$

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=c,\
N(t)=2k+1\\}=\frac{(2k+1)!}{t^{2k+1}}\beta\sum_{j=1}^{k}\frac{(t-s)^{2k+1-2j}}{j!(j-1)!(2k+1-2j)!}\frac{(c^{2}s^{2}-\beta^{2})^{j-1}}{(2c)^{2j-1}}\mathop{}\\!\mathrm{d}s$
(4.9)

and

$P\Big{\\{}F_{\beta}=\frac{\beta}{c}\ \Big{|}\ V(0)=c,\
N(t)=2k+1\Big{\\}}=\Bigl{(}1-\frac{\beta}{ct}\Bigr{)}^{2k+1}.$

We point out that the distribution for $s>t$ is not displayed. It is
interesting to observe that for $\beta\downarrow 0$, $F_{\beta}\longrightarrow
0\,a.s.$, conditionally on $V(0)=c$ and the number of switches. Note that if
$N(t)=1$, the absolutely continuous component (4.9) is null.

###### Proof.

Let $k\in\mathbb{N},\ 0<\frac{\beta}{c}<s\leq t$. By considering that
$P\big{\\{}F_{\beta}=\frac{\beta}{c}\ \big{|}\ V(0)=c,\
N(t)=2k\big{\\}}=P\big{\\{}N\bigl{(}\frac{\beta}{c}\bigr{)}=0\ \big{|}\
V(0)=c,\ N(t)=2k\big{\\}}$ we obtain result (4.1).

Throughout this proof we modify notation (1.7) by considering the Poisson
process at time $s>0$, $N(s)$, instead of at time $t\geq s$, therefore

$P_{n}^{\pm}\\{\ \cdot\ \\}=P\\{\ \cdot\ |\ V(0)=\pm c,\ N(s)=n\\}.$ (4.10)

To prove (4.8) we start by studying the following distribution of the maximum.
For integer $k\geq 1,\ 0<\frac{\beta}{c}<s\leq t$,

$\displaystyle P\\{M(s)\leq\beta\ |\ V(0)=c,\ N(t)=2k\\}$
$\displaystyle=\sum_{j=0}^{2k}P\\{M(s)\leq\beta,\ N(s)=j\ |\ V(0)=c,\
N(t)=2k\\}$
$\displaystyle=\sum_{j=0}^{2k}P_{j}^{+}\\{M(s)\leq\beta\\}\,\frac{P\\{N(s)=j\\}\,P\\{N(t-s)=2k-j\\}}{P\\{N(t)=2k\\}}$
$\displaystyle=\frac{1}{t^{2k}}\sum_{j=0}^{2k}P_{j}^{+}\\{M(s)\leq\beta\\}\binom{2k}{j}s^{j}(t-s)^{2k-j},$
(4.11)

where in the second equality we used that $M(s)$ and $N(t)$ are conditionally
independent with respect to $N(s)=j$. We can now can study the distribution of
the first passage time.

$\displaystyle P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=$
$\displaystyle c,\
N(t)=2k\\}/\mathop{}\\!\mathrm{d}s=-\frac{\partial}{\partial
s}P\\{M(s)\leq\beta\ |\ V(0)=c,\ N(t)=2k\\}$
$\displaystyle\begin{array}[]{l}\displaystyle=\frac{(2k)!}{t^{2k}}\>\sum_{j=0}^{2k}\frac{1}{j!(2k-j)!}\Biggl{[}\frac{\beta}{s\,\mathop{}\\!\mathrm{d}\beta}P_{j}^{+}\\{M(s)\in\mathop{}\\!\mathrm{d}\beta\\}\,s^{j}(t-s)^{2k-j}\
+\\\ \ \ \
-P_{j}^{+}\\{M(s)\leq\beta\\}\biggl{(}js^{j-1}(t-s)^{2k-j}-(2k-j)s^{j}(t-s)^{2k-j-1}\biggr{)}\Biggr{]},\end{array}$
(4.14)

where in the last equality we used (4.11) and (4.1).
We can write the first term of (4.14) as follows

