A conformal Skorokhod embedding

TL;DR

This paper introduces a new complex-analytic method for the Skorokhod embedding theorem, allowing the construction of barriers that produce Brownian motion hitting points with any prescribed centered distribution with finite variance.

Contribution

It provides a constructive, explicit approach to the Skorokhod embedding problem using complex analysis, expanding the toolkit for stochastic process embedding techniques.

Findings

Constructed barriers for prescribed distributions

Provided explicit descriptions of barriers

Offered a new proof of the Skorokhod embedding theorem

Abstract

Start a planar Brownian motion and let it run until it hits some given barrier. We show that the barrier may be crafted so that the x coordinate at the hitting time has any prescribed centered distribution with finite variance. This provides a new, complex-analytic proof of the Skorokhod embedding theorem. Our method is constructive and can give an explicit description of the barrier.

Equations

$$T\left(X_t,\Omega \right)=\inf \left\{t>0\mid X_t\notin \Omega \right\}.$$

$$G\left(y\right)=\inf \left\{x\in \mathbb{R}\mid F\left(x\right)\ge y\right\}.$$

$${\int }_0^1{G}^2\left(y\right)dy$$

$$\phi \left(\theta \right)=G\left(\frac{\left|\theta \right|}{\pi }\right).$$

$$\psi \left(z\right)=\sum _{n=0}^{\infty }\hat{\phi }\left(n\right)z^n.$$

$$\mathrm{Re}\psi \left(z\right)$$

$$=\mathrm{Re}\sum _{n=0}^{\infty }\hat{\phi }\left(n\right)e^{i\theta n}$$

$$=\sum _{n=0}^{\infty }\hat{\phi }\left(n\right)\cos \theta n.$$

$$\left|\psi \left(z\right)-{\psi }_k\left(z\right)\right|$$

$$\le \sum _{n=0}^{\infty }\left|\left(\hat{\phi }\left(n\right)-{\hat{\phi }}_k\left(n\right)\right)r^n{e}^{in\theta }\right|$$

$$\le \underset{n}{\sup }\left|\left(\hat{\phi }\left(n\right)-{\hat{\phi }}_k\left(n\right)\right)\right|\sum _{n=0}^{\infty }{r}^n.$$

$$\left|\hat{\phi }\left(n\right)-{\hat{\phi }}_k\left(n\right)\right|$$

$$\le \frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left|\phi \left(\theta \right)-{\phi }_k\left(\theta \right)\right|d\theta ,$$

$$f\left(X_t\right)=Y_{\zeta \left(t\right)}$$

$$\zeta \left(t\right)={\int }_0^t{\left|f^{\prime }\left(X_s\right)\right|}^{2}ds.$$

$${\mathbb{E}}_z\left(T\left(Y_t,\Omega \right)\right)$$

$$\mathbb{E}{\left(Y_{T\left(Y_t,\Omega \right)}^{\left(1\right)}\right)}^2-\mathbb{E}T\left(Y_t,\Omega \right)=0,$$

$$\mathrm{Re}\psi \left(e^{i\theta }\right)$$

$$\mathrm{Im}\psi \left(e^{i\theta }\right)$$

$$\left(Hu\right)\left(x\right)=\frac{1}{\pi }PV{\int }_{-\infty }^{\infty }\frac{u\left(\tau \right)}{t-\tau }d\tau ,$$

$$\mu \left(x\right)=\sum _i{p}_i{\delta }_{x_i}\left(x\right).$$

$$F\left(x\right)=\sum _i{p}_i{1}_{x\ge x_i},$$

$$\phi \left(\theta \right)=\sum _i{\alpha }_i{1}_{|\theta |\ge {\theta }_i}$$

$$y_i=\frac{1}{\pi }\sum _j{\alpha }_j\log \left|\frac{\sin \left(\frac{{\theta }_j}{2}-\frac{x}{2}\right)}{\sin \left(\frac{{\theta }_j}{2}+\frac{x}{2}\right)}\right|,$$

