Random walks are determined by their trace on the positive half-line

TL;DR

This paper proves that the distribution of a random walk is uniquely determined by the distributions of its positive parts, confirming a recent conjecture and utilizing complex-analytic methods.

Contribution

It establishes that the law of a random walk is uniquely determined by its positive half-line distributions, resolving a recent conjecture.

Findings

Law of a random walk determined by positive parts

Confirmation of recent conjecture by Chaumont and Doney

Use of complex-analytic methods

Abstract

We prove that the law of a random walk $X_n$ is determined by the one-dimensional distributions of $\max(X_n, 0)$ for $n = 1, 2, \ldots$, as conjectured recently by Lo\"ic Chaumont and Ron Doney. Equivalently, the law of $X_n$ is determined by its upward space-time Wiener-Hopf factor. Our methods are complex-analytic.

Equations

$$\hat{\mu }(z)$$

$$({\mu }^{\ast n}{)}_{+}$$

$$({\mu }_{+}^{\ast n}\ast {\mu }_{-}^{\ast k}{)}_{+}$$

$$0$$

$$=\sum _{j=0}^{N+1}\left(\genfrac{}{}{0ex}{}{N+1}{j}\right)({\mu }_{+}^{\ast j}\ast {\mu }_{-}^{\ast N+1-j}-{\nu }_{+}^{\ast j}\ast {\nu }_{-}^{\ast N+1-j}{)}_{+}.$$

$$0$$

$$({\mu }_{+}^N\ast {\mu }_{-}^{\ast K+1}{)}_{+}$$

$$=(({\nu }_{+}^N\ast {\nu }_{-}^{\ast K}{)}_{+}\ast {\mu }_{-}{)}_{+}=({\nu }_{+}^N\ast {\nu }_{-}^{\ast K}\ast {\mu }_{-}{)}_{+}$$

$$=({\mu }_{+}^N\ast {\mu }_{-}\ast {\nu }_{-}^{\ast K}{)}_{+}=(({\mu }_{+}^N\ast {\mu }_{-}{)}_{+}\ast {\nu }_{-}^{\ast K}{)}_{+}$$

$$=(({\nu }_{+}^N\ast {\nu }_{-}{)}_{+}\ast {\nu }_{-}^{\ast K}{)}_{+}=({\nu }_{+}^N\ast {\nu }_{-}\ast {\nu }_{-}^{\ast K}{)}_{+}=({\nu }_{+}^N\ast {\nu }_{-}^{\ast K+1}{)}_{+};$$

$$f(z)$$

$$f_{\mathrm{b}}(z)$$

$$|f_{\mathrm{o}}(z)|$$

$$|f_{\mathrm{s}}(z)|$$

$$D$$

$$A$$

$$\phi (z)$$

$${\phi }^n$$

$$|{\phi }_{n,\mathrm{o}}(z)|$$

$$=\exp \left(\frac{1}{\pi }{\int }_{-\infty }^{\infty }\frac{-\mathrm{Im}z}{|z-x{|}^2}\log |f(x){|}^{n}dx\right).$$

$$|{\phi }_{n,\mathrm{s}}(z)|$$

Full text

Work supported by the Polish National Science Centre (NCN) grant no. 2015/19/B/ST1/01457

1. Introduction and main result

In this note we give an affirmative answer to the question posed by Loïc Chaumont and Ron Doney in [1], inspired by Vincent Vigon’s conjecture in [9]. The main result was previously stated without proof in a more general form in [6], and an erroneous proof was given in [8].

A random walk $X_{n}$ is said to be non-degenerate if $\mathds{P}(X_{n}>0)\neq 0$. Similarly, a finite signed Borel measure $\mu$ on $\mathds{R}$ is said to be non-degenerate if the restriction of $\mu$ to $(0,\infty)$ is a non-zero measure.

