TITLE: Is the force Exerted by a ball thrown at a wall regardless of the angle always perpendicular to the wall? Why?
QUESTION [1 upvotes]: The observation of the ball going of in a different direction says but why does this happen?

REPLY [2 votes]: Sometimes to visualise where forces apply and what in direction the forces apply it is useful to think about the momentum of the objects in question. Loosely speaking:

Momentum is conserved, so if two objects interact and one loses momentum, the other must gain it. (This is very similar to the idea in Newton's 3rd law - every action has an equal and opposite reaction).
When the momentum of an object changes, the rate of change of the momentum is the force applied to the object. (This is Newton's 2nd law - force equals mass times acceleration).

Throw a ball directly (i.e. perpendicular) at a wall, the ball's momentum will change direction back towards the thrower, when it hits the wall. Therefore, there will be a force applied, from the wall to the ball, in the direction back towards the thrower. The wall also has a change in momentum due to statement 1 and therefore the ball also applies a force to the wall due to statement 2.
When throwing the ball at an angle to the wall, imagine the momentum of the ball resolved into motion perpendicular and tangential to the point of impact on the wall. We've already looked at the perpendicular motion so now look at the motion of the ball tangential to the point of impact on the wall. If the momentum of the ball changes tangentially then you can identify a force in the direction of the changing momentum. This will only happen if the ball interacts with the wall because it's not perfectly smooth (i.e. friction) or the wall's not straight. (You can add the parallel and tangential forces by vector addition into a single force which won't necessarily be perpendicular to the wall if you like)
This approach also helps with another question commonly asked when people first encounter Newton's third law. 'What happens if instead of hitting the wall the ball hits and breaks a window and falls through the other side. Where is the equal and opposite reaction then? Tracing through the changes in momentum will again answer this question and if you can do that you will have got the concept.


text:
{\bf Problem.} In any month with five Mondays, a school declares the fifth Monday a Marvelous Monday. School begins on Monday, September 4. What is the date of the first Marvelous Monday after school starts? (If you abbreviate, do not use a dot in the abbreviation.)
{\bf Level.} Level 5
{\bf Type.} Number Theory
{\bf Solution.} September has 30 days.  September 4 is a Monday, so September 9 is a Saturday.  Since September 30 is exactly 21 days later (or 3 weeks), September 30 is also a Saturday.

Then October 1 is a Sunday, and October 2 is a Monday.  Then October 2, 9, 16, 23, and 30 are all Mondays, so the first Marvelous Monday is $\boxed{\text{October 30}}$.

set_name: MATH


TITLE: limits using $ \epsilon - \delta $ to prove two variable function
QUESTION [6 upvotes]: I'm trying to use the $ \epsilon - \delta $ argument to prove $\lim_{(x,y) \rightarrow (1.1)}  \frac{2xy}{x^2+y^2} =1$. I know that I need to show that  $\forall  \epsilon>0,  \exists \delta>0$ s.t. for all (x,y) in the domain of f,  $| \frac{2xy}{x^2+y^2} -1| <  \epsilon$  whenever $0<  \sqrt{(x-1)^2 + (y-1)^2} <\delta$.
I've made $| \frac{2xy}{x^2+y^2} -1|$ to $\frac{(x-y)^2}{x^2 + y^2} $, but I can't get it to look like $\sqrt{(x-1)^2 + (y-1)^2}$..... what should I do from now? :(

REPLY [1 votes]: Hint:
With $t:=\dfrac yx$,$$f(x,y):=\frac{2xy}{x^2+y^2}=\frac{2t}{1+t^2}.$$ As this is a continuous function, it suffices to show that
$$\lim_{(x,y)\to(1,1)}\frac yx=1$$ and compute $f(1,1).$
Now you have to establish
$$\left|\frac yx-1\right|=\frac{|y-x|}{x}<\epsilon,$$ which is true when $x,y$ are sufficiently close to $1$.


TITLE: Image of the product of a Matrix and its transpose
QUESTION [2 upvotes]: If $A$ is an $n\times m$ matrix, is it necessarily true that $\text{im}(A)=\text{im}(AA^T)$ where $A^T$ is the transpose of $A$.

REPLY [4 votes]: The following holds for matrices with real entries.
We first claim that $\ker(AA^T)=\ker(A^T)$. Indeed, if $v\in\ker(AA^T)$, then$$\langle AA^Tv,v\rangle=0$$yields$$\langle A^Tv,A^Tv\rangle=0,$$and hence$$A^Tv=0.$$It now follows that $rank(AA^T)=rank(A^T)=rank(A).$ Since clearly $im(AA^T)\subset im(A),$ we obtain the desired equality.


TITLE: Magnetic moment derivation from Dirac equation
QUESTION [2 upvotes]: I am reading a text book where they show the electron has spin 1/2 using Dirac's equation. At one point in the derivation they define $\pi=P-qA/c$ where $P$ is the momentum operator and A is the vector potential. They then claim that $\pi\times \pi=iq\hbar B / c$ where B is the magnetic field. Apparently $\nabla\times A=B$ as we are assuming the scalar potential is static.
My question is what happened to the $A\times P$ term in $\pi\times\pi$, why is that set to zero?

REPLY [3 votes]: $A\times P$ – more precisely, an expression proportional to $A\times P + P\times A$ – wasn't set to zero. It was properly evaluated and the result gave the $iq\hbar B/c$ term.
Note that if $\pi$ were a vector of $c$-numbers rather than operators, $\pi\times \pi$ would be equal to zero. That's how the cross product behaves. So any term in the cross product $\pi\times \pi$ that is nonzero must be proportional to the nonzero commutators between the components of $\pi$. Now, all three components of $P$ commute with each other; and all three components of $A$ (which depend on the vector $x$) commute with each other. So all the terms in $\pi\times \pi$ must arise from the commutators of components of $P$, essentially a derivative with respect to $x$, and components of the vector potential $A$. By rotational symmetry, it's clear that one must get a multiple of $\nabla\times A$ in this way. And by the way, $B = \nabla\times A$ holds exactly even if there is a time-dependent scalar potential!
Let me write the calculation here:
$$ \pi\times\pi = \epsilon_{ijk} \pi_j\pi_k = \frac 12\epsilon_{ijk} [\pi_j,\pi_k]=\dots  $$
Here, I could have replaced the product $\pi_j\pi_k$ by one-half of the commutator because it's multiplied by a $jk$-antisymmetric epsilon symbol, anyway. Continue:
$$ \dots = \epsilon_{ijk} [P_j,-qA_k/c] = \dots  $$
Here, I used the distributive law for the commutator, realized that $[P_j,P_k]=0$ and $[A_j,A_k]=0$, so only the mixed commutators contribute something that is nonzero and these mixed terms are there twice, $[P,A]$ and $[A,P]$ with the opposite sign (cancelled by the opposite sign of the epsilon symbol), so it's enough to write one of them and erase the factor of $1/2$ again.
Now, $[P_j,Y]\equiv -i\hbar \partial_j Y$ so we have
$$\dots = -i\hbar \epsilon_{ijk} \partial_j(-qA_k/c) = \frac{iq\hbar}{c} \epsilon_{ijk}\partial_j A_k = \frac{iq\hbar}{c} B_i. $$
QED.


