# A collection of results relating the geometry of plane domains and the exit
time of planar Brownian motion.

Maher Boudabra, Andrew Buttigieg, and Greg Markowsky
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
Department of Mathematics, Monash University, Australia

###### Abstract

We prove a number of results relating exit times of planar Brownian with the
geometric properties of the domains in question. Included are proofs of the
conformal invariance of moduli of rectangles and annuli using Brownian motion;
similarly probabilistic proofs of some recent results of Karafyllia on
harmonic measure on starlike domains; examples of domains and their
complements which are simultaneously large when measured by the moments of
exit time of Brownian motion, and examples of domains and their complements
which are simultaneously small; and proofs of several identities involving the
Cauchy distribution using the optional stopping theorem.

##  1 Introduction

In this paper, we prove a series of loosely connected results relating the
geometry of domains in the plane with the distribution of the exit time of
planar Brownian motion from those domains. Results of this type seem to have
originated in such seminal papers as [10] and [6], and continue to attract
current researchers (see the references contained in [20] for a large list).
We first informally describe our results.

In the next section, we show how Brownian motion can be used to prove a staple
of complex analysis, which is the conformal invariance of the modulus of a
rectangle, or of an annulus. In other words, conformal maps between rectangles
which fix the vertices cannot change the ratio of the side lengths, and
conformal maps between annuli cannot change the ratio of the radii of the
annuli. In the following section, we show how Brownian motion can be used to
prove some recent analytic results on harmonic measure on starlike domains due
to Karafyllia. Next we discuss the relationship between moments of Brownian
exit time and the Hardy norm of domains. We discuss how these quantities can
indicate the size of a domain at infinity, but also discuss potentially
surprising examples of domains and their complements which are simultaneously
large when measured by the moments of exit time of Brownian motion, and
examples where domains and their complements are simultaneously small. In the
final section, we show how the optional stopping theorem can be used to prove
a number of results on the Cauchy distribution which are proved in the
literature by other methods.

We now fix notation. $B_{t}$ will always denote a planar Brownian motion, and
given a domain $U$ let $\tau(U)$ denote the exit time of planar Brownian
motion from $U$; that is, $\tau(U)=\inf\\{t\geq 0:B_{t}\notin U\\}$. The next
theorem is the core of planar Brownian motion theory.

######  Theorem 1.1.

If $f$ is a non constant analytic function and $B_{t}$ is a planar Brownian
motion starting at a then $f(B_{t})$ is a time-changed Brownian motion
starting at $f(a)$, where the time change rate is governed by

$\sigma(t):=\int_{0}^{t}\lvert f^{\prime}(B_{s})\rvert^{2}ds.$

In other words, the process $f(B_{\sigma^{-1}(t)})$ is a planar Brownian
motion.

Lévy’s Theorem is often referred to as the conformal invariance principle,
although a more accurate name might be the analytic invariance principle,
since injectivity plays no role in the theorem. This result will play the
major role in everything that is to follow.

##  2 Exit distributions of Brownian motion from rectangles and annuli, and
applications to conformal maps

###  2.1 Applications of $\psi$ and probabilistic proofs for known results.

We will use the following notation for a rectangle (with $a,b>0$).

$R_{a,b}:=(-a,a)\times(-b,b).$

Fix a rectangle $R_{a,b}$, and label the vertices clockwise as
$z_{1},\ldots,z_{4}$, starting with $z_{1}=a+bi$. Now suppose that we have a
similarly defined rectangle $R_{a^{\prime},b^{\prime}}$ with vertices
$z^{\prime}_{1},\ldots,z^{\prime}_{4}$, and that there is a conformal map $f$
between them such that $f(z_{j})=z^{\prime}_{j}$ for $j=1,\ldots,4$; note that
in making this last statement we are implicitly using Caratheodory’s Theorem
(see [17]), which implies that such a conformal map between Jordan domains
extends to a homeomorphism between the boundaries. The following is a standard
result in complex analysis.

######  Theorem 2.1.

Given the setup described above, we have
$\frac{a}{b}=\frac{a^{\prime}}{b^{\prime}}$, and furthermore $f(z)=cz$, where
$c$ is a positive constant.

There are at least two known proofs of this. The standard one uses the concept
of extremal length (see [1], [12], or [16]). The second, which is not as
widely known, is to repeatedly apply Schwarz’s reflection principle in order
to extend $f$ to a conformal self-map of $\mathbb{C}$. We present here a
third, using the conformal invariance of Brownian motion.

###### Proof.

Label the right side of $R_{a,b}$ as $S_{1}$, and then label the remaining
sides $S_{2},S_{3},S_{4}$ clockwise. Label the sides of
$R_{a^{\prime},b^{\prime}}$ similarly as
$S^{\prime}_{1},\ldots,S^{\prime}_{4}$. Let

$W=\\{z\in R_{a,b}:P_{z}(B_{\tau(R_{a,b})}\in
S_{1})=P_{z}(B_{\tau(R_{a,b})}\in S_{3})\\}.$

It is clear by symmetry that $i\mathbb{R}\cap R_{a,b}\subseteq W$, where
$i\mathbb{R}$ denotes the imaginary axis. It is also not hard to see that
$W\subseteq i\mathbb{R}\cap R_{a,b}$, since a Brownian motion starting from
some point not on $i\mathbb{R}\cap R_{a,b}$ has a positive probability to exit
on the nearest vertical side before hitting $i\mathbb{R}\cap R_{a,b}$, but if
it hits $i\mathbb{R}\cap R_{a,b}$ before hitting the nearest vertical side it
has equal probabilities of exiting on $S_{1}$ and $S_{3}$ after that by
symmetry and the strong Markov property. We conclude that $W=i\mathbb{R}\cap
R_{a,b}$. By the conformal invariance of Brownian motion, for $z\in W$ we must
have $P_{f(z)}(B_{\tau(R_{a^{\prime},b^{\prime}})}\in
S^{\prime}_{1})=P_{f(z)}(B_{\tau(R_{a^{\prime},b^{\prime}})}\in
S^{\prime}_{3})$, and this implies that $f(i\mathbb{R}\cap
R_{a,b})=i\mathbb{R}\cap R_{a^{\prime},b^{\prime}}$. The same argument,
applied to the horizontal sides, shows that $f(\mathbb{R}\cap
R_{a,b})=\mathbb{R}\cap R_{a^{\prime},b^{\prime}}$. In particular, we see that
$f(0)=0$. Furthermore, $R_{a,b}$ and $R_{a^{\prime},b^{\prime}}$ are divided
into four smaller rectangles by the real and imaginary axes, and we have shown
that $f$ maps each of these rectangles in $R_{a,b}$ onto the corresponding one
in $R_{a^{\prime},b^{\prime}}$. Applying the same argument to each of these
smaller rectangles, and then iterating the argument shows that
$f(\frac{j}{2^{n}}a+\frac{k}{2^{n}}bi)=\frac{j}{2^{n}}a^{\prime}+\frac{k}{2^{n}}b^{\prime}i$,
where $n$ is any positive integer and $j,k$ are any odd integers with
$|j|,|k|<2^{n}$. These points form a dense set in $R_{a,b}$ and $f$ is
continuous, so $f$ must be the map
$(x,y)\to(\frac{a^{\prime}}{a}x,\frac{b^{\prime}}{b}y)$. However, the Cauchy-
Riemann equations now imply that $\frac{a^{\prime}}{a}=\frac{b^{\prime}}{b}$,
and the result follows. ∎