Theorem 1**.**

If $X_{n}$ and $Y_{n}$ are non-degenerate random walks such that $\max(X_{n},0)$ and $\max(Y_{n},0)$ are equal in distribution for all $n=1,2,\ldots$ , then $X_{n}$ and $Y_{n}$ are equal in distribution for $n=1,2,\ldots$

More generally, if $\mu$ and $\nu$ are non-degenerate finite signed Borel measures and their $n$-fold convolutions $\mu^{*n}$ and $\nu^{*n}$ agree on $(0,\infty)$ for $n=1,2,\ldots$ , then $\mu=\nu$.

Following [1], we remark that various reformulations of the above result are possible. A non-degenerate random walk $X_{n}$ is determined by any of the following objects:

• •

The law of the ascending ladder process $(T_{k},S_{k})$; here $S_{k}=X_{T_{k}}$ is the $k$-th running maximum of the random walk.

• •

The upward space-time Wiener–Hopf factor $\Phi_{+}(q,\xi)$, that is, the characteristic function of $(T_{1},S_{1})$.

• •

The distributions of the running maxima $\max(0,X_{1},X_{2},\ldots,X_{n})$ for $n=1,2,\ldots$

Theorem 1 clearly implies that a non-degenerate Lévy process $X_{t}$ is determined by any of the following objects:

• •

The distributions of $\max(X_{t},0)$ for all $t>0$ (or even for $t=1,2,\ldots$).

• •

The law of the ascending ladder process $(T_{t},S_{t})$.

• •

The upward space-time Wiener–Hopf factor $\kappa_{+}(q,\xi)$, that is, the characteristic exponent of $(T_{t},S_{t})$.

• •

The distributions of the running suprema $\sup\{X_{s}:s\in[0,t]\}$ for all $t>0$.

For further discussion, we again refer to [1], where Theorem 1 was proved under various relatively mild additional conditions. For related research, see [1, 3, 5, 6, 7, 8] and the references therein.

Theorem 1 was given without proof in [6] in a more general form: Theorem 4 therein claims that $\mu=\nu$ if $\mu$ and $\nu$ are non-degenerate finite Borel measures on $\mathds{R}$ and the restrictions of $\mu^{*n_{k}}$ and $\nu^{*n_{k}}$ to $(0,\infty)$ are equal for $k=1,2,\ldots$ , where $n_{1}=1$ and $n_{2}-1,n_{3}-1,\ldots$ are distinct and have no common divisor other than $1$. Noteworthy, this result is stated for measures on the Euclidean space of arbitrary dimension, and their restrictions to the half-space. A proof is given in [8] under the additional condition $n_{2}=2$, and only in dimension one. However, the argument in [8] contains a gap, that we describe at the end of this article.

2. Proof

All measures considered below are finite, signed Borel measures. For a measure $\mu$ on $\mathds{R}$, we denote the restrictions of $\mu$ to $(0,\infty)$ and $(-\infty,0]$ by $\mu_{+}=\operatorname{\mathds{1}}_{(0,\infty)}\mu$ and $\mu_{-}=\operatorname{\mathds{1}}_{(-\infty,0]}\mu$. This should not be confused with the Hahn decomposition of $\mu$ into the positive and negative part. By $\mu^{*n}$ we denote the $n$-fold convolution of $\mu$, and we define $\mu^{*0}$ to be the Dirac measure $\delta_{0}$. For brevity, we write $\mu_{\pm}^{*n}=(\mu_{\pm})^{*n}$, as opposed to $(\mu^{*n})_{\pm}$. We denote the characteristic function of a measure $\mu$ by $\hat{\mu}$:

[TABLE]

for $z\in\mathds{R}$, and also for those $z\in\mathds{C}$ for which the integral converges. We recall that $\hat{\mu}_{+}$ is a bounded holomorphic function in the upper complex half-plane $\mathds{C}_{+}=\{z\in\mathds{C}:\operatorname{Im}z>0\}$, continuous on the boundary. Similarly, $\hat{\mu}_{-}$ is a bounded holomorphic function on the lower complex half-plane $\mathds{C}_{-}=\{z\in\mathds{C}:\operatorname{Im}z<0\}$.

