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\begin{abstract}
In this note, we compute the Mellin transform of a Poissonian exponential functional, the underlying process being a simple continuous time random walk. It shows that the Poissonian functional can be expressed in term of the inverse of a $q$-gamma random variable.

The result interpolates between two known results. When the random walk has only positive increments, we retrieve a theorem due to Bertoin, Biane and Yor. In the Brownian limit ($q \rightarrow 1^-$), one recovers Dufresne's identity involving an inverse gamma random variable. Hence, one can see it as a $q$-deformed Dufresne identity.
\end{abstract}
{\bf MSC 2010 subject classifications:} 33D05, 60J27, 60J65\\
{\bf Keywords:} $q$-calculus, $q$-gamma random variable, exponential functionals of compound Poisson process, $q$-analogue of Dufresne's identity for exponential functionals of Brownian motion, Wiener-Hopf factorization.

\section{Introduction}

Let $P^+$ and $P^-$ be two standard Poisson processes with respective intensities $1$ and $z = q^\mu \in \left[0, 1\right)$. In all the following, we assume $0 < q < 1$ and $\mu>0$. Then $\zeta_t := P_t^+ - P_t^-$ is a compound Poisson process with jumps $+1$ or $-1$, and drifting to $+\infty$. We investigate the law of the perpetuity
\begin{align}
\label{eq:def_perpetuity}
I^{(q, z)} := & \int_0^\infty q^{\zeta_s} ds \ ,
\end{align}
where $0<q<1$. We can also rewrite $I^{(q, z)}$ as
\begin{align}
\label{eq:def_perpetuity2}
I^{(q, z)} = & \frac{1}{1+z}\sum_{n=0}^\infty q^{B_n} \varepsilon_n \ ,
\end{align}
where $\left( \varepsilon_n; n \geq 1 \right)$ are i.i.d exponential random variables and $B_n$ is a standard random walk drifting to $+\infty$. More precisely
$$ \P\left( B_{n+1}-B_n = 1 \right) = 1-\P\left( B_{n+1}-B_n = -1 \right) = \frac{1}{1+z} > \half \ .$$

In the denomination of \cite{bib:by}, the random variable of interest is an exponential functional associated to the L\'evy process $ -\log(q) \zeta_t $ .
We have:
$$ \E\left( e^{-s \log(q) \zeta_t} \right) = \exp\left( t \psi(s) \right) $$
where the L\'evy-Khintchine exponent is:
\begin{align}
\label{eq:levy_khintchine_exp}
\psi(s) = & \left( q^{-s} - 1 \right) + z\left( q^s - 1 \right) \ .
\end{align}

The subject of exponential functionals has been extensively studied (\cite{bib:by}, \cite{bib:cpy97}, \cite{bib:bl13}) with a focus on many possible aspects such as their continuity properties \cite{bib:ls11} and their Mellin transforms \cite{bib:hy13}. From a theoretical perspective, it is known since Kesten \cite{bib:kesten} that such perpetuities appear naturally when analyzing products of random matrices. More precisely, perpetuities appear as matrix coefficients of random walks in lower (or upper) triangular matrices. The study of such objects is in fact our original motivation. In insurance and mathematical finance, the perpetuity \eqref{eq:def_perpetuity} can be understood as the total discounted cash-flow from a perpetual bond, when interest rates are random. Also, the discussion in \cite{bib:bby} shows a connection to models in DNA replication and transmission protocols. With such applications in mind, corollary \ref{corollary:cv_gamma} allows to simulate $I^{(q, z)}$ with little effort.

Our main theorem \ref{thm:main} gives an explicit formula for the Mellin transform of $I^{(q, z)}$ and its density. This allows to recognize in corollary \ref{corollary:cv_gamma} a Wiener-Hopf factorization: $I^{(q, z)}$ has the same law as the product of two independent random variables, one of which is the inverse of a $q$-gamma random variable - to be defined later in the text.

On the one hand, notice that if we specialize $z$ to zero, we obtain
$$ I^{(q)} := I^{(q, z=0)} = \sum_{n=0}^\infty q^{n} \varepsilon_n \ ,$$
which is exactly the object studied by Bertoin, Biane and Yor in \cite{bib:bby}. In that respect, our result is a refinement of their theorem 1. A purely analytic description of that case has been given by \cite{bib:berg}.

On the other hand, consider the compound Poisson process $\zeta_{t}$ properly rescaled in time and space thanks to the parameter $q$. Thanks to Donsker's invariance principle, it will converge weakly to a Brownian motion as $q \rightarrow 1$, in the uniform topology. The exponential functionals will naturally converge too, although the claim requires some work because the exponential functionals are not continuous on the sample path space. This way, one recovers Dufresne's identity involving the inverse of the usual gamma random variable.

In the light of what has been just said, corollary \ref{corollary:cv_gamma} is a $q$-deformed Dufresne identity. As we will explain, the definition of the $q$-gamma distribution from $q$-calculus can be subject to controversy since many candidates qualify as $q$-deformations. Therefore, our result shows that the $q$-gamma distribution appears in relation to the random walk exactly in the same way as the gamma distribution appears in relation to Brownian motion. For possible multi-dimensional generalisations using certain random walks on solvable Lie groups, see \cite{bib:chh14b}.\\

It is worth mentioning that the L\'evy-Khintchine exponent \eqref{eq:levy_khintchine_exp} has the following Wiener-Hopf factorization (see Chapter 6 in \cite{bib:kyprianou})
\begin{align}
\label{eq:wiener_hopf}
\psi(s) = -\phi_+(s) \phi_-(-s) & \textrm{ with } \phi_+(s) = q^{-s} - 1, \ \phi_-(s) = z q^{-s} - 1 \ .
\end{align}
The functions $\phi_+$ and $\phi_-$ are exponents of subordinators and more precisely the ascending and descending ladder height processes associated to $-\zeta \log q$. One recognizes immediately that $\phi_+$ is the exponent of a rescaled standard Poisson process while $\phi_-$ is the exponent of a rescaled Poisson process with rate $z$ and killing rate $1-z$. The rescaling in space is $-\log q$.

Let $\xi$ be a L\'evy process and let $I_\xi := \int_0^\infty e^{-\xi_s} ds$ be its exponential functional. Under fairly general assumption on $\xi$, \cite{bib:PPS12} and \cite{bib:PS12} prove that $I_\xi$ enjoys a remarkable factorization into the product of two independent exponential functionals $I_{H}$ and $I_Y$. Since this factorization \emph{in law} is a shadow of the Wiener-Hopf factorization \eqref{eq:wiener_hopf} of the Lévy exponent, it is referred to by the same denomination. In their construction, $H$ is the ascending ladder height process and $Y$ is a L\'evy process with only negative jumps, constructed from the descending ladder height process $H^-$. One of their requirements on the descending ladder height process is to have a non-decreasing L\'evy density. In the more recent paper \cite{bib:PS13}, Patie and Savov removed the hypothesis of a non-decreasing density. Indeed, they have announced in their theorem 2.1 that such Mellin transforms always take the form of a ratio of generalized Weierstrass products. In our case, the relationship to $q$-calculus arises from the fact that the generalized Weierstrass products at hand are $q$-Pochhammer symbols.

The compound Poisson process $\zeta \log q$, although simple, is a very degenerate case which we make entirely explicit, along with its relations to $q$-calculus. Moreover, our methodology of proof, which hinges on the complex-analytic Carleson's theorem, can be easily generalized.

\subsection*{Notations}
If $E$ is a topological space, we will denote by $Bor(E)$ the $\sigma$-algebra of its Borel sets. And if $X$ is a random variable valued in a measurable space $\left( E, \Ec \right)$, $\Lc\left( X \right) = \P_X$ is the probability measure on $\left( E, \Ec \right)$ known as the distribution of $X$. Equality in law is denoted by $\eqlaw$.

We will also make use of the Vinogradov symbol:
$$ f \ll g \Leftrightarrow f = \Oc(g) \ .$$

\subsection*{Structure of the paper}
Right away, we start by stating the main theorem \ref{thm:main}, as the prerequisites for such an analytic result are minimal. Then, we justify and give the definition of the $q$-gamma random variable in subsection \ref{subsection:def_q_gamma}. This allows to recognize, hidden in the expression of $s \mapsto \E\left( \left( I^{(q, z)} \right)^s \right)$, the Mellin transform of an inverse $q$-gamma random variable. As such, the main theorem is reinterpreted in a probabilistic fashion as corollary \ref{corollary:cv_gamma}: $I^{(q, z)}$ factors as the product of two independent random variables. Finally, in subsection \ref{subsection:scaling_limit}, we formulate the weak convergence of $I^{(q, z)}$ as $q \rightarrow 1^-$, jointly with the scaling of the random walk to Brownian motion.

The proofs of theorem \ref{thm:main} and proposition \ref{proposition:cv_bm} are given separately in section \ref{section:proofs}. There, we start by explaining why the approach of \cite{bib:bby} cannot be adapted to our setting.