There is a well-known analogue for an annulus (again, see [1], [12], or [16]).
This states that if there is a conformal map between two annuli then the
ratios of their inner and outer radii must be the same, and that the map must
be either a linear map or an inversion. Two standard proofs are well-known
which parallel the ones for the rectangle, one using extremal length and the
other using Schwarz reflection. We will offer here a third using Brownian
motion, and will note that our proof yields a more general result which
applies to analytic functions which are not necessarily univalent. To state
this generalisation, we need the definition of a proper map.

######  Definition 2.2.

We say that an analytic function $f:U\rightarrow W$ is proper if for any
compact set $K$, $f^{-1}(K)$ is also compact.

Definition 2.2 extends to any continuous function between two topological
spaces [23]. The interpretation of properness is the following : If
$(z_{n})_{n}$ is a sequence of $U$ that converges to $\partial U$, i.e the
complement of any compact set of $U$ contains all but a finite number of
$(z_{n})_{n}$, then $(f(z_{n}))_{n}$ converges necessarily to $\partial W$.
Often, we express that as $f$ maps boundary to boundary. Note that conformal
maps are automatically proper, since $f^{-1}$ is itself conformal.

We will use the following notation for an annulus

$\displaystyle A_{h,r}$ $\displaystyle:=\\{z\mid r<|z|<R\\}.$

Our generalisation is as follows.

######  Theorem 2.3.

Suppose $f:A_{r,R}\rightarrow A_{r^{\prime},R^{\prime}}$ is a proper analytic
function. Then there is a positive integer $n$ such that
$\frac{R^{n}}{r^{n}}=\frac{R^{\prime}}{r^{\prime}}$, and $f$ is of the form
$f(z)=\xi z^{n}$ or $f(z)=\frac{\xi}{z^{n}}$ for some constant $\xi$.

###### Proof.

We will consider a Brownian motion $B_{t}$ running in $A_{r,R}$, and the image
time-changed Brownian motion $B^{\prime}_{t}$ running in
$A_{r^{\prime},R^{\prime}}$. Since $f$ is proper, it is easy to see that
$\sigma(\tau(A_{r,R}))=\tau^{\prime}(A_{r^{\prime},R^{\prime}})$, where
$\tau^{\prime}(A_{r^{\prime},R^{\prime}})$ denotes the exit time of the
Brownian motion $B^{\prime}_{t}=f(B_{\sigma^{-1}(t)})$. Fix $a\in A_{r,R}$ and
choose $r^{\prime\prime},R^{\prime\prime}$ such that
$r^{\prime}<r^{\prime\prime}<|f(a)|<R^{\prime\prime}<R^{\prime}$. Since $f$ is
proper, $f^{-1}(cl(A_{r^{\prime\prime},R^{\prime\prime}}))$ is compact (here
$cl$ denotes the closure), and we may therefore find $\tilde{r}>r$ and
$\tilde{R}<R$ such that
$f^{-1}(cl(A_{r^{\prime\prime},R^{\prime\prime}}))\subseteq
A_{\tilde{r},\tilde{R}}$. The sets $f(\\{|z|=\tilde{r}\\})$ and
$f(\\{|z|=\tilde{R}\\})$, must lie in the complement of
$A_{r^{\prime\prime},R^{\prime\prime}}$, however their union must separate
$f(a)$ from the boundary of $A_{r^{\prime},R^{\prime}}$; to see this, note
that the set $\\{B_{t}:0\leq t\leq\tau(A_{r,R})\\}$ must intersect
$\\{|z|=\tilde{r}\\})\cup\\{|z|=\tilde{R}\\}$, and thus
$\\{B^{\prime}_{t}:0\leq t\leq\tau^{\prime}(A_{r^{\prime},R^{\prime}})\\}$
must intersect $f(\\{|z|=\tilde{r}\\})\cup f(\\{|z|=\tilde{R}\\})$. By
continuity, each of $f(\\{|z|=\tilde{r}\\})$ and $f(\\{|z|=\tilde{R}\\})$ must
lie in just one of $\\{|z|<r^{\prime\prime}\\}$ and
$\\{|z|>R^{\prime\prime}\\}$. We can then, if necessary, replace $f$ by
$\frac{Rr}{f}$ and assume that
$f(\\{|z|=\tilde{r}\\})\subseteq\\{|z|<r^{\prime\prime}\\}$ and
$f(\\{|z|=\tilde{R}\\})\subseteq\\{|z|>R^{\prime\prime}\\}$. Now let
$\tilde{r}\searrow r$ and $\tilde{R}\nearrow R^{\prime}$; as $f$ is proper,
the distances from the images $f(\\{|z|=\tilde{r}\\})$ and
$f(\\{|z|=\tilde{R}\\})$ to the boundary of $A_{r^{\prime},R^{\prime}}$ must
go to 0, and we conclude the following equality of events:

$\displaystyle\\{|B_{\tau(A_{r,R})}|=r\\}=\\{|B^{\prime}_{\tau^{\prime}(A_{r^{\prime},R^{\prime}})}|=r^{\prime}\\}$
$\displaystyle\\{|B_{\tau(A_{r,R})}|=R\\}=\\{|B^{\prime}_{\tau^{\prime}(A_{r^{\prime},R^{\prime}})}|=R^{\prime}\\}$