TITLE: Microcanonical ensemble energy relation to kT
QUESTION [0 upvotes]: So today I derived the Boltzmann distribution and I stumbled across a particular assertion whose correctness I am wondering about:
In the limit of the microcanonical ensemble (where the energy range is infinitesimally small for microstates), is the following true?
$$ dE = kT(dN) $$
Where I am denoting E as total system energy, N as total particles/modes, k as Boltzmann constant and T as temperature.
For reference, here is my derivation:
We take the case of a microcanonical ensemble, a closed system with N particles, W microstates, each with equal probability, P, and fixed energy, E. We can represent W by the combinations of particles, N, divided by the combinations of states (i) that contain particles $ n_i $:
$$ W = \frac{N!}{n_0!n_1!n_2!\space...} $$
Each state energy is defined as $ \epsilon_i $, therefore we can assert that $ E = \sum_{i=0}^{\infty}n_i\epsilon_i $ and $ N = \sum_{i=0}^{\infty}n_i $. Since each microstate has equal probability, the following is also true:
$$ WP = 1$$
Basically we want to maximize W here and after some unfortunate math involving taking the derivative of a factorial:
$$ dlnW = -\sum_{i}n_i\space dn_i = 0 $$
At this point we introduce two more terms involving dN and dE, which are both zero as well:
$$ dN = \sum_{i}dn_i = 0$$
$$ dE = \sum_{i}\epsilon_i \space dn_i = 0 $$
This results in:
$$ 0 = -\sum_{i}(\alpha + \beta \epsilon_i + ln\space n_i)dn_i $$
Which after a bit more math, ultimately gives:
$$ n_i = \frac{N}{\sum_{i}e^{-\beta \epsilon_i}}e^{-\beta \epsilon_i} $$
And the probability of being in state i is $ p_i = n_i/N $. This is the desired result and is the Boltzmann Distribution.
We also know that when analyzing with the laws of thermodynamics, $ \beta = \frac{1}{kT} $, which also implies the following:
$$ \beta \space dE = \frac{1}{kT}dE = \frac{1}{kT}\sum_{i}\epsilon_i \space dn_i = 0 $$
But I am reminded of $ E = kTN $ which implies $ dE = kT(dN)$ that can only be true if:
$$ \epsilon_i = kT $$
and therefore
$$ \frac{1}{kT}dE = \frac{1}{kT}\sum_{i}kT \space dn_i = \sum_{i}dn_i = dN $$
Which means that the energy per state is kT, which at first made sense to me. But this makes the Boltzmann Distribution fall apart, going to zero, in fact.
My deduction can't be valid, please correct me where I am wrong.

REPLY [1 votes]: The equipartition theorem says that the total energy in equilibrium will be $E = \frac{1}{2} NkT$ where N is the number of degrees of freedom (number of particles if each has only 1 d.o.f.).
The i index in your equations runs from 0 to $\infty$ because there are an infinite number of possible states in the system.  Their energy levels are represented by the $\epsilon_i$'s.  Because there are only N particles, only a finite number of these can be occupied.  In other words, there are an infinite number of states $i$ for which $n_i = 0$ and only a finite number where $n_i \neq 0$.
If $\epsilon_i$ were equal to $\frac{1}{2}kT$ for all i, then this would mean all infinitely many states were occupied, leading to obvious problems.  For one thing, it would mean $\sum n_i = \infty$ instead of $\sum n_i = N$.  So obviously, this is not the case.
Because the total energy of a system in equilibrium is $\frac{1}{2} NkT$, that implies that the average energy per particle is $\frac{1}{2} kT$.  However, that does not mean that the energy of each particle is exactly $\frac{1}{2}kT$, just that this is the average of all of them.
So it's not true that either $\epsilon_i = \frac{1}{2} kT$ or even that $\epsilon_i n_i = \frac{1}{2} kT$, it's just that $\frac{1}{N}\sum n_i \epsilon_i = \frac{1}{2} kT$.  In words, the average energy per degree of freedom is $\frac{1}{2} kT$.
You write $dE = kT dN$ and then jump to the conclusion that $\epsilon_i = kT$.  I'm not sure where this logical leap comes from but I have a guess.  With differential equations, one has to be very careful to state what is being held fixed and what is changing, otherwise they could have multiple interpretations.  $dE = \frac{1}{2} kT dN$ is correct as long as by that you mean "if I hold the temperature of a system fixed, each additional particle I add to it will add $\frac{1}{2} kT$ amount of energy to the system."  But keep in mind that if you are holding the temperature fixed but allowing the energy and number of particles to change, all of the $n_i$'s will change along with that when you add a new particle.  In other words, adding 1 more particle causes the rest of the particles to rearrange in whatever way will keep the temperature fixed... some occupation numbers of states may increase or decrease.  So it's not that the new additional particle you added ends up having an energy of exactly $\frac{1}{2} kT$.
Instead of holding the temperature of the system fixed, you could choose to hold the energy fixed.  Fixing the temperature in terms of the physics means allowing the system to be in contact with a large heat bath at temperature T (the environment).  If instead you hold the energy fixed, this means isolating the system from the environment so that no heat is allowed to flow in or out.  If you do that, then adding a new particle on top of the N you already have would decrease the temperature.  Writing that in terms of an equation looks like $kdT = -2EN^{-2}dN$, which comes from taking the derivative of both sides of $kT = 2E/N$.
Some of this ambiguity can be resolved by using partial derivatives and by indicating what variables are being held fixed.


text:
\begin{document}
\maketitle

\section{Introduction}

We will in this paper suggest how the mathematical concept of
hyperstructures may be a useful tool in the study of the higher,
hierachical structure of languages.

By a language $\L$ we will mean a spoken, written, formal or geometric
language.  We will assume that in the context of a language there are
some basic units: sounds, letters, geometric forms, signs, etc.  They
represent the ``atoms'' of the language.  We suggest that
Hyperstructures \cite{Cog, HAM, NSCS, NS, SO, HOA, HOS, MHS, PHS}
represent a useful framework in which to represent a language
structure, and suggest the following procedure.

\section{Hyperstructures}

Let us denote the collection or set of basic linguistic units by
$X_0$.  Let $\P(X_0)$ stand for all (finite) subsets or a
subcollection.  We will then have an assignment (technically a
presheaf in many cases):
\begin{equation*}
  \Omega_0 \colon \P(X_0) \to \sets,
\end{equation*}
for example ``giving some meaning'' to a collection of sounds (or
letters).  This is what we call the Observation part (property,
state).

