# Planar random motions in a vortex

Enzo Orsingher
Department of Statistical Sciences
Sapienza University of Rome
<EMAIL_ADDRESS>
&Manfred Marvin Marchione
Department of Statistical Sciences
Sapienza University of Rome
<EMAIL_ADDRESS>
###### Abstract

We study a planar random motion $\big{(}X(t),Y(t)\big{)}$ with orthogonal
directions which can turn clockwise, counterclockwise and reverse its
direction each with a different probability. The support of the process is
given by a time-varying square and the singular distributions on the boundary
and the diagonals of the square are obtained. In the interior of the support,
we study the hydrodynamic limit of the distribution. We then investigate the
time $T(t)$ spent by the process moving vertically and the joint distribution
of $\big{(}T(t),Y(t)\big{)}$. We prove that, in the hydrodynamic limit, the
process $\big{(}X(t),Y(t)\big{)}$ spends half the time moving vertically.

_Keywords_ Telegraph process $\cdot$ Bessel functions $\cdot$ Hyperbolic
equations

##  1 Introduction

Finite-velocity planar random motions have been investigated by several
authors in the literature since the Eighties. These motions can be classified
according to different criteria. One such classification considers the number
of possible directions the motion can take, distinguishing between minimal (Di
Crescenzo [4]), orthogonal (Orsingher [12]) and motions with infinite
directions (Orsingher and De Gregorio [13]). A categorization is also possible
based on the distribution of the waiting time before a change of direction
occurs. While the classical problem involves direction changes at Poisson
times, variants such as the Polya process (Crimaldi et al. [3]), processes
with gamma-distributed intertimes (Martinucci et al. [9]) and the geometric
counting process (Di Crescenzo et al. [5]) have been explored in the
univariate case. In the bivariate case, Cinque and Orsingher [2] examined the
case in which the direction changes are governed by a non-homogeneous Poisson
process. Fractional extensions of finite-velocity planar random motions have
been investigated by Masoliver and Lindenberg [11] and Masoliver [10].
A relevant criteria for classifying planar random motion considers the
mechanism for determining the direction taken after a change occurs. Orsingher
[12] investigated a motion with orthogonal directions which turns clockwise or
counter-clockwise with equal probabilities. Di Crescenzo [4] examined a
minimal cyclic random motion, which can only turn counter-clockwise when the
change of direction occurs. A cyclic planar random motion with orthogonal
directions was examined by Orsingher et al. [14]. Kolesnik and Orsingher [6]
and Cinque and Orsingher [2] discussed a variant of the planar motion which
can turn clockwise, counter-clockwise and can reverse its direction, each with
an equal probability of $\frac{1}{3}$. The term reflection was used by the
authors to denote the direction reversal, and they showed that expressing the
distribution of the process explicitly becomes a challenging task when
reflection is present.
Multivariate extensions have been proposed, for minimal cyclic random motions,
by Samoilenko [15] and Lachal et al. [8]. Lachal [7] studied multidimensional
random motions with an arbitrary number of directions, while three-dimensional
random motions with orthogonal directions were investigated by Cinque and
Orsingher [1].
In this paper, we study a finite-velocity planar random motion for which
orthogonal changes of direction and reflection occur with different
probabilities. We consider a planar random motion $\big{(}X(t),Y(t)\big{)}$
which moves along four directions

$d_{j}=\Big{(}\cos\left(\frac{\pi}{2}j\right),\;\sin\left(\frac{\pi}{2}j\right)\Big{)},\qquad
j=0,1,2,3.$

Of course, we have that $d_{j}=d_{j+4n}$ for integer values of $n$. We assume
that the random vector $\big{(}X(t),Y(t)\big{)}$ lies at the origin of the
Cartesian plane at the initial time $t=0$, and it starts moving along one of
the four possible directions with equal probabilities. The changes of
direction are modeled as Poisson arrivals and we denote by $N(t),\;t>0$, the
total number of changes. The intensity of $N(t)$ is assumed to be constant and
we denote it by $\lambda$. Moreover, we assume that at each Poisson event the
new direction is taken according to the following rule. With probability $p$
the direction changes by a counter-clockwise turn, therefore passing from
$d_{j}$ to $d_{j+1}$. With probability $q$ the direction changes by a
clockwise turn, therefore passing from $d_{j+1}$ to $d_{j}$. A reflection
occurs wih probability $1-p-q$, changing the direction from $d_{j}$ to
$d_{j+2}$. Thus, reflection is possible only if $p+q<1$. In the special case
$p=q=\frac{1}{2}$, the process we study reduces to that studied by Kolesnik
and Orsingher [6] for which an explicit representation of the distribution can
be obtained. If $p\neq q$, a vorticity effect is introduced in the sense that
the process tends to perform clockwise turns with higher probability if $p<q$,
while counter-clockwise turns are more likely to occur if $p>q$. In the
limiting cases $p=1$ or $q=1$, the cyclic random motion investigated by
Orsingher et al. [14] is obtained. It is clear that the support of the motion
is time-dependent and coincides with the square $S_{ct}$ defined as

$S_{ct}=\left\\{(x,y)\in\mathbb{R}^{2}:\;|x+y|\leq ct,\;|x-y|\leq
ct\right\\}.$

Moreover, the distribution of $\big{(}X(t),Y(t)\big{)}$ has a singular
component on the boundary $\partial S_{ct}$ of the domain $S_{ct}$. The
particle with position $\big{(}X(t),Y(t)\big{)}$ lies on $\partial S_{ct}$ if
no changes of direction occur or if clockwise and counter-clockwise turns
alternate. In particular, if no changes of direction occur the particle lies
on the vertices of the square $S_{ct}$. A sketch of some sample paths is
provided in figure 1.

Figure 1: some sample paths of the process $\big{(}X(t),Y(t)\big{)}$. The
motion takes place in the time-varying square $S_{ct}$ and a degenerate
component of the distribution is present on the boundary and the diagonals of
the square. In particular, the process lies on $\partial S_{ct}$ if the path
changes direction by always alternating two contiguous directions. This
behaviour is displayed, for instance, by the red path in the figure. The green
and blue sample paths lead the particle to the interior of the support. While
the green path only exhibits clockwise and counterclockwise changes of
direction, in the blue path a reflection has occurred and the particle has
reversed its direction.

Our work starts with the presentation of some results concerning the
distribution of the process $\big{(}X(t),Y(t)\big{)}$ on the boundary
$\partial S_{ct}$. In the general case where reflection is admitted, the
probability of the particle lying on the boundary is given by

$\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in\partial S_{ct}\Big{)}=e^{-\lambda
t}\left\\{e^{\lambda
t\sqrt{pq}}\left(1+\frac{p+q}{2\sqrt{pq}}\right)+e^{-\lambda
t\sqrt{pq}}\left(1-\frac{p+q}{2\sqrt{pq}}\right)-1\right\\}.$

On each side of $\partial S_{ct}$, the distribution of
$\big{(}X(t),Y(t)\big{)}$ admits a continuous component which is explicitly
obtained in this paper. In particular, we prove that

$\displaystyle\mathbb{P}\Big{(}X(t)+$ $\displaystyle
Y(t)=ct,\;X(t)-Y(t)\in\mathop{d\eta}\Big{)}/\mathop{d\eta}$
$\displaystyle=\frac{e^{-\lambda
t}}{4c}\left[\frac{\lambda(p+q)}{2}\;I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)+\;\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)\right],\qquad\lvert\eta\lvert<ct$

with characeristic function

$\displaystyle\mathbb{E}\left[e^{i\alpha\left(X(t)-Y(t)\right)}\;\mathds{1}_{\\{X(t)+Y(t)=ct\\}}\right]=\frac{e^{-\lambda
t}}{4}$
$\displaystyle\left[\left(1+\frac{\lambda(p+q)}{2\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right)e^{t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right.$
$\displaystyle+$
$\displaystyle\left.\left(1-\frac{\lambda(p+q)}{2\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right)e^{-t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right].$

We then proceed with our analysis by studying the distribution of
$\big{(}X(t),Y(t)\big{)}$ on the diagonals of $S_{ct}$, that is on the set

$Q_{ct}=\left\\{(x,y)\in S_{ct}:\;x=0\vee y=0\right\\}.$

We first show that

$\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in
Q_{ct}\Big{)}=e^{-\lambda(p+q)t}.$ (1)

Formula (1) above highlights that if $p+q=1$, that is if reflection is not
admitted, the probability of the particle lying on the diagonals of the square
$S_{ct}$ coincides with the probability of no Poisson events occurring, which
implies that no direction changes occur and therefore the particle must lie on
one of the vertices of $S_{ct}$. In other words, if reflection is not admitted
the particle lies on $Q_{ct}$ if and only if it lies on one of the vertices of
$S_{ct}$. Conversely, if reflection is possible, the distribution of
$\big{(}X(t),Y(t)\big{)}$ on $Q_{ct}$ admits a continuous component. In fact,
we show that, if $p+q<1$,

$\displaystyle\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)=0\Big{)}/\mathop{dx}=\frac{e^{-\lambda
t}}{4c}\Bigg{[}\lambda(1-p-q)$
$\displaystyle\;I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)$
$\displaystyle+$ $\displaystyle\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)\Bigg{]},\qquad\lvert
x\lvert<ct$

and we obtain the corresponding characteristic function.
We then study the distribution of $\big{(}X(t),Y(t)\big{)}$ in the interior of
its support $S_{ct}$. We show that the probability density function

$u(x,y,t)\mathop{dx}\mathop{dy}=\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)\in\mathop{dy}\Big{)},\qquad\lvert
x\lvert+\lvert y\lvert<ct$ (2)

is the solution to a complicated fourth-order partial differential equation
for which a general solution is difficult to give in an explicit form. In the
special case in which reflection is not admitted, we derive the characteristic
function of $\big{(}X(t),Y(t)\big{)}$ in the interior of $S_{ct}$ explicitly.
In the general case, we are able to study the hydrodynamic limit of the
distribution for $\lambda,c\to+\infty$, with $\frac{\lambda}{c^{2}}\to 1$, and
we prove that the heat equation

$\frac{\partial u}{\partial
t}=\frac{1}{4}\;\frac{(1-p)+(1-q)}{(1-p)^{2}+(1-q)^{2}}\;\Delta u$ (3)

is satisfied in the limiting case. The interpretation of equation (3) is that
the process $\big{(}X(t),Y(t)\big{)}$ converges to a planar Brownian motion
with independent components and diffusion coefficient depending on $p$ and
$q$. It is interesting to observe that the diffusion coefficient is maximized
for $p=q=\frac{1}{2}$.

The final part of our work is devoted to the study of the time spent by
$\big{(}X(t),Y(t)\big{)}$ moving vertically, in parallel to the $y$-axis.
Thus, we study the random process

$T(t)=\int_{0}^{t}\mathds{1}_{\\{D(\tau)\in\\{d_{1},d_{3}\\}\\}}\mathop{d\tau},\qquad
t>0.$

We obtain the exact distribution of $T(t)$ and its characteristic function. In
particular, we prove that

$\displaystyle\mathbb{P}\Big{(}T(t)\in\mathop{ds}\Big{)}/ds=e^{-\lambda(p+q)t}\Bigg{[}\lambda(p+$
$\displaystyle q)\;I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)$
$\displaystyle+\frac{\partial}{\partial
t}I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)\Bigg{]},\qquad s\in(0,t).$ (4)

We then analyze the joint distribution of the vector
$\big{(}T(t),Y(t)\big{)}$. The support of such random vector is given by the
triangle

$R_{ct}=\left\\{(s,y)\in\mathbb{R}^{2}:\;\lvert y\lvert\leq
cs,\,s\in[0,t]\right\\}.$

The process $\big{(}T(t),Y(t)\big{)}$ can move along two directions which are
parallel to the oblique sides of $R_{ct}$. Moreover, as we will discuss in the
remainder of the paper, the process can arrest its movement at some points and
stay still for some time. In particular, this occurs when the process
$\big{(}X(t),Y(t)\big{)}$ is moving horizontally in $S_{ct}$. Some sample
paths of the process $\big{(}T(t),Y(t)\big{)}$ are exemplified in figure 2.

Figure 2: some sample paths of the process $\big{(}T(t),Y(t)\big{)}$, whose
support coincides with the time-varying triangle $R_{ct}$. The particle with
position $\big{(}T(t),Y(t)\big{)}$ can move along two diffrent directions,
parallel to the oblique side of the triangle, and can arrest its movement at
some points and stay still for some time. For each path in the figure, we
highlighted with a dot the points in which the particle has stopped. In
particular, the particle which moved along the blue path has never arrested
its movement and therefore it lies on the verical side of $R_{ct}$. The
particle lies on one of the oblique sides of $R_{ct}$ if it always moves along
the same direction, possibly stopping its movement sometimes. This is the case
of the red path in the figure.