Fix $\theta\in(0,1)$, and let $W=\\{a:P_{a}(|B_{\tau(A_{r,R})}|=r)=\theta\\}$.
It is well known that $W$ is the circle $\\{\ln|z|=\ln
R-\theta\ln\frac{R}{r}\\}$; see for example [21, Thm. 3.17]. By conformal
invariance, we must have $f(W)\subseteq W^{\prime}$, where
$W^{\prime}=\\{a:P_{a}(|B^{\prime}_{\tau^{\prime}(A_{r^{\prime},R^{\prime}})}|=r^{\prime})=\theta\\}=\\{\ln|z|=\ln
R^{\prime}-\theta\ln\frac{R^{\prime}}{r^{\prime}}\\}$. Rearranging and
dividing yields

$\frac{\ln|f(z)/R^{\prime}|}{\ln|z/R|}=\frac{\ln|R^{\prime}/r|}{\ln|R/r|}.$

Setting $\eta:=\frac{\ln|R^{\prime}/r|}{\ln|R/r|}$, we obtain

$|f(z)|=\frac{R^{\prime}}{R^{\eta}}|z|^{\eta}.$

We want to now consider $\frac{f(z)}{z^{\eta}}$ as an analytic function,
however it can not be defined to be analytic on the entire annulus unless
$\eta$ is an integer. Since we do not, as of yet, know that it is an integer,
we solve this problem by removing the slit $S=\\{Im(z)=0,-R<Re(z)<-r\\}$, from
the annulus, and choose a branch of the function $z\to z^{\eta}$ which is
analytic on $A_{r,R}\backslash S$. On this domain, the function
$\frac{f(z)}{z^{\eta}}$ is indeed analytic, and as it has constant absolute
value it must be a constant by the open mapping theorem for analytic
functions. Thus, $f(z)=\xi z^{\eta}$ for some constant $\xi$. However, $f$ is
analytic on all of $A_{r,R}$, and the result now follows upon recalling again
that $z\to z^{\eta}$ can only be analytic on the annulus if $\eta$ is an
integer $n$. ∎

######  Remark.

The ratio of the two dimensions of a rectangle (resp. inner and outer radii of
an annulus) is referred to as the modulus of the rectangle (resp. annulus),
and these results imply that it is a conformal invariant. Interestingly, a
different probabilistic interpretation of modulus has recently been given in
[2]. The approach taken there is quite different from ours, and makes use of
discretization.

##  3 Brownian Motion and Starlike Domains

###  3.1 Exit of Brownian motion from a Starlike Domain

The main purpose of this section is to examine a recent result of Karafyllia
[13] on the harmonic measure of starlike domains. We will show how the result
can be extended, and will give a proof using Brownian motion. First, we need
some definitions.

######  Definition 3.1.

(Harmonic Measure)

For a simply connected domain $D$ and a subset $A\subset\partial D$, we define
the harmonic measure as

$\omega_{D}(a,A):=\mathbb{P}_{a}(B_{\tau(D)}\in A)$

Remark: The harmonic measure can also be defined analytically, i.e. without
mentioning Brownian motion. This is the definition used by Karafyllia in [13],
but the two definitions are equivalent.

######  Definition 3.2.

(Starlike Domain)

An open set $D$ on $\mathbb{C}$ is starlike (with respect to $z_{0}$ if there
exists some $z_{0}\in D:\\{(1-t)z_{0}+tz:t\in[0,1]\\}\in D$ for all $z\in D$.

######  Definition 3.3.

($\triangle-$Starlike Domain)

A domain $U$ is $\triangle-$starlike if, given any $z\in U$, the horizontal
ray

$\triangle(z)=\\{w:\Im(w)=\Im(z),\Re(w)\leq\Re(z)\\}$

lies entirely in $U$.

Figure 1: All rays going left lie inside U

The following result was proved analytically for starlike domains by
Karafyllia [13], in order to answer a question posed by Betsakos in [4].

######  Theorem 3.4.

Let $D$ be a domain in $\mathbb{C}$ which is starlike with respect to $0$.
Then for every $R>0$,

$\hat{\nu}_{D}(0,R)\leq 2\nu_{0,D}(R),$

where $\nu_{D}(0,R):=\omega_{D}(0,\partial D\cap\\{|z|>R\\})$,
$\hat{\nu}_{D}(0,r):=\omega_{D\cap\\{|z|<R\\}}(0,D\cap\\{|z|=R\\})$. The
constant $2$ is the best possible.

Now a generalisation is offered and proved using the reflection principle of
Brownian motion. Our result pertains to $\Delta$-starlike domains, and to see
that it really is a generalisation of Karafyllia’s result we note that any
domain starlike with respect to 0 corresponds to a $\triangle-$starlike domain
with periodic boundary of period $2\pi i$ through the complex exponential map.
The image of this map will be the starlike domain with the origin removed, but
the removal of the origin is irrelevant to the Brownian motion since planar
Brownian motion doesn’t see points (see [20]), provided that the Brownian
motion doesn’t start at the origin. To include the origin, simply take a point
near the origin, conclude the theorem for that starting point, and then let
the point approach the origin. To summarize, our result can be translated to
Karafyllia’s by the exponential map. The following is our result.

######  Theorem 3.5.

Let U be a $\triangle-$starlike Domain in $\mathbb{C}$. Then for every
$r\in\mathbb{R},$ $a\in\\{\Re(z)<r\\}$, we have

$\hat{\nu}_{U}(a,r)\leq 2\nu_{U}(a,r),$

where $\nu_{U}(a,r):=\omega_{U}(a,\partial U\cap\\{\Re(z)>r\\})$ and
$\hat{\nu}_{U}(a,r):=\omega_{U\cap\\{\Re(z)<r\\}}(a,U\cap\\{\Re(z)=r\\})$. The
constant 2 is the best possible.

Remark: In addition to transferring the result to $\Delta$-starlike domains,
we also show that the harmonic measure can be taken with respect to any point
in $a\in\\{|z|<R\\}$, rather than just $0$. That is, we prove

$\hat{\nu}_{D}(a,R)\leq 2\nu_{a,D}(R),$

for any $a\in\\{|z|<R\\}$. In terms of the Brownian motion interpretation of
harmonic measure, this means that the Brownian motion can start at any
$a\in\\{|z|<R\\}$ and the result still holds.

