# Occupation times and areas derived from random sampling

Frank Aurzada  Department of Mathematics, Technical University of Darmstadt
Leif Döring  Mathematics Institute, University of Mannheim  Helmut H. Pitters
Mathematics Institute, University of Mannheim

###### Abstract

We consider the occupation area of spherical (fractional) Brownian motion,
i.e. the area where the process is positive, and show that it is uniformly
distributed. For the proof, we introduce a new simple combinatorial view on
occupation times of stochastic processes that turns out to be surprisingly
effective. A sampling method is used to relate the moments of occupation times
to persistence probabilities of random walks that again relate to
combinatorial factors in the moments of beta distributions. Our approach also
yields a new and completely elementary proof of Lévy’s second arcsine law for
Brownian motion. Further, combined with Spitzer’s formula and the use of Bell
polynomials, we give a characterisation of the distribution of the occupation
times for all Lévy processes.

Keywords— Bell polynomials; fluctuation theory for random walks; Lévy process;
occupation time; spherical fractional Brownian motion

##  1 Introduction and main results

Consider a measure space $(I,\mathcal{I},\alpha)$, where $\alpha$ is a finite
measure with total mass $|\alpha|=\alpha(I)$, and a stochastic process
$X=\\{X_{t},t\in I\\}$ with index set $I$ whose state space $\mathscr{X}$ is
endowed with some sigma algebra $\mathcal{X}$. We do not assume $I$ to be an
ordered set. For a real-valued, non-negative, measurable function
$f:\mathscr{X}\to[0,\infty)$ consider the path integral

$\int_{I}f(X_{s})\alpha(ds).$

Path integrals for diverse stochastic processes have a rich history in several
areas of probability theory. In the present article, we deal with occupation
times $\int_{I}{\mathbf{1}}_{\\{X_{s}\in S\\}}\alpha(ds)$ for some measurable
set $S$. For $I=[0,t]$, $\alpha$ the Lebesgue measure, and $S$ measurable,
this is the portion of time that the process spends in the set $S$. Most
classically, the occupation time of the non-negative half-line $S=[0,\infty)$
during $[0,1]$ by a Brownian motion is well-known to be arcsine distributed,
i.e. it has the density $\pi^{-1}(x(1-x))^{-1/2}$ on $(0,1)$. This result goes
back to Paul Lévy [34] and is sometimes referred to as the second arcsine law
for Brownian motion, cf. [37]. Since Lévy’s seminal work, many proofs for this
result have been found (e.g. Kac’s derivation via the Feynman-Kac formula as
expounded in [37, application of Theorem 7.43], or via approximation by
(simple) random walks, cf. [37, Theorem 5.28]). Further, various
generalizations to other processes have been considered (see for instance [30,
11, 7, 21, 24, 32, 8, 31, 19, 38, 36]).

While the one-dimensional stochastic process setting is well-understood, many
open problems remain for multi-dimensional processes and processes with
general index sets. Most prominently, characterising the distribution of
occupation times of planar (and higher dimensional) Brownian motion (random
walks) in cones are open problems to this day (except for cases that may
trivially be reduced to one-dimensional problems). The major focus of this
paper is on random fields, i.e. on processes with multidimensional index sets.
Here, the Brownian sheet is a natural object to consider, and [31] derives
asymptotic bounds, but the exact distribution of the occupation ‘area’ of the
Brownian sheet remains unknown. For the Brownian pillow we refer to [25]. In
this paper we compute the distribution of the occupation area for the
fractional generalisation of Lévy’s spherical Brownian motion. This is our
main result.

To motivate our approach let us recall some attempts towards the occupation
times of planar Brownian motion using moments. Note that occupation times are
bounded random variables and as such are uniquely determined by their moments.
As a specific example consider the time that planar Brownian motion spends in
some fixed cone $C$. The problem to characterise the distribution of this time
was put forth in [8] and is still open. The authors were able to derive the
first three moments of this occupation time if $C$ is taken to be a quadrant.
Motivated by this work, [19] studied the time that planar Brownian motion
spends in the ‘hourglass’, i.e. the union of the first and third quadrant, and
rephrased this problem in the language of Kontorovich-Lebedev transforms.
Desbois [14] generalized the quadrant problem to wedges with apex at the
origin and some angle $\theta>0$. Employing methods from physics, the author
computed the first three moments in the case of a wedge with angle $\theta$,
the fourth moment in the quadrant case ($\theta=\pi/2$), and derived a general
formula for second moments in high-dimensional orthants. We follow these
research efforts and attack occupation time distributions through their
integer moments, introducing a simple sampling method.

Suppose that we were to ‘guess’ the proportion of time that the process $X$
spends in some set $S$ during $[0,t]$, and to this end we were allowed to
sample $X$ at $m$ instances chosen according to our liking. It seems rather
natural to choose the times $U_{1},\ldots,U_{m}$ independently (and
independent of $X$) and uniformly at random in $[0,t],$ and to take the
empirical probability $\texttt{\\#}\\{1\leq k\leq m\colon X_{U_{k}}\in S\\}/m$
as an estimator of said proportion. In fact, it turns out that the probability
that $X$ is in $S$ at all times $U_{1},\ldots,U_{m}$ agrees with the $m$-th
moment of the occupation time of $S$ (up to the factor $t^{m}$), a
generalization of which we will see in Proposition 1. Sampling a stochastic
process at random times is by no means a new idea, and has been employed in
various other contexts. For instance, the random tree may be constructed from
broken lines derived from Brownian excursion sampled at independent uniform
times [1, 33]. In [39] the author studied Brownian motion, bridge, excursion
and meander by sampling at i.i.d. uniform times, and the convex hull of
multidimensional Brownian motion was studied in [18] by sampling at the points
of an independent Poisson process.

A surprising consequence of the computation of moments by means of sampling at
random times is a completely elementary proof for the arcsine law of the
Brownian motion, the uniform distribution of the occupation time of Lévy
bridges, and also a new characterisation of the occupation times for all Lévy
processes. Our approach combines occupation time moments with random walk
probabilities and elementary combinatorics. The use of combinatorics is not
surprising as beta distributions often appear as occupation time distributions
have explicit moment expressions involving elementary combinatorial factors.
For example, the $m$-th moments of the arcsine distribution are
$2^{-2m}\binom{2m}{m}$, combinatorial factors that appear in many
combinatorial problems, in particular in persistence probabilities of random
walks. This suggests to ask if the $m$-th moments of occupation times are
inherently related to combinatorial terms. Our answer is yes. The main insight
of this article is to realise that the following simple sampling formula is a
surprisingly effective link to relate occupation times, random walks, and,
depending on the situation, beta distributions.

######  Proposition 1.

Consider a stochastic process $(X_{t})_{t\in I}$ indexed by a measure space
$(I,\mathcal{I},\alpha)$ that attains values in a measurable space
$(\mathscr{X},\mathcal{X})$. Let $S\in\mathcal{X}$ and $m\in\mathbb{N}$. Then

$\displaystyle\mathbb{E}\left[\left(\int_{I}{\mathbf{1}}_{\\{X_{t}\in
S\\}}\alpha(dt)\right)^{m}\right]=\lvert\alpha\rvert^{m}\mathbb{P}\left\\{X_{U_{1}}\in
S,\ldots,X_{U_{m}}\in S\right\\},\quad m\in\mathbb{N},$ (1)

where $U_{1},U_{2},\ldots$ is an i.i.d. sequence independent of $X$ such that
$U_{1}$ has distribution $\alpha/\lvert\alpha\rvert$.

###### Proof.

Set $f(x):={\mathbf{1}}_{\\{x\in S\\}}$. Re-writing the expectation w.r.t. to
the distribution $\alpha/|\alpha|$ (independent of $X$) as integrals, we
obtain

$\mathbb{E}\big{[}f(X_{U_{1}})\cdots
f(X_{U_{m}})\big{]}=\mathbb{E}\left[\int_{I}\cdots\int_{I}f(X_{u_{1}})\cdots
f(X_{u_{m}})\frac{\alpha(du_{1})}{|\alpha|}\cdots\frac{\alpha(du_{m})}{|\alpha|}\right].$

Multiplying by $|\alpha|^{m}$, noticing that all the integrals are identical,
and inserting $f(x):={\mathbf{1}}_{\\{x\in S\\}}$ shows the claim. ∎

In order to discuss the use of this result, let us consider the example
$I=[0,t]$, $\alpha$ the Lebesgue measure, and $S=[0,\infty)$. Then Proposition
1 shows that the occupation time of continuous-time processes $(X_{t})$ can be
characterised through the persistence probabilities
$\mathbb{P}\left\\{X_{U_{1}}\geq 0,...,X_{U_{m}}\geq 0\right\\}$. In many
situations these persistence probabilities may be reduced to persistence
probabilities $\mathbb{P}\left\\{S_{1}\geq 0,...,S_{m}\geq 0\right\\}$ for a
well-understood discrete-time process $(S_{n})$. For example for random walks,
there is a vast literature going back to seminal works of Spitzer [44] and
Sparre Andersen [42], see also the exposition in [28, Section 1.3] for more
recent results, where such probabilities were computed under different
assumptions on the set $S$. Little suprisingly, the moments of arcsine
distributions appear naturally in persistence probabilities. While there is a
long tradition of deriving arcsine laws for continuous-time processes from
discrete-time processes using Donsker-type limiting arguments, the simple
connection between moments of occupation times and persistence probabilities
seems to be new.

######  Remark 1.

The sampling approach also shows that the $m$-th moment of the occupation time
of $d$-dimensional Brownian motion in some cone $C$ is equal to the
probability that a $d$-dimensional random walk stays in the cone $C$ up to
time $m.$ The exit time from a cone of a multi-dimensional random walk has
received great interest in mathematical research (cf. e.g. [23]), not least
because this quantity has connections to many areas such as representation
theory [5, 6], conditioned random walks [5, 6], random matrices [16], non-
colliding random walks [13, 17], and enumerative combinatorics [10, 20, 27].
We leave for future research whether this direct link between continuous-time
occupation times and discrete-time exit probabilities may help to solve open
problems for the planar and multidimensional Brownian motion.