$\displaystyle\frac{\beta}{s\,\mathop{}\\!\mathrm{d}\beta}\sum_{j=1}^{2k}$
$\displaystyle\frac{s^{j}(t-s)^{2k-j}}{j!(2k-j)!}P_{j}^{+}\\{M(s)\in\mathop{}\\!\mathrm{d}\beta\\}=\frac{\beta}{\mathop{}\\!\mathrm{d}\beta}\sum_{l=0}^{2k-1}\frac{s^{l}(t-s)^{2k-1-l}}{(l+1)!(2k-1-l)!}P_{l+1}^{+}\\{M(s)\in\mathop{}\\!\mathrm{d}\beta\\}$
(4.15)
$\displaystyle=\frac{\beta}{cs}\sum_{h=0}^{k-1}\frac{(t-s)^{2k-1-2h}}{(2k-1-2h)!}\frac{(c^{2}s^{2}-\beta^{2})^{h}}{h!^{2}(2c)^{2h}}\>+\>\frac{\beta}{\mathop{}\\!\mathrm{d}\beta}\,\sum_{h=0}^{k-1}\frac{s^{2h+1}(t-s)^{2k-2-2h}}{(2h+2)!(2k-2-2h)!}P_{2h+2}^{+}\\{M(s)\in\mathop{}\\!\mathrm{d}\beta\\},$

where the two sums concern the indexes $l$ even and odd respectively. We
simplify the second term of (4.14) as follows (note that, from (4.2), if $j$
odd, $P_{j}^{+}\\{M(s)\leq\beta\\}=P_{j+1}^{+}\\{M(s)\leq\beta\\}$ )

$\displaystyle\sum_{j=0}^{2k}\frac{P_{j}^{+}\\{M(s)\leq\beta\\}}{j!(2k-j)!}$
$\displaystyle\biggl{[}(2k-j)s^{j}(t-s)^{2k-j-1}-js^{j-1}(t-s)^{2k-j}\biggr{]}=$
$\displaystyle=\sum_{j=0}^{2k-1}P_{j}^{+}\\{M(s)\leq\beta\\}\frac{s^{j}(t-s)^{2k-1-j}}{j!(2k-1-j)!}-\sum_{l=0}^{2k-1}P_{l+1}^{+}\\{M(s)\leq\beta\\}\frac{s^{l}(t-s)^{2k-1-l}}{l!(2k-1-l)!}$
$\displaystyle=\sum_{j=0}^{k-1}\frac{s^{2j}(t-s)^{2k-1-2j}}{(2j)!(2k-1-2j)!}\Biggl{(}P_{2j}^{+}\\{M(s)\leq\beta\\}-P_{2j+1}^{+}\\{M(s)\leq\beta\\}\Biggr{)}$
$\displaystyle=-\frac{\beta}{cs}\sum_{j=0}^{k-1}\frac{(t-s)^{2k-1-2j}}{j!^{2}(2k-1-2j)!}\frac{(c^{2}s^{2}-\beta^{2})^{j}}{(2c)^{2j}},$
(4.16)

where in the last equation we used (4.2) and compute some trivial
calculations. Finally, by applying (4.15) and (4.16) in formula (4.14) we
obtain the claimed result (4.8).

The proof for the odd case work in the same way and therefore it is omitted. ∎

It is important to notice that, for natural $k$ and $0<\frac{\beta}{c}<t$

$\displaystyle P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=c,\
N(t)=2k\\}$
$\displaystyle=\frac{2(2k-1)!}{(k-1)!^{2}}\frac{(c^{2}t^{2}-\beta^{2})^{k-1}}{(2ct)^{2k-1}}\frac{\beta}{t}\mathop{}\\!\mathrm{d}t$
(4.17)
$\displaystyle=\frac{\beta\,\mathop{}\\!\mathrm{d}t}{t\,\mathop{}\\!\mathrm{d}\beta}P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta\
|\ V(0)=c,\ N(t)=2k\\}.$ (4.18)

Formula (4.18) connect the density of the first passage time across level
$\beta$ with the density of the maximum in a fascinating way which resembles
the relationship between the two distributions in the case of Brownian motion.

######  Theorem 4.2.

Let $F_{\beta}$ be the first passage time of a symmetric telegraph process
across level $\beta>0$. For $0<\frac{\beta}{c}<t$,

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\
V(0)=c\\}=\frac{\lambda\beta\,e^{-\lambda
t}}{\sqrt{c^{2}t^{2}-\beta^{2}}}I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}t$
(4.19)

and

$P\Big{\\{}F_{\beta}=\frac{\beta}{c}\ \Big{|}\
V(0)=c\Big{\\}}=e^{-\lambda\frac{\beta}{c}}.$ (4.20)

