The expected signature of Brownian motion stopped on the boundary of a circle has finite radius of convergence

TL;DR

This paper investigates the properties of the expected signature of Brownian motion stopped at a boundary, providing a counterexample to a previously posed question about its moment conditions and convergence.

Contribution

It presents the first example showing that the expected signature of Brownian motion may not satisfy the moment condition, challenging prior assumptions.

Findings

Expected signature has finite radius of convergence

Counterexample to the moment condition question

Challenges previous beliefs about expected signatures

Abstract

The expected signature is an analogue of the Laplace transform for rough paths. Chevyrev and Lyons showed that, under certain moment conditions, the expected signature determines the laws of signatures. Lyons and Ni posed the question of whether the expected signature of Brownian motion up to the exit time of a domain satisfies Chevyrev and Lyons' moment condition. We provide the first example where the answer is negative.

Equations

$$\begin{array}{rl}S(X{)}_{0,T}& :=1+{\int }_0^{T}d{X}_s+{\int }_0^T{\int }_0^{r_2}d{X}_{r_1}\otimes d{X}_{r_2}\\ & +{\int }_0^T{\int }_0^{r_3}{\int }_0^{r_2}d{X}_{r_1}\otimes d{X}_{r_2}\otimes d{X}_{r_3}+\cdots \in T(({\mathbb{R}}^d)):=\prod _{n=0}^{\infty }({\mathbb{R}}^d{)}^{\otimes n}.\end{array}$$

$$f(X_t)=f(X_0)+\sum _{i=1}^d{\int }_0^t{\partial }_{x_i}f(X_s)d{X}_s^i.$$

$$\mathbb{E}\left[\Vert {\int }_0^T{\int }_0^{r_n}\cdots {\int }_0^{r_2}d{X}_{r_1}\otimes \cdots \otimes d{X}_{r_n}\Vert \right]<+\infty .$$

$$\mathsf{ExpSig}(X{)}_T$$

$$:=\sum _{n=0}^{\infty }\mathbb{E}\left[{\int }_0^T{\int }_0^{r_n}\cdots {\int }_0^{r_2}d{X}_{r_1}\otimes \cdots \otimes d{X}_{r_n}\right]\in T(({\mathbb{R}}^d)).$$

$$\sum _{n\ge 0}\Vert {\mathrm{proj}}_n\mathsf{ExpSig}(X{)}_T\Vert {\lambda }^n<+\infty ,$$

$$a_n:=\int x^n\mu (dx).$$

$$S(X{)}_{0,T}=1+TZ+\frac{T^2}{2!}{Z}^2+\frac{T^3}{3!}{Z}^3+\dots$$

$$\mathsf{ExpSig}(X{)}_T=1+T{a}_1+\frac{T^2}{2!}{a}_2+\frac{T^3}{3!}{a}_3+\dots ,$$

$$\mathsf{ExpSig}(B{)}_T=\exp \left(\frac{T}{2}\sum _{i=1}^d{e}_i\otimes e_i\right).$$

$$\Vert {\mathrm{proj}}_{2n}\mathsf{ExpSig}(B)\Vert =\Vert \frac{T^n}{2^{n}n!}{\left(\sum _{i=1}^d{e}_i\otimes e_i\right)}^n\Vert \le \frac{d^n{T}^n}{2^{n}n!},$$

$$\sum \Vert {\mathrm{proj}}_n\mathsf{ExpSig}(B{)}_T\Vert {\lambda }^n<+\infty ,$$

$$\tau :=\inf \left\{t\ge 0:|B_t^z|\in \partial \mathbb{D}\right\}.$$

$$\Phi (z):=\mathsf{ExpSig}(X^z{)}_{\infty },$$

$$\Delta \left({\mathrm{proj}}_n\left(\Phi (z)\right)\right)=-2\sum _{i=1}^d{e}_i\otimes \frac{\partial {\mathrm{proj}}_{n-1}\left(\Phi (z)\right)}{\partial z_i}-\left(\sum _{i=1}^d{e}_i\otimes e_i\right)\otimes {\mathrm{proj}}_{n-2}\left(\Phi (z)\right),$$

$${\mathrm{proj}}_n\left(\Phi (z)\right)=\{\begin{array}{ll}0,& \text{if}n\ge 1\\ 1,& \text{if }n=0.\end{array}$$

$$M\left(v_1\otimes \cdots \otimes v_k\right)=M\left(v_1\right)\cdots M\left(v_k\right),$$

$$M:\left(\begin{array}{c}x\\ y\end{array}\right)\mapsto \left(\begin{array}{ccc}0& 0& x\\ 0& 0& y\\ x& y& 0\end{array}\right).$$