\section{Main theorem and consequences}

The $n$-th $q$-Pochhammer symbol, for $n \in \N \sqcup \{ \infty \}$ and $a \in \C$, is given by:
$$ \left(a; q \right)_{n} = \prod_{j=0}^n\left( 1 - a q^j \right) \ ,$$
and the $q$-gamma function (\cite{bib:gasper_rahman}  eq. (1.10.1)) is defined for $x>0$ by:
$$ \Gamma_q\left( x \right) := \left( 1 - q \right)^{1-x} \frac{\left(q; q \right)_{\infty}}{\left(q^x; q \right)_{\infty}} \stackrel{q \rightarrow 1^-}{\longrightarrow} \Gamma(x) \ .$$

The main result, whose proof is postponed to section \ref{section:proofs}, is the following:
\begin{thm}
\label{thm:main}
The Mellin transform of $I^{(q, z)}$ is analytic for $\Re(s)<\mu$ and is given by:
\begin{align*}
    \E\left( \left( I^{(q, z)} \right)^s \right)
= & \Gamma\left(1 + s\right) \frac{\left( q^{1+s}; q \right)_\infty}
                                  {\left( q; q \right)_\infty}
                             \frac{ \left( z; q \right)_\infty }
                                  { \left( z q^{-s}; q \right)_\infty } \\
= & \left(1-q\right)^{-2s} \frac{ \Gamma\left( 1 + s \right) }{ \Gamma_q\left( 1+s \right) }
                           \frac{ \Gamma_q\left(\mu-s\right) }{ \Gamma_q\left( \mu \right) } \ ,
\end{align*}
where $z = q^\mu$ with $\mu>0$. Moreover, the density of the perpetuity \eqref{eq:def_perpetuity} is:
$$ \forall x \geq 0, \ i_{q,z}(x) :=  \frac{\left( z; q \right)_\infty}{\left( q; q \right)_\infty} \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(qz)^m}{ \left( q; q \right)_m } \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -x q^{m-n} } \ . $$
\end{thm}

\subsection{On the \texorpdfstring{$q$}{q}-gamma distribution}
\label{subsection:def_q_gamma}

Let $G_a$ denote a geometric random variable with parameter $0 \leq a < 1$.  Following the notations in \cite{bib:bby}, if $\left( G_{a}, G_{aq}, G_{aq^2}, \dots \right)$ are independent geometric random variables with the specified parameters, we define the random variable $R^{(q)}_a$ as:
\begin{align}
\label{eq:def_q_gamma}
R^{(q)}_a := & \left( 1- q \right)^{-1} q^{\sum_{n=0}^\infty G_{a q^n}} \ .
\end{align}
The distribution of $R^{(q)}_a$ is the $q$-analogue of the gamma distribution. Such a denomination might appear arbitrary at this point. We hope that the reader will be convinced by the following useful properties and expressions.

Remarkably, the Mellin transform of $R^{(q)}_a$ for $\Re(s) \geq 0$ is expressed thanks to $q$-exponentials. If $a=q^\kappa$, $\kappa>0$:
\begin{align}
\label{eq:mellin_q_gamma}
\E\left( \left( R^{(q)}_a \right)^s \right) = & \left( 1- q \right)^{-s} \frac{ \left( a; q \right)_\infty }{\left( a q^s; q \right)_\infty}
                                            = \frac{ \Gamma_q\left( s + \kappa \right) }{ \Gamma_q\left( \kappa \right) } \ .
\end{align}
By taking $q \rightarrow 1^-$ in equation \eqref{eq:mellin_q_gamma}, one sees the convergence in law of $R^{(q)}_{q^\kappa}$ to a gamma random variable with parameter $\kappa$. Moreover, the $q$-binomial theorem (\cite{bib:gasper_rahman} eq. (1.3.2)):
\begin{align}
\label{eq:q_binomial_thm}
\forall x \in \C, |q|<1, |z|<1, \ \frac{\left(xz; q \right)_\infty}{\left(z; q \right)_\infty} = & \sum_{n=0}^\infty \frac{\left(x; q \right)_n}{\left(q; q \right)_n} z^n
\end{align}
yields for $x=0$ and $z=a q^s$ that:
$$ \E\left( \left( R^{(q)}_a \right)^s \right) = \left( 1- q \right)^{-s} \left( a; q \right)_\infty \sum_{n=0}^\infty \frac{a^n q^{sn}}{\left(q; q \right)_n}$$
which implies, via injectivity of the Mellin transform, that:
\begin{align}
\label{eq:q_gamma_proba}
\forall n \in \N, \ \P\left( R^{(q)}_a = \left(1-q\right)^{-1} q^n \right) = & \frac{\left(a; q \right)_\infty}{\left(q; q \right)_n} a^n \ .
\end{align}
To the best of the author's knowledge, the $q$-gamma distribution was first defined by Pakes (\cite{bib:pakes} section 5), motivated by length biasing. Equation \eqref{eq:q_gamma_proba} matches indeed with the last equation of page 844 \cite{bib:pakes}. So far, we preferred a naive point of view that avoids discussing why this definition yields the ``correct'' $q$-deformation. After all, from the point of view of $q$-calculus, a $q$-gamma distribution can be defined as soon as we have a $q$-deformation of the Gamma integral given for $x>0$:
$$ \Gamma(x) = \int_{\R_+} t^{x-1} e^{-t} dt \ ,$$
which serves to define the gamma distribution. Now, the $q$-integral of a function $f: \R_+ \rightarrow \R$ is given by (1.11 in \cite{bib:gasper_rahman}):
\begin{align}
\label{eq:def_q_integral}
\forall u \in \R_+, \ \int_0^u f(t) \ d_q t & := (1-q) \sum_{n \geq 0} q^n u f\left(q^n u \right)
\end{align}
and we have:
\begin{align}
\label{eq:q_integral}
\forall x>0, \Gamma_q(x) = \left( 1 - q \right)^{1-x} \frac{\left(q; q \right)_{\infty}}{\left(q^x; q \right)_{\infty}}
                         = \int_{0}^{\frac{1}{1-q}} t^{x-1} E_q(-t) d_q t
\end{align}
where
$$ \forall t \in \C, \ E_q(t) := \left( t(1-q); q \right)_{\infty}$$
is the $q$-exponential. In order to see that, take $a = q^x$ in \eqref{eq:q_gamma_proba} and using the fact that probability weights sum up to $1$:
\begin{align*}
\left( 1 - q \right)^{1-x} \frac{\left(q; q \right)_{\infty}}{\left(q^x; q \right)_{\infty}}
\quad = \quad & \left( 1 - q \right)^{1-x} \frac{\left(q; q \right)_{\infty}}{\left(q^x; q \right)_{\infty}} \sum_{n \geq 0} \P\left( R^{(q)}_{q^x} = \left(1-q\right)^{-1} q^n \right) \\
\quad = \quad & \left( 1 - q \right)^{1-x} \sum_{n \geq 0} q^{nx} \left(q^n; q \right)_{\infty}\\
\quad = \quad & \sum_{n \geq 0} q^n \left( \frac{q^{n}}{1-q} \right)^{x-1} \left(q^n; q \right)_{\infty}\\
\stackrel{eq. \eqref{eq:def_q_integral}}{=} & \int_{0}^{\frac{1}{1-q}} t^{x-1} E_q(-t) d_q t
\end{align*}
De Sole and Kac \cite{bib:desole_kac05} convincingly argue that \eqref{eq:q_integral} is indeed the ``correct'' expression among other possible candidates because of its compatibility with other formulae from $q$-calculus. As mentioned in the introduction, our approach can be seen as another argument in favor of De Sole and Kac's claim, with a probabilistic point of view.

\subsection{Wiener-Hopf factorization into independent random variables}

The random variable $I^{(q, z)}$ can be factored into the product of two independent random variables, one of which is an inverse $q$-gamma:
\begin{corollary}
\label{corollary:cv_gamma}
The following equality in law holds:
$$ I^{(q, z)} \stackrel{\Lc}{=} q^{-\sum_{n=0}^\infty G_{z q^n}} I^{(q)} = \frac{\left( 1- q \right)^{-1} I^{(q)}}{ R^{(q)}_z}$$
where $I^{(q)}$ and $R^{(q)}_z$ are independent random variables.

In particular, as $q \rightarrow 1$, $\left(1-q\right)^2 I^{(q, q^\mu)}$ converges in law to $\frac{1}{\gamma_\mu}$ with $\gamma_\mu$ a gamma random variable with parameter $\mu$.
\end{corollary}
\begin{proof}
For the equality in law, Mellin transforms of both sides are equal.

Since we already know that $R^{(q)}_{q^\mu}$ converges in law to $\gamma_\mu$ it is sufficient to prove that $(1-q) I^{(q)}$ converges to $1$ in probability, which is implied by:
$$ \E\left( (1-q) I^{(q)} \right) \stackrel{eq. \eqref{eq:def_perpetuity2}}{=}
   \sum_{n=0}^\infty (1-q) q^n \E( \varepsilon_n ) = 1 \ .$$
$$ \Var\left( (1-q) I^{(q)} \right) \stackrel{eq. \eqref{eq:def_perpetuity2}}{=}
   \sum_{n=0}^\infty (1-q)^2 q^{2n} \Var( \varepsilon_n ) = \half \frac{1-q}{1+q} \stackrel{q \rightarrow 1}{\longrightarrow} 0 \ .$$
\end{proof}

The Wiener-Hopf factorization is a general phenomenon for exponential functionals of L\'evy processes. As in the work of \cite{bib:PPS12} and \cite{bib:PS12}, the term $I^{(q)}$ appearing in the previous factorization is the exponential functional associated to the descending ladder height process of $\zeta \log q$. In this case, $I^{(q)} = I^{(q,z=0)}$ corresponds indeed to the exponential functional if the random walk $B$ only goes upward.

However, the other term i.e the inverse $q$-gamma variable is not the exponential functional of a L\'evy process. Among the possible arguments for that claim, the $q$-gamma variable is discrete, while exponential functionals are known to have a density - very degenerate cases aside (theorem 2.2 in \cite{bib:blr08}). This result provides a good example to Corollary 2.2 in \cite{bib:PS13}, where Patie and Savov give a necessary and sufficient condition for this second term to be an exponential functional.

\begin{rmk}
This Wiener-Hopf factorization in law is, to the author, quite puzzling. Indeed, the perpetuity on the left-hand side involves adding  both positive and negative powers of $q$ randomly. On the right-hand side, we see that this is tantamount \emph{in law} to paying the price of certain negative power of $q$ that is a sum of geometric random variables, then multiplying by the exponential functional associated to the descending ladder height process of $\zeta_t \log q$.