Next we consider collections of basic units with properties
\begin{equation*}
  (S_0,\omega_0)
\end{equation*}
and we want to bind these together to meaningful entities
\begin{equation*}
  \Gamma_1 = \{(S_0,\omega_0) \mid S_0 \subseteq X_0, \omega_0 \in
  \Omega_0(S_0)\}.
\end{equation*}

$B_0 \colon \Gamma_1 \to \sets$ is another assignment (presheaf) such
that $B_0(S_0,\omega_0)$ is the set of bonds, ways of gluing the
element in $S_0$ together to meaningful units, like words.

This is the process we will iterate in order to get a
hyperstructure. We think of
\begin{itemize}
\item[] $\Omega_0$ as the semantic assignment, ``meaning''
\end{itemize}
and
\begin{itemize}
\item[] $B_0$ as the syntactic (``grammatical'') structure assignment.
  Admissible generation of words.
\end{itemize}

In hyperstructures we can also define composition of bonds, by a
``gluing'' process (see \cite{HOA}) which is a huge
generalization of concatenation of letters for example.  From bonds
$b_1,b_2,\ldots,b_k$ we can generate new bonds:
\begin{equation*}
  b = b_1 \boks b_2 \boks \cdots \boks b_k.
\end{equation*}
These assignments give a ``grammar'' both for syntax and semantics.
For example Chomsky grammars and their generative rules will be
covered by bond composition in the form of concatenation. We may also
consider higher dimensional geometric alphabets and higher dimensional
gluing. The semantic part may also play a role in the composition.

Now we can proceed to produce the higher tiers of the language:

Put
\begin{equation*}
  X_1 = \{ b_0 \mid b_0 \in B_0(S_0,\omega_0)\}.
\end{equation*}
Then we choose or construct (a ``presheaf''), specific for each
language:
\begin{equation*}
  \Omega_1 \colon \P(X_1) \to \sets
\end{equation*}
from
\begin{equation*}
  \Gamma_1 = \{(S_1,\omega_1)\}
\end{equation*}
and
\begin{equation*}
  B_1 \colon \Gamma_1 \to \sets.
\end{equation*}
This means that $B_1(S_1,\omega_1)$ represent bonds of bonds, for
example words of words bound together in meaningful admissable
sentences given by the bond presheaves.

Then we can construct arbitrarily many tiers getting a formal
hyperstructure $\H(\L)$.

We also have ``boundary'' maps connecting the tiers:
\begin{equation*}
  \partial_i \colon X_{i + 1} \to \P(X_i)
\end{equation*}
which dissolve the bonds --- or put differently: ``a bond knows what
it binds'' or ``a sentence knows its words'' (linguistic version).

Starting with a top bond and iteratively applying the $\partial_i$'s
is like given a sentence and constructing a parsing tree:
\begin{center}
  \begin{tikzpicture}
    \begin{scope}
      \draw[midarrow] (0,0) -- (1,-1);
      \draw[midarrow] (0,0) -- (-1,-1);
      \node[above] at (0,0){$C_n$};
      \node[below] at (0,-1){$\{C_{n - 1}^j\}$};
      \node at (1.75,-0.5){$\partial_{n - 1}$};
    \end{scope}

    \begin{scope}[yshift=-2cm]
      \draw[midarrow] (0,0) -- (1,-1);
      \draw[midarrow] (0,0) -- (-1,-1);
      \node[below] at (0,-1){$\{C_{n - 2}^k\}$};
      \node at (1.75,-0.5){$\partial_{n - 2}$};
    \end{scope}

    \begin{scope}[yshift=-4cm]
      \draw[midarrow] (0,0) -- (1,-1);
      \draw[midarrow] (0,0) -- (-1,-1);
      \node[below] at (0,-1){$\vdots$};
      \node at (1.75,-0.5){$\partial_{n - 3}$};
    \end{scope}
  \end{tikzpicture}
\end{center}
i.e., describing $n$-levels of generalized ``morphemes''.

\emph{We suggest that this is the basic general structure of language,
and that any reasonable language is governed by a Hyperstructure, and
it may be very useful to identify it.}

Hyperstructures are really organizing tools, and so are languages for
communication, thought and learning.

\section{Consequences}

Notice that even starting with a finite set of basic units, infinite
numbers of bonds and meanings can be generated for any tier.  Our
approach covers semiotic systems as well.

Another advantage with hyperstructures is the possibility of passing
from local to global via an extended Grothendieck type topology (see
\cite{HOA} where this is introduced).

This means that we consider tiers of bonds
\begin{center}
  \begin{tikzpicture}
    \diagram{d}{1em}{2.5em}{
      B_n & \sets\\
      B_{n - 1} & \sets\\
      \vdots &\\
      B_0 & \sets\\
    };

    \path[->,font=\scriptsize,above,midway]
      (d-1-1.east)
      edge[decorate,decoration={snake,amplitude=.4mm,segment
        length=3mm,post length=1mm}] node{$\Lambda_n$} (d-1-2.west)
      (d-2-1.east)
      edge[decorate,decoration={snake,amplitude=.4mm,segment
        length=3mm,post length=1mm}] node{$\Lambda_{n - 1}$}
        (d-2-2.west)
      (d-4-1.east)
      edge[decorate,decoration={snake,amplitude=.4mm,segment
        length=3mm,post length=1mm}] node{$\Lambda_0$} (d-4-2.west);
  \end{tikzpicture}
\end{center}
where the $\Lambda_i$'s are ``meaning'' functions or assignments at
each level. We have developed rules for when we from these
\emph{levelwise} meanings can deduce a global meaning.  This is done
by what we call a globalizer (generalizing sections in sheaves), and
shows how global meanings can be integrated and also that in many
situations you cannot just pass from local to global.  Tiers or levels
are needed.  This extends the linguistic notion of \emph{``duality of
  patterning''}.  But duality is just for two levels, in general many
levels are needed in an efficient language.  Our approach is
\emph{``multiplicity of patterning''}!

In general: Data, Stimuli, Inputs, Outputs may be unstructured ---
have no immediate meaning.

Hence put a Hyperstructure on the situation (which may be inherent in
the living brain or a synthetic one) and via a suitable ``globalizer''
structure and meaning will appear (unless random!).

This is a way to handle the ``Generalized Binding Problem'' (see
\cite{HOS}).  If two languages are $\H$-structured communication
will (should) take place in an $\H$-structured way and understanding
represents some kind of structural compatibility or resonance.

Finally, these structures also relate to memory and learning.  For
example a \emph{machine} organized as an $\H$-structure will take data
(stimuli), process them and adjust according to a global goal state or
meaning.

This is \emph{automatic} learning.  Learning in general requires or at
least profits from a Hyperstructure.  Hyperstructured libraries of
events or data are important in learning and memory and
$\H$-structures facilitates the process.

We hope that this note will initiate a further study of
hyperstructures in languages.