###### Proof.

It clearly suffices to assume $r=0$. Let $\mathbb{I}:=\\{z:\Re(z)=0\\}$,
$\tau(U):=\inf\\{t:B_{t}\notin U\\}$,
$\tau_{0}:=\inf\\{t:B_{t}\notin\\{\Re(B_{t})<0\\}\cap U\\}$. Define the
coupling

$\hat{B}_{t}=\begin{cases}B_{t}&t\leq\tau_{0}\\\
-\bar{B}_{t}&t>\tau_{0}\end{cases}.$

By the reflection principle for planar Brownian motion, $\hat{B}_{t}$ is a
Brownian motion. It is evident that, as a consequence of
$\triangle-$starlikeness,

$\\{\Re(\hat{B}_{\tau(U)})<0,\tau(U)>\tau_{0}\\}\subseteq\\{\Re(B_{\tau(U)})>0,\tau(U)>\tau_{0}\\}.$

Figure 2 captures this idea, that once the Brownian motion hits $\mathbb{I}$
it will be at least as likely to exit to the right of this line than to the
left.

Figure 2: On the left, if $B_{\tau(U)}$ is left of $\mathbb{I}$,
$\hat{B}_{\tau(U)}$ is guaranteed to lie on the right of $\mathbb{I}$. On the
other hand if $B_{\tau(U)}$ lies to the right of $\mathbb{I}$, then
$\hat{B}_{t}$ may have lived past that time, and may or may not exit to the
left of $\mathbb{I}$.

Note also that, up to a set of probability 0, $\\{\tau(U)>\tau_{0}\\}$ is the
disjoint union of $\\{\Re(B_{\tau(U)})<0,\tau(U)>\tau_{0}\\}$ and
$\\{\Re(B_{\tau(U)})>0,\tau(U)>\tau_{0}\\}$. We see that

$P(\Re(B_{\tau(U)})>0,\tau(U)>\tau_{0})\geq\frac{1}{2}P(\tau(U)>\tau_{0}).$

The result now follows upon noting that

$P(\Re(B_{\tau(U)})>0,\tau(U)>\tau_{0})=P(\Re(B_{\tau(U)})>0)=\nu_{U}(a,0)$

and

$P(\tau(U)>\tau_{0})=\hat{\nu}_{U}(a,0)$

∎

######  Remark.

As an alternative to the proof given here, one may obtain Karafyllia’s result
directly by reflecting over the circle of radius $R$. Such a reflection does
preserve Brownian motion but also introduces a time change, however that is
not relevant for our problem since we are only interested in the distribution
of the Brownian motion at time $\tau$, rather than the distribution of $\tau$
itself. The constant $2$ in our result is optimal, and occurs when the
$\triangle-$starlike domain is a horizontal strip or half-plane with
horizontal boundary. We note that the constant 2 is also optimal in
Karafyllia’s result as well, but in that case the inequality is always strict;
that is, the constant 2 is never obtained for a given domain. This is because
the starlike domain can never actually equal its reflection over the circle,
since $0$ is in the starlike domain but $\infty$ is in its complement.

##  4 Domains with constraints on the exit moments

###  4.1 Hardy Spaces and $p$-th moments of exit times

The theory of Hardy spaces provides another point of intersection between the
theory of Brownian motion and complex analysis. Hardy spaces provide an
analytic way of putting a measure on the size of the domain as seen by
Brownian motion through its exit time out of the domain. Moreover, Hardy
spaces play a major role when it comes to the problem of finiteness of the
moments of the exit times. To this end, let $f$ be a analytic function on the
unit disc and for any $p>0$ and $0\leq r<1$ set

$N_{p,r}(f):=\left\\{\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^{p}d\theta\right\\}^{\frac{1}{p}}.$

The quantity $N_{p,r}(f)$ can be interpreted as the $L^{p}$ norm 111The word
norm is an abuse of language as $N_{p,r}(f)$ is not a true norm when $p<1$. of
the function $\theta\mapsto f(re^{\theta i})$. It can be shown, using harmonic
analysis techniques, that $N_{p,r}(f)$ is non decreasing in terms of $r$ [23].
Hence, we are ready now to define the $p^{th}$-Hardy norm.

######  Definition 4.1.

For any analytic function $f$ on the unit disc, the $p^{th}$-Hardy norm of $f$
is defined by

$H_{p}(f):=\sup_{0\leq r<1}N_{p,r}(f)=\sup_{0\leq
r<1}\left\\{\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{\theta
i})|^{p}d\theta\right\\}^{\frac{1}{p}}.$

The set of analytic functions whose $p^{th}$-Hardy norm is finite is denoted
by $\mathcal{H}^{p}$ and called Hardy space (of index $p$). A crucial result
about Hardy norms is that, if $H_{p}(f)$ is finite then $f$ has a radial
extension to the boundary. More precisely $f^{\ast}(z):=\lim_{r\rightarrow
1}f(rz)$ exists a.e for all $z\in\partial\mathbb{D}$ and it belongs to $L^{p}$
as well. More details about the topic can be found in [11]. A consequence of
Hölder’s inequality is the inclusion $\mathcal{H}^{q}\subseteq\mathcal{H}^{p}$
whenever $0<p\leq q$. This leads to the following definition.

######  Definition 4.2.

(Hardy Number)

Given a simply connected domain $W$ and conformal $f:\mathbb{D\rightarrow}W$,
the Hardy number, $H(W)$, of $W$ is defined as

$H(W):=\sup\\{p>0\mid H_{p}(f)<\infty\\}.$

Although the definition is in terms of $f$ but it can be shown that the Hardy
number does not depend of the functional mapping $\mathbb{D}$ onto $U$ $f$ and
so it is well defined. We will be interested in the Hardy numbers of a
particular type of domain, namely Jordan* domains, which we now define.

######  Definition 4.3.

(Jordan curve)

A Jordan curve is a closed curve in $\mathbb{C}$ which is the homeomorphic
image of a circle.

######  Definition 4.4.

(Jordan domain)

A domain whose boundary is a Jordan curve is called a Jordan domain.

Note that Jordan domains are automatically simply connected. The question of
calculating the Hardy number of a Jordan domain is not interesting, because a
Jordan domain is bounded and it follows easily that its Hardy number is
infinite. In order to make the question more interesting, we describe another
type of domain, the Jordan* domains, which were previously considered in [18].

######  Definition 4.5.

(Jordan* Domain)

A Jordan* domain is a domain in $\mathbb{C}$ which is the image of Jordan
domain $U$ under a Mobius transformation which takes a boundary point of $U$
to $\infty$.

Essentially we are modifying the class of Jordan domains to require $\infty$
to be a boundary point, and the Hardy number of Jordan* domains is no longer
trivial. Alternatively, Jordan* domains could be defined to be domains whose
boundary contains $\infty$ and is homeomorphic (as a set in the Riemann
sphere) to a circle. An important paper which relates the theory of Hardy
spaces to the exit time of Brownian motion is that of Burkholder [6] and we
will implement from it without proof the following two results.