Lemma 2**.**

Suppose that $\mu,\nu$ are measures on $\mathds{R}$ satisfying

[TABLE]

Then $\mu_{+}=\nu_{+}$ and

[TABLE]

Proof.

We proceed by induction with respect to $N$. For $N=1$ the result is trivial: we have $\mu_{+}=(\mu^{*1})_{+}=(\nu^{*1})_{+}=\nu_{+}$. Suppose that the assertion of the lemma holds true for some $N$, and suppose that $(\mu^{*n})_{+}=(\nu^{*n})_{+}$ for $n=1,2,\ldots,N,N+1$. By the induction hypothesis, formula (1) holds for $n=0,1,\ldots,N-1$ and $k=1,2,\ldots$ , and we have $\mu_{+}=\nu_{+}$. Therefore, we only need to prove (1) for $n=N$ and $k=1,2,\ldots$

By the binomial theorem,

[TABLE]

We already know that $\mu_{+}^{*N+1}=\nu_{+}^{*N+1}$ and $(\mu_{+}^{*j}*\mu_{-}^{*N+1-j})_{+}=(\nu_{+}^{*j}*\nu_{-}^{*N+1-j})_{+}$ for $j=1,2,\ldots,N-1$. Furthermore, $(\mu_{-}^{*N+1})_{+}=0=(\nu_{-}^{*N+1})_{+}$. It follows that all terms corresponding to $j\neq N$ in the above sum are zero. Thus,

[TABLE]

which proves (1) for $n=N$ and $k=1$. The proof for $n=N$ and $k>1$ proceeds again by induction: if (1) holds for $n=N$ and $k=1,2,\ldots,K$, then

[TABLE]

we used the identity $(\pi*\sigma_{-})_{+}=(\pi_{+}*\sigma_{-})_{+}$ in second, fourth, sixth and eighth equalities, (1) for $n=N$ and $k=K$ in the third one, (1) for $n=N$ and $k=1$ in the seventh one, and $\mu_{+}=\nu_{+}$ in the fifth one. We conclude that (1) holds for $n=N$ and all $k=1,2,\ldots$ , and the proof is complete. ∎

A holomorphic function $f$ on $\mathds{C}_{-}$ is said to be of bounded type (or belong to the Nevanlinna class) if $\log|f(x)|$ has a harmonic majorant on $\mathds{C}_{-}$. Equivalently, $f$ is of bounded type if it is a ratio of two bounded holomorphic functions on $\mathds{C}_{-}$. We recall the following fundamental factorisation theorem for holomorphic functions on $\mathds{C}_{-}$ which are bounded or of bounded type, and we refer to [2] for further details.

Theorem 3** (Theorem II.5.5 and Corollary II.5.7 in [2]).**

Let $f$ be a holomorphic function of bounded type on the lower complex half-plane, and suppose that $f$ is not identically zero. Let $\alpha_{0}$ be the multiplicity of the zero of $f$ at $z=-i$ (possibly $n_{0}=0$), and let $z_{1},z_{2},\ldots$ be the (finite or infinite) sequence of all zeros of $f$ in the lower complex half-plane, with corresponding multiplicities $\alpha_{1},\alpha_{2},\ldots$  Then $f$ admits a factorisation

[TABLE]

(unique, up to multiplication of $f_{\mathrm{o}}$ and $f_{\mathrm{s}}$ by a constant of modulus $1$), with the folllowing factors. The function $f_{\mathrm{b}}$ is a Blaschke product, determined uniquely by the zeros of $f$:

[TABLE]

The function $f_{\mathrm{o}}$ is an outer function, a holomorphic function determined uniquely up to multiplication by a constant of modulus $1$ by the formula:

[TABLE]

Finally, the function $f_{\mathrm{s}}$ is a singular inner function, a holomorphic function determined uniquely up to multiplication by a constant of modulus $1$ by the expression:

[TABLE]

where $a\in\mathds{R}$ is a constant and $\lambda$ is a signed measure, singular with respect to the Lebesgue measure.