\subsection*{Acknowledgements}
I would like to thank M.~Thaule for his kind technical assistance in
preparing the manuscript.

config: arxiv


TITLE: Example of truncated expectation which gives a disconnected diagram. Where am I wrong?
QUESTION [0 upvotes]: My notes introduce the truncated expectation in the following way:
given $S_0$ a quadratic form, consider a generating function
$$e^{W(J)} = \int \prod_x d\varphi_x e^{-S_0+ JO} = \int P(d\varphi) e^{JO}. $$
Let define the truncated expectation of $O$ as
$$ E^T(O;n) = \frac{\partial^n}{\partial J^n} W(J) \bigg|_{J=0}=  \frac{\partial^n}{\partial J^n} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} $$
And so for example $E^T(O;1)=\frac{\partial}{\partial J} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} =\frac{\int P(d\varphi) O}{\int P(d\varphi) }=\langle O\rangle $.
Now, using the Wick rule, I calculate $$E^T(\varphi_1 \varphi_2 \varphi_3 \varphi_4;1)=\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle = \langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle  +\langle \varphi_1 \varphi_4\rangle \langle \varphi_2 \varphi_3\rangle +\langle \varphi_1 \varphi_3\rangle \langle \varphi_2 \varphi_4\rangle. $$
The rules for Feynman diagrams are: construct a Feynman diagram drawing a segment to each field and then join the different segments if there is a contraction between the field, that is if I have the expectation between the field. Then, following this rule I have that $\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle $ is composed by disconnected diagram, because, for example for the first term that comes from  the Wick rule, $\langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle $ I have one segment that connects the vertices $1-2$ and one segment for $3-4$.
Where am I wrong?

REPLY [0 votes]: $$
Z=\int P(d\varphi)e^{JO}
$$
is the partition function of $\varphi^4$ type model. You can rigorously prove in the sense of power series in $\mathbb{C}[[J]]$ that $Z$ is a sum over all graphs with degree 4 vertices (this is math i.e. graph theory terminology) corresponding to the $O$'s, connected or not. When you take the log $W$ you get a similar sum where you only keep connected graphs. In the example above, graphs look like the number 8. You have one central vertex with 4 half edges emanating from it, with two loops (math terminology), i.e., tadpoles (physics terminology).


TITLE: How to estimate the experimental acceptance of a particle as a function of its life-time?
QUESTION [2 upvotes]: My lecturer stated that we can calculate the experimental acceptance of the Higgs boson as a function of the life-time and also as a function of the  mass by reading the following article: Davier and Ngoc, “An unambiguous search for a light Higgs boson, Physics Letters B, 1989.

What mathematical equations allow us to do that?

How does one read/interpret the graph in figure 6 (shown below)? Does it help in any way to get the acceptance?

In the article, it is stated that the Higgs' life-time is given $(1)$
$$ \tag{1} \tau^{-1}_H = \frac{1}{2} \alpha_H m_H \left( 1 - \frac{4m^2}{m^2_H} \right)^{3/2} $$
where $\alpha_H = \frac{m^2 G_F \sqrt{2}}{4\pi}$ , $m$ is the electron mass and $m_H$ is the Higgs' mass. I presume this is useful, but I don't know how.

How do I read the following figure? Can this be interpreted into knowing the experimental acceptance based on the life-time?

REPLY [1 votes]: Compare with the previous figure from the same paper (click to embiggen):

This says that they observed a few hundred, perhaps a thousand, total events in their detector, all with energies well below 1 GeV.  A hypothetical 20 MeV Higgs, interacting as described in the paper and its references, would have produced a few dozen additional events with energies between 1–2 GeV. Since there were no such events (not even background!) that low-mass Higgs is “ruled out.” From your excerpted Figure 4, you can read that the 20 MeV Higgs is only ruled out if its lifetime is in the cross-hatched “excluded range” from $10^{-11}$ to $10^{-9}$ seconds, presumably because a shorter- or longer-lived particle would have decayed mostly before or mostly after the detectors.  A previous experiment ruled out
a teardrop-shaped blob on the lower left.
I think (but it would be useful for you to verify) that your lifetime-versus-mass relation is plotted with the dot-dashed line labeled “SM.” The result of the paper, then, is that nearly all of this curve lies in the excluded region.
Acceptance/exclusion plots are terrifically confusing to read, especially in old papers where they are printed in black-and-white with only one line width available.
(Evidence: I misread this one at least twice.)
Modern exclusion plots show colored regions, which makes them slightly easier to understand, but because they are easier to understand people tend to put dozens of different colored regions on them.


TITLE: Help with Bayes Rule and Probability
QUESTION [1 upvotes]: I've seen many similar problems to this one, but I'm still a bit confused as to how to solve it:
Two buckets:
bucket $1$ has $3$ black balls, $7$ blue balls
bucket $2$ has $7$ black balls, $3$ blue balls
We draw randomly $12$ times from one of the buckets with replacement. This yielded $4$ black and $8$ blue.
What's the probability that the draw was made from the first bucket?

REPLY [3 votes]: Hints:
Since you're just learning Bayes' Theorem, you need to break up the problem into parts. Bayes' Theorem for this problem will look like this (please verify you understand how I got this):
Let $J_i$ be the event that you are drawing from jar $i$. Let $D$ be the event that you drew 4 black and 8 blue balls.
$$P(J_i|D)=\frac{P(J_i)P(D|J_i)}{\sum_{i=1}^2 P(J_i)P(D|J_i)}$$
Hint: All you are doing is "renormalizing" your probabilities to 1, assuming you are in the part of the Venn diagram where $D$ has occurred. Draw the Venn Diagram for this problem and you'll see why.

What is the implicit assumption about how you pick jars? Is each jar equally likely to be chosen?
Can you derive the probability of getting $D$ assuming you are drawing from the first jar? Derive this for the second jar too.
Now, can you calculate the probability of $D$ regardless of which jar you choose (unconditional probability)?

Ok, you now have all the ingredients to solve this problem. Draw your Venn Diagram, study the above formula and Baye's Theorem and solve it. :-)

REPLY [1 votes]: Let $F$ be the event of "getting$4$ black and $8$ blue",and let $E$ be the event "the selection is from bucket $1$". Thus we have:
$P\left(E\mid F\right) = \dfrac{P\left(E \cap F\right)}{P\left(F\right)}= \dfrac{P\left(E \cap F\right)}{P\left(E \cap F\right) + P\left(E^c \cap F\right)}= \dfrac{\binom{12}{4}(0.3)^4(0.7)^8}{\binom{12}{4}(0.3)^4(0.7)^8+\binom{12}{4}(0.7)^4(0.3)^8}=...$


text:
\begin{document}

\baselineskip=18pt
\setcounter{page}{1}

\vglue50pt

\centerline{\large\bf The stable Derrida--Retaux system at criticality}

\bigskip
\bigskip

\centerline{\footnotesize Xinxing Chen\footnote{\scriptsize School of Mathematical Sciences, Shanghai Jiaotong University, 200240 Shanghai, China, {\tt chenxinx@sjtu.edu.cn} $\;$Partially supported by NSFC grant Nos.  11771286 and 11531001.}
and
Zhan Shi\footnote{\scriptsize LPSM, Sorbonne Universit\'e Paris VI, 4 place Jussieu, F-75252 Paris Cedex 05, France, {\tt zhan.shi@upmc.fr} $\;$Partially supported by ANR MALIN.}
}

