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\title[Random walks are determined by their trace on the positive half-line]{Random walks are determined by their trace on~the positive half-line}
\alttitle{Les marches aléatoires sont déterminées par leur trace sur la demi-droite positive}

\subjclass{60G50, 60G51, 45E10, 30H15}
\keywords{Random walk, Lévy process, Wiener--Hopf factorisation, Nevanlinna class}

%[\initial{M.} \lastname{Kwa{\'s}nicki}]
\author{\firstname{Mateusz} \lastname{Kwa{\'s}nicki}}
\address{Department of Pure Mathematics,\\ Wrocław University of Science and Technology,\\
Wybrzeże Wyspiańskiego 27,\\
50-370 Wrocław (Poland)}
\email{mateusz.kwasnicki@pwr.edu.pl}
\thanks{Work supported by the Polish National Science Centre (NCN) grant no.\@ 2015/19/B/ST1/01457}

\editor{L. Chaumont}


\begin{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.
\end{abstract}

\begin{altabstract}
Nous démontrons que la loi d'une marche aléatoire $X_n$ est déterminée par les distributions de $\max(X_n, 0)$ pour $n = 1, 2, \ldots\,$, comme l'avaient conjecturé récemment Lo\"ic Chaumont et Ron Doney. De manière équivalente, la loi de $X_n$ est déterminée par son facteur de Wiener--Hopf espace-temps ascendant. Nos méthodes relèvent de l'analyse complexe.
\end{altabstract}


\datereceived{2019-03-13}
\dateaccepted{2020-02-19}


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\begin{document}

\maketitle


\section{Introduction and main result}

In this note we give an affirmative answer to a question posed by Lo\"ic Chaumont and Ron Doney in~\cite{bib:cd17}, inspired by Vincent Vigon's conjecture in~\cite{bib:v02}. The main result was previously stated without proof in a more general form in~\cite{bib:ou90}, and an erroneous proof was given in~\cite{bib:u94}.

A random walk $X_n$ is said to be \emph{non-degenerate} if $\pr(X_n > 0) \ne 0$. Similarly, a finite signed Borel measure $\mu$ on $\R$ is said to be \emph{non-degenerate} if the restriction of $\mu$ to $(0, \infty)$ is a non-zero measure.

\begin{theo}\label{thm:wh}
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 $\mu = \nu$.
\end{theo}

Following~\cite{bib:cd17}, 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:
\begin{itemize}
\item The law of the ascending ladder process $(T_k, S_k)$; here $S_k = X_{T_k}$ is the $k^{\rm th}$ running maximum of the random walk.
\item The upward space-time Wiener--Hopf factor $\Phi_+(q, \xi)$, that is, the characteristic function of $(T_1, S_1)$.
\item The distributions of the running maxima $\max(0, X_1, X_2,\,\ldots,\,X_n)$ for all\linebreak$n = 1, 2,\,\ldots$
\end{itemize}
Theorem~\ref{thm:wh} clearly implies that a~non-degenerate Lévy process $X_t$ is determined by any of the following objects:
\begin{itemize}
\item The distributions of $\max(X_t, 0)$ for all $t > 0$ (or even for $t = 1, 2,\,\ldots$).
\item The law of the ascending ladder process $(T_t, S_t)$.
\item The upward space-time Wiener--Hopf factor $\kappa_+(q, \xi)$, that is, the characteristic exponent of $(T_t, S_t)$.
\item The distributions of the running suprema $\sup\{X_s : s \in [0, t]\}$ for all $t > 0$.
\end{itemize}
For further discussion, we again refer to~\cite{bib:cd17}, where Theorem~\ref{thm:wh} was proved under various relatively mild additional conditions. For related research, see~\cite{bib:cd17,bib:lms76,bib:o85,bib:ou90,bib:u90,bib:u94} and the references therein.

Theorem~\ref{thm:wh} was given without proof in~\cite{bib:ou90} in a more general form: Theorem~4 therein claims that $\mu = \nu$ if $\mu$ and $\nu$ are non-degenerate finite Borel measures on $\R$ 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~\cite{bib:u94} under the additional condition $n_2 = 2$, and only in dimension one. However, the argument in~\cite{bib:u94} contains a gap, that we describe at the end of this article.


