# Martin boundary of a space-time Brownian motion with drift
killed at the boundary of a moving cone

Sandro Franceschi  Télécom SudParis, Institut Polytechnique de Paris

###### Abstract

We study a space-time Brownian motion with drift
$B(t)=(t_{0}+t,y_{0}+W(t)+\gamma t)$ killed at the moving boundary of the cone
$\\{(t,x):0\leqslant x\leqslant t\\}$. This article determines the parabolic
Martin boundary and all harmonic functions associated with this process. To
that end, the asymptotics of Green’s functions are determined along all
directions. We also find the exit probabilities at the edges, the probability
of remaining in the cone forever and the laws of the exit point and exit time.
From this, we derive an explicit formula for the transition kernel of the
process. These results arise from two different methods initially introduced
to study random walks. An analytical approach, developed in the 1970s by
Malyshev and based on the steepest descent method on a Riemann surface, is
used to determine the asymptotics of the Green’s functions. A recursive
compensation approach, inspired by the method developed in the 1990s by Adan,
Wessels and Zijm, is used to determine the harmonic functions.

##  1 Introduction

#### Main process and overview of the results

Let $W(t)$ be a standard Brownian motion, $\gamma\in(0,1)$ a drift,
$(t_{0},y_{0})$ a starting point such that $0<t_{0}<y_{0}$. We define the
space-time Brownian motion by

$B(t):=(t_{0}+t,y_{0}+W(t)+\gamma t).$ (1)

Let the cone

$C:=\\{(t^{\prime},y):0<y<t^{\prime}\\}$ (2)

which defines a two-sided moving boundary, and $T$ the first exit time of the
cone

$T:=\inf\\{t\geqslant 0:B(t)\notin C\\}.$ (3)

We also define the exit times on each edge of the cone

$T_{1}:=\inf\\{t\geqslant 0:B(t)=(t_{0}+t,t_{0}+t),\
t>0\\}\quad\text{and}\quad T_{2}:=\inf\\{t\geqslant 0:B(t)=(t_{0}+t,0),\
t>0\\}$ (4)

and we have $T=T_{1}\wedge T_{2}$.

The purpose of the present work is to determine the parabolic Martin boundary
associated with the space-time Brownian motion killed at the boundary of the
cone $C$ (Theorem 10). This result derives from the asymptotics of Green’s
functions (Theorem 4). An explicit expression is given for all the associated
harmonic functions (Theorem 13), that is the positive functions $u$ satisfying

$\begin{cases}(\partial^{2}_{y}+\gamma\partial_{y}+\partial_{t})u(t,y)=0&\text{for
all }(t,y)\in C,\\\ u(t,y)=0&\text{for all }(t,y)\in\partial C.\end{cases}$
(5)

We find the persistence probability $\mathbb{P}^{\alpha}(T=\infty)$
(Proposition 22) and the probabilities of exit on an edge
$\mathbb{P}^{\alpha}(T_{1}<T_{2})$ and $\mathbb{P}^{\alpha}(T_{2}<T_{1})$
(Proposition 15) for the process conditioned to drift in the direction
$\alpha$ via Doob’s $h$-transform. We also compute the laws of $T$, $T_{1}$,
$T_{2}$ and of the law of the exit point of the process $B(T)$ when $T<\infty$
(Theorem 18). Finally, this gives a new original approach to obtain the
transition kernel of the process killed at time $T$ (Corollary 20), which we
note

$p^{k,C}_{(t_{0},y_{0})}(t,y)\mathrm{d}y:=\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y),T>t).$
(6)

#### Proof strategy and structure of the article

Section 2 sets out some fundamental analytical preliminaries for our study.
For practical reasons we first transform via a simple linear application, the
space-time Brownian motion in the cone $C$ into a degenerate Brownian motion
in the quadrant $\mathbb{R}_{+}^{2}$. Then, we find a kernel functional
equation (Proposition 10) connecting the Laplace transform of the Green’s
functions of this new process and the Laplace transform of the exit densities
on the axes. Section 3 is devoted to the asymptotics of Green’s functions. In
this section, the key element of the proofs is the application of the steepest
descent method on a Riemann surface generated by the kernel. Section 4 gives a
probabilist interpretation of all the harmonic functions in the cone in terms
of the persistence probabilities of some conditioned processes via Doob’s
$h$-transform. Then, the full Martin boundary is determined. Section 5 is
based on a recursive compensation approach used to determine explicit
expressions for the harmonic functions. The minimal Martin boundary is shown
to be homeomorphic to a portion of a certain parabola defined by the kernel.
Persistence and exit probabilities are computed. Finally, Section 6 is
dedicated to the inversion of the Laplace transforms to obtain the law of the
exit point and the transition kernel of the process.

#### Related literature

The literature on Brownian motion and moving boundaries is very rich, from the
1960s to the present day. The study of the crossing probability of the
boundary is at the heart of many of these articles. The boundary can be one-
sided or two-sided, depending on the problems considered. We cannot claim to
be exhaustive here and will simply cite a few articles related to our problem.
In a seminal paper published in 1960 [2], Anderson considers a Brownian motion
between two non-parallel straight lines and determines the probability of
crossing one of the two straight lines before the other for a finite or
infinite time horizon. In 1964 [39], see [22, page 306-315] for an English
version, Skorokhod founds the probability density of the escape location of a
Lévy process that does not cross two parallel straight lines, which provides
the law of the process killed at the boundary. In 1967 [38] Sheep considered
for the first time in the continuous setting a square-root boundary. This type
of boundary will subsequently be the subject of numerous studies [31]. In the
1970s until recently Novikov solved many interesting related issues [36, 37].

There are also links between moving boundary results and Brownian motion in
Weil chambers and reflection groups. For example, Biane, Bougerol, and
O’Connell [4] studied the probability that a Brownian motion with drift stays
forever in a Weyl chamber, also called the persistence probability. Recently,
Defosseux [8] studied space-time Brownian motion in an affine Weyl chamber.
Some of the results of this last paper are found again in the present article
using other methods.

Since the seminal work of Martin [34], many books [11] and articles have
studied Brownian motion and random walks in cones, the asymptotics of their
Green’s functions, their Martin boundary and the associated harmonic
functions. In the discrete setting, examples include the work of Malyshev,
Kourkova et al. on the Martin boundary of random walks in a quarter plane [33,
29, 30, 28] using a saddle point method on a Riemann surface generated by the
kernel of a functional equation. This powerful technique used in the present
paper has recently been developed with success in the continuous setting for
reflected Brownian motion in wedges or half-planes [17, 18, 13].

In the higher dimensional case, the famous article of Ney and Spitzer [35]
computes the asymptotics of the Green’s function of random walks with drift in
$\mathbb{Z}^{d}$ and shows that the Martin boundary is homeomorphic to the
unit sphere. Many other interesting articles by Ignatiouk et al. deals with
the Martin boundary of random walks in half-spaces and orthants [27, 26, 25].
Denisov and Wachtel study the tail asymptotics for the exit time of a
multidimensional random walk in a cone [9]. Duraj et al. determine the
asymptotic of the Green function for random walks without drift confined to
multidimensional convex cones [12]. Garbit and Raschel study the survival
probability of multidimensional random walks in pyramids [21]. We also mention
the article of Hoang et al. on the construction of harmonic functions in
wedges using Boundary value problems [23].

In the continuous setting, several papers study Brownian motion in cones in
higher dimensions and we mention a few of them here. De Blassie determines the
distributions of first exit times of Brownian motion in cones and computes
asymptotics [7]. Bañuelos and Smits study the asymptotic behavior of Brownian
motion in cones and express their results with infinite series involving its
transition density [3]. Garbit and Raschel study the asymptotics of the tail
distribution of the first exit time of Brownian motion with drift for a large
class of cones [20].

Finally, we mention the compensation approach developed by Adan, Wessels and
Zijm [1] which is an inspiration for this paper. The recent article of Hoang
et al. [24] uses this method to compute discrete harmonic functions in the
quadrant. The paper of Ichiba et al. [19] also uses this approach to compute
the invariant measure of a degenerate competing three-particle system.

##  2 Analytical preliminaries

#### Killed degenerate Brownian motion in the quadrant

A simple linear transformation maps the space-time Brownian motion $B(t)$
defined in (2) in the cone $C$ onto a degenerate Brownian motion with drift
$Z(t)$ in the quadrant $\mathbb{R}_{+}^{2}$. To that purpose, we define the
linear transform $\ell$ by

$\ell(t^{\prime},y):=(t^{\prime}-y,y)\quad\text{and}\quad\ell^{-1}(x,y):=(x+y,y).$

Thanks to $\ell$ we perform the change of variable

$t^{\prime}=t_{0}+t=x+y.$

We denote $B(t)=(t_{0}+t,y_{0}+W(t)+\gamma t)=(t^{\prime},Y(t))$ the
coordinates of the space-time Brownian motion. We define the drifted
degenerate Brownian motion by

$\displaystyle Z(t)=(X(t),Y(t))$ $\displaystyle:=\ell(B(t))$ $\displaystyle\
=(t_{0}-y_{0}-W(t)+(1-\gamma)t,y_{0}+W(t)+\gamma t)$ $\displaystyle\
=z_{0}+\mathbf{W}(t)+\mu t$

where $\mathbf{W}(t):=(-W(t),W(t))$ is a degenerate Brownian motion in
$\mathbb{R}^{2}$ of covariance matrix $\begin{pmatrix}1&-1\\\
-1&1\end{pmatrix}$ and we denote $\mu:=(1-\gamma,\gamma)\in\mathbb{R}_{+}^{2}$
the positive drift and
$z_{0}=(x_{0},y_{0}):=(t_{0}-y_{0},y_{0})\in\mathbb{R}_{+}^{2}$ the starting
point.

We have $\ell(C)=(\mathbb{R}_{+}^{*})^{2}$ where $C$ is defined in (2). The
stopping time $T$ defined in (3) is equal to

$T=\inf\\{t\geqslant 0:Z(t)\notin(\mathbb{R}_{+}^{*})^{2}\\}.$

It is the first exit time of the quadrant for the process $Z(t)$. The stopping
times on each of the boundaries defined in (4) are equal to

$T_{1}=\inf\\{t\geqslant
0:Z(t)\in\\{0\\}\times\mathbb{R}_{+}\\}\quad\text{and}\quad
T_{2}=\inf\\{t\geqslant 0:Z(t)\in\mathbb{R}_{+}\times\\{0\\}\\}$

and we still have $T=T_{1}\wedge T_{2}$.