Organisation of the article: In the following sections we illustrate the power
of this simple approach. The paper is structured as follows. In Section 1.1 we
give a very simple proof of the second arcsine law of Brownian motion. In
Section 1.2 we discuss the main result of this paper, i.e. we determine the
distribution of the occupation ‘area’ of Lévy’s Brownian motion on the sphere.
In Section 1.3 we characterise all occupation times of one-dimensional Lévy
processes using combinatorial expressions. The proofs are given in Section 2.

###  1.1 An elementary proof of Lévy’s arcsine law

As a first illustration of our line of attack we give a new, very elementary
proof of Lévy’s second arcsine law for Brownian motion.

######  Theorem 1 (Lévy [34]).

If $B$ is a standard Brownian motion, then
$t^{-1}\int_{0}^{t}{\mathbf{1}}_{\\{B_{s}>0\\}}ds$ is arcsine distributed.

In contrast to other proofs of the second arcsine law of Brownian motion, our
proof is completely elementary and in particular does not require any limiting
procedure nor does it employ analytic computations or excursion theory, as
Lévy’s original proof. At first sight our argument might resemble proofs that
approximate Brownian motion using discrete-time random walks. However,
instead, we use an entirely different connection between Brownian motion and
the so-called Laplace random walk. Instead of discretising $(B_{t})$ and
studying the same problem for random walks, the sampling method relates the
moments of the occupation time of continuous Brownian motion to discrete
persistence probabilities.

###### A simple proof of Theorem 1.

W.l.o.g. we may assume $t=1$, by the self-similiarity of Brownian motion. Fix
$m\in\mathbb{N}$. The sampling formula (1) gives

$\mathbb{E}\left[\left(\int_{0}^{1}{\mathbf{1}}_{\\{B_{t}>0\\}}dt\right)^{m}\right]=\mathbb{P}\left\\{B_{U_{1}}>0,\ldots,B_{U_{m}}>0\right\\}=\mathbb{P}\left\\{B_{U_{m:1}}>0,\ldots,B_{U_{m:m}}>0\right\\},$
(2)

where $(U_{i})$ are i.i.d. uniform in $[0,1]$ independent of the Brownian
motion and $(U_{m:i})$ is the corresponding order statistics. Further, let
$(E_{i})$ be i.i.d. standard exponential random variables independent of the
$(U_{i})$ and of the Brownian motion and set $T_{k}:=\sum_{i=1}^{k}E_{i}$,
$k=0,1,2,\ldots$. Conditioning on $T_{m+1}$ and on the $(U_{i})$ (which are
independent of the Brownian motion $B$), we can use the self-similarity of
Brownian motion, $(B_{s})_{s\geq 0}=_{d}(T_{m+1}^{-1/2}B_{T_{m+1}s})_{s\geq
0}$, to see that the probability in (2) equals

$\mathbb{P}\left\\{B_{T_{m+1}U_{m:1}}>0,\ldots,B_{T_{m+1}U_{m:m}}>0\right\\}=\mathbb{P}\left\\{B_{T_{1}}>0,\ldots,B_{T_{m}}>0\right\\},$
(3)

where we used the independence of $(U_{i})$ and $(E_{i})$ from the Brownian
motion and the fact that the vector $(T_{m+1}U_{m:1},\ldots,T_{m+1}U_{m:m})$
has the same distribution as $(T_{1},\ldots,T_{m})$, see e.g. Theorem V.2.2 in
[15].

Thus, the moments of the occupation time of Brownian motion on the left-hand
side in (2) are given by the persistence probabilities on the right-hand side
in (3). We note that these are the persistence probabilities of the Laplace
random walk $R_{i}:=B_{T_{i}}$, $i=0,1,2,\ldots$. It is well-known that the
probabilities on the right-hand side in (3) are equal to
$2^{-2m}\binom{2m}{m}$, which are – in turn – the moments of the arcsine
distribution. Since the occupation times are bounded the proof of the second
arcsine law of Brownian motion is complete.

To keep the proof self-contained let us also give an elementary argument for
the persistence probabilities in (3). Define
$\tau:=\min\\{j\in\\{0,\ldots,m\\}:R_{j}=\max_{k\in\\{0,\ldots,m\\}}R_{k}\\}$
to be the first (and only) index where the maximum of $(R_{k})_{k=0}^{m}$ is
attained. Since $\tau\in\\{0,\ldots,m\\}$ by construction, we must have (using
the continuity of the distribution of the $R_{k}$ in the second step):

$\displaystyle 1=$
$\displaystyle\sum_{j=0}^{m}\mathbb{P}\left\\{\tau=j\right\\}=\sum_{j=0}^{m}\mathbb{P}\left\\{R_{k}<R_{j},k=0,\ldots,j-1,j+1,\ldots,m\right\\}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{m}\mathbb{P}\left\\{R_{k}<R_{j},k=0,\ldots,j-1\right\\}\cdot\mathbb{P}\left\\{R_{k}<R_{j},k=j+1,\ldots,m\right\\}$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{m}\mathbb{P}\left\\{R_{k}>0,k=1,\ldots,j\right\\}\cdot\mathbb{P}\left\\{R_{k}>0,k=1,\ldots,m-j\right\\},$

where we used the independence of increments of $(R_{k})$ in the third step
and the stationarity and the symmetry of the increments of $(R_{k})$ in the
fourth step. It is again elementary to show that the unique solution of this
recursive equation is given by
$\mathbb{P}\left\\{R_{k}>0,k=1,\ldots,j\right\\}=\frac{(2j-1)!!}{(2j)!!}=2^{-2j}\binom{2j}{j}$
for all $j=0,\ldots,m$. To see the latter, multiply the recursion by
$x\in[0,1)$, sum in $m$, and the generating function of the probabilities in
question is found to be $(1-x)^{-1/2}$, cf. [12] for similar arguments. ∎

The proof does not fully use the Brownian properties, in particular,
continuity does not play a role in the sampling. Actually, exactly the same
argument works for symmetric strictly stable Lévy processes, recovering the
arcsine law first derived by Kac [29]. Below we also provide a simple proof
for the occupation time of a Brownian bridge to be $\mathcal{U}([0,1])$ but we
do so directly in the more general setting of Lévy bridges, cf. Theorem 4.

###  1.2 Spherical fractional Brownian motion

We now come to the main result of this article, the occupation ‘area’ law for
Lévy’s spherical Brownian motion and the fractional generalisation. Fix
$H\in(0,1/2]$ and $d\in\mathbb{N}$, $d\geq 2$, and let $\lVert
x\rVert\coloneqq\sqrt{x_{1}^{2}+\cdots+x_{d}^{2}}$ denote the Euclidean norm
of $x\in\mathbb{R}^{d}$. Recall that spherical fractional Brownian motion
(spherical fBM) $X\coloneqq(X_{t})_{t\in\mathbb{S}^{d-1}}$ is a centred
Gaussian process on the unit $(d-1)$-sphere
$\mathbb{S}^{d-1}\coloneqq\\{x\in\mathbb{R}^{d}\colon\lVert x\rVert=1\\}$ such
that $X_{O}=0$ a.s. for some arbitrary fixed point $O\in\mathbb{S}^{d-1}$ with

$\displaystyle\mathbb{E}[(X_{s}-X_{t})^{2}]=(d(s,t))^{2H},\qquad
s,t\in\mathbb{S}^{d-1},$ (4)

where $d(s,t)$ denotes the geodesic distance between two points $s,t$ on
$\mathbb{S}^{d-1}$. The special case $H=1/2$ was first studied by Paul Lévy
[35] and is sometimes referred to as Lévy’s spherical Brownian motion. Istas
[26] showed that there exists a Gaussian process indexed by $\mathbb{S}^{d-1}$
with covariance structure as in (4) if and only if $H\leq 1/2$. Let

$\displaystyle
A\coloneqq\int_{\mathbb{S}^{d}}{\mathbf{1}}_{\\{X_{s}>0\\}}\sigma^{d-1}(ds)$

denote the ‘area’ that $X$ spends positive, or rather the measure of the area
on $\mathbb{S}^{d-1}$ on which $X$ is positive as measured by the surface
measure $\sigma^{d-1}$.

######  Theorem 2 (Occupation time of spherical fractional Brownian motion).

Let $(X_{t})_{t\in\mathbb{S}^{d-1}}$ be a spherical fractional Brownian motion
$X$ with Hurst parameter $H\in(0,1/2]$. Then

$\lvert\sigma^{d-1}\rvert^{-1}\int_{\mathbf{S}^{d-1}}{\mathbf{1}}_{\\{X_{s}>0\\}}\sigma^{d-1}(ds),$

i.e. the ‘area’ that $X$ spends positive, is uniformly distributed on $(0,1)$,
where
$\lvert\sigma^{d-1}\rvert=\sigma^{d-1}(\mathbb{S}^{d-1})=2\pi^{\frac{d}{2}}/\Gamma(\frac{d}{2})$
is the surface area of the unit $(d-1)$-sphere.

