%\documentclass[10pt]{article} \documentclass[11pt,draft]{article} \voffset -24.5mm \hoffset -5mm \textwidth 165mm \textheight 245mm \oddsidemargin=6mm \evensidemargin=6mm \usepackage[intlimits]{amsmath} \usepackage{amssymb} \usepackage{amsthm} \usepackage[mathscr]{eucal} %\usepackage{cite} %\usepackage{dsfont} %\usepackage[dvips]{graphicx} \usepackage{indentfirst} \allowdisplaybreaks[4] \renewcommand{\baselinestretch}{1.05} %********************************************* \renewcommand{\theequation}{\arabic{section}.\arabic{equation}} %********************************************* \makeatletter \@addtoreset{equation}{section} \makeatother \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}{Lemma}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{remark}{Remark}[section] %new \newtheorem{proposition}{Proposition}[section] \newtheorem{example}{Example}[section] \begin{document} \author{V.Nekrutkin\thanks{Supported by the RFFI grant 129755 ${\rm N^o}$ 3072507.}} \title{P\'{o}ly{a} processes and sequences: the martingale characterization} \date{} \maketitle \begin{abstract} It is proved that a mixed Poisson process $\xi_t$ is either P\'{o}ly{a} or pure Poisson process iff there exists a non-degenerate linear transformation $\xi_t\mapsto \eta_t = a(t)\xi_t+b(t)$ such that $\eta_t$ is a martingale. Analogous result is demonstrated for P\'{o}ly{a} sequences. \end{abstract} \bigskip \noindent {\small {\bf 2000 Mathematics Subject Classification}: primary 60G55, secondary 60G44, 91B42.}\\ {\small {\bf Keywords}:\, Mixed Poisson processes, P\'{o}lya processes, martingale characterization.} \section{Introduction} Let $\eta_{\centerdot}=\{\eta_i, i\geq 1\}$ be a sequence of i.i.d. random variables such that $\mathcal L(\eta_i)=\mathrm{EXP}(1)$, where $\mathcal L(\xi)$ stands for the distribution of the random variable $\xi$ and $\mathrm{EXP}(\lambda)$ is the exponential distribution with the parameter $\lambda$. Consider a non-negative %\marginpar{?} random variable $\tau$ with the distribution ${\mathcal P}=\mathcal L(\tau)$ and assume that $\tau$ is independent of $\eta_\centerdot$. %and denote ${\mathcal P}=\mathcal L(\tau)$. Then the random process $N_{\centerdot}=\{N_t,\,t\geq 0\}$ defined by the equality \begin{gather} \label{eq:MP_exp} N_t=\max\{n: \eta_1+\ldots+\eta_n0$ ($k$ is the {\em shape parameter} and $\mu$ the {\em scale parameter}). Then $N_{\centerdot}$ is called {\em the P\'{o}lya process} %(the other term is {\em gamma-Poison process}) with the parameters $k,\mu$. The natural problem is to characterize P\'{o}lya processes in the class of all mixed Poisson processes. Several characterization results can be found in \cite{Gr97}. For example (see \cite[th.\!\! 2.2]{Gr97}), if $\mathbb{E}\tau\!<\!\infty$ and %there exists such $t_0\!>\!0$ that $\mathbb{E}(\tau\,|\,N_{t_0}=n)\!=\!a_{t_0}n\!+\!b_{t_0}$ for some $t_0\!