A conformal Skorokhod embedding

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.


Introduction
The Skorokhod embedding problem asks the following: Given a Brownian motion X t and a probability distribution µ with expectation zero and finite variance, find a stopping time T so that X t ∼ µ and ET < ∞.
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 (t, X t ) of the Brownian motion hits some barrier Ω ⊆ R + × R. Root's solution does not use any additional randomness. Finding out the barrier Ω, 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 Ω ⊆ R 2 and a planar Brownian motion X t = X (1) t , X (2) t with X 0 ∈ Ω, let T (X t , Ω) be the first time that X t exits the domain Ω, Theorem 1. Let µ be a probability distribution on R with zero expectation and finite variance. There exists a simply connected domain Ω ⊆ R 2 containing the origin such that if Y t is a standard planar Brownian, then Y T (Yt,Ω) has distribution µ.
The proof of Theorem 1 is given in the next section. Section 3 gives properties and examples of Ω. Finally, in Section 4 we exhibit some open questions.

Proof of theorem
Proof. We identify R 2 with the complex plane C. Denote the open unit disc by D = {z ∈ C | |z| < 1} and the unit circumference by ∂D = {z ∈ C | |z| = 1}.
Denote by F : R → [0, 1] the cumulative distribution function of µ, and by G : (0, 1) → R its generalized inverse: Observe that both F and G are monotone increasing. Despite the fact that G may diverge at 0 and 1, it is still in L 2 : If X is a uniform random variable on [0, 1] then G (X) distributes as µ, and so Define the function ϕ : (−π, π) \ {0} → R by The function ϕ may be viewed as a periodic function on R with period 2π, with the possibility of it diverging at integer multiples of π. It is an even function in L 2 , and thus has a Fourier series representation containing only cosines. Denote the n-th Fourier coefficient of ϕ byφ (n), and recall that by Carleson's theorem [2] for an L 2 function, the Fourier representation ∞ n=0φ (n) cos (θn) agrees with ϕ almost everywhere.
Let ψ : D → C be the complex function defined by The function ψ 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 iθ with θ ∈ [−π, π], then So on the unit circle, the real part of ψ e iθ agrees with ϕ (θ) almost everywhere.
Since lim n→∞φ (n) = 0 by the Riemann-Lebesgue lemma, the function ψ is analytic (and non constant) in the open disc D, and so the image Ω = ψ (D) is some connected domain in C. As all the coefficients of ψ are real, this domain is symmetric to conjugation: ψ (z) = ψ (z), i.e it is symmetric to reflection about the x axis.
Proposition 2. ψ 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]: k=1 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 ψ k which converge to ψ uniformly on every compact subset of D. Let {G k : [0, 1] → R} ∞ k=1 be a sequence of bounded, twice differentiable, strictly increasing functions satisfying 2. The first and second derivatives of G (s) Such a sequence may be found, for example, by taking step-function approximations of G on increasingly finer partitions of (0, 1), smoothingly interpolating the jump discontinuities, and adding some small strictly increasing smooth function.
For each G k , define the corresponding ϕ k and ψ k as in equations (1) and (2). By properties (2) and (3) above, the ϕ k are also smooth and their derivatives are all in L 1 . Using the fact that if a function f is s-times differentiable and f (s) ∈ L 1 then f (n) ≤ f (s) 1 /n s , we get that the Fourier coefficients {φ k (n)} ∞ n=0 are absolutely convergent, and so ψ k can be extended to the closed disc D. This allows us to look at the image of the unit circle ∂D under ψ k .
All that remains is to show that {ψ k } ∞ k=1 converge uniformly to ψ on every compact subset A ⊆ D . To see this, note that A is contained in some closed disc of radius ρ < 1. Denoting The sum ∞ n=0 r n converges, and the supremum converges to 0 uniformly in k since for each n ∈ N we have and this integral converges to 0 since ϕ k converges to ϕ in L 1 .
Having established that ψ is one-to-one, we can invoke the conformal invariance of Brownian motion, which can be found in [7,Theorem 7.20]: Thus, the process ψ (X t ) is a time-changed Brownian motion, i.e there exists a planar Brownian motion Y t and a monotone function The domain Ω has the properties that we are looking for: 1. Upon hitting the boundary, Y  2. Y t hits the boundary ∂Ω in finite time. This is an immediate consequence of the following lemma: Lemma 5 (Lemma 1.1 in [1]). Suppose that f (w) = ∞ n=0 a n w n is a conformal mapping from the unit disc D onto a domain Ω with f (0) = z. Then Since ϕ is in L 2 , the right hand side of equation (3) is finite. Since Y (1) t 2 − t is a martingale with expectation 0, we get by the optional stopping theorem that 3. The process Y t starts at 0, i.e Y 0 = ψ (0) = 0: The imaginary component is 0 since ψ is symmetric to complex conjugation, and the real component is 0 since Y t is a martingale and ReY t distributes as µ, so that ReY 0 = R xdµ = 0. 4. Ω is simply connected by Brouwer's invariance of domain theorem, which states that: Theorem 6 (Theorem 1 in chapter 1 in [10]). If U ⊆ R n is open and f : U → R n is one-to-one and continuous, then f is a homeomorphism.