######  Theorem 4.6.

Suppose $f$ is a conformal map taking the unit disc to a simply connected
domain $R$. Then for all $x\in R$,

$f\in
H^{2p}\iff\mathbb{E}_{x}[\tau(R)^{p}]<\infty\iff\mathbb{E}_{x}[|B_{\tau(R)}|^{2p}]<\infty.$

The $p$-Hardy norm can be seen as the expectation of the $p^{th}$ moment of
$f(B_{T(\mathbb{D})})$ as $B_{T(\mathbb{D})}$ is uniformly distributed on the
circle. Thus, the equivalence of the finiteness of the $p^{th}$ moment of the
stopped Brownian motion and the $2p^{th}$ Hardy norm of a conformal map can be
seen as a consequence of the conformal invariance of Brownian motion. Another
crucial fact about Theorem 4.6 is that the finiteness of
$\mathbb{E}_{x}[\tau(R)^{p}]$ and $\mathbb{E}_{x}[|B_{\tau(R)}|^{2p}]$ does
not depend on the starting point $x$ [6]. Burkholder was also able to give the
following general result about the $p^{th}$ moment of the exit time of a
Brownian motion exiting any simply connected domain, which is a consequence of
the corresponding result for Hardy norms (see [11]).

######  Theorem 4.7.

If $R$ is simply connected and not the whole complex plane then, for all $z\in
R$, $\mathbb{E}_{z}[\tau(R)^{p}]<\infty$ for any $p<\frac{1}{4}$.

In other words, the minimum Hardy number for any simply connected domain is
$\frac{1}{2}$. This is therefore considered an extremal case with regards to
the size of a simply connected domain as seen by Brownian motion. The Koebe
domain, $K:=\mathbb{C}\backslash(-\infty,-\frac{1}{4}]$ is considered in many
ways extremal among simply connected domains, and it has infinite $1/4^{th}$
moment of $\tau_{K}$ and thus a Hardy number of $\frac{1}{2}$ (this is proved
in [11], and can also be proved using Burkholder’s result and calculations
similar to Example 4.8 below). There are examples of smaller domains which
also have infinite $1/4^{th}$ moment. One of these will be important in what
follows, so we take the time to describe it.

######  Example 4.8.

Consider a half-strip, for example $\\{\Re(z)<0,-1<\Im(z)<1\\}$ and label the
complement of this half strip $V$. Now consider a slightly smaller domain $W$,
which is the complement of the parabola given by the equation
$x=1-\frac{y^{2}}{4}$. Also consider the conformal map from the unit disk to
$W$, $f(z)=\frac{4}{(1+z)^{2}}.$ We see

$\displaystyle H_{p}(f)^{p}$
$\displaystyle={\textstyle\frac{1}{2\pi}}\int_{0}^{2\pi}|f(e^{ti})|^{2p}dt$
$\displaystyle={\textstyle\frac{1}{2\pi}}\int_{0}^{2\pi}\Bigl{|}\frac{4}{(1+e^{it})^{2}}\Bigr{|}^{2p}dt.$

The singularity at $t=\pi$ is only contained when $p<\frac{1}{4}$, thus the
$2p^{th}$-Hardy norm is infinite for $p=1/4$. By Theorem 4.6 we have $f\notin
H^{\frac{1}{2}}$, meaning $\mathbb{E}[\tau_{W}^{\frac{1}{4}}]=\infty$ and
since $W\subset V$, $\mathbb{E}[\tau_{V}^{\frac{1}{4}}]=\infty$ hence
$H(V)=\frac{1}{2}$. We may translate the parabola as required such that any
half strip may be fit inside, meaning any half strip will have infinite
$1/4^{th}$ moment of its exit time out of the domain and thus a Hardy number
of $\frac{1}{2}$.

As another application of Theorem 4.6 (applied to the conformal map
$f(z)=(\frac{1-z}{1+z})^{\frac{\theta}{\pi}}$), let us consider the following
result (which can also be deduced from the explicit formula for the
distribution of $\tau(W)$ given in [19]).

######  Proposition 4.9.

[6] Let $W=\\{-\frac{\theta}{2}<Arg(z)<\frac{\theta}{2}\\}$ be a wedge of
aperture $\theta\in(0,2\pi]$. Then

$\mathbb{E}(\tau(W)^{\frac{p}{2}})<+\infty\Longleftrightarrow
p<\frac{\pi}{\theta}.$

Note that the Koebe domain can be considered as a wedge of size $2\pi$ with
$\theta=\pi$ giving the required result of

$\mathbb{E}(\tau^{\frac{1}{4}})<+\infty\Longleftrightarrow p<\frac{1}{2},$

resulting in a Hardy number of $\frac{1}{2}$ as expected.

We will now give an intuitive discussion to motivate a natural conjecture that
we have considered. Proposition 4.9 indicates that the aperture of a domain at
$\infty$ essentially decides whether the domain is viewed as "small" or
"large" by Brownian motion; we will continue to use these terms in quotation
marks to indicate their informal nature. Consider now a wedge $V$ of size
$\theta\in(0,2\pi)$ and its complement $W$ of size $\alpha=2\pi-\theta$.
Suppose that $\mathbb{E}(\tau_{V}^{\frac{p}{2}})<+\infty$ for some $p$. This
means that $V$ is "small", so $W$ must be "large". To be precise, suppose
$\frac{1}{p}+\frac{1}{q}\leq 2$; we will show that
$\mathbb{E}(\tau_{W}^{\frac{q}{2}})=+\infty$. We know $p<\frac{\pi}{\theta}$,
so

$2>\frac{1}{q}+\frac{\theta}{\pi}=\frac{1}{q}+\frac{2\pi-\alpha}{\pi},$

hence $q>\frac{\pi}{\alpha}$ and thus
$\mathbb{E}(\tau_{W}^{\frac{q}{2}})=+\infty$. On the other hand, if we suppose
$\mathbb{E}(\tau_{V}^{\frac{p}{2}})=+\infty$ (i.e. $V$ is "large") and
$\frac{1}{p}+\frac{1}{q}>2$ then we can show
$\mathbb{E}(\tau_{W}^{\frac{q}{2}})<+\infty$ (i.e. $W$ is "large"). To see
this, note that $p\geq\frac{\pi}{\theta}$, so

$2<\frac{1}{q}+\frac{\theta}{\pi}=\frac{1}{q}+\frac{2\pi-\alpha}{\pi},$

hence $q<\frac{\pi}{\alpha}$ and $\mathbb{E}(\tau_{W}^{\frac{q}{2}})<+\infty$.
These observations have led us to an attempt at generalization, and the
Jordan* domains appear to be ideally suited for this purpose, since the Jordan
Curve Theorem implies that the interior of the complement of a Jordan* domain
is again a Jordan* domain. The following conjecture is therefore natural.