We first study the distribution of $\big{(}T(t),Y(t)\big{)}$ in the interior
of $R_{ct}$ and we show that the probability density function

$\bar{u}(s,y,t)\mathop{ds}\mathop{dy}=\mathbb{P}\Big{(}T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy}\Big{)},\qquad(s,y)\in
R_{ct}.$ (5)

satisfies a third-order partial differential equation. By studying the
hydrodynamic limit of the equation, we prove that, for $\lambda,c\to+\infty$
with $\frac{\lambda}{c^{2}}\to 1$, the partial differential equation

$\frac{\partial u}{\partial t}=-\frac{1}{2}\frac{\partial u}{\partial
s}+\frac{1}{4}\;\frac{(1-p)+(1-q)}{(1-p)^{2}+(1-q)^{2}}\frac{\partial^{2}u}{\partial
y^{2}.}$ (6)

holds. The interpretation of equation (6) is that, by taking the hydrodynamic
limit, the asymptotic distribution is

$\lim_{\lambda,c\to+\infty}\mathbb{P}\Big{(}T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy}\Big{)}/\left(\mathop{ds}\mathop{dy}\right)=\sqrt{\frac{(1-p)^{2}+(1-q)^{2}}{(1-p)+(1-q)}}\cdot\frac{e^{-\frac{(1-p)^{2}+(1-q)^{2}}{(1-p)+(1-q)}\,\frac{y^{2}}{t}}}{\sqrt{\pi
t}}\cdot\delta\left(s-\frac{t}{2}\right)$

where the symbol $\delta(\cdot)$ denotes the Dirac delta. Therefore, while in
the hydrodynamic limit the variable $Y(t)$ becomes a Brownian motion with
variance depending on $p$ and $q$, the process $T(t)$ becomes deterministic
and its limiting value is equal to $\frac{t}{2}$. Thus, in the limit, the
process $\big{(}X(t),Y(t)\big{)}$ spends half of the time moving vertically.
The joint distribution of $\big{(}T(t),Y(t)\big{)}$ is also studied on the
sides of the triangle $R_{ct}$. As for the oblique side, on which the particle
lies if $Y(t)=cT(t)$, the distribution satisfies a third-order partial
differential equation. In the special case $p+q=1$, we are able to study such
equation in detail. We first calculate the probability of the particle
$\big{(}T(t),Y(t)\big{)}$ lying on the oblique side of the triangle, which
reads

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{)}=\frac{e^{-\lambda
t}}{8}\Bigg{\\{}$
$\displaystyle\left(1+\frac{1}{\sqrt{2p(1-p)}}\right)^{2}e^{\lambda
t\sqrt{2p(1-p)}}$
$\displaystyle+\left(1-\frac{1}{\sqrt{2p(1-p)}}\right)^{2}e^{-\lambda
t\sqrt{2p(1-p)}}-\frac{(2p-1)^{2}}{p(1-p)}\Bigg{\\}}.$

We then investigate the exact distribution of $\big{(}T(t),Y(t)\big{)}$ when
$Y(t)=cT(t)$. If $p+q=1$, we show that

$\displaystyle\mathbb{P}\Big{(}T(t)\in\mathop{ds},\,Y(t)=c\,T(t)\Big{)}/\mathop{ds}=$
$\displaystyle\frac{\lambda}{2\sqrt{2}}\;I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)}$
$\displaystyle+\frac{1}{4}\left(1+\frac{1}{2p(1-p)}\right)\frac{\partial}{\partial
t}I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)},\qquad\lvert
s\lvert<t.$

We then discuss a special case in which the partial differential equation
governing the distribution of $\big{(}T(t),Y(t)\big{)}$ on the oblique side of
$R_{ct}$ reduces to a second-order equation, namely the case $p=q$ in which
clockwise and counterclockwise changes of direction are equiprobable. While we
are not able to find the exact distribution, for $p=q$ we obtain the
probability of the particle lying on the oblique side of $R_{ct}$, that is

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{)}=$
$\displaystyle\frac{3}{8}\left(1+\frac{(1+6p)}{3\sqrt{12p^{2}-4p+1}}\right)\;e^{\frac{\lambda
t}{2}\left(-(1+2p)+\sqrt{12p^{2}-4p+1}\right)}$
$\displaystyle\;\;+\frac{3}{8}\left(1-\frac{(1+6p)}{3\sqrt{12p^{2}-4p+1}}\right)\;e^{\frac{\lambda
t}{2}\left(-(1+2p)-\sqrt{12p^{2}-4p+1}\right)}.$

Finally, we study the distribution of $\big{(}T(t),Y(t)\big{)}$ on the
vertical side of $R_{ct}$ and we show that it perfecly resembles the
distribution of the process $\big{(}X(t),Y(t)\big{)}$ on the diagonals
$Q_{ct}$ of the square $S_{ct}$. Thus, it holds that

$\displaystyle\mathbb{P}\Big{(}Y(t)\in\mathop{dy},\,T(t)=t\Big{)}/\mathop{dy}=\frac{e^{-\lambda
t}}{4c}\Bigg{[}\lambda$
$\displaystyle(1-p-q)\;I_{0}\left(\frac{\lambda}{c}\,(1-p-q)\,\sqrt{c^{2}t^{2}-y^{2}}\right)$
$\displaystyle+\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}\,(1-p-q)\,\sqrt{c^{2}t^{2}-y^{2}}\right)\Bigg{]},\qquad\lvert
y\lvert<ct$

The paper is structured in the following manner. In section 2 we study the
distribution of the process $\big{(}X(t),Y(t)\big{)}$ on the boundary
$\partial S_{ct}$ of its support, while in section 3 we study the distribution
on the diagonals $Q_{ct}$. Section 4 is devoted to the study of the
distribution in the interior of the square $S_{ct}$. In the final section, the
distribution of the time spent moving vertically $T(t)$ and the joint
distribution of the vector $\big{(}T(t),Y(t)\big{)}$ are studied.

##  2 Distribution on the boundary

In this section, we study the behaviour of the particle
$\big{(}X(t),Y(t)\big{)}$ on the boundary of the square $S_{ct}$. We first
calculate the probability of the particle lying on $\partial S_{ct}$ and we
then obtain the exact distribution of the position of the particle on the
boundary. We examine the general case in which reflection is possible. Our
analysis starts by observing that, regardless of the initial direction of the
motion, the boundary can be reached if the particle always moves along two
contiguous directions by alternating clockwise and counterclockwise changes of
direction. Of course, if no changes of direction occur, the motion always
takes place on $\partial S_{ct}$ because the particle lies on the vertices of
the square. Therefore, we distinguish three cases in which the particle is on
$\partial S_{ct}$:

  * •

the particle is on the boundary because no changes of direction occur

  * •

an even number of alternating changes of direction occurs, in which case the
first change of direction can be either clockwise or counterclockwise

  * •

an odd number of alternating changes of direction occurs, in which case the
first change of direction can be either clockwise or counterclockwise.

Thus, we have the following result.

######  Theorem 1.

It holds that

$\displaystyle\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in\partial
S_{ct}\Big{)}=e^{-\lambda t}\left\\{e^{\lambda
t\sqrt{pq}}\left(1+\frac{p+q}{2\sqrt{pq}}\right)+e^{-\lambda
t\sqrt{pq}}\left(1-\frac{p+q}{2\sqrt{pq}}\right)-1\right\\}.$ (7)

###### Proof.

By treating separately the cases in which the process performs no changes of
direction, an even number of changes and an odd number of changes, we can
write that

$\displaystyle\mathbb{P}\Big{(}\big{(}X$
$\displaystyle(t),Y(t)\big{)}\in\partial S_{ct}\Big{)}$ $\displaystyle=$
$\displaystyle\mathbb{P}\big{(}N(t)=0\big{)}+\sum_{k=1}^{\infty}\mathbb{P}\big{(}N(t)=2k\big{)}\left(p^{k}q^{k}+q^{k}p^{k}\right)+\sum_{k=0}^{\infty}\mathbb{P}\big{(}N(t)=2k+1\big{)}\left(p^{k+1}q^{k}+q^{k+1}p^{k}\right)$
$\displaystyle=$ $\displaystyle e^{-\lambda t}+2e^{-\lambda
t}\sum_{k=1}^{\infty}\frac{(\lambda t\sqrt{pq})^{2k}}{(2k)!}+e^{-\lambda
t}\frac{(p+q)}{\sqrt{pq}}\sum_{k=0}^{\infty}\frac{(\lambda
t\sqrt{pq})^{2k+1}}{(2k+1)!}$ $\displaystyle=$ $\displaystyle e^{-\lambda
t}+2e^{-\lambda t}\big{(}\operatorname{cosh}\left(\lambda
t\sqrt{pq}\right)-1\big{)}+e^{-\lambda
t}\frac{(p+q)}{\sqrt{pq}}\operatorname{sinh}\left(\lambda t\sqrt{pq}\right)$
$\displaystyle=$ $\displaystyle e^{-\lambda t}\left\\{e^{\lambda
t\sqrt{pq}}\left(1+\frac{p+q}{2\sqrt{pq}}\right)+e^{-\lambda
t\sqrt{pq}}\left(1-\frac{p+q}{2\sqrt{pq}}\right)\right\\}-e^{-\lambda t}$

which completes the proof. ∎

We emphasize that formula (7) represents the probability of the particle being
at any point on the entire boundary of the support, encompassing all four
sides and the vertices of the square. In order to calculate the probability of
the particle lying in the interior of a specific side of $\partial S_{ct}$,
the probability mass on the vertices must be subtracted. The result must then
be divided by four in order to focus on just one side. For instance, the
probability of the process $\big{(}X(t),Y(t)\big{)}$ belonging to the interior
of the side of $\partial S_{ct}$ which belongs to the first quadrant of the
Cartesian plane is given by

$\displaystyle\mathbb{P}\Big{(}X(t)+Y(t)=c$ $\displaystyle t,\;\left\lvert
X(t)-Y(t)\right\lvert<ct\Big{)}=\frac{1}{4}\left[\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in\partial
S_{ct}\Big{)}-\mathbb{P}\big{(}N(t)=0\big{)}\right]$ $\displaystyle=$
$\displaystyle\frac{e^{-\lambda t}}{4}\left\\{e^{\lambda
t\sqrt{pq}}\left(1+\frac{p+q}{2\sqrt{pq}}\right)+e^{-\lambda
t\sqrt{pq}}\left(1-\frac{p+q}{2\sqrt{pq}}\right)-2\right\\}.$ (8)

We now highlight some interesting special cases of formula (7). First of all,
if reflection is not admitted, that is if $p+q=1$, we have that

$\displaystyle\mathbb{P}\Big{(}\big{(}X$
$\displaystyle(t),Y(t)\big{)}\in\partial S_{ct}\Big{)}$
$\displaystyle\;\;=e^{-\lambda t}\left\\{e^{\lambda
t\sqrt{p(1-p)}}\left(1+\frac{1}{2\sqrt{p(1-p)}}\right)+e^{-\lambda
t\sqrt{p(1-p)}}\left(1-\frac{1}{2\sqrt{p(1-p)}}\right)-1\right\\}.$ (9)

Moreover, a noticeable simplification of formula (7) occurs for $p=q$, in
which case we can write

$\displaystyle\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in\partial
S_{ct}\Big{)}=2e^{-\lambda t(1-p)}-e^{-\lambda t}.$ (10)

For $p=q=\frac{1}{3}$, formula (10) is consistent with the results obtained by
Cinque and Orsingher [2]. In the special case $p=\frac{1}{2}$, both formulas
(9) and (10) reduce to

$\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in\partial
S_{ct}\Big{)}=2e^{-\frac{\lambda t}{2}}-e^{-\lambda t}$

which coincides with the result obtained by Orsingher [12].

We now study the distribution of $\big{(}X(t),Y(t)\big{)}$ on a side of
$\partial S_{ct}$. In particular, we consider the side belonging to the first
quadrant of the Cartesian plane. Therefore, we want to determine the
probability density function

$f(\eta,t)=\mathbb{P}\Big{(}X(t)+Y(t)=ct,\;X(t)-Y(t)\in\mathop{d\eta}\Big{)}/\mathop{d\eta},\qquad-
ct<\eta<ct.$ (11)

We start by considering, for $j=0,1$, the density functions

$f_{j}(\eta,t)=\mathbb{P}\Big{(}X(t)+Y(t)=ct,\;X(t)-Y(t)\in
d\eta,\;D(t)=d_{0}\Big{)}/\mathop{d\eta}.$

Clearly, we have that

$f(\eta,t)=f_{0}(\eta,t)+f_{1}(\eta,t).$ (12)

By means of standard methods it can be shown that

$\begin{dcases}f_{0}(\eta,t+\mathop{dt})=f_{0}(\eta-c\mathop{dt},t)(1-\lambda\mathop{dt})+f_{1}(\eta,t)\lambda
q\mathop{dt}+o(\mathop{dt})\\\
f_{1}(\eta,t+\mathop{dt})=f_{1}(\eta+c\mathop{dt},t)(1-\lambda\mathop{dt})+f_{0}(\eta,t)\lambda
p\mathop{dt}+o(\mathop{dt})\end{dcases}$ (13)

Therefore, by performing a first-order Taylor expansion of the equations in
formula (13), we obtain that $f_{0}$ and $f_{1}$ satisfy the system of partial
differential equations

$\begin{dcases}\frac{\partial f_{0}}{\partial t}=-c\frac{\partial
f_{0}}{\partial\eta}+\lambda qf_{1}-\lambda f_{0}\\\ \frac{\partial
f_{1}}{\partial t}=c\frac{\partial f_{1}}{\partial\eta}+\lambda pf_{0}-\lambda
f_{1}\end{dcases}$ (14)

with initial condition
$f_{0}(\eta,0)=f_{1}(\eta,0)=\frac{1}{4}\,\delta(\eta).$ The relationship (12)
together with the system of differential equations (14) implies that the
probability density function $f$ satisfies, for $-ct<\eta<ct$, the partial
differential equation

$\left(\frac{\partial^{2}}{\partial t^{2}}+2\lambda\frac{\partial}{\partial
t}-c^{2}\frac{\partial^{2}}{\partial\eta^{2}}+\lambda^{2}\left(1-pq\right)\right)f=0$
(15)

Therefore, the continuous component of the distribution (11) is given in the
following theorem.