###  1.3 Lévy processes and bridges

In this section, we apply the sampling formula to compute all moments of
occupation times of one-dimensional Lévy processes, i.e. stochastic processes
with independent and stationary increments. Let $(X_{t})_{t\geq 0}$ be a Lévy
process. We characterize the distribution of the random variable

$A_{t}\coloneqq\int_{0}^{t}{\mathbf{1}}_{\\{X_{s}>0\\}}\,ds$ (5)

by working out explicitly all its moments. In order to state the result, let
us introduce some further notation. A partition of a set $S$ is a set, $\rho$
say, of nonempty pairwise disjoint subsets of $S$ whose union is $S$. The
members of $\rho$ are also called the blocks of $\rho$. Let $\texttt{\\#}S$
denote the cardinality of $S$ and for some natural number $n$ let
${\mathscr{P}}_{n}$ denote the set of all partitions of $\\{1,\ldots,n\\}$.
Further, we recall that $(f\ast g)(t):=\int_{0}^{t}f(t-s)g(s)ds$ is the
convolution of two functions $f,g:[0,\infty)\to\mathbb{R}$. Sampling the
occupations at Poisson times in combination with Spitzer’s identity and a Bell
polynomial trick yields the following moment formula:

######  Theorem 3 (Occupation time of a Lévy process).

Fix $m\geq 1$ arbitrarily. The $m$-th moment of the occupation time $A_{t}$ of
the real-valued Lévy process $X$ in the set $(0,\infty)$ is given by

$\displaystyle\mathbb{E}[A_{t}^{m}]$
$\displaystyle=\sum_{\rho\in{{\mathscr{P}}}_{m}}\int_{0}^{t}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}\left(u^{\texttt{\\#}B-1}\mathbb{P}\left\\{X_{u}>0\right\\}\right)(s)ds.$
(6)

In particular, the first two moments of $A_{t}$ are given by

$\displaystyle\mathbb{E}[A_{t}]$
$\displaystyle=\int_{0}^{t}\mathbb{P}\left\\{X_{s}>0\right\\}ds,$ (7)
$\displaystyle\mathbb{E}[A_{t}^{2}]$
$\displaystyle=\int_{0}^{t}s\mathbb{P}\left\\{X_{s}>0\right\\}ds+\int_{0}^{t}\int_{0}^{s}\mathbb{P}\left\\{X_{u}>0\right\\}\mathbb{P}\left\\{X_{s-u}>0\right\\}duds.$

Equations (6) and (7) still hold when their (strict) inequalities together
with the (strict) inequality in the definition of the occupation time (5) are
replaced by weak inequalities.

Theorem 3 shows how to work out explicitly the moments of the distribution of
the occupation time above zero of a Lévy process $X$. In particular, the
formula shows that the distribution of $A_{t}$ is completely determined by the
positivity function $s\mapsto\mathbb{P}\left\\{X_{s}>0\right\\}$. In fact, the
only ingredient coming from the Lévy process in the moment formula (6) is the
positivity function. Equivalently, the theorem shows that the first moment of
the occupation times already determines their entire distribution.

There are a few situations in which the moments formulas can be used to
compute the occupation time distributions. One example, that could not be
treated in the literature before, is the $\frac{1}{2}$-stable subordinator
with negative drift $\mu$ for which the positivity function is known as
$\mathbb{P}\left\\{X_{t}>0\right\\}=\text{erf}(\sqrt{t/(4\mu)})$. The slightly
tedious computations will be presented in an accompanying article. A more
common situation is that of constant positivity, i.e.
$\mathbb{P}\left\\{X_{t}>0\right\\}=c$ for all $t>0$, which occurs for
instance in the case of strictly stable Lévy processes. Inserting into (6)
leaves us with a simple combinatorial expression for the moments of $A_{t}$. A
short computation shows that those expressions are precisely those of the
generalised arcsine distributions, i.e. a beta distribution with parameters
$(a,b)=(c,1-c)$ for some $c\in(0,1)$.

######  Corollary 1 (cf. [24]).

Fix $c\in(0,1)$. The following two statements are equivalent:

  1. 1.

We have $\mathbb{P}\left\\{X_{t}>0\right\\}=c$ for all $t>0$.

  2. 2.

The occupation time
$t^{-1}A_{t}=t^{-1}\int_{0}^{t}{\mathbf{1}}_{\\{X_{s}>0\\}}ds$ is generalised
arcsine distributed with parameter $c\in(0,1)$ for all $t>0$.

The symmetric case $c=\frac{1}{2}$ thus recovers the classical arcsine law.
The corollary can be deduced from Theorem 3 with a short combinatorial
computation because the moments of generalised arcsine distributions have the
combinatorial form

$\displaystyle\frac{\Gamma(m+c)}{\Gamma(m+1)\Gamma(c)}=\frac{c^{\overline{m}}}{m!},$

where $x^{\overline{m}}\coloneqq x(x+1)\cdots(x+m-1)$ denotes the $m$-the
rising factorial power of $x\in\mathbb{R}$, and the last identity is easily
seen by induction. The corollary was already proved by Getoor and Sharpe [24]
by guessing of Laplace transforms. Our proof highlights once more the
combinatorial nature behind occupation times seen through their moments.

Finally, we use the sampling method to provide a simple proof of the
uniformity of occupation times for Lévy bridges.

######  Theorem 4 (Occupation time of a Lévy bridge; cf. [22] and [32]).

Let $X$ denote a Lévy process and consider the stochastic process
$\mathring{X}\coloneqq(\mathring{X}_{t})_{t\in[0,1]}$ defined by
$\mathring{X}_{t}\coloneqq X_{t}-tX_{1}$, that we refer to as the Lévy bridge
induced by $X$. Provided that the distribution of $X_{1}$ has no atoms, the
occupation time
$\mathring{A}\coloneqq\int_{0}^{1}{\mathbf{1}}_{\\{\mathring{X}_{t}>0\\}}dt$
of the Lévy bridge $\mathring{X}$ is uniformly distributed on $(0,1)$.

The result in Theorem 4 essentially goes back to Fitzsimmons and Getoor [22]
and Knight [32]. In fact, Knight [32, Theorem 2.1(a)] provides a complete
characterization of Lévy bridges with uniform occupation times. However, we
consider our derivation interesting in its own right, as the sampling approach
yields significantly simpler proofs.

##  2 Proofs of the theorems

###  2.1 Proofs for the spherical fractional Brownian motion result

Before we start with the proof of our results on spherical fBM let us first
examine its index set. Looking at $\mathbb{S}^{d-1}$ through the glasses of
Cartesian coordinates, there seems to be no natural way to order its elements
that suits our purposes. Instead, the spherical coordinates naturally suggest
an order on the sphere that is very useful. Let us recall the definition of
_spherical coordinates._ For any point $x\in\mathbb{R}^{d}$ with Euclidean
norm $r\coloneqq r(x)\coloneqq\lVert x\rVert$ its angles
$(\varphi_{1},\ldots,\varphi_{d-2},\theta)\coloneqq(\varphi_{1}(x),\ldots,\varphi_{d-2}(x),\theta(x))\in[0,\pi)^{d-2}\times[0,2\pi)$
are defined (cf. [9]) implicitly by

$\displaystyle\begin{split}x_{k}&=r\cos\varphi_{k}\prod_{j=1}^{k-1}\sin\varphi_{j},\qquad
1\leq k\leq d-2,\\\ x_{d-1}&=r\sin\theta\prod_{j=1}^{d-2}\sin\varphi_{j},\\\
\end{split}$ (8) which then implies $\displaystyle x_{d}$
$\displaystyle=r\cos\theta\prod_{j=1}^{d-2}\sin\varphi_{j}.$

We refer to $(r(x),\varphi_{1}(x),\ldots,\varphi_{d-2}(x),\theta(x))$ as the
spherical coordinates of $x$. At times we allow ourselves to slightly misuse
terminology and refer to the angles
$(\varphi_{1}(x),\ldots,\varphi_{d-2}(x),\theta(x))$ as the spherical
coordinates of $x$, in particular if $r(x)=1$. In what follows we agree on
using the following (lexicographic) order $\leq$ on $\mathbb{S}^{d-1}$. For
$x,x^{\prime}\in\mathbb{S}^{d-1}$ we set $x\leq x^{\prime}$ if one of the
following (mutually exclusive) conditions holds:

  1. i)

$x=x^{\prime}$,

  2. ii)

$\theta(x)<\theta(x^{\prime})$,

  3. iii)

$\theta(x)=\theta(x^{\prime})$, and there exists $1\leq k\leq d-2$ such that
$\varphi_{j}(x)=\varphi_{j}(x^{\prime})$ for all $1\leq j\leq k-1$, and
$\varphi_{k}(x)<\varphi_{k}(x^{\prime})$.

In what follows we will deal with a finite number $U_{1},\ldots,U_{m}$ of,
say, i.i.d. r.v.s sampled according to some continuous distribution with
support $\mathbb{S}^{d-1}$. Consequently, for any pair $U_{i},U_{j}$ all their
angles are distinct a.s. Therefore, their order, i.e. whether $U_{i}\leq
U_{j}$ or $U_{j}\leq U_{i}$, is completely determined by $\theta(U_{i})$ and
$\theta(U_{j})$. This means that the order statistics $U_{m:1}\leq\cdots\leq
U_{m:m}$ again only depends on the angles $\theta(U_{1}),\ldots,\theta(U_{m})$
a.s.

At the heart of our proof of Theorem 2 lies the following proposition on the
increments of spherical fBM that we consider of interest in its own right. We
call a finite permutation $\pi$ a cyclic permutation if there is a
decomposition of $\pi$ into one cycle only. For $m\in\mathbb{N}$ we denote by
$\operatorname{Cyc}(m)$ the set of all cyclic permutations of
$\\{1,\ldots,m\\}$. A finite sequence $(Y_{1},\ldots,Y_{m})$ of r.v.s is
called cyclically exchangeable if for any cyclic permutation
$\pi\in\operatorname{Cyc}(m)$ the random vectors $(Y_{1},\ldots,Y_{m})$ and
$(Y_{\pi(1)},\ldots,Y_{\pi(m)})$ have the same distribution. Intuition
suggests that the increments of spherical fractional Brownian motion $X$
induced by the order statistics of $m$ i.i.d. points $U_{1},\ldots,U_{m}$
sampled from $\mathbb{S}^{d-1}$ uniformly at random should be cyclically
exchangeable. (This is most easily seen first in the special case $d=2$.) Our
next proposition shows that this intuition is in fact true.