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\centerline{\it Dedicated to the memory of Vladas Sidoravicius}

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{\leftskip=1.5truecm \rightskip=1.5truecm \baselineskip=15pt \small

\noindent{\slshape\bfseries Summary.} The Derrida--Retaux recursive system was investigated by Derrida and Retaux~\cite{derrida-retaux} as a hierarchical renormalization model in statistical physics. A prediction of \cite{derrida-retaux} on the free energy has recently been rigorously proved (\cite{bmvxyz_conjecture_DR}), confirming the Berezinskii--Kosterlitz--Thouless-type phase transition in the system. Interestingly, it has been established in \cite{bmvxyz_conjecture_DR} that the prediction is valid only under a certain integrability assumption on the initial distribution, and a new type of universality result has been shown when this integrability assumption is not satisfied. We present a unified approach for systems satisfying a certain domination condition, and give an upper bound for derivatives of all orders of the moment generating function. When the integrability assumption is not satisfied, our result allows to identify the large-time order of magnitude of the product of the moment generating functions at criticality, confirming and completing a previous result in \cite{collet-eckmann-glaser-martin}.

\bigskip

\noindent{\slshape\bfseries Keywords.} Derrida--Retaux recursive system, moment generating function.

\bigskip

\noindent{\slshape\bfseries 2020 Mathematics Subject Classification.} 60J80, 82B27.

}

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\bigskip

\section{Introduction}
\label{s:intro}

Fix an integer $m\ge 2$. Let $X_0$ be a random variable taking values in $\z_+ := \{ 0, \, 1, \, 2, \ldots\}$. To avoid trivial discussion, it is assumed, throughout the paper, that $\P ( X_0 \ge 2) > 0$. Let us consider the Derrida--Retaux recursive system $(X_n, \, n\ge 0)$ defined as follows: for all $n\ge 0$,
\begin{equation}
    X_{n+1}
    =
    (X_{n,1} + \cdots + X_{n,m} -1)^+ ,
    \label{iteration}
\end{equation}

\noindent where $X_{n,i}$, $i\ge 1$, are independent copies of $X_n$. This was investigated by Derrida and Retaux~\cite{derrida-retaux} as a toy model to study depinning in presence of impurities \cite{luck,derrida-hakim-vannimenus,tang-chate,giacomin,giacomin-toninelli,dglt,monthus}. We refer to \cite{bz_dobrushin} for an overview on rigorous results and predictions about the Derrida--Retaux system.

Assuming $\E(X_0)<\infty$, it is immediate from \eqref{iteration} that $\E(X_{n+1}) \le m\, \E(X_n)$, so the free energy
$$
F_\infty := \lim_{n\to \infty} \downarrow \frac{\E(X_n)}{m^n} \in [0, \, \infty),
$$

\noindent is well-defined. A remarkable result by Collet et al.~\cite{collet-eckmann-glaser-martin} tells us that assuming $\E(X_0 \, m^{X_0})<\infty$ (which we take for granted throughout the paper) and writing $\eta := (m-1) \E(X_0 \, m^{X_0}) - \E(m^{X_0})$, then $F_\infty >0$ if $\eta >0$, and $F_\infty =0$ if $\eta \le 0$.

As such, it is natural to say that the system $(X_n, \, n\ge 0)$ is supercritical if $\eta>0$, is critical if $\eta=0$, and is subcritical if $\eta<0$.

It has been conjectured by Derrida and Retaux~\cite{derrida-retaux} that if $\eta >0$, then we would have
\begin{equation}
    F_\infty = \exp \Big( - \frac{C+o(1)}{\eta^{1/2}} \Big), \qquad \eta \to 0+ \, ,
    \label{conj_DR}
\end{equation}

\noindent for some constant $C\in (0, \, \infty)$ possibly depending on the law of $X_0$. A (somehow weak) result has been proved in \cite{bmvxyz_conjecture_DR}: assuming $\E(X_0^3 m^{X_0})<\infty$,
$$
F_\infty = \exp \Big( - \frac{1}{\eta^{1/2+o(1)}} \Big), \qquad \eta \to 0+ \, .
$$

\noindent This confirms that the Derrida--Retaux system has a Berezinskii--Kosterlitz--Thouless-type phase transition of infinite order. The integrability assumption $\E(X_0^3 m^{X_0})<\infty$ might look exotic, but it is optimal. [We believe that there should be a change-of-measures argument, and that the assumption is equivalent to saying that $X_0$ has a finite second moment under a new probability measure; however, we have not succeeded in making this idea into a rigorous argument.] In fact, it has also been proved in \cite{bmvxyz_conjecture_DR} that if $\P(X_0 = k) \sim c \, m^{-k} k^{-\alpha}$, $k\to \infty$, for some $2<\alpha<4$ and $c>0$,\footnote{Notation: $a_k \sim b_k$, $k\to \infty$, means $\lim_{k\to \infty} \frac{a_k}{b_k} =1$.} then
\begin{equation}
    F_\infty = \exp \Big( - \frac{1}{\eta^{\nu +o(1)}} \Big), \qquad \eta \to 0+ \, ,
    \label{energie_cas_stable}
\end{equation}

\noindent where $\nu = \nu(\alpha) := \frac{1}{\alpha-2}$. In other words, \eqref{conj_DR} predicts only a small part of universalities, under the assumption $\E(X_0^3 m^{X_0})<\infty$, while other universality phenomena are described by \eqref{energie_cas_stable}. We expect many other universality results in the latter setting (for example, corresponding to those in \cite{4authors} for an analogous continuous-time model); unfortunately, they are currently only on a heuristic level.

It is well-known that sum of i.i.d.\ random variables, after an appropriate normalization, converges to a Gaussian limiting law under the condition of finiteness of second moment, and to a stable limiting law under a weaker integrability condition. We say that the Derrida--Retaux system has a ``finite variance" if $\E(X_0^3 m^{X_0})<\infty$, and that it is a stable system if integrability condition holds for lower orders. In this paper, we are interested in the stable system when it is critical, i.e., when $(m-1) \E(X_0 \, m^{X_0}) = \E(m^{X_0})$. [We are going to see in Section \ref{s:domination}, quite easily, that this implies $(m-1) \E(X_n \, m^{X_n}) = \E(m^{X_n})$ for all $n\ge 0$.] We write $(Y_n, \, n\ge 0)$ instead of $(X_n, \, n\ge 0)$ in order to insist on criticality. {F}rom now on, we assume $(Y_n, \, n\ge 0)$ to be a Derrida--Retaux system satisfying $(m-1) \E(Y_0 \, m^{Y_0}) = \E(m^{Y_0}) <\infty$, such that
\begin{equation}
    \P(Y_0 = k) \sim c_0 \, m^{-k} k^{-\alpha},
    \qquad
    k\to \infty ,
    \label{alpha}
\end{equation}

\noindent for some $2<\alpha<4$ and $c_0>0$. We intend to prove the following result.