$$\left|\left({\alpha }_1,{\alpha }_2\right)\right|=\left|{\alpha }_1\right|+\left|{\alpha }_2\right|,D^{\left({\alpha }_1,{\alpha }_2\right)}u(z)=\frac{{\partial }^{{\alpha }_1+{\alpha }_2}u}{\partial z_1^{{\alpha }_1}\partial z_2^{{\alpha }_2}}(z),\left({\alpha }_1,{\alpha }_2\right)\in {\mathbb{N}}^2.$$

$$\left(\lambda M\right)\Phi (z)=\sum _{n=0}^{\infty }{\lambda }^{n}M{\mathrm{proj}}_n\left(\Phi (z)\right).$$

$$\Vert D^{\alpha }{\mathrm{proj}}_n\left(\Phi (z)\right)\Vert \le C^n.$$

$$z\mapsto \sum _{n=0}^{\infty }{\lambda }^{n}M{\mathrm{proj}}_n\left(\Phi (z)\right)$$

$$D^{\alpha }\sum _{n=0}^{\infty }{\lambda }^{n}M{\mathrm{proj}}_n\left(\Phi (z)\right)=\sum _{n=0}^{\infty }{\lambda }^n{D}^{\alpha }M{\mathrm{proj}}_n\left(\Phi (z)\right).$$

$$\Vert {\mathrm{proj}}_n\left(\Phi (z)\right){\Vert }_{W^{m,2}\left(\mathbb{D}\right)}\le C^n,$$

$$\Vert u{\Vert }_{W^{m,2}\left(\mathbb{D}\right)}=\underset{|\alpha |=m}{\max }\Vert D^{\alpha }u{\Vert }_{L^2\left(\mathbb{D}\right)}.$$

$$\left|D^{\alpha }{\mathrm{proj}}_n\left(\Phi (z)\right)\right|$$

$$\le \tilde{C}(2)\Vert {\mathrm{proj}}_n\left(\Phi (z)\right){\Vert }_{W^{4,2}\left(\mathbb{D}\right)}$$

$$\le \tilde{C}(2)C(4{)}^n.$$

$$\left|D^{\alpha }M{\mathrm{proj}}_n\left(\Phi (z)\right)\right|\le c^n.$$

$$\sum _{n=0}^{\infty }{\lambda }^n{D}^{\alpha }M{\mathrm{proj}}_n\left(\Phi (z)\right)$$

$$\sum _{n=0}^{\infty }{\lambda }^{n}M{\mathrm{proj}}_n\left(\Phi (z)\right)$$

$$F_{\lambda }(z)=(\lambda M)\Phi (z)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)=\sum _{n=0}^{\infty }{\lambda }^{n}M{\mathrm{proj}}_n\left(\Phi (z)\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$$

Full text

1. 1 Introduction
2. 2 Differentiability of the development of expected signature
3. 3 A polar decomposition for the development
4. 4 ODE for $A_{\lambda},B_{\lambda},C_{\lambda}$
5. 5 Solving the ODE for $A_{\lambda},B_{\lambda},C_{\lambda}$
6. 6 Concluding
7. 7 Appendix

1 Introduction

Let a probability measure on a subset of the real line have moments of all orders. Under which conditions do these moments pin down the probability measure uniquely? This is the well-studied moment problem. When the subset is compact, the answer is always affirmative. In the noncompact case uniqueness is more delicate (see [Sch2017]).

In stochastic analysis one is usually concerned with measures on some space of paths, the prime example being Wiener measure on the space of continuous functions. It turns out that for many purposes a good replacement for “monomials” in this setting are the iterated integrals of paths. The collection of all of these integrals is called the iterated-integrals signature.

For smooth paths $X:[0,T]\to\mathbb{R}^{d}$ (and with respect to a time horizon $T>0$), it is defined, using Riemann-Stieltjes integration, as

[TABLE]

It is well-known (see [BGLY2016] and references therein) that

• •

$S(X)_{0,T}\in G$, where $G\subset T((\mathbb{R}^{d}))$ is the group of grouplike elements,

• •

$S(X)_{0,T}$ completely characterizes the path $X$ up to reparametrization and up to tree-likeness.

Let $X:\Omega\times[0,T]\to\mathbb{R}^{d}$ now be a stochastic process. For fixed $\omega\in\Omega$, $t\mapsto X_{t}(\omega)$ is usually not smooth, so that we have to assume that the stochastic process posesses a “reasonable” integration theory. In particular assume that integrals of the form $\int g(X_{s})dX_{s}$, exist for a large class of functions $g\in C(\mathbb{R}^{d},L(\mathbb{R}^{d},\mathbb{R}^{n}))$ and that the fundamental theorem of calculus holds,

[TABLE]

An important example is Brownian motion with Stratonovich integration. Other examples include: the Young integration against a fractional Brownian motion with Hurst parameter strictly larger than $1/2$, …(see [FV10]).