Figure 1: Space time Brownian motion in $C$ and degenerate Brownian motion in
$\mathbb{R}_{+}^{2}$

#### Green’s functions and exit densities

For a set $S\subset(\mathbb{R}_{+}^{*})^{2}$ we define the Green’s measure of
the process killed at the boundary by

$G_{z_{0}}(S):=\int_{0}^{\infty}\mathbb{P}(Z(t)\in S,t<T)\mathrm{d}t.$

Let us remark that for $S\subset(\mathbb{R}_{+}^{*})^{2}$ we have
$\mathbb{P}(Z(t)\in S,t<T)=\mathbb{P}(Z(t\wedge T)\in S)$. We assume that this
measure has a density denoted

$g_{z_{0}}(z)\mathrm{d}z:=G_{z_{0}}(\mathrm{d}z).$

Let us recall that in (6) we noted $p^{k,C}_{(t_{0},y_{0})}(t,y)$ the
transition kernel of the space time Brownian motion killed in the cone $C$. It
is important to remark that, up to the following change of variables, the
Green’s function of the process $Z$ killed at the boundary of the quadrant is
equal to the transition kernel of $B$ the space time Brownian motion killed at
the boundary of $C$ :

$p^{k,C}_{(t_{0},y_{0})}(t,y)=g_{z_{0}}(z)\quad\text{ where
}z_{0}=(t_{0}-y_{0},y_{0})\text{ and }z=(t_{0}+t-y,y).$ (7)

One may consult the famous Doob’s book [11, Chap IX §17] which explains how
Brownian Motion transition kernel may be seen as Green functions. The exit
measures on the boundaries are defined for $S\subset\mathbb{R}_{+}^{2}$ by

$\displaystyle A_{1}(S)$ $\displaystyle:=\mathbb{P}(Z(T_{1})\in
S,T_{1}<T_{2})=\mathbb{E}(\mathds{1}_{Z(T_{1})\in
S}\mathds{1}_{T_{1}<T_{2}}),$ $\displaystyle A_{2}(S)$
$\displaystyle:=\mathbb{P}(Z(T_{2})\in
S,T_{2}<T_{1})=\mathbb{E}(\mathds{1}_{Z(T_{2})\in
S}\mathds{1}_{T_{2}<T_{1}}),$

and the support of the measure $A_{1}$ (resp. $A_{2}$) lies on
$\\{0\\}\times\mathbb{R}_{+}$ (resp. $\mathbb{R}_{+}\times\\{0\\}$). We denote
their densities $f_{1}$ and $f_{2}$ defined as follows

$f_{1}(y)\mathrm{d}y\delta_{0}(\mathrm{d}x):=A_{1}(\mathrm{d}x,\mathrm{d}y),\quad
f_{2}(x)\mathrm{d}x\delta_{0}(\mathrm{d}y):=A_{2}(\mathrm{d}x,\mathrm{d}y),$

where $\delta_{0}$ is the Dirac measure in $0$. One may notice that
$(t_{0}+T)\mathds{1}_{T<\infty}=(X(T)+Y(T))\mathds{1}_{T<\infty}=Y(T_{1})\mathds{1}_{T_{1}<T_{2}}+X(T_{2})\mathds{1}_{T_{2}<T_{1}}$
and we deduce that the cumulative distribution function of $T$ is equal to
$F_{T}(t):=\mathbb{P}(T<t)=A_{1}(\\{0\\}\times[0,t_{0}+t])+A_{2}([0,t_{0}+t]\times\\{0\\})$
and then $f_{T}$ the density of $T$ is equal to

$f_{T}(t)=f_{1}(t_{0}+t)+f_{2}(t_{0}+t).$ (8)

Note that to lighten the notation we have not noted the dependence at $z_{0}$
for $A_{1}$, $A_{2}$, $f_{1}$, $f_{2}$ and $f$.

#### Functional equation

Let us define the Laplace transform of the Green’s function for
$(p,q)\in\mathbb{C}^{2}$ by

$L(p,q):=\iint_{\mathbb{R}_{+}^{2}}g_{z_{0}}(x,y)e^{px+qy}\mathrm{d}x\mathrm{d}y=\mathbb{E}\left(\int_{0}^{\infty}e^{(p,q)\cdot
Z_{t}}\mathds{1}_{t<T}\mathrm{d}t\right).$

Remembering that $Z=(X,Y)$ the Laplace transform of the exit densities are
defined by

$L_{1}(q):=\int_{0}^{\infty}f_{1}(y)e^{qy}\mathrm{d}y=\mathbb{E}\left(e^{qY({T_{1}})}\mathds{1}_{T_{1}<T_{2}}\right),\quad
L_{2}(p):=\int_{0}^{\infty}f_{2}(x)e^{px}\mathrm{d}x=\mathbb{E}\left(e^{pX({T_{2}})}\mathds{1}_{T_{2}<T_{1}}\right).$

Once again, we omit to note the dependence at $z_{0}$ for $L$, $L_{1}$ and
$L_{2}$. One may remark that $L(0,0)=\mathbb{E}[T]=\infty$ since
$\mathbb{P}(T=\infty)>0$ and

$L_{1}(0)=\mathbb{P}(T_{1}<T_{2}),\quad L_{2}(0)=\mathbb{P}(T_{2}<T_{1}),\quad
L_{1}(0)+L_{2}(0)=\mathbb{P}(T<\infty).$

######  Proposition 1 (Functional equation).

For $(p,q)\in\mathbb{C}^{2}$ such that $\Re p<0$ and $\Re q<0$ the Laplace
transforms converge and we have

$K(p,q)L(p,q)=L_{1}(q)+L_{2}(p)-e^{px_{0}+qy_{0}}$ (9)

where the kernel $K$ is defined by

$K(p,q):=\frac{1}{2}(p-q)^{2}+(1-\gamma)p+\gamma q.$ (10)

###### Proof.

We denote

$\mathcal{G}:=\frac{1}{2}\left(\frac{\partial}{\partial
x}-\frac{\partial}{\partial y}\right)^{2}+(1-\gamma)\frac{\partial}{\partial
x}+\gamma\frac{\partial}{\partial y}$ (11)

the infinitesimal generator of $Z$. We apply Dynkin’s formula, for $f$ twice
differentiable we obtain

$\mathbb{E}f(Z(t\wedge T))=f(z_{0})+\mathbb{E}\int_{0}^{t\wedge
T}\mathcal{G}f(Z(s))\mathrm{d}s$

which can be rewritten

$\mathbb{E}(f(Z(t\wedge
T_{1}))\mathds{1}_{T_{1}<T_{2}})+\mathbb{E}(f(Z(t\wedge
T_{2}))\mathds{1}_{T_{2}<T_{1}})=f(z_{0})+\mathbb{E}\int_{0}^{t}\mathcal{G}f(Z(s))\mathds{1}_{s<T}\mathrm{d}s.$

By taking $f(x,y)=e^{px+qy}$ we obtain

$\mathbb{E}\left(e^{qY({t\wedge
T_{1}})}\mathds{1}_{T_{1}<T_{2}}\right)+\mathbb{E}\left(e^{pX({t\wedge
T_{2}})}\mathds{1}_{T_{2}<T_{1}}\right)=e^{px_{0}+qy_{0}}+K(p,q)\mathbb{E}\left(\int_{0}^{t}e^{(p,q)\cdot
Z_{s}}\mathds{1}_{s<T}\mathrm{d}s\right).$

The two expectations of the left-hand side are bounded by $1$ since $\Re p<0$
and $\Re q<0$ and then the expectation of the right-hand side is also finite.
By making $t$ tend towards infinity and by definition of $L$, $L_{1}$ and
$L_{2}$ we obtain the functional equation (9). ∎

#### Study of the kernel and analytic continuation

The kernel $K(p,q)$ defined in (10) leads us to introduce four algebraic
functions $P^{+}$, $P^{-}$, $Q^{+}$ and $Q^{-}$ satisfying

$K(P^{\pm}(q),q)=K(p,Q^{\pm}(p))=0,$

analytic on $\mathbb{C}\setminus[(1-\gamma)^{2}/2,\infty)$ for $P^{\pm}$, on
$\mathbb{C}\setminus[\gamma^{2}/2,\infty)$ for $Q^{\pm}$, and defined by

$P^{\pm}(q):=\gamma-1+q\pm\sqrt{(1-\gamma)^{2}-2q}\quad\text{and}\quad
Q^{\pm}(p):=-\gamma+p\pm\sqrt{\gamma^{2}-2p}.$ (12)

Thanks to the functional equation (9) we can now continue meromorphically
$L_{1}$ to the domain $\\{q\in\mathbb{C}:\Re q<(1-\gamma)^{2}/2\\}$ by the
formula

$L_{1}(q)=e^{P^{-}(q)x_{0}+qy_{0}}-L_{2}(P^{-}(q)).$

We just have to verify that $\Re P^{-}(q)<0$ when $\Re q<(1-\gamma)^{2}/2$ and
to remember that $L_{2}$ is analytic in the complex half-plane with negative
real part. In the same way, we extend $L_{2}$ analytically to the domain
$\\{p\in\mathbb{C}:\Re p<\gamma^{2}/2\\}$.

#### PDE and derivative of the Green’s functions

Let $z=(x,y)\in\mathbb{R}^{2}$. Let us recall that we defined in (11) the
infinitesimal generator $\mathcal{G}$. We now define its dual
$\mathcal{G}^{*}$ by

$\mathcal{G}^{*}:=\frac{1}{2}\left(\frac{\partial}{\partial
x}-\frac{\partial}{\partial y}\right)^{2}-(1-\gamma)\frac{\partial}{\partial
x}-\gamma\frac{\partial}{\partial y}.$

The Green’s function $g_{z_{0}}$ satisfies the following classical parabolic
partial differential equation [11, Chap IX §17], denoting $\delta_{z_{0}}$ the
Dirac measure in $z_{0}\in\mathbb{R}^{2}$ we have

$\begin{cases}\mathcal{G}^{*}g_{z_{0}}=-\delta_{z_{0}}&\text{on
}(\mathbb{R}_{+}^{*})^{2},\\\ g_{z_{0}}=0&\text{on
}\mathbb{R}_{+}\times\\{0\\}\cup\\{0\\}\times\mathbb{R}_{+}.\end{cases}$ (13)

######  Proposition 2 (Exit densities seen as derivative of the Green’s
function).

We have the following links between the Green’s density and the exit
densities:

$f_{1}(y)=\frac{1}{2}\frac{\partial g_{z_{0}}}{\partial x}(0,y),\quad
f_{2}(x)=\frac{1}{2}\frac{\partial g_{z_{0}}}{\partial y}(x,0).$

###### Proof.

A weak formulation of the partial differential equation (13) applied to an
exponential test function leads to a functional equation linking the Laplace
transforms of the Green’s function and their derivatives. To do this, it is
enough to integrate the partial differential equation (13) and to make
successive integration by parts on the integral

$\iint_{\mathbb{R}_{+}^{2}}\mathcal{G}^{*}g(z_{0},z)e^{z\cdot(p,q)}\mathrm{d}z=-e^{z_{0}\cdot(p,q)}.$

Then, defining $\widehat{L}_{1}(q):=\frac{1}{2}\int_{0}^{\infty}\frac{\partial
g}{\partial x}(0,y)e^{qy}\mathrm{d}y$ and symmetrically $\widehat{L}_{2}(p)$
we obtain

$K(p,q)L(p,q)=\widehat{L}_{1}(q)+\widehat{L}_{2}(p)-e^{px_{0}+qy_{0}}.$

Comparing this equation with the functional equation (9), we obtain by
identification that $\widehat{L}_{1}=L_{1}$ and $\widehat{L}_{2}=L_{2}$ and
then we can conclude. ∎

######  Remark 3 (Harmonic functions).

A function $u(t,y)$ satisfies (5) if and only if $h(x,y):=u(x+y,y)$ satisfies
$\mathcal{G}h=0$ on $(\mathbb{R}_{+}^{*})^{2}$ and $h=0$ on the boundaries
$\mathbb{R}_{+}\times\\{0\\}\cup\\{0\\}\times\mathbb{R}_{+}$. We say that $h$
(resp. $u$) is harmonic for the killed degenerate Brownian motion $Z$ (resp.
is harmonic for the killed space-time Brownian motion $B$).