>\!0$ and any $n>0$, then $N_{\centerdot}$ is either a Poisson process or a P\'{o}lya process. Other characterizations of P\'{o}lya/Poisson processes are given in terms of their transition intensities and in terms of time-transformation of the corresponding mixed Poison processes (see \cite[th. 4.1 and th. 4.3]{Gr97}). The aim of this paper is to characterize P\'{o}lya processes in martingale terms. The interest to martingale characterizations can be explained as follows. By definition, the process $\xi_{\centerdot}$ is the martingale relative to the {\em natural filtration} ${\mathcal F}^{\,\xi}_t\!=\!\sigma(\xi_s, 0\leq s\leq t)$, if $\mathbb{E}|\xi_t|\!<\!\infty$ for any $t\geq 0$ and \begin{gather} \label{eq:mart} \mathbb{E}(\xi_T\,|{\mathcal F}^{\,\xi}_t)=\xi_t \qquad a.s. \end{gather} for any $T\geq t\geq0$. If additionally $\mathbb{E}\xi_t^2<\infty$ for any $t\geq 0$, then the left side of \eqref{eq:mart} has the interpretation in terms of prediction: $\mathbb{E}(\xi_T\,|{\mathcal F}^{\,\xi}_t)$ is the best least-square approximation of $\xi_T$ in the class $L^2_{{\mathcal F}^{\,\xi}_t}$ of all random variables $\zeta$ such that a) $\mathbb{E}\zeta^2\!<\!\infty$ and b) $\zeta$ depends only on the trajectory $\{\xi_s, \, 0\!\leq\! s\!\leq\! t\}$ of the process $\xi_{\centerdot}$\,. In other words, \eqref{eq:mart} means that $\xi_t$ is the best (least-square) prediction of $\xi_T$ for any $T>t$, if the approximation is based only on the knowledge of the trajectory of the process $\xi_{\centerdot}$ up to time $t$. Evidently, such a prediction seems to be somewhat degenerate. Wiener process is the most well-known example of this situation. A bit less trivial case occurs if some non-degenerate linear trans\-for\-ma\-tion of $\xi_{\centerdot}$ is a martingale (rather than the process $\xi_{\centerdot}$ itself). Indeed, assume that $\mathbb{E}\xi_t^2<\infty$ for any $t\geq 0$ and consider the process $\zeta_t=b(t)(\xi_t+a(t))$ with $b(t)\neq 0$ for any $t\geq 0$. Evidently, ${\mathcal F}^{\,\zeta}_t \stackrel{\rm def}=\sigma(\zeta_s, 0\leq s\leq t)= {\mathcal F}^{\,\xi}_t \stackrel{\rm def}=\sigma(\xi_s, 0\leq s\leq t)$. If $\zeta_{\centerdot}$ is the martingale relative to the natural filtration ${\mathcal F}^{\,\zeta}_t$, then for any $T>t$ \begin{gather} \mathbb{E}\big(\xi_T\,|\,{\mathcal F}^{\,\xi}_t\big)= %\frac{1}{b(T)}\,\mathbb{E}(\zeta_T\,|\,{\mathcal F}^{\,\zeta}_t)-a(T)= %\nonumber %\\ %=\frac{1}{b(T)}\,b(t)(\xi_t+a(t))-a(T)= \frac{b(t)}{b(T)}\,\xi_t+\frac{b(t)a(t)}{b(T)}-a(T). \label{eq:mart_lin} \end{gather} (Of course, this equality is valid only almost surely. Starting from this point, we ignore these technicalities.) Therefore, the best least-square prediction of $\xi_T$ is equal to the right side of \eqref{eq:mart_lin}. The good example of such a situation is the Poisson process with the parameter $\lambda$: then $b(t)\!\equiv\!1$, $a(t)=-\lambda t$ and the best prediction of $\xi_T$ is $\xi_t+\lambda(T-t)$. Other processes with independent increments and finite second moments produce analogous examples. %Below\marginpar{?