The iterated-integrals signature $S(X)_{0,T}$ defined by the expression (1) — now, using the given integration theory — is then a random variable. Let us assume that we can take its expectation level-by-level, i.e. for all $n\geq 1$ (we postpone the discussion of the choice of norm on $(\mathbb{R}^{d})^{\otimes n}$ to later)

[TABLE]

We can then define expected signature level-by-level

[TABLE]

where $\mathbb{E}_{\mathsf{p}}$ denotes the expectation level-by-level ($\mathsf{p}$ for “projective”) of a $G$-valued random variable. The question arises:

Does $\mathsf{ExpSig}(X)$ completely characterize the law of $X$?

As we have seen above, the computation of $S(X)_{0,T}$ already incurs a loss of information: the parametrization of $X$ and any tree-like parts are lost. The relevant question is hence

Does $\mathsf{ExpSig}(X)_{T}$ completely characterize the law of $X$, up to parametrization and tree-likeness?

Since this formulation is a bit awkard, and since the (deterministic) step $S(X)\mapsto X$ is completely understood, we can instead focus on

Does $\mathsf{ExpSig}(X)_{T}$ completely characterize the law of $S(X)_{0,T}$ on $G$?

A sufficient condition for this to be the case is given in [ChevyrevLyons]: if $\mathsf{ExpSig}(X)_{T}$ has infinite radius of convergence, that is

[TABLE]

for all $\lambda>0$ then the law of $S(X)_{0,T}$ on $G$ is the unique law with this (projective) expected value. Here $\operatorname{proj}_{n}:T((\mathbb{R}^{d}))\to(\mathbb{R}^{d})^{\otimes n}$ denotes projection onto tensors of length $n$.

Let us give two examples. Let $\mu$ be a probability measure on $\mathbb{R}$ having all moments and define

[TABLE]

Consider the stochastic process $X_{t}:=tZ$, where $Z$ is distributed according to $\mu$. Since $X$ is smooth, its signature is well-defined and actually has the simple form

[TABLE]

Then

[TABLE]

and a sufficient condition for $\sum_{n}a_{n}T^{n}\lambda^{n}/n!$ to have infinite radius of convergence is $|a_{n}|\leq C^{n}$, for some $C>0$.111The condition $|a_{n}|\leq C^{n}$ is of course more than enough in the classical moment problem to have uniqueness for the law $\mu$ on $\mathbb{R}$ ([SR75, Example X.6.4]). Then [ChevyrevLyons, Proposition 6.1] applies, and the law of $S(X)_{0,T}$ on $G$ is uniquely determined by these moments.

Consider now the expected signature of a standard Brownian motion $B$ calculated up to some fixed time $T>0$. It is known (see for example [LV2004, Proposition 4.10]) that

[TABLE]

It follows that

[TABLE]

and hence

[TABLE]

for any $\lambda>0$. Again, by [ChevyrevLyons, Proposition 6.1], the law of $S(B)_{0,T}$ is uniquely determined by $\mathsf{ExpSig}(B)_{T}$.

The decay rate of expected signature of the stochastic process up to the exit time from a bounded domain is a very challenging problem, even for the simple Brownian motion case. The literature on the decay rate of the expected signature focuses on the case for the fixed time interval, e.g. [Passeggeri2016] and [FrizRiedel2014], which heavily relies on the Gaussian tail of the increment. However, the increment of stopped processes would violate this assumption. The question on the finiteness of the convergence radius of the expected signature of the Brownian motion was firstly proposed in [LyonsNi] in 2015. It has remained open until our paper provides the first negative example, i.e. a 2-dimensional Brownian motion up to the unit disk, which implies the lower bound of the decay rate of the expected signature in this case. In [LyonsNi, Theorem 3.6], using Sobolev estimates, under certain smoothness and boundedness condition of the domain, geometric upper bounds for the decay rate of the expected signature of stopped Brownian is established (see also [ChevyrevLyons, Example 6.20] for a probabilistic approach). It is even more challenging to establish a non-trivial lower bound of the decay rate of the expected signature in this case. Our work may shed some light on how to use the PDE approach to derive the lower bound of the decay rate of the stopped diffusion processes.

Concretely, we consider the Brownian motion $B^{z}$ in $\mathbb{R}^{2}$ started at some point $z$ in the unit circle $\mathbb{D}:=\{z\in\mathbb{R}^{2}:|z|\leq 1\}$, and stopped at hitting the boundary, that is

[TABLE]

In the notation introduced above, we are interested in

[TABLE]

where $X^{z}_{t}:=B^{z}_{t\wedge\tau}$. In [LyonsNi] it was shown that for every $n\in\mathbb{N}$ and $n\geq 2$, the $n$th term of $\Phi$ satisfies the following PDE:

[TABLE]

with the boundary condition that for each $|z|=1$,

[TABLE]

Additionally, one has $\operatorname{proj}_{0}\left(\Phi(z)\right)=1$ and $\operatorname{proj}_{1}\left(\Phi(z)\right)=0$ for all $z\in\mathbb{D}$. Using this, they were able to obtain the bound $\left\|\operatorname{proj}_{n}\left(\Phi(x)\right)\right\|\leq C^{n}$ for some $C>0$ ([LyonsNi, Theorem 3.6]). This bound is not enough to decide whether the radius of convergence for $\mathsf{ExpSig}(X^{z})_{\infty}$ is infinite or not, but it is enough to deduce that $\mathsf{ExpSig}(X^{z})_{\infty}$ has radius of converge strictly larger than [math]. In this work we show that the radius of convergence is indeed finite.