Furthermore, for almost all $x\in\mathds{R}$ with respect to both the Lebesgue measure and the measure $\lambda$, the limit $f(x)$ of $f(x+iy)$ as $y\to 0^{-}$ exists. This boundary limit $f(x)$ is non-zero almost everywhere with respect to the Lebesgue measure and zero almost everywhere with respect to $\lambda$. The symbol $f(x)$ used in the definition of the outer function $f_{\mathrm{o}}$ refers precisely to this boundary limit. Additionally, we have $\sum_{j}\alpha_{j}|\operatorname{Im}z_{j}|(1+|z_{j}|^{2})^{-1}<\infty$, $\int_{-\infty}^{\infty}(1+x^{2})^{-1}|\log|f(x)||dx<\infty$ and $\int_{\mathds{R}}(1+x^{2})^{-1}|\lambda|(dx)<\infty$, and any parameters $\alpha_{j}$, $z_{j}$, $a$, $\lambda$ and boundary values $|f(x)|$, $x\in\mathds{R}$, which satisfy these conditions, correspond to some function $f$ of bounded type.

Finally, $f$ is a bounded holomorphic function in the lower complex half-plane if and only if $a\geqslant 0$, $\lambda$ is a non-negative measure and the boundary values $|f(x)|$ are bounded for $x\in\mathds{R}$.

Lemma 4**.**

Suppose that $\mu$ is a measure on $\mathds{R}$ such that $\mu_{-}$ is a non-zero measure and $(\mu_{+}*\mu_{-})_{+}=0$. Then $\hat{\mu}_{+}$ has a holomorphic extension $\varphi$ to the the connected open set

[TABLE]

and $\varphi$ is a meromorphic function on $\mathds{C}\setminus\{z\in\mathds{R}:\hat{\mu}_{-}(z)=0\}$. Furthermore, $\varphi\hat{\mu}_{-}$ extends to a function which is holomorphic on $\mathds{C}_{-}$ and continuous on $\mathds{C}_{-}\cup\mathds{R}$, namely, the characteristic function of $\mu_{+}*\mu_{-}$.

Proof.

Denote $\nu=\mu_{+}*\mu_{-}$; by the assumption, $\nu=\nu_{-}$. Let $f=\hat{\mu}_{+}$, $g=\hat{\mu}_{-}$ and $h=\hat{\nu}=\hat{\nu}_{-}$. Clearly, $h(z)=f(z)g(z)$ for $z\in\mathds{R}$. Let

[TABLE]

so that $D=\mathds{C}\setminus(A\cup B)$.

We note basic properties of $A$ and $B$. By continuity of $g$, $A$ and $A\cup B$ are closed sets, and $D$ is an open set. Since $g$ is holomorphic on $\mathds{C}_{-}$ (and not identically zero), $B$ is a countable (possibly finite) set with no accumulation points on $\mathds{C}_{-}$. By Theorem 3, $A$ has zero Lebesgue measure (as a subset of $\mathds{R}$). In particular, $D$ is connected.

We define a function $\varphi$ on $D$ by the formula

[TABLE]

By definition, $\varphi$ is holomorphic both on $\mathds{C}_{+}$ and on $\mathds{C}_{-}\setminus B$, as well as meromorphic on $\mathds{C}_{-}$. Furthermore, $\varphi$ is continuous at each point $z\in\mathds{R}\setminus A$, because both $f$ (defined on $\mathds{C}_{+}\cup\mathds{R}$) and $h/g$ (defined on $(\mathds{C}_{-}\setminus B)\cup(\mathds{R}\setminus A)$) are continuous at $z$ and $f(z)=h(z)/g(z)$. By a standard application of Morera’s theorem, $\varphi$ is holomorphic in $D$. It remains to note that $\varphi(z)g(z)=h(z)$ for $z\in\mathds{C}_{-}\setminus B$. ∎

Lemma 5**.**

If $\mu$ is a measure on $\mathds{R}$ such that $(\mu_{+}^{*n}*\mu_{-})_{+}=0$ for all $n=1,2,\ldots$ , then either $\mu_{+}$ or $\mu_{-}$ is a zero measure.