\medskip

\begin{theorem}
\label{t:main}

 Let $(Y_n, \, n\ge 0)$ be such that $(m-1) \E(Y_0 \, m^{Y_0}) = \E(m^{Y_0}) <\infty$. Under assumption \eqref{alpha}, there exist constants $c_2\ge c_1 >0$ such that for all $n\ge 1$,
 $$
 c_1 \, n^{\alpha-2} \le \prod_{i=0}^{n-1} [\E(m^{Y_i})]^{m-1} \le c_2 \, n^{\alpha-2} .
 $$

\end{theorem}

\medskip

When the system is of ``finite-variance" (i.e., $\E(Y_0^3 m^{Y_0})<\infty$), the analogue of Theorem \ref{t:main} was known (\cite{collet-eckmann-glaser-martin}, \cite{bmxyz_questions}), and has played an important role in the study of the asymptotics of $\P(Y_n >0)$ and $\E(Y_n)$ in \cite{bmvxyz_sustainability}. It would be tempting to believe that Theorem \ref{t:main} could play an equally important role in the study of the same problems for the stable system.

Just like the usual random walk has a nice continuous-time analogue which is Brownian motion, the Derrida--Retaux system has analogues in continuous time (Derrida and Retaux~\cite{derrida-retaux}, Hu, Mallein and Pain~\cite{HMP}), defined via appropriate integro-differential equations. For the continuous-time analogue of the stable Derrida--Retaux system, see~\cite{4authors}. These continuous-time models have been studied in depth in \cite{derrida-retaux}, \cite{HMP} and \cite{4authors}, while most of the corresponding problems remain open for the original Derrida--Retaux system.

With the exception of the case $\alpha=3$, Theorem \ref{t:main} was already stated in Collet et al.~\cite{collet-eckmann-glaser-martin}: its proof in case $2<\alpha<3$ was indicated, whereas the proof in case $3<\alpha<4$ was only summarized in a ``very succinct account". By means of the notion of dominability (see the forthcoming Definition \ref{d:dominable}), we give a unified approach to the system in both situations, i.e., either it is stable (no need for discussions separately on the cases $2<\alpha<3$ and $3<\alpha<4$) or is of ``finite variance". Concretely, in both situations, we use a truncating argument by considering a bounded random variable defined by
$$
Z_0 = Z_0(M) := Y_0 \, {\bf 1}_{\{ Y_0 \le a(M)\} } ,
$$

\noindent where $a(M) \in [1, \, \infty]$ can be possibly infinite (in which case there is no need for truncation), whose value depends on an integer parameter $M\ge 1$. Consider the Derrida--Retaux system $(Z_n, \, n\ge 0)$ whose initial distribution is given by $Z_0$.\footnote{Strictly speaking, it is a {\it sequence} of Derrida--Retaux systems, indexed by $M$.} We prove, in Theorem \ref{t:dominable}, that in both situations, it is possible to choose a convenient value of $a(M)$ such that the new system $(Z_n, \, n\ge 0)$ is dominable (in the sense of Definition \ref{d:dominable}), while it is possible to connect the moment generating functions of $Y_n$ and $Z_n$. In Theorem \ref{t:H_nkupper}, we give an upper bound for the moment generating function of any dominable system $(Z_n, \, n\ge 0)$. As such, a combined application of Theorems \ref{t:dominable} and \ref{t:H_nkupper} will yield information for the moment generating function of the original Derrida--Retaux system, in both situations. In the stable case, it will yield Theorem \ref{t:main}, whereas in the case of ``finite variance", under a stronger integrability assumption on the law of $Y_0$, it will give the following result:

\medskip

\begin{theorem}
 \label{t:moment_ub}

 Let $(Y_n, \, n\ge 0)$ be such that $(m-1) \E(Y_0 \, m^{Y_0}) = \E(m^{Y_0})$. If $\E(s^{Y_0})<\infty$ for some $s>m$, then there exists a constant $c_3 >0$ such that for all integers $n\ge 1$ and $k \ge 1$,
 \begin{equation}
     \frac{\mathrm{d}^k}{\mathrm{d} u^k}\, \E(u^{Y_n}) \Big|_{u=m}
     \le
     k!\, \ee^{c_3 k} \, n^{k-1} \, .
     \label{eq_t:moment_ub}
 \end{equation}

\end{theorem}

\medskip

In the ``finite-variance" case $\E(Y_0^3 m^{Y_0})<\infty$, \eqref{eq_t:moment_ub} for $k\in \{ 1, \, 2, \, 3\}$ was known: the case $k=1$ is simple because by criticality,  $\E(Y_n m^{Y_n-1}) = \frac{1}{m(m-1)} \, \E(m^{Y_n})$ which is bounded in $n$ (\cite{collet-eckmann-glaser-martin}, \cite{bmxyz_questions}), the case $k=3$ was proved in \cite{bmxyz_questions}, and the case $k=2$, stated in \cite{bmvxyz_conjecture_DR}, follows immediately from the cases $k=1$ and $k=3$ by means of the Cauchy--Schwarz inequality. More generally, if $\E(Y_0^\ell m^{Y_0})<\infty$ for some integer $\ell \ge 1$, then for all $k\in [1, \, \ell] \cap \z$, it is quite easy to prove (\cite{bmvxyz_sustainability}) by induction in $k$, using the recursion \eqref{iteration}, that there exists a constant $c>0$ such that for all integer $n\ge 1$,
$$
\frac{\mathrm{d}^k}{\mathrm{d} u^k}\, \E(u^{Y_n}) \Big|_{u=m}
\le
c \, n^{k-1} \, .
$$

\noindent Theorem \ref{t:moment_ub} gives information about the dependence in $k$ of the constant $c$, under the integrability assumption $\E(s^{Y_0})<\infty$ for some $s>m$.

The rest of the paper is organized as follows. In Section \ref{s:domination}, we introduce the notion of dominable systems. Theorem \ref{t:H_nkupper}, which gives an upper bound for the moment generating function of dominable systems, is the main technical result of the paper. The brief Section \ref{s:moment_s>m} is devoted to the proof of Theorem \ref{t:moment_ub}, obtained as a simple consequence of Theorem \ref{t:H_nkupper}. In Section \ref{s:truncating}, for both ``finite-variance" and stable systems, we construct a dominable system $(Z_n, \, n\ge 0)$ such that $Z_0$ is obtained from an appropriate truncation of $Y_0$. Finally, Theorem \ref{t:main} is proved in Section \ref{s:pf_thm}, also as a consequence of Theorem \ref{t:H_nkupper}.