Conjecture: If $V$ and $W$ are complementary Jordan* domains (that is, they
share the same boundary), then they cannot both be "small" or both be "large"
as viewed by Brownian motion.

We have not attempted to make this statement more rigorous, since we have
found example domains that show that virtually any formulation of it will be
false. We will now take the time to describe them; in particular, we will
prove the following theorem.

######  Theorem 4.10.

  1. 1.

There exist complementary Jordan* domains $V$ and $W$ such that
$\mathbb{E}[\tau(V)^{1/4}]=\mathbb{E}[\tau(W)^{1/4}]=\infty$.

  2. 2.

There exist complementary Jordan* domains $V$ and $W$ such that
$\mathbb{E}[\tau(V)^{p}]<\infty$ and $\mathbb{E}[\tau(W)^{p}]<\infty$ for all
$p>0$.

Part 1 of the theorem provides an example of complementary Jordan* domains
which are both "large" as viewed by Brownian motion, and part 2 does the same
with both "small". We remind the reader that any Jordan* domain is simply
connected and therefore has Hardy number at least $1/2$. This implies that
part 1 of the theorem cannot be improved. The remainder of this subsection
will provide the proof of part 1, while part 2 will be proved in the next
subsection.

###### Proof.

We will utilise the following result for domains, which is a straightforward
consequence of the monotone convergence theorem.

######  Theorem 4.11.

If $\Omega_{n}$ is an increasing sequence of domains (i.e.
$\Omega_{n}\subseteq\Omega_{n+1}$) and $\Omega=\cup_{n=1}^{\infty}\Omega_{n},$
then
$\mathbb{E}[\tau(\Omega_{n})^{p}]\rightarrow\mathbb{E}[\tau(\Omega)^{p}].$

We will construct a sequence of domains $(W_{n})_{n\in\mathbb{N}_{0}}$ and a
sequence of each respective complement $(V_{n})_{n\in\mathbb{N}_{0}}$ in which
the limiting domains of each sequence will both have infinite $1/4^{th}$
moment. In each $W_{n}$ we start a Brownian motion $B_{t}$ at $-1$ and in each
$V_{n}$ we start a Brownian motion $\hat{B}_{t}$ at 1.

We begin with an extension of Example 4.8, in which we restrict the complement
of the half strip by the lines shown in Figure 3

$\Re(z)=b_{1},\quad\Im(z)=a_{1}\quad\Im(z)=-a_{1},$

where $a_{1}>a_{0}:=1$ $V_{1}$ is a subset of the complement of a half strip
and so, if we take $a_{1}\rightarrow\infty$ and $b_{1}\rightarrow-\infty$,
then $W_{1}$ becomes a half strip, so $V_{1}$ becomes the complement of a half
strip and by (4.11), $\mathbb{E}[\tau(V_{1})^{\frac{1}{4}}]\rightarrow\infty$.
Thus we may take $a_{1},b_{1}$ large enough such that
$\mathbb{E}[\tau(V_{1})^{\frac{1}{4}}]>1$, and may also ensure that $a_{1}\geq
a_{0}+1$. Since the complement of a half strip can fit inside $W_{1}$, we have
that $\mathbb{E}[\tau(W_{1})^{\frac{1}{4}}]=\infty$.

Figure 3: $1^{st}$ iteration of domain

We can then restrict $W_{1}$ by the lines shown in Figure 4

$\Re(z)=b_{2},\quad\Im(z)=a_{2}\quad\Im(z)=-a_{2}.$

As $a_{2}\rightarrow\infty$ and $b_{2}\rightarrow\infty$, $W_{2}\rightarrow
W_{1}$ and we have that $W_{2}\subset W_{1},$ meaning by Theorem 4.11,

$\mathbb{E}[\tau(W_{2})^{\frac{1}{4}}]\rightarrow\mathbb{E}[\tau(W_{1})^{\frac{1}{4}}]=\infty.$

We may then take these limits large enough such that
$\mathbb{E}[\tau(W_{2})^{\frac{1}{4}}]>2$ while again ensuring $a_{2}\geq
a_{1}+1$. Since the complement of a half strip can fit inside $V_{2}$, we also
have that $\mathbb{E}[\tau(W_{1})^{\frac{1}{4}}]=\infty$.

Figure 4: $2^{nd}$ iteration of domain

We may continue iterating this process. In the odd iterations, for all
$n\in\mathbb{N}_{0}$, as $a_{2n+1}\rightarrow\infty$,
$b_{2n+1}\rightarrow-\infty$, then $V_{2n+1}\rightarrow V_{2n}$ and since
$V_{2n+1}\subset V_{2n}$, by Theorem 4.11,

$\mathbb{E}[\tau(V_{2n+1})^{\frac{1}{4}}]\rightarrow\mathbb{E}[\tau(V_{2n})^{\frac{1}{4}}]=\infty.$

Thus, the limits may be taken large enough such that
$\mathbb{E}[\tau(V_{2n+1})^{\frac{1}{4}}]>2n+1$ while once again ensuring that
$a_{2n+1}\geq a_{2n}+1$. The complement of a half strip can fit inside
$W_{2n+1}$ meaning $\mathbb{E}[\tau(W_{2n+1})^{\frac{1}{4}}]=\infty$.

Following the same logic with the as the odd iterations we construct the even
iterations such that $\mathbb{E}[\tau(W_{2n+2})^{\frac{1}{4}}]>2n+2$ and
$\mathbb{E}[\tau(V_{2n})^{\frac{1}{4}}]=\infty$, with $a_{2n}\geq a_{2n-1}+1$
.

We show three more iterations in Figure 5.