##  3 Asymptotics via Malyshev’s analytical approach

###  3.1 Asymptotics of Green’s functions

By using the functional equation (9), inverting the Laplace transform and
applying the steepest descent method, it is possible to compute the
asymptotics of Green’s function.

######  Theorem 4 (Asymptotics of Green’s function and exit densities).

Let $\alpha\in(0,\pi/2)$ and $z=(x,y)\in\mathbb{R}_{+}^{2}$. When $z\to\infty$
and $y/x\to\tan\alpha$ we have

$g_{z_{0}}(z)\sim\frac{h^{\alpha}(z_{0})}{\sqrt{|z|}}\frac{e^{-z\cdot(p(\alpha),q(\alpha))}}{\sqrt{2\pi(\cos\alpha+\sin\alpha)}}$
(14)

where the so-called saddle point $(p(\alpha),q(\alpha))$ is defined by

$\displaystyle(p(\alpha),q(\alpha))$
$\displaystyle:=\mathrm{argmax}\\{p\cos\alpha+q\sin\alpha:(p,q)\in\mathbb{R}^{2},K(p,q)=0\\}$
(15)
$\displaystyle=\left(\frac{\gamma^{2}}{2}-\frac{1}{2(1+\frac{1}{\tan\alpha})^{2}},\frac{(1-\gamma)^{2}}{2}-\frac{1}{2(1+{\tan\alpha})^{2}}\right)$
(16)

and $h^{\alpha}$ is a harmonic function for the process (i..e
$\mathcal{G}h^{\alpha}=0$ on $\mathbb{R}_{+}^{2}$ and $h^{\alpha}=0$ on
$\mathbb{R}\times\\{0\\}\cup\\{0\\}\times\mathbb{R}$) defined for
$\alpha\in(0,\pi/2)$ by

$h^{\alpha}(z_{0}):=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}-L_{1}(q(\alpha))-L_{2}(p(\alpha)).$
(17)

We also have the asymptotics of the exit densities on the boundaries

$f_{1}(y)\underset{y\to\infty}{\sim}\frac{h^{\pi/2}(z_{0})}{y^{3/2}}\frac{e^{-yq(\pi/2)}}{\sqrt{2\pi}}\quad\text{and}\quad
f_{2}(x)\underset{x\to\infty}{\sim}\frac{h^{0}(z_{0})}{x^{3/2}}\frac{e^{-xp(0)}}{\sqrt{2\pi}}$
(18)

where we define

$h^{\pi/2}(z_{0}):=x_{0}e^{x_{0}p(\pi/2)+y_{0}q(\pi/2)}-L_{2}^{\prime}(p(\pi/2))\quad\text{and}\quad
h^{0}(z_{0}):=y_{0}e^{x_{0}p(0)+y_{0}q(0)}-L_{1}^{\prime}(q(0)).$

The proof is done below in Section 3.2.

######  Remark 5 ($h^{0}$ and $h^{\pi/2}$ as derivatives).

It should be noted that for $\alpha>0$ we have $h^{\alpha}\to 0$ when
$\alpha\to 0$ (it derives from Proposition 22 below). That is why we have
defined $h^{0}$ differently. We can interpret $h^{0}$ as the $q$-derivative of
$h^{\alpha}$ when $\alpha>0$ evaluated in $q(0)$, i.e.
$h^{0}=\left.\frac{\partial h^{\alpha}}{\partial q}\right|_{q(0)}$. A similar
remark holds for $h^{\pi/2}$.

Figure 2: The parabola
$\mathcal{P}:=\\{(p,q)\in\mathbb{R}_{+}^{2}:K(p,q)=0\\}$ in blue and the arc
of parabola
$\overline{\mathcal{A}}:=\\{(p(\alpha),q(\alpha)):\alpha\in[0,\pi/2]\\}\subset\mathcal{P}$
which parametrizes the Martin’s boundary in red

###  3.2 Proof of Theorem 4

The proof of Theorem 4 follows the same steps as in [13] and [18]. We will
first show asymptotics for the Green functions $g_{z_{0}}$ (14) and then
asymptotics for the exit densities $f_{1}$ and $f_{2}$ (18).

#### From double to simple integral via the residue theorem

The Laplace inversion formula, see [10, Theorem 24.3 and 24.4] and [5],
implies that for $\epsilon>0$ we have

$g_{z_{0}}(z)=\frac{1}{(2i\pi)^{2}}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}L(p,q)e^{-px-
qy}\mathrm{d}p\mathrm{d}q.$

We are now going to use the functional equation (9) and the residue theorem to
transform the double integral into a sum of simple integrals.

######  Lemma 6 (Reduction to simple integrals).

The Green’s function satisfies $g_{z_{0}}(z)=I_{1}+I_{2}+I_{3}$ where

$\begin{cases}I_{1}:=\frac{-1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{1}(q)}{\sqrt{(1-\gamma)^{2}-2q}}e^{-P^{+}(q)x-qy}\mathrm{d}q,\\\
I_{2}:=\frac{-1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{2}(p)}{\sqrt{\gamma^{2}-2p}}e^{-px-Q^{+}(p)y}\mathrm{d}p,\\\
I_{3}:=\frac{1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{e^{px_{0}+Q^{+}(p)y_{0}}}{\sqrt{\gamma^{2}-2p}}e^{-px-Q^{+}(p)y}\mathrm{d}p.\end{cases}$
(19)

###### Proof.

The following equalities are explained below.

$\displaystyle g(z_{0},z)$
$\displaystyle=\frac{1}{(2i\pi)^{2}}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{1}(q)+L_{2}(p)-e^{px_{0}+qy_{0}}}{K(p,q)}e^{-px-
qy}\mathrm{d}p\mathrm{d}q$
$\displaystyle=\frac{1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{1}(q)}{\frac{1}{2}(P^{+}(q)-P^{-}(q))}e^{-P^{+}(q)x-qy}\mathrm{d}q+\frac{1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{2}(p)}{\frac{1}{2}(Q^{+}(p)-Q^{-}(p))}e^{-px-Q^{+}(p)y}\mathrm{d}p$
$\displaystyle-\frac{1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{-e^{px_{0}+Q^{+}(p)y_{0}}}{\frac{1}{2}(Q^{+}(p)-Q^{-}(p))}e^{-px-Q^{+}(p)y}\mathrm{d}p$
$\displaystyle=I_{1}+I_{2}+I_{3}$

The first equality above comes from the functional equation (9). The second
equality above comes from the residue theorem applied to an integration
contour $\mathcal{C}_{R}$ represented in Figure 3 to obtain $I_{1}$ (and a
similar contour to obtain $I_{2}$ and $I_{3}$). In a classic way, the residue
theorem implies that
$\frac{1}{2i\pi}\int_{\mathcal{C}_{R}}\frac{L_{1}(q)}{K(p,q)}e^{-px-
qy}\mathrm{d}p=\frac{L_{1}(q)}{\frac{1}{2}(P^{+}(q)-P^{-}(q))}e^{-P^{+}(q)x-qy}$
since $P^{+}(q)$ is the only pole of the integrand inside $\mathcal{C}_{R}$.
Furthermore, the integral on the red half circle and the blue segments tends
to $0$ when $R\to\infty$ and we obtain that
$-\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{1}(q)}{K(p,q)}e^{-px-
qy}\mathrm{d}p=\lim_{R\to\infty}\int_{\mathcal{C}_{R}}\frac{L_{1}(q)}{K(p,q)}e^{-px-
qy}\mathrm{d}p$ taking into account the orientation of the contour. Details
are omitted and one can refer to [18, Lemma 4.1] or [13, Proposition 5.] which
perform similar calculations. The third equality just comes from the
definition of $P^{\pm}$ and $Q^{\pm}$ in (12).

Figure 3: Integration contour $\mathcal{C}_{R}$ and $Q^{+}(p)$ the only pole
of the integrand inside $\mathcal{C}_{R}$

∎

#### Steepest descent method

We are going to denote by $(r,\alpha)\in\mathbb{R}_{+}\times[0,\pi/2]$ the
polar coordinate of the point $z=(x,y)\in\mathbb{R}_{+}^{2}$, i.e.
$(x,y)=(r\cos\alpha,r\sin\alpha)$ and
$(r,\alpha)=(\sqrt{x^{2}+y^{2}},\arctan(y/x))$. Let us define the functions

$F(p,\alpha):=p\cos\alpha+Q^{+}(p)\sin\alpha\quad\text{and}\quad
G(q,\alpha):=P^{+}(q)\cos\alpha+q\sin\alpha.$

The point $(p(\alpha),q(\alpha))$ defined in (15) is a saddle point for these
functions. Indeed, for $\alpha\in(0,\pi/2)$ the first derivatives (according
to $p$ for $F$ and $q$ for $G$) satisfy

$F^{\prime}_{p}(p(\alpha),\alpha)=G^{\prime}_{q}(q(\alpha),\alpha)=0,$

and the second derivatives satisfy

$F^{\prime\prime}_{p}(p(\alpha),\alpha)=-\frac{(\cos\alpha+\sin\alpha)^{3}}{\sin^{2}\alpha}<0,\quad
G^{\prime\prime}_{q}(q(\alpha),\alpha)=-\frac{(\cos\alpha+\sin\alpha)^{3}}{\cos^{2}\alpha}<0.$
(20)

######  Lemma 7 (Asymptotics of $I_{1}$).

When $r\to\infty$ and $\alpha\to\alpha_{0}$ the following asymptotics holds:

$I_{1}=-L_{1}(q(\alpha_{0}))\frac{e^{-z\cdot(p(\alpha_{0}),q(\alpha_{0}))}}{\sqrt{2\pi
r(\cos\alpha_{0}+\sin\alpha_{0})}}+o\left(\frac{1}{\sqrt{r}}e^{-z\cdot(p(\alpha_{0}),q(\alpha_{0}))}\right).$

Similar asymptotics hold for $I_{2}$ and $I_{3}$.