$$\sum _i{\alpha }_i\left(\cot \left(\frac{{\theta }_i}{2}-\frac{x}{2}\right)+\cot \left(\frac{{\theta }_i}{2}+\frac{x}{2}\right)\right)=0.$$

$$u\left(x\right)=\{\begin{array}{ll}0& 0<\left|x\right|<{\theta }_0\\ 1& {\theta }_0\le \left|x\right|\le \pi \end{array}$$

$$\left(Hu\right)\left(x\right)=\frac{1}{\pi }\log \left|\frac{\sin \left(\frac{{\theta }_0}{2}-\frac{x}{2}\right)}{\sin \left(\frac{{\theta }_0}{2}+\frac{x}{2}\right)}\right|.$$

$${\sqcap }_{\alpha }\left(x\right)=\{\begin{array}{ll}1& \left|x\right|<\alpha \\ 0& o.w.\end{array}.$$

$$u\left(x\right)=\sum _{k=-\infty }^{\infty }{\sqcap }_{\pi -{\theta }_0}\left(x+\pi +2k\pi \right).$$

$$\left(H{\sqcap }_{\alpha }\right)\left(x\right)=\frac{1}{\pi }\log \left|\frac{\alpha +x}{\alpha -x}\right|.$$

$$\left(Hu\right)\left(x\right)$$

$$=\frac{1}{\pi }\log \prod _{k=-\infty }^{\infty }\left|1+\frac{\frac{1}{2\pi }\left(2\pi -2{\theta }_0\right)}{k+\frac{1}{2\pi }\left({\theta }_0+x\right)}\right|.$$

$$\prod _{k=-\infty }^{\infty }\left|1+\frac{b}{k-a}\right|,$$

$$\frac{\sin \left(\pi \left(a-b\right)\right)}{\sin \left(\pi a\right)}$$

Full text

1 Introduction

The Skorokhod embedding problem asks the following: Given a Brownian motion $X_{t}$ and a probability distribution $\mu$ with expectation zero and finite variance, find a stopping time $T$ so that $X_{t}\sim\mu$ and $\mathbb{E}T<\infty$.

There have been numerous solutions to this formulation and to several variations and generalizations over the years. See Obłój’s extensive survey [8] for a detailed account of the problem, its characteristics and applications.

One important solution is given by Root [9], who sets $T$ as the first time that the graph $\left(t,X_{t}\right)$ of the Brownian motion hits some barrier $\Omega\subseteq\mathbb{R}^{+}\times\mathbb{R}$. Root’s solution does not use any additional randomness. Finding out the barrier $\Omega$, however, is often a difficult task, and not many explicit solutions are known (but see [4] for constructions relying on solutions to PDEs).

In this paper, we present a new solution to the Skorokhod embedding problem. Our method is similar to Root’s, in that the stopping time is the first hitting time of some barrier by a Brownian motion. The method requires additional randomness in the form of another independent Brownian motion, but can offer tractable analytic expressions for calculating the shape of the barrier explicitly.

For a domain $\Omega\subseteq\mathbb{R}^{2}$ and a planar Brownian motion $X_{t}=\left(X_{t}^{\left(1\right)},X_{t}^{\left(2\right)}\right)$ with $X_{0}\in\Omega$, let $T\left(X_{t},\Omega\right)$ be the first time that $X_{t}$ exits the domain $\Omega$,

[TABLE]

Theorem 1**.**

Let $\mu$ be a probability distribution on $\mathbb{R}$ with zero expectation and finite variance. There exists a simply connected domain $\Omega\subseteq\mathbb{R}^{2}$ containing the origin such that if $Y_{t}$ is a standard planar Brownian, then $Y_{T\left(Y_{t},\Omega\right)}^{\left(1\right)}$ has distribution $\mu$.

The proof of Theorem 1 is given in the next section. Section 3 gives properties and examples of $\Omega$. Finally, in Section 4 we exhibit some open questions.

2 Proof of theorem

Proof.

We identify $\mathbb{R}^{2}$ with the complex plane $\mathbb{C}$. Denote the open unit disc by $D=\left\{z\in\mathbb{C}\mid\left|z\right|<1\right\}$ and the unit circumference by $\partial D=\left\{z\in\mathbb{C}\mid\left|z\right|=1\right\}$.