Figure 5: $3^{rd},$$4^{th}$, and $5^{th}$ iterations of domain

Notice $V_{2n+1}\subset V_{2n+3}$ and $W_{2n+2}\subset W_{2n+4}$ for all
$n\in\mathbb{N}_{0}$. Thus we define the required domains

$V_{\infty}=\bigcup_{n=0}^{\infty}V_{2n+1}\qquad
W_{\infty}=\bigcup_{n=1}W_{2n+2}.$

By construction, the $a_{n}$’s go to $+\infty$, so $V_{\infty}$ and
$W_{\infty}$ are both Jordan* domains. Since $(V_{2n+1})_{n=0}^{\infty}$ is an
increasing sequence of domains, by Theorem 4.11 we have

$\mathbb{E}[\tau(V_{2n+1})^{\frac{1}{4}}]\rightarrow\mathbb{E}[\tau(V_{\infty})^{\frac{1}{4}}]=\infty.$

The same is true of $W_{\infty}$ and thus in the limit we have a domain and
its complement who both have infinite $1/4^{th}$ moment of Brownian exit time,
as required.

∎

###  4.2 Hyperbolic Geometry and Hardy Numbers

We now construct a second counterexample to the conjecture in which the domain
and its complement both have finite $p^{th}$ moment of the exit time for any
$p>0$. To construct this second counterexample we must recall theory from
hyperbolic geometry ([3], [15]). We first define hyperbolic distance on the
unit disc, which is induced by the hyperbolic density.

######  Definition 4.12.

(Hyperbolic density)

The hyperbolic density on the unit disc $\mathbb{D}$ is defined as

$\lambda_{\mathbb{D}}(z)|dz|=\frac{2|dz|}{1-|z|^{2}}$

The definition of hyperbolic density is motivated by the following isometric
property

$\lambda_{\mathbb{D}}(\phi(z))|\phi^{\prime}(z)|=\lambda_{\mathbb{D}}(z)$

for a conformal automorphism $\phi$. The hyperbolic density induces hyperbolic
length and hyperbolic distance in the following way.

######  Definition 4.13.

(Hyperbolic length)

For any two points, $z,w\in\mathbb{D}$, we define the hyperbolic length
$\ell_{\mathbb{D}}$ as

$\ell_{\mathbb{D}}(\gamma)=\int_{\gamma}\lambda_{\mathbb{D}}(z)|dz|$

As such, we define the hyperbolic distance as follows:

######  Definition 4.14.

(Hyperbolic distance)

The hyperbolic distance on the unit disc $\mathbb{D}$, for $z,w\in\mathbb{D}$
is defined by

$d_{\mathbb{D}}(z,w)=\inf_{\gamma}\ell_{\mathbb{D}}(\gamma)$

More generally, for any simply connected domain $U\neq\mathbb{C}$, any
conformal $\phi:\mathbb{D}\rightarrow U$, and $z,w\in U$,

$d_{U}(z,w)=d_{\mathbb{D}}(\phi^{-1}(z),\phi^{-1}(w))$

and for any $E\subset U$,

$d_{U}(z,E)=\inf\\{d_{U}(z,w):w\in E\\}.$

We may also a define a related quantity used to estimate the hyperbolic
distance.

######  Definition 4.15.

(Quasi-hyperbolic distance)

For any two points, $z,w$, in a simply connected domain $U$, the quasi-
hyperbolic distance is defined as

$\delta_{U}(z,w)=\inf_{\gamma:z\rightarrow
w}\int_{\gamma}\frac{|ds|}{|s-\partial U|}.$

Where $|.|$ is the Euclidean norm and the infinimum is taken over all paths
$\gamma$ which connect $z$ to $w$.

The quasi-hyperbolic distance estimates the hyperbolic distance by the
following result, which is a consequence of the celebrated Kobe$-\frac{1}{4}$
theorem [7].

######  Theorem 4.16.

$\frac{1}{2}\delta_{U}\leq d_{U}\leq 2\delta_{U}$.

We may apply this theory to the theory of Hardy domains to construct the
second counterexample using the following theorem given in [14].

######  Theorem 4.17.

Let $U$ be a simply connected domain, $a\in U$, $W_{R}=\\{|z-a|=R\\}$, and
$F_{R}=W_{R}\cap U$ for $R>0$, then

$H(U)=\liminf_{R\rightarrow\infty}\frac{d_{U}(a,F_{R})}{\ln(R)}.$

######  Remark 4.18.

In [14], $F_{R}$ was defined as $R\partial\mathbb{D}\cap U$, but here we
require the natural extension to circles centered at arbitrary values $a$ in
$U$.

Finally we will implement the following lemma.

######  Lemma 4.19.

If $U$ is simply connected and $|z-\partial U|<K$ for all $z\in U$ for some
$K\in\mathbb{R^{+}}$, then $H(U)=\infty$.

###### Proof.

By definition and assumption we have

$\delta_{U}(a,z)=\inf_{\gamma:a\rightarrow
z}\int_{\gamma}\frac{|ds|}{|s-\partial U|}>\inf_{\gamma:a\rightarrow
z}\int_{\gamma}\frac{|ds|}{K}>\frac{|z-a|}{K}.$

From (4.17),

$d_{U}(a,z)\geq\frac{1}{2}\delta_{U}(a,z)>\frac{|z-a|}{2K}.$

Hence

$H(U)=\liminf_{R\rightarrow\infty}\frac{d_{U}(a,F_{R})}{\ln(R)}\geq\liminf_{R\rightarrow\infty}\frac{R}{2K\text{$\ln(R)$}}=\infty.$

∎

######  Example 4.20.

We will construct a domain and its complement which will both have with finite
$p^{th}$ moment for any $p$ for the exit time of a Brownian motion exiting the
domains.

Let $\gamma_{1}=\\{t\geq 0:te^{it}\\}$ shown in (6) in orange and
$\gamma_{2}=\\{t:te^{i(t-\pi)}\\}$ shown in blue. Concatenating these curves
we split the complex plane into $U$ and $\mathbb{C}\backslash U$. Any point
$a$ in $U$ or $b$ in $\mathbb{C}\backslash U$ will be within $\pi$ of
$\gamma_{1}$ or $\gamma_{2}$. Hence by (4.17), $H(U)=\infty$ and
$H(\mathbb{C}\backslash U)=\infty$ meaning in turn both have exit times with
finite $p^{th}$ moment for any $p$.

Figure 6: Domain and complement with infinite Hardy numbers

##  5 Characterisations of the Cauchy Distribution

In this short final section, we examine a method of deducing identities
related to planar Brownian motion by applying the optional stopping theorem to
complex-valued martingales. This method has been used previously in [8] to
deduce the exit distributions of Brownian motion from a half-plane and from a
strip, as well as the identity

$\mathbb{E}[e^{i\lambda\frac{2}{\pi}\ln|C_{1}|}]=\frac{1}{\cosh\lambda},$

with $C_{1}$ a standard Cauchy, which was proved in [5] by other methods
(however, unbeknownst to the authors of [8] at the time, the arguments for the
strip and half-plane were already present in Exercises 2.18 and 2.19 in [21]).