\section{Proof}

All measures considered below are finite, signed Borel measures. For a measure $\mu$ on $\R$, we denote the restrictions of $\mu$ to $(0, \infty)$ and $(-\infty, 0]$ by $\mu_+ = \ind_{(0, \infty)} \mu$ and $\mu_- = \ind_{(-\infty, 0]} \mu$. 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 record the following elementary identities: $(\delta_0 * \sigma_-)_+ = (\sigma_-)_+ = 0$, $(\pi_- * \sigma_-)_+ = 0$, and $(\pi * \sigma_-)_+ = (\pi_+ * \sigma_-)_+$.

We denote the characteristic function of a measure $\mu$ by $\hat\mu$:
\[
\hat\mu(z) = \int_\R e^{i z x} \mu(dx) 
\]
for $z \in \R$, and also for those $z \in \CC$ for which the integral converges. We recall that $\hat\mu_+$ is a bounded holomorphic function in the upper complex half-plane $\CC_+ =\linebreak\{z \in \CC : \im z > 0\}$, continuous on the boundary. Similarly, $\hat\mu_-$ is a bounded holomorphic function on the lower complex half-plane $\CC_- = \{z \in \CC : \im z < 0\}$.

\begin{lemm}\label{lem:mupm}
Suppose that $\mu, \nu$ are measures on $\R$ satisfying
\[
(\mu^{*n})_+ = (\nu^{*n})_+\quad\text{for }n = 1, 2,\,\ldots,\,N.
\]
Then $\mu_+ = \nu_+$ and
\begin{equation}\label{eq:mupm}
\left(\mu_+^{*n} * \mu_-^{*k}\right)_+ = \left(\nu_+^{*n} * \nu_-^{*k}\right)_+ \quad\text{for }n = 0, 1,\,\ldots,\,N - 1\text{ and }k = 1, 2,\,\ldots
\end{equation}
\end{lemm}

\begin{proof}
We proceed by induction with respect to $N$. For $N = 1$ the result is trivial: we have $\mu_+ = (\mu^{*1})_+ = (\nu^{*1})_+ = \nu_+$ and $(\mu_+^{*0} * \mu_-^{*k})_+ = (\delta_0 * \mu_-^{*k})_+ = 0\linebreak= (\delta_0 * \nu_-^{*k})_+ = (\nu_+^{*0} * \nu_-^{*k})_+$ for $k = 1, 2,\,\ldots\,$ Suppose that the assertion of Lemma~\ref{lem:mupm} holds for some $N$, and suppose that $(\mu^{*n})_+ = (\nu^{*n})_+$ for $n = 1, 2,\,\ldots,\,N,\linebreak N + 1$. By the induction hypothesis, formula~\eqref{eq:mupm} holds for $n = 0, 1,\,\ldots,\,N - 1$ and $k = 1, 2, \ldots\,$, and we have $\mu_+ = \nu_+$. Therefore, we only need to prove~\eqref{eq:mupm} for $n = N$ and $k = 1, 2, \ldots$