Recall from [ChevyrevLyons, Proposition 6.1] that if $A,B$ are $G$-valued random variables such that $\mathbb{E}_{\mathsf{p}}[A]=\mathbb{E}_{\mathsf{p}}[B]$ and $\mathbb{E}_{\mathsf{p}}[A]$ has an infinite radius of convergence, then $A\overset{\mathcal{D}}{=}B$. Our main theorem, proven in Section 6, is the following.

Theorem**.**

The expected signature $\Phi(0)=\mathsf{ExpSig}(X^{0})_{\infty}=\mathbb{E}_{\mathsf{p}}[S(X^{0})_{0,\infty}]$ of a two-dimensional Brownian motion stopped upon exiting the unit disk has a finite radius of convergence.

The condition of $\mathbb{E}_{\mathsf{p}}[A]$ having an infinite radius of convergence is equivalent to $\mathbb{E}_{\mathsf{p}}[A]$ lying in $E$. Here $E$ is defined as the closure of $T((\mathbb{R}^{2}))$ under the coarsest topology such that for all normed algebras $\mathcal{A}$ and all $M\in L(\mathbb{R}^{2},\mathcal{A})$, the extension $M:T((\mathbb{R}^{2}))\to\mathcal{A}$ is continuous. Recall that for $M\in L\left(\mathbb{R}^{2},A\right)$, we may define $M$ firstly as a map on the $k$-times algebraic tensor product $(\mathbb{R}^{2})^{\otimes_{a}k}$, by the relation

[TABLE]

and then extended it to $T((\mathbb{R}^{2}))$ by linearity.

We want to show that $\Phi(z)$ does not lie in the space $E$. It is sufficient to show that there exists $\lambda^{*}\in\mathbb{R}$, and $M\in L\left(\mathbb{R}^{2},M_{3\times 3}\left(\mathbb{R}\right)\right)$, such that $\left(\lambda M\right)\left(\Phi(z)\right)$ diverges as $\lambda$ tends to a finite number $\lambda^{*}$. In fact, we will choose $M$ to be

[TABLE]

Such a map $M$ first appeared in [HamblyLyons] to study the signature of bounded variation paths and is also subsequently used in [LyonsXu].

We proceed as follows. In Section 2, for $\lambda>0$ we let $\lambda M$ act on $\Phi(z)$. For $\lambda$ small enough, we show that resulting linear map in $L(\mathbb{R}^{3},\mathbb{R}^{3})$, evaluated at $(0,0,1)\in\mathbb{R}^{3}$ is smooth in $z$ and solves a certain PDE. Using rotational invariance of Brownian motion, in Section 3 we rewrite said PDE solution in polar coordinates. In Sections 4 and 5, we obtain an explicit solution for the PDE (still, for $\lambda$ small enough) in terms of Bessel functions. Finally, in Section 6 we show that the solution blows up as $\lambda\to\tilde{\lambda}$ for some $\tilde{\lambda}<+\infty$, proving our main theorem. The Appendix, Section 7, contains some auxiliary results on PDEs.

2 Differentiability of the development of expected signature

We first need two technical lemmas which assert that the development of the expected signature is twice differentiable, and satisfies the PDE we expect it to. In the lemma below we will adopt the multi-index notation

[TABLE]

Observe that

[TABLE]

Let $\|\cdot\|$ be the projective norm.

Lemma 1**.**

The function $z\mapsto\operatorname{proj}_{n}\left(\Phi(z)\right)$ is twice continuously differentiable. There exists a constant $C>0$ such that for all $n\in\mathbb{N}$, all $z\in\mathbb{D}$ and all $\alpha\in\mathbb{N}^{2}$ satisfying $|\alpha|\leq 2$, one has the bound

[TABLE]

Moreover, there exists $\lambda^{*}>0$ such that for all $\lambda<\lambda^{*}$

[TABLE]

is twice differentiable in $z$ and if $|\alpha|\leq 2$, then

[TABLE]

Proof.