} Our aim is to prove that a non-degenerate linear transformation of a mixed Poisson process is the martingale iff this process is either a Poisson process or a P\'{o}lya process. (Note that a P\'{o}lya process has by no means independent increments.) Analogous proposition is demonstrated for the so-called P\'{o}lya sequences, common in some popular models of consumer behavior. \section{Characterization of P\'{o}lya processes} Let us start with the simple lemma. \begin{lemma} \label{lem:MPOS} Let $N_\centerdot$ be a mixed Poison process with the structure distribution $\mathcal P$. Denote ${\mathcal F}^{\,N}_t=\sigma(N_s, 0\leq s\leq t)$. Then \begin{gather} \label{eq:cond_exp} \mathbb{E}(N_{T}\,|\,{\mathcal F}^{\,N}_t)=N_t+(T-t)\,\frac{\int_{(0,\infty)} \lambda^{N_t+1} e^{-\lambda t}{\mathcal P}(d\lambda)} {\int_{(0,\infty)}\lambda^{N_t} e^{-\lambda t}{\mathcal P}(d\lambda)} \end{gather} for any $T>t\geq0$. \end{lemma} \begin{proof} It is easy to show that \begin{gather*} \mathbb{P}(N_{t_1}=m_1,\ldots,N_{t_n}=m_n)= \prod_{i=1}^{n-1}\frac{(t_{i+1}-t_i)^{m_{i+1}-m_i}}{(m_{i+1}-m_i)!} \int_{(0,\infty)}\lambda^{m_n}e^{-\lambda t_n}{\mathcal P}(d\lambda) %\label{eq:mult_distr_cp} \end{gather*} for any $n\geq 1$, $0=t_0t>0$. Analogously, if $t=0$ and $m\geq 0$, then \begin{gather} \label{eq:distr} \mathbb{P}(N_{T}=m\,|\,N_0=0)=\mathbb{P}(N_{T}=m)= \frac{T^m}{m!}\,\int_{(0,\infty)}\lambda^{m}e^{-\lambda T}{\mathcal P}(d\lambda). \end{gather} %It follows from the\marginpar{?} % Markov property of the process %$N_\centerdot$ that In view of \eqref{eq:tr_func} and \eqref{eq:distr}, we get \eqref{eq:cond_exp}. \end{proof} \begin{proposition} 1. Consider a P\'{o}lya process $N_\centerdot$ with the parameters $k,\mu$ and denote $\zeta_t=(N_t+k)/(t+\mu)$. Then $\zeta_\centerdot$ is the martingale relative to the natural filtration ${\mathcal F}^{\,\zeta}_t$.\\ 2. Let $N_\centerdot$ be a mixed Poisson process with the structure distribution $\mathcal P$. Suppose that there exist functions $b(\cdot)$ and $a(\cdot)$ such that the process $\zeta_t=b(t)(N_t+a(t))$ is the martingale relative to the natural filtration ${\mathcal F}^{\,\zeta}_t$. Suppose additionally that $b(t)\neq 0$ for any $t\!\geq\! 0$. Then either ${\mathcal P}=\delta_{\alpha_0}$ for some $\alpha_0>0$ or ${\mathcal P}={\rm G}(k,\mu)$ with certain $k,\mu>0$. \end{proposition} \begin{proof} 1. As it was already mentioned, ${\mathcal F}^{\,\zeta}_t={\mathcal F}^{\,N}_t$. Since ${\mathcal P}={\rm G}(k,\mu)$, then \begin{gather} \label{eq:gamma_mom_exp} \int_{(0,\infty)}\lambda^{l} e^{-\lambda t}{\mathcal P}(d\lambda)= %\frac{\mu^k}{\Gamma(k)}\int_{0}^\infty \lambda^{k+l-1}e^{-(\mu+t)\lambda}d\lambda= \frac{\mu^k}{\Gamma(k)}\,\frac{\Gamma(k+l)}{(\mu+t)^{k+l}} \end{gather} for any $l\geq 0$. In view of \eqref{eq:gamma_mom_exp}, we obtain from \eqref{eq:cond_exp} that \begin{gather*} \mathbb{E}(N_{T}\,|\,{\mathcal F}^{\,N}_t) =N_t+(T-t)\,\frac{k+N_t}{\mu+t}=\frac{N_t(\mu+T)+k(T-t)}{\mu+t}\ . \end{gather*} Therefore, \begin{gather*} \mathbb{E}(\zeta_T\,|\,{\mathcal F}^{\,\zeta}_t)= \frac{\mathbb{E}(N_T\,|\,{\mathcal F}^{\,N}_t)+k}{\mu+T}+\frac{k}{\mu+T}= \frac{N_t}{\mu+t}+\frac{k(T-t)}{(\mu+t)(\mu+T)}+\frac{k}{\mu+T}=\zeta_t \end{gather*} and the first assertion is proved.