By the binomial theorem,
\begin{align*}
0 & = \left(\mu^{*N + 1} - \nu^{*N + 1}\right)_+ = \left((\mu_+ + \mu_-)^{*N + 1} - (\nu_+ + \nu_-)^{*N + 1}\right)_+ \\
& = \sum_{j = 0}^{N + 1} \binom{N + 1}{j} \left(\mu_+^{*j} * \mu_-^{*N + 1 - j} - \nu_+^{*j} * \nu_-^{*N + 1 - j}\right)_+. 
\end{align*} 
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 $j \ne N$ in the above sum are zero. Thus,
\[
0 = \binom{N + 1}{N} \left(\mu_+^{*N} * \mu_- - \nu_+^{*N} * \nu_-\right)_+,
\]
which proves~\eqref{eq:mupm} for $n = N$ and $k = 1$. The proof for $n = N$ and $k > 1$ proceeds again by induction. Suppose that~\eqref{eq:mupm} holds for $n = N$ and $k = 1, 2, \ldots, K$. By the identity $(\pi * \sigma_-)_+ = (\pi_+ * \sigma_-)_+$,
\[
\left(\mu_+^{*N} * \mu_-^{*K + 1}\right)_+ = \left(\mu_+^{*N} * \mu_-^{*K} * \mu_-\right)_+ = \left(\left(\mu_+^{*N} * \mu_-^{*K}\right)_+ * \mu_-\right)_+.
\]
Applying~\eqref{eq:mupm}, with $n = N$ and $k = K$, and then again the identity $(\pi * \sigma_-)_+\linebreak= (\pi_+ * \sigma_-)_+$, we find that
\[
\left(\left(\mu_+^{*N} * \mu_-^{*K}\right)_+ * \mu_-\right)_+ = \left(\left(\nu_+^{*N} * \nu_-^{*K}\right)_+ * \mu_-\right)_+ = \left(\nu_+^{*N} * \nu_-^{*K} * \mu_-\right)_+.
\]
Recall that $\mu_+ = \nu_+$, so that
\[
\left(\nu_+^{*N} * \nu_-^{*K} * \mu_-\right)_+ = \left(\mu_+^{*N} * \mu_- * \nu_-^{*K}\right)_+.
\]
We use the identity $(\pi * \sigma_-)_+ = (\pi_+ * \sigma_-)_+$ for the third time, and then we again apply~\eqref{eq:mupm}, with $n = N$ and $k = 1$:
\[
\left(\mu_+^{*N} * \mu_- * \nu_-^{*K}\right)_+ = \left(\left(\mu_+^{*N} * \mu_-\right)_+ * \nu_-^{*K}\right)_+ = \left(\left(\nu_+^{*N} * \nu_-\right)_+ * \nu_-^{*K}\right)_+.
\]
Finally, once again we apply the identity $(\pi * \sigma_-)_+ = (\pi_+ * \sigma_-)_+$:
\[
\left(\left(\nu_+^{*N} * \nu_-\right)_+ * \nu_-^{*K}\right)_+ = \left(\nu_+^{*N} * \nu_- * \nu_-^{*K}\right)_+ = \left(\nu_+^{*N} * \nu_-^{*K + 1}\right)_+.
\]
The above chain of equalities implies that $(\mu_+^{*N} * \mu_-^{*K + 1})_+ = (\nu_+^{*N} * \nu_-^{*K + 1})_+$, which is just~\eqref{eq:mupm} with $n = N$ and $k = K + 1$. We conclude that~\eqref{eq:mupm} holds for $n = N$ and every $k = 1, 2,\,\ldots\,$, and the proof of Lemma~\ref{lem:mupm} is complete.
\end{proof}

A holomorphic function $f$ on $\CC_-$ is said to be of \emph{bounded type} (or belong to the \emph{Nevanlinna class}) if $\log |f(x)|$ has a harmonic majorant on $\CC_-$. Equivalently, $f$ is of bounded type if it is a ratio of two bounded holomorphic functions on $\CC_-$. We recall the following fundamental factorisation theorem for holomorphic functions on $\CC_-$ which are bounded or of bounded type, and we refer to~\cite{bib:g07,bib:m09} for further details.

\begin{theo}{\cite[Theorem~II.5.5 and Corollary~II.5.7]{bib:g07}; \cite[Theorem~13.15]{bib:m09}}\label{thm:fact}

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 $\alpha_0 = 0$), 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
\begin{equation}\label{eq:fact}
f(z) = f_\bb(z) f_\oo(z) f_\si(z)
\end{equation}
(unique, up to multiplication of $f_\oo$ and $f_\si$ by a constant of modulus~$1$), with the following factors. The function $f_\bb$ is a \emph{Blaschke product}, determined uniquely by the zeros of $f$:
\begin{equation}\label{eq:bb}
f_\bb(z) = \biggl(\frac{z + i}{z - i} \biggr)^{\!\alpha_0} \prod_j \biggl(\frac{|1 + z_j^2|}{1 + z_j^2} \, \frac{z - z_j}{z - \bar{z}_j} \biggr)^{\!\alpha_j}.
\end{equation}
The function $f_\oo$ is an \emph{outer function}, a holomorphic function determined uniquely up to multiplication by a constant of modulus~$1$ by the formula:
\begin{equation}\label{eq:oo}
|f_\oo(z)| = \exp \biggl(\frac{1}{\pi} \int_{-\infty}^\infty \frac{-\im z}{|z - x|^2} \, \log |f(x)| \, dx \biggr).
\end{equation}
Finally, the function $f_\si$ is a \emph{singular inner function}, a holomorphic function determined uniquely up to multiplication by a constant of modulus~$1$ by the expression:
\begin{equation}\label{eq:si}
|f_\si(z)| = \exp \biggl(a \im z - \frac{1}{\pi} \int_\R \frac{-\im z}{|z - x|^2} \, \sigma(dx) \biggr),
\end{equation}
where $a \in \R$ is a constant and $\sigma$ is a signed measure, singular with respect to the Lebesgue measure.