######  Theorem 2.

The probability density function (11) is

$f(\eta,t)=\frac{e^{-\lambda
t}}{4c}\left[\frac{\lambda(p+q)}{2}\;I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)+\;\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)\right],\qquad\lvert\eta\lvert<ct.$
(16)

###### Proof.

In view of equation (15), the function

$\widetilde{f}(\eta,t)=e^{\lambda t}f(\eta,t)$ (17)

satisfies the partial differential equation

$\frac{\partial^{2}\widetilde{f}}{\partial
t^{2}}-c^{2}\frac{\partial^{2}\widetilde{f}}{\partial\eta^{2}}=\lambda^{2}pq\widetilde{f}.$
(18)

The change of variables

$z=\sqrt{c^{2}t^{2}-\eta^{2}}$ (19)

reduces equation (18) to the Bessel equation

$\frac{d^{2}\widetilde{f}}{z^{2}}+\frac{1}{z}\frac{d\widetilde{f}}{dz}-\frac{\lambda^{2}}{c^{2}}pq\widetilde{f}=0$

whose general solution reads

$\widetilde{f}(z)=A\;I_{0}\left(\frac{\lambda}{c}\sqrt{pq}z\right)+B\;K_{0}\left(\frac{\lambda}{c}\sqrt{pq}z\right).$
(20)

We disregard the term involving the modified Bessel function of the second
kind $K_{0}(\cdot)$ because it would make the density function $f(\eta,t)$
non-integrable in proximity of the endpoints $\eta=\pm ct$. Therefore, by
inverting the transformations (17) and (19), we express the solution to
equation (15) in the form

$f(\eta,t)=e^{-\lambda
t}\left[A\;I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)+B\;\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}\sqrt{pq}\sqrt{c^{2}t^{2}-\eta^{2}}\right)\right]$
(21)

Comparing formulas (20) and (21), observe that we added a term involving the
time derivative of the Bessel function in order to add flexibility to the
general solution to equation (15). The introduction of the additional term can
be performed because equation (15) is homogeneous with respect to the time
variable. We now have to determine the coefficients $A$ and $B$ in the
expression (21). For this purpose, we use the well-known relation

$\int_{-ct}^{ct}I_{0}\left(K\;\sqrt{c^{2}t^{2}-\eta^{2}}\right)\mathop{d\eta}=\frac{1}{K}\left(e^{Kct}-e^{-Kct}\right),\qquad
c,t>0$ (22)

from which we also obtain

$\int_{-ct}^{ct}\frac{\partial}{\partial
t}I_{0}\left(K\;\sqrt{c^{2}t^{2}-\eta^{2}}\right)\mathop{d\eta}=c\left(e^{Kct}+e^{-Kct}-2\right),\qquad
c,t>0.$ (23)

Using the above formulas and integrating the expression (21), we obtain that

$\int_{-ct}^{ct}f(\eta,t)\mathop{d\eta}=e^{-\lambda
t}\left\\{\left(Bc+\frac{Ac}{\lambda\sqrt{pq}}\right)e^{\lambda
t\sqrt{pq}}+\left(Bc-\frac{Ac}{\lambda\sqrt{pq}}\right)e^{-\lambda
t\sqrt{pq}}-2Bc\right\\}.$ (24)

Comparing formulas (8) and (24) yields

$A=\frac{\lambda(p+q)}{8c},\qquad B=\frac{1}{4c}.$

which completes the proof. ∎

In the following theorem, we derive an explicit expression for the
characteristic function of $\big{(}X(t),Y(t)\big{)}$ on the boundary of the
support.

######  Theorem 3.

The characteristic function of $\big{(}X(t),Y(t)\big{)}$ on $\partial S_{ct}$
is

$\displaystyle\mathbb{E}\left[e^{i\alpha\left(X(t)-Y(t)\right)}\;\mathds{1}_{\\{X(t)+Y(t)=ct\\}}\right]=\frac{e^{-\lambda
t}}{4}$
$\displaystyle\left[\left(1+\frac{\lambda(p+q)}{2\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right)e^{t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right.$
$\displaystyle+$
$\displaystyle\left.\left(1-\frac{\lambda(p+q)}{2\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right)e^{-t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}\right]$
(25)

###### Proof.

By using the notation

$\widehat{f}(\alpha,t)=\mathbb{E}\left[e^{i\alpha\left(X(t)-Y(t)\right)}\;\mathds{1}_{\\{X(t)+Y(t)=ct\\}}\right]$

equation (15) implies that the following ordinary differential equation is
satisfied

$\begin{dcases}\frac{d^{2}\widehat{f}}{dt^{2}}+2\lambda\frac{d\widehat{f}}{dt}+\left[c^{2}\alpha^{2}+\lambda^{2}(1-pq)\right]\widehat{f}=0\\\
\widehat{f}(\alpha,0)=\frac{1}{2}\\\
\frac{d}{dt}\left.\widehat{f}(\alpha,t)\right\lvert_{t=0}=-\frac{\lambda(2-p-q)}{4}\end{dcases}$
(26)

where the initial conditions have been determined from the linear system (14).
The general solution to equation (26) reads

$\widehat{f}(\alpha,t)=k_{0}(\alpha)\;e^{-\lambda
t+t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}+k_{1}(\alpha)\;e^{-\lambda
t-t\sqrt{\lambda^{2}pq-\alpha^{2}c^{2}}}.$

The coefficients $k_{0}$ and $k_{1}$ can be determined by using the initial
conditions. ∎

We conclude this section by verifying that the results presented so far are
consistent with each other. We start by observing that, for $\alpha=0$,
formula (3) reduces to

$\displaystyle\mathbb{P}\Big{(}X(t)+Y(t)=ct\Big{)}=\frac{e^{-\lambda
t}}{4}\left\\{e^{\lambda
t\sqrt{pq}}\left(1+\frac{p+q}{2\sqrt{pq}}\right)+e^{-\lambda
t\sqrt{pq}}\left(1-\frac{p+q}{2\sqrt{pq}}\right)\right\\}$

which is consistent with formula (8). Moreover, a direct calculation of the
characteristic function (3) can be performed by observing that

$\displaystyle\mathbb{E}$
$\displaystyle\Big{[}e^{i\alpha\left(X(t)-Y(t)\right)}\;\mathds{1}_{\\{X(t)+Y(t)=ct\\}}\Big{]}$
$\displaystyle\qquad=e^{i\alpha
ct}\cdot\mathbb{P}\Big{(}X(t)=ct,\,Y(t)=0\Big{)}+e^{-i\alpha
ct}\cdot\mathbb{P}\Big{(}X(t)=0,\,Y(t)=ct\Big{)}+\int_{-ct}^{ct}e^{i\alpha\eta}f(\eta,\,t)\mathop{d\eta}$
$\displaystyle\qquad=\frac{e^{-\lambda t}}{2}\cos\left(\alpha
ct\right)+\int_{-ct}^{ct}e^{i\alpha\eta}f(\eta,\,t)\mathop{d\eta}.$ (27)

By using the expression (12) for the density $f$ and the integral formula

$\int_{-ct}^{ct}e^{i\alpha\eta}\,I_{0}\left(\frac{\lambda}{c}\sqrt{c^{2}t^{2}-\eta^{2}}\right)\mathop{d\eta}=\frac{c}{\sqrt{\lambda^{2}-\alpha^{2}c^{2}}}\left[e^{t\sqrt{\lambda^{2}-\alpha^{2}c^{2}}}-e^{-t\sqrt{\lambda^{2}-\alpha^{2}c^{2}}}\right]$
(28)

it is a matter of straightforward calculation to prove that formula (27)
coincides with the characteristic function (3).

##  3 Distribution on the diagonals

If reflection is admitted, a degenerate component of the distribution of
$\big{(}X(t),Y(t)\big{)}$ emerges on the diagonals of $S_{ct}$. In order to
verify this, note that the process must start moving along a diagonal at time
$t=0$. If the process only performs reflections, it is clear that it continues
to move along the same diagonal as it only changes the travel orientation.
Therefore, recalling the notation

$Q_{ct}=\left\\{(x,y)\in S_{ct}:\;x=0\vee y=0\right\\},$

we have that

$\displaystyle\mathbb{P}\Big{(}\big{(}X(t),Y$ $\displaystyle(t)\big{)}\in
Q_{ct}\Big{)}=\sum_{k=0}^{\infty}\mathbb{P}\Big{(}\left(X(t),Y(t)\right)\in
Q_{ct}\Big{\lvert}N(t)=k\Big{)}\,\mathbb{P}\left(N(t)=k\right)$
$\displaystyle=$ $\displaystyle e^{-\lambda
t}\sum_{k=0}^{\infty}\frac{(1-p-q)^{k}(\lambda
t)^{k}}{k!}=e^{-\lambda(p+q)t}.$ (29)

Observe that, if $p+q=1$, the expression (29) corresponds to the probability
of no changes of direction occurring. In other words, if reflection is not
admitted the particle lies on the diagonals if and only if it never changes
direction, in which case it must lie in one of the vertices of $S_{ct}$. We
also emphasize that formula (29) gives the probability of the particle
belonging to whole diagonals, including the extremal points. In order to only
consider the interior of the diagonals, the probability mass on the vertices
of $S_{ct}$ must be subtracted. For instance, is we consider the horizontal
diagonal, we have that

$\displaystyle\mathbb{P}\Big{(}Y(t)=0,\;-ct<X(t)<ct\Big{)}=\frac{1}{2}\left[\mathbb{P}\Big{(}\big{(}X(t),Y(t)\big{)}\in
Q_{ct}\Big{)}-\mathbb{P}\big{(}N(t)=0\big{)}\right]=\frac{e^{-\lambda(p+q)t}-e^{-\lambda
t}}{2}.$ (30)

We now want to determine the exact distribution of the process on the
diagonals $Q_{ct}$. Without loss of generality, we only consider the
horizontal diagonal. Therefore we study the probability density function

$g(x,t)=\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)=0\Big{)}/\mathop{dx},\qquad\lvert
x\lvert<ct.$ (31)

By defining, for $j=0,2$, the densities

$g_{j}(x,t)=\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)=0,\;D(t)=d_{j}\Big{)}/\mathop{dx},\qquad\lvert
x\lvert<ct$

it is clear that

$g(x,t)=g_{0}(x,t)+g_{2}(x,t).$ (32)

Moreover, we can write

$\begin{cases}g_{0}(x,t+\mathop{dt})=g_{0}(x-c\mathop{dt},t)(1-\lambda\mathop{dt})+g_{2}(x,t)\lambda(1-p-q)\mathop{dt}+o(\mathop{dt})\\\
g_{2}(x,t+\mathop{dt})=g_{2}(x+c\mathop{dt},t)(1-\lambda\mathop{dt})+g_{0}(x,t)\lambda(1-p-q)\mathop{dt}+o(\mathop{dt})\end{cases}$
(33)

which implies that the system of partial differential equations

$\begin{dcases}\frac{\partial g_{0}}{\partial t}=-c\frac{\partial
g_{0}}{\partial x}+\lambda(1-p-q)g_{2}-\lambda g_{0}\\\ \frac{\partial
g_{2}}{\partial t}=c\frac{\partial g_{2}}{\partial
x}+\lambda(1-p-q)g_{0}-\lambda g_{2}\end{dcases}$ (34)

is satisfied with initial conditions
$g_{0}(x,0)=g_{2}(x,0)=\frac{1}{4}\delta(x)$. The system (34) implies, in view
of equation (32), that

$\left(\frac{\partial^{2}}{\partial t^{2}}+2\lambda\frac{\partial}{\partial
t}-c^{2}\frac{\partial^{2}}{\partial
x^{2}}+\lambda^{2}(p+q)\left(2-p-q\right)\right)g=0$ (35)

Therefore, we obtain an explicit expression for the density $g(x,t)$ in the
following theorem.

######  Theorem 4.

If $p+q<1$, the probability density function (31) is

$\displaystyle g(x,t)=\frac{e^{-\lambda t}}{4c}\Bigg{[}\lambda(1-p-q)$
$\displaystyle\;I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)$
$\displaystyle+$ $\displaystyle\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)\Bigg{]},\qquad|x\lvert<ct.$

###### Proof.

We start by defining the auxiliary function

$\widetilde{g}(x,t)=e^{\lambda t}g(x,t).$

Equation (35) implies that $\widetilde{g}(x,t)$ satisifes the partial
differential equation satisfies the partial differential equation

$\frac{\partial^{2}\widetilde{g}}{\partial
t^{2}}-c^{2}\frac{\partial^{2}\widetilde{g}}{\partial
x^{2}}=\lambda^{2}(1-p-q)^{2}\widetilde{g}.$ (36)

The change of variables

$z=\sqrt{c^{2}t^{2}-x^{2}}$ (37)

transforms equation (36) into the Bessel equation

$\frac{d^{2}\widetilde{g}}{z^{2}}+\frac{1}{z}\frac{d\widetilde{g}}{dz}-\frac{\lambda^{2}}{c^{2}}(1-p-q)^{2}\widetilde{g}=0$

whose general solution reads

$\widetilde{g}(z)=A\;I_{0}\left(\frac{\lambda}{c}(1-p-q)z\right)+B\;K_{0}\left(\frac{\lambda}{c}(1-p-q)z\right).$
(38)

Similarly to theorem 2, we disregard the term involving the modified Bessel
function of the second kind $K_{0}(\cdot)$ and, since equation (35) is
homogeneous with respect to the time variable, we express its solution in the
form

$g(x,t)=e^{-\lambda
t}\left[A\;I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)+B\;\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}(1-p-q)\sqrt{c^{2}t^{2}-x^{2}}\right)\right]$
(39)

In order to determine the coefficients $A$ and $B$ in formula (39), we use
formulas (22) and (23) and we obtain

$\int_{-ct}^{ct}g(x,t)\mathop{dx}=e^{-\lambda
t}\left\\{\left(Bc+\frac{Ac}{\lambda(1-p-q)}\right)e^{\lambda
t(1-p-q)}+\left(Bc-\frac{Ac}{\lambda(1-p-q)}\right)e^{-\lambda
t(1-p-q)}-2Bc\right\\}.$ (40)

Comparing formulas (30) and (40) yields

$A=\frac{\lambda(1-p-q)}{4c},\qquad B=\frac{1}{4c}.$

which completes the proof. ∎

We now derive the characteristic function of the process on on the horizontal
diagonal of the support.