\\ 2. Since $\zeta_\centerdot$ is the martingale, \begin{gather} \label{eq:Mart} \mathbb{E}(\zeta_{t+\Delta}\,|\,{\mathcal F}^\zeta_t)= b(t+\Delta)\Big(\mathbb{E}(N_{t+\Delta}\,|{\mathcal F}^N_t)+a(t+\Delta)\Big)=\zeta_t=b(t)(N_t+a(t)) \end{gather} for any $t,\Delta\geq 0$. In view of Markov property of the process $N_\centerdot$\,, we obtain from \eqref{eq:Mart} that \begin{gather} \mathbb{E}(N_{t+\Delta}\,|N_t=l)=l\,\frac{b(t)}{b(t+\Delta)}+\frac{b(t)a(t)}{b(t+\Delta)}-a(t+\Delta) \label{eq:cond_exp_hyp} \end{gather} for any $t>0$ and integer $l\geq 0$. Denote \begin{gather*} f_1(t,\Delta)=\frac{b(t)-b(t+\Delta)}{b(t+\Delta)}\quad \mathrm{and} \quad f_2(t,\Delta)=\frac{b(t)a(t)}{b(t+\Delta)}-a(t+\Delta). \end{gather*} It follows from (\ref{eq:cond_exp}) and (\ref{eq:cond_exp_hyp}) that functions $f_1(t,\Delta)$ and $f_2(t,\Delta)$ are both proportional to $\Delta$: $f_1(t,\Delta)=c_1(t)\Delta$ and $f_2(t,\Delta)=c_2(t)\Delta$ with $c_2(t)>0$ and $c_1(t)\geq 0$. Moreover, if $c_1(t_0)= 0$ for some $t_0>0$, then $c_1(t)= 0$ for any $t>0$. As the result, we come to the equality \begin{gather} \label{eq:mom_recc} \frac{\int_{(0,\infty)}\lambda^{l+1} e^{-\lambda t}{\mathcal P}(d\lambda)} {\int_{(0,\infty)}\lambda^{l} e^{-\lambda t}{\mathcal P}(d\lambda)}=c_1(t)l+c_2(t), \quad t>0, \quad l\geq 0. \end{gather} Let us fix $t_0>0$ and denote $c_1=c_1(t_0)$ and $c_2=c_2(t_0)$ for brevity. Additionally, denote \begin{gather*} \gamma=\int_{(0,\infty)}e^{-\lambda t_0}{\mathcal P}(d\lambda) \end{gather*} and ${\mathcal Q}(d\lambda)\!=\!e^{-\lambda t_0}{\mathcal P}(d\lambda)/\gamma$. Evidently, $Q$ is the probability distribution. Now the equality \eqref{eq:mom_recc} obtains the form \begin{gather} \label{eq:mom_recc1} \frac{\int_{(0,\infty)}\lambda^{l+1} {\mathcal Q}(d\lambda)} {\int_{(0,\infty)}\lambda^{l} {\mathcal Q}(d\lambda)}=c_1l+c_2. \end{gather} Let $c_1 =0$. Then ${\int_{(0,\infty)}\lambda^{l} {\mathcal Q}(d\lambda)}=c_2^{\,l}$ for any integer $l\geq 0$ and ${\mathcal Q}=\delta_{c_2}$. Therefore ${\mathcal Q}=\delta_{\alpha}$ with $\alpha=c_2$ and $N_\centerdot$ is the Poisson process with the parameter $\alpha$. Suppose that $c_1>0$. If $\mathcal R$ stands for the gamma-distribution $\mathrm{G}(k',\mu')$, then \begin{gather*} \frac{\int_{(0,\infty)}\lambda^{l+1} {\mathcal R}(d\lambda)} {\int_{(0,\infty)}\lambda^{l} {\mathcal R}(d\lambda)}=(l+k')/\mu' \end{gather*} in view of \eqref{eq:gamma_mom_exp}. The latter equality is of the same form as \eqref{eq:mom_recc1}. Since any gamma-distribution is determined by its moments, then ${\mathcal Q}={\rm G}(k_1,\mu_1)$ with $k_1=c_2/c_1$ and $\mu_1=1/c_1$. Therefore, the distribution ${\mathcal P}$ has the density \begin{gather*} p(\lambda)=\gamma\frac{\mu_1^{k_1}}{\Gamma(k_1)}\lambda^{k_1-1}e^{-(\mu_1-t_0)\lambda}\,, \end{gather*} where $\lambda>0$ and $\mu_1>t_0$. In other words, $\mathcal P={\rm G}(k,\mu)$ with $k=k_1$ and $\mu=\mu_1-t_0$. Since these parameters do not depend on $t_0$, we get the explicit expressions for functions $c_1(\,\cdot\,)$, $c_2(\,\cdot\,)$: $c_1(t)=1/(\mu+t)$ and $c_2(t)=k/(\mu+t)$. \end{proof} \section{Characterization of P\'{o}lya sequences} %\label{sect:POLYA_seq} The so-called ``NBD-Dirichlet'' model of consumer behavior (see \cite{Ehr88} for the detailed dis\-cus\-sion) uses a P\'{o}lya process $N_t$ to describe the number of purchase occasions up to time $t$. % moments\marginpar{?} of a random buyer. The second part of the model deals with a so-called P\'{o}lya sequence of $d$-dimensional random vectors. This sequence explains how a buyer makes a choice of one among $d$ interchangeable goods. To be short, we restrict ourselves to the case $d=2$. Consider the sequence $\varepsilon_\centerdot=\{\varepsilon_j,\,j\geq 1\}$ of random variables such that $\varepsilon_j\in \{0,1\}$. Assume that the sequence $\varepsilon_\centerdot$ is permutable. This means that for any $n$ all permutations of random variables $\varepsilon_1,\ldots,\varepsilon_n$ have identical $n$-dimensional distributions. Then (e.g., \cite[ch. 7]{F84}) there exists a {\em structure distribution} $\mathcal P$ such that $\mathcal P([0,1])=1$ and \begin{gather} \label{eq:def_ssym} \mathbb{P}(\varepsilon_1=\delta_1,\ldots,\varepsilon_n=\delta_n)=\int_{[0,1]}z^m(1-z)^{n-m}{\mathcal P}(dz) \quad {\rm with} \quad m=\delta_1+\ldots+\delta_n \end{gather} %with $m=\delta_1+\ldots+\delta_n$ for any $n\geq 1$ and any $\delta_i\in \{0,1\}$. If $\mathcal P$ is a beta-distribution with some parameters $(\nu_1,\nu_2)$, then the sequence $\varepsilon_\centerdot$ can be described in terms of P\'{o}lya urn models. Therefore it is natural to call $\varepsilon_\centerdot$ as the two-dimensional {\em P\'{o}lya sequence} with parameters $(\nu_1,\nu_2)$. (Various results on P\'{o}lya urn models can be found in \cite{JK77}.) Our aim is to give a martingale characterization of P\'{o}lya sequences among all permutable random sequences $\varepsilon_\centerdot=\{\varepsilon_j,\,j\geq 1\}$. More precisely, we consider random variables $S_n=\varepsilon_1+\ldots+ \varepsilon_n$ and their non-degenerate linear transformations. Evidently, it is sufficient to restrict ourselves to the case ${\mathcal P}((0,1))=1$. Let us start with the proposition analogous to Lemma \ref{lem:MPOS}. \begin{lemma} \label{lem:Msymm} Suppose that the distribution of the sequence $\varepsilon_\centerdot=\{\varepsilon_j,\,j\geq 1\}$ is defined by \eqref{eq:def_ssym} with ${\mathcal P}((0,1))=1$. Denote $S_n=\varepsilon_1+\ldots+\varepsilon_n$. Then \begin{gather} \label{eq:cond_seq} \mathbb{E}(S_{N}\,|\, S_1,\ldots,S_n)=S_n+(N-n)\,\frac{\int_{(0,1)}z^{S_n+1}(1-z)^{n-S_n}{\mathcal P}(dz)} {\int_{(0,1)}z^{S_n}(1-z)^{n-S_n}{\mathcal P}(dz)} \end{gather} for any $N>n\geq0$. \end{lemma} \begin{proof} Let $\delta_j\in \{0,1\}$ and $m_j=\delta_1+\ldots+\delta_j$. It follows from \eqref{eq:def_ssym} that \begin{gather*} \mathbb{P}(S_{n+1}=m_{n+1}\,|\,S_1=m_1,\ldots,S_n=m_n)=\frac{\int_{(0,1)}z^{m_{n+1}}(1-z)^{n+1-m_{n+1}}{\mathcal P}(dz)} {\int_{(0,1)}z^{m_{n}}(1-z)^{n-m_{n}}{\mathcal P}(dz)}\,. \end{gather*} Therefore, the sequence $\{S_n,\, n\geq 0\}$ is the Markov chain with the transition function \begin{gather*} \mathbb{P}(S_{n+1}=m+\delta\,|\,S_n=m)=\frac{\int_{(0,1)}z^{m+\delta}(1-z)^{n+1-m-\delta}{\mathcal P}(dz)} {\int_{(0,1)}z^{m}(1-z)^{n-m}{\mathcal P}(dz)}\,, \end{gather*} where $\delta\in \{0,1\}$ and $m\in \{0,\ldots,n\}$. As the result, $\mathbb{E}(S_{N}\,|\, S_1,\ldots,S_n)= \mathbb{E}(S_{N}\,|\,S_n)$. Analogously, \begin{gather} \label{eq:cond_SN} \mathbb{P}(S_{N}=l+m\,|\, S_n=l)=\frac{{\rm C}^m_{N-n}\int_{(0,1)}z^{l+m}(1-z)^{N-l-m}{\mathcal P}(dz)} {\int_{(0,1)}z^{l}(1-z)^{n-l}{\mathcal P}(dz)} \end{gather} for $N>n\geq 0$, $l\in \{0,\ldots,n\}$ and $m\in \{0,\ldots,N-n\}$, Using \eqref{eq:cond_SN}, we obtain that \begin{gather} \label{eq:cond_l} \mathbb{E}(S_{N}\,|\, S_n=l)=l+(N-n)\,\frac{\int_{(0,1)}z^{l+1}(1-z)^{n-l}{\mathcal P}(dz)} {\int_{(0,1)}z^{l}(1-z)^{n-l}{\mathcal P}(dz)} %=l+\Delta A(n,l). \end{gather} and the proof is complete. %\ebox \end{proof} \begin{proposition} 1. Let $\varepsilon_{\centerdot}=\{\varepsilon_j,\,j\geq 1\}$ be a {\em P\'{o}lya sequence} with parameters $(\nu_1,\nu_2)$. Denote $\zeta_n=(S_n+\nu_1)/(n+\nu_1+\nu_2)$. Then $\{\zeta_n,\,n\geq 1\}$ is the martingale relative to the natural filtration ${\mathcal F}^\zeta_n=\sigma(\zeta_1,\ldots,\zeta_n)$.\\ 2. Let the conditions of Lemma \ref{lem:Msymm} be fulfilled. Suppose that there exist sequences $b_n\neq 0$ and $a_n$ such that the random sequence $\zeta_n=b_n(S_n+a_n)$ is a martingale relative to the natural filtration ${\mathcal F}^{\,\zeta}_n=\sigma(\zeta_1,\ldots,\zeta_n)$. Then either ${\mathcal P}=\delta_{\alpha}$ for some $\alpha\in(0,1)$ or ${\mathcal P}$ is a beta-distribution with a certain parameters $(\nu_1,\nu_2)$. \end{proposition} \begin{proof} 1. Evidently, ${\mathcal F}^\zeta_n={\mathcal F}^S_n\stackrel{\rm def}=\sigma(S_1,\ldots,S_n)$. Besides, if $\mathcal P$ is the beta-distribution with parameters $(\nu_1,\nu_2)$, then \begin{gather} \label{eq:beta_mom} \int_{(0,1)}z^{k}(1-z)^{m}{\mathcal P}(dz)=\frac{\Gamma(\nu_1+\nu_2)}{\Gamma(\nu_1)\Gamma(\nu_2)}\, \frac{\Gamma(\nu_1+k)\Gamma(\nu_1+m)}{\Gamma(\nu_1+\nu_2+k+m)} \end{gather} for $k>-\nu_1$ and $m>-\nu_2$. Using \eqref{eq:beta_mom} and \eqref{eq:cond_seq}, we see that \begin{gather*} \mathbb{E}(S_{n+1}\,|\,{\mathcal F}^S_n)=S_n+\frac{S_n+\nu_1}{\nu_1+\nu_2+n}\,. \end{gather*} Therefore, \begin{gather*} \mathbb{E}(\zeta_{n+1}\,|\,{\mathcal F}^\zeta_n)=\frac{\mathbb{E}(S_{n+1}\,|\,{\mathcal F}^S_n)+\nu_1}{n+1+\nu_1+\nu_2}= (S_n+\nu_1)\frac{1+1/(\nu_1+\nu_2+n)}{n+1+\nu_1+\nu_2}=\zeta_n. \end{gather*} 2. Without loss of generality we can take $b_0=1$. Let $N\geq n\geq 0$ and denote $\Delta=N-n$. Since $\{\zeta_n,\,n\geq 0\}$ is a