Denote by $F:\mathbb{R}\to\left[0,1\right]$ the cumulative distribution function of $\mu$, and by $G:\left(0,1\right)\to\mathbb{R}$ its generalized inverse:

[TABLE]

Observe that both $F$ and $G$ are monotone increasing. Despite the fact that $G$ may diverge at [math] and $1$, it is still in $\mathrm{L}^{2}$: If $X$ is a uniform random variable on $\left[0,1\right]$ then $G\left(X\right)$ distributes as $\mu$, and so

[TABLE]

Define the function $\varphi:\left(-\pi,\pi\right)\backslash\left\{0\right\}\to\mathbb{R}$ by

[TABLE]

The function $\varphi$ may be viewed as a periodic function on $\mathbb{R}$ with period $2\pi$, with the possibility of it diverging at integer multiples of $\pi$. It is an even function in $\mathrm{L}^{2}$, and thus has a Fourier series representation containing only cosines. Denote the $n$-th Fourier coefficient of $\varphi$ by $\hat{\varphi}\left(n\right)$, and recall that by Carleson’s theorem [2] for an $\mathrm{L}^{2}$ function, the Fourier representation $\sum_{n=0}^{\infty}\hat{\varphi}\left(n\right)\cos\left(\theta n\right)$ agrees with $\varphi$ almost everywhere.

Let $\psi:D\to\mathbb{C}$ be the complex function defined by

[TABLE]

The function $\psi$ may not necessarily be defined for points on the unit circle, but its real part is well-defined for almost all such points. Indeed, let $z=e^{i\theta}$ with $\theta\in\left[-\pi,\pi\right]$, then

[TABLE]

So on the unit circle, the real part of $\psi\left(e^{i\theta}\right)$ agrees with $\varphi\left(\theta\right)$ almost everywhere.

Since $\lim_{n\to\infty}\hat{\varphi}\left(n\right)=0$ by the Riemann-Lebesgue lemma, the function $\psi$ is analytic (and non constant) in the open disc $D$, and so the image $\Omega=\psi\left(D\right)$ is some connected domain in $\mathbb{C}$. As all the coefficients of $\psi$ are real, this domain is symmetric to conjugation: $\psi\left(\overline{z}\right)=\overline{\psi\left(z\right)}$, i.e it is symmetric to reflection about the $x$ axis.

Proposition 2**.**

$\psi$* is one-to-one in the unit disc $D$.*

Proof.

The proof relies on the following theorem, which can be found in [3, chapter VIII.3]:

Theorem 3**.**

Let $\left\{f_{k}\left(z\right)\right\}_{k=1}^{\infty}$ be a sequence of one-to-one analytic functions on a domain $D$ that converge uniformly on every compact subset of $D$ to a function $f$. Then $f$ is either one-to-one or constant.

All we have to do then is find a sequence of one-to-one functions $\psi_{k}$ which converge to $\psi$ uniformly on every compact subset of $D$. Let $\left\{G_{k}:\left[0,1\right]\to\mathbb{R}\right\}_{k=1}^{\infty}$ be a sequence of bounded, twice differentiable, strictly increasing functions satisfying

$G_{k}\to G$ in $\mathrm{L}^{1}$. 2. 2.

The first and second derivatives of $G_{k}^{\left(s\right)}$ are in $\mathrm{L}^{1}$. 3. 3.

$G_{k}^{\prime}\left(0\right)=G_{k}^{\prime}\left(1\right)=0$ and $G_{k}^{\prime\prime}\left(0\right)=G_{k}^{\prime\prime}\left(1\right)=0$.

Such a sequence may be found, for example, by taking step-function approximations of $G$ on increasingly finer partitions of $\left(0,1\right)$, smoothingly interpolating the jump discontinuities, and adding some small strictly increasing smooth function.