Let $m\in\mathbb{N}$. By Theorem 16 in the appendix, the function $z\mapsto\operatorname{proj}_{n}\left(\Phi(z)\right)$ is twice continuously differentiable (it is in fact infinitely differentiable on $\mathbb{D}$). By Lemma 15 in the Appendix, there exists $C>0$ such that for all $n\in\mathbb{N}$

[TABLE]

where the norm $\|\cdot\|_{W^{m,2}\left(\mathbb{D}\right)}$ is the Sobolev norm on the unit disc $\mathbb{D}$ with respect to the variable $z$,

[TABLE]

By Theorem 2.2 in [LyonsNi], which bounds the values of a function $u$ in terms of the Sobolev norm of $u$, there is some constant $\tilde{C}(2)$ such that for all $z\in\mathbb{D}$ and $\left|\alpha\right|\leq 2$,

[TABLE]

Since $M\operatorname{proj}_{n}\left(\Phi(z)\right)$ is a linear image of $\operatorname{proj}_{n}\left(\Phi(z)\right)$, the function $M\operatorname{proj}_{n}\left(\Phi(z)\right)$ is twice continuously differentiable in $z$, and moreover, there exists $c>0$ such that for all $z\in\mathbb{D}$,

[TABLE]

This bound also allows us to deduce that for $\left|\alpha\right|=2$, the series

[TABLE]

converges uniformly and hence the series

[TABLE]

is twice continuously differentiable and the derivatives can be taken inside the infinite summation. ∎

Lemma 2**.**

There exists $\lambda^{*}>0$ such that if $\lambda<\lambda^{*}$, the function $F_{\lambda}$ defined by

[TABLE]

is twice continuously differentiable on $\mathbb{D}$, and satisfies

[TABLE]

with $F_{\lambda}(z)=(0,0,1)$ for $z\in\partial\mathbb{D}$. Here $(e_{1},e_{2})$ denotes the canonical basis of $\mathbb{R}^{2}$.

Proof.

If we apply the linear map $M$ to the PDE (5), then we have a matrix-valued PDE

[TABLE]

together with the boundary condition

[TABLE]

We may multiply both sides with $\lambda^{n}$, sum to infinity and apply to the vector $(0,0,1)$ to get

[TABLE]

By Lemma 1, each of the infinite sums converges and we may take the derivatives outside the infinite sum. ∎

3 A polar decomposition for the development

Let $x=\left(x_{1},x_{2}\right)^{t}\in\mathbb{R}^{2}$. Recall that

[TABLE]

We may consider $M(x)$ as a linear endomorphism of $\mathbb{R}^{2}\oplus\mathbb{R}$ mapping $\left(v,\alpha\right)$ to $\left(\alpha x,\left\langle x,v\right\rangle\right)$.

Lemma 3**.**

For any linear map $R:\mathbb{R}^{2}\to\mathbb{R}^{2}$,

[TABLE]

where $R^{*}$ is the transpose of $R$.

Proof.

Note that

[TABLE]

In what follows, we will use the notation

[TABLE]

Corollary 4**.**

Let $R:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ be the rotation map

[TABLE]

Then

[TABLE]

Proof.

Brownian motion $B^{R(z)}$ starting at $R(z)$ has the same distribution as the rotated Brownian motion $R\left(B^{z}\right)$, where $B^{z}$ starts from $z$. Let $\circ\mathrm{d}$ denote the Stratonovich differential. Then

[TABLE]

By Lemma 3

[TABLE]

As $R$ is orthogonal, we have

[TABLE]

Therefore,

[TABLE]

Corollary 5**.**

Define the functions $A_{\lambda},B_{\lambda},C_{\lambda}:\left[0,1\right]\rightarrow\mathbb{R}$ by

[TABLE]

where $F_{\lambda}$ is the function defined by (9). In polar coordinates, the expression of $F_{\lambda}$ reads

[TABLE]

Additionally, there exists $\lambda^{*}>0$ such that if $\lambda<\lambda^{*}$, then $A_{\lambda},B_{\lambda},C_{\lambda}$ are twice continuously differentiable functions in the variable $r$ for all $r\in\left[0,1\right]$.

Proof.

The functions $A_{\lambda},B_{\lambda},C_{\lambda}$ are twice continuously differentiable because $F_{\lambda}(r,0)$ is twice continuously differentiable by Lemma 2. Let $R:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be the rotation of angle $\theta$. For $z=(r\cos\theta,r\sin\theta)=R(r,0)$, the definition of $F_{\lambda}$ gives

[TABLE]

By Corollary 4, one has

[TABLE]

where

[TABLE]

4 ODE for $A_{\lambda},B_{\lambda},C_{\lambda}$

Lemma 6**.**

There exists $\lambda^{*}>0$ such that for $\lambda<\lambda^{*}$, the functions $A_{\lambda},B_{\lambda},C_{\lambda}$ defined in Lemma 5 satisfy

[TABLE]

and

[TABLE]

Proof.