martingale, then \begin{gather} \label{eq:mart_P} \mathbb{E}(S_{n+\Delta}\,|S_n=l)=\frac{b_n(l+a_n)}{b_{n+\Delta}}-a_{n+\Delta} \end{gather} for any $\Delta,n\geq 0$ and $0\leq l\leq n$. Denote \begin{gather*} A(n,l)=\frac{\int_{(0,1)}z^{l+1}(1-z)^{n-l}{\mathcal P}(dz)} {\int_{(0,1)}z^{l}(1-z)^{n-l}{\mathcal P}(dz)}\,. \end{gather*} Then (see Lemma \ref{lem:Msymm}) the equality \eqref{eq:mart_P} can be rewritten in the form \begin{gather} \label{eq:ANL} \Delta A(n,l)=\frac{b_n-b_{n+\Delta}}{b_{n+\Delta}}\,l+\frac{b_n-b_{n+\Delta}}{b_{n+\Delta}}\,a_n+a_n-a_{n+\Delta}. \end{gather} In other terms, \begin{gather*} \Delta A(n,l)=\frac{b_n-b_{n+\Delta}}{b_{n+\Delta}}\,l+\Delta A(n,0). \end{gather*} %It follows from the latter equality This means that $(b_n-b_{n+\Delta})/b_{n+\Delta}$ is proportional to $\Delta$: $(b_n-b_{n+\Delta})/b_{n+\Delta}=c_1(n)\Delta$. In other words, $b_{n+\Delta}={b_n}/({1+c_1(n)\Delta})$ and $b_n={1}/{(1+c_1\,n)}$ with $c_1=c_1(0)$. Besides, \begin{gather} (b_n-b_{n+\Delta})/(b_{n+\Delta})= {c_1\Delta}/(1+c_1 n). \label{eq:delta_bn} \end{gather} Analogously, $a_{n+\Delta}-a_n$ is proportional to $\Delta$: $a_{n+\Delta}-a_n=c_2(n)\Delta$. Therefore \begin{gather} \label{eq:delta_an} a_n=a_0+c_2 n \quad {\mbox and} \quad a_{n+\Delta}-a_n=c_2\Delta \end{gather} with $c_2=c_2(0)$. Applying \eqref{eq:ANL}, \eqref{eq:delta_bn} and \eqref{eq:delta_an}, we obtain that \begin{gather} \label{eq:mom2} \frac{\int_{(0,1)}z^{l+1}(1-z)^{n-l}{\mathcal P}(dz)} {\int_{(0,1)}z^{l}(1-z)^{n-l}{\mathcal P}(dz)}= \frac{c_1 l}{1+c_1 n}+\frac{c_1(a_0+c_2 n)}{1+c_1 n}-c_2=\frac{c_1l+c_1 a_0-c_2}{1+c_1 n} \end{gather} and %Under the choice $n=l$ \begin{gather} \label{eq:eq_main} \frac{\int_{(0,1)}z^{l+1}{\mathcal P}(dz)} {\int_{(0,1)}z^l{\mathcal P}(dz)} =\frac{c_1 a_0-c_2+c_1 l}{1+c_1 l}\,. \end{gather} %$b_3=b_2+1$. Suppose now that $c_1=0$. Then the right side of \eqref{eq:eq_main} is a constant $\alpha=-c_2\in (0,1)$ and $\mathcal P=\delta_\alpha$. Consider the case $c_1\neq0$. If we take $l=0$ and $l=1$ in \eqref{eq:eq_main}, then under denotation $x=c_1 a_0-c_2$ \begin{gather*} 0<\int_{(0,1)}z{\mathcal P}(dz)=x<1 \quad \mbox{and} \quad \int_{(0,1)}z^2{\mathcal P}(dz)=x\,\frac{x+c_1}{1+c_1}\,. \end{gather*} Since $\mathcal P\neq\delta_\alpha$, then $x(x+c_1)/(1+c_1)>x^2$ and either $c_1>0$ or $c_1<-1$. On the other hand, if we choose $l=0$ in \eqref{eq:mom2}, we see that the supposition $c_1<-1$ leads to the contradiction: in this case the left side of \eqref{eq:mom2} is negative. Therefore $c_1>0$. If $\mathcal R$ stands for the beta-distribution with parameters $(\nu_1,\nu_2)$, then \begin{gather} \label{eq:beta_mom1} \frac{\int_{(0,1)}z^{l+1}{\mathcal R}(dz)} {\int_{(0,1)}z^l{\mathcal R}(dz)}= \frac{\nu_1+l}{\nu_1+\nu_2+l}\,. \end{gather} in view of \eqref{eq:beta_mom}. Let us denote $\nu_1\!=\!(c_1 a_0-c_2)/c_1$ and $\nu_2\!=\!1/c_1-\nu_1$. Then \eqref{eq:eq_main} has the same form as \eqref{eq:beta_mom1}. 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