###### Proof.

The integral $I_{1}$ is typical to apply the classical saddle point method. A
famous reference is Fedoryuk’s book [15, §4]. Starting from (19) we then
obtain

$\displaystyle I_{1}$
$\displaystyle=\frac{-1}{2i\pi}\int_{-\epsilon-i\infty}^{-\epsilon+i\infty}\frac{L_{1}(q)}{\sqrt{(1-\gamma)^{2}-2q}}e^{-rG(q,\alpha)}\mathrm{d}q$
$\displaystyle\underset{r\to\infty\atop\alpha\to\alpha_{0}}{=}\frac{-1}{2i\pi}\frac{L_{1}(q(\alpha_{0}))}{\sqrt{(1-\gamma)^{2}-2q(\alpha_{0})}}i\sqrt{\frac{-2\pi}{rG^{\prime\prime}_{q}(q(\alpha_{0}))}}e^{-rG(q(\alpha_{0}),\alpha_{0})}+o\left(\frac{1}{\sqrt{r}}e^{-rG(q(\alpha_{0}),\alpha_{0})}\right).$

Let us give some details about the application of this method. To obtain the
asymptotics we shift the integration contour onto the steepest descent line
which passes through the saddle point $q(\alpha)$. This is possible since the
integrand is a meromorphic function and we do not cross the only pole since
$q(\alpha)<(1-\gamma)^{2}/2$ for $\alpha\in(0,\pi/2)$, see (15). Then, the
asymptotics derives from the classical Laplace’s method, see for example [13,
Lemma 16]. To handle the convergence in $\alpha\to\alpha_{0}$ we just apply a
simple parameter-dependent Morse lemma, see [18, Appendix A] which details
this technique. To conclude, a straightforward calculation using (16) and (20)
shows that

$\frac{1}{2\pi}\frac{1}{\sqrt{(1-\gamma)^{2}-2q(\alpha)}}\sqrt{\frac{-2\pi}{G^{\prime\prime}_{q}(q(\alpha))}}=\frac{1}{\sqrt{2\pi(\cos\alpha+\sin\alpha)}}$

and it just remains to remark that
$rG(q(\alpha),\alpha)=z\cdot(p(\alpha),q(\alpha))$. ∎

###### Proof of (14).

The main asymptotics (14) of Theorem 4 directly derives from Lemmas 19 and 7.
One will compute more explicitly
$h^{\alpha}(z_{0})=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}-L_{1}(q(\alpha))-L_{2}(p(\alpha))$
in the following section which will show that this function is non-zero for
$\alpha\in(0,\pi/2)$. ∎

#### Asymptotics on the boundaries via Tauberian lemmas

######  Lemma 8 (Branching point behaviour).

When $q\to q^{+}:=q(\pi/2)=(1-\gamma)^{2}/2$ we have

$L_{1}(q)=L_{1}(q^{+})-(x_{0}e^{P^{-}(q^{+})x_{0}+q^{+}y_{0}}-L_{2}^{\prime}(P^{-}(q^{+})))\sqrt{2(q^{+}-q)}+o(\sqrt{q^{+}-q})$

and a symmetrical result holds for $L_{2}$.

###### Proof.

Evaluating (9) at $(P^{-}(q),q)$ and making a Taylor series expansion when
$q\to q^{+}$ we obtain

$\displaystyle L_{1}(q)$
$\displaystyle=e^{P^{-}(q)x_{0}+qy_{0}}-L_{2}(P^{-}(q))$
$\displaystyle=L_{1}(q^{+})+(x_{0}e^{P^{-}(q^{+})x_{0}+q^{+}y_{0}}-L_{2}^{\prime}(P^{-}(q^{+})))(P^{-}(q))-P^{-}(q^{+}))+o(P^{-}(q))-P^{-}(q^{+}))$

and it just remains to notice that
$P^{-}(q))-P^{-}(q^{+})=-\sqrt{2(q^{+}-q)}$. ∎

###### Proof of (18).

Lemma 8 and classical Tauberian inversion lemmas [10, Theorem 37.1], [6, Lemma
C.2], imply that

$f_{1}(y)\underset{y\to\infty}{\sim}(x_{0}e^{P^{-}(q^{+})x_{0}+q^{+}y_{0}}-L_{2}^{\prime}(P^{-}(q^{+})))\frac{-\sqrt{2}}{\Gamma(-1/2)}y^{-3/2}e^{-q^{+}y}$

and we deduce the asymptotics (18) of Theorem 4 noticing that
$\Gamma(-1/2)=-2\sqrt{\pi}$. ∎

##  4 Martin boundary and persistence probabilities

###  4.1 Doob’s $h$-transform and persistence

In this short section, we give a probabilist interpretation of the harmonic
functions $h^{\alpha}$, found in Theorem 4, in term of the persistence
probabilities of the process conditioned to drift in a given direction.

For $\alpha\in[0,\pi/2]$ we consider the probability $\mathbb{P}^{\alpha}$
defined by the Doob’s $h$-transform for the harmonic function
$e^{xp(\alpha)+yq(\alpha)}$ associated to the process
$Z(t)=z_{0}+\mathbf{W}(t)+\mu t$. We have

$\mathbb{P}^{\alpha}(Z(t)\in
S)=\mathbb{E}(e^{(p(\alpha),q(\alpha))\cdot(\mathbf{W}(t)+\mu
t)}\mathds{1}_{Z(t)\in S})$

for a set $S\subset\mathbb{R}^{2}$. Under $\mathbb{P}^{\alpha}$ the process
$Z(t)$ is a Markov process of transition kernel

$\widetilde{P}_{t}(z_{0},z)=e^{(x-x_{0})p(\alpha)+(y-y_{0})q(\alpha)}P_{t}(z_{0},z)$

where $P_{t}$ is the transition kernel of $Z(t)$ under $\mathbb{P}$. Under
$\mathbb{P}^{\alpha}$ the process $Z(t)$ is conditioned to drift in the
direction $\alpha$ (in the sense of the Doob’s $h$-transform).

We introduce the corresponding modified Green’s measure, for
$S\subset\mathbb{R}_{+}^{2}$ we set

$G^{\alpha}_{z_{0}}(S):=\int_{0}^{\infty}\mathbb{P}^{\alpha}(Z(t)\in
S,t<T)\mathrm{d}t$

and its density

$g^{\alpha}_{z_{0}}(z):=e^{(x-x_{0})p(\alpha)+(y-y_{0})q(\alpha)}g_{z_{0}}(z).$

We define $L^{\alpha}$ the Laplace transform relative to these modified
Green’s functions and we have

$L^{\alpha}(p,q)=L(p+p(\alpha),q+q(\alpha))e^{-x_{0}p(\alpha)-y_{0}q(\alpha)}.$

On the same way, we introduce the modified exit measures

$A_{1}^{\alpha}(S):=\mathbb{P}^{\alpha}(Z(T_{1})\in
S,T_{1}<T_{2})\quad\text{and}\quad
A_{2}^{\alpha}(S):=\mathbb{P}^{\alpha}(Z(T_{2})\in S,T_{2}<T_{1})$

and their densities

$f_{1}^{\alpha}(y):=e^{-x_{0}p(\alpha)+(y-y_{0})q(\alpha)}f_{1}(y)\quad\text{and}\quad
f_{2}^{\alpha}(x):=e^{(x-x_{0})p(\alpha)-y_{0}q(\alpha)}f_{2}(x).$

We denote their Laplace transforms $L_{1}^{\alpha}$ and $L_{2}^{\alpha}$ and
we have

$L_{1}^{\alpha}(q)=L_{1}(q+q(\alpha))e^{-x_{0}p(\alpha)-y_{0}q(\alpha)},\quad
L_{2}^{\alpha}(p)=L_{2}(p+p(\alpha))e^{-x_{0}p(\alpha)-y_{0}q(\alpha)}.$

One may remark that

$L_{1}^{\alpha}(0)=\mathbb{P}^{\alpha}(T_{1}<T_{2}),\quad
L_{2}^{\alpha}(0)=\mathbb{P}^{\alpha}(T_{2}<T_{1}),\quad
L_{1}^{\alpha}(0)+L_{2}^{\alpha}(0)=\mathbb{P}^{\alpha}(T<\infty).$

######  Proposition 9 (Conditioned persistence probabilities).

Let $z_{0}=(x_{0},y_{0})$, for $\alpha\in(0,\pi)$ we have

$h^{\alpha}(z_{0})=e^{x_{0}p(\alpha)+y_{0}q(\alpha)}\mathbb{P}^{\alpha}(T=\infty)=e^{x_{0}p(\alpha)+y_{0}q(\alpha)}-\mathbb{E}(e^{X_{T}p(\alpha)+Y_{T}q(\alpha)}\mathds{1}_{T<\infty})$
(21)

and

$h^{0}(z_{0})=y_{0}e^{x_{0}p(0)+y_{0}q(0)}-\mathbb{E}\left(Y_{T_{1}}e^{q(0)Y_{T_{1}}}\mathds{1}_{T_{1}<T_{2}}\right),\quad
h^{\pi/2}(z_{0})=x_{0}e^{x_{0}p(\pi/2)+y_{0}q(\pi/2)}-\mathbb{E}\left(X_{T_{2}}e^{p(\pi/2)X_{T_{2}}}\mathds{1}_{T_{2}<T_{1}}\right).$

When $x_{0}\to\infty$ and $y_{0}\to\infty$ we obtain

$h^{\alpha}(z_{0})\sim e^{x_{0}p(\alpha)+y_{0}q(\alpha)},\quad
h^{0}(z_{0})\sim y_{0}e^{x_{0}p(0)+y_{0}q(0)}\quad\text{and}\quad
h^{\pi/2}(z_{0})\sim x_{0}e^{x_{0}p(\pi/2)}.$ (22)

###### Proof.

By Theorem 4, we have

$\displaystyle h^{\alpha}(z_{0})$
$\displaystyle=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}-L_{1}(q(\alpha))-L_{2}(p(\alpha))$
$\displaystyle=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}(1-L_{1}^{\alpha}(0)-L_{2}^{\alpha}(0))$
$\displaystyle=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}(1-\mathbb{P}^{\alpha}(T_{1}<T_{2})-\mathbb{P}^{\alpha}(T_{2}<T_{1}))$
$\displaystyle=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}\mathbb{P}^{\alpha}(T=\infty)$
$\displaystyle=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}-e^{p(\alpha)x_{0}+q(\alpha)y_{0}}\mathbb{P}^{\alpha}(T<\infty)$
$\displaystyle=e^{x_{0}p(\alpha)+y_{0}q(\alpha)}-\mathbb{E}(e^{X_{T}p(\alpha)+Y_{T}q(\alpha)}\mathds{1}_{T<\infty}),$

and also

$\displaystyle h^{0}(z_{0})$
$\displaystyle=y_{0}e^{x_{0}p(0)+y_{0}q(0)}-L_{1}^{\prime}(q(0))$
$\displaystyle=y_{0}e^{x_{0}p(0)+y_{0}q(0)}-\mathbb{E}\left(Y_{T_{1}}e^{q(0)Y_{T_{1}}}\mathds{1}_{T_{1}<T_{2}}\right).$

We now assume that $x_{0}\to\infty$ and $y_{0}\to\infty$. Noticing that
$\mathbb{P}^{\alpha}(T=\infty){\to}1$ we obtain the asymptotics for
$h^{\alpha}$ when $\alpha\in(0,\pi/2)$. Remembering that $p(0)>0$, $q(0)<0$
and that $Y_{T_{1}}>y_{0}$, we see that
$\mathbb{E}\left(Y_{T_{1}}e^{q(0)Y_{T_{1}}}\mathds{1}_{T_{1}<T_{2}}\right)=o(y_{0}e^{x_{0}p(0)+y_{0}q(0)})$
and the asymptotics of $h^{0}$ follows. A symmetrical proof holds for
$h^{\pi/2}$. ∎

These asymptotics will be useful in the proof of Theorem 13 to show that the
Martin boundary is minimal.

###  4.2 Martin boundary

We refer to the classic book by Doob [11] which gives a comprehensive
presentation of Martin boundary theory.