Proof.

Let $\mu$ be such a measure, and suppose that both $\mu_{+}$ and $\mu_{-}$ are non-zero measures. Let $\varphi,f,g,h,A,B,D$ be as in the proof of Lemma 4. Clearly, $\varphi^{n}$ is the holomorphic extension of $f^{n}$, the characteristic function of $\mu_{+}^{*n}$. An application of Lemma 4 to the measure $\mu_{+}^{*n}+\mu_{-}$ implies that for all $n=1,2,\ldots$ , the function $\varphi^{n}g$ extends from $\mathds{C}_{-}\setminus B$ to a function $h_{n}$ which is bounded and holomorphic on $\mathds{C}_{-}$ and continuous on $\mathds{C}_{-}\cup\mathds{R}$, namely, $h_{n}$ is the characteristic function of $\mu_{+}^{*n}*\mu_{-}$.

Consider the factorisations $g=g_{\mathrm{b}}g_{\mathrm{o}}g_{\mathrm{s}}$ and $h_{n}=h_{n,{\mathrm{b}}}h_{n,{\mathrm{o}}}h_{n,{\mathrm{s}}}$ given in Theorem 3, and let $\lambda_{g}$, $a_{g}$ and $\lambda_{h,n}$, $a_{h,n}$ denote the corresponding non-negative measures $\lambda$ and constants $a$ for $g$ and $h_{n}$, respectively. Note that Theorem 3 applies both to $g$ and to $h_{n}=\varphi^{n}g$, as these functions are not identically zero: $f$ and $g$ are characteristic functions of non-zero measures $\mu_{+}$ and $\mu_{-}$, while $h_{n}$ is the product of $g$ and the holomorphic extension of $f^{n}$.

Recall that $\varphi^{n}=h_{n}/g$ on $\mathds{C}_{-}\setminus B$. It follows that if $\varphi_{n,{\mathrm{b}}}=h_{n,{\mathrm{b}}}/g_{\mathrm{b}}$, $\varphi_{n,{\mathrm{o}}}=h_{n,{\mathrm{o}}}/g_{\mathrm{o}}$ and $\varphi_{n,{\mathrm{s}}}=h_{n,{\mathrm{s}}}/g$, then

[TABLE]

on $\mathds{C}_{-}\setminus B$. Let us examine the above factors in more detail.

By definition, $\varphi_{n,{\mathrm{o}}}$ and $\varphi_{n,{\mathrm{s}}}$ have no zeros in $\mathds{C}_{-}$. This means that if $z_{0}\in\mathds{C}_{-}$ is a pole of $\varphi$ of order $\alpha_{0}$, then $z_{0}$ is a pole of $\varphi_{n,{\mathrm{b}}}=h_{n,{\mathrm{b}}}/g_{\mathrm{b}}$ of order $n\alpha_{0}$, and therefore $g_{\mathrm{b}}$ has a zero at $z_{0}$ of multiplicity at least $n\alpha_{0}$ for all $n=1,2,\ldots$  Since all zeroes of $g_{\mathrm{b}}$ have finite multiplicity, $\varphi$ has no poles in $\mathds{C}_{-}$. In particular, $\varphi$ extends to a holomorphic function on $\mathds{C}\setminus A$, which will be denoted again by $\varphi$, and $\varphi_{n,{\mathrm{b}}}=h_{n,{\mathrm{b}}}/g_{\mathrm{b}}$ has no poles in $\mathds{C}_{-}$. Therefore, the zeros of $h_{n,{\mathrm{b}}}$ must cancel the zeros of $g_{\mathrm{b}}$, and $\varphi_{n,{\mathrm{b}}}$ is a Blaschke product.