######  Proposition 2.

Fix $H\in(0,1/2]$, and let $(X_{t})_{t\in\mathbb{S}^{d-1}}$ denote spherical
fBM with Hurst index $H$ as defined by (4) with the property that $X_{O}=0$
a.s. for some fixed (deterministic) $O\in\mathbb{S}^{d-1}$ with $\theta(O)=0$.
Let $U_{1},U_{2},\ldots$ denote a sequence of i.i.d. r.v.s uniformly
distributed on $\mathbb{S}^{d-1}$. Fix $m\in\mathbb{N}$. Then the sequence of
increments

$\displaystyle(X_{U_{m:k}}-X_{U_{{m:k-1}}})_{k=1}^{m+1}$ (9)

is cyclically exchangeable, where we set $U_{m:0}\coloneqq U_{m:m+1}\coloneqq
O$.

Before we turn to the proof of Proposition 2 we make some further
observations.

######  Lemma 1.

Let $U$ be a point sampled uniformly at random from $\mathbb{S}^{d-1}$. Then
$(\varphi_{1}(U),\ldots,\varphi_{d-2}(U))$ and $\theta(U)$ are independent,
and $\theta(U)$ is uniformly distributed on $(0,2\pi)$.

###### Proof.

Fix some arbitrary $x\in\mathbb{R}^{d}$. Notice from the definition of
spherical coordinates in Equations (8) that the angles of $x$ and $cx$ agree
for any $c>0$, i.e.

$\displaystyle\varphi_{k}(x)$ $\displaystyle=\varphi_{k}\left(cx\right),\qquad
1\leq k\leq d-2,\quad\text{and}\quad\theta(x)=\theta\left(cx\right).$

Moreover, the angles $\varphi_{1},\ldots,\varphi_{d-2}$ depend on
$x_{1},\ldots,x_{d-2}$, but not on $x_{d-1},x_{d}$. The projection of $x$ onto
the hyperplane $x_{1}=\cdots=x_{d-2}=0$ has distance
$\sqrt{x_{d-1}^{2}+x_{d}^{2}}=r\prod_{j=1}^{d-2}\sin\varphi_{j}$ from the
(Euclidian) origin by the Pythagorean identity
$\sin^{2}\varphi+\cos^{2}\varphi=1$, and since $\sin\varphi\geq 0$ for
$\varphi\in[0,\pi)$. Consequently,
$\sin\theta=x_{d-1}/\sqrt{x_{d-1}^{2}+x_{d}^{2}}$, and therefore $\theta$ only
depends on $x_{d-1}$ and $x_{d}$. Recall now that $U=_{d}X/\lVert X\rVert$
with $X=(X_{1},\ldots,X_{d})$ having i.i.d. standard Gaussian coordinates.
This shows that
$(\varphi_{1}(U),\ldots,\varphi_{d-2}(U))=_{d}(\varphi_{1}(X),\ldots,\varphi_{d-2}(X))$
and $\theta(U)=_{d}\theta(X)$ are independent. Moreover, $\theta$ is the angle
enclosed by the positive $x_{d}$-axis and the line through the origin and the
projection $(0,\ldots,0,x_{d-1},x_{d})$ of $x$ onto the hyperplane
$x_{1}=x_{2}=\cdots=x_{d-2}=0$. Since the distribution of $(X_{d-1},X_{d})$ is
invariant under rotations in the plane, $\theta(X)$ is uniformly distributed
on $(0,2\pi)$. ∎

The last lemma allows us to show that the (geodesic) distances between
consecutively ordered i.i.d uniformly distributed points on $\mathbb{S}^{d-1}$
are exchangeable. Recall that a finite sequence $(Y_{1},\ldots,Y_{m})$ of
r.v.s is called exchangeable if for any permutation $\pi$ the random vectors
$(Y_{1},\ldots,Y_{m})$ and $(Y_{\pi(1)},\ldots,Y_{\pi(m)})$ have the same
distribution. Clearly, if $(Y_{1},\ldots,Y_{m})$ is exchangeable than it is
also cyclically exchangeable.

######  Proposition 3.

Fix $m\in\mathbb{N}$. Let $U_{1},\ldots,U_{m}$ be a sequence of i.i.d. r.v.s
with uniform distribution on $\mathbb{S}^{d-1}$. Then the random vector of
geodesic distances

$\displaystyle\left(d(U_{m:k},U_{m:k-1})\right)_{k=1}^{m+1}$

between consecutive order statistics $U_{m:0},U_{m:1},\ldots,U_{m:m+1}$ is
exchangeable, where $U_{m:0}\coloneqq U_{m:m+1}\coloneqq O\in\mathbb{S}^{d-1}$
is a fixed (deterministic) point with $\theta(O)=0$.

###### Proof.

We make use of the fact that the geodesic distance $d(x,x^{\prime})$ between
two points $x,x^{\prime}\in\mathbb{S}^{d-1}$ satisfies $\cos
d(x,x^{\prime})=x\cdot x^{\prime}$, where $x\cdot
x^{\prime}\coloneqq\sum_{k=1}^{d}x_{k}x^{\prime}_{k}$ denotes the scalar
product of $x$ and $x^{\prime}$, cf. [3, p. 141–142]. Thus it suffices to show
that $(U_{m:k}\cdot U_{m:k-1})_{k=1}^{m+1}$ is exchangeable. Now, denoting by
$(x)_{\ell}$ the $\ell$-th component of $x$, by definition of the scalar
product,

$\displaystyle U_{m:k}\cdot U_{m:k-1}=$
$\displaystyle\sum_{\ell=1}^{d}(U_{m:k})_{\ell}(U_{m:k-1})_{\ell},$

and by the implicit definition of spherical coordinates, Equations (8), the
last term equals

$\displaystyle=$
$\displaystyle\sum_{\ell=1}^{d-2}\cos\varphi_{\ell}(U_{m:k})\cos\varphi_{\ell}(U_{m:k-1})\prod_{j=1}^{\ell-1}\sin\varphi_{j}(U_{m:k})\sin\varphi_{j}(U_{m:k-1})$
$\displaystyle+\left(\sin\theta(U_{m:k})\sin\theta(U_{m:k-1})+\cos\theta(U_{m:k})\cos\theta(U_{m:k-1})\right)\prod_{j=1}^{d-2}\sin\varphi_{j}(U_{m:k})\sin\varphi_{j}(U_{m:k-1})$
$\displaystyle=$
$\displaystyle\sum_{\ell=1}^{d-2}\cos\varphi_{\ell}(U_{m:k})\cos\varphi_{\ell}(U_{m:k-1})\prod_{j=1}^{\ell-1}\sin\varphi_{j}(U_{m:k})\sin\varphi_{j}(U_{m:k-1})$
$\displaystyle+\cos(\theta(U_{m:k})-\theta(U_{m:k-1}))\prod_{j=1}^{d-2}\sin\varphi_{j}(U_{m:k})\sin\varphi_{j}(U_{m:k-1}),$

where we used the identity
$\cos(\varphi-\varphi^{\prime})=\cos\varphi\cos\varphi^{\prime}+\sin\varphi\sin\varphi^{\prime}$
in the last equation. By Lemma 1 $(\theta(U_{k}))_{k=1}^{m}$ is an i.i.d.
sequence of uniform $(0,2\pi)$ r.v.s. Consequently, the gaps
$(\theta(U_{m:k})-\theta(U_{m:k-1}))_{k=1}^{m+1}$ rescaled by $1/(2\pi)$ obey
a Dirichlet distribution with all parameters equal to one, cf. Theorem V.2.2
in [15]. In particular, the gaps are exchangeable. Since
$(\varphi_{1}(U_{m:k}),\ldots,\varphi_{d-2}(U_{m:k}))_{k=1}^{m}$ is an i.i.d.
sequence of random variables and again by Lemma 1 and the expansion of the dot
product in the last display we see that $(U_{m:k}\cdot U_{m:k-1})_{k=1}^{m}$
is exchangeable. ∎

We are now ready to show Proposition 2.

###### Proof of Proposition 2.

Set $d_{H}(s,t)\coloneqq(d(s,t))^{2H}$, and define the function
$\tilde{c}\colon(\mathbb{S}^{d-1})^{4}\to\mathbb{R}$ by

$\displaystyle\tilde{c}(s,s^{\prime},t,t^{\prime})$
$\displaystyle\coloneqq\operatorname{Cov}(X_{s^{\prime}}-X_{s},X_{t^{\prime}}-X_{t})$
$\displaystyle=\operatorname{Cov}(X_{s^{\prime}},X_{t^{\prime}})-\operatorname{Cov}(X_{s^{\prime}},X_{t})-\operatorname{Cov}(X_{s},X_{t^{\prime}})+\operatorname{Cov}(X_{s},X_{t})$
$\displaystyle=c(s^{\prime},t^{\prime})-c(s^{\prime},t)-c(s,t^{\prime})+c(s,t)$
$\displaystyle=\frac{1}{2}\left(d_{H}(s^{\prime},t)+d_{H}(s,t^{\prime})-d_{H}(s^{\prime},t^{\prime})-d_{H}(s,t)\right),$

where the covariance function of $X$,

$\displaystyle c(s,t)$
$\displaystyle=\frac{1}{2}(d_{H}(O,s)+d_{H}(O,t)-d_{H}(s,t)),\qquad
s,t\in\mathbb{S}^{d-1},$

can be computed from (4). Conditionally given $U\coloneqq(U_{1},\ldots,U_{m})$
the random vector of increments $(X_{U_{m:k}}-X_{U_{m:k-1}})_{k=1}^{m+1}$ has
characteristic function