######  Theorem 10 (Martin boundary).

For $\alpha\in[0,\pi/2]$, the function $z_{0}\mapsto h^{\alpha}(z_{0})$ is
harmonic for the process $Z$, i.e. $\mathcal{G}h^{\alpha}=0$ inside the
quarter plane and $h^{\alpha}=0$ on its boundaries. The Martin boundary of the
process $Z$ killed outside of the quarter plane is denoted
$\partial_{M}^{\mathbb{R}_{+}^{2}}Z$ and is homeomorphic to $[0,\pi/2]$
through the map $\Phi$ given by

$\begin{array}[]{ccccc}\Phi:[0,\frac{\pi}{2}]&\longrightarrow&\partial_{M}^{\mathbb{R}_{+}^{2}}Z\\\
\alpha&\longmapsto&\displaystyle\frac{h^{\alpha}(\cdot)}{h^{\alpha}(1,1)}.\\\
\end{array}$

###### Proof.

We define the Martin’s kernel for the arbitrary point
$(1,1)\in\left(\mathbb{R}_{+}^{*}\right)^{2}$ by

$k(z_{0},z):=\frac{g_{z_{0}}(z)}{g_{(1,1)}(z)}.$

By the Martin’s boundary theory in a parabolic context [11, Chap XIX] we have
to show that for $\alpha\in[0,\pi/2]$, $z=(x,y)$, $z\to\infty$ and
$y/x\to\tan\alpha$ we have

$k(z_{0},z)\rightarrow
k^{\alpha}(z_{0}):=\frac{h^{\alpha}(z_{0})}{h^{\alpha}(1,1)}.$

For $\alpha\in(0,\pi/2)$ it directly derives from the asymptotics of the
Green’s function $g_{z_{0}}$ (14) found in Theorem 4. For $\alpha=0$ or
$\pi/2$ it is enough to use formula (18) of Theorem 4 and Proposition 2 and to
apply l’Hôpital’s rule. The continuity of $\Phi$ is clear on $(0,\pi/2)$. The
continuity at $0$ derives from Remark 5, noticing that $p^{\prime}(0)=0$, with
a simple Taylor approximation we obtain that $h^{\alpha}=\alpha
q^{\prime}(0)h^{0}+o(\alpha)$ when $\alpha\to 0$ and then $k^{\alpha}\to
k^{0}$ when $\alpha\to 0$. The same reasoning holds for the continuity at
$\pi/2$. The injectivity of $\Phi$ follows from the asymptotics of
$h^{\alpha}$ when $x_{0}\to\infty$ and $y_{0}\to\infty$ which are all
different for different values of $\alpha$, see (22) of Proposition 22. Since
$[0,\pi/2]$ is compact and $\Phi$ is a continuous bijection, $\Phi$ is an
homeomorphism. ∎

In Theorem 13, we will deduce from the asymptotics (22) that the Martin
boundary is minimal.

##  5 Harmonic functions via the compensation approach

###  5.1 Minimal harmonic functions

The harmonic function $h^{\alpha}(z_{0})$ defined in (17) has been expressed
in terms of the Laplace transforms $L_{1}$ and $L_{2}$ and then interpreted in
(21) as a persistence probability of a conditioned process. But this is not a
very tractable expression since these Laplace transforms and these
probabilities are not known explicitly. In this section, we compute explicitly
all the harmonic functions $h^{\alpha}(z_{0})$ thanks to a recursive
compensation approach.

######  Proposition 11 (Harmonic functions via compensation approach,
$\alpha\in(0,\pi/2)$).

For $z=(x,y)\in\mathbb{R}_{+}^{2}$, we define the functions
$\widetilde{h}^{(p_{0},q_{0})}$ by

$\widetilde{h}^{(p_{0},q_{0})}(x,y):=\sum_{n\in\mathbb{Z}}(-1)^{n}e^{xp_{n}+yq_{n}}$
(23)

where $(p_{n},q_{n})$ is a recursive sequence of points on the parabola
$\mathcal{P}:=\\{(p,q)\in\mathbb{R}_{+}^{2}:K(p,q)=0\\}$ starting from
$(p_{0},q_{0})\in\mathcal{A}:=\\{(p(\alpha),q(\alpha)):\alpha\in(0,\pi/2)\\}$
the red arc of the parabola, see Figures 2 and 4, and defined by

$\begin{cases}p_{2n}:=p_{0}+2n(p_{0}-q_{0})-2n(n+\gamma)\\\
q_{2n}:=q_{0}+2n(p_{0}-q_{0})-2n(n+\gamma-1)\end{cases}\quad\text{and}\quad(p_{2n+1},q_{2n+1})=(p_{2n},q_{2n+2}).$
(24)

Then, the functions $\widetilde{h}^{(p_{0},q_{0})}$ are positive and harmonic
for the process $Z$ killed at the boundary of the quadrant, that is

$\begin{cases}\mathcal{G}\widetilde{h}^{(p_{0},q_{0})}=0&\text{on
}(\mathbb{R}_{+}^{*})^{2}\\\ \widetilde{h}^{(p_{0},q_{0})}=0&\text{on
}\mathbb{R}\times\\{0\\}\cup\\{0\\}\times\mathbb{R}.\end{cases}$ (25)

###### Proof.

This sum is typical of the compensation approach. First, we may notice that
the convergence of the sum is trivial since the sequence $(p_{n},q_{n})$ tends
to minus infinity quadratically. Let us notice that a function
$f(z)=e^{z\cdot(p,q)}$ satisfies $\mathcal{G}f=0$ if and only if $K(p,q)=0$,
i.e. $(p,q)\in\mathcal{P}$. It is easy to verify that for all $n\in\mathbb{Z}$
we have $(p_{n},q_{n})\in\mathcal{P}$, see Figure 4. Therefore, we deduce that
$\mathcal{G}\widetilde{h}^{(p_{0},q_{0})}=0$ for
$z=(x,y)\in\mathbb{R}_{+}^{2}$. Furthermore, using
$(p_{2n+1},q_{2n+1})=(p_{2n},q_{2n+2})$ we have the following recursive
compensations

$\widetilde{h}^{(p_{0},q_{0})}(x,y)=\overunderbraces{&\br{4}{=0\text{ when
}x=0}\br{4}{=0\text{ when
}x=0}}{\cdots+&e^{p_{-2}x+q_{-2}y}-e^{p_{-1}x+q_{-1}y}+e^{p_{0}x+q_{0}y}-e^{p_{1}x+q_{1}y}+e^{p_{2}x+q_{2}y}+\cdots}{&\br{3}{=0\text{
when }y=0}\br{3}{=0\text{ when }y=0}}$

and we deduce that $\widetilde{h}^{(p_{0},q_{0})}$ satisfies the boundary
conditions of (25).

It only remains to show the positivity. First, there exists $t>0$ large enough
such that when $x+y>t$ we have $\widetilde{h}^{(p_{0},q_{0})}(z)>0$, indeed

$\displaystyle\widetilde{h}^{(p_{0},q_{0})}(x,y)=\overbrace{e^{p_{0}x+q_{0}y}-e^{p_{1}x+q_{1}y}-e^{p_{-1}x+q_{-1}y}}^{\geqslant
0\text{ when }x+y\text{ large enough}}$
$\displaystyle+\sum_{\mathbb{N}^{*}}\overbrace{e^{p_{2n}x+q_{2n}y}-e^{p_{2n+1}x+q_{2n+1}y}}^{\geqslant
0\text{ always}}$
$\displaystyle+\sum_{\mathbb{N}^{*}}\overbrace{e^{p_{-2n}x+q_{-2n}y}-e^{p_{-2n-1}x+q_{-2n-1}y}.}^{\geqslant
0\text{ always}}$

For all starting point $z_{0}=(x_{0},y_{0})\in\mathbb{R}_{+}^{2}$ of the
process $Z=(X,Y)$, remembering that $X(t)+Y(t)=t_{0}+t$ and since
$\widetilde{h}^{(p_{0},q_{0})}$ is harmonic, we deduce that

$\widetilde{h}^{(p_{0},q_{0})}(x_{0},y_{0})=\mathbb{E}[\widetilde{h}^{(p_{0},q_{0})}(Z(t))]\geqslant
0.$

∎

Figure 4: Sequel of the points $(p_{n},q_{n})$ on the parabola
$\mathcal{P}:=\\{(p,q)\in\mathbb{R}^{2}:K(p,q)=0\\}$. On the left
$(p_{0},q_{0})\in\mathcal{A}:=\\{(p(\alpha),q(\alpha)):\alpha\in(0,\pi/2)\\}$,
on the right, $(p_{0},q_{0})=(p(0),q(0))\in\partial\mathcal{A}$.

It remains to define the two harmonic functions associated with the extremal
points of the arc
$\partial{\mathcal{A}}=\\{(p(0),q(0)),(p(\pi/2),q(\pi/2))\\}$. Indeed, when
$(p_{0},q_{0})\to(p(0),q(0))$ or $(p(\pi/2),q(\pi/2))$ we have
$\widetilde{h}^{(p_{0},q_{0})}\to 0$. That is why we have to define
$\widetilde{h}^{(p(0),q(0))}$ and $\widetilde{h}^{(p(\pi/2),q(\pi/2))}$
differently.

######  Proposition 12 (Harmonic functions via compensation approach,
$\alpha=0$ or $\pi/2$).

For $z=(x,y)\in\mathbb{R}_{+}^{2}$, we define the function
$\widetilde{h}^{(p(0),q(0))}$ by

$\widetilde{h}^{(p(0),q(0))}(x,y):=\sum_{n\in\mathbb{Z}}(-2nx+(1-2n)y)e^{xp_{2n}+yq_{2n}}$
(26)

where $(p_{2n},q_{2n})$ is a recursive sequence of points on the parabola
$\mathcal{P}$ starting from
$(p_{0},q_{0})=(p(0),q(0))=(\gamma^{2}/2,-\gamma+\gamma^{2}/2)\in\partial\mathcal{A}$,
and defined by (24), which means in this case

$\begin{cases}p_{2n}:=\gamma^{2}/2-2n^{2},\\\
q_{2n}:=\gamma^{2}/2-\gamma-2n(n-1).\end{cases}$ (27)

Then, the function $\widetilde{h}^{(p_{0},q_{0})}$ is positive and harmonic
for the process $Z$ killed at the boundary of the quadrant, that is satisfying
(25). A symmetric expression holds for $\widetilde{h}^{(p(\pi/2),q(\pi/2))}$.