Furthermore, for almost all $x \in \R$ with respect to both the Lebesgue measure and the measure $\sigma$, the limit $f(x)$ of $f(x + i y)$ 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 $\sigma$. The symbol $f(x)$ used in the definition of the outer function $f_\oo$ refers precisely to this boundary limit. Additionally, we have \[
\sum_j \alpha_j |\im z_j| \left(1 + |z_j|^2\right)^{-1} < \infty, \int_{-\infty}^\infty \left(1 + x^2\right)^{-1} |\log |f(x)|| dx < \infty
\] 
and
\[
\int_\R \left(1 + x^2\right)^{-1} |\sigma|(dx) < \infty\,,
\]
 and any parameters $\alpha_j$, $z_j$, $a$, $\sigma$ and boundary values $|f(x)|$, $x \in \R$, 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 $a \ge 0$, $\sigma$ is a non-negative measure and the boundary values $|f(x)|$ are bounded for $x \in \R$.
\end{theo}

\begin{lemm}\label{lem:ext}
Suppose that $\mu$ is a measure on $\R$ such that $\mu_-$ is a non-zero measure and $(\mu_+ * \mu_-)_+ = 0$. Then $\hat\mu_+$ has a holomorphic extension $\ph$ to the connected open set
\[
D = \CC \setminus \left\{z \in \CC_- \cup \R : \hat\mu_-(z) = 0\right\},
\]
and $\ph$ is a meromorphic function on $\CC \setminus \{z \in \R : \hat\mu_-(z) = 0\}$. Furthermore, $\ph \hat\mu_-$ extends to a function which is holomorphic on $\CC_-$ and continuous on $\CC_- \cup \R$, namely, the characteristic function of $\mu_+ * \mu_-$.
\end{lemm}

\begin{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 $z \in \R$. Let
\[
A = \{ z \in \R : g(z) = 0 \},\quad B = \{ z \in \CC_- : g(z) = 0 \},
\]
so that $D = \CC \setminus (A \cup B)$.

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 $\CC_-$ (and not identically zero), $B$ is a countable (possibly finite) set with no accumulation points on $\CC_-$. By Theorem~\ref{thm:fact}, $A$ has zero Lebesgue measure (as a subset of $\R$). In particular, $D$ is connected. Indeed: the sets $D \cap \CC_+ = \CC_+$ and $D \cap \CC_- = \CC_- \setminus B$ are clearly path-connected, the set $D \cap \R = \R \setminus A$ is non-empty, and since $D$ is open, each point of $D \cap \R$ is path-connected with points from both $D \cap \CC_+$ and $D \cap \CC_-$.

We define a function $\ph$ on $D$ by the formula
\[
\ph(z) =
\begin{cases}
f(z) & \text{if }z \in \CC_+ \cup (\R \setminus A), \\[0.5em]
\dfrac{h(z)}{g(z)} & \text{if }z \in \CC_- \setminus B\,.
\end{cases}
\]
By definition, $\ph$ is holomorphic both on $\CC_+$ and on $\CC_- \setminus B$, as well as meromorphic on $\CC_-$. Furthermore, $\ph$ is continuous at each point $z \in \R \setminus A$, because both $f$ (defined on $\CC_+ \cup \R$) and $h / g$ (defined on $(\CC_- \setminus B) \cup (\R \setminus A)$) are continuous at\linebreak$z$ and $f(z) = h(z) / g(z)$. By a standard application of Morera's theorem (see~\cite[Theorem~IV.5.10 and Exercise~IV.5.9]{bib:c73}, or~\cite[Exercise~II.12]{bib:g07}), $\ph$ is holomorphic in $D$. It remains to note that $\ph(z) g(z) = h(z)$ for $z \in \CC_- \setminus B$.
\end{proof}

\begin{lemm}\label{lem:wh}
If $\mu$ is a measure on $\R$ such that $(\mu_+^{*n} * \mu_-)_+ = 0$ for all\linebreak$n = 1, 2,\,\ldots\,$, then either $\mu_+$ or $\mu_-$ is a zero measure.
\end{lemm}