$\displaystyle\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}s_{k}\left(X_{U_{m:k}}-X_{U_{m:k-1}}\right)\right)\Big{|}\,U\right]$
$\displaystyle=\exp\left(-\frac{1}{2}s^{\intercal}Rs\right)\qquad(s\in\mathbb{R}^{m+1}),$
(10)

where $R=(R_{ij})$ is the $(m+1)\times(m+1)$ covariance matrix (a random
matrix, as it depends on $(U_{1},\ldots,U_{m})$) defined by
$R_{ij}\coloneqq\tilde{c}(U_{m:i},U_{m:i-1},U_{m:j},U_{m:j-1})$. Let
$\pi\in\operatorname{Cyc}(m+1)$ be an arbitrary but fixed cyclic permutation.
If we can show that the matrices $(R_{ij})_{i,j\in\\{1,\ldots,m+1\\}}$ and
$(R_{\pi(i)\pi(j)})_{i,j\in\\{1,\ldots,m+1\\}}$ have the same distribution,
the claim is proved. Assume without loss of generality that $i\leq j-1$.
Notice that by definition of $\tilde{c}$

$\displaystyle\quad R_{ij}$
$\displaystyle=\frac{1}{2}\left(d_{H}(U_{m:i-1},U_{m:j})+d_{H}(U_{m:i},U_{m:j-1})-d_{H}(U_{m:i-1},U_{m:j-1})-d_{H}(U_{m:i},U_{m:j})\right)$
$\displaystyle=_{d}\frac{1}{2}\bigg{(}d_{H}(U_{m:\pi(i)-1},U_{m:\pi(j)})+d_{H}(U_{m:\pi(i)},U_{m:\pi(j)-1})$
$\displaystyle\qquad-
d_{H}(U_{m:\pi(i)-1},U_{m:\pi(j)-1})-d_{H}(U_{m:\pi(i)},U_{m:\pi(j)})\bigg{)}$
$\displaystyle=R_{\pi(i)\pi(j)},$

where we applied Proposition 3 in the second equality. The identity in the
last math display implies that $(R_{ij})_{i,j\in\\{1,\ldots,m+1\\}}$ has the
same distribution as $(R_{\pi(i)\pi(j)})_{i,j\in\\{1,\ldots,m+1\\}}$ and thus
the claim follows. ∎

The last ingredient in our derivation of the uniform distribution of the
occupation time of spherical fBM, Theorem 2, is a fluctuation result on random
walk bridges. We construct a random walk bridge
$S_{0}=0,S_{1},\ldots,S_{m},S_{m+1}=0$ from the (cyclically exchangeable)
increments in (9) by setting

$\displaystyle S_{k}$
$\displaystyle\coloneqq\sum_{\ell=1}^{k}(X_{U_{m:\ell}}-X_{U_{m:\ell-1}})=X_{U_{m:k}},\qquad
1\leq k\leq m+1.$

The event
$\\{X_{U_{1}}>0,\ldots,X_{U_{m}}>0\\}=\\{X_{U_{m:1}}>0,\ldots,X_{U_{m:m}}>0\\}$
may now be viewed as the event $\\{S_{1}>0,\ldots,S_{m}>0\\}$, i.e. that
$(S_{k})_{k=0}^{m+1}$ is positive (except for its two endpoints
$S_{0}=S_{m+1}=0$). The fluctuation result on random walk bridges (with
cyclically exchangeable increments) will be formulated quite generally and may
be of independent interest. We stress that a similar result by Sparre Andersen
[42], cf. the exposition in [28, Section 1.3], is not sufficient for our
purposes, as it assumes exchangeable increments rather than only cyclically
exchangeable increments. We refer to [4] for similar results, but they do not
apply to bridges, as needed in our case. Our notation partly follows the
exposition in [28].

Fix $m\in\mathbb{N}$. Let $\xi_{1},\ldots,\xi_{m}$ be real r.v.s. Define the
partial sums $(S_{k})_{k=0}^{m}$ by

$\displaystyle S_{0}\coloneqq 0,\qquad S_{k}$
$\displaystyle\coloneqq\xi_{1}+\cdots+\xi_{k},\qquad 1\leq k\leq m.$

We impose the following assumptions on the increments
$\xi_{1},\ldots,\xi_{m}$:

  1. i)

Bridge property: $S_{m}=\xi_{1}+\ldots+\xi_{m}=0$ a.s.

  2. ii)

Cyclic exchangeability: For every cyclic permutation
$\pi\in\operatorname{Cyc}(m)$ of $\\{1,\ldots,m\\}$ we have the distributional
identity

$\displaystyle(\xi_{1},\ldots,\xi_{m})=_{d}(\xi_{\pi(1)},\ldots,\xi_{\pi(m)}).$

  3. iii)

For any $1\leq k\leq m-1$ the distribution of $S_{k}$ has no atoms.

We call $(S_{k})_{k=0}^{m}$ a random walk bridge with cyclically exchangeable
increments.

######  Proposition 4.

Fix $m\in\mathbb{N}$. Let $(S_{k})_{k=0}^{m}$ be a random walk bridge with
cyclically exchangeable increments such that $S_{k}$ has no atoms for any
$1\leq k\leq m$, then

$\displaystyle\mathbb{P}\left\\{S_{1}>0,\ldots,S_{m-1}>0\right\\}$
$\displaystyle=\frac{1}{m}.$

###### Proof.

Define
$\tau:=\min\\{j\in\\{0,\ldots,m-1\\}:S_{j}=\min_{k\in\\{0,\ldots,m-1\\}}S_{k}\\}$
to be the index where the minimum of $S_{0}=0,S_{1},\ldots,S_{m-1}$ is
attained. We show that
$\mathbb{P}\left\\{\tau=j\right\\}=\mathbb{P}\left\\{\tau=0\right\\}$ for all
$j=0,1,\ldots,m-1$. Since
$1=\sum_{j=0}^{m-1}\mathbb{P}\left\\{\tau=j\right\\}$, this will imply
$\mathbb{P}\left\\{\tau=0\right\\}=\frac{1}{m}$. Noting further that
$\mathbb{P}\left\\{\tau=0\right\\}=\mathbb{P}\left\\{S_{1}>0,\ldots,S_{m-1}>0\right\\}$
we will have proved our claim. In order to prove
$\mathbb{P}\left\\{\tau=j\right\\}=\mathbb{P}\left\\{\tau=0\right\\}$ we will
use the cyclic permutation $\pi$ given by

$\pi(i):=\begin{cases}i+j&:i=1,\ldots,m-j,\\\
i-m+j&:i=m-j+1,\ldots,m.\end{cases}$

Note that by cyclic exchangeability

$\mathbb{P}\left\\{\tau=0\right\\}=\mathbb{P}\left\\{0<\sum_{i=1}^{k}\xi_{i},k=1,\ldots,m-1\right\\}=\mathbb{P}\left\\{0<\sum_{i=1}^{k}\xi_{\pi(i)},k=1,\ldots,m-1\right\\}.$

We are going to analyse the conditions $0<\sum_{i=1}^{k}\xi_{\pi(i)}$,
$k=1,\ldots,m-1$, and see that they are equivalent to the event
$\\{\tau=j\\}$. Indeed, first note that for $k=1,\ldots,m-j-1$

$0<\sum_{i=1}^{k}\xi_{\pi(i)}=\sum_{i=1}^{k}\xi_{i+j}=\sum_{i=j+1}^{k+j}\xi_{i}=S_{k+j}-S_{j}.$

This means that $S_{j}<S_{\ell}$ for all $\ell=j+1,\ldots,m-1$. Second, note
that for $k=m-j,\ldots,m-1$

$\displaystyle 0$ $\displaystyle<$
$\displaystyle\sum_{i=1}^{k}\xi_{\pi(i)}=\sum_{i=1}^{m-j}\xi_{\pi(i)}+\sum_{i=m-j+1}^{k}\xi_{\pi(i)}=\sum_{i=1}^{m-j}\xi_{i+j}+\sum_{i=m-j+1}^{k}\xi_{i-m+j}$
$\displaystyle=$
$\displaystyle\sum_{i=j+1}^{m}\xi_{i}+\sum_{i=1}^{k-m+j}\xi_{i}=S_{m}-S_{j}+S_{k-m+j}=0-S_{j}+S_{k-m+j}.$

This means that $S_{j}<S_{\ell}$ for all $\ell=0,\ldots,j-1$. ∎

We are now ready to prove Theorem 2.

###### Proof of Theorem 2.

Recall that the uniform distribution on $(0,1)$ has moment sequence
$\int_{0}^{1}x^{m}dx=\frac{1}{m+1}$, $m\in\mathbb{N}$. For this reason,
according to Proposition 1 it is sufficient to show that
$\mathbb{P}\left\\{X_{U_{1}}>0,\ldots,X_{U_{m}}>0\right\\}=\frac{1}{m+1}$ for
$m\geq 1$. From Proposition 2 we know that we can view
$\\{X_{U_{1}}>0,\ldots,X_{U_{m}}>0\\}=\\{X_{U_{m:1}}>0,\ldots,X_{U_{m:m}}>0\\}$
as the event $\\{S_{1}>0,\ldots,S_{m}>0\\}$, where $S_{1},\ldots,S_{m+1}$ is a
random walk bridge with exchangeable increments defined in (9). The claim thus
follows from Proposition 4. ∎

###  2.2 Proofs for the results on Lévy processes and Lévy bridges

####  2.2.1 Lévy processes

For the proof of Theorem 3 we rely on some well-known results of Spitzer and
some basic facts on Bell polynomials that we now recall. Let
$\xi_{1},\xi_{2},\ldots$ denote a sequence of i.i.d. real-valued random
variables with partial sums $S_{n}\coloneqq\sum_{k=1}^{n}\xi_{k}$,
$n\in\mathbb{N}$. As a consequence of what is now called Spitzer’s identity,
he obtains the following fact.

######  Corollary 2 (Corollary 2 in [44], Theorem 1 in [43]).

The survival probabilities of the partial sums $S_{1},S_{2},\ldots$ have
generating function

$\displaystyle\sum_{k=0}^{\infty}t^{k}\mathbb{P}\left\\{S_{1}\geq
0,\ldots,S_{k}\geq 0\right\\}$
$\displaystyle=\exp\left(\sum_{k=1}^{\infty}\frac{t^{k}}{k}\mathbb{P}\left\\{S_{k}\geq
0\right\\}\right),\qquad|t|<1.$ (11)

This identity still holds when the inequalities in (11) are replaced by strict
inequalities.