######  Theorem 5.

The characteristic function of $\big{(}X(t),Y(t)\big{)}$ on the horizontal
diagonal of the square $S_{ct}$ is

$\displaystyle\mathbb{E}\left[e^{i\alpha
X(t)}\;\mathds{1}_{\\{Y(t)=0\\}}\right]=\frac{e^{-\lambda t}}{4}$
$\displaystyle\left[\left(1+\frac{\lambda(1-p-q)}{\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}\right)e^{t\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}\right.$
$\displaystyle+$
$\displaystyle\left.\left(1-\frac{\lambda(1-p-q)}{\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}\right)e^{-t\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}\right].$
(41)

###### Proof.

In view of equation (35), the characteristic function

$\widehat{g}(\alpha,t)=\mathbb{E}\left[e^{i\alpha
X(t)}\;\mathds{1}_{\\{Y(t)=0\\}}\right]$

satisfies the ordinary differential equation

$\begin{dcases}\frac{d^{2}\widehat{g}}{dt^{2}}+2\lambda\frac{d\widehat{g}}{dt}+\left[\lambda^{2}(p+q)\left(2-p-q\right)+\alpha^{2}c^{2}\right]\widehat{g}=0\\\
\widehat{g}(\alpha,0)=\frac{1}{2}\\\
\left.\frac{d}{dt}\widehat{g}(\alpha,t)\right\lvert_{t=0}=-\frac{\lambda(p+q)}{2}.\end{dcases}$
(42)

The general solution to equation (26) is

$\widehat{g}(\alpha,t)=k_{0}(\alpha)\;e^{-\lambda
t+t\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}+k_{1}(\alpha)\;e^{-\lambda
t-t\sqrt{\lambda^{2}(1-p-q)^{2}-\alpha^{2}c^{2}}}.$

The coefficients $k_{0}$ and $k_{1}$ can be obtained by using the initial
conditions. ∎

We emphasize that the consistency of the results presented in this section can
be proved by using a similar approach to that of section 2. In particular,
setting $\alpha=0$ in formula (41) yields the probability (29) while the
characteristic function (41) can be obtained direcly by using theorem 4 and
the integral (28).

##  4 Distribution in the interior of the domain

In this section, we study the distribution of the process
$\big{(}X(t),Y(t)\big{)}$ in the interior of its domain, that is the interior
of the set

$S_{ct}=\left\\{(x,y)\in\mathbb{R}^{2}:\;\lvert x\lvert+\lvert y\lvert\leq
ct\right\\}.$

Therefore, we are interested in studying the function $u(x,y,t)$ defined by
the relationship

$u(x,y,t)\mathop{dx}\mathop{dy}=\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)\in\mathop{dy}\Big{)},\qquad\lvert
x\lvert+\lvert y\lvert<ct.$ (43)

We emphasize that when the process admits reflection, that is when $p+q<1$,
the distribution of $\big{(}X(t),Y(t)\big{)}$ has a degenerate component on
the diagonals of the square $S_{ct}$. Therefore, if reflection is possible we
study the function $u(x,y,t)$ under the implicit assumption that $x\neq 0$ and
$y\neq 0$ since the definition (43) does not make sense otherwise. In order to
study the distribution (43) we consider, for $j=0,1,2,3$, the probability
density functions $u_{j}(x,y,t)$ defined as

$u_{j}(x,y,t)\mathop{dx}\mathop{dy}=\mathbb{P}\Big{(}X(t)\in\mathop{dx},\;Y(t)\in\mathop{dy},\;D(t)=d_{j}\Big{)}.$

By means of standard techniques it can be shown that the following linear
system of equations is satisfied by $u_{j}(x,y,t),\;j=0,1,2,3$

$\begin{dcases}\dfrac{\partial u_{0}}{\partial t}=-c\dfrac{\partial
u_{0}}{\partial x}+\lambda qu_{1}+\lambda(1-p-q)u_{2}+\lambda pu_{3}-\lambda
u_{0}\\\ \dfrac{\partial u_{1}}{\partial t}=-c\dfrac{\partial u_{1}}{\partial
y}+\lambda pu_{0}+\lambda qu_{2}+\lambda(1-p-q)u_{3}-\lambda u_{1}\\\
\dfrac{\partial u_{2}}{\partial t}=c\dfrac{\partial u_{2}}{\partial
x}+\lambda(1-p-q)u_{0}+\lambda pu_{1}+\lambda qu_{3}-\lambda u_{2}\\\
\dfrac{\partial u_{3}}{\partial t}=c\dfrac{\partial u_{3}}{\partial y}+\lambda
qu_{0}+\lambda(1-p-q)u_{1}+\lambda pu_{2}-\lambda u_{3}\end{dcases}$ (44)

with initial conditions $u_{j}(x,y,0)=\frac{1}{4}\,\delta(x)\,\delta(y)$ for
$j=0,1,2,3$. Moreover, since $u(x,y,t)=\sum_{j=0}^{3}u_{j}(x,y,t)$, the system
of equations (44) implies that $u(x,y,t)$ satisfies a partial differential
equation which can be represented in the form

$\displaystyle\Bigg{\\{}\Bigg{(}\left(\frac{\partial}{\partial
t}+\lambda\right)$ ${}^{2}-c^{2}\frac{\partial^{2}}{\partial
x^{2}}\Bigg{)}\left(\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}-c^{2}\frac{\partial^{2}}{\partial
y^{2}}\right)-4\lambda^{2}pq\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}$ $\displaystyle-$
$\displaystyle\lambda^{2}(1-p-q)^{2}\left(2\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}-c^{2}\Delta\right)-4\lambda^{3}(p^{2}+q^{2})(1-p-q)\left(\frac{\partial}{\partial
t}+\lambda\right)$
$\displaystyle\qquad\quad+\lambda^{4}\left[(1-p-q)^{2}-2pq\right]^{2}-\lambda^{4}(p^{2}+q^{2})^{2}\Bigg{\\}}u=0.$
(45)

######  Theorem 6.

For $\lambda,c\to+\infty$ with $\frac{c^{2}}{\lambda}\to 1$, the partial
differential equation (45) becomes

$\frac{\partial u}{\partial
t}=\frac{1}{4}\;\frac{(1-p)+(1-q)}{(1-p)^{2}+(1-q)^{2}}\;\Delta u$ (46)

where $\Delta=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial
y^{2}}$ is the Laplacian.

###### Proof.

The theorem is proved by dividing equation (45) by $\lambda^{3}$, simplifying
the expression for eliminating all the terms which grow as a multiple of
$\lambda$ and taking the limit. ∎

The interpretation of theorem 6 is that the process $\big{(}X(t),Y(t)\big{)}$
becomes, in the hydrodynamic limit, a planar Brownian motion with independent
components and variance depending on $p$ and $q$. By taking into account the
natural constraints to which $p$ and $q$ are subject, a constrained
maximization problem can be solved in order to prove that the diffusion
coefficient of equation (46) is maximized for $p=q=\frac{1}{2}$. Intuitively,
this is due to the fact that both reflection and the vorticity effect induced
by the condition $p\neq q$ tend to prevent the particle from moving too far
from the starting point. The case $p=q=\frac{1}{2}$ is the only case in which
both reflection and the vorticity effect are absent, which implies that the
diffusion coefficient is maximized.
We now examine the special case in which the process $\big{(}X(t),Y(t)\big{)}$
does not admit reflection. By setting $q=1-p$, it can be easily verified that
the partial differential equation (45) reduces to

$\left(\left(\frac{\partial}{\partial
t}+\lambda\right)^{4}-\left(c^{2}\Delta+4\lambda^{2}p(1-p)\right)\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}+c^{2}\frac{\partial^{4}}{\partial x^{2}\partial
y^{2}}\right)u=\lambda^{4}(2p-1)^{2}u.$ (47)

Even in this simpler case, finding the distribution of
$\big{(}X(t),Y(t)\big{)}$ in closed form is a difficult task. However, we are
able to obtain an explicit form of the characteristic function as shown in the
following theorem.

######  Theorem 7.

If $p+q=1$, the characteristic function of $\big{(}X(t),Y(t)\big{)}$ in the
interior of $\partial S_{ct}$ is

$\displaystyle\mathbb{E}\Big{[}$ $\displaystyle e^{i\left(\alpha X(t)+\beta
Y(t)\right)}\Big{]}$ $\displaystyle\;\;\;\;=e^{-\lambda
t}\Bigg{\\{}k_{0}(\alpha,\beta)\,\operatorname{cosh}\Big{(}A(\alpha,\beta)\mathop{t}+B(\alpha,\beta)\mathop{t}\Big{)}+k_{1}(\alpha,\beta)\,\operatorname{sinh}\Big{(}A(\alpha,\beta)\mathop{t}+B(\alpha,\beta)\mathop{t}\Big{)}$
$\displaystyle\qquad\;\;\;\;-k_{2}(\alpha,\beta)\,\operatorname{cosh}\Big{(}A(\alpha,\beta)\mathop{t}-B(\alpha,\beta)\mathop{t}\Big{)}-k_{3}(\alpha,\beta)\,\operatorname{sinh}\Big{(}A(\alpha,\beta)\mathop{t}-B(\alpha,\beta)\mathop{t}\Big{)}\Bigg{\\}}$
(48)

where

$A(\alpha,\beta)=\frac{1}{2}\sqrt{4\lambda^{2}p(1-p)-c^{2}(\alpha^{2}+\beta^{2})+2\sqrt{c^{4}\alpha^{2}\beta^{2}-\lambda^{4}(2p-1)^{2}}}$
(49)
$B(\alpha,\beta)=\frac{1}{2}\sqrt{4\lambda^{2}p(1-p)-c^{2}(\alpha^{2}+\beta^{2})-2\sqrt{c^{4}\alpha^{2}\beta^{2}-\lambda^{4}(2p-1)^{2}}}$
(50)

and

$\displaystyle k_{0}(\alpha,\beta)=$
$\displaystyle\frac{2\lambda^{2}-c^{2}(\alpha^{2}+\beta^{2})-2\left(A(\alpha,\beta)-B(\alpha,\beta)\right)^{2}}{8A(\alpha,\beta)B(\alpha,\beta)}$
$\displaystyle k_{1}(\alpha,\beta)=$ $\displaystyle\frac{\lambda^{3}-\lambda
c^{2}(\alpha^{2}+\beta^{2})-\lambda\left(A(\alpha,\beta)-B(\alpha,\beta)\right)^{2}}{4A(\alpha,\beta)B(\alpha,\beta)\left(A(\alpha,\beta)+B(\alpha,\beta)\right)}$
$\displaystyle k_{2}(\alpha,\beta)=$
$\displaystyle\frac{2\lambda^{2}-c^{2}(\alpha^{2}+\beta^{2})-2\left(A(\alpha,\beta)+B(\alpha,\beta)\right)^{2}}{8A(\alpha,\beta)B(\alpha,\beta)}$
$\displaystyle k_{3}(\alpha,\beta)=$ $\displaystyle\frac{\lambda^{3}-\lambda
c^{2}(\alpha^{2}+\beta^{2})-\lambda\left(A(\alpha,\beta)+B(\alpha,\beta)\right)^{2}}{4A(\alpha,\beta)B(\alpha,\beta)\left(A(\alpha,\beta)-B(\alpha,\beta)\right)}.$

###### Proof.

We adopt the notation

$\widehat{u}(\alpha,\beta,t)=\mathbb{E}\Big{[}e^{i\left(\alpha X(t)+\beta
Y(t)\right)}\Big{]}.$ (51)

In view of equation (47), the characteristic function
$\widehat{u}(\alpha,\beta,t)$ satisfies the ordinary differential equation

$\left(\left(\frac{d}{dt}+\lambda\right)^{4}+\left(c^{2}(\alpha^{2}+\beta^{2})-4\lambda^{2}p(1-p)\right)\left(\frac{d}{dt}+\lambda\right)^{2}+c^{2}\alpha^{2}\beta^{2}\right)\widehat{u}=\lambda^{4}(2p-1)^{2}\widehat{u}.$
(52)

We now establish the initial conditions for equation (52). For this purpose we
define the functions

$\widehat{u}_{j}(\alpha,\beta,t)=\mathbb{E}\Big{[}e^{i\left(\alpha X(t)+\beta
Y(t)\right)}\;\mathds{1}_{\\{D(t)=d_{j}\\}}\Big{]},\qquad j=0,1,2,3$

and, in view of the system of equations (44), we observe that for all
$n\in\mathbb{N}$

$\frac{d^{n}}{dt^{n}}\begin{pmatrix}\widehat{u}_{0}\\\ \widehat{u}_{1}\\\
\widehat{u}_{2}\\\
\widehat{u}_{3}\end{pmatrix}=\text{L}^{n}\cdot\begin{pmatrix}\widehat{u}_{0}\\\
\widehat{u}_{1}\\\ \widehat{u}_{2}\\\ \widehat{u}_{3}\end{pmatrix}$ (53)

where

$\text{L}=\begin{pmatrix}i\alpha c-\lambda&\lambda(1-p)&0&\lambda p\\\ \lambda
p&i\beta c-\lambda&\lambda(1-p)&0\\\ 0&\lambda p&-i\alpha
c-\lambda&\lambda(1-p)&\\\ \lambda(1-p)&0&\lambda p&-i\beta
c-\lambda\end{pmatrix}.$

Equation (53) can be proved for $n=1$ by taking the Fourier transforms of the
equations in the system (44) and can be extended to general values of $n$ by
induction. Moreover, formula (53) implies that

$\frac{d^{n}}{dt^{n}}\left.\begin{pmatrix}\widehat{u}_{0}\\\
\widehat{u}_{1}\\\ \widehat{u}_{2}\\\
\widehat{u}_{3}\end{pmatrix}\right\rvert_{t=0}=\frac{1}{4}\;\text{L}^{n}\cdot\begin{pmatrix}1\\\
1\\\ 1\\\ 1\end{pmatrix}$ (54)

where we have used the fact that, for $j=0,1,2,3$, the condition
$u_{j}(x,y,0)=\frac{1}{4}\delta(x)\delta(y)$ implies that
$\widehat{u}_{j}(\alpha,\beta,0)=\frac{1}{4}$. By using formula (54) and by
taking into account that
$\widehat{u}(\alpha,\beta,t)=\sum_{j=0}^{3}\widehat{u}_{j}(\alpha,\beta,t)$,
we obtain the following initial conditions for equation (52):