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

Consider the factorisations $g = g_\bb\,g_\oo\,g_\si$ and $h_n = h_{n,\bb}\,h_{n,\oo}\, h_{n,\si}$ given in Theorem~\ref{thm:fact}, and let $\sigma_g$, $a_g$ and $\sigma_{h,n}$, $a_{h,n}$ denote the corresponding non-negative measures $\sigma$ and constants $a$ for $g$ and $h_n$, respectively. Note that Theorem~\ref{thm:fact} applies both to $g$ and to $h_n = \ph^n g$, 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 $\ph^n = h_n / g$ on $\CC_- \setminus B$. It follows that if $\ph_{n,\bb} = h_{n,\bb} / g_\bb$, $\ph_{n,\oo} = h_{n,\oo} / g_\oo$ and $\ph_{n,\si} = h_{n,\si} / g$, then
\[
\ph^n = \ph_{n,\bb}\,\ph_{n,\oo}\,\ph_{n,\si} \]
on $\CC_- \setminus B$. Let us examine the above factors in more detail.

By definition, $\ph_{n,\oo}$ and $\ph_{n,\si}$ have no zeros in $\CC_-$. This means that if $z_0 \in \CC_-$ is a pole of $\ph$ of order $\alpha_0$, then $z_0$ is a pole of $\ph_{n,\bb} = h_{n,\bb} / g_\bb$ of order $n \alpha_0$, and therefore $g_\bb$ has a zero at $z_0$ of multiplicity at least $n \alpha_0$ for all $n = 1, 2, \ldots$\, Since all zeroes of $g_\bb$ have finite multiplicity, $\ph$ has no poles in $\CC_-$. In particular, $\ph$ extends to a~holomorphic function on $\CC \setminus A$, which will be denoted again by $\ph$, and $\ph_{n,\bb} = h_{n,\bb} / g_\bb$ has no poles in $\CC_-$. Therefore, the zeros of $h_{n,\bb}$ must cancel the zeros of $g_\bb$, and $\ph_{n,\bb}$ is a~Blaschke product.

Since $h_n(x) / g(x) = (f(x))^n$ for $x \in \R \setminus A$ and $A$ has Lebesgue measure zero, we have
\begin{align*}
|\ph_{n,\oo}(z)| & = \exp \biggl(\frac{1}{\pi} \int_{-\infty}^\infty \frac{-\im z}{|z - x|^2} \, (\log |h_n(x)| - \log |g(x)|) \, dx \biggr) \\
& = \exp \biggl(\frac{1}{\pi} \int_{-\infty}^\infty \frac{-\im z}{|z - x|^2} \, \log |f(x)|^n \, dx \biggr). 
\end{align*}
In particular, $\ph_{n,\oo}$ 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 $\ph_{n,\si}$ is the ratio of two singular inner functions, and hence a singular inner function. If we denote $a_{\ph,n} = a_{h,n} - a_g$ and $\sigma_{\ph,n} = \sigma_{h,n} - \sigma_g$, then
\[
|\ph_{n,\si}(z)| = \exp \biggl(-a_{\ph,n} \im z - \frac{1}{\pi} \int_\R \frac{-\im z}{|z - x|^2} \, \sigma_{\ph,n}(dx) \biggr).
\]

The above properties imply that $\ph^n$ is of bounded type, and therefore the factors $\ph_{n,\bb}$, $\ph_{n,\oo}$, $\ph_{n,\si}$, the signed measure $\sigma_{\ph,n}$ and the constant $a_{\ph,n} \in \R$ are uniquely determined (up to multiplication by a constant of modulus $1$ in case of $\ph_{n,\oo}$ and~$\ph_{n,\si}$).