Let us rewrite this identity in a more combinatorial form that is better
suited for our purposes. To this end we utilize the Bell polynomials. For any
two sequences of real numbers $v_{\bullet}=(v_{k})_{k\in\mathbb{N}}$ and
$w_{\bullet}=(w_{k})_{k\in\mathbb{N}}$ let

$B_{k}(v_{\bullet},w_{\bullet})\coloneqq\sum_{\ell=1}^{k}v_{\ell}B_{k,\ell}(w_{\bullet}),\quad
k\in\mathbb{N},$

denote the $k$-th complete Bell polynomial (associated with
$(v_{\bullet},w_{\bullet})$), where

$B_{k,\ell}(w_{\bullet})\coloneqq\sum_{\rho\in{\mathscr{P}}_{k,\ell}}\prod_{B\in\rho}w_{\texttt{\\#}B},\quad
1\leq\ell\leq k,$

denotes the $(i,\ell)$-th partial Bell polynomial (associated with
$w_{\bullet}$) and ${\mathscr{P}}_{i,\ell}$ denotes the set of all partitions
of $\\{1,\ldots,i\\}$ that contain exactly $\ell$ blocks. We use the well
known fact, cf. [40, Equation (1.11)], that for any two sequences
$v_{\bullet},w_{\bullet}$ the exponential generating function of the
associated complete Bell polynomials is given by

$\displaystyle\sum_{k=1}^{\infty}B_{k}(v_{\bullet},w_{\bullet})\frac{x^{k}}{k!}=v(w(x)),$
(12)

whenever either of these quantities is well-defined and where
$v(x)\coloneqq\sum_{k\geq 1}v_{k}\frac{x^{k}}{k!}$ and
$w(y)\coloneqq\sum_{k\geq 1}w_{k}\frac{y^{k}}{k!}$ denote the exponential
generating function of $v_{\bullet}=(v_{k})_{k\in\mathbb{N}}$ and
$w_{\bullet}=(w_{k})_{k\in\mathbb{N}}$, respectively. For more information on
Bell polynomials, the interested reader is referred to the lecture notes [40].

We will work with the following reformulation of Spitzer’s result:

######  Corollary 3.

For the survival probability of the sequence of partial sums we obtain

$\displaystyle\mathbb{P}\left\\{S_{1}\geq 0,\ldots,S_{k}\geq 0\right\\}$
$\displaystyle=\frac{1}{k!}\,\sum_{\rho\in{\mathscr{P}}_{k}}\prod_{B\in\rho}(\texttt{\\#}B-1)!\,\mathbb{P}\left\\{S_{\texttt{\\#}B}\geq
0\right\\};$ (13)

and the identity in (13) still holds when the inequalities are replaced by
strict inequalities.

###### Proof.

Define the sequences $v_{\bullet}\coloneqq(v_{k})_{k\in\mathbb{N}}$ and
$w_{\bullet}\coloneqq(w_{k})_{k\in\mathbb{N}}$ by setting

$\displaystyle v_{k}=1\quad\text{and}\quad
w_{k}=(k-1)!\,\mathbb{P}\left\\{S_{k}\geq 0\right\\},\quad k\geq 1.$

With this particular choice for $v_{\bullet},w_{\bullet}$, we find that
$v(x)=e^{x}-1$ and
$w(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k}\mathbb{P}\left\\{S_{k}\geq
0\right\\}$ and we can observe that the right hand side in (11) equals
$v(w(t))+1$. For $v(w(t))$ we can use the expansion (12). Comparing this to
the left hand side of (11) we find that

$\displaystyle\mathbb{P}\left\\{S_{1}\geq 0,\ldots,S_{k}\geq 0\right\\}$
$\displaystyle=\frac{1}{k!}\,B_{k}(v_{\bullet},w_{\bullet})$
$\displaystyle=\frac{1}{k!}\,\sum_{\ell=1}^{k}B_{k,\ell}(w_{\bullet})$
$\displaystyle=\frac{1}{k!}\,\sum_{\ell=1}^{k}\sum_{\rho\in{\mathscr{P}}_{k,\ell}}\prod_{B\in\rho}w_{\texttt{\\#}B}$
$\displaystyle=\frac{1}{k!}\,\sum_{\rho\in{\mathscr{P}}_{k}}\prod_{B\in\rho}(\texttt{\\#}B-1)!\,\mathbb{P}\left\\{S_{\texttt{\\#}B}\geq
0\right\\},$

as claimed, where we used that ${\mathscr{P}}_{k}$ is the disjoint union of
the ${\mathscr{P}}_{k,\ell}$ for $\ell=1,\ldots,k$. The claim for strict
inequalities follows from the same proof using strict inequalities at all
places. ∎

We are now prepared to prove Theorem 3. The proof idea is similar to our proof
of the Brownian (or stable Lévy) arcsine law, Theorem 1. The moments are
rewritten as persistence probabilities at random times that come from the
normalised jump-times of an independent Poisson process. As we are not
assuming the scaling property we cannot scale out the terminal time. This
forces us to work with a Lévy process extension of the sampling formula at
Poisson times. Combined with the above variant of Spitzer’s identity the claim
can be deduced.

###### Proof of Theorem 3.

For the proof it is convenient to turn to a variant of the sampling method.
Specifically, instead of starting with the $m$-th moment of the sojourn time
of $X$ up until time $t$, we focus instead on its Laplace transform. We use
the Poisson sampling formula

$\displaystyle F(q)$
$\displaystyle\coloneqq\int_{0}^{\infty}e^{-qt}\mathbb{E}[A_{t}^{m}]dt=\frac{m!}{q^{m+1}}\mathbb{P}\left\\{X_{T_{1}^{(q)}}>0,\ldots,X_{T_{m}^{(q)}}>0\right\\},\qquad
m\in\mathbb{N},q>0,$ (14)

where $(T_{k}^{(q)})$ denotes the sequence of waiting times in a standard
Poisson process of intensity $q>0$ independent of $X$. The formula was for
instance used by [24] but most certainly also appeared elsewhere. Here is a
quick proof for completeness. First note that the Markov property of the
random walk $X_{T_{1}^{(q)}},X_{T_{2}^{(q)}},\ldots$ yields

$\displaystyle\mathbb{P}\left\\{X_{T_{1}^{(q)}}>0,\ldots,X_{T_{m}^{(q)}}>0\right\\}$
$\displaystyle=\int_{x_{1}>0}\cdots\int_{x_{m}>0}P(dx_{1})\cdots
P(dx_{m}-x_{m-1})$
$\displaystyle=q^{m}\int_{x_{1}>0}\cdots\int_{x_{m}>0}U^{q}(dx_{1})\cdots
U^{q}(dx_{m}-x_{m-1}),$

where $P(A):=\mathbb{P}\left\\{X_{T_{1}^{(q)}}\in A\right\\}$ is the jump
distribution of the random walk and the quantity
$U^{q}(A):=\int_{0}^{\infty}e^{-qt}\mathbb{P}\left\\{X_{t}\in A\right\\}dt$ is
the so-called $q$-potential measure of $X$. Here we used that $P=qU^{q}$. To
rewrite $F$ denote by $\bar{X}$ the killed process, i.e. the Lévy process
killed at an independent exponential time with parameter $q$. By $\bar{p}$
denote the transition kernel of the killed process. Then Fubini, monotone
convergence, the Markov property, and symmetry of the integrand yield

$\displaystyle\quad\int_{0}^{\infty}e^{-qt}\mathbb{E}[A_{t}^{m}]dt$
$\displaystyle=\frac{1}{q}\int_{0}^{\infty}\cdots\int_{0}^{\infty}\mathbb{E}[\mathbf{1}_{\bar{X}_{s_{1}}>0}\cdots\mathbf{1}_{\bar{X}_{s_{m}>0}}]ds_{m}\cdots
ds_{s}$
$\displaystyle=\frac{1}{q}\lim_{N_{1},...,N_{m}\to\infty}\int_{0}^{N_{1}}\cdots\int_{0}^{N_{m}}\mathbb{E}[\mathbf{1}_{\bar{X}_{s_{1}}>0}\cdots\mathbf{1}_{\bar{X}_{s_{m}>0}}]ds_{m}\cdots
ds_{1}$
$\displaystyle=\frac{m!}{q}\lim_{N_{1},...,N_{m}\to\infty}\int_{0}^{N_{1}}\int_{s_{1}}^{N_{2}}\cdots\int_{s_{m-1}}^{N_{m}}\mathbb{E}[\mathbf{1}_{\bar{X}_{s_{1}}>0}\cdots\mathbf{1}_{\bar{X}_{s_{m}>0}}]ds_{m}\cdots
ds_{1}$
$\displaystyle=\frac{m!}{q}\lim_{N_{1},...,N_{m}\to\infty}\int_{0}^{N_{1}}\int_{s_{1}}^{N_{2}}\cdots\int_{s_{m-1}}^{N_{m}}$
$\displaystyle\qquad\times\int_{x_{1}>0}\cdots\int_{x_{m}>0}\bar{p}_{s_{1}}(dx_{1})\cdots\bar{p}_{s_{m}-s_{m-1}}(dx_{m}-x_{m-1})ds_{m}\cdots
ds_{1}$
$\displaystyle=\frac{m!}{q}\lim_{N_{1},...,N_{m}\to\infty}\int_{0}^{N_{1}}\int_{0}^{N_{2}-s_{1}}\cdots\int_{0}^{N_{m}-s_{m-1}}$
$\displaystyle\qquad\times\int_{x_{1}>0}\cdots\int_{x_{m}>0}\bar{p}_{s_{1}}(dx_{1})\cdots\bar{p}_{s_{m}}(dx_{m}-x_{m-1})ds_{m}\cdots
ds_{1}$
$\displaystyle=\frac{m!}{q}\int_{x_{1}>0}\cdots\int_{x_{m}>0}U^{q}(dx_{1})\cdots
U^{q}(dx_{m}-x_{m-1}).$