$\displaystyle\widehat{u}$ $\displaystyle(\alpha,\beta,0)=1,\qquad\qquad$
$\displaystyle\;\frac{d}{dt}\widehat{u}(\alpha,\beta,t)\Big{\rvert}_{t=0}=0,$
(55) $\displaystyle\frac{d^{2}}{dt^{2}}\widehat{u}$
$\displaystyle(\alpha,\beta,t)\Big{\rvert}_{t=0}=-\frac{c^{2}(\alpha^{2}+\beta^{2})}{2},\qquad\qquad$
$\displaystyle\frac{d^{3}}{dt^{3}}\widehat{u}(\alpha,\beta,t)\Big{\rvert}_{t=0}=\frac{\lambda
c^{2}(\alpha^{2}+\beta^{2})}{2}.$

We now define the function

$\widetilde{u}(\alpha,\beta,t)=e^{\lambda t}\;\widehat{u}(\alpha,\beta,t)$
(56)

and we observe that, in view of formulas (52) and (55), the following ordinary
differential equation is satisfied by $\widetilde{u}(\alpha,\beta,t)$:

$\begin{dcases}\dfrac{d^{4}\widetilde{u}}{dt^{4}}+\left(c^{2}(\alpha^{2}+\beta^{2})-4\lambda^{2}p(1-p)\right)\dfrac{d^{2}\widetilde{u}}{dt^{2}}+\left(c^{2}\alpha^{2}\beta^{2}-\lambda^{4}(2p-1)^{2}\right)\widetilde{u}=0\\\
\widetilde{u}(\alpha,\beta,0)=1\\\
\dfrac{d}{dt}\widetilde{u}(\alpha,\beta,t)\Big{\rvert}_{t=0}=\lambda\\\
\dfrac{d^{2}}{dt^{2}}\widetilde{u}(\alpha,\beta,t)\Big{\rvert}_{t=0}=\lambda^{2}-\frac{c^{2}(\alpha^{2}+\beta^{2})}{2}\\\
\dfrac{d^{3}}{dt^{3}}\widetilde{u}(\alpha,\beta,t)\Big{\rvert}_{t=0}=\lambda^{3}-\lambda
c^{2}(\alpha^{2}+\beta^{2}).\end{dcases}$ (57)

Equation (57) is an Euler-type ordinary differential equation which can be
solved by finding the roots of the associated algebraic equation

$r^{4}+\left(c^{2}(\alpha^{2}+\beta^{2})-4\lambda^{2}p(1-p)\right)r^{2}+\left(c^{2}\alpha^{2}\beta^{2}-\lambda^{4}(2p-1)^{2}\right)=0.$

Since the above algebraic equation is biquadratic, its roots can be easily
found and it can be shown that the general solution to equation (57) can be
expressed in the form

$\displaystyle\widetilde{u}(\alpha,\beta,t)=\varphi_{0}($
$\displaystyle\alpha,\beta)\,e^{A(\alpha,\beta)t+B(\alpha,\beta)t}+\varphi_{1}(\alpha,\beta)\,e^{A(\alpha,\beta)t-B(\alpha,\beta)t}$
$\displaystyle\;\qquad+$
$\displaystyle\varphi_{2}(\alpha,\beta)\,e^{-A(\alpha,\beta)t-B(\alpha,\beta)t}+\varphi_{3}(\alpha,\beta)\,e^{-A(\alpha,\beta)t+B(\alpha,\beta)t}$
(58)

where $A(\alpha,\beta)$ and $B(\alpha,\beta)$ are defined in formulas (49) and
(50). By taking the derivatives of the general solution (58) and using the
initial conditions of equation (57), it can be proved that the coefficients
$\varphi_{j}(\alpha,\beta),\;j=0,1,2,3$, satisfy the linear system of
equations

$\begin{pmatrix}1&1&1&1\\\ A+B&A-B&-A-B&-A+B\\\
(A+B)^{2}&(A-B)^{2}&(A+B)^{2}&(A-B)^{2}\\\
(A+B)^{3}&(A-B)^{3}&-(A+B)^{3}&-(A-B)^{3}\\\
\end{pmatrix}\begin{pmatrix}\varphi_{0}\\\ \varphi_{1}\\\ \varphi_{2}\\\
\varphi_{3}\end{pmatrix}=\begin{pmatrix}1\\\ \lambda\\\
\lambda^{2}-\frac{c^{2}(\alpha^{2}+\beta^{2})}{2}\\\\[2.84544pt]
\lambda^{3}-\lambda c^{2}(\alpha^{2}+\beta^{2})\end{pmatrix}$ (59)

where we omitted, for simplicity, the dependence of
$A,B,\varphi_{0},\varphi_{1},\varphi_{2}$ and $\varphi_{3}$ on $\alpha$ and
$\beta$. The linear system (59) can be solved by inverting the coefficient
matrix. Thus, we have that

$\begin{pmatrix}\varphi_{0}\\\ \varphi_{1}\\\ \varphi_{2}\\\
\varphi_{3}\end{pmatrix}=\begin{pmatrix}-\frac{(A-B)^{2}}{8AB}&-\frac{(A-B)^{2}}{8AB(A+B)}&\frac{1}{8AB}&\frac{1}{8AB(A+B)}\\\\[2.84544pt]
\frac{(A+B)^{2}}{8AB}&\frac{(A+B)^{2}}{8AB(A-B)}&-\frac{1}{8AB}&-\frac{1}{8AB(A-B)}\\\\[2.84544pt]
-\frac{(A-B)^{2}}{8AB}&\frac{(A-B)^{2}}{8AB(A+B)}&\frac{1}{8AB}&-\frac{1}{8AB(A+B)}\\\\[2.84544pt]
\frac{(A+B)^{2}}{8AB}&-\frac{(A+B)^{2}}{8AB(A-B)}&-\frac{1}{8AB}&\frac{1}{8AB(A-B)}\end{pmatrix}\begin{pmatrix}1\\\
\lambda\\\ \lambda^{2}-\frac{c^{2}(\alpha^{2}+\beta^{2})}{2}\\\\[2.84544pt]
\lambda^{3}-\lambda c^{2}(\alpha^{2}+\beta^{2})\end{pmatrix}$ (60)

from which the coefficients $\varphi_{j}(\alpha,\beta),\;j=0,1,2,3$, are
immediately obtained. In particular, it can be shown that

$\displaystyle\varphi_{0}(\alpha,\beta)=\frac{k_{0}(\alpha,\beta)+k_{1}(\alpha,\beta)}{2},\qquad\varphi_{1}(\alpha,\beta)=-\frac{k_{2}(\alpha,\beta)+k_{3}(\alpha,\beta)}{2},$
$\displaystyle\varphi_{2}(\alpha,\beta)=\frac{k_{0}(\alpha,\beta)-k_{1}(\alpha,\beta)}{2},\qquad\varphi_{3}(\alpha,\beta)=-\frac{k_{2}(\alpha,\beta)-k_{3}(\alpha,\beta)}{2},$

where the coefficients $k_{j}(\alpha,\beta),\;j=0,1,2,3$, are defined in the
statement of the theorem. After substituting the expressions of
$\varphi_{j}(\alpha,\beta),\;j=0,1,2,3$, into formula (58), the proof is
completed by inverting the transformation (56) and properly collecting the
terms of the final expression. ∎

We observe that, although the characteristic function (48) has a cumbersome
representation, it is consistent with the existing literature since it
coincides with the results obtained by Orsingher [12] for $p=q=\frac{1}{2}$.
Moreover, we emphasize that the key step for obtaining the characteristic
function was to solve the fourth-order algebraic equation (57). Such equation
is easy to solve because it is biquadratic in the case $p+q=1$. In principle,
the characteristic function of $\big{(}X(t),Y(t)\big{)}$ could also be
obtained in the case in which reflection is admitted by solving a more general
fourth-order equation. However, since the equation is not biquadratc for
$p+q<1$, the resulting characteristic function would have an extremely
complicated form.

##  5 Time spent in vertical direction

We conclude our analysis by studying the time spent by the process
$\big{(}X(t),Y(t)\big{)}$ moving vertically, parallel to the $y$-axis.
Formally, we define the process

$T(t)=\int_{0}^{t}\mathds{1}_{\\{D(\tau)\in\\{d_{1},d_{3}\\}\\}}\mathop{d\tau},\qquad
t>0.$

For fixed $t>0$, the support of the random variable $T(t)$ coincides with the
interval $[0,t]$. Moreover, it is clear that the distribution of $T(t)$ has a
degenerate component on the extrema of its support. In particular, we have
that $T(t)=0$ if the process $\big{(}X(t),Y(t)\big{)}$ starts moving in
horizontal direction at time 0 and it never changes direction or it changes
direction by only performing reflections until time $t$. Similarly, it holds
that $T(t)=t$ if the process $\big{(}X(t),Y(t)\big{)}$ starts moving
vertically at time 0 and it never changes direction or it only performs
reflections until time $t$. Therefore, we have that

$\mathbb{P}\left(T(t)=0\right)=\mathbb{P}\left(T(t)=t\right)=\frac{1}{2}\;e^{-\lambda
t}\sum_{k=0}^{\infty}\frac{(\lambda
t)^{k}(1-p-q)^{k}}{k!}=\frac{1}{2}\;e^{-\lambda t(p+q)}.$ (61)

In the interior of its support, the random variable $T(t)$ has a continuous
distribution. We denote the corresponding probability density function by

$h(s,t)=\mathbb{P}\Big{(}T(t)\in\mathop{ds}\Big{)}/ds,\qquad s\in(0,t).$ (62)

In order to determine the density (62), we start by defining the functions
$h_{j}(s,t),\;j=0,1,$

$h_{j}(s,t)=\mathbb{P}\Big{(}T(t)\in\mathop{ds},\;D(t)\in\\{d_{j},\;d_{j+2}\\}\Big{)}/ds,\qquad
s\in(0,t).$

It is clear that

$\begin{dcases}h_{0}(s,t+\mathop{dt})=h_{0}(s,t)(1-\lambda(p+q)\mathop{dt})+h_{1}(s-\mathop{ds},t)\lambda(p+q)\mathop{dt}+o(dt)\\\
h_{1}(s,t+\mathop{dt})=h_{1}(s-\mathop{ds},t)(1-\lambda(p+q)\mathop{dt})+h_{0}(s,t)\lambda(p+q)\mathop{dt}+o(dt)\end{dcases}$

which implies that the system of differential equations

$\begin{dcases}\dfrac{\partial h_{0}}{\partial
t}=\lambda(p+q)\left(h_{1}-h_{0}\right)\\\ \dfrac{\partial h_{1}}{\partial
t}=-\dfrac{\partial h_{1}}{\partial s}+\lambda(p+q)\left(h_{0}-h_{1}\right)\\\
\end{dcases}$ (63)

is satisfied with initial conditions
$h_{0}(s,0)=h_{1}(s,0)=\frac{1}{2}\;\delta(s)$. In view of the system (63) and
taking into account that

$h(s,t)=h_{0}(s,t)+h_{1}(s,t)$

the density function $h(s,t)$ satisfies the partial differential equation

$\left(\frac{\partial^{2}}{\partial t^{2}}+\frac{\partial^{2}}{\partial
s\;\partial t}+2\lambda(p+q)\frac{\partial}{\partial
t}+\lambda(p+q)\frac{\partial}{\partial s}\right)h=0.$ (64)

Thus, we are able to prove the following result.

######  Theorem 8.

The probability density function (62) reads

$h(s,t)=e^{-\lambda(p+q)t}\left[\lambda(p+q)\;I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)+\frac{\partial}{\partial
t}I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)\right],\qquad s\in(0,t).$ (65)

###### Proof.

In light of equation (64), the function

$\widetilde{h}(s,t)=e^{\lambda(p+q)t}\;h(s,t)$

satisfies the partial differential equation

$\left(\frac{\partial^{2}}{\partial t^{2}}+\frac{\partial^{2}}{\partial
s\partial t}-\lambda^{2}(p+q)^{2}\right)\widetilde{h}=0.$ (66)

The change of variables

$z=\sqrt{s(t-s)}$

transforms equation (66) into the Bessel equation

$\frac{d^{2}\widetilde{h}}{z^{2}}+\frac{1}{z}\frac{d\widetilde{h}}{dz}-4\lambda^{2}(p+q)^{2}\widetilde{h}=0$

whose general solution reads

$\widetilde{h}(z)=A\;I_{0}\Big{(}2\lambda(p+q)z\Big{)}+B\;K_{0}\Big{(}2\lambda(p+q)z\Big{)}.$

Therefore, usual arguments permit us to express the solution to equation (64)
in the form

$h(s,t)=e^{-\lambda(p+q)t}\left[A\;I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)+B\;\frac{\partial}{\partial
t}I_{0}\left(2\lambda(p+q)\sqrt{s(t-s)}\right)\right].$ (67)

In order for formula (67) to be consistent with (61), we must have that

$\int_{0}^{t}h(s,t)\mathop{ds}=1-e^{-\lambda(p+q)t}.$

By using the relation

$\int_{0}^{t}I_{0}\Big{(}K\;\sqrt{s(t-s)}\Big{)}\mathop{ds}=\frac{1}{K}\left(e^{\frac{Kt}{2}}-e^{-\frac{Kt}{2}}\right),\qquad
t>0$ (68)

we obtain

$A=\lambda(p+q),\qquad B=1$

which completes the proof. ∎

The corresponding characteristic function is given in the following theorem.