For each $G_{k}$, define the corresponding $\varphi_{k}$ and $\psi_{k}$ as in equations (1) and (2). By properties (2) and (3) above, the $\varphi_{k}$ are also smooth and their derivatives are all in $\mathrm{L}^{1}$. Using the fact that if a function $f$ is $s$-times differentiable and $f^{\left(s\right)}\in\mathrm{L}^{1}$ then $\left|\hat{f}\left(n\right)\right|\leq\left\|f^{\left(s\right)}\right\|_{1}/n^{s}$, we get that the Fourier coefficients $\left\{\hat{\varphi}_{k}\left(n\right)\right\}_{n=0}^{\infty}$ are absolutely convergent, and so $\psi_{k}$ can be extended to the closed disc $\overline{D}$. This allows us to look at the image of the unit circle $\partial D$ under $\psi_{k}$.

Parameterize the circle by $\gamma\left(\theta\right)=e^{i\theta}$ for $\theta\in\left[-\pi,\pi\right]$. As $\theta$ increases from $-\pi$ to [math], $\mathrm{Re}\psi_{k}\left(\gamma\left(\theta\right)\right)=\varphi_{k}\left(\theta\right)$ strictly decreases from $G_{k}\left(1\right)$ to $G_{k}\left(0\right)$, and as $\theta$ increases from [math] to $\pi$, $\mathrm{Re}\psi_{k}\left(\gamma\left(\theta\right)\right)$ strictly increases from $G_{k}\left(0\right)$ to $G_{k}\left(1\right)$. Further, using the Hilbert transform equation (4) in Section 3, it can be directly calculated that $\mathrm{Im}\psi_{k}\left(\gamma\left(\theta\right)\right)$ is always positive for $\theta\in\left(-\pi,0\right)$ and always negative for $\theta\in\left(0,\pi\right)$. Thus, the image $\psi_{k}\left(\partial D\right)$ is a simple loop. Since the preimage of $\partial\psi_{k}\left(D\right)$ is a subset of $\partial D$ (this is a consequence of the maximum principle: If an interior point $z\in D$ were mapped to $\partial\psi_{k}\left(D\right)$, we could obtain a local maximum for $\left|\psi_{k}\right|$), we have that $\psi_{k}\left(\partial D\right)=\partial\psi_{k}\left(D\right)$. So for every $x\in\psi_{k}\left(D\right)$, $\psi_{k}\left(\partial D\right)$ winds around $x$ exactly once. By Cauchy’s argument principle, this means that $x$ has exactly one preimage in $D$ under $\psi_{k}$, i.e $\psi_{k}$ is one-to-one.

All that remains is to show that $\left\{\psi_{k}\right\}_{k=1}^{\infty}$ converge uniformly to $\psi$ on every compact subset $A\subseteq D$ . To see this, note that $A$ is contained in some closed disc of radius $\rho<1$. Denoting $z=re^{i\theta}\in A$ with $0\leq r\leq\rho$, we have

[TABLE]

The sum $\sum_{n=0}^{\infty}r^{n}$ converges, and the supremum converges to [math] uniformly in $k$ since for each $n\in\mathbb{N}$ we have

[TABLE]

and this integral converges to [math] since $\varphi_{k}$ converges to $\varphi$ in $\mathrm{L}^{1}$. ∎

Having established that $\psi$ is one-to-one, we can invoke the conformal invariance of Brownian motion, which can be found in [7, Theorem 7.20]:

Theorem 4**.**

Let $U$ be a domain in the complex plain, $x\in U$ and $f:U\to\Omega$ be analytic. Let $\left\{X_{t}\mid t\geq 0\right\}$ be a planar Brownian motion started in $x$. Then the process $\left\{f\left(X_{t}\right)\mid 0\leq t\leq T\left(X_{t},U\right)\right\}$ is a time-changed Brownian motion, i.e there exists a planar Brownian motion $\left\{Y_{t}\mid t\geq 0\right\}$ such that for any $t\in\left[0,T\left(X_{t},U\right)\right)$,

[TABLE]

where

[TABLE]

If $f$ is one-to-one then $\zeta\left(T\left(X_{t},U\right)\right)$ is the first exit time from $\Omega$ by $\left\{Y\mid t\geq 0\right\}$.