Properties and examples
In this section, we will see examples of domains Ω for various distributions. Numeric approximations of Ω 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 Ω can immediately be gathered from µ: 3. If the support of µ is infinite, then the projection of Ω onto the x axis is unbounded.
One can describe Ω by looking at its boundary ∂Ω, and in general it is useful to do this by considering how the curve γ (θ) = ψ e iθ behaves as θ traverses from −π to π. Of course, for some distributions both the real part and the imaginary part of ψ e iθ can be infinite, in the latter case even countably many times, so γ (θ) really sits in the Riemann sphere. When calculating ψ e iθ , the real part is already known -it is just equal to ϕ (θ) -so all that is left to do is to find the imaginary part. The relation between the real part and the imaginary part of ψ e iθ is given by replacing all cosines with sines in the series expansion of ψ: Alternatively, the imaginary part is obtained from the real part by the Hilbert transform operator H, which, for a function u : R → R, is given by where P V 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 Imψ. The first is to inspect the series ∞ n=1φ (n) sin θn directly, and the second is to consider the Hilbert transform of ϕ. In the next two subsections, we will see both approaches.

Discrete distributions
Let µ be an atomic distribution, supported on a (possibly infinite) discrete set of values {x i } i , which obtains x i with probability p i : The cumulative distribution function is then just a sum of step functions, and so is the inverse G (y) and the symmetric ϕ µ (θ). We write for some weights α i and thresholds θ 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)  As the y i 's give a complete characterization of Ω, calculating them may be of interest.
Theorem 7. Each y i is given by where x is a solution to the equation Proof. Although Imψ e iθ may be discontinuous and even infinite, it is differentiable almost everywhere; in fact, by section III.2.10 in [5], if Reψ e iθ is constant in an interval, then Imψ e iθ is analytic there. This means that Imψ e iθ is piecewise analytic, since Reψ e iθ is piecewise constant except for a discrete set of jumps at {x i } i . The y i 's can therefore be calculated by finding the local minima and maxima of Imψ e iθ , and these in turn are obtained by differentiation.
Our first step is to calculate Imψ e iθ . As noted at the beginning of the section, the relation between Reψ e iθ and Imψ e iθ 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 (x) be a periodic step function, i.e u (x) = 0 0 < |x| < θ 0 1 θ 0 ≤ |x| ≤ π for some 0 ≤ θ 0 ≤ π, with u (x) = u (x + 2π). Then Proof. Denote by α (x) the square pulse of half-width α around the origin: Then u (x) is the sum of infinitely many square pulses, each with half-width π − θ 0 : The Hilbert transform of a single square pulse α can readily be calculated to be The Hilbert transform commutes with shifts, and so . This is an expression of the form form with b = 1 2π (2π − 2θ 0 ) and a = − 1 2π (θ 0 + x). It can be shown that a product in the form of equation (6) is equal to sin (π (a − b)) sin (πa) (to see this, recall Euler's infinite product identity for the sine function, sin (x) = x ∞ n=1 1 − x 2 n 2 π 2 ) . We then have Differentiating Equation (5) in Lemma 8, each y i is therefore given by (Hϕ) (x) where x is a zero of the derivative of Hϕ:

The uniform distribution
Let µ be the uniform distribution on [−1, 1], so that the inverse function is G (y) = 2y − 1, and ϕ (θ) = 2 |θ| π − 1. The Fourier series of ϕ µ is given by which gives The boundary ∂Ω is given in Figure 1b. The y component of γ (θ) = ψ e iθ = (x (θ) , y (θ)) is then given by Differentiating by θ, we get If we denote by gd −1 (x) = tanh −1 tan x 2 the inverse Gudermannian function, then a short calculation reveals that Thus the domain Ω is bounded by the parametric curve This integral may be solved with the assistance of computational software; the antiderivative of tanh −1 tan π 4 − x 2 is given by where Li n (x) is the polylogarithm function, Li n (z) = ∞ k=1 z k k n .

Other directions and open questions
For a given measure µ, the domain Ω in Theorem 1 is not unique, even if we require Ω 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 Ω from properties of µ? For example, For what measures µ is the domain Ω 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 µ on R n can be generated this way: For example, high dimensional Brownian motion does not hit points, so any µ containing atoms cannot be obtained. What can we say about measures µ for which we already know that there is a Skorokhod embedding?
Question 10. Let µ be a distribution on R n for which there exists a Skorokhod embedding. Is there a domain Ω ⊆ R n+m for some m so that Y

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