By Corollary 5,

[TABLE]

As $A_{\lambda},B_{\lambda},C_{\lambda}$ are twice continuously differentiable for $\lambda<\lambda^{*}$, we may substitute (4) for $F_{\lambda}$ into the equation in Lemma 2, which gives for $r>0$

[TABLE]

Using the identities

[TABLE]

the right hand side of the PDE in Lemma 2 is

[TABLE]

Equating (13) and (14) gives the first equation as

[TABLE]

As this holds for all $\theta$, we may equate the coefficients of $\sin\theta$ and $\cos\theta$ to obtain

[TABLE]

and from the second equation,

[TABLE]

and therefore,

[TABLE]

Combining (15) and (16) and multiplying the equations throughout by $r^{2}$, we have

[TABLE]

for all $r>0$. By continuity of the second derivatives of $A_{\lambda},B_{\lambda},C_{\lambda}$ (see Lemma 5), the equations hold for all $r\geq 0$. Again using the continuity of the second derivatives of $A_{\lambda},B_{\lambda},C_{\lambda}$, we may substitute $r=0$ into (17) to get

[TABLE]

Using the boundary conditions for $z\in\partial\mathbb{D}$ in Lemma 2, we have

[TABLE]

5 Solving the ODE for $A_{\lambda},B_{\lambda},C_{\lambda}$

Lemma 7**.**

Let

[TABLE]

where $J_{0},J_{1}$ are the Bessel functions of the first kind. Fix $\lambda>0$ such that $d(\lambda)J_{1}(\lambda)\neq 0$. Then the real-valued functions defined for all $r>0$ by222Note that the two determinations of the square root in the definition of $\zeta$ yield the yield the same $A_{\lambda}$, $B_{\lambda}$ and $C_{\lambda}$.

[TABLE]

are the unique solution of the differential system (12) satisfying the boundary conditions stated in Lemma 6.

Proof.

Recall that, for $\nu=0,1$, Bessel’s differential equation

[TABLE]

has a canonical basis of solutions consisting of the ordinary Bessel functions $J_{\nu}(x)$ and $Y_{\nu}(x)$ [DLMF, §10.2]. The function $J_{\nu}$ is entire, while $Y_{\nu}$ is analytic on $\mathbb{C}\setminus\mathbb{R}_{-}$, and $Y_{\nu}(x)$ diverges to $-\infty$ as $x\rightarrow 0$ along the positive reals.

Let $\mathcal{C}_{\nu}$ denote a cylinder function, i.e., a linear combination $aJ_{\nu}+bY_{\nu}$ with coefficients $a,b$ that do not depend on $\nu$ [DLMF, §10.2(ii)]. One has [DLMF, (10.6.2), (10.6.3)]

[TABLE]

The equation for $B_{\lambda}$ is exactly Bessel’s equation with $\nu=1$ and $x=\lambda r$; its general solution for $r>0$ is hence $B_{\lambda}(r)=\mathcal{C}_{1}(\lambda r)$. Since $Y_{1}(x)$ diverges as $x\rightarrow 0$ and $\lambda$ is nonzero, the initial condition at [math] forces $b=0$. Similary, the initial condition at $1$ implies $a=0$ and therefore $B_{\lambda}=0$, unless $J_{1}(\lambda)=0$, in which case any $B_{\lambda}(r)=aJ_{1}(\lambda r)$ is a solution.

Let us turn to the coupled equations for $A_{\lambda}$ and $C_{\lambda}$. Make the ansatz

[TABLE]

where $\alpha,\zeta$ are yet unspecified complex numbers. The equation involving $C_{\lambda}^{\prime\prime}$ becomes

[TABLE]

The change of variable passing from $\mathcal{C}_{0}$ to $f_{0}$ transforms Bessel’s equation into

[TABLE]

and the relations (21) yield $rf_{1}^{\prime}(r)+f_{1}(r)=\lambda\zeta rf_{0}(r)$, so (23) holds when

[TABLE]

i.e., when $\zeta^{2}=2(1+\alpha\zeta)$.

Similarly, the equation involving $A_{\lambda}^{\prime\prime}$ rewrites as

[TABLE]

and the last term on the left-hand side is equal to $2\lambda^{2}\zeta r^{2}f_{1}(r)$ by (21). Thus, (24) reduces to Bessel’s equation provided that $\zeta^{2}=1-2\alpha^{-1}\zeta$.

In summary, the functions (22) define a solution of (12) for any choice of $a,b$ in the definition of $\mathcal{C}_{\nu}$ and $\alpha$, $\zeta$ such that $\zeta^{2}=2(1+\alpha\zeta)=1-2\alpha^{-1}\zeta$. The latter condition is equivalent to

[TABLE]

Letting $\zeta$ now denote a fixed root of $\zeta^{4}+\zeta^{2}+2$, say the one in (18), the choices

[TABLE]

provide us with four linearly independent333This follows, for instance, from the expressions [DLMF, (10.2.2), (10.8.1)] and the fact that $(\lambda\zeta)^{2}\neq 1$. solutions, which hence form a basis of the solution space of the system of two linear differential equation of order two.