Thus, the process $\psi\left(X_{t}\right)$ is a time-changed Brownian motion, i.e there exists a planar Brownian motion $Y_{t}$ and a monotone function $\zeta\left(t\right)$ such that $\psi\left(X_{t}\right)=Y_{\zeta\left(t\right)}$. The domain $\Omega$ has the properties that we are looking for:

Upon hitting the boundary, $Y_{t}^{\left(1\right)}$ distributes as $Y_{T\left(Y_{t},\Omega\right)}^{\left(1\right)}\sim\mu$: The position of $X_{T\left(X_{t},D\right)}$ is uniform on the unit circle and the real part of $\psi\left(\partial D\right)$ is made of two (reflected) copies of $G=F^{-1}$, so by the inverse transform sampling method we get that $\mathrm{Re}Y_{T\left(Y_{t},\Omega\right)}$ distributes as $\mu$. 2. 2.

$Y_{t}$ hits the boundary $\partial\Omega$ in finite time. This is an immediate consequence of the following lemma:

Lemma 5** (Lemma 1.1 in [1]).**

Suppose that $f\left(w\right)=\sum_{n=0}^{\infty}a_{n}w^{n}$ is a conformal mapping from the unit disc $D$ onto a domain $\Omega$ with $f\left(0\right)=z$. Then

[TABLE]

Since $\varphi$ is in $\mathrm{L}^{2}$, the right hand side of equation (3) is finite. Since $\left(Y_{t}^{\left(1\right)}\right)^{2}-t$ is a martingale with expectation [math], we get by the optional stopping theorem that

[TABLE]

implying $\mathbb{E}T\left(Y_{t},\Omega\right)=\text{Var}\mu$. 3. 3.

The process $Y_{t}$ starts at [math], i.e $Y_{0}=\psi\left(0\right)=0$: The imaginary component is [math] since $\psi$ is symmetric to complex conjugation, and the real component is [math] since $Y_{t}$ is a martingale and $\mathrm{Re}Y_{t}$ distributes as $\mu$, so that $\mathrm{Re}Y_{0}=\int_{\mathbb{R}}xd\mu=0$. 4. 4.

$\Omega$ is simply connected by Brouwer’s invariance of domain theorem, which states that:

Theorem 6** (Theorem 1 in chapter 1 in [10]).**

If $U\subseteq\mathbb{R}^{n}$ is open and $f:U\to\mathbb{R}^{n}$ is one-to-one and continuous, then f is a homeomorphism.

∎

3 Properties and examples

In this section, we will see examples of domains $\Omega$ for various distributions. Numeric approximations of $\Omega$ for Bernoulli, Gaussian and Cantor distributions, as well as some distribution on the natural numbers, can be found in Figure 1.

Some easy properties of $\Omega$ can immediately be gathered from $\mu$:

If $x$ is an atom of $\mu$, i.e $\mu\left(\left\{x\right\}\right)>0$, then $\partial\Omega$ contains a straight line segment at $x$, i.e $\left\{x\right\}\times\left[a,b\right]\subseteq\partial\Omega$ for some numbers $a,b$. 2. 2.

If $\left(a,b\right)$ is an interval with $\mu\left(\left(a,b\right)\right)=0$ then $\Omega$ contains the infinite rectangle $\left(a,b\right)\times\left(-\infty,\infty\right)$. 3. 3.

If the support of $\mu$ is infinite, then the projection of $\Omega$ onto the $x$ axis is unbounded.

One can describe $\Omega$ by looking at its boundary $\partial\Omega$, and in general it is useful to do this by considering how the curve $\gamma\left(\theta\right)=\psi\left(e^{i\theta}\right)$ behaves as $\theta$ traverses from $-\pi$ to $\pi$. Of course, for some distributions both the real part and the imaginary part of $\psi\left(e^{i\theta}\right)$ can be infinite, in the latter case even countably many times, so $\gamma\left(\theta\right)$ really sits in the Riemann sphere.