We will focus the method on two identities of the Cauchy distribution given in
[22] and proved via the residue theorem. Here alternate probabilistic proofs
are offered.

######  Proposition 5.1.

Let C be distributed $\textnormal{Cauchy}(a,b)$ and
$\gamma=a+bi\in\mathbb{C}$. If $\alpha\in\mathbb{C}$ and $\Im(\alpha)>0$ then
$\mathbb{E}[\frac{C-\alpha}{C-\bar{\alpha}}]=\frac{\gamma-\alpha}{\gamma-\bar{\alpha}}$.

###### Proof.

Start a Brownian motion at $\gamma$ in the upper half plane and stop it at
$T(\mathbb{H})$. As is well-known (see [8]),
$B_{T(\mathbb{H})}\sim\textnormal{Cauchy}(a,b)$ so
$B_{T(\mathbb{H})}\overset{d}{=}C$. We can transform this distribution by
$\phi(z)=\frac{z-\alpha}{z-\bar{\alpha}}$ which maps $\mathbb{H}$ to
$\mathbb{D}$ and $\alpha$ to $0$. Since $\phi$ maps to $\mathbb{D}$, it is
bounded and since it is analytic, $\phi(B_{t})$ is a martingale. Thus we may
utilise the optional stopping theorem.

$\mathbb{E}[\frac{C-\alpha}{C-\bar{\alpha}}]=\mathbb{E}[\frac{B_{\tau}-\alpha}{B_{\tau}-\bar{\alpha}}]=\mathbb{E}[\frac{B_{0}-\alpha}{B_{0}-\bar{\alpha}}]=\frac{\gamma-\alpha}{\gamma-\bar{\alpha}}.$

∎

In proving the second Cauchy distribution identity, the following results from
Burkholder [6] will be utilised.

######  Theorem 5.2.

If $0<p<\infty$, $B_{t}$ is a planar Brownian motion and $\tau$ is a stopping
time such that $\mathbb{E}[\textnormal{Log}(1+\tau)]<\infty$, then

$\mathbb{E}(\sup_{0\leq t<\infty}|B_{\tau\wedge t}|)^{p}\leq
c_{p}\mathbb{E}|B_{\tau}|^{p}.$

Remark: Clearly, if $\mathbb{E}[\tau^{p}]<\infty$ for some $p>0$, then
$\mathbb{E}[\textnormal{$\ln$}(1+\tau)]<\infty$ and Theorem 5.2 may be
applied.

We will also use the following convergence theorem given in [9].

######  Lemma 5.3.

If $X=(X_{n},\mathcal{F}_{n})$ is a discrete time submartingale and, for some
$p>1$,

$\sup_{n}\mathbb{E}|X_{n}|^{p}<\infty$

then there is an integrable random variable $X_{\infty}$ such that

$X_{n}\rightarrow X_{\infty}\ a.s.\qquad
X_{n}\overset{L^{1}}{\rightarrow}X_{\infty}.$

We thus show a probabilistic proof for the following Cauchy distribution
identity.

######  Proposition 5.4.

Let C be distributed $\textnormal{Cauchy}(a,b)$ and
$\gamma=a+bi\in\mathbb{C}$. Let $\phi(z)=z^{\alpha}$ be defined for
$z\in\mathbb{C}$ and $\alpha\in(0,1)$, using a principal branch of the
logarithm; that is, $z^{\alpha}=e^{\alpha\textnormal{Log}(z)}$, where
$\textnormal{Log}(z)=\text{$\ln|z|+i\textup{Arg}(z)$}$ and
$-\pi<\textnormal{{Arg}}(z)\leq\pi$. Then

$\mathbb{E}[C^{\alpha}]=\gamma^{\alpha}.$

###### Proof.

Start a Brownian motion at $\gamma\in\mathbb{H}$ and stop it at
$T(\mathbb{H})$. $\phi$ is defined to be analytic, thus $B_{t}^{\alpha}=M_{t}$
is a martingale. Since $\alpha\in(0,1)$, $\phi(\mathbb{H})\subset\mathbb{H}$
and is a wedge of size $\alpha\pi$. Define $\tau=T(\phi(\mathbb{H}))$. Thus,
using (4.7) for $p<\frac{1}{\alpha}$,
$\mathbb{E}(\tau^{\frac{p}{2}})<+\infty$, and so
$\mathbb{E}[\textnormal{Log}(1+\tau)]<\infty$.

Thus, by (5.2), for any $0<q<\infty$

$\mathbb{E}_{x}[(\sup_{0\leq t<\infty}|B_{\tau\wedge t}|)^{q}]\leq
c_{q}\mathbb{E}|B_{\tau}|^{q}.$

We define $1<\beta<\frac{1}{\alpha}$ so that $\alpha\beta<1$. Setting
$q=\alpha\beta$ we have, for all $t$,

$\displaystyle\sup_{0\leq t<\infty}\mathbb{E}_{x}[|M_{\tau\wedge
t}|{}^{\beta}]$ $\displaystyle\leq\mathbb{E}_{x}[(\sup_{0\leq
t<\infty}|M_{\tau\wedge t}|)^{\beta}]$ $\displaystyle\leq
c_{p}\mathbb{E}|M_{\tau}|^{\beta}=D<\infty,$

where $D=\mathbb{E}|C|^{\alpha\beta}$ is a finite constant since the Cauchy
distribution has finite absolute fractional moments. By discretising we have
$\sup_{n}\mathbb{E}_{x}[|M_{\tau\wedge n}|{}^{\beta}]<\infty$ and so by (5.3),
as $n\rightarrow\infty$, $M_{\tau\wedge n}\rightarrow M_{\tau}$ in $L^{1}$ and
almost surely and thus in turn $\mathbb{E}[M_{\tau\wedge
n}]\rightarrow\mathbb{E}[M_{\tau}]$. However since $\tau\wedge n$ is bounded
almost surely, we can apply the optional stopping theorem to the stopped
martingale $M_{\tau\wedge n}$ hence achieving the result

$\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_{0}]=\mathbb{E}[M_{\tau}].$

Hence we arrive at the result

$\mathbb{E}[C^{\alpha}]=\mathbb{E}[B_{\tau}^{\alpha}]=\mathbb{E}[B_{0}^{\alpha}]=\gamma^{\alpha}.$

∎

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