Combining the two previous displays yields (14). Combined with Corollary 3 we
obtain

$\displaystyle F(q)$
$\displaystyle=\frac{1}{q^{m+1}}\sum_{\rho\in{\mathscr{P}}_{m}}\prod_{B\in\rho}(\texttt{\\#}B-1)!\,\mathbb{P}\left\\{S_{\texttt{\\#}B}>0\right\\},$

with $S_{k}\coloneqq X_{T_{k}^{(q)}}$. Since $T_{k}^{(q)}$ is gamma
distributed with parameters $k,q$; i.e. with density $s\mapsto
q^{k}s^{k-1}e^{-qs}/(k-1)!\mathbf{1}\\{s>0\\}$, and setting
$p_{s}\coloneqq\mathbb{P}\left\\{X_{s}>0\right\\}$ we conclude

$\displaystyle F(q)$
$\displaystyle=\frac{1}{q}\sum_{\rho\in{\mathscr{P}}_{m}}\prod_{B\in\rho}\int_{0}^{\infty}s^{\texttt{\\#}B-1}p_{s}e^{-qs}ds.$

The integral on the right-hand side is a Laplace transform, and we will denote
the Laplace transform of a function $f:[0,\infty)\to\mathbb{R}$ by
$(\mathcal{L}f)(q):=\int_{0}^{\infty}f(s)e^{-qs}ds$. Using basic properties of
Laplace transforms we can thus write

$\displaystyle F(q)$
$\displaystyle=\frac{1}{q}\sum_{\rho\in{\mathscr{P}}_{m}}\prod_{B\in\rho}\mathcal{L}\left(s^{\texttt{\\#}B-1}p_{s}\right)(q)$
$\displaystyle=\frac{1}{q}\mathcal{L}\left(\sum_{\rho\in{\mathscr{P}}_{m}}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}s^{\texttt{\\#}B-1}p_{s}\right)(q)$
$\displaystyle=\mathcal{L}\left(1\ast\sum_{\rho\in{\mathscr{P}}_{m}}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}s^{\texttt{\\#}B-1}p_{s}\right)(q).$

From this calculation we find

$\displaystyle\mathbb{E}[A_{t}^{m}]$
$\displaystyle=\left(1\ast\sum_{\rho\in{\mathscr{P}}_{m}}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}s^{\texttt{\\#}B-1}p_{s}\right)(t),$

which is the claim. ∎

Before we turn to the proof of Corollary 1 we provide a helpful lemma.

######  Lemma 2.

Fix $m\in\mathbb{N}$ and positive real numbers $a_{1},\ldots,a_{m}>0.$ Then,
for $t>0$,

$\displaystyle\left(\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{k=1}^{m}\left(s^{a_{k}-1}\right)\right)(t)=\Gamma\left(\sum_{k=1}^{m}a_{k}\right)^{-1}\prod_{k=1}^{m}\Gamma(a_{k})\cdot
t^{\sum_{k=1}^{m}a_{k}-1}.$

###### Proof.

We show the claim by induction on $m$. It is clear that the claim holds for
$m=1.$ Assume now the claim is true for some positive integer
$m\in\mathbb{N}.$ Then, using the induction hypothesis in the second equality,

$\displaystyle\left(\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{k=1}^{m+1}\left(s^{a_{k}-1}\right)\right)(t)$
$\displaystyle=\left(\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{k=1}^{m}\left(s^{a_{k}-1}\right)\ast
s^{a_{m+1}-1}\right)(t)$
$\displaystyle=\left(\Gamma\left(\sum_{k=1}^{m}a_{k}\right)^{-1}\prod_{k=1}^{m}\Gamma(a_{k})\cdot
s^{\sum_{k=1}^{m}a_{k}-1}\ast s^{a_{m+1}-1}\right)(t)$
$\displaystyle=\Gamma\left(\sum_{k=1}^{m}a_{k}\right)^{-1}\prod_{k=1}^{m}\Gamma(a_{k})\int_{0}^{t}(t-s)^{\sum_{k=1}^{m}a_{k}-1}s^{a_{m+1}-1}ds$
$\displaystyle=\Gamma\left(\sum_{k=1}^{m}a_{k}\right)^{-1}\prod_{k=1}^{m}\Gamma(a_{k})\cdot
t^{\sum_{k=1}^{m+1}a_{k}-1}\int_{0}^{1}(1-s)^{\sum_{k=1}^{m}a_{k}-1}s^{a_{m+1}-1}ds$
$\displaystyle=\Gamma\left(\sum_{k=1}^{m}a_{k}\right)^{-1}\prod_{k=1}^{m}\Gamma(a_{k})\cdot
t^{\sum_{k=1}^{m+1}a_{k}-1}\frac{\Gamma(\sum_{k=1}^{m}a_{k})\Gamma(a_{m+1})}{\Gamma(\sum_{k=1}^{m+1}a_{k})}$
$\displaystyle=\Gamma\left(\sum_{k=1}^{m+1}a_{k}\right)^{-1}\prod_{k=1}^{m+1}\Gamma(a_{k})\cdot
t^{\sum_{k=1}^{m+1}a_{k}-1},$

where we transformed coordinates in the integral in the fourth equality and
used the beta integral in the fifth equality. ∎

###### Proof of Corollary 1.

Assume that $\mathbb{P}\left\\{X_{t}>0\right\\}=c\in(0,1)$ for all $t>0$. By
(6) we have

$\displaystyle\mathbb{E}[A_{t}^{m}]=\sum_{\rho\in{\mathscr{P}}_{m}}\int_{0}^{t}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}\left(u^{\texttt{\\#}B-1}\mathbb{P}\left\\{X_{u}>0\right\\}\right)(s)ds=\sum_{\rho\in{\mathscr{P}}_{m}}c^{\texttt{\\#}\rho}\int_{0}^{t}\mathop{\scalebox{1.5}{\raisebox{-0.77498pt}{$\ast$}}}_{B\in\rho}\left(u^{\texttt{\\#}B-1}\right)(s)ds$

and applying Lemma 2 the right hand side equals

$\displaystyle\quad\frac{1}{\Gamma(m)}\sum_{\rho\in{\mathscr{P}}_{m}}c^{\texttt{\\#}\rho}\prod_{B\in\rho}(\texttt{\\#}B-1)!\cdot\int_{0}^{t}s^{m-1}ds$
$\displaystyle=\frac{t^{m}}{m!}\sum_{\rho\in{\mathscr{P}}_{m}}c^{\texttt{\\#}\rho}\prod_{B\in\rho}(\texttt{\\#}B-1)!$
$\displaystyle=\frac{t^{m}}{m!}\sum_{b=1}^{m}c^{b}\sum_{\rho\in{\mathscr{P}}_{m,b}}\prod_{B\in\rho}(\texttt{\\#}B-1)!.$

Notice that $(k-1)!$ is the number of cyclic permutations of $k$ elements,
thus $\sum_{\rho\in{\mathscr{P}}_{m,b}}\prod_{B\in\rho}(\texttt{\\#}B-1)!$ is
the number of permutations of $\\{1,\ldots,m\\}$ with $b$ blocks, also known
as the $(n,b)$-th unsigned Stirling number, which we denote by
$\genfrac{[}{]}{0.0pt}{}{n}{k}$. Recall that the unsigned Stirling numbers may
also be written as

$\displaystyle
c^{\overline{m}}=\sum_{b=0}^{m}\genfrac{[}{]}{0.0pt}{}{m}{k}c^{b},$

where $c^{\overline{m}}\coloneqq c(c+1)\cdots(c+m-1)$ and we recall that
$\genfrac{[}{]}{0.0pt}{}{m}{0}=0$ if $m>0$. Putting everything together, we
conclude that $\mathbb{E}[A_{t}^{m}]=t^{m}c^{\overline{m}}/m!$, which is the
$m$-th moment of $tA$ where the distribution of $A$ is the arcsine law on
$(0,1)$ with parameter $c$. Since this distribution is uniquely determined by
its moments, we are done with the proof of the first implication.

To see the opposite implication, assume that $t^{-1}A_{t}$ is arcsine
distributed with parameter $c$. In particular, the first moment has to have
the form $\mathbb{E}[A_{t}]=ct$ for all $t>0$. Using the first moment formula
(7), we know that the positivity function of the Lévy process is constant. ∎

####  2.2.2 Lévy bridges

We now offer an elementary derivation of the occupation time distribution of
Lévy bridges. In some sense the proof is a simpler version of our proof for
the theorem on the occupation time of spherical Brownian motion. Here the
proof can rely on exchangeability whereas the spherical situation is more
subtle and requires to work with cyclic exchangeability only. The proof is
based on the so-called Baxter’s combinatorial lemma for permutations of
vectors. Fix $n\in\mathbb{N}$ and $z_{1},\ldots,z_{n}\in\mathbb{R}^{2}$. For
any permutation $\pi\in\mathfrak{\operatorname{Cyc}}(m)$ define the
corresponding partial sums We now offer an elementary derivation of the
occupation time distribution of Lévy bridges. In some sense the proof is a
simpler version of our proof for the theorem on the occupation time of
spherical Brownian motion. Here the proof can rely on exchangeability whereas
the spherical situation is more subtle and requires to work with cyclic
exchangeability only. The proof is based on the so-called Baxter’s
combinatorial lemma for permutations of vectors. Fix $n\in\mathbb{N}$ and
$z_{1},\ldots,z_{n}\in\mathbb{R}^{2}$. For any permutation
$\pi\in\mathfrak{\operatorname{Cyc}}(m)$ define the corresponding partial sums