When calculating $\psi\left(e^{i\theta}\right)$, the real part is already known - it is just equal to $\varphi\left(\theta\right)$

• so all that is left to do is to find the imaginary part. The relation between the real part and the imaginary part of $\psi\left(e^{i\theta}\right)$ is given by replacing all cosines with sines in the series expansion of $\psi$:

[TABLE]

Alternatively, the imaginary part is obtained from the real part by the Hilbert transform operator $H$, which, for a function $u:\mathbb{R}\to\mathbb{R}$, is given by

[TABLE]

where $PV$ is Cauchy’s principal value. One way to see that this is true is to note that the Hilbert transform is linear and maps the function $\cos nx$ to $\sin nx$ for positive integers $n$. For a comprehensive source on the Hilbert transform, see [6].

There are therefore two straightforward ways to go about studying $\mathrm{Im}\psi$. The first is to inspect the series $\sum_{n=1}^{\infty}\hat{\varphi}\left(n\right)\sin\theta n$ directly, and the second is to consider the Hilbert transform of $\varphi$. In the next two subsections, we will see both approaches.

3.1 Discrete distributions

Let $\mu$ be an atomic distribution, supported on a (possibly infinite) discrete set of values $\left\{x_{i}\right\}_{i}$, which obtains $x_{i}$ with probability $p_{i}$:

[TABLE]

The cumulative distribution function is then just a sum of step functions,

[TABLE]

and so is the inverse $G\left(y\right)$ and the symmetric $\varphi_{\mu}\left(\theta\right)$. We write

[TABLE]

for some weights $\alpha_{i}$ and thresholds $\theta_{i}$ (these can be calculated explicitly from $p_{i}$ and $x_{i}$, but we omit the calculations for brevity).

As noted in items (1) and (2) above, the boundary $\partial\Omega$ consists of infinite rays of the form $\left\{x_{i}\right\}\times\left(y_{i},\infty\right]$ and $\left\{x_{i}\right\}\times\left(-y_{i},\infty\right]$ for some values $y_{i}\geq 0$. If there is an extremal value $x=\min_{i}\left\{x_{i}\right\}$ or $x=\max_{i}\left\{x_{i}\right\}$ among the $x_{i}$’s, then clearly $\partial\Omega$ contains the infinite line $\left\{x\right\}\times\left(-\infty,\infty\right)$.

As the $y_{i}$’s give a complete characterization of $\Omega$, calculating them may be of interest.

Theorem 7**.**

Each $y_{i}$ is given by

[TABLE]

where $x$ is a solution to the equation

[TABLE]

Proof.

Although $\mathrm{Im}\psi\left(e^{i\theta}\right)$ may be discontinuous and even infinite, it is differentiable almost everywhere; in fact, by section III.2.10 in [5], if $\mathrm{Re}\psi\left(e^{i\theta}\right)$ is constant in an interval, then $\mathrm{Im}\psi\left(e^{i\theta}\right)$ is analytic there. This means that $\mathrm{Im}\psi\left(e^{i\theta}\right)$ is piecewise analytic, since $\mathrm{Re}\psi\left(e^{i\theta}\right)$ is piecewise constant except for a discrete set of jumps at $\left\{x_{i}\right\}_{i}$.

The $y_{i}$’s can therefore be calculated by finding the local minima and maxima of $\mathrm{Im}\psi\left(e^{i\theta}\right)$, and these in turn are obtained by differentiation.

Our first step is to calculate $\mathrm{Im}\psi\left(e^{i\theta}\right)$. As noted at the beginning of the section, the relation between $\mathrm{Re}\psi\left(e^{i\theta}\right)$ and $\mathrm{Im}\psi\left(e^{i\theta}\right)$ is given by the Hilbert transform $H$. Since $H$ is a linear operator, we first compute the Hilbert transform of a single step function.

Lemma 8**.**

Let $u\left(x\right)$ be a periodic step function, i.e

[TABLE]

for some $0\leq\theta_{0}\leq\pi$, with $u\left(x\right)=u\left(x+2\pi\right)$. Then

[TABLE]

Proof.