A trajectorial explanation of such an equality would certainly be desirable. The same question is raised in \cite{bib:PS13} for the Wiener-Hopf factorizations of exponential functionals associated to more general L\'evy processes.
\end{rmk}

For general L\'evy processes, the distribution of the remainder random variable $R_z^{(q)}$ is extracted from the L\'evy measure of the descending ladder process $H^-$ (see the expressions in Theorem 1.9 \cite{bib:alili}).

\subsection{Scaling limit}
\label{subsection:scaling_limit}
The thorough reference for weak convergence in sample path spaces is \cite{bib:billingsley}. The acronym càdlàg stands for ``continu à droite, limité à gauche'' in French and means right-continuous with left-limits. Let $D_\infty$ be the Skorohod space of functions on $\R$:
$$ D_\infty := \left\{ f: \R_+ \rightarrow \R \ | \ f \textrm{ is càdlàg } \right\} \ .$$
Let $d$ be the following distance which metrizes the topology of uniform convergence on compact sets:
$$ \forall (x,y) \in D_\infty \times D_\infty, \ d(x,y) := \sum_{n=1}^\infty \frac{1 \wedge d_n(x,y)}{2^n}$$
with
$$ d_n(x,y) := \sup_{0 \leq t \leq n} |x(t)-y(t)| \ .$$
As such, we will refer to $d$ as the local uniform distance. The induced topology is the (locally) uniform topology and is finer than the usual Skorohod topology on $D_\infty$. However it is not separable. There is no essential need for defining the usual Skorohod topology, but the reader is referred to section 12 in \cite{bib:billingsley} for completeness. $\Dc$ is the ball $\sigma$-field for the local uniform distance - or equivalently the Borel $\sigma$-field for the usual Skorohod topology ( See equation (15.2) in \cite{bib:billingsley}).

We are concerned with the weak convergence of the following sequence of compound Poisson processes indexed by $q$:
\begin{align}
\label{eq:def_W_q}
\left( W_t^{(q, \mu)} ; t \geq 0 \right) := & \left( \frac{-\log q}{2} \ \zeta_{ \frac{2 t}{(1-q)^2} } ; t \geq 0 \right) \ ,
\end{align}
jointly with their exponential functionals $I\left( W^{(q, \mu)} \right)$ where:
 $$\begin{array}{cccc}
    I: & D_\infty & \rightarrow & \R \sqcup \{ \infty \} \\
       &     x    & \mapsto     & 2 \int_0^\infty e^{-2 x(s)} ds \ .
   \end{array}$$

The following theorem deals with the sequence of probability measures
$$ \Lc\left( W^{(q, \mu)} , I\left( W^{(q, \mu)} \right) \right)$$
on $\left( D_\infty \times \R_+, \Dc \otimes Bor\left( \R_+ \right) \right)$. There are three reasons why we choose to work in the locally uniform topology on the Skorohod space. First, the functional $I$ is better controlled under variations of its argument's uniform norm. In the usual Skorohod topology, one would have to deal with a troublesome time rescaling as well. Secondly, our result in this setting is stronger as the uniform topology has less open sets. And thirdly, our presentation is improved by not having to define the Skorohod topology.

\begin{proposition}
\label{proposition:cv_bm}
$$\left( 1 - q \right)^2 I^{(q,z)} = I\left( W^{(q, \mu)} \right) \ .$$
and we have the weak convergence on $\left( D_\infty \times \R_+, \Dc \otimes Bor\left( \R_+ \right) \right)$:
$$ \left( W^{(q, \mu)}, I\left( W^{(q, \mu)} \right) \right) \stackrel{q \rightarrow 1^-}{\longrightarrow}
   \left( W^{(\mu)}   , I\left( W^{(   \mu)} \right) \right) \ .
$$
\end{proposition}

Putting together proposition \ref{proposition:cv_bm} and corollary \ref{corollary:cv_gamma} recovers Dufresne's identity in law \cite{bib:Dufresne90}:
$$ 2 \int_0^\infty e^{- 2 W_s^{(\mu)}} ds \eqlaw \frac{1}{\gamma_\mu} \ .$$

If one is only interested in proving that $I\left( W^{(q, \mu)} \right)$ converges weakly to $I\left( W^{(   \mu)} \right)$, which is the convergence of the second marginal in the previous theorem, one can invoke general approximation theorems such as theorem 7.3 in \cite{bib:bl13} or lemma 4.8 in \cite{bib:PPS12}. As mentioned in their text, the latter result was specifically tailored to by-pass the issue that exponential functionals are not continuous on $D_\infty$. One cannot a priori invoke the mapping theorem 2.7 in \cite{bib:billingsley}. The problem arises from the infinite time horizon: two paths may be close to each other in the locally uniform topology and yet be very different in the long run, causing their exponential functionals to be also very different. Of course, these unfavorable events occur with small probability. Not only our proof gives joint convergence, but it also shows that the mapping theorem can be applied, upon controlling such unfavorable events. The argument is robust and simple enough to generalize to other Lévy processes, although we have not explored that possibility.

\section{Proofs}
\label{section:proofs}

\subsection{Proof of theorem \ref{thm:main}}
Let us comment on the proof of \cite{bib:bby} which handles the case of $z=0$ in theorem \ref{thm:main}. It goes as follows. Consider two independent random variables $R^{(q)}_q$ and $I^{(q)}$. After computing the entire moments of $I^{(q)}$, one can notice that those of $(1-q) R^{(q)}_q I^{(q)}$ match the moments of a standard exponential random variable $\varepsilon$. Since the exponential distribution is characterized by its moments, we can deduce $(1-q) R^{(q)}_q I^{(q)} \eqlaw \varepsilon$. One concludes by taking the Mellin transform on both sides of the previous equality in law. In the proof of theorem \ref{thm:main} though, there is a small catch. If the random variable $I^{(q)}$ has entire moments, $I^{(q, z)}$ does not for $z>0$. Therefore, we will have to take a different path - through complex analysis. The crux of the argument in computing the Mellin transform is Carlson's theorem (see 9.2.1 in \cite{bib:boas}). It allows under certain conditions to identify two entire functions that match on the positive integers.

\paragraph{Mellin transform:} Recall that
$$ \E\left( e^{-s \log(q) \zeta_t} \right) = \exp\left( t \psi(s) \right)$$
where the L\'evy-Khintchine exponent is:
$$\psi(s) = \left( q^{-s} - 1 \right) + z\left( q^s - 1 \right) = \left( q^{-s} - 1 \right) \left(1-z q^s \right) \ .$$
It satisfies (8) and (9) in \cite{bib:by}, thereby confirming the existence of negative moments for $I^{(q,z)}$ (theorem 3 in \cite{bib:by}). Since a Mellin transform is always holomorphic on the interior of the vertical strip where it converges absolutely, $s \mapsto \E\left( \left( I^{(q,z)} \right)^{s} \right)$ is holomorphic on the left half of the complex plane.

Let $s = -r$, with $\Re(r)>0$. One has the recurrence formula:
\begin{align}
\label{eq:mellin_recurrence}
\E\left( \left( I^{(q,z)} \right)^{-r} \right)
= & \frac{r}{\left( q^{-r} - 1 \right) \left(1-q^{\mu+r} \right)} \E\left( \left( I^{(q,z)} \right)^{-(r+1)} \right) \ .
\end{align}
Indeed, from equation (11) in \cite{bib:by}:
\begin{align*}
\E\left( \left( I^{(q,z)} \right)^{-r} \right)
& = \frac{r}{\psi(r)} \E\left( \left( I^{(q,z)} \right)^{-(r+1)} \right)\\
& = \frac{r}{\left( q^{-r} - 1 \right) \left(1-z q^r \right)} \E\left( \left( I^{(q,z)} \right)^{-(r+1)} \right)\\
& = \frac{r}{\left( q^{-r} - 1 \right) \left(1-q^{\mu+r} \right)} \E\left( \left( I^{(q,z)} \right)^{-(r+1)} \right) \ .
\end{align*}
The idea is rather simple: Equation \eqref{eq:mellin_recurrence} identifies the Mellin transform on integers, as a meromorphic function. Then, Carleson's theorem enters the picture in order to discriminate between the different possibilities. More precisely, the recurrence allows to extend the map $r \mapsto \E\left( \left( I^{(q,z)} \right)^{-r} \right)$ to all of $\C$ except when $r \in -\mu - \N$. These are precisely the poles and the Mellin transform admits a meromorphic extension to all of $\C$. Moreover, it is holomorphic on $\{ \Re(r)>-\mu \}$.

Let us define the meromorphic function $\varphi$ on the right half plane:
$$\varphi(r) := -1 +
                \frac{ \E\left( \left( I^{(q,z)} \right)^{-r} \right) }
                     { \Gamma\left( 1 - r \right) }
                \frac{\left( q; q \right)_\infty}
                     {\left( q^{1-r}; q \right)_\infty}
                \frac{ \left( z q^{r}; q \right)_\infty }
                     { \left( z; q \right)_\infty } \ .$$
We are done upon proving that $\varphi$ vanishes identically. Using the recurrence \eqref{eq:mellin_recurrence}, we find that:
$$ \varphi(0) = 0 , \varphi(r) = \varphi(r+1) \ ,$$
for $r$ outside of poles. Now, by examining its constituents, the function $\varphi$ is holomorphic on the strip $-\mu < \Re(r) < 1$. Moreover, this strip is of length larger than one. Because the function $\varphi$ is $1$-periodic, it is an entire function that vanishes on the integers. In order to see it vanishes identically, we invoke Carlson's theorem as announced (theorem 9.2.1 in \cite{bib:boas}); which requires from $\varphi$ not to grow too fast on vertical strips. In order to exclude maps such as $z \mapsto \frac{\sin\left( \pi z \right)}{\pi z}$, we need to check that $\varphi$ is a function of exponential type at most $\pi$. This is handled by the following lemma.