###### Proof.

First, we again notice that the convergence of the sum is trivial since the
sequence $(p_{2n},q_{2n})$ tends to minus infinity quadratically. Here we
cannot compensate in the same way that in the previous proposition because we
would obtain the null function since $(p_{2n},q_{2n})=(p_{-2n+1},q_{-2n+1})$.
Therefore, we compensate using linear polynomials in front of the exponential
terms. With a straightforward computation we notice that a function
$f(x,y)=(\lambda_{1}x+\lambda_{2}y)e^{xp+yq}$ satisfies $\mathcal{G}f=0$ if
and only if $K(p,q)=0$ and
$\lambda_{1}((1-\gamma)+p-q)+\lambda_{2}(\gamma+q-p)=0$. It is easy to verify
that $K(p_{2n},q_{2n})=0$ and
$-2n((1-\gamma)+p_{2n}-q_{2n})+(1-2n)(\gamma+q_{2n}-p_{2n})=0$. Therefore we
deduce that $\mathcal{G}\widetilde{h}^{(p_{0},q_{0})}=0$ for
$z=(x,y)\in\mathbb{R}_{+}^{2}$. Furthermore, using the fact that
$(p_{2n},q_{2n})=(p_{-2n},q_{-2n+2})$ we have the following recursive
compensations, we obtain that $\widetilde{h}^{(p(0),q(0))}(x,y)$ is equal to

$\overunderbraces{&\br{1}{=0\text{ when }y=0}&\br{4}{=0\text{ when
}y=0}&\br{4}{=0\text{ when
}y=0}}{&ye^{p_{0}x+q_{0}y}&-(2x+y)&e^{p_{2}x+q_{2}y}&+&(2x+3y)e^{p_{-2}x-q_{-2}y}&-(4x+3y)&e^{p_{4}x+q_{4}y}&+(4x+5y)&e^{p_{-4}x+q_{-4}y}&+\cdots}{&\br{3}{=0\text{
when }x=0}&&\br{3}{=0\text{ when }x=0}}$

and we deduce that $\widetilde{h}^{(p(0),q(0))}$ satisfies the boundary
conditions of (25). Positivity is proved in the same way as in the proof of
Proposition 25. It is easy to see that the function
$\widetilde{h}^{(p(0),q(0))}$ is positive for $x+y$ large enough and we
conclude in the same way thanks to the harmonic character of the function. ∎

The following proposition shows that the compensation approach actually allows
to construct all positive minimal harmonic functions. This implies a
correspondence between minimal positive harmonic functions and the arc of
parabola $\overline{\mathcal{A}}$.

######  Theorem 13 (Minimal harmonic functions).

For all $\alpha\in[0,\pi/2]$ we have

$\widetilde{h}^{(p(\alpha),q(\alpha))}=h^{\alpha}$ (28)

where $(p(\alpha),q(\alpha))$ is the saddle point defined in (15). The Martin
boundary given in Theorem 10 is minimal. It is homeomorphic to the arc of
parabola $\overline{\mathcal{A}}$ through the map

$\begin{array}[]{ccccc}\overline{\mathcal{A}}&\longrightarrow&\partial_{M}^{\mathbb{R}_{+}^{2}}Z\\\
(p(\alpha),q(\alpha))&\longmapsto&\displaystyle\frac{\widetilde{h}^{(p(\alpha),q(\alpha))}(z_{0})}{\widetilde{h}^{(p(\alpha),q(\alpha))}(1,1)}.\\\
\end{array}$

See Figures 2 and 4.

###### Proof.

Recall that the Martin boundary $\partial_{M}^{\mathbb{R}_{+}^{2}}Z$ is
determined in Theorem 10 and is homeomorphic to $[0,\pi/2]$. In Propositions
25 and 12 we have seen that the functions
$\widetilde{h}^{(p(\alpha),q(\alpha))}$ are harmonic for $\alpha\in[0,\pi/2]$.
By Martin boundary theory [11], we have

$\widetilde{h}^{(p(\alpha),q(\alpha))}(z_{0})=\int_{0}^{\pi/2}\frac{h^{\beta}(z_{0})}{h^{\beta}(1,1)}\mathrm{d}m_{\alpha}(\beta)$
(29)

for some measure $m_{\alpha}$ on $[0,\pi/2]$. Let $z_{0}=(x_{0},y_{0})$ and
$\alpha\in(0,\pi/2)$, when $x_{0}\to\infty$ and $y_{0}\to\infty$ we have
$\widetilde{h}^{(p(\alpha),q(\alpha))}(z_{0})\sim
e^{x_{0}p(\alpha)+y_{0}q(\alpha)}$ by (23) and $h^{\beta}(z_{0})\sim
e^{x_{0}p(\beta)+y_{0}q(\beta)}$ by Proposition 22. For $\alpha=0$, we have
$\widetilde{h}^{(p(0),q(0))}(z_{0})\sim y_{0}e^{x_{0}p(0)+y_{0}q(0)}$ by (26)
and $h^{\beta}(z_{0})\sim y_{0}e^{x_{0}p(\beta)+y_{0}q(\beta)}$ and a
symetrical result holds for $\alpha=\pi/2$. Then, from these asymptotics and
(29) we can deduce that $m_{\alpha}=h^{\alpha}(1,1)\delta_{\alpha}$ (where
$\delta_{\alpha}$ is the dirac measure at $\alpha$) and therefore
$\widetilde{h}^{(p(\alpha),q(\alpha))}=h^{\alpha}$. Indeed, it is not possible
to create an exponential asymptotic of a certain parameter by averaging
exponential asymptotics of other parameters. We deduce that these harmonic
functions are minimal since there is a unique measure on the Martin boundary
which satisfies (29). The Martin boundary is then equal to the minimal Martin
boundary and is homeomorphic to the arc of parabola $\overline{\mathcal{A}}$
through the composition between the map $\Phi$ given in Theorem 10 and the
homeomorphism
$[0,\pi/2]\to\overline{\mathcal{A}}:\alpha\mapsto(p(\alpha),q(\alpha))$. ∎

The same kind of phenomenon appears for non-singular random walks [27] and
singular one [24].

###  5.2 Persistence and exit probabilities

From (21), (23) and (28) we find explicitly the persistence probability of the
conditioned process:
$\mathbb{P}^{\alpha}(T=\infty)=e^{-x_{0}p(\alpha)-y_{0}q(\alpha)}h^{\alpha}(z_{0})$.
The next remark focuses on the non-conditioned process.

######  Remark 14 (Persistence probability).

Let $\alpha_{\gamma}\in(0,\pi/2)$ such that
$\tan\alpha_{\gamma}=\gamma/(1-\gamma)$, we have
$(p(\alpha_{\gamma}),q(\alpha_{\gamma}))=(0,0)$ and then
$\mathbb{P}=\mathbb{P}^{\alpha_{\gamma}}$. Then, the probability that the
process stays in the cone forever is equal to

$\mathbb{P}_{(t_{0},y_{0})}(\forall t>0:X_{t}\in
C)=\mathbb{P}(T=\infty)=\widetilde{h}^{(0,0)}(z_{0})=h^{\alpha_{\gamma}}(z_{0}).$

For $z_{0}=(x_{0},y_{0})=(t_{0}-y_{0},y_{0})$ and remembering equation (24) we
obtain

$\displaystyle\mathbb{P}(T=\infty)$
$\displaystyle=\sum_{n\in\mathbb{Z}}(-1)^{n}e^{p_{n}x_{0}+q_{n}y_{0}}=\sum_{n\in\mathbb{Z}}e^{p_{2n}x_{0}+q_{2n}y_{0}}-\sum_{n\in\mathbb{Z}}e^{\overbrace{p_{2n+1}}^{p_{2n}}x_{0}+\overbrace{q_{2n+1}}^{q_{2n+2}}y_{0}}$
$\displaystyle=\sum_{n\in\mathbb{Z}}2\sinh\left(\frac{q_{2n}-q_{2n+2}}{2}y_{0}\right)e^{p_{2n}x_{0}+\frac{q_{2n}+q_{2n+2}}{2}y_{0}}$
$\displaystyle=2\sum_{n\in\mathbb{Z}}\sinh((2n+\gamma)y_{0})e^{-2n(n+\gamma)t_{0}-\gamma
y_{0}}.$

We can also express this quantity in another way to match Defosseux’s result
[8, Proposition 2.3]. By noting that
$p_{-2n}x_{0}+q_{-2n}y_{0}=-2n^{2}t_{0}+2n\gamma t_{0}-2ny_{0}$ and
${p_{2n}}x_{0}+{q_{2n+2}}y_{0}=-2n^{2}t_{0}-2n\gamma t_{0}-2\gamma
y_{0}-2ny_{0}$ and with a simple change of variable in the sum we obtain

$\displaystyle\mathbb{P}(T=\infty)$
$\displaystyle=\sum_{n\in\mathbb{Z}}e^{p_{-2n}x_{0}+q_{-2n}y_{0}}-\sum_{n\in\mathbb{Z}}e^{{p_{2n}}x_{0}+{q_{2n+2}}y_{0}}$
$\displaystyle=2\sum_{n\in\mathbb{Z}}\sinh(\gamma(y_{0}+2nt_{0}))e^{-2(ny_{0}+n^{2}t_{0})-\gamma
y_{0}}.$

We are now interested in $\mathbb{P}^{\alpha}(T_{1}<T_{2})$ and
$\mathbb{P}^{\alpha}(T_{2}<T_{1})$ which are the exit probabilities on each
side of the cone of the conditioned process.

######  Proposition 15 (Conditioned probabilities of exit on an edge).

For $\alpha\in[0,\pi]$ we have

$\displaystyle
L_{1}(q(\alpha))=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}\mathbb{P}^{\alpha}(T_{1}<T_{2})=\sum_{n\in-\mathbb{N}^{*}}(-1)^{n+1}e^{p_{n}x_{0}+q_{n}y_{0}}$

and

$\displaystyle
L_{2}(p(\alpha))=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}\mathbb{P}^{\alpha}(T_{2}<T_{1})=\sum_{n\in+\mathbb{N}^{*}}(-1)^{n+1}e^{p_{n}x_{0}+q_{n}y_{0}}$

where $(p_{0},q_{0})=(p(\alpha),q(\alpha))$ and the sequence
$(p_{n},q_{n})_{n\in\mathbb{Z}}$ is defined in (24).