Denote by $\sqcap_{\alpha}\left(x\right)$ the square pulse of half-width $\alpha$ around the origin:

[TABLE]

Then $u\left(x\right)$ is the sum of infinitely many square pulses, each with half-width $\pi-\theta_{0}$:

[TABLE]

The Hilbert transform of a single square pulse $\sqcap_{\alpha}$ can readily be calculated to be

[TABLE]

The Hilbert transform commutes with shifts, and so

[TABLE]

This is an expression of the form form

[TABLE]

with $b=\frac{1}{2\pi}\left(2\pi-2\theta_{0}\right)$ and $a=-\frac{1}{2\pi}\left(\theta_{0}+x\right)$. It can be shown that a product in the form of equation (6) is equal to

[TABLE]

(to see this, recall Euler’s infinite product identity for the sine function, $\sin\left(x\right)=x\prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)$) . We then have

[TABLE]

∎

Differentiating Equation (5) in Lemma 8, each $y_{i}$ is therefore given by $\left(H\varphi\right)\left(x\right)$ where $x$ is a zero of the derivative of $H\varphi$:

[TABLE]

∎

3.2 The uniform distribution

Let $\mu$ be the uniform distribution on $\left[-1,1\right]$, so that

[TABLE]

the inverse function is $G\left(y\right)=2y-1$, and $\varphi\left(\theta\right)=2\frac{\left|\theta\right|}{\pi}-1$. The Fourier series of $\varphi_{\mu}$ is given by

[TABLE]

which gives

[TABLE]

The boundary $\partial\Omega$ is given in Figure 1b. The $y$ component of $\gamma\left(\theta\right)=\psi\left(e^{i\theta}\right)=\left(x\left(\theta\right),y\left(\theta\right)\right)$ is then given by

[TABLE]

Differentiating by $\theta$, we get

[TABLE]

If we denote by $\mathrm{gd}^{-1}\left(x\right)=\tanh^{-1}\left(\tan\left(\frac{x}{2}\right)\right)$ the inverse Gudermannian function, then a short calculation reveals that

[TABLE]

Thus the domain $\Omega$ is bounded by the parametric curve

[TABLE]

This integral may be solved with the assistance of computational software; the antiderivative of $\tanh^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)$ is given by

[TABLE]

where $\mathrm{Li}_{n}\left(x\right)$ is the polylogarithm function,

[TABLE]

4 Other directions and open questions

For a given measure $\mu$, the domain $\Omega$ in Theorem 1 is not unique, even if we require $\Omega$ to be simply connected and symmetric to conjugation (see Figure 2 for two different domains giving the same distribution). Is there any distinguishing trait to the construction given above? Can we deduce properties of $\Omega$ from properties of $\mu$? For example,

Question 9**.**

For what measures $\mu$ is the domain $\Omega$ in Theorem 1 convex?

Theorem 1 utilizes the conformal invariance of Brownian motion and uses complex-analytic tools, and so relies on the fact that the Brownian motion is planar. However, we may ask similar hitting-time questions concerning the marginal distribution of the first $n$ coordinates of a stopped Brownian motion in higher dimensions. It is already well known that not every distribution $\mu$ on $\mathbb{R}^{n}$ can be generated this way: For example, high dimensional Brownian motion does not hit points, so any $\mu$ containing atoms cannot be obtained. What can we say about measures $\mu$ for which we already know that there is a Skorokhod embedding?

Question 10**.**

Let $\mu$ be a distribution on $\mathbb{R}^{n}$ for which there exists a Skorokhod embedding. Is there a domain $\Omega\subseteq\mathbb{R}^{n+m}$ for some $m$ so that $\left(Y_{T\left(X_{t},\Omega\right)}^{\left(1\right)},\ldots,Y_{T\left(X_{t},\Omega\right)}^{\left(n\right)}\right)$ is a Skorokhod embedding for $\mu$?

5 Acknowledgments

The author thanks Itai Benjamini, Krzysztof Burdzy, Ronen Eldan, Boa’z Klartag and Dan Mikulincer for their insightful comments and suggestions.