\begin{lemma}
$\varphi$ is a function of type $\half \pi$ i.e:
$$ \forall \varepsilon>0, \exists M_\varepsilon > 0, \forall r \in \C, |\varphi(r)| \leq M_\varepsilon e^{ (\frac{\pi}{2} + \varepsilon)|\Im(r)| } \ .$$
\end{lemma}
\begin{proof}
Given that $\varphi$ is $1$-periodic, it suffices to prove the result for $\Re(r) \in \left[ \sigma_- ; \sigma_+ \right] \subset \left(-\mu; 1 \right)$, with $\left[ \sigma_- ; \sigma_+ \right]$ having a length larger than $1$. In that domain:
$$ \left| \E\left( \left( I^{(q,z)} \right)^{-r} \right) \right| \leq  \E\left( \left( I^{(q,z)} \right)^{-\sigma_-}  + \left( I^{(q,z)} \right)^{-\sigma_+} \right) \ll 1 \ ,$$
$$ \left| \left( z q^{r}; q \right)_\infty \right| \leq \left( -q^{\mu+\sigma_-}; q \right)_\infty \ll 1 \ ,$$
$$ \left| \frac{1}{\left( q^{1-r}; q \right)_\infty} \right| \leq \frac{1}{\left( q^{1-\sigma_+}; q \right)_\infty} \ll 1 \ .$$
For the two previous dominations, we respectively used the fact that the $q$-Pochhammer symbol $\left( x; q \right)_\infty$ satisfies $\left|\left( x; q \right)_\infty \right| \leq \left( -|x|; q \right)_\infty$ for $x \in \C$ and that it is decreasing for $x \in (-\infty, 1) \subset \R$. Thus:
$$   |\varphi(r)|
\ll  1 + \left| \frac{1}{\Gamma\left( 1 - r \right)} \right| \ .$$
To conclude, we apply the complex Stirling formula ( e.g Proposition IV.1.4 in \cite{bib:freitag_busam}) with the variable going to infinity within a vertical strip. As $t \rightarrow \infty$ with $\sigma$ in a fixed compact $K$, we have:
$$ \Gamma\left( \sigma + it \right) \sim \sqrt{2 \pi} \left( i t\right)^{\sigma - \half} e^{-\frac{\pi}{2} |t|}
               \left( \frac{|t|}{e} \right)^{it} \left( 1 + O_K(\frac{1}{|t|}) \right) \ .$$
In particular:
$$ \forall \varepsilon>0, \exists M_\varepsilon>0, \forall \Re(r) \in \left[ \sigma_- ; \sigma_+ \right], |\varphi(r)| \leq M_\varepsilon e^{ (\frac{\pi}{2} + \varepsilon)|\Im(r)| } \ .$$
\end{proof}

\paragraph{Density:}
The expression for the density is obtained via inversion of the Mellin transform
\begin{align}
\label{eq:mellin_transform}
 \E\left( \left( I^{(q, z)} \right)^s \right)
= & \Gamma\left(1 + s\right) \frac{\left( q^{1+s}; q \right)_\infty}
                                  {\left( q; q \right)_\infty}
                             \frac{ \left( z; q \right)_\infty }
                                  { \left( z q^{-s}; q \right)_\infty }.
\end{align}
In the following, all series are absolutely convergent for $0<z<1$, $0< q < 1$ and $\Re(s)<\mu$, making the use of Fubini's theorem possible in order to exchange summations and integrals. As in \cite{bib:bby}, start by the Euler formula:
$$ \left( q^{1+s}; q \right)_\infty = \sum_{n=0}^\infty \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} q^{n(1+s)}$$
together with
$$ q^{n(1+s)} \Gamma(1+s) = \int_0^\infty x^s e^{ -x/q^n } dx$$
in order to obtain the first part of the right-hand side in equation \eqref{eq:mellin_transform}
\begin{align}
\label{eq:density_part1}
\Gamma(1+s) \frac{\left( q^{1+s}; q \right)_\infty}{\left( q; q \right)_\infty} & = \frac{1}{\left( q; q \right)_\infty} \int_0^\infty dx \ x^s \sum_{n=0}^\infty \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -x/q^n } \ .
\end{align}
Now, for the second part of the expression, an application of the $q$-binomial theorem (eq. \eqref{eq:q_binomial_thm}) yields
\begin{align}
\label{eq:density_part2}
\frac{ \left( z; q \right)_\infty }{ \left( z q^{-s}; q \right)_\infty } = &
\sum_{m=0}^\infty \left( z; q \right)_\infty \frac{z^m q^{-ms}}{ \left( q; q \right)_m } \ .
\end{align}
Forming the product of \eqref{eq:density_part1} and \eqref{eq:density_part2} gives:
\begin{align*}
  & \E\left( \left( I^{(q, z)} \right)^s \right) \\
= \quad & \Gamma\left(1 + s\right) \frac{\left( q^{1+s}; q \right)_\infty}
                                  {\left( q; q \right)_\infty}
                             \frac{ \left( z; q \right)_\infty }
                                  { \left( z q^{-s}; q \right)_\infty }\\
= \quad & \frac{\left( z; q \right)_\infty}{\left( q; q \right)_\infty}
          \left[ \sum_{m=0}^\infty \frac{z^m q^{-ms}}{ \left( q; q \right)_m } \right]
          \left[ \int_0^\infty dx \ x^s \sum_{n=0}^\infty \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -x/q^n } \right]\\
= \quad & \frac{\left( z; q \right)_\infty}{\left( q; q \right)_\infty} \sum_{m=0}^\infty \frac{z^m}{ \left( q; q \right)_m } \int_0^\infty dx \ \left( x q^{-m} \right)^s \sum_{n=0}^\infty \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -x/q^n }\\
\stackrel{(y=x q^{-m})}{=} & \frac{\left( z; q \right)_\infty}{\left( q; q \right)_\infty} \sum_{m=0}^\infty \frac{(qz)^m}{ \left( q; q \right)_m } \int_0^\infty dy \ y^s \sum_{n=0}^\infty \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -y q^{m-n} }\\
\stackrel{(Fubini)}{=} & \int_0^\infty dy \ y^s \left[ \frac{\left( z; q \right)_\infty}{\left( q; q \right)_\infty} \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(qz)^m}{ \left( q; q \right)_m } \frac{(-1)^n q^{ \binom{n}{2} } }{\left(q;q\right)_n} e^{ -y q^{m-n} } \right] \ ,
\end{align*}
hence the result for the density.

\subsection{Proof of theorem \ref{proposition:cv_bm}}
To recognize the random variable $\left( 1 - q \right)^2 I^{(q,z)}$ as nothing but $I\left( W^{(q, \mu)} \right)$, we perform the change of variable $s = \frac{2 u}{(1-q)^2}$ in the time parameter:
$$
  \left( 1 - q \right)^2 I^{(q,z)}
= \left( 1 - q \right)^2 \int_0^\infty q^{\zeta_s} ds
= 2 \int_0^\infty q^{ \zeta_{ \frac{2 u}{(1-q)^2} } } du
= 2 \int_0^\infty e^{ -2 W_s^{(q, \mu)}} ds \ .
$$

And, in order to prove that for any bounded continuous functional $F: D_\infty \times \R \rightarrow \R$:
$$ \E\left( F\left( W^{(q, \mu)}, 2 \int_0^\infty e^{ -2 W_s^{(q, \mu)}} ds \right) \right) \stackrel{q \rightarrow 1^-}{\longrightarrow}
   \E\left( F\left( W^{(\mu)}, 2 \int_0^\infty e^{ -2 W_s^{(\mu)}} ds \right) \right), $$
we will approximate the exponential functional $I(\cdot)$ by the exponential functional $I_T(\cdot)$ up to a finite horizon $T>0$:
 $$\begin{array}{cccc}
    I_T: & D_\infty & \rightarrow & \R \\
       &     x    & \mapsto     & 2 \int_0^T e^{-2 x(s)} ds
   \end{array}$$
which is continuous for the locally uniform topology. Also, we adopt the convention that $W^{(q, \mu)}$ for $q=1$ refers to a Brownian motion with drift $\mu$. The rest of the proof is broken down into three steps.

\paragraph{ \bf Step 1: Weak convergence in the uniform topology on the Skorohod space}\mbox{}\\
Using Wald's identity:
$$ \E\left(   W_t^{(q, \mu)} \right) = -\frac{\log q}{2 (1-q)^2} (1+z) t \E\left( B_1 \right) = -\frac{\left( 1-q^\mu \right) \log q}{(1-q)^2} t \stackrel{q \rightarrow 1^-}{\longrightarrow} \mu t \ ,$$
$$ \Var\left( W_t^{(q, \mu)} \right) = \half \log^2 q \ (1+z) t(1-q)^{-2} \E\left( B_1^2 \right) \stackrel{q \rightarrow 1^-}{\longrightarrow} t \ .$$
Thanks to Donsker's invariance principle, we have weak convergence to $W^{(\mu)}$ in the usual Skorohod topology. Because the statement is usually phrased for the rescaled discrete random walk, one can adapt theorem 14.1 in \cite{bib:billingsley} to the case of a compound Poisson process: the two previous estimates on mean and variance give tightness and because of the independence of increments, the convergence of finite dimensional marginals is obtained from the convergence of Lévy exponents. We record the expression of the Lévy exponent for later use:
$$ \E\left( e^{s W_t^{(q, \mu)}} \right) = e^{t \psi_q(s)}$$
where
\begin{align}
\label{eq:levy_khintchine_exp_q}
\psi_q(s) = & - 2\frac{1-q^{-\half s}}{1-q} \frac{1-q^{\mu+\half s}}{1-q} \stackrel{q \rightarrow 1^-}{\longrightarrow} \half s^2 + s \mu \ .
\end{align}

At this point, we need to upgrade convergence to the finer (locally uniform) topology. This is achieved by applying theorem 6.6 in \cite{bib:billingsley}. Indeed, the Wiener measure is supported on the space of continuous functions; which is separable for the locally uniform topology. Moreover convergence in the usual Skorohod topology to a continuous function implies local uniform convergence.