\section{Dominable systems}
\label{s:domination}

We introduce the notion of dominable systems and prove a general upper bound for the moment generating function of such systems (Theorem \ref{t:H_nkupper}). As before, we talk about the Derrida--Retaux system $(Z_n, \, n\ge 0)$, while it is, in fact, a sequence of Derrida--Retaux systems $(Z_n(M), \, n\ge 0)$ indexed by the integer-valued parameter $M$.

\medskip

\begin{definition}
\label{d:dominable}

 Let $\gamma>0$. The system $(Z_n, \, n\ge 0)$ is said to be $\gamma$-dominable if for all sufficiently large integer $M$, say $M\ge M_0$, $Z_0=Z_0(M)$ is bounded, and there exists a constant $\vartheta(M) \ge 1$ such that $M \mapsto \vartheta(M)$ is non-decreasing in $M\ge M_0$, and that$\,$\footnote{In \eqref{assump_H0k} and \eqref{assump_Qn}, $k\ge 3$ and $n\ge 1$ are integers. Notation: $a\vee b := \max\{ a, \, b\}$.}
\begin{eqnarray}
    \E(Z_0^k m^{Z_0})
 &\le& M^{k-3} (\vartheta(M) + k!) ,
    \qquad
    k\ge 3,
    \label{assump_H0k}
    \\
    \vartheta (n\vee M) \prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1}
 &\le& \gamma \, (n\vee M)^2 ,
    \qquad
    n\ge 1.
    \label{assump_Qn}
\end{eqnarray}

\end{definition}

\medskip

\begin{remark}
\label{r:dominable=>subcritical}

Let $(Z_n, \, n\ge 0)$ be $\gamma$-dominable. By \eqref{assump_Qn} and the trivial inequality $\E(m^{Z_i}) \ge 1$ ($\forall i\ge 0$), we have $\E(Z_n) \le \E(m^{Z_n}) \le [\gamma \, (n\vee M)^2]^{1/(m-1)}$, so the free energy $F_\infty := \lim_{n\to \infty} \frac{\E(Z_n)}{m^n}$ vanishes; by the criterion of Collet et al.~\cite{collet-eckmann-glaser-martin} recalled in the introduction, the system $(Z_n, \, n\ge 0)$ is subcritical or critical for all $M\ge M_0$: we have $(m-1)\E(Z_0 \, m^{Z_0}) \le \E(m^{Z_0})$.\qed

\end{remark}

\medskip

\begin{theorem}
 \label{t:H_nkupper}

 Let $\gamma>0$. Let $(Z_n, \, n\ge 0)$ be a $\gamma$-dominable system, and let $H_n(u) := \E(u^{Z_n})$. There exists a constant $c_4\ge 1$, depending only on $(m, \, \gamma)$, such that for all integers $k\ge 3$, $n\ge 1$ and $M\ge M_0$,
 $$
 H_n^{(k)}(m)\le k!\, c_4^k\, (n\vee M)^{k-1}\, ,
 $$

 \noindent where $H_n^{(k)}(\cdot)$ stands for the $k$-th derivative of $H_n(\cdot)$.

\end{theorem}

\medskip

\begin{corollary}
 \label{p:sXsub}

 Let $\gamma>0$. Let $(Z_n, \, n\ge 0)$ be a $\gamma$-dominable system.
 There exists a constant $c_5>0$, depending only on $(m, \, \gamma)$, such that for $M\ge M_0$, $n\ge 1$ and $v := m + \frac{1}{2c_4 (n\vee M)}$,
 \begin{equation}
     \E(Z_n^2 v^{Z_n})
     \le
     c_5 [\vartheta (n\vee M)]^{1/2} \prod_{i=0}^{n-1} [\E(m^{Z_i})]^{(m-1)/2} ;
     \label{cor1}
 \end{equation}
 in particular, we have, with $c_6 := \gamma^{1/2}c_5$,
 \begin{equation}
     \E(Z_n^2 v^{Z_n}) \le c_6 (n\vee M).
     \label{e:922745}
 \end{equation}

\end{corollary}

\medskip

The rest of the section is devoted to the proof of Theorem \ref{t:H_nkupper} and Corollary \ref{p:sXsub}. We start by mentioning a general technique going back to Collet et al.~\cite{collet-eckmann-glaser-martin}. Let $(X_n, \, n\ge 0)$ denote a Derrida--Retaux system satisfying $\E(X_0 \, m^{X_0})<\infty$. Let
$$
G_n(u) := \E(u^{X_n}) , \qquad n\ge 0.
$$

\noindent The iteration formula \eqref{iteration} is equivalent to:
\begin{equation}
    G_{n+1}(u) = \frac1u \, [G_n(u)]^m + (1-\frac1u) \, [G_n(0)]^m , \qquad n\ge 0.
    \label{iteration_Gn}
\end{equation}

\noindent A useful trick of Collet et al.~\cite{collet-eckmann-glaser-martin} consists in observing that this yields
$$
(u-1)u\, G_{n+1}'(u) - G_{n+1}(u)
=
[m(u-1)\, G_n'(u)-G_n(u)] \, [G_n(u)]^{m-1} \, .
$$

\noindent In particular, taking $u=m$ yields that
$$
G_{n+1}(m) - (m-1)m\, G_{n+1}'(m)
=
[G_n(m) - (m-1)m \, G_n'(m)] \, [G_n(m)]^{m-1} \, .
$$

\noindent Iterating this formula gives that for $n\ge 1$,
\begin{equation}
    G_n(m) - (m-1)m \, G_n'(m)
    =
    [G_0(m) - (m-1)m \, G_0'(m)] \prod_{i=0}^{n-1} [G_i(m)]^{m-1} \, .
    \label{iteration_Collet}
\end{equation}

\noindent In particular, \eqref{iteration_Collet} tells us that the sign of $\E(m^{X_n})-(m-1) \E(X_n\, m^{X_n})$ remains identical for all $n\ge 0$: it either is always positive (meaning that the system is supercritical), or is always negative (subcritical), or vanishes identically (critical).

A couple of known results which we are going to use for the subcritical or critical system: if $(X_n, \, n\ge 0)$ is a Derrida--Retaux satisfying $\E(m^{X_0}) \le (m-1) \E(X_0 \, m^{X_0}) < \infty$, then for all $n\ge 0$,
\begin{eqnarray}
    (m-1) \E(X_n \, m^{X_n})
 &\le& \E(m^{X_n})
    \le m^{1/(m-1)} ,
    \qquad n\ge 0,
    \label{Gn(m)<}
    \\
    \prod_{i=0}^{n-1} [\E(m^{X_i})]^{m-1}
 &\le& c_7 \, n^2,
    \qquad n\ge 1,
    \label{prod_Gi(m)<}
\end{eqnarray}

\noindent where $c_7>0$ is a constant depending on $m$ and on the law of $X_0$. See \cite[(3.11)]{bmvxyz_conjecture_DR} for the second inequality in \eqref{Gn(m)<} (proved in \cite{bmvxyz_conjecture_DR} for critical systems, and the same proof valid for subcritical systems as well), and \cite[Proposition 1]{bmxyz_questions} for \eqref{prod_Gi(m)<}.