Since $h_{n}(x)/g(x)=(f(x))^{n}$ for $x\in\mathds{R}\setminus A$ and $A$ has Lebesgue measure zero, we have

[TABLE]

In particular, $\varphi_{n,{\mathrm{o}}}$ is a bounded outer function, namely, the outer function in the factorisation of the bounded holomorphic function $(\overline{f(\bar{z})})^{n}$ on the lower complex half-plane.

Finally $\varphi_{n,{\mathrm{s}}}$ is the ratio of two singular inner functions, and hence a singular inner function. If we denote $a_{\varphi,n}=a_{h,n}-a_{g}$ and $\lambda_{\varphi,n}=\lambda_{h,n}-\lambda_{g}$, then

[TABLE]

The above properties imply that $\varphi^{n}$ is of bounded type, and therefore the factors $\varphi_{n,{\mathrm{b}}}$, $\varphi_{n,{\mathrm{o}}}$, $\varphi_{n,{\mathrm{s}}}$, the signed measure $\lambda_{\varphi,n}$ and the constant $a_{\varphi,n}\in\mathds{R}$ are uniquely determined (up to multiplication by a constant of modulus $1$ in case of $\varphi_{n,{\mathrm{o}}}$ and $\varphi_{n,\ss}$).

By comparing the factorisations of $\varphi$ and $\varphi^{n}$, we find that $\varphi_{n,{\mathrm{s}}}=c_{n}(\varphi_{1,{\mathrm{s}}})^{n}$ for some constant $c_{n}$ with modulus $1$. It follows that $a_{\varphi,n}=na_{\varphi,1}$ and $\lambda_{\varphi,n}=n\lambda_{\varphi,1}$. This, however, implies that $a_{\varphi,1}=\tfrac{1}{n}a_{\varphi,n}\geqslant-\tfrac{1}{n}a_{g}$ for all $n=1,2,\ldots$ , and so $a_{\varphi,1}\geqslant 0$. Similarly, the negative part of $\lambda_{\varphi,1}=\tfrac{1}{n}\lambda_{\varphi,n}$ is dominated by $\tfrac{1}{n}\lambda_{g}$ for any $n=1,2,\ldots$  This is not possible if the negative part of $\lambda_{\varphi,1}$ is non-zero, and therefore $\lambda_{\varphi,1}$ is a non-negative measure. We conclude that $\varphi=\varphi_{1,{\mathrm{b}}}\varphi_{1,{\mathrm{o}}}\varphi_{1,{\mathrm{s}}}$ is a bounded holomorphic function on $\mathds{C}_{-}$.

Since $\varphi=f$ on $\mathds{C}_{+}$ and $f$ is a bounded holomorphic function on $\mathds{C}_{+}$, we have proved that $\varphi$ is a bounded holomorphic function on $\mathds{C}\setminus A$. However, $A$ has zero Lebesgue measure (as a subset of $\mathds{R}$). By Painlevé’s theorem (see Theorem 2.7 in [10]), $\varphi$ extends to a bounded holomorphic function on $\mathds{C}$. This, in turn, implies that $\varphi$ is constant, and so $\hat{\mu}_{+}$ is constant, contradicting the assumption that $\mu_{+}$ is a non-zero measure on $(0,\infty)$. ∎

Proof of Theorem 1.