Now by applying the mapping theorem (theorem 2.7 in \cite{bib:billingsley}), we have that for all $T>0$:
\begin{align}
\label{eq:weak_cv_finite_T}
 \E\left( F\left( W^{(q, \mu)}, I_T\left( W^{(q, \mu)} \right) \right) \right) & \stackrel{q \rightarrow 1^-}{\longrightarrow}
 \E\left( F\left( W^{(   \mu)}, I_T\left( W^{(   \mu)} \right) \right) \right) \ .
\end{align}

\paragraph{ \bf Step 2: Uniform control of the error in $q$}\mbox{}\\
We prove that there is a $0 < q_0 < 1$ such that for all $\delta>0$:
\begin{align}
\label{eq:uniform_control}
\limsup_{T  \rightarrow \infty} \sup_{ q_0 \leq q \leq 1}
\P\left( \left| I\left( W^{(q, \mu)} \right) - I_T\left( W^{(q, \mu)} \right) \right| \geq \delta \right) & = 0 \ .
\end{align}

As $W^{(q, \mu)}$ has independent increments:
$$ I\left( W^{(q, \mu)} \right) = I_T\left( W^{(q, \mu)} \right) + e^{-2 W^{(q, \mu)}_T } I\left( \widetilde{ W^{(q, \mu)} } \right)$$
with $\widetilde{ W^{(q, \mu)} } := W^{(q, \mu)}_{T+\cdot} - W^{(q, \mu)}_{T}$ independent from $W^{(q, \mu)}_T$. By the Markov inequality, for $0 < \nu < \mu$:
\begin{align*}
         \P\left( \left| I\left( W^{(q, \mu)} \right) - I_T\left( W^{(q, \mu)} \right) \right| \geq \delta \right)
\leq \ & \E\left[ \left(\frac{1}{\delta}\right)^\nu e^{-2 \nu W^{(q, \mu)}_T } I\left( \widetilde{ W^{(q, \mu)} } \right)^\nu \right]\\
=    \ & \left(\frac{1}{\delta}\right)^\nu e^{T \psi_q(-2 \nu) } \E\left[ I\left( W^{(q, \mu)} \right)^\nu \right] \ .
\end{align*}
Let us examine the behavior of this bound as $q \rightarrow 1$. From the Mellin transform's explicit expression in theorem \ref{thm:main}, it is clear that $\E\left( I\left( W^{(q, \mu)} \right)^\nu \right)$ is uniformly bounded as $q \rightarrow 1$. Also, from \eqref{eq:levy_khintchine_exp_q}, $\psi_q(-2 \nu) \stackrel{q \rightarrow 1^-}{\longrightarrow} 2 \nu(\nu-\mu) < 0$. Hence the existence of a $0 < q_0 < 1$ such that for all $q_0  \leq q \leq 1$, $-\psi_q(-2 \nu) \geq \nu(\mu-\nu) > 0$. Therefore:
$$   \sup_{ q_0  \leq q \leq 1} \P\left( \left| I\left( W^{(q, \mu)} \right) - I_T\left( W^{(q, \mu)} \right) \right| \geq \delta \right)
 \leq \frac{e^{ - T \nu(\mu-\nu) }}{\delta^\nu} \sup_{ q_0  \leq q \leq 1} \E\left( I\left( W^{(q, \mu)} \right)^\nu \right),
$$
which is a finite quantity. This proves the claim \eqref{eq:uniform_control} upon taking $T \rightarrow \infty$.

\paragraph{ \bf Step 3: Conclusion}\mbox{}\\
Let $\varepsilon>0$. Without loss of generality, we can assume that $F$ is uniformly continuous in the second variable, thanks to the Portmanteau theorem. Let $\delta = \delta(\varepsilon)$ be the modulus of continuity of $F$ associated to $\varepsilon>0$:
$$ \forall w \in D_\infty, \ \left( |x-y| \leq \delta \Rightarrow \left| F(w,x) - F(w,y) \right| \leq \varepsilon \right) \ .$$
We have:
\begin{align}
\label{eq:inequality_control_delta}
\forall w \in D_\infty, \ \left| F\left( w, I(w) \right) - F\left( w, I_T(w) \right) \right| &
\leq \varepsilon + 2 |F|_\infty \mathds{1}_{\left\{ |I(x)-I_T(x)| \geq \delta \right\}}  \ .
\end{align}

By combining the triangular inequality with \eqref{eq:inequality_control_delta}, we have for all $q_0 \leq q \leq 1$ and $T>0$:
\begin{align*}
     & \left| \E\left( F\left( W^{(q, \mu)}, I\left( W^{(q, \mu)} \right) \right) \right)
           -  \E\left( F\left( W^{(   \mu)}, I\left( W^{(   \mu)} \right) \right) \right) \right|\\

\leq & \left| \E\left( F\left( W^{(q, \mu)}, I_T\left( W^{(q, \mu)} \right) \right) \right)
           -  \E\left( F\left( W^{(   \mu)}, I_T\left( W^{(   \mu)} \right) \right) \right) \right|\\
     & + 2 \varepsilon + 2 |F|_\infty \sup_{q_0 \leq q' \leq 1} \P\left( \left| I\left( W^{(q', \mu)} \right) - I_T\left( W^{(q', \mu)} \right) \right| \geq \delta \right) \ .
\end{align*}
Now, thanks to the weak convergence \eqref{eq:weak_cv_finite_T}, the following holds for all $T>0$:
\begin{align*}
 & \limsup_{q \rightarrow 1^-}
   \left| \E\left( F\left( W^{(q, \mu)}, I\left( W^{(q, \mu)} \right) \right) \right)
       -  \E\left( F\left( W^{(   \mu)}, I\left( W^{(   \mu)} \right) \right) \right) \right|\\
\leq \ & 2 \varepsilon + 2 |F|_\infty \sup_{q_0 \leq q' \leq 1} \P\left( \left| I\left( W^{(q', \mu)} \right) - I_T\left( W^{(q', \mu)} \right) \right| \geq \delta \right) \ .
\end{align*}
Using the uniform control estimate \eqref{eq:uniform_control} and taking $T \rightarrow \infty$ in the previous expression:
$$ \limsup_{q \rightarrow 1^-}
   \left| \E\left( F\left( W^{(q, \mu)}, I\left( W^{(q, \mu)} \right) \right) \right)
       -  \E\left( F\left( W^{(   \mu)}, I\left( W^{(   \mu)} \right) \right) \right) \right|
\leq 2 \varepsilon \ .
$$
We are done as $\varepsilon>0$ is arbitrary.

\section*{Acknowledgments}
The author would like to thank, one the one hand, Alexander Watson and Juan Carlos Pardo for encouraging him to write this note; and on the other hand Mladen Savov for pointing out references on Wiener-Hopf factorizations.

\small
\bibliographystyle{halpha}
\bibliography{Bib_q_functionals}

\end{document}

config: arxiv


TITLE: Uncountable family of infinite subsets with pairwise finite intersections
QUESTION [8 upvotes]: I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family $(X_\alpha)_{\alpha \in A}$ of infinite subsets of $X$ such that $X_\alpha \cap X_\beta$ is finite whenever $\alpha \neq \beta$.  The way I know how to prove this statement is as follows.
First, it suffices to prove the case when $X$ is countable.  Thus we can choose a bijection between $X$ and $\mathbb{Q} \cap [0,1]$.  To save notation we can tacitly assume that $X = \mathbb{Q} \cap [0,1]$.
Let the index set be $A = [0,1] \setminus X$, i.e. all the irrationals in $[0,1]$.  For each $\alpha \in A$, choose a sequence  $(x_{\alpha 1},x_{\alpha 2},\dots)$ of elements of $X$ such that $x_{\alpha n} \to \alpha$ as $n \to \infty$, and let $X_\alpha = \{ x_{\alpha_n} \mid n \in \mathbb{N} \}$.
Since $\alpha$ is irrational, the sequence $(x_{\alpha n})$ cannot be eventually constant, so $X_\alpha$ is infinite.  And if $\alpha \neq \beta$ then the sequences $(x_{\alpha n})$ and $(x_{\beta n})$ can have only finitely many terms in common since they have different limits, so $X_\alpha \cap X_\beta$ is finite.
Is it possible to do this in a more constructive way?  I know very little about set theory and logic, so I apologize if this question is too elementary.  Also, I wasn't sure about any relevant tags other than set-theory, so please feel free to add appropriate tags.
Edit: to clarify, I didn't have a clear notion of what I meant by "constructive" here.  What I didn't like about the proof I gave above was that it required a choice of sequence of rationals converging to each irrational.  The answers so far all address this concern adequately.

REPLY [12 votes]: Here is a transcription of Todd Eisworth's answer into constructive mathematics. His construction is more or less constructive, he just uneccessarily says things which ruin the constructivity such as:

In constructive mathematics only decidable subsets of $\mathbb{N}$ have characteristic functions. But we can avoid the problem by speaking only about the characteristic functions.
Not every countable set can be put in bijective correspondence with the natural numbers. This can be avoided, because we don't need such a bijection, just an injection from the natural numbers into the set (which is the definition of "infinite").
In constructive mathematics cardinals don't work too well, so it is better not to mention $2^{\aleph_0}$. This can be avoidded by talking explicitly about a set being uncountable (i.e., there is no surjection from $\mathbb{N}$ onto the set).

And here is the proof:
The set $2^\ast$ of finite sequences of $0$'s and $1$'s is in bijective correspondence with $\mathbb{N}$, therefore it clearly suffices to find an uncountable collection of subsets of $2^\ast$ such that any two of them have only a finite intersection. Once such a collection is found, it can be embeded into $X$.
The set $2^\mathbb{N}$ is uncountable by the usual Cantor's diagonal argument (which is constructive!). Given any $f \in 2^\mathbb{N}$, let $S_f \subseteq 2^\ast$ be the set of finite prefixes of $f$. The assignment $f \mapsto S_f$ is injective, therefore the $S_f$'s form an uncountable family. If $f$ and $g$ are distinct, then there is a smallest $n \in \mathbb{N}$ such that $f(n) \neq g(n)$, hence $S_f \cap S_g$ is finite, as it contains precisely the first $n$ prefixes of $f$.