###### Proof.

The proof is again based on a compensation method. Let us define

$\widetilde{L}_{1}^{\alpha}(z_{0}):=\sum_{n\in-\mathbb{N}^{*}}(-1)^{n+1}e^{p_{n}x_{0}+q_{n}y_{0}}.$

Let us verify that the function $\widetilde{L}_{1}^{\alpha}$ is harmonic and
satisfies the following conditions:

$\begin{cases}(\mathcal{G}\widetilde{L}_{1}^{\alpha})(z_{0})=0&\text{for
}z_{0}\in(\mathbb{R}_{+}^{*})^{2},\\\
\widetilde{L}_{1}^{\alpha}(z_{0})=0&\text{for
}z_{0}=(x_{0},0)\in\mathbb{R}\times\\{0\\},\\\
\widetilde{L}_{1}^{\alpha}(z_{0})=e^{q(\alpha)y_{0}}&\text{for
}z_{0}=(0,y_{0})\in\\{0\\}\times\mathbb{R}.\end{cases}$ (30)

The first condition is trivial by the construction of the sequel
$(p_{n},q_{n})\in\mathcal{P}$, see Proposition 25. The boundary conditions can
be seen by summing in packets of two (taking into account the first term which
cannot be compensated on one side) as follows

$\widetilde{L}_{1}^{\alpha}(z_{0})=\overunderbraces{&\br{1}{=e^{q(\alpha)y_{0}}\text{
when }x_{0}=0}\br{4}{=0\text{ when }x_{0}=0}\br{4}{=0\text{ when
}x_{0}=0}}{&e^{p_{-1}x_{0}+q_{-1}y_{0}}-e^{p_{-2}x_{0}+q_{-2}y_{0}}+e^{p_{-3}x_{0}-q_{-3}y_{0}}-e^{p_{-4}x_{0}+q_{-4}y_{0}}+e^{p_{-5}x_{0}+q_{-5}y_{0}}+\cdots}{&\br{3}{=0\text{
when }y_{0}=0}\br{3}{=0\text{ when }y_{0}=0}}.$

With a classical martingale argument we remark that the function $z_{0}\mapsto
L_{1}(q(\alpha))$ seen as a function of $z_{0}$ is also harmonic, i.e.
$\mathcal{G}_{z_{0}}L_{1}(q(\alpha))=0$. This function also satisfies the same
boundary conditions, to verify it, it is enough to remark that via the Doob’s
$h$-transform we have

$L_{1}(q(\alpha))=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}L_{1}^{\alpha}(0)=e^{p(\alpha)x_{0}+q(\alpha)y_{0}}\mathbb{P}_{z_{0}}^{\alpha}(T_{1}<T_{2})$

and of course $\mathbb{P}^{\alpha}(T_{1}<T_{2})=0$ if
$z_{0}\in\mathbb{R}\times\\{0\\}$ and $\mathbb{P}^{\alpha}(T_{1}<T_{2})=1$ if
$z_{0}\in\\{0\\}\times\mathbb{R}$. The functions $\widetilde{L}_{1}^{\alpha}$
and $L_{1}(q(\alpha))$ seen as a function of $z_{0}$ both satisfy (30) and
tend to $0$ at infinity. We deduce by the maximum principle that these
functions are equals, i.e. $L_{1}(q(\alpha))=\widetilde{L}_{1}^{\alpha}$. ∎

######  Remark 16 (Probability of exit on an edge).

For $\alpha_{\gamma}\in(0,\pi/2)$ such that
$\tan\alpha_{\gamma}=\gamma/(1-\gamma)$ we have
$(p(\alpha_{\gamma}),q(\alpha_{\gamma}))=(0,0)$ and we find that the
probability $\mathbb{P}(T_{1}<T_{2})$ that the process exit of the cone by the
vertical axis is equal to

$\displaystyle\mathbb{P}(T_{2}<T_{1})=L_{2}(0)$
$\displaystyle=\sum_{n\in\mathbb{N}^{*}}(-1)^{n+1}e^{p_{n}x_{0}+q_{n}y_{0}}=\sum_{n\in\mathbb{N}}e^{\overbrace{p_{2n+1}}^{p_{2n}}x_{0}+\overbrace{q_{2n+1}}^{q_{2n+2}}y_{0}}-\sum_{n\in\mathbb{N}}e^{p_{2n+2}x_{0}+{q_{2n+2}}y_{0}}$
$\displaystyle=\sum_{n\in\mathbb{N}}2\sinh\left(\frac{p_{2n}-p_{2n+2}}{2}x_{0}\right)e^{\frac{p_{2n}+p_{2n+2}}{2}x_{0}+q_{2n+2}y_{0}}$
$\displaystyle=2\sum_{n\in\mathbb{N}}\sinh((2n+1+\gamma)x_{0})e^{-(2n+2)(n+\gamma)t_{0}-(\gamma+1)x_{0}}$

where $(p_{0},q_{0})=(0,0)$. Similarly, we obtain

$\mathbb{P}(T_{1}<T_{2})=2\sum_{n\in\mathbb{N}}\sinh((2n+2-\gamma)y_{0})e^{-(2n+2)(n+1-\gamma)t_{0}-(2-\gamma)y_{0}}.$

This result matches with Anderson’s result [2, Thm 4.1].

######  Corollary 17 (Laplace transform on the boundary).

The Laplace transform $L_{1}$ can be extended analytically on
$\mathbb{C}\setminus[q^{+},\infty)$ by the formula

$\displaystyle L_{1}(q)$
$\displaystyle=\sum_{n\in-\mathbb{N}^{*}}2\sinh\left((2n-p_{0}-q_{0}+\gamma+1)x_{0}\right)e^{(p_{0}+(2n+1)(p_{0}-q_{0})-2n(n+\gamma+1)-\gamma-1)x_{0}+(q+(2n+2)(P^{+}(q)-q)-(2n+2)(n+\gamma))y_{0}}$
$\displaystyle=\sum_{n\in-\mathbb{N}^{*}}e^{(p_{0}+2n(p_{0}-q_{0})-2n(n+\gamma))x_{0}+(q_{0}+(2n+2)(p_{0}-q_{0})-(2n+2)(n+\gamma))y_{0}}$
$\displaystyle\quad-e^{(p_{0}+(2n+2)(p_{0}-q_{0})-(2n+2)(n+1+\gamma))x_{0}+(q_{0}+(2n+2)(p_{0}-q_{0})-(2n+2)(n+\gamma))y_{0}}$

where $(p_{0},q_{0})=(P^{+}(q),q)$ and

$\begin{cases}p_{2n}=P^{+}(q)+2n(P^{+}(q)-q)-2n(n+\gamma)\\\
q_{2n}=q+2n(P^{+}(q)-q)-2n(n+\gamma-1)\end{cases}\quad\text{and}\quad(p_{2n+1},q_{2n+1})=(p_{2n},q_{2n+2}).$
(31)

###### Proof.

The result directly derives from Proposition 15, the fact that
$(p(\alpha),q(\alpha))=(P^{+}(q(\alpha)),q(\alpha))$ and the analyticity of
$P^{+}$ on $\mathbb{C}\setminus[q^{+},\infty)$. Taking
$(p_{0},q_{0})=(P^{+}(q),q)$ we obtain that

$\displaystyle L_{1}(q)$
$\displaystyle=\sum_{n\in-\mathbb{N}^{*}}(-1)^{n+1}e^{p_{n}x_{0}+q_{n}y_{0}}=\sum_{n\in-\mathbb{N}^{*}}(e^{p_{2n}x_{0}}-e^{p_{2n+2}x_{0}})e^{q_{2n+2}y_{0}}$
$\displaystyle=\sum_{n\in-\mathbb{N}^{*}}2\sinh\left(\frac{p_{2n}-p_{2n+2}}{2}x_{0}\right)e^{\frac{p_{2n}+p_{2n+2}}{2}x_{0}+q_{2n+2}y_{0}}$

and we conclude noticing that (24) implies (31). ∎

##  6 Exit densities and transition kernel

This last section aims at computing the inverse Laplace transforms of $L_{1}$
and $L_{2}$ to obtain an explicit expression for the exit densities $f_{1}$
and $f_{2}$. Finally, these expressions are used to calculate the kernel
$p_{(t_{0},y_{0})}^{k,C}(t,y)$.

######  Theorem 18 (Exit densities on the boundaries).

Let $z_{0}=(x_{0},y_{0})$ the starting point and $t_{0}=x_{0}+y_{0}$. Exit
densities are given by the following formulas:

$\displaystyle
f_{1}(y)=\frac{e^{-t_{0}(1-\gamma)-y_{0}\gamma}}{\sqrt{2\pi}}\frac{e^{-\frac{1}{2}(1-\gamma)^{2}(y-t_{0})}}{\sqrt{(y-t_{0})^{3}}}\sum_{n\in\mathbb{Z}}((2n+1)t_{0}-y_{0})e^{-\frac{1}{2}\frac{((2n+1)t_{0}-y_{0})^{2}}{y-t_{0}}+(2n+1)y_{0}-2n(n+1)t_{0}}$
$\displaystyle f_{2}(x)=\frac{e^{-t_{0}\gamma-
x_{0}(1-\gamma)}}{\sqrt{2\pi}}\frac{e^{-\frac{1}{2}\gamma^{2}(x-t_{0})}}{\sqrt{(x-t_{0})^{3}}}\sum_{n\in\mathbb{Z}}((2n+1)t_{0}-x_{0})e^{-\frac{1}{2}\frac{((2n+1)t_{0}-x_{0})^{2}}{x-t_{0}}+(2n+1)x_{0}-2n(n+1)t_{0}}$

###### Proof.

We seek to inverse the Laplace transform of Corollary 31. We have

$\displaystyle L_{1}(q)$
$\displaystyle=\sum_{n\in\mathbb{N}}(e^{p_{-2n-2}x_{0}+q_{-2n}y_{0}}-e^{p_{-2n-2}x_{0}+q_{-2n-2}y_{0}})$
$\displaystyle=\sum_{n\in\mathbb{N}}e^{-(P^{+}(q)-q)((2n+1)t_{0}-y_{0})+qt_{0}-(2n+2)(n+1-\gamma)t_{0}+2(n+1-\gamma)y_{0}}$
$\displaystyle\quad-e^{-(P^{+}(q)-q)((2n+1)t_{0}+y_{0})+qt_{0}-(2n+2)(n+1-\gamma)t_{0}-(2n+2)y_{0}}$
$\displaystyle=\sum_{n\in\mathbb{N}}e^{-((2n+1)t_{0}-y_{0})\sqrt{(1-\gamma)^{2}-2q}+qt_{0}-t_{0}(2n(n+1)+1-\gamma)+y_{0}(2n+1-\gamma)}$
$\displaystyle\quad-e^{-((2n+1)t_{0}+y_{0})\sqrt{(1-\gamma)^{2}-2q}+qt_{0}-t_{0}(2n(n+1)+1-\gamma)-y_{0}(2n+1+\gamma)}$

We denote $\mathcal{L}^{-1}_{q}$ the inverse Laplace operator related to the
$q$-variable. For $a>0$, $b>0$ and $y>t_{0}$ we have

$\mathcal{L}^{-1}_{q}(e^{-b\sqrt{a-2q}+qt_{0}})(y)=\frac{b}{\sqrt{2\pi}}\frac{1}{\sqrt{(y-t_{0})^{3}}}e^{-\frac{1}{2}(a(y-t_{0})+\frac{b^{2}}{y-t_{0}})}.$

Taking the inverse Laplace transform of the sum term by term, it reads

$\displaystyle f_{1}(y)$
$\displaystyle=\frac{e^{-t_{0}(1-\gamma)-y_{0}\gamma}}{\sqrt{2\pi}}\frac{1}{\sqrt{(y-t_{0})^{3}}}$
$\displaystyle\sum_{n\in\mathbb{N}}((2n+1)t_{0}-y_{0})e^{-\frac{1}{2}((1-\gamma)^{2}(y-t_{0})+\frac{((2n+1)t_{0}-y_{0})^{2}}{y-t_{0}})}e^{-2n(n+1)t_{0}+(2n+1)y_{0}}$
$\displaystyle\quad-((2n+1)t_{0}+y_{0})e^{-\frac{1}{2}((1-\gamma)^{2}(y-t_{0})+\frac{((2n+1)t_{0}+y_{0})^{2}}{y-t_{0}})}e^{-2n(n+1)t_{0}-(2n+1)y_{0}}$

which coincides with the stated result summing over $\mathbb{Z}$. The formula
for $f_{2}(x)$ is obtained replacing $1-\gamma$ by $\gamma$ and $y$ by $x$. ∎

With some tedious computations, one may verify that this result matches with
Anderson’s result [2, Thm 4.3] by deriving its formula (4.32).