Suppose that $(\mu^{*n})_{+}=(\nu^{*n})_{+}$ for $n=1,2,\ldots$ for some measures $\mu$ and $\nu$ such that $\mu_{+}$ and $\nu_{+}$ are non-zero measures. By Lemma 2, $\mu_{+}=\nu_{+}$ and $(\mu_{+}^{*n}*\mu_{-})_{+}=(\nu_{+}^{*n}*\nu_{-})_{+}$ for $n=1,2,\ldots$  Let $\eta=\mu_{+}+\mu_{-}-\nu_{-}$, so that $\eta_{+}=\mu_{+}=\mu_{-}$ and $\eta_{-}=\mu_{-}-\nu_{-}$. Then $(\eta_{+}^{*n}*\eta_{-})_{+}=0$ for $n=1,2,\ldots$ , and therefore, by Lemma 5, either $\eta_{+}$ or $\eta_{-}$ is a zero measure. Since $\eta_{+}=\mu_{+}$ is a non-zero measure, we must have $\eta_{-}=0$, that is, $\mu_{-}=\nu_{-}$. ∎

3. An error in [8]

In [8] an analogue of Theorem 1 is given, with equality of $\mu^{*n}$ and $\nu^{*n}$ on $(-\infty,0)$ rather than on $(0,\infty)$. In page 3001, line 16 of [8], it is claimed that the measures $\mu$ and $\nu$ satisfy condition (B) of Theorem A in [8], as a consequence of the results of Section 11.2 in [4]. This reasoning would have been correct if the holomorphic extensions of $\hat{\mu}$ and $\hat{\nu}$ to the upper complex half-plane had been known to be continuous on the boundary. However, this is not verified in [8].

More precisely, it is observed in [8] that $\hat{\mu}=(\hat{\chi}_{2}-(\hat{\chi}_{1})^{2})/(2\hat{\chi}_{1})$ almost everywhere on $\mathds{R}$, where $\chi_{1}=\mu-\nu$ and $\chi_{2}=\mu^{*2}-\nu^{*2}$ are measures concentrated on $(0,\infty)$. Since $\hat{\chi}_{1}$ and $\hat{\chi}_{2}$ extend to holomorphic functions on $\mathds{C}_{+}$, $\hat{\mu}$ extends to a meromorphic function on $\mathds{C}_{+}$. Equality of $\mu^{*n}$ and $\nu^{*n}$ on $(-\infty,0)$ for $n\geqslant 3$ is used only to show that the extension of $\hat{\mu}$ has no poles in $\mathds{C}_{+}$. However, the extension of $\hat{\mu}$ can have singularities near $\mathds{R}$ and thus fail to satisfy condition (B) of Theorem A in [8].

To be specific, observe that $\hat{\mu}(z)=z^{2}(z+i)^{-4}\exp(i/z)$ is the characteristic function of a measure $\mu$ on $\mathds{R}$. Namely, $\mu$ is the convolution of $\tfrac{1}{6}x^{3}e^{-x}\operatorname{\mathds{1}}_{(0,\infty)}(x)dx$ and $\tfrac{1}{6}\,{{}_{0}F_{1}}(4;x)\operatorname{\mathds{1}}_{(-\infty,0)}(x)dx-\tfrac{1}{2}\delta_{0}(dx)-\delta_{0}^{\prime}(dx)-\delta_{0}^{\prime\prime}(dx)$ (in the sense of distributions; ${{}_{0}F_{1}}$ is the hypergeometric function; we omit the details). Clearly, $\hat{\mu}$ extends holomorphically to the upper complex half-plane, but this extension is not continuous on the boundary, and thus $\mu$ does not satisfy condition (B) of Theorem A in [8]. Furthermore, $\hat{\mu}(z)$ is the ratio of two characteristic functions of finite measures supported in $[0,\infty)$: $z^{4}/(z+i)^{8}$ and $z^{2}(z+i)^{-4}\exp(-i/z)$.

The author of the present article was not able to correct the error in [8]. The proof given above uses a related, but essentially different idea.

Acknowledgments

I thank Loïc Chaumont for numerous discussions on the subject of the article and encouragement. I thank Jan Rosiński for letting me know about reference [6]. I thank Alexander Ulanovskiĭ for discussions on references [6, 8]. The main part of this article was written during the 39th Conference on Stochastic Processes and their Applications in Moscow.