REPLY [3 votes]: There are several constructions, some of which are quite nice and visual, and which you are therefore more likely to consider "constructive". These examples were shown to me by Imre Leader.

Your example: take your infinite set to be $\mathbf{Q}$ and look at approximations to reals.
Take your infinite set to be the nodes of an infinite binary tree, and take your sets to be the paths from the root to infinity. Every two such paths will go separate ways after finitely many steps.
Take your infinite set to be the quarter-lattice $\mathbf{N}^2$, and take your sets, indexed by $\theta\in[0,\pi/2)$, to consist of the lattice points between $y=x \tan(\theta)$ and $y=x\tan(\theta) + 1$. Any two such slivers have finite intersection.

Complementary question: Find an uncountable totally ordered chain of subsets of $\mathbf{N}$.


TITLE: How to prove union and intersection of classes if sets?
QUESTION [2 upvotes]: Im working on some book exercises and struggle with a proof.
If $B_i$ and $C_j$ are two classes of sets such that $B_i \subseteq  C_j $.
How can I show that $\bigcup\limits_i B_{i}  \subseteq \bigcup\limits_{j} C_{j}$ and $\bigcap\limits_j C_{j} \subseteq \bigcap\limits_{i} B_{i}$
So far I have tried for $\bigcup\limits_i B_{i}  \subseteq \bigcup\limits_{j} C_{j}$:
Let $x  \in \bigcup\limits_i B_{i} \subseteq \bigcup\limits_j C_{j} \iff x \in B_i \cup C_j$
By the definition of union, it follows that $x \in B_i$ or  $x \in C_j$
By definition of subset it follows that $x \in B_i \implies x \in C_j$
So far I have tried for $\bigcap\limits_j C_{j} \subseteq \bigcap\limits_{i} B_{i}$:
Take any subset $x \subseteq B_i \cap C_j$.
$x \subseteq B_i \wedge x \subseteq C_j$, meaning that $x \in \bigcup\limits_i B_{i} \wedge x \in \bigcup\limits_j C_{j}$.
However not sure whether that is correct. Appreciate any help.

REPLY [1 votes]: You misstated the problem for  Simmons Introduction to Topology and Analysis.
The problem was:
Let $\{A_i\}, \{B_j\}$ be to classes of sets. so that $\{A_i\} \subset \{B_i\}$ (in other words it is the classes of sets that are subsets of each other and NOT the individual sets)
Now show that $\cup_i A_i \subset \cup_j B_j$ and $\cap B_j \subset \cap_i A_i$.
Again, my advice is element chase:
Let $x \cup_i A_i$ then $x \in A_i$ for some $A_i$.  Now $\{A_i\} \subset \{B_i\}$ so $A_i = B_k$ for some $B_k \in \{B_j\}$ so $x \in B_k$. so $x \in \cup_j B_j$.
So $\cup_i A_i \subset \cup_j B_j$.
If $x \in \cap_j B_j$ then $x \in B_j$ for all the $B_j$.  Let $A_i$ be any set in $\{A_i\} \subset \{B_j\}$.  Then $A_i = B_k$ for $B_k \in \{B_j\}$.  So $x \in B_k = A_i$ because $x$ is in all $B_j$.  Thus it is in all the $A_i \in \{A_i\} \subset \{B_j\}$ so it is in $\cap_i A_i$.
$\cap B_j \subset \cap_i A_i$.


TITLE: Solution of the quasilinear equation $p - z q + z = 0$ with initial data curve
QUESTION [0 upvotes]: Hello ı cant write like maths notation hear sorry for that.
Find the solution of the initial value problem for the quasilinear equation
$p - z q + z = 0$ for the initial data curve
$\Gamma: x_{0} = 0, y_{0} = s, z_{0} = -2s, −\infty \leq s \leq \infty $
I found $c_{1} = (e^x).z $
But ı dont know is it true or not and ı couldn't continue from this part. Can u help me ?

REPLY [1 votes]: $$z_x-zz_y=-z$$
You correctly found a first characteristic equation :
$$ze^x=c_1$$
A second characteristic equation is :
$$z-y=c_2$$
The general solution is :
$$ze^x=F(z-y)\quad\implies\quad z=e^{-x}F(z-y)$$
Condition :
$$z(0,y)=-2y=e^0F(-2y-y)=F(-3y)$$
The function $F$ is determined :
$$F(X)=\frac23 X$$
We put in into the general solution where $X=z-y$
$$ze^x=\frac23 (z-y)$$
Solving for $z$ :
$$z(x,y)=\frac{-2y}{3e^x-2}$$


TITLE: Can a mass matrix be asymmetric?
QUESTION [8 upvotes]: I am developing a mathematical model of a mechanical device consisting basically of coupled harmonic oscillators. It turns out that the system mass matrix is asymmetric. I seem to read somewhere that a mass matrix has to be symmetric, but I am not sure. So I would like to know whether it is possible for a mass matrix in this case to be asymmetric. If it can't, what are the physical implications of an asymmetric mass matrix in this case?

REPLY [6 votes]: I agree with the answer of Lubos (and ja72 is right that there is also a positivity criterion). However, since you repeated the question again to ja72, let me explain the answer of Lubos more explicitly and with more details.
Suppose that in your system of harmonic oscillators the kinetic term is defined with a mass matrix $M$, i.e. (as in the answer of ja72):
$$K=\frac{1}{2} \dot{q}^T M \dot{q}$$
Now define the following to matrices $M_s=\frac{1}{2}(M+M^T)$ and $M_a=\frac{1}{2}(M-M^T)$. It is very easy to verify that $M=M_s+M_a$, $M_s^T=M_s$ and $M_a^T=-M_a$. Because of these properties, $M_s$ is called the symmetric part of $M$, while $M_a$ is called the antisymmetric part (this is also the terms Lubos used).
Now, let us notice that since $M_a$ is antisymmetric the identity $v^T M_a v=0$ holds for any vector $v$. Writing up the kinetic term again, we obtain
$$K=\frac{1}{2} \dot{q}^T M \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q} + \frac{1}{2} \dot{q}^T M_a \dot{q}= \frac{1}{2} \dot{q}^T M_s \dot{q}$$
Hence $M_s$ defines exactly the same kinetic term as $M$! This means that without loss of generality we can assume that the mass matrix is symmetric.
Or in other words: whenever you write down a kinetic energy with a a non-symmetric mass matrix $M$, you can just forget about the antisymmetric part, and write down the same kinetic energy with $M_s$ - it simply leads to the same system.
I hope that this more detailed explanation helps in understanding the answer (which was also previously given). Don't hesitate to ask, if something is still not clear.
Best,
Zoltan


TITLE: Functions that can be computed faster simultaneously than expected
QUESTION [6 upvotes]: The following is an elementary question about circuit complexity. It is different from the kind of thing I have seen discussed, so I would be interested in any work that has been done on this kind of question. My apologies if this is easy or well known.
By a Boolean function, I mean a function from $2^n\rightarrow2^m$ where $2$  is the set with two elements $0,1$. If $f:2^n\rightarrow2^m$ and $g:2^n\rightarrow2^k$ are Boolean functions, I denote by $\langle f,g\rangle$ the function $2^n\rightarrow2^{m+k}$ whose output is the output of $f$ followed by the output of $g$. Finally, let $|f|$ be the size (i.e. number of internal nodes) of the minimal circuit computing $f$. My question is the following: Suppose $| \langle f, \langle g,h \rangle \rangle | < |f|+|g|+|h|$ . Does it follow that $|\langle x,y \rangle | < |x|+|y|$ for some pair $x,y$ chosen from $f,g,h $ ? Naively, the idea is that we can build a circuit computing $\langle f,g \rangle$ from one computing $f$ and one computing $g$, and this has size $|f|+|g|$. But if $f$ and $g$ have something "in common", perhaps we can do better. Then the question is, suppose $f,g,h$ "have something in common". Does this mean two of these functions already "have something in common"?
I would also be interested in any variations on this question, for example for considering only formulas or bounded circuits.

REPLY [6 votes]: A specific example that has a negative answer comes from Karatsuba multiplication in which one multiplies two polynomials
$$(a_0 + a_1x)(b_0 + b_1x) =: c_0 + c_1x + c_2x^2$$
$$= (a_0b_0) + (a_0b_1 + a_1b_0)x + (a_1b_1)x^2$$
$$= (a_0b_0) + (a_0b_0 + (a_0+a_1)(b_0 + b_1) + a_1b_1)x + (a_1b_1)x^2$$
If we take $\mathbb{F}_2$-coefficients (or $2$-coefficients, in your notation), this can be viewed as three Boolean functions $2^4 \to 2^1$, or their concatenation a single function $2^4 \to 2^3$.
Karatsuba's trick allows one to turn the four multiplications and one addition into three multiplications and four additions. If you're counting all operations, and not just multiplications, it looks like you have increased from five to seven operations. But if you use this trick recursively, or take the $a_i,b_j$ to be polynomials themselves, then the reduction in multiplication wins out, and you can save overall operations.
I don't have a proof that two of the coefficients cannot be optimized without the third, but the reason I believe it is the case is because the Karatsuba trick exploits the fact that our coefficients of interest live in a rank-3 subspace of a 4-dimensional space.