$\displaystyle s_{0}[\pi]\coloneqq
0,\,\,\,s_{k}[\pi]\coloneqq\sum_{\ell=1}^{k}z_{\pi(\ell)},\quad 1\leq k\leq
n.$

For the sake of brevity we set $s_{k}\coloneqq
s_{k}[id_{n}]=\sum_{\ell=1}^{k}z_{\ell}$ for $0\leq k\leq n$, where
$id_{n}\colon[n]\to[n]$ denotes the identity permutation $id_{n}(k)=k$. Notice
that $s_{n}[\pi]=\sum_{\ell=1}^{n}z_{\pi(\ell)}=s_{n}$ does not depend on
$\pi$. Moreover, for any subset $M\subseteq\\{1,\ldots,n\\}$ let
$s_{M}\coloneqq\sum_{k\in M}z_{k}$. Following Baxter, we call
$z_{1},\ldots,z_{n}$ skew if the fact that $z_{M}$ and $z_{M^{\prime}}$ lie on
a common line (i.e. there exists a real $c\neq 0$ such that
$z_{M}=cz_{M^{\prime}}$) implies $M=M^{\prime}$. Any point
$z\in\mathbb{R}^{2}\setminus\\{0\\}$ together with the origin
$0\in\mathbb{R}^{2}$ defines a line in the plane containing $0$ and $z$ that
divides the plane into two half spaces. We call these the left and right half
space (in clockwise orientation when the direction of the line is induced by
moving from $0$ to $z$) induced by $z$. Let $H(z)$ denote the left half space
induced by $z$ and including the line containing $z$. Then Baxter’s
combinatorial lemma may be stated as follows.

######  Lemma 3 (Baxter’s combinatorial lemma, cf. Lemma 1 in [2]).

Fix $n\in\mathbb{N}$ and assume that $z_{1},\ldots,z_{n}\in\mathbb{R}^{2}$
that are skew. Then

$\displaystyle\texttt{\\#}\left\\{\pi\in\operatorname{Cyc}(n)\colon\\{s_{k}[\pi]\\}_{k=1}^{n}\subseteq
H(s_{n})\right\\}$ $\displaystyle=1.$

In words, there is precisely one cyclic permutation $\pi$ of
$z_{1},\ldots,z_{n}$ such that the corresponding partial sums
$s_{1}[\pi],\ldots,s_{n}[\pi]$ all lie in the left half space $H(s_{n})$ of
$s_{n}$.

We note that Baxter’s lemma is rather elementary to prove, the proof is a
clever few line computation. The way in which we will apply Baxter’s lemma is
the following. If the partial sums $(s_{k})_{k=1}^{n}$ induced by skew points
$z_{1},\ldots,z_{n}\in\mathbb{R}^{2}$ are such that $s_{n}$ lies on the
positive $x$-axis, then $H(s_{n})$ is the upper half-plane. Thus Baxter’s
lemma is well suited to approach persistence probabilities from a
combinatorial perspective.

Moreover, we make the following observation. Consider a real function
$f:\mathbb{R}\to\mathbb{R}$ such that $f(0)=0$ and fix
$0=t_{0}<t_{1}<\cdots<t_{n}<t_{n+1}=1$. Define the function
$\mathring{f}\colon[0,1]\to\mathbb{R}$ by setting $\mathring{f}(t)\coloneqq
f(t)-tf(1)$, and call $\mathring{f}$ the bridge induced by $f$.

######  Lemma 4.

We have $\mathring{f}(t_{k})>0$ for all $1\leq k\leq n$ if and only if the
points $(t_{k},f(t_{k}))$, $1\leq k\leq n+1$, all lie in the left half plane
induced by $(1,f(1))$ (which is $H((1,f(1))$).

###### Proof.

The line through the origin containing $(1,f(1))$ may be parameterised as
$\\{(t,tf(1))\colon t\in\mathbb{R}\\}$. Inserting the arguments $t_{k}$ gives
the claim. ∎

We are now ready to prove Theorem 4. Proposition 1 reduces the problem to
persistence probabilities of Lévy bridges, which is then reformulated with the
help of Lemma 4. Baxter’s lemma then simplifies the expressions to the moments
of the uniform distribution.

###### Proof of Theorem 4.

Recall that the uniform distribution on $(0,1)$ is uniquely identified by its
moment sequence $\int_{0}^{1}x^{m}dx=\frac{1}{m+1}$, $m\geq 1$. By Proposition
1 it suffices to show that, for any $m\in\mathbb{N}$,

$\displaystyle\mathbb{P}\left\\{\mathring{X}_{U_{m:1}}>0,\ldots,\mathring{X}_{U_{m:m}}>0\right\\}=\frac{1}{m+1},$

where $U_{1},U_{2},\ldots$ is an i.i.d. sequence of uniform $(0,1)$ random
variables independent of $(X_{t})$ and $U_{m:1}\leq\ldots\leq U_{m:m}$ is the
corresponding order statistics. We further set $U_{m:0}:=0$ and
$U_{m:m+1}:=1$.

Step 1: We first show that the random vector
$(X_{U_{m:k}}-X_{U_{m:k-1}})_{k=1}^{m+1}$ is exchangeable. For this it
suffices to show that for any permutation $\pi$ of $\\{1,\ldots,m+1\\}$ and
for any $t_{1},\ldots,t_{m+1}\in\mathbb{R}$

$\displaystyle\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}t_{k}(X_{U_{m:k}}-X_{U_{m:k-1}})\right)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}t_{k}(X_{U_{m:\pi(k)}}-X_{U_{m:\pi(k)-1}})\right)\right].$
(15)

Let $B\coloneqq\\{x\in\mathbb{R}\colon\lVert x\rVert\leq 1\\}$ denote the unit
ball in $\mathbb{R}$, and let $(a,\gamma,\nu)$ denote the generating triplet
of the law of $X_{1}$, where $a\geq 0$, $\gamma\in\mathbb{R}$, and $\nu$ is a
measure on $\mathbb{R}$ with $\nu(\\{0\\})=0$ and $\int_{\mathbb{R}}(\lvert
x\rvert\wedge 1)\nu(dx)<\infty$. Conditionally given
$U\coloneqq(U_{1},\ldots,U_{m})$, and using the fact that $X$ has independent
increments, we have

$\displaystyle\quad\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}t_{k}(X_{U_{m:k}}-X_{U_{m:k-1}})\right)\middle|U\right]$
$\displaystyle=\prod_{k=1}^{m+1}\mathbb{E}\left[\exp(it_{k}(X_{U_{m:k}}-X_{U_{m:k-1}}))\middle|U\right]$
$\displaystyle=\prod_{k=1}^{m+1}\exp\left((U_{m:k}-U_{m:k-1})\left(-\frac{1}{2}t_{k}^{2}a+i\gamma
t_{k}+\int_{\mathbb{R}}(e^{it_{k}x}-1-izx\mathbf{1}_{B}(x))\nu(dx)\right)\right),$
(16)

where in the last step we used the well-known Lévy-Khinchine representation of
an infinitely divisible distribution, cf. [41, Theorem 8.1]. Using the fact
that the $m+1$ gaps
$(U_{m:1}-U_{m:0},U_{m:2}-U_{m:1},\ldots,U_{m:m+1}-U_{m:m})$ induced by
$U_{1},\ldots,U_{m}$ obey a Dirichlet distribution with parameters
$1,\ldots,1$, and thus constitute an exchangeable random vector, we obtain
from (16) that

$\displaystyle\mathbb{E}\,\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}t_{k}(X_{U_{m:k}}-X_{U_{m:k-1}})\right)\middle|U\right]$
$\displaystyle=\mathbb{E}\,\mathbb{E}\left[\exp\left(i\sum_{k=1}^{m+1}t_{k}(X_{U_{m:\pi(k)}}-X_{U_{m:\pi(k)-1}})\right)\middle|U\right].$

By Fubini’s theorem, this shows (15). We can now come to the main argument of
the proof.

Step 2: Set

$S_{k}:=\left(U_{m:k},X_{U_{m:k}}\right)=\sum_{i=1}^{k}\left(U_{m:i}-U_{m:i-1},X_{U_{m:i}}-X_{U_{m:i-1}}\right),\quad
k=0,\ldots,m+1.$

Note that the events $\\{(U_{m:1},X_{U_{m:1}}),\ldots,(U_{m:m},X_{U_{m:m}})\in
H((1,X_{1}))\\}=\\{S_{1},\ldots,S_{m}\in H((1,X_{1}))\\}$ and
$\\{\mathring{X}_{U_{m:1}}>0,\ldots,\mathring{X}_{U_{m:1}}>0\\}$ are equal by
Lemma 4. Using the cyclic exchangeability in this first step (note that
$X_{1}$ is not altered by the permutations), we obtain

$\displaystyle\mathbb{P}\left\\{S_{1},\ldots,S_{m}\in H((1,X_{1}))\right\\}$
$\displaystyle=\frac{1}{m+1}\sum_{\pi\in\operatorname{Cyc}(m+1)}\mathbb{P}\left\\{S_{1}[\pi],\ldots,S_{m}[\pi]\in
H((1,X_{1}))\right\\}$
$\displaystyle=\mathbb{E}\Big{[}\frac{1}{m+1}\sum_{\pi\in\operatorname{Cyc}(m+1)}\mathbf{1}\\{S_{1}[\pi],\ldots,S_{m}[\pi]\in
H((1,X_{1}))\\}\Big{]}$ $\displaystyle=\frac{1}{m+1},$

where in the second to last line the sum equals one a.s. by Baxter’s
combinatorial lemma. Here, we used that the points
$(U_{m:k}-U_{m:k-1},X_{U_{m:k}}-X_{m:k-1})$ are almost surely skew in the
application of Baxter’s combinatorial lemma, which is due to the assumption
that $X_{1}$ has no atoms.∎

######  Remark 2.

Note that the main argument applies to all stochastic processes whose
increments over gaps induced by i.i.d. sampled times are exchangeable.

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