By comparing the factorisations of $\ph$ and $\ph^n$, we find that $\ph_{n,\si} = c_n (\ph_{1,\si})^n$ for some constant $c_n$ with modulus $1$. It follows that $a_{\ph,n} = n a_{\ph,1}$ and $\sigma_{\ph,n} = n \sigma_{\ph,1}$. This, however, implies that $a_{\ph,1} = \tfrac{1}{n} a_{\ph,n} \ge - \tfrac{1}{n} a_g$ for all $n = 1, 2, \ldots$\,, and so $a_{\ph,1} \ge 0$. Similarly, the negative part of $\sigma_{\ph,1} = \tfrac{1}{n} \sigma_{\ph,n}$ is dominated by $\tfrac{1}{n} \sigma_g$ for any $n = 1, 2, \ldots\,$This is not possible if the negative part of $\sigma_{\ph,1}$ is non-zero, and therefore $\sigma_{\ph,1}$ is a~non-negative measure. We conclude that $\ph = \ph_{1,\bb}\,\ph_{1,\oo}\,\ph_{1,\si}$ is a bounded holomorphic function on $\CC_-$.

Since $\ph = f$ on $\CC_+$ and $f$ is a bounded holomorphic function on $\CC_+$, we have proved that $\ph$ is a bounded holomorphic function on $\CC \setminus A$. However, $A$ has zero Lebesgue measure (as a subset of $\R$). By Painlevé's theorem (see~\cite[Theorem~2.7]{bib:y15}),\linebreak$\ph$ extends to a bounded holomorphic function on $\CC$. This, in turn, implies that $\ph$ is constant, and so $\hat\mu_+$ is constant, contradicting the assumption that $\mu_+$ is a non-zero measure on $(0, \infty)$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:wh}]
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~\ref{lem:mupm}, $\mu_+ = \nu_+$ 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~\ref{lem:wh}, either $\eta_+$ or $\eta_-$ is a~zero measure. Since $\eta_+ = \mu_+$ is a~non-zero measure, we must have $\eta_- = 0$, that is, $\mu_- = \nu_-$.
\end{proof}


\section{An error in~\cite{bib:u94}}

In~\cite{bib:u94} an analogue of Theorem~\ref{thm:wh} is given, with equality of $\mu^{*n}$ and $\nu^{*n}$ on $(-\infty, 0)$ rather than on $(0, \infty)$. In~\cite[Page~3001, line~16]{bib:u94}, it is claimed that the measures $\mu$ and $\nu$ satisfy~\cite[condition~(B) of Theorem~A]{bib:u94}, as a consequence of the results of~\cite[Section~11.2]{bib:lo77}. 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~\cite{bib:u94}.

More precisely, it is observed in~\cite{bib:u94} that $\hat{\mu} = (\hat{\chi}_2 - (\hat{\chi}_1)^2) / (2 \hat{\chi}_1)$ almost everywhere on~$\R$, 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 $\CC_+$, $\hat{\mu}$ extends to a meromorphic function on $\CC_+$. Equality of $\mu^{*n}$ and $\nu^{*n}$ on $(-\infty, 0)$ for $n \ge 3$ is used only to show that the extension of $\hat{\mu}$ has no poles in $\CC_+$. However, the extension of $\hat{\mu}$ can have singularities near $\R$ and thus fail to satisfy~\cite[condition~(B) of Theorem~A]{bib:u94}.

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 $\R$. Namely, $\mu$ is the convolution of $\tfrac{1}{6} x^3 e^{-x} \ind_{(0, \infty)}(x) dx$ and $\tfrac{1}{6} \, {_0F_1}(4; x) \ind_{(-\infty, 0)}(x) dx - \tfrac{1}{2} \delta_0(dx) - \delta_0'(dx) - \delta_0''(dx)$ (in the sense of distributions; ${_0F_1}$ 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~\cite[condition~(B) of Theorem~A]{bib:u94}. 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~\cite{bib:u94}. The proof given above uses a related, but essentially different idea.


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\subsection*{Acknowledgments}
I thank Lo\"ic Chaumont for numerous discussions on the subject of the article and encouragement. I thank Jan Rosiński for letting me know about reference~\cite{bib:ou90}.\linebreak I thank Alexander Ulanovski\u{\i} for discussions on references~\cite{bib:ou90,bib:u94}. The main part of this article was written during the \emph{$39^{\rm th}$ Conference on Stochastic Processes and their Applications} in Moscow.



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