---
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 Loïc 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.'
address: |
    Mateusz Kwaśnicki\
    Faculty of Pure and Applied Mathematics\
    Wrocław University of Science and Technology\
    ul. Wybrzeże Wyspiańskiego 27\
    50-370 Wrocław, Poland
author:
- Mateusz Kwaśnicki
title: |
    Random walks are determined\
    by their trace on the positive half-line
---

[^1]

Introduction and main result
============================

In this note we give an affirmative answer to the question posed by Loïc Chaumont and Ron Doney in [@bib:cd17], inspired by Vincent Vigon’s conjecture in [@bib:v02]. The main result was previously stated without proof in a more general form in [@bib:ou90], and an erroneous proof was given in [@bib:u94].

A random walk $X_n$ is said to be *non-degenerate* if ${\mathds{P}}(X_n > 0) \ne 0$. 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.

\[thm:wh\] 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 $\mu = \nu$.

Following [@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:

-   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 \[thm:wh\] 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 [@bib:cd17], where Theorem \[thm:wh\] was proved under various relatively mild additional conditions. For related research, see [@bib:cd17; @bib:lms76; @bib:o85; @bib:ou90; @bib:u90; @bib:u94] and the references therein.

Theorem \[thm:wh\] was given without proof in [@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 ${\mathds{R}}$ 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 [@bib:u94] under the additional condition $n_2 = 2$, and only in dimension one. However, the argument in [@bib:u94] contains a gap, that we describe at the end of this article.

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 $\hat\mu$: [ ]{} 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 ${\mathds{C}}_- = \{z \in {\mathds{C}}: \operatorname{Im}z < 0\}$.

\[lem:mupm\] Suppose that $\mu, \nu$ are measures on ${\mathds{R}}$ satisfying [ ]{} Then $\mu_+ = \nu_+$ and [ ]{}

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  holds for $n = 0, 1, \ldots, N - 1$ and $k = 1, 2, \ldots$, and we have $\mu_+ = \nu_+$. Therefore, we only need to prove  for $n = N$ and $k = 1, 2, \ldots$

By the binomial theorem, [ ]{} 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, [ ]{} which proves  for $n = N$ and $k = 1$. The proof for $n = N$ and $k > 1$ proceeds again by induction: if  holds for $n = N$ and $k = 1, 2, \ldots, K$, then [ ]{} we used the identity $(\pi * \sigma_-)_+ = (\pi_+ * \sigma_-)_+$ in second, fourth, sixth and eighth equalities, for $n = N$ and $k = K$ in the third one, for $n = N$ and $k = 1$ in the seventh one, and $\mu_+ = \nu_+$ in the fifth one. We conclude that  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 [@bib:g07] for further details.

\[thm:fact\] 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 $n_0 = 0$), 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 [ ]{} (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 $f$: [ ]{} 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: [ ]{} 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: [ ]{} 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 $f(x + i y)$ 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 $\lambda$. 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 $x \in {\mathds{R}}$.

\[lem:ext\] 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 [ ]{} 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 $\mu_+ * \mu_-$.

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 [ ]{} 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 \[thm:fact\], $A$ 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 [ ]{} 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 $z \in {\mathds{C}}_- \setminus B$.

\[lem:wh\] 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.

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 \[lem:ext\]. Clearly, ${\varphi}^n$ is the holomorphic extension of $f^n$, the characteristic function of $\mu_+^{*n}$. An application of Lemma \[lem:ext\] 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 \[thm:fact\], 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 \[thm:fact\] 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 [ ]{} 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 [ ]{} 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 [ ]{}

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 $a_{{\varphi},n} = n a_{{\varphi},1}$ 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 [@bib:y15]), ${\varphi}$ 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 $(0, \infty)$.

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 \[lem:mupm\], $\mu_+ = \nu_+$ 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 \[lem:wh\], either $\eta_+$ or $\eta_-$ is a zero measure. Since $\eta_+ = \mu_+$ is a non-zero measure, we must have $\eta_- = 0$, that is, $\mu_- = \nu_-$.

An error in [@bib:u94]
======================

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

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

Acknowledgments {#acknowledgments .unnumbered}
---------------

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 [@bib:ou90]. I thank Alexander Ulanovskiĭ for discussions on references [@bib:ou90; @bib:u94]. The main part of this article was written during the *39th Conference on Stochastic Processes and their Applications* in Moscow.

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[^1]: Work supported by the Polish National Science Centre (NCN) grant no. 2015/19/B/ST1/01457