The asymptotic behaviour of $Y_{1}$ at the origin, $Y_{1}(\lambda\zeta r)\sim-2(\pi\lambda\zeta r)^{-1}$ [DLMF, (10.7.4)], shows that linear combinations involving any of the last two solutions (25) are incompatible with the conditions $A_{\lambda}(0)=C_{\lambda}^{\prime}(0)=0$. Therefore, one has

[TABLE]

for some $u,v\in\mathbb{C}$. The conditions $A_{\lambda}(1)=0$, $C_{\lambda}(1)=1$ translate into a linear system for $u,v$ of determinant

[TABLE]

(where we have used the fact that $J_{\nu}(\bar{z})=\overline{J_{\nu}(z)}$ [DLMF, (10.11.9)]). When $d(\lambda)\neq 0$, the unique solution is $u=-\bar{v}=\bar{\alpha}J_{1}(\lambda\bar{\zeta})$. Since $|\alpha|^{2}=2\sqrt{2}$, this leads to the expressions (19). ∎

6 Concluding

Lemma 8**.**

In the notation of Lemma 7, there exists $\tilde{\lambda}>0$ such that $C(0)$, viewed as a function of $\lambda$, has a pole at $\tilde{\lambda}$.

Proof.

Let us first show that $d(\lambda)$ has a zero lying in the interval $(2.5,3)$. Consider the series expansions [DLMF, (10.2.2)]

[TABLE]

For $x\in\mathbb{C}$ and $n\in\mathbb{N}$ such that $|x|<2(n+1)$, the remainders starting at index $n$ of both series are bounded by

[TABLE]

In particular, for $x=\lambda\zeta$ or $x=\lambda\bar{\zeta}$ with $0<\lambda\leqslant 3$, we have $|x/2|<1.784$. For $n=5$, the quantity (27) is bounded by $0.025$. By replacing $J_{0}$ and $J_{1}$ by the first five terms of the series (26) in the expression of $d(\lambda)$ and propagating this bound by the triangle inequality, one can check that $d(2.5)<-0.06$. A similar calculation shows that $d(3)>0.03$. Since $d(\lambda)$ is a continuous function of $\lambda$, it follows that $d(\tilde{\lambda})$ vanishes for some $\tilde{\lambda}\in(2.5,3)$.

We still need to check that the numerator of $C_{\lambda}(0)$ in (19) does not vanish at $\tilde{\lambda}$. One has $J_{0}(0)=1$. Taking $n=3$ in (27) yields an expression of the form

[TABLE]

where one can check that $\operatorname{Im}(c_{0})<-0.33$, $\operatorname{Im}(c_{1})<-0.12$, $\operatorname{Im}(c_{2})<-0.001$. For all $\lambda\geqslant 2.5$, this implies

[TABLE]

The claim follows. ∎

Remark 9**.**

Instead of doing the calculation sketched in the proof manually, one can easily prove the result using a computer implementation of Bessel functions that provides rigorous error bounds. For example, using the interval arithmetic library Arb [Johansson2017] via SageMath, the check that $d(\lambda)$ has a zero goes as follows. The quantities of the form [x.xxx +/- eps] appearing in the output are guaranteed to be rigorous enclosures of the corresponding real quantities. We check the presence of a zero in the interval $[2.82,2.83]$ instead of $[2.5,3.0]$ because having a tighter estimate simplifies the second step.

sage: zeta = CBF(sqrt((-1+Isqrt(7))/2)) sage: alpha = zeta^3/2 + zeta sage: lb, ub = CBF(282/100), CBF(283/100) sage: (alpha.conjugate()(lbzeta).bessel_J(0) ....: (lbzeta.conjugate()).bessel_J(1)) [-13.208370024264 +/- 4.16e-13] + [-0.003639973760 +/- 4.63e-13]I sage: (alpha.conjugate()(ubzeta).bessel_J(0) ....: (ubzeta.conjugate()).bessel_J(1)) [-13.424373315124 +/- 4.75e-13] + [0.005782411521 +/- 4.38e-13]*I

One can then verify as follows that the image by the function $\lambda\mapsto\bar{\alpha}J_{1}(\lambda\bar{\zeta})$ of the interval $[2.82,2.83]$ only contains elements of negative imaginary part.

sage: crit = lb.union(ub); crit # convex hull (real interval) [2.8 +/- 0.0301] sage: alpha.conjugate()(critzeta.conjugate()).bessel_J(1) [+/- 0.0707] + [-4e+0 +/- 0.303]*I

Theorem 10**.**

The series expansion with respect to $\lambda$ of $\Phi(0)$ has a finite radius of convergence.