######  Theorem 9.

The characteristic function of $T(t)$ is

$\displaystyle\mathbb{E}\left[e^{i\alpha T(t)}\right]=$
$\displaystyle\frac{1}{2}\left(1+\frac{2\lambda(p+q)}{\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}}\right)e^{i\frac{\alpha}{2}t-\lambda(p+q)t+\frac{1}{2}\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}\;t}$
$\displaystyle+\frac{1}{2}\left(1-\frac{2\lambda(p+q)}{\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}}\right)e^{i\frac{\alpha}{2}t-\lambda(p+q)t-\frac{1}{2}\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}\;t}.$
(69)

###### Proof.

The function

$\widehat{h}(\alpha,t)=\mathbb{E}\left[e^{i\alpha T(t)}\right]$

satisfies, in view of equation (64), the ordinary differential equation

$\begin{dcases}\frac{d^{2}\widehat{h}}{dt^{2}}+\left[2\lambda(p+q)-i\alpha\right]\frac{d\widehat{h}}{dt}-i\lambda(p+q)\alpha\widehat{h}=0\\\
\widehat{h}(\alpha,0)=1\\\
\frac{d}{dt}\left.\widehat{h}(\alpha,t)\right\lvert_{t=0}=\frac{i\alpha}{2}.\end{dcases}$
(70)

The general solution to equation (70) can be expressed in the form

$\widehat{h}(\alpha,t)=k_{0}(\alpha)\;e^{i\frac{\alpha}{2}t-\lambda(p+q)t+\frac{1}{2}\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}\;t}+k_{1}(\alpha)\;e^{i\frac{\alpha}{2}t-\lambda(p+q)t-\frac{1}{2}\sqrt{4\lambda^{2}(p+q)^{2}-\alpha^{2}}\;t}.$

The coefficients $k_{0}$ and $k_{1}$ can be determined by using the initial
conditions. ∎

Observe that the characteristic function (69) can be computed directly from
the density function (65). By writing that

$\mathbb{E}\left[e^{i\alpha T(t)}\right]=\int_{0}^{t}e^{i\alpha
s}\,h(s,t)\mathop{ds}$

the integral formula

$\int_{0}^{t}e^{i\alpha
s}\,I_{0}\left(\beta\sqrt{s(t-s)}\right)\mathop{ds}=\frac{e^{\frac{i\alpha
t}{2}}}{\beta^{2}-\alpha^{2}}\left[e^{\frac{t}{2}\sqrt{\beta^{2}-\alpha^{2}}}-e^{-\frac{t}{2}\sqrt{\beta^{2}-\alpha^{2}}}\right]$

yields the result.

We conclcude our analysis by discussing some aspects of the joint distribution
of the time spent moving vertically $T(t)$ and the component $Y(t)$ of the
planar random motion studied so far. Clearly, the support of the bivariate
process $\big{(}T(t),\;Y(t)\big{)}$ coincides with the triangle

$R_{ct}=\left\\{(s,y)\in\mathbb{R}^{2}:\;0\leq s\leq t,\;-ct\leq y\leq
ct\right\\}.$ (71)

We consider the probability density function $\bar{u}(s,y,t)$ defined by the
relationship

$\bar{u}(s,y,t)\mathop{ds}\mathop{dy}=\mathbb{P}\Big{(}T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy}\Big{)},\qquad(s,y)\in
R_{ct}.$ (72)

At the initial time $t=0$, the process $\big{(}T(t),\;Y(t)\big{)}$ starts at
the vertex $(0,0)$ of the triangle $R_{ct}$ and it moves along two directions
which are parallel to the oblique sides of $R_{ct}$. In particular, the
normalized vectors which describe the directions along which
$\big{(}T(t),\;Y(t)\big{)}$ can move are given by

$\theta_{+}=\left(\frac{1}{\sqrt{1+c^{2}}},\;\frac{c}{\sqrt{1+c^{2}}}\right),\qquad\theta_{-}=\left(\frac{1}{\sqrt{1+c^{2}}},\;-\frac{c}{\sqrt{1+c^{2}}}\right).$

Clearly, in both cases the particle with position $\big{(}T(t),\;Y(t)\big{)}$
moves rightwards since the process $T(t)$ is non-decreasing. We denote by
$\Theta(t)$ the direction of $\big{(}T(t),\;Y(t)\big{)}$ at time $t>0$. In
order to understand how the motion of $\big{(}X(t),\;Y(t)\big{)}$ on $S_{ct}$
and that of $\big{(}T(t),\;Y(t)\big{)}$ on $R_{ct}$ relate to each other, we
first observe that $\Theta(t)=\theta_{+}$ if and only if $D(t)=d_{1}$ and
$\Theta(t)=\theta_{-}$ if and only if $D(t)=d_{3}$. It is now clear that when
$\big{(}X(t),\;Y(t)\big{)}$ moves horizontally, neither $T(t)$ increases nor
$Y(t)$, so that the particle in $R_{ct}$ remains still. If the process
$\big{(}T(t),\;Y(t)\big{)}$ is not moving at time $t$ we say that
$\Theta(t)=0$. Therefore, we have that $\Theta(t)=0$ if and only if
$D(t)\in\\{d_{0},d_{2}\\}$.
We now study the rule which determines the changes of the direction
$\Theta(t)$ for the particle moving in the triangle $R_{ct}$. For this
purpose, we denote by $E_{t}$ the event of a change of direction occurring
during the time interval $(t,t+\mathop{dt}]$. This enables us to write that

$\Theta(t+dt)\Big{\lvert}\left\\{E_{t},\;\Theta(t)=\theta_{+}\right\\}\;=\begin{cases}0&\text{with
prob.}\;p+q\\\ \theta_{-}&\text{with prob.}\;1-p-q.\end{cases}$ (73)

Formula (73) can be interpreted in the following manner. Assume that at time
$t$ the particle $\big{(}T(t),\;Y(t)\big{)}$ is moving with direction
$\Theta(t)=\theta_{+}$ and that the event $E_{t}$ occurs. Since
$\Theta(t)=\theta_{+}$, we have that $D(t)=d_{1}$ and thus we can distinghuish
between two cases. In the first case reflection occurs with probability
$1-p-q$, which implies that $D(t+dt)=d_{3}$ and thus $\Theta(t)=\theta_{-}$.
If reflection does not occur, which happens with probability $p+q$, we have
that $D(t)\in\\{d_{0},d_{2}\\}$ and therefore $\Theta(t+dt)=0$. Similarly, we
have that

$\Theta(t+dt)\Big{\lvert}\left\\{E_{t},\;\Theta(t)=\theta_{-}\right\\}\;=\begin{cases}0&\text{with
prob.}\;p+q\\\ \theta_{+}&\text{with prob.}\;1-p-q.\end{cases}$

We now emphasize that difficulties emerge when conditioning the distribution
of $\Theta(t+dt)$ on the event $\Theta(t)=0$. It can be easily verified that

$\Theta(t+dt)\Big{\lvert}\left\\{E_{t},\;\Theta(t)=0,\;D(t)=d_{0}\right\\}\;=\begin{cases}0&\text{with
prob.}\;1-p-q\\\ \theta_{+}&\text{with prob.}\;p\\\ \theta_{-}&\text{with
prob.}\;q.\end{cases}$ (74)

and analogously

$\Theta(t+dt)\Big{\lvert}\left\\{E_{t},\;\Theta(t)=0,\;D(t)=d_{2}\right\\}\;=\begin{cases}0&\text{with
prob.}\;1-p-q\\\ \theta_{+}&\text{with prob.}\;q\\\ \theta_{-}&\text{with
prob.}\;p.\end{cases}$ (75)

The difference between formulas (74) and (75) shows that the distribution of
$\Theta(t+dt)$, conditional on the fact that $\Theta(t)=0$ and that a change
of direction occurs, cannot be determined exactly unless the additional
information on $D(t)$ is given. The interesting fact about this phenomenon is
that it only occurs if $p\neq q$. For $p=q$, the right-hand sides of formulas
(74) and (75) are equal and we can write

$\Theta(t+dt)\Big{\lvert}\left\\{E_{t},\;\Theta(t)=0\right\\}\;=\begin{cases}0&\text{with
prob.}\;1-2p\\\ \theta_{+}&\text{with prob.}\;p\\\ \theta_{-}&\text{with
prob.}\;p\end{cases}$

which noticeably simplifies the joint distribution of the couple
$\big{(}T(t),\;Y(t)\big{)}$. In order to understand how the study of the
density function $\bar{u}(s,y,t)$ is simplified for $p=q$, we define, for
$j=0,1,2,3$, the auxiliary density functions

$\bar{u}_{j}(s,y,t)\mathop{ds}\mathop{dy}=\mathbb{P}\left(T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy},\;D(t)=d_{j}\right),\qquad(s,y)\in
R_{ct}.$

and we observe that the following linear system of partial differential
equations is satisfied:

$\begin{dcases}\dfrac{\partial\bar{u}_{0}}{\partial
t}=-\lambda\bar{u}_{0}+\lambda q\bar{u}_{1}+\lambda(1-p-q)\bar{u}_{2}+\lambda
p\bar{u}_{3}\\\ \dfrac{\partial\bar{u}_{1}}{\partial
t}=-\dfrac{\partial\bar{u}_{1}}{\partial
s}-c\dfrac{\partial\bar{u}_{1}}{\partial y}+\lambda p\bar{u}_{0}+\lambda
q\bar{u}_{2}+\lambda(1-p-q)\bar{u}_{3}-\lambda\bar{u}_{1}\\\
\dfrac{\partial\bar{u}_{2}}{\partial
t}=-\lambda\bar{u}_{2}+\lambda(1-p-q)\bar{u}_{0}+\lambda p\bar{u}_{1}+\lambda
q\bar{u}_{3}\\\ \dfrac{\partial\bar{u}_{3}}{\partial
t}=-\dfrac{\partial\bar{u}_{3}}{\partial
s}+c\dfrac{\partial\bar{u}_{3}}{\partial y}+\lambda
q\bar{u}_{0}+\lambda(1-p-q)\bar{u}_{1}+\lambda
p\bar{u}_{2}-\lambda\bar{u}_{3}.\end{dcases}$ (76)

The system (76) implies that $\bar{u}(s,y,t)$ satisfies the cumbersome fourth-
order partial differential equation

$\displaystyle\Bigg{\\{}\left[\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}-\lambda^{2}(1-p-q)^{2}\right]\left[\left(\frac{\partial}{\partial
t}+\lambda+\frac{\partial}{\partial
s}\right)^{2}-c^{2}\frac{\partial^{2}}{\partial
y^{2}}\right]-4\lambda^{2}pq\left(\frac{\partial}{\partial
t}+\lambda\right)\left(\frac{\partial}{\partial
t}+\lambda+\frac{\partial}{\partial s}\right)$
$\displaystyle\qquad-4\lambda^{3}(1-p-q)(p^{2}+q^{2})\left(\frac{\partial}{\partial
t}+\lambda+\frac{1}{2}\frac{\partial}{\partial
s}\right)-\lambda^{2}(1-p-q)^{2}\left(\frac{\partial}{\partial
t}+\lambda\right)^{2}-\lambda^{4}(p^{2}-q^{2})^{2}$
$\displaystyle\qquad-4\lambda^{4}pq(1-p-q)^{2}+\lambda^{4}(1-p-q)^{4}\Bigg{\\}}\bar{u}=0.$
(77)

We now show that the system (76) and equation (77) simplify in the case $p=q$.
We define the function

$\bar{u}_{02}(s,y,t)\mathop{ds}\mathop{dy}=\mathbb{P}\left(T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy},\;D(t)\in\\{d_{0},\;d_{2}\\}\right),\qquad
s\in[0,t],\;y\in[-ct,ct].$

Of course, we have that

$\bar{u}_{02}(s,y,t)=\bar{u}_{0}(s,y,t)+\bar{u}_{2}(s,y,t).$

Therefore, the first and the third equations of the system (76) imply that, if
$p=q$, the ordinary differential equation

$\dfrac{\partial\bar{u}_{02}}{\partial t}=-2\lambda p\,\bar{u}_{02}+2\lambda
p(\bar{u}_{1}+\bar{u}_{3})$ (78)

holds which permits to simplify the system (76) into

$\begin{dcases}\dfrac{\partial\bar{u}_{02}}{\partial t}=-2\lambda
p\,\bar{u}_{02}+2\lambda p(\bar{u}_{1}+\bar{u}_{3})\\\
\dfrac{\partial\bar{u}_{1}}{\partial t}=-\dfrac{\partial\bar{u}_{1}}{\partial
s}-c\dfrac{\partial\bar{u}_{1}}{\partial y}+\lambda
p\bar{u}_{02}+\lambda(1-2p)\bar{u}_{3}-\lambda\bar{u}_{1}\\\
\dfrac{\partial\bar{u}_{3}}{\partial t}=-\dfrac{\partial\bar{u}_{3}}{\partial
s}+c\dfrac{\partial\bar{u}_{3}}{\partial y}+\lambda
p\bar{u}_{02}+\lambda(1-2p)\bar{u}_{1}-\lambda\bar{u}_{3}.\end{dcases}$ (79)

Observe that equation (78) is not satisfied if $p\neq q$. Since

$\bar{u}(s,y,t)=\bar{u}_{02}(s,y,t)+\bar{u}_{1}(s,y,t)+\bar{u}_{3}(s,y,t)$

the linear system (79) implies that, if $p=q$, equation (77) reduces to the
third-order partial differential equation

$\displaystyle\Bigg{\\{}\left(\frac{\partial}{\partial t}+2\lambda
p\right)\left[\left(\frac{\partial}{\partial
t}+\lambda+\frac{\partial}{\partial
s}\right)-c^{2}\frac{\partial^{2}}{\partial
y^{2}}\right]-4\lambda^{2}p^{2}\left(\frac{\partial}{\partial
t}+\lambda+\frac{\partial}{\partial s}\right)$
$\displaystyle\qquad\qquad-\lambda^{2}(1-2p)^{2}\left(\frac{\partial}{\partial
t}+2\lambda p\right)-4\lambda^{3}p^{2}(1-2p)\Bigg{\\}}\bar{u}=0.$ (80)

Unfortunately, finding an explicit solution to equation (77) seems a difficult
task even in the special case (80). However, in the following theorem we are
able to obtain the limiting behaviour of $\big{(}T(t),\;Y(t)\big{)}$ when both
the velocity of the process and the rate of the direction changes are
infinite.