TITLE: Real Analysis - Uniform Convergence Problem
QUESTION [1 upvotes]: So I screwed up a problem on my exam. I know that now. But pure mathematics is as difficult and terrifying as it is rewarding for me, and I can't let this go! If someone could tell me if the following is right or wrong, that would be much appreciated.
I was asked to prove the uniform convergence of the series $f_n(x)=\frac{x}{x+n}$ on $[1,\infty)$.
So I've taken $$f=\lim_{n\to \infty} \frac{x}{x+n}=0$$ and $$|f_n-f|=|\frac{x}{x+n}-0|=\frac{x}{x+n}<1, \forall x\in[1,\infty), \forall n\in\Bbb{N}$$
Then, I simply took $N>0$ and $n \ge N$, concluding that $f_n(x)$ converged uniformly to $1$.
This seems too easy to be right (so did the solution I put on my exam, God help me!). Is it?
Thanks!

REPLY [3 votes]: The sequence $f_n$ does not converge uniformly to $f$ since
$$
\sup_{x\ge 1}\biggl|\frac x{x+n}\biggr|=1
$$
for each $n\ge1$.

REPLY [2 votes]: Not really, sorry. You need to show, for given $\varepsilon>0$, that there is $N\in \mathbb{N}$, such that $$\frac{x}{x+n}< \varepsilon$$
for all $n\ge N$, independently of $x$. Which, I'm afraid, is not even true, because if you have any $N$ you can choose $x $ very large (larger than $N$) such that the fraction becomes $>\frac{1}{2}$, say. What can be shown is uniform convergence on compact sets, then you'd start out with an upper bound on $x$


TITLE: find vector x so that vector: diag(x).C.x has all components equal, where C is positive-definite
QUESTION [2 upvotes]: PROBLEM: I am trying to find closed form solutions or provable general properties of solutions for the solution $x$ of the following:
Find $\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_N \end{bmatrix}$ where:
 $$diag(x) C x == \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$$
...(or more generally equals $\begin{bmatrix} t \\ \vdots \\ t \end{bmatrix}$ for any positive constant $t>0$).
The solution, $x$, is an $N$-by-1 (column) vector.  The right hand side, the column vector of ones, is as well.
$C$ is an $N$-by-$N$ correlation matrix: it is symmetric, the diagonal elements are exactly $1$, and the off-diagonal elements are in the range $-1 < C(j,k) < 1$.  Actually for my problem we can assume $0 < C(j,k) < 1$, for all $j \ne k$.
We should assume $N>3$ and $C(j,k)$ is distinct for every distinct combination $(j,k)$ where $j \ne k$.  I have found easy solutions for $N=2$, and where all the off-diagonal elements have the same value.
I am using the convention $M=diag(x)$ to refer to the diagonal square matrix $M$ with whose diagonal is formed by the vector $x$, that is, $M_{j,j} = x_j$.
This problem involves (a) simultaneous polynomial equations, and (b) positive definite matrices (more specifically nonsingular correlation matrices).  I am trying to find either a closed form solution (which may not exist) or at least to find if this family of equations has “a name” and some well-known properties within some sub field of algebra.  It started out as a signal / data viz problem I got obsessed with after working on it in 2 & 3 variable problems, and realized there wasn't (?) a solution for higher dimensions that was easy/obvious (to me at least).
Anything you got is appreciated.

REPLY [1 votes]: No closed form: basically isomorphic to finding a general closed-form solution for the roots of polynomials of arbitrarily high degree.  It is however frequently solved by numerical optimization in applications such as the determination of so-called "risk parity" portfolios in finance.


TITLE: What is the necessary condition for the process of "proceeding to the limit" in Whitehead and Russell's PM?
QUESTION [1 upvotes]: I read this from Introduction of the 1st edition of Principia Mathematica by Whitehead and Russell:

Since the orders of functions are only defined step by step, there can be no process of "proceeding to the limit," and functions of infinite order cannot occur.

Here is a summary of orders of functions:

An nth-order function is one that takes (n-1)th and lower order functions as arguments.
A first order function is one that takes only "individuals" as arguments.
A second order function is one that take first order function and "individuals" as arguments.
By "individuals" we mean constituent objects that are neither a function or a proposition and will not disappear like a class or a set after analysis.

At this point, I think an nth-order function is perfectly eligible for mathematical induction. I can't see the connection between something defined step by step and its inability to "proceed to the limit." It seems that the authors assume the readers are familiar with some well-known rules that specify the necessary condition for the process of "proceeding to the limit." I wonder what these rules are.

REPLY [0 votes]: Short answer:
Mathematical induction can never proceed to infinity. The necessary condition for mathematical induction is that the property in question is possessed by an inductive number and is hereditary with respect to the relation $+_c1$.

Long answer:
This book is like building a tower from ground up without scaffolding,
  crane or any outside support. Everything that touches the ground must
  be explicitly spelt out and every brick must be supported by the ones
  from beneath not above. In other words, primitive ideas and primitive
  propositions are all the foundations this book has.* And every
  proposition is either a primitive one or must be proved by numbers
  less than itself. The formulae of induction are not reached until
  ✳91·17·171·373. The hierarchy of types is introduced in ✳12 which precedes the formulae of induction.
✳91·17 $\vdash:. P \in Potid‘R:\phi S. \supset_S. \phi(S|R):\phi(I {\restriction C‘R}): \supset \phi P $
Which states that if the property $\phi$ is hereditary with respect to $|R$, then if $\phi$ belongs to $I \restriction C‘R$ it belongs to any member of $Potid‘R$
Based on summary of Part III Section C, mathematical induction is a
  definition rather than a principle. In other words, some cardinals are
  inductive, some other cardinals are non-inductive. The defining
  property of inductive cardinals is this: An inductive cardinal is one
  which obeys mathematical induction starting from $0$, i.e. it is one
  which possesses every property possessed by $0$ and by the numbers
  obtained by adding $1$ to numbers possessing the property. In other
  words:
$\vdash :: \alpha \in NC induct .≡:. \xi \in \mu .\supset_\xi. \xi +_c
 1 \in \mu : 0 \in \mu \supset_\mu. \alpha \in \mu $
An inductive number by definition is one which can be reached from $0$
  by successive additions of $1$. Such a process can never reach
  infinity because ✳123 shows that the smallest infinite number $\aleph_0$ is not an inductive number.

*Due to the speculative nature of philosophy, the authors admit that "there must always be some elements of doubt, since it is hard to be
 sure that one never uses some principle unconsciously."  --see Part I.
 Section A, on first page of theory of deduction.


TITLE: Quotients and exact sequences of algebraic group schemes
QUESTION [5 upvotes]: Now I study group schemes to understand fundamental properties about semi abelian varieties and generalized jacobian varieties.
Let $G$ be an algebraic group scheme over a field $k$ ($=$ $k$ group schemes of finite type), $H$ an algebraic group sub-scheme.
In Milne's online note, the author defines that the quotient $G/H$ is the representable algebraic scheme of the sheaf associated to the presheaf $S \mapsto G(S)/H(S)$, in the "faithfully flat finite type site". (not fppf.)
But in Conrad's "semistable reduction for abelian varieties", the author uses the different definition.
(It seems for me that he uses the fppf site.)
And so these 2 references define exact sequences in the different way.
The former says that a sequence $1 \to G' \xrightarrow{f} G \xrightarrow{g} G'' \to 1$ is exact if $g$ is faithfully flat and $f$ identifies $G'$ to $\ker g$.
(The author shows that $g$ is faithfully flat $\iff$ $g$ is surjective $\iff$ $g$ induces $G / \ker g \cong G''$.)
Are these two definitions same?
Next, In Serre's "algebraic groups and class fields", the author says that if $k$ is algebraically closed, a sequence of group varieties (= algebraic groups that are varieties = smooth algebraic groups) $1 \to G' \xrightarrow{f} G \xrightarrow{g} G'' \to 1$ is exact iff this is exact on the rational points as abstract groups and it induces the exact sequence of the tangent spaces at $1$.
Is this true?
I don't know the Weil's foundation, so I don't understand whether this is true for schemes.
And please suggest me some references.
Thank you very much!

REPLY [2 votes]: I don't have Milne on me, but I'm pretty sure that "faithfully flat finite type" is just fppf. Recall that fppf is just french for faithfully flat and finite presentation, but our objects are Noetherian.
As to your second point, the only question is whether or not if $f:G\to H$ is faithfully flat then $H$ is isomorphic to the fppf sheafification of $G/\ker f$. It sounds like you know that the fppf sheafification is representable by some group scheme $Q$. Note moreover that we get a map $Q\to H$ of groups since we have a map of presheaves $(G/\ker f)^\mathrm{pre}\to H$ and thus we get a map by the definition of sheafification since $H$ is a sheaf. We want to show that this map of groups is an isomorphism. The fact that it's injective is clear. To show surjectivity it suffices to show that $f:G\to H$ is a surjective map of fppf sheaves. But, this almost follows by definition/the fact that we have effective fppf descent for morphisms.
For your final point, if I'm understanding you correctly. There are two points here:

If $f:G\to H$ is a map of finite type algebraic groups over $k$, then $\ker f$ is trivial iff $(\ker f)(\overline{k})=1$ and the map $(df)_e:T_e G\to T_e H$ is injective. Moreover, both of these are equivalent to the fact that $f$ is a closed immersion.
If $f:G\to H$ is a map of smooth algebraic groups over $k$, then $f$ is fatithfully flat if and only if $f(\overline{k}):G(\overline{k})\to H(\overline{k})$ is a surjection.
If $f:G\to H$ is a surjective map of smooth groups, and $\ker f$ is smooth, then the induced maps on tangent spaces is a short exact sequence.

The first fact can be found here as proposition 10.2.1 with proof in subsequent pages. The second fact follows from miracle flatness and basic facts about surjectivity of maps of schemes. The last part is clear by noting that since $\ker f$ is smooth that $f$ itself is smooth by fppf descent (since $G\times_H G\cong G\times_{\mathrm{Spec}(k)}\ker f$) and thus surjectivity on the tangetn spaces is clear, and the short exact sequence part is an exercise I leave to you.