Remark 11**.**

Since the condition of [ChevyrevLyons] of uniqueness of laws is only sufficient, the questions remains on whether there exists another law on $G$ having the same moments as $S(X^{0})_{0,T}$.

Proof.

Assume for contradiction that $\Phi(0)$ has an infinite radius of convergence. Then $F_{\lambda}(0)$ is an entire function in $\lambda$. We also know from Corollary 5 that there exists $\lambda^{*}>0$ such that for real $\lambda<\lambda^{*}$

[TABLE]

where $A_{\lambda},B_{\lambda},C_{\lambda}$ are defined by Lemma 7. By the Identity theorem, $A_{\lambda},B_{\lambda},C_{\lambda}$ are entire functions and (28) holds for all $\lambda$. This contradicts Lemma 8, and therefore $\Phi(0)$ has a finite radius of convergence. ∎

7 Appendix

Let $\Gamma$ be a domain in $\mathbb{R}^{d}$.

Definition 12**.**

Let $u$ be a locally integrable function in $\Gamma$ and $\alpha$ be a multi-index. Then a locally integrable function $r_{\alpha}u$ such that for every $g\in C_{c}^{\infty}(\Gamma)$,

[TABLE]

will be called weak derivative of $u$ and $r_{\alpha}$ is denoted by $D^{\alpha}u$. By convention, $D^{\alpha}u=u$ if $|\alpha|=0$.

Definition 13**.**

The Sobolev space $W^{k,p}(\Gamma)$ for $p,k\in\mathbb{N}$ is defined to be the set of all $\mathbb{R}^{\tilde{d}}$-valued functions $u\in L^{p}(\Gamma)$ such that for every multi-index $\alpha$ with $|\alpha|\leq k$, the weak partial derivative $D^{\alpha}u$ belongs to $L^{p}(\Gamma)$, i.e.

[TABLE]

It is endowed with the Sobolev norm defined as follows:

[TABLE]

When $k=0$, this norm coincides with the $L^{p}(\Gamma)$-norm, i.e.

[TABLE]

Theorem 14**.**

Let $M$ be a second order differential operator with coefficients $\{a^{i,j}\}$. Let $u$ be a weak solution of

[TABLE]

Suppose that the ellipticity condition holds. Let $f\in W^{k,2}(\Gamma)$, $g\in W^{k+2,2}(\Gamma)$. Let $\Gamma$ be a bounded domain of class $C^{k+2}$ and let the coefficients of $M$ be of class $C^{k+1}(\bar{\Gamma})$. Then

[TABLE]

with $c$ depending on $\lambda,d$, $\Gamma$ and on the $C^{k+1}$-norms for the $a^{i,j}$.

Proof.

It is proved by using Theorem 8.13 in [gilbarg2015elliptic] and setting the boundary condition $\varphi=0$. ∎

In the following we prove Lemma 15 for $m\geq 2$, which is a generalization of Lemma 3.11 for the case $m=\lfloor\frac{d}{2}\rfloor$ in [LyonsNi].

Lemma 15**.**

Let $\Gamma$ be a bounded domain of class $C^{m}$ in $\mathbb{R}^{d}$, where $m\geq 2$. Then there exists a constant $C$ only depending on $\Gamma$ and $d$, such that for every positive integer $n\geq 2$,

[TABLE]

Proof.

The proof of Lemma 3.11 in [LyonsNi] can be applied here directly, except for that we need to check that $\operatorname{proj}_{n}(\Phi)\in W^{m,2}$, which is proved in the following theorem 16. ∎

Theorem 16**.**

Suppose that $\Gamma$ is a non-empty bounded domain in $E$. It follows $\Phi$ is infinitely differentiable in componentwise sense, i.e. for all index $I$, $\operatorname{proj}_{n}\circ\Phi$ is infinitely differentiable for all $n$.

Proof.

Based on Theorem 3.2 in [LyonsNi], it shows that

[TABLE]

where $K_{\varepsilon}(r)$ is a smooth distribution with compact support $[0,\frac{\varepsilon}{2}]$, $*$ is the convolution, and $G_{\varepsilon}$ be a map from $\mathbb{R}^{d}$ to $T((\mathbb{R}^{d}))$ defined by:

[TABLE]

Since $\Psi$ is smooth (in polynomial form) and $K_{\varepsilon}$ is a smooth function with compact support, $G_{\varepsilon}$ is a smooth function with compact support. It is easy to show that for any partial derivative $D^{\alpha}G_{\varepsilon}$ is $L_{1}$ integrable.

[TABLE]

On the other hand, $\Phi\in L^{1}$ as well, and so we have

[TABLE]

Thus $G_{\varepsilon}*\Phi$ is infinitely differentiable, since $D^{\alpha}\Phi=(D^{\alpha}G_{\varepsilon})*\Phi\in L^{1}$. ∎