######  Theorem 10.

For $\lambda,\,c\to+\infty$, with $\frac{\lambda}{c^{2}}\to 1$, the partial
differential equation (77) becomes

$\frac{\partial u}{\partial t}=-\frac{1}{2}\frac{\partial u}{\partial
s}+\frac{1}{4}\;\frac{(1-p)+(1-q)}{(1-p)^{2}+(1-q)^{2}}\frac{\partial^{2}u}{\partial
y^{2}.}$ (81)

###### Proof.

The proof can be performed by straightfoward calculation after dividing
equation (77) by $\lambda^{3}$. Simplifying the expression and taking the
limit yields the result. ∎

In order to interpret theorem 10 observe that, by taking the Fourier transform
of both sides of equation (81), it can be shown that the characteristic
function of $\big{(}T(t),\;Y(t)\big{)}$ becomes, in the hydrodynamic limit,

$\lim_{\lambda,c\to+\infty}\mathbb{E}\left[e^{i\left(\alpha T(t)+\beta
Y(y)\right)}\right]=e^{i\frac{\alpha
t}{2}-\frac{(1-p)+(1-q)}{(1-p)^{2}+(1-q)^{2}}\frac{\beta^{2}t}{4}}.$ (82)

Taking the inverse Fourier transform of the expression (82) yields

$\lim_{\lambda,c\to+\infty}\mathbb{P}\Big{(}T(t)\in\mathop{ds},\;Y(t)\in\mathop{dy}\Big{)}/\left(\mathop{ds}\mathop{dy}\right)=\sqrt{\frac{(1-p)^{2}+(1-q)^{2}}{(1-p)+(1-q)}}\cdot\frac{e^{-\frac{(1-p)^{2}+(1-q)^{2}}{(1-p)+(1-q)}\,\frac{y^{2}}{t}}}{\sqrt{\pi
t}}\cdot\delta\left(s-\frac{t}{2}\right).$ (83)

Equation (83) implies that, while $Y(t)$ converges to a Brownian motion, the
process $T(t)$ becomes deterministic and its limiting value is equal to
$\frac{t}{2}$. Thus, in the hydrodynamic limit, the process
$\big{(}X(t),\;Y(t)\big{)}$ spends half of the time moving vertically.

We are now interested in studying the behaviour of $\big{(}T(t),\;Y(t)\big{)}$
on the boundary of $R_{ct}$. We start by considering the oblique side with
equation $y=cs$ and we define the probability density function

$\bar{f}(s,t)=\mathbb{P}\Big{(}T(t)\in\mathop{ds},\,Y(t)=c\,T(t)\Big{)}/\mathop{ds},\qquad
s\in(0,t).$ (84)

By setting, for $j=0,1,2$,

$\bar{f}_{j}(s,t)=\mathbb{P}\Big{(}T(t)\in\mathop{ds},\,Y(t)=c\,T(t),\,D(t)=d_{j}\Big{)}/\mathop{ds},\qquad
s\in(0,t)$

we clearly have that

$\begin{dcases}\bar{f}_{0}(s,t+dt)=\bar{f}_{0}(s,t)(1-\lambda\mathop{dt})+\bar{f}_{1}(s,t)\lambda
q\mathop{dt}+\bar{f}_{2}(s,t)\lambda(1-p-q)\mathop{dt}+o(dt)\\\
\bar{f}_{1}(s,t+dt)=\bar{f}_{1}(s-\mathop{ds},t)(1-\lambda\mathop{dt})+\bar{f}_{0}(s,t)\lambda
p\mathop{dt}+\bar{f}_{2}(s,t)\lambda q\mathop{dt}+o(dt)\\\
\bar{f}_{2}(s,t+dt)=\bar{f}_{2}(s,t)(1-\lambda\mathop{dt})+\bar{f}_{0}(s,t)\lambda(1-p-q)\mathop{dt}+\bar{f}_{1}(s,t)\lambda
p\mathop{dt}+o(dt)\end{dcases}$

which implies that

$\begin{dcases}\frac{\partial\bar{f}_{0}}{\partial t}=\lambda
q\bar{f}_{1}+\lambda(1-p-q)\bar{f}_{2}-\lambda\bar{f}_{0}\\\
\frac{\partial\bar{f}_{1}}{\partial t}=-\frac{\partial\bar{f}_{1}}{\partial
s}+\lambda p\bar{f}_{0}+\lambda q\bar{f}_{2}-\lambda\bar{f}_{1}\\\
\frac{\partial\bar{f}_{2}}{\partial t}=\lambda(1-p-q)\bar{f}_{0}+\lambda
p\bar{f}_{1}-\lambda\bar{f}_{2}\end{dcases}$ (85)

with initial conditions $\bar{f}_{j}(s,0)=\frac{1}{4}\,\delta(s)$ for
$j=0,1,2$. Thus, the probability density function (84) satisfies the third-
order partial differential equation

$\displaystyle\bigg{\\{}\left(\frac{\partial}{\partial
t}+\lambda\right)^{3}+\left(\frac{\partial}{\partial t}+\lambda\right)^{2}$
$\displaystyle\frac{\partial}{\partial
s}-\lambda^{2}\left[2pq+(1-p-q)^{2}\right]\left(\frac{\partial}{\partial
t}+\lambda\right)$
$\displaystyle-\lambda^{2}(1-p-q)^{2}\frac{\partial}{\partial
s}-\lambda^{3}(1-p-q)(p^{2}+q^{2})\bigg{\\}}\bar{f}=0.$ (86)

While finding a general solution to equation (86) is a difficult task, we
observe that a noticeable simplification of the equation occurs if the process
does not admit reversion, that is if $p+q=1$. In this case equation (86) can
be expressed in the form

$\left(\frac{\partial}{\partial
t}+\lambda\right)\Bigg{(}\frac{\partial^{2}}{\partial
t^{2}}+\left(\frac{\partial}{\partial
s}+2\lambda\right)\frac{\partial}{\partial
t}+\lambda^{2}\left[1-2p(1-p)\right]+\lambda\frac{\partial}{\partial
s}\Bigg{)}\bar{f}=0.$ (87)

Thus, if reflection is not admitted, we are able to study the exact
distribution of $\big{(}T(t),Y(t)\big{)}$ on the oblique sides of $R_{ct}$. We
start by calculating the probability of the particle lying on the upper
oblique side.

######  Theorem 11.

If the process $\big{(}X(t),Y(t)\big{)}$ does not admit reflection, that is if
$p+q=1$, it holds that

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{)}=\frac{e^{-\lambda
t}}{8}\Bigg{\\{}$
$\displaystyle\left(1+\frac{1}{\sqrt{2p(1-p)}}\right)^{2}e^{\lambda
t\sqrt{2p(1-p)}}$
$\displaystyle+\left(1-\frac{1}{\sqrt{2p(1-p)}}\right)^{2}e^{-\lambda
t\sqrt{2p(1-p)}}-\frac{(2p-1)^{2}}{p(1-p)}\Bigg{\\}}.$ (88)

###### Proof.

We start by studying the probability of $\big{(}T(t),Y(t)\big{)}$ lying on the
oblique side of $R_{ct}$ conditional on the initial direction of
$\big{(}X(t),Y(t)\big{)}$ being $d_{1}$. Clearly, we can write that

$\displaystyle\mathbb{P}\big{(}Y(t)$
$\displaystyle=c\,T(t)\big{\lvert}D(0)=d_{1}\big{)}=\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t),\;N(t)=k\big{\lvert}D(0)=d_{1}\big{)}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t),\;N(t)=2k\big{\lvert}D(0)=d_{1}\big{)}+\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t),\;N(t)=2k+1\big{\lvert}D(0)=d_{1}\big{)}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k\big{)}\,\mathbb{P}\left(N(t)=2k\right)$
$\displaystyle\;\;+\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k+1\big{)}\,\mathbb{P}\left(N(t)=2k+1\right)$
$\displaystyle=$ $\displaystyle e^{-\lambda
t}\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k\big{)}\,\frac{(\lambda
t)^{2k}}{(2k)!}$ $\displaystyle\;\;+e^{-\lambda
t}\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k+1\big{)}\,\frac{(\lambda
t)^{2k+1}}{(2k+1)!}.$ (89)

We now observe that

$\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k\big{)}=\left(2p(1-p)\right)^{k}.$
(90)

In order to clarify the formula above, observe that the event
$\\{Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k\\}$ occurs if the process
alternates the upward direction $d_{1}$ with horizontal directions, which can
be both $d_{0}$ and $d_{2}$. In order for $N(t)$ to be equal to $2k$, a change
of direction from $d_{1}$ to horizontal and then back to $d_{1}$ must occur
$k$ times, and it always occurs with probability $p(1-p)$ no matter whether
the horizontal direction is $d_{0}$ or $d_{2}$. Therefore, a path having
initial direction $d_{1}$ and involving $2k$ changes of directions which
alternate $d_{1}$ and horizontal directions can occur with probability
$\left(p(1-p)\right)^{k}$. Moreover, there are $2^{k}$ equiprobable ways of
combining all the possible choices of the horizontal directions of the
sequence, since every horizontal direction must belong to the set
$\\{d_{0},\,d_{2}\\}$.
Similar arguments permit to show that

$\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{1},\;N(t)=2k+1\big{)}=\left(2p(1-p)\right)^{k}.$
(91)

Formula (91) holds because, in order to perform $2k+1$ changes of direction,
the process must first perform $2k$ changes of direction with probability
$\left(2p(1-p)\right)^{k}$ and then perform an additional change. Since the
initial direction is $d_{1}$, the first $2k$ changes lead to the direction
$d_{1}$ again. The event $\\{Y(t)=c\,T(t)\\}$ is then true if and only if the
additional change of direction occurs towards a horizontal direction, which
occurs with probability 1 because reflection is not admitted by hypothesis.
By now substituting formulas (90) and (91) into (89), we obtain

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}$ $\displaystyle
D(0)=d_{1}\big{)}=\sum_{k=0}^{\infty}\mathbb{P}\big{(}Y(t)=c\,T(t),\;N(t)=k\big{\lvert}D(0)=d_{1}\big{)}$
$\displaystyle=$ $\displaystyle e^{-\lambda
t}\sum_{k=0}^{\infty}\frac{(\sqrt{2p(1-p)}\lambda
t)^{2k}}{(2k)!}+\frac{e^{-\lambda
t}}{\sqrt{2p(1-p)}}\sum_{k=0}^{\infty}\frac{(\sqrt{2p(1-p)}\lambda
t)^{2k+1}}{(2k+1)!}$ $\displaystyle=$ $\displaystyle e^{-\lambda
t}\Bigg{\\{}\operatorname{cosh}\left(\lambda
t\sqrt{2p(1-p)}\right)+\frac{1}{\sqrt{2p(1-p)}}\,\operatorname{sinh}\left(\lambda
t\sqrt{2p(1-p)}\right)\Bigg{\\}}$ $\displaystyle=$
$\displaystyle\frac{e^{-\lambda
t}}{2}\Bigg{\\{}\left(1+\frac{1}{\sqrt{2p(1-p)}}\right)e^{\lambda
t\sqrt{2p(1-p)}}$
$\displaystyle\;\;\;\;\;\;\;+\left(1-\frac{1}{\sqrt{2p(1-p)}}\right)e^{-\lambda
t\sqrt{2p(1-p)}}\Bigg{\\}}.$

Similar arguments permit to obtain the probability of
$\big{(}T(t),Y(t)\big{)}$ lying on the oblique side of $R_{ct}$ conditional on
the initial direction being horizontal. In particular, we have that

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{0}\big{)}=e^{-\lambda
t}\Bigg{\\{}$
$\displaystyle\frac{p}{2}\left(\frac{1}{2p(1-p)}+\frac{1}{\sqrt{2p(1-p)}}\right)e^{\lambda
t\sqrt{2p(1-p)}}$ $\displaystyle+$
$\displaystyle\frac{p}{2}\left(\frac{1}{2p(1-p)}-\frac{1}{\sqrt{2p(1-p)}}\right)e^{-\lambda
t\sqrt{2p(1-p)}}+1-\frac{1}{2(1-p)}\Bigg{\\}}$

and

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{2}\big{)}=e^{-\lambda
t}\Bigg{\\{}$
$\displaystyle\frac{1-p}{2}\left(\frac{1}{2p(1-p)}+\frac{1}{\sqrt{2p(1-p)}}\right)e^{\lambda
t\sqrt{2p(1-p)}}$ $\displaystyle+$
$\displaystyle\frac{1-p}{2}\left(\frac{1}{2p(1-p)}-\frac{1}{\sqrt{2p(1-p)}}\right)e^{-\lambda
t\sqrt{2p(1-p)}}+1-\frac{1}{2p}\Bigg{\\}}.$

The theorem is finally proved by observing that

$\mathbb{P}\big{(}Y(t)=c\,T(t)\big{)}=\frac{1}{4}\,\sum_{j=0}^{2}\mathbb{P}\big{(}Y(t)=c\,T(t)\big{\lvert}D(0)=d_{j}\big{)}.$

∎

We are now able to obtain the distribution (84) in explicit form.

######  Theorem 12.

If $p+q=1$, the probability density function (84) is

$\displaystyle\bar{f}(s,t)=\frac{\lambda}{2\sqrt{2}}\;$ $\displaystyle
I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)}$
$\displaystyle+\frac{1}{4}\left(1+\frac{1}{2p(1-p)}\right)\frac{\partial}{\partial
t}I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)},\qquad\lvert
s\in(0,t).$

###### Proof.

In order to find a solution to equation (87), we solve the equation

$\Bigg{(}\frac{\partial^{2}}{\partial t^{2}}+\left(\frac{\partial}{\partial
s}+2\lambda\right)\frac{\partial}{\partial
t}+\lambda^{2}\left[1-2p(1-p)\right]+\lambda\frac{\partial}{\partial
s}\Bigg{)}\bar{f}=0.$

By setting

$\widetilde{\bar{f}}(s,t)=e^{\lambda t}\bar{f}(s,t)$

the equation is transformed into

$\Bigg{(}\frac{\partial^{2}}{\partial t^{2}}+\frac{\partial^{2}}{\partial
s\partial t}-2\lambda^{2}p(1-p)\Bigg{)}\widetilde{\bar{f}}=0.$ (92)

The change of variables $z=\sqrt{s(t-s)}$ transforms equation (92) into the
Bessel equation

$\frac{d^{2}\widetilde{\bar{f}}}{z^{2}}+\frac{1}{z}\frac{d\widetilde{\bar{f}}}{dz}-8\lambda^{2}p(1-p)\widetilde{\bar{f}}=0$

which admits general solution in the form

$\widetilde{\bar{f}}(z)=A\;I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,z\Big{)}+B\;K_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,z\Big{)}.$

By discarding the second kind modified Bessel function $K_{0}$ we write that

$\bar{f}(s,t)=e^{-\lambda
t}\left[A\;I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)}+B\;\frac{\partial}{\partial
t}I_{0}\Big{(}2\lambda\sqrt{2p(1-p)}\,\sqrt{s(t-s)}\Big{)}\right].$ (93)

In view of the integral (68), the coefficients $A$ and $B$ which make formula
(93) consistent with formula (88) are

$A=\frac{\lambda}{2\sqrt{2}},\qquad
B=\frac{1}{4}\left(1+\frac{1}{2p(1-p)}\right).$

∎

We conclude our study of the distribution (84) in the case $p+q=1$ by
obtaining, in the following theorem, the characteristic function of
$\big{(}T(t),Y(t)\big{)}$ on the oblique side of $R_{ct}$.