###### Proof.

By means of (4.18) we have

$\displaystyle P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=c\\}$
$\displaystyle=\frac{\beta\,\mathop{}\\!\mathrm{d}t}{t\,\mathop{}\\!\mathrm{d}\beta}\sum_{k=1}^{\infty}P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\
N(t)=2k\ |\ V(0)=c\\}$
$\displaystyle=\frac{\beta\,\mathop{}\\!\mathrm{d}t}{t\,\mathop{}\\!\mathrm{d}\beta}P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,N(t)\
\text{even}\ |\ V(0)=c\\},$

and probability (4.19) easily follows by considering that

$P\\{M(t)\in\mathop{}\\!\mathrm{d}\beta,\ N(t)\ \text{even}\ |\
V(0)=c\\}=\frac{\lambda t\,e^{-\lambda
t}}{\sqrt{c^{2}t^{2}-\beta^{2}}}I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}\mathop{}\\!\mathrm{d}\beta,$

that is formula (3.23) of [3]. ∎

Clearly, for $\beta\downarrow 0$, $F_{\beta}\longrightarrow 0\,a.s.$, if we
know that $V(0)=c$. Moreover, distribution (4.19) converges to the
distribution of the first passage time of Brownian motion under Kac’s
conditions, $\lambda,c\longrightarrow\infty$ such that
$\frac{\lambda}{c^{2}}\longrightarrow 1$. The reader can prove this assertion
by considering the asymptotic estimate of the Bessel functions, when
$x\longrightarrow\infty$, $I_{r}(x)\sim\frac{e^{x}}{\sqrt{2\pi x}}$ for
$r=0,1$.

###  4.2 First passage time: motion starting with negative velocity

In the case of the telegraph process starting with negative speed, the first
passage time across a level $\beta>0$ can occur at time $t>\frac{\beta}{c}$
and only if $N(t)$ is odd. In this case the unconditional distribution has no
singularity in $\frac{\beta}{c}$ and $T_{\beta}>\beta/c\ a.s.$ under any
condition.

######  Theorem 4.3.

Let $F_{\beta}$ be the first passage time of a symmetric telegraph process
across level $\beta>0$. For $0<\frac{\beta}{c}<s\leq t$ and
$k\in\mathbb{N}_{0}$

$\displaystyle P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ $ $\displaystyle
V(0)=-c,\ N(t)=2k+1\\}$ (4.21)
$\displaystyle=\frac{(2k+1)!}{t^{2k+1}}\sum_{j=0}^{k}\frac{(t-s)^{2k-2j}}{j!(j+1)!(2k-2j)!}\frac{(c^{2}s^{2}-\beta^{2})^{j-1}(cs-\beta)\bigl{[}cs+(2j+1)\beta\bigr{]}}{2^{2j+1}c^{2j}}\mathop{}\\!\mathrm{d}s$

and, for $k\in\mathbb{N}$

$\displaystyle P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}s\ |\ $ $\displaystyle
V(0)=-c,\ N(t)=2k\\}$ (4.22)
$\displaystyle=\frac{(2k)!}{t^{2k}}\sum_{j=0}^{k}\frac{(t-s)^{2k-1-2j}}{j!(j+1)!(2k-1-2j)!}\frac{(c^{2}s^{2}-\beta^{2})^{j-1}(cs-\beta)\bigl{[}cs+(2j+1)\beta\bigr{]}}{2^{2j+1}c^{2j}}\mathop{}\\!\mathrm{d}s$

We point out that the distribution for $s>t$ is not displayed. Note that for
$\beta\downarrow 0$, probability (4.21) reduces to
$\frac{(2k+1)!}{2t^{2k+1}}\sum_{j=0}^{k}\frac{(t-s)^{2k-2j}s^{2j}}{2^{2j}j!(j+1)!(2k-2j)!}\mathop{}\\!\mathrm{d}s$,
which does not depend on the velocity of the motion (an equivalent result
follows from (4.22) ).

###### Proof.

To prove the theorem we must proceed as we showed in the proof of Theorem 4.1.
In this case, the difference is that, considering notation (4.10), for integer
$k\geq 0$, we have

$P_{2k+1}^{-}\\{M(s)\leq\beta\\}-P_{2k+2}^{-}\\{M(s)\leq\beta\\}=-2\beta\binom{2k+1}{k}\frac{(c^{2}s^{2}-\beta^{2})^{k}}{(2cs)^{2k+2}}(cs-\beta),$
(4.23)

which follow from (4.4) and (4.6). Note that in the case of $V(0)=c$ this
difference is null. ∎

It is important to notice that, for $k\in\mathbb{N}_{0},\ 0<\frac{\beta}{c}<t$

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=-c,\
N(t)=2k+1\\}=\binom{2k+1}{k}\frac{(c^{2}t^{2}-\beta^{2})^{k-1}(ct-\beta)\bigl{[}ct+(2k+1)\beta\bigr{]}}{c^{2k}(2t)^{2k+1}}\mathop{}\\!\mathrm{d}t.$
(4.24)

######  Theorem 4.4.

Let $F_{\beta}$ be the first passage time of a symmetric telegraph process
across level $\beta>0$. For $0<\frac{\beta}{c}<t$