######  Remark 19 (Density of the first exit time and persistence
asymptotics).

The previous result gives an explicit expression for the density of $T$ since
we have seen in (8) that $f_{T}(t)=f_{1}(t_{0}+t)+f_{2}(t_{0}+t)$. The
persistence probability of the process after a time $t$ is given by

$\displaystyle\mathbb{P}(T>t$
$\displaystyle)=\mathbb{P}(T=\infty)+\int_{t}^{\infty}f_{T}(s)\mathrm{d}s$
$\displaystyle\underset{t\to\infty}{=}h^{\alpha_{\gamma}}(z_{0})+\left(\frac{h^{0}(z_{0})}{\sqrt{2\pi}}\frac{e^{-\frac{\gamma^{2}}{2}(t+t_{0})}}{t^{3/2}}+\frac{h^{\pi/2}(z_{0})}{\sqrt{2\pi}}\frac{e^{-\frac{(1-\gamma)^{2}}{2}(t+t_{0})}}{t^{3/2}}\right)(1+o(1))$

where the asymptotics derives from (18) and $\alpha_{\gamma}$ is defined in
Remark 14.

We now use the previous result about the exit densities to obtain an explicit
formula for the Green’s function of the process $Z$, and then for the
transition density of the space-time Brownian motion $B$, see (7).

######  Corollary 20 (Transition density of the killed space-time Brownian
motion in a cone).

For $(t_{0}+t,y)\in C$, the transition density defined in (6) is equal to

$\displaystyle p_{(t_{0},y_{0})}^{k,C}(t,y)=\frac{1}{\sqrt{2\pi
t}}e^{-\frac{(y-y_{0}-\gamma
t)^{2}}{2t}}-\mathbb{E}_{(t_{0},y_{0})}\left(\frac{1}{\sqrt{2\pi(t-T)}}e^{-\frac{(y-Y_{T}-\gamma(t-T))^{2}}{2(t-T)}}\mathds{1}_{T\leqslant
t}\right)$ (32) $\displaystyle=\frac{1}{\sqrt{2\pi
t}}\sum_{n\in\mathbb{Z}}\left(e^{-\frac{(y-2nt_{0}-y_{0}-\gamma
t)^{2}}{2t}+\gamma(2nt_{0}+y_{0})}-e^{-\frac{(y+2nt_{0}+y_{0}-\gamma
t)^{2}}{2t}-\gamma(2nt_{0}+y_{0})}\right)e^{-\gamma
y_{0}-2n^{2}t_{0}-2ny_{0}}$ (33)

###### Proof.

Let us define $p_{t}^{\gamma}(y):=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(y-\gamma
t)^{2}}{2t}}$ the transition kernel of the free Brownian motion with drift
$\gamma$ starting from $0$ and recall that we note
$p^{k,C}_{(t_{0},y_{0})}(t,y)\mathrm{d}y=\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y),T>t)$
the transition kernel of the killed space-time Brownian motion in $C$. A
direct calculus based on the Markov property gives

$\displaystyle p^{k,C}_{(t_{0},y_{0})}(t,y)\mathrm{d}y$
$\displaystyle=\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y))-\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y),T\leqslant
t)$
$\displaystyle=\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y))-\int_{0}^{t}\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y)|T=T_{2}=u)f_{2}(t_{0}+u)\mathrm{d}u$
$\displaystyle-\int_{0}^{t}\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y)|T=T_{1}=v)f_{1}(t_{0}+v)\mathrm{d}v$
$\displaystyle=\mathbb{P}_{(t_{0},y_{0})}(B(t)=(t_{0}+t,\mathrm{d}y))-\int_{0}^{t}\mathbb{P}_{(t_{0}+u,0)}(B(t)=(t_{0}+t,\mathrm{d}y))f_{2}(t_{0}+u)\mathrm{d}u$
$\displaystyle-\int_{0}^{t}\mathbb{P}_{(t_{0}+v,t_{0}+v)}(B(t)=(t_{0}+t,\mathrm{d}y))f_{1}(t_{0}+v)\mathrm{d}v$
$\displaystyle=\left(p_{t}^{\gamma}(y-y_{0})-\underbrace{\int_{0}^{t}p_{t-u}^{\gamma}(y)f_{2}(t_{0}+u)\mathrm{d}u}_{=:I_{2}(t)}-\underbrace{\int_{0}^{t}p_{t-v}^{\gamma}(y-t_{0}-v)f_{1}(t_{0}+v)\mathrm{d}v}_{=:I_{1}(t)}\right)\mathrm{d}y$

and equality (32) follows. We defined in the previous formula the integrals
$I_{1}(t)$ and $I_{2}(t)$ and we are going to compute these integrals by
computing their Laplace transforms. We denote $\mathcal{L}_{t}$ the Laplace
transform operator related to the $t$ variable and $\star$ the convolution
operation. We have

$\displaystyle\mathcal{L}_{t}(I_{2}(t))(p)$
$\displaystyle=\mathcal{L}_{t}\left(\left(p_{\cdot}^{\gamma}(y)\star
f_{2}(t_{0}+\cdot)\right)(t)\right)(p)$
$\displaystyle=\mathcal{L}_{t}\left(p_{t}^{\gamma}(y)\right)(p)\times\mathcal{L}_{t}\left(f_{2}(t_{0}+t)\right)(p)$
(34)

One may compute that

$\mathcal{L}_{t}\left(p_{t}^{\gamma}(y)\right)(p)=\frac{1}{\sqrt{2(p+\gamma^{2}/2)}}e^{\gamma
y-y\sqrt{2(p+\gamma^{2}/2)}}.$ (35)

Recalling that Theorem 18 gives

$f_{2}(t_{0}+u)=\frac{e^{-t_{0}\gamma-
x_{0}(1-\gamma)}}{\sqrt{2\pi}}\frac{e^{-\frac{1}{2}\gamma^{2}u}}{{u^{3/2}}}\sum_{n\in\mathbb{Z}}((2n+1)t_{0}-x_{0})e^{-\frac{1}{2}\frac{((2n+1)t_{0}-x_{0})^{2}}{u}+(2n+1)x_{0}-2n(n+1)t_{0}}$

we compute the following Laplace transform (we note $sgn$ the sign function)

$\mathcal{L}_{t}\left(\frac{((2n+1)t_{0}-x_{0})}{\sqrt{2\pi}{t^{3/2}}}e^{-\frac{1}{2}\gamma^{2}t-\frac{1}{2}\frac{((2n+1)t_{0}-x_{0})^{2}}{t}}\right)(p)=sgn(n)e^{-|(2n+1)t_{0}-x_{0}|\sqrt{2(p+\gamma^{2}/2)}}$

and we obtain

$\mathcal{L}_{t}\left(f_{2}(t_{0}+t)\right)(p)={e^{-t_{0}\gamma-
x_{0}(1-\gamma)}}\sum_{n\in\mathbb{Z}}sgn(n)e^{-sgn(n)((2n+1)t_{0}-x_{0})\sqrt{2(p+\gamma^{2}/2)}+(2n+1)x_{0}-2n(n+1)t_{0}}.$
(36)

Using (34), (35) and (36) we obtain the Laplace transform of $I_{2}$. By
inverting this Laplace transform we compute

$\displaystyle I_{2}(t)$ $\displaystyle={e^{-t_{0}\gamma-
x_{0}(1-\gamma)}}\sum_{n\in\mathbb{Z}}sgn(n)p_{t}^{\gamma}(y+sgn(n)((2n+1)t_{0}-x_{0}))e^{-sgn(n)\gamma((2n+1)t_{0}-x_{0})+(2n+1)x_{0}-2n(n+1)t_{0}}$
$\displaystyle={e^{-\gamma
y_{0}}}\sum_{n\in\mathbb{Z}}sgn(n)p_{t}^{\gamma}(y+sgn(n)(2nt_{0}+y_{0}))e^{-2n^{2}t_{0}-2ny_{0}-sgn(n)\gamma(2nt_{0}+y_{0})}$
$\displaystyle={e^{-\gamma
y_{0}}}\sum_{n\in\mathbb{N}}p_{t}^{\gamma}(y+2nt_{0}+y_{0})e^{-2n^{2}t_{0}-2ny_{0}-\gamma(2nt_{0}+y_{0})}$
$\displaystyle-{e^{-\gamma
y_{0}}}\sum_{n\in-\mathbb{N}^{*}}p_{t}^{\gamma}(y-2nt_{0}-y_{0})e^{-2n^{2}t_{0}-2ny_{0}+\gamma(2nt_{0}+y_{0})}$

In the same way, we show that for $x+y=t_{0}+t$ we have

$\displaystyle I_{1}(t)$
$\displaystyle={e^{-(1-\gamma)x_{0}}}\sum_{n\in\mathbb{N}}p_{t}^{1-\gamma}(x+2nt_{0}+x_{0})e^{-2n^{2}t_{0}-2nx_{0}-(1-\gamma)(2nt_{0}+x_{0})}$
$\displaystyle-{e^{-(1-\gamma)x_{0}}}\sum_{n\in-\mathbb{N}^{*}}p_{t}^{1-\gamma}(x-2nt_{0}-x_{0})e^{-2n^{2}t_{0}-2nx_{0}+(1-\gamma)(2nt_{0}+x_{0})}$
$\displaystyle={e^{-\gamma
y_{0}}}\sum_{n\in\mathbb{N}}p_{t}^{\gamma}(y-2(n+1)t_{0}+y_{0})e^{-2(n+1)^{2}t_{0}+2(n+1)y_{0}+\gamma(2(n+1)t_{0}-y_{0})}$
$\displaystyle-{e^{-\gamma
y_{0}}}\sum_{n\in-\mathbb{N}^{*}}p_{t}^{\gamma}(y+2nt_{0}-y_{0})e^{-2n^{2}t_{0}+2ny_{0}-\gamma(2nt_{0}-y_{0})}$
$\displaystyle={e^{-\gamma
y_{0}}}\sum_{n\in-\mathbb{N}^{*}}p_{t}^{\gamma}(y+2nt_{0}+y_{0})e^{-2n^{2}t_{0}-2ny_{0}-\gamma(2nt_{0}+y_{0})}$
$\displaystyle-{e^{-\gamma
y_{0}}}\sum_{n\in\mathbb{N}^{*}}p_{t}^{\gamma}(y-2nt_{0}-y_{0})e^{-2n^{2}t_{0}-2ny_{0}+\gamma(2nt_{0}+y_{0})}$

Since $p^{k,C}_{(t_{0},y_{0})}(t,y)=p_{t}^{\gamma}(y-y_{0})-I_{1}(t)-I_{2}(t)$
we obtain formula (33) which concludes the proof. ∎

We find again the result obtained by Defosseux [8, Proposition 2.2].

The approach developed in this article which combines the steepest descent
method and a recursive compensation method allowed us to determine the Martin
boundary and all the harmonic functions associated with the space-time
Brownian motion. By using these two methods together, we also developed a new
and original approach which enabled us to recover some interesting results.

### Acknowledgments

This project has received funding from Agence Nationale de la Recherche, ANR
JCJC programme under the Grant Agreement RESYST ANR-22-CE40-0002. The author
would like to thank Manon Defosseux, Kilian Raschel and Pierre Tarrago for
interesting discussions related to space-time Brownian motion, the
compensation approach and Martin’s boundary theory.

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