######  Theorem 13.

If $p+q=1$, the characteristic function of $\big{(}T(t),Y(t)\big{)}$ on the
oblique side of $R_{ct}$ reads

$\displaystyle\mathbb{E}$ $\displaystyle\left[e^{i\alpha
T(t)}\;\mathds{1}_{\\{Y(t)=c\,T(t)\\}}\right]$
$\displaystyle=\frac{1}{16}\left(\frac{1+2p(1-p)}{p(1-p)}+\frac{4p(1-p)(2\lambda+i\alpha)-i\alpha(1+2p(1-p))}{p(1-p)\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}}\right)e^{-\lambda
t+\frac{t}{2}\left(i\alpha+\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}\right)}$
$\displaystyle\;\;+\frac{1}{16}\left(\frac{1+2p(1-p)}{p(1-p)}-\frac{4p(1-p)(2\lambda+i\alpha)-i\alpha(1+2p(1-p))}{p(1-p)\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}}\right)e^{-\lambda
t+\frac{t}{2}\left(i\alpha-\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}\right)}$
$\displaystyle\;\;\;\;\;-\frac{(2p-1)^{2}}{8p(1-p)}\;e^{-\lambda t}.$ (94)

###### Proof.

We set

$\widehat{\bar{f}}(s,t)=\mathbb{E}\left[e^{i\alpha
T(t)}\;\mathds{1}_{\\{Y(t)=c\,T(t)\\}}\right].$

By taking into account equation (87), the characteristic function
$\widehat{\bar{f}}(s,t)$ can be obtained by finding the solution to the
ordinary differential equation

$\begin{dcases}\frac{d^{3}\widehat{\bar{f}}}{dt^{3}}+\left[3\lambda-i\alpha\right]\frac{d^{2}\widehat{\bar{f}}}{dt^{2}}+\lambda\left[\lambda\left(3-2p(1-p)\right)-2i\alpha\right]\frac{d\widehat{\bar{f}}}{dt}+\lambda^{2}\left[\lambda\left(1-2p(1-p)\right)-i\alpha\right]\widehat{\bar{f}}=0\\\
\widehat{\bar{f}}(\alpha,0)=\frac{3}{4}\\\
\left.\frac{d}{dt}\widehat{\bar{f}}(\alpha,t)\right\lvert_{t=0}=\frac{i\alpha-\lambda}{4}\\\
\left.\frac{d^{2}}{dt^{2}}\widehat{\bar{f}}(\alpha,t)\right\lvert_{t=0}=\frac{2\lambda^{2}p(1-p)-\alpha^{2}}{4}.\end{dcases}$
(95)

The general solution to equation (95) reads

$\widehat{\bar{f}}(\alpha,t)=k_{0}(\alpha)\;e^{-\lambda
t+\frac{t}{2}\left(i\alpha+\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}\right)}+k_{1}(\alpha)\;e^{-\lambda
t+\frac{t}{2}\left(i\alpha-\sqrt{8\lambda^{2}p(1-p)-\alpha^{2}}\right)}+k_{2}(\alpha)\;e^{-\lambda
t}.$

and the coefficients $k_{0},\;k_{1}$ and $k_{2}$ can be obtained by using the
initial conditions. ∎

So far, we have studied the distribution of $\big{(}T(t),Y(t)\big{)}$ on the
oblique side of the triangle $R_{ct}$ and we observed that, while finding the
general distribution is a difficult problem, an explicit representation for
the distribution can be obtained if reflection is not admitted. We now discuss
another special case in which equation (86) can be replaced by a simpler
second-order equation, namely the case in which $p=q$. By defining

$\bar{f}_{02}(s,t)=\mathbb{P}\big{(}T(t)\in\mathop{ds},\,Y(t)=cs,\,D(t)\in\\{d_{0},\,d_{2}\\}\big{)}/\mathop{ds},\qquad
s\in(0,t)$

it is clear that, under the assumption that $p=q$, the system of equations
(85) can be written in the form

$\begin{dcases}\frac{\partial\bar{f}_{02}}{\partial t}=-2\lambda
p\bar{f}_{02}+2\lambda p\bar{f}_{1}\\\ \frac{\partial\bar{f}_{1}}{\partial
t}=-\frac{\partial\bar{f}_{1}}{\partial s}+\lambda
p\bar{f}_{02}-\lambda\bar{f}_{1}\end{dcases}$ (96)

which implies that the probability density function (84) satisfies the second-
order partial differential equation

$\left(\frac{\partial^{2}}{\partial t^{2}}+\frac{\partial^{2}}{\partial
s\,\partial t}+\lambda(1+2p)\frac{\partial}{\partial t}+2\lambda
p\frac{\partial}{\partial s}+2\lambda^{2}p(1-p)\right)\bar{f}=0.$ (97)

In principle, we would start the analysis of the density function $\bar{f}$ by
first studying the probability of the particle with position
$\big{(}T(t),Y(t)\big{)}$ being on the oblique side of $R_{ct}$.
Unfortunately, a combinatorial approach in the fashion of theorem 11 is not
trivial if $p=q$ because reversion is admitted, except for the special case
$p=q=\frac{1}{2}$. Therefore, for $p=q$, we start by studying the
characteristic function in the following theorem.

######  Theorem 14.

If $p=q$, the characteristic function of $\big{(}T(t),Y(t)\big{)}$ on the
oblique side of $R_{ct}$ reads

$\displaystyle\mathbb{E}$ $\displaystyle\left[e^{i\alpha
T(t)}\;\mathds{1}_{\\{Y(t)=c\,T(t)\\}}\right]$ $\displaystyle=$
$\displaystyle\frac{3}{8}\left(1+\frac{\lambda-i\alpha+6\lambda
p}{3\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}}\right)\;e^{\frac{t}{2}\left(i\alpha-\lambda(1+2p)+\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}\right)}$
$\displaystyle\;\;+\frac{3}{8}\left(1-\frac{\lambda-i\alpha+6\lambda
p}{3\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}}\right)\;e^{\frac{t}{2}\left(i\alpha-\lambda(1+2p)-\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}\right)}.$
(98)

###### Proof.

Similarly to theorem 13, we set

$\widehat{\bar{f}}(s,t)=\mathbb{E}\left[e^{i\alpha
T(t)}\;\mathds{1}_{\\{Y(t)=c\,T(t)\\}}\right].$

Since $p=q$, equation (97) implies that the characteristic function
$\widehat{\bar{f}}(s,t)$ is the solution to the ordinary differential equation

$\begin{dcases}\frac{d^{2}\widehat{\bar{f}}}{dt^{2}}+\left[\lambda\left(1+2p\right)-i\alpha\right]\frac{d\widehat{\bar{f}}}{dt}+\left[2\lambda^{2}p(1-p)-2i\alpha\lambda
p\right]\widehat{\bar{f}}=0\\\ \widehat{\bar{f}}(\alpha,0)=\frac{3}{4}\\\
\left.\frac{d}{dt}\widehat{\bar{f}}(\alpha,t)\right\lvert_{t=0}=\frac{i\alpha-\lambda}{4}.\end{dcases}$
(99)

The general solution to equation (99) reads

$\displaystyle\widehat{\bar{f}}(\alpha,t)=k_{0}(\alpha)\;$ $\displaystyle
e^{\frac{t}{2}\left(i\alpha-\lambda(1+2p)+\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}\right)}$
$\displaystyle\qquad+k_{1}(\alpha)\;e^{\frac{t}{2}\left(i\alpha-\lambda(1+2p)-\sqrt{\lambda^{2}(12p^{2}-4p+1)-\alpha^{2}-2i\alpha\lambda(1-2p)}\right)}.$
(100)

and the coefficients $k_{0},\;k_{1}$ and $k_{2}$ are obtained by using the
initial conditions. ∎

We are now able to calculate the probability of the process
$\big{(}T(t),Y(t)\big{)}$ lying on the oblique side of $R_{ct}$.

######  Theorem 15.

If $p=q$, it holds that

$\displaystyle\mathbb{P}\big{(}Y(t)=c\,T(t)\big{)}=$
$\displaystyle\frac{3}{8}\left(1+\frac{(1+6p)}{3\sqrt{12p^{2}-4p+1}}\right)\;e^{\frac{\lambda
t}{2}\left(-(1+2p)+\sqrt{12p^{2}-4p+1}\right)}$
$\displaystyle\;\;+\frac{3}{8}\left(1-\frac{(1+6p)}{3\sqrt{12p^{2}-4p+1}}\right)\;e^{\frac{\lambda
t}{2}\left(-(1+2p)-\sqrt{12p^{2}-4p+1}\right)}.$ (101)

###### Proof.

Formula (15) follows immediately from formula (14) by setting $\alpha=0$.∎

It is interesting to observe that formulas (88) and (15) coincide if
$p=q=\frac{1}{2}$.
Unfortunately, the inverse Fourier transform of the characteristic function
(14) is difficult to find. Therefore, in the case $p=q$ we are not able to
obtain the exact distribution of the process on the oblique side of $R_{ct}$.

We conclude our work by studying the distribution of the process
$\big{(}T(t),Y(t)\big{)}$ on the vertical side of $R_{ct}$. Of course, the
particle with position $\big{(}T(t),Y(t)\big{)}$ lies on the vertical side if
and only if $T(t)=t$, that is if the process $\big{(}X(t),Y(t)\big{)}$ has
always moved vertically. Recalling formula (61), we have that

$\mathbb{P}\left(T(t)=t\right)=\frac{1}{2}\;e^{-\lambda t(p+q)}.$ (102)

We are now interested in studying the probability density function

$\bar{g}(y,t)\mathop{dy}=\mathbb{P}\Big{(}Y(t)\in\mathop{dy},\,T(t)=t\Big{)},\qquad|y|<ct.$
(103)

Moreover, for $j=1,3$, we define the functions

$\bar{g}_{j}(y,t)\mathop{dy}=\mathbb{P}\Big{(}Y(t)\in\mathop{dy},\,T(t)=t,\,D(t)=d_{j}\Big{)},\qquad|y|<ct.$

We have that

$\begin{dcases}\frac{\partial\bar{g}_{1}}{\partial
t}=-c\frac{\partial\bar{g}_{1}}{\partial
y}+\lambda(1-p-q)\bar{g}_{3}-\lambda\bar{g}_{1}\\\
\frac{\partial\bar{g}_{3}}{\partial t}=c\frac{\partial\bar{g}_{3}}{\partial
y}+\lambda(1-p-q)\bar{g}_{1}-\lambda\bar{g}_{3}\end{dcases}$ (104)

which implies that the partial differential equation

$\left(\frac{\partial^{2}}{\partial t^{2}}+2\lambda\frac{\partial}{\partial
t}-c^{2}\frac{\partial^{2}}{\partial
y^{2}}+\lambda^{2}(p+q)(2-p-q)\right)\bar{g}=0.$ (105)

is satisfied. We now observe that the system of equations (104) and equation
(105) perfectly resemble equations (34) and (35), where the variable $x$ is
replaced by $y$. Therefore, the motion on the diagonal of the square $S_{ct}$
is identical in distribution to the process moving on the vertical side of the
triangle $R_{ct}$. The reason for which this relationship holds is clear. Both
the motions take place at costant velocity $c$ on the segment $[-ct,\,ct]$ and
the changes of direction occur at Poisson times with constant intensity
$\lambda(1-p-q)$, that is when the process $\big{(}X(t),Y(t)\big{)}$ performs
a reflection. Thus, by using the ideas of theorems 4 and 5, the exact
distribution of $\big{(}T(t),Y(t)\big{)}$ on the vertical side of $R_{ct}$ can
be obtained in explicit form. In particular, if $p+q<1$, we have that

$\bar{g}(y,t)=\frac{e^{-\lambda
t}}{4c}\left[\lambda(1-p-q)\;I_{0}\left(\frac{\lambda}{c}\,(1-p-q)\,\sqrt{c^{2}t^{2}-y^{2}}\right)+\frac{\partial}{\partial
t}I_{0}\left(\frac{\lambda}{c}\,(1-p-q)\,\sqrt{c^{2}t^{2}-y^{2}}\right)\right]$

and

$\displaystyle\mathbb{E}\left[e^{i\beta
Y(t)}\;\mathds{1}_{\\{T(t)=t\\}}\right]=\frac{e^{-\lambda t}}{4}$
$\displaystyle\left[\left(1+\frac{\lambda(1-p-q)}{\sqrt{\lambda^{2}(1-p-q)^{2}-\beta^{2}c^{2}}}\right)e^{t\sqrt{\lambda^{2}(1-p-q)^{2}-\beta^{2}c^{2}}}\right.$
$\displaystyle+$
$\displaystyle\left.\left(1-\frac{\lambda(1-p-q)}{\sqrt{\lambda^{2}(1-p-q)^{2}-\beta^{2}c^{2}}}\right)e^{-t\sqrt{\lambda^{2}(1-p-q)^{2}-\beta^{2}c^{2}}}\right].$

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