TITLE: how can I find maximum value of this function?
QUESTION [1 upvotes]: Can any one help me to calculate this function :
$$f(y)=\max\limits_{\mu>0}[\exp(\frac{-n\mu^{2}}{\sigma^{2}})\exp(\frac{2\mu}{\sigma^{2}}\sum_{k=1}^{n}y_{k})]$$
where $y_{k}$ is random variable with normal distribution. $$$$
Thank you in advance.
Sorry, I had forgotten to put the second power of $\mu$ in first exponential(I modified it).

REPLY [2 votes]: To maximize $\exp(g(\mu)/\sigma^2)$ with $g(\mu)=-n\mu^2+2\mu s$ and $s=\sum\limits_{k=1}^ny_k$, one should maximise $g(\mu)$. Since $g'(\mu)=-2n\mu+2s$ is positive for $\mu\lt s/n$ and negative for $\mu\gt s/n$, $g(\mu)$ is maximal at $\mu=s/n$ and $f(y)=\exp(g(s/n)/\sigma^2)$. Since $g(s/n)=s^2/n$, $f(y)=\exp(s^2/(n\sigma^2))$.


TITLE: Determining Degeneracies of Operator (Quantum Mechanics / Linear Algebra)
QUESTION [3 upvotes]: Hello all! Above is my question. I am fine all the way up to the final part about the degeneracy. I find counting degeneracies quite difficult, and this is no exception! I have really no idea why the trace is as that power series, and don't know why the degeneracies are as given.
By the standard power series / binomial expansion, I get
$$ d_n = {(-1)^n \over n!} \prod_{m=0}^{n-1}(-D-m) = {1 \over n!} \prod_{m=0}^{n-1}(D+m).$$
(Empty sum is identity, as usual.) I can't work out why this is the degeneracy (I know it is related to $n \choose r$, but not sure how to use that...). I can't see any reason why this should be the coefficients in the power series expansion.
Also, while I'm on the topic of degeneracies, how can I show that the degeneracy of the $n-$th excited state of the 3D harmonic oscillator ($E_{n_1,n_2,n_3} = \hbar \omega (n_1 + n_2 + n_3 + 3/2))$ is $(n+1)(n+2)/2$ (which I know is equal to $\sum_{r=0}^{n+1}r$). I guess it's just the number of ways that three (positive) integers can add up to $n$, but I don't know how to prove that.
Thank you in advance for your help!

(Please note: I'm going away shortly, so if I don't respond to your answer, then I'm not ignoring it, I simply haven't seen it yet; once I see any responses, I shall reply asap! Thanks!)

(Oh, and please no-one try to move this to physics.SE - this is a maths questions; I am not asking what the physical significance of this is, just the maths behind it - it's linear algebra really! Sorry to be blunt, but I had someone move one over incorrectly before - rather annoying! Thanks!)

REPLY [2 votes]: When you have $D$ oscillators the state will be $| n\rangle = | n_1, n_2, \cdots, n_D\rangle$ and the trace you want:
$$
Tr(x^{\mathcal{N}}) = \sum_{n} \langle n|x^{\mathcal{N}}|n\rangle = \sum_{n_1, \cdots, n_D =0}^{\infty} \langle n_1, \cdots, n_D|x^{\mathcal{N}}| n_1,\cdots, n_D \rangle
$$
The second sum is actually (all number operators commute with each other)
$$
\sum_{n_1, \cdots, n_D = 0}^{\infty} x^{n_1 + \cdots +n_D } = \sum_{n_1 =0}^{\infty}x^{n_1} \sum_{n_2=0}^{\infty}x^{n_2} \cdots \sum_{n_D =0}^{\infty}x^{n_D} = (1-x)^{-D}
$$
In the last step one uses the result derived in (a).
The degeneracy of the operator $\mathcal{N}$ is the amount of all such partitions of the integer $n$ into the sum $n_1 +n_2 + \cdots + n_D$, this is a standard combinatioral problem (solved usually by introducing "separations"), the answer is
$$
d_n = \binom{n+D-1}{n}
$$
One can see this number is exactly the coefficent in the trace expansion by noting that the one can recast the sum into the form
$$
\sum_{n_1, \cdots, n_D = 0}^{\infty} x^{n_1 + \cdots +n_D } = \sum_{n=0}^{\infty} \sum_{n_1 + \cdots + n_D =n}x^{n} = \sum_{n=0}^{\infty} d_n x^n
$$
(all we have done is classify the terms of the sum according to their powers)
Your approach seems to make use of the generating function for $d_n$, which should also work.
For the 3D harmonic oscillator you have $D=3$ so you get
$$
d_3 = \binom{n+3-1}{n} = \binom{n+2}{n} = \frac{(n+1)(n+2)}{2}
$$


TITLE: Are all primes the same?
QUESTION [4 upvotes]: Are there properties "intrinsic"* to primes that set them apart?

Example 1: The fields $\mathbb{F}_p$ do not all have the same structure. If $p=1 \mod 4$ is a prime then there is a number $1<x<p$ where $x^{-1}=-x\mod p$, i.e. in the field $\mathbb{F}_p$, its multiplicative inverse is also its additive inverse. Moreover, this happens always and only when $p=1\mod 4$ and not when $p=3\mod 4$. Thus the odd primes are neatly divided into two classes with distinctly different properties.
Example 2: The fields $\mathbb{F}_p$ yield good expander graphs, but some are better than others. For a prime $p$, construct an undirected graph on the vertices $\{0,1,\ldots,p-1\}$ as follows: node $n$ is connected to $n-1,n+1$ and $n^{-1}\mod p$. Let $\lambda_p$ be the spectral gap of the graph thus obtained (the difference between the two largest eigenvalues of the adjacency matrix). It follows from Selberg's 3/16 theorem that the resulting graph is a Ramanujan graph: it has an almost maximal spectral gap, hence these graphs are good expanders, and their diameter will be very small: only $\mathcal{O}(\log p)$. But some primes yield better expanders than others. I'm looking at a plot for the spectral gaps of primes up to 727, and there are no discernable local trends, although it looks like the spectral gap might converge to the expected value of $3-\sqrt{8}$.
Are there more examples that show that not all primes are equal? This question is born of curiosity, so feel free to bring surprising results from any area of mathematics!

*Of course there are twin primes, Sophie Germain primes and Mersenne primes, but these labels only show where a prime is located with respect to other numbers. For the purposes of this question, I am interested in properties "intrinsic" to a prime that set it apart from other primes.

REPLY [2 votes]: You can divide the primes into two types: those primes $p$ such that $p+1$ has exactly one prime factor and the primes $p$ such that $p+1$ has more than one prime factor. Of course, the primes of the first type are exactly the Mersenne primes.


TITLE: Moments of products of independent random variables: $E[X^mY^n]$ Part II
QUESTION [2 upvotes]: This is a follow up question from here.
Let $X$ and $Y$ denote two real-valued bounded random variables. Then all joint moments exist and uniquely define their joint probability $P(X,Y)$.
Given for all $m,n\in\mathbb{N}\cup\{0\}$ we have $E[X^m Y^n]=a_m b_n$ with $a_m, b_n \in \mathbb{R}$. Does it follow that $X$ and $Y$ are independent random variables and, thus, $a_m=E[X^m]$, $b_n=E[Y^n]$ and $E[X^m Y^n]=E[X^m]E[Y^n]$?
If yes, is it a straight forward result?
Thanks for your comments / suggestions.

REPLY [2 votes]: Assuming that the equation $EX^{m}Y^{n}=a_mb_n$ holds for all $n, m \geq 0$ and $a_0b_0=1$ the answer is YES. In this case $a_mb_0=EX^{m}$ and $b_na_0=EY^{n}$ so we get  $EX^{m}Y^{n}=EX^{m}EY^{n}$.
From the fact that the moments uniquely determine the joint distribution it follows that $X$ and $Y$ are independent. The reason is if $U$ and $V$ are independent with the same distributions as $X$ and $Y$ respectively then  $(U,V)$ and $(X,Y)$ have the same moments. The uniqueness of the  joint distribution now implies that $(U,V)$ has same distribution as $(X,Y)$ which implies that $X$ and $Y$ are independent.


TITLE: Show this inequality $\frac{n}{a_1 - a_0} + \frac{n - 1}{a_2 - a_1} + \cdots + \frac{1}{a_n - a_{n-1}} \ge \sum_{k=1}^n \frac{k^2}{a_k}$
QUESTION [0 upvotes]: For $a_1, \ldots , a_n \in \mathbb{R}, a_1 < a_2 < \cdots <a_n$ and $a_i \ne 0$, show that
$\dfrac{n}{a_1 - a_0} + \dfrac{n - 1}{a_2 - a_1} + \cdots + \dfrac{1}{a_n - a_{n-1}} \ge \sum_{k=1}^n \dfrac{k^2}{a_k}$
where $a_0 = 0$.
I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.
The inequality mentioned in the chapters are
Cauchy-Schwarz Inequality
Weierstrass's Inequality
Tchebychev's Inequality
I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.

REPLY [2 votes]: We proceed by induction on $n$. For $n = 1$ we have
    $$
  \frac{1}{a_{1} - a_{0}} \geq \frac{1}{a_{1}},
 $$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
    \begin{align*}
  \sum_{k = 1}^{n + 1}\frac{n + 2 - k}{a_{k} - a_{k - 1}} &= \sum_{k = 1}^{n}\frac{n + 1 - k}{a_{k} - a_{k - 1}} + \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}}  \\
             & \geq \sum_{k = 1}^{n} \frac{k^{2}}{a_{k}} + \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}},
 \end{align*}
    by the induction hypothesis. It is therefore sufficient to prove that
    $$
  \sum_{k = 1}^{n + 1}\frac{1}{a_{k} - a_{k - 1}} \geq \frac{(n + 1)^{2}}{a_{n + 1}}.
 $$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, \ldots, a_n - a_{n - 1}$.