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=-c\\}=\frac{e^{-\lambda
t}}{ct+\beta}\Biggr{[}\lambda\beta
I_{0}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}+c\sqrt{\frac{ct-\beta}{ct+\beta}}I_{1}\Bigl{(}\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\beta^{2}}\Bigr{)}\Biggr{]}\mathop{}\\!\mathrm{d}t.$
(4.25)

###### Proof.

$P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\
V(0)=-c\\}=\sum_{k=0}^{\infty}P\\{F_{\beta}\in\mathop{}\\!\mathrm{d}t\ |\
V(0)=-c,\ N(t)=2k+1\\}P\\{N(t)=2k+1\\}$

and the result (4.25) follows by means of (4.24) (we suggest to rewrite its
last factor considering that
$\bigl{[}ct+(2k+1)\beta\bigr{]}=(2k+2)\beta+(ct-\beta)$ ). ∎

Under Kac’s conditions, distribution (4.25) converges to the distribution of
the first passage time of Brownian motion. Finally, if $\beta\downarrow 0$,
density (4.25) reads $e^{-\lambda t}/t\,I_{1}(\lambda t)$ and as above, this
quantity does not depend on $c$.

We want to recall that distributions (4.17), (4.19), (4.24) and (4.25) were
first proved in [11] by means of the Darlin-Siegert’s relationship. The reader
can also find more general expression in Theorem 3.8 of [15].

###  4.3 Return time to the origin

The results on the first passage time easily yield the distribution of the
first returning time to the origin for the telegraph process,
$F_{0}=\inf\\{s>0\ :\ \mathcal{T}(s)=0\\}$. It is clear that the probabilities
of the first returning time conditioned on $V(0)=c$ coincide with those
conditioned on $V(0)=-c$.

######  Theorem 4.5.

Let $F_{0}$ be the first returning time to the origin of a symmetric telegraph
process. Let $v=\pm c$. For $0<s\leq t$

$P\\{F_{0}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=v,\
N(t)=1\\}=\frac{\mathop{}\\!\mathrm{d}s}{2t},$ (4.26)

and, for $k\in\mathbb{N}$

$\displaystyle P\\{F_{0}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=v,\
N(t)=2k+1\\}=\frac{(2k+1)!}{t^{2k+1}}\sum_{j=1}^{k}\frac{(t-s)^{2k-2j}\,s^{2j}}{j!(j+1)!(2k-2j)!\,2^{2j+1}}\mathop{}\\!\mathrm{d}s,$
(4.27) $\displaystyle P\\{F_{0}\in\mathop{}\\!\mathrm{d}s\ |\ V(0)=v,\
N(t)=2k\\}=\frac{(2k)!}{t^{2k}}\sum_{j=1}^{k-1}\frac{(t-s)^{2k-1-2j}\,s^{2j}}{j!(j+1)!(2k-1-2j)!\,2^{2j+1}}\mathop{}\\!\mathrm{d}s.$
(4.28)

Furthermore, for $t>0$

$P\\{F_{0}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=v\\}=\frac{e^{-\lambda
t}}{t}I_{1}(\lambda t)\mathop{}\\!\mathrm{d}t.$ (4.29)

Note that all these results are independent of $c$.

###### Proof.

Without any loss of generality we consider $V(0)=c$.

Probability (4.26) follows by observing that if $N(t)=1$ it is necessary that
the change of direction occurs at time $s/2$ in order to return to the origin
at time $s$.

In order to return in $0$ at time $s$, the motion need at least one change of
direction before time $s/2$. Now, with (4.7) at hand, we immediately obtain

$\displaystyle P\\{F_{0}\in\mathop{}\\!\mathrm{d}s\ |\ $ $\displaystyle
V(0)=c,\ N(t)=n\\}$
$\displaystyle=\int_{0}^{\frac{s}{2}}P\\{F_{ct_{1}}\in\mathop{}\\!\mathrm{d}s-t_{1}\
|\ V(0)=c,\ N(t-t_{1})=n-1\\}P\\{T_{1}\in\mathop{}\\!\mathrm{d}t_{1}\ |\
N(t)=n\\}$

and results (4.27) and (4.28) follows by using Theorem 4.1.

If $s=t$, we can reach the $0$ only if $V(t)=-c$ (we are assuming $V(0)=c$),
meaning that an odd number of switches occur up to time $t$. Now, for
$k\in\mathbb{N}_{0}$, we have that

$P\\{F_{0}\in\mathop{}\\!\mathrm{d}t\ |\ V(0)=v,\
N(t)=2k+1\\}=\binom{2k+1}{k}\frac{\mathop{}\\!\mathrm{d}t}{t\,2^{2k+1}}=P^{-}_{2k+1}\\{M(t)=0\\}\mathop{}\\!\mathrm{d}t/t.$
(4.30)

Hence, for $t>0$, we immediately obtain (4.29). ∎

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