The proof of Theorem \ref{t:H_nkupper} relies on the following preliminary result. Recall that $H_n(u) := \E(u^{Z_n})$.

\medskip

\begin{lemma}
\label{l:Pi_n<}

 Let $(Z_n, \, n\ge 0)$ be a $\gamma$-dominable system for some $\gamma>0$, in the sense of Definition $\ref{d:dominable}$.
 There exist constants $c_8>0$ and $c_9>0$, depending only on $m$, such that for $M\ge M_0$ and all integer $n\ge 1$,
 \begin{eqnarray}
     H_n''(m)
  &\le& c_8 \, [\vartheta (M)]^{1/2} \prod_{i=0}^{n-1} H_i(m)^{(m-1)/2},
     \label{Hn_2nd_derivative<}
     \\
     H_n'''(m)
  &\le& c_9 \, \vartheta (M) \prod_{i=0}^{n-1} H_i(m)^{m-1} .
     \label{Hn_3rd_derivative<}
 \end{eqnarray}

\end{lemma}

\medskip

\noindent {\it Proof.} Let
$$
D_n(u) := (m-1)u^3 H_n'''(u) + (4m-5)u^2 H_n''(u) + 2(m-2) uH_n'(u),
\qquad
n\ge 0.
$$

\noindent Recall from Remark \ref{r:dominable=>subcritical} that $m(m-1)H_0'(m) \le H_0(m)$. By \cite[Equation (19)]{bmxyz_questions}, this yields $D_{n+1}(m) \le D_n(m) H_n(m)^{m-1}$ for all $n\ge 0$. [In \cite{bmxyz_questions}, it was proved that if $m(m-1)H_0'(m) = H_0(m)$, then $D_{n+1}(m) = D_n(m) H_n(m)^{m-1}$, but the proof is valid for inequalities in place of equalities.] Accordingly, for all $n\ge 1$,
$$
D_n(m)
\le
D_0(m) \prod_{i=0}^{n-1} H_i(m)^{m-1} .
$$

\noindent By definition, $D_n(m) \ge m^3(m-1) H_n'''(m)$, so
\begin{equation}
    H_n'''(m) \le \frac{D_0(m)}{m^3(m-1)} \, \prod_{i=0}^{n-1} H_i(m)^{m-1}\, .
    \label{pf_Hn_2nd_derivative<_eq1}
\end{equation}

For any $j\ge 0$ and $\ell \in \{ 1, \, 2, \, 3\}$, writing $Z_j^3  m^{Z_j} \le \frac{9m^3}{2} \, Z_j(Z_j-1)(Z_j-2) m^{Z_j-3} \, {\bf 1}_{\{ Z_j \ge 3\}} + 8m^2$, and $Z_j^\ell \le Z_j^3$, we obtain
\begin{equation}
    \max_{\ell \in \{ 1, \, 2, \, 3\}} \E(Z_j^\ell  m^{Z_j}) \le \frac{9m^3}{2}\, H_j'''(m) + 8m^2\, .
    \label{pf_Hn_2nd_derivative<_eq2}
\end{equation}

\noindent In particular, with $j=0$ and $\ell \in \{ 1, \, 2\}$, this gives $\max\{ H_0'(m), \, H_0''(m)\} \le \frac{9m^3}{2}\, H_0'''(m) + 8m^2$. Consequently, with $c_{10} := m^3 (m-1)$, $c_{11} := m^2 (4m-5)$ and $c_{12} := 2m(m-2)$,
$$
D_0(m)
=
c_{10}\, H_0'''(m) + c_{11}\, H_0''(m) + c_{12}\, H_0'(m)
\le
c_{13} \, H_0'''(m) + c_{14}\, ,
$$

\noindent where $c_{13}:= c_{10} + \frac{9m^3}{2}(c_{11}+c_{12})$ and $c_{14}:=8m^2(c_{11}+c_{12})$. Since $H_0'''(m) \le \vartheta (M) +6 \le 7\vartheta (M)$ by assumption \eqref{assump_H0k} (applied to $k=3$; recalling that $\vartheta (M) \ge 1$), we get $D_0(m) \le 7c_{13} \, \vartheta (M) + c_{14} \le c_{15} \, \vartheta (M)$ with $c_{15} := 7c_{13} + c_{14}$. Going back to \eqref{pf_Hn_2nd_derivative<_eq1}, we have
$$
H_n'''(m) \le \frac{c_{15}}{m^3(m-1)} \, \vartheta (M) \prod_{i=0}^{n-1} H_i(m)^{m-1} ,
$$

\noindent proving \eqref{Hn_3rd_derivative<} with $c_9 := \frac{c_{15}}{m^3(m-1)}$.

It remains to prove \eqref{Hn_2nd_derivative<}. By \eqref{pf_Hn_2nd_derivative<_eq2} (applied to $j=n$),
$$
\E(Z_n^3 m^{Z_n}) \le \frac{9m^3}{2}\, H_n'''(m) + 8m^2\, ,
$$

\noindent whereas by \eqref{Gn(m)<}, $\E(Z_n m^{Z_n}) \le \frac{m^{1/(m-1)}}{m-1} =: c_{16}$, it follows from the Cauchy--Schwarz inequality that
$$
\E(Z_n^2 m^{Z_n}) \le c_{16}^{1/2} \Big( \frac{9m^3}{2}\, H_n'''(m) + 8m^2 \Big)^{\! 1/2}.
$$

\noindent Recall from \eqref{Hn_3rd_derivative<}, which we have just proved, that $H_n'''(m) \le c_9 \vartheta (M) \prod_{i=0}^{n-1} H_i(m)^{m-1}$. Writing $8m^2 \le 8m^2 \vartheta (M) \prod_{i=0}^{n-1} H_i(m)^{m-1}$ (because $\vartheta (M)\ge 1$ and $H_i(m) \ge 1$), this yields
$$
\E(Z_n^2 m^{Z_n})
\le
c_{17} \, [\vartheta (M)]^{1/2} \prod_{i=0}^{n-1} H_i(m)^{(m-1)/2},
$$

\noindent with $c_{17}:= c_{16}^{1/2} (\frac{9m^3}{2}\, c_9 + 8m^2)^{1/2}$. Since $H_n''(m) \le \E(Z_n^2 m^{Z_n})$, we obtain \eqref{Hn_2nd_derivative<} with $c_8 := c_{17}$.\qed

\medskip

\begin{remark}

 We often use the following inequalities for dominable systems:
 \begin{eqnarray}
     H_n''(m)
  &\le& c_8 \, [\vartheta (n\vee M)]^{1/2} \prod_{i=0}^{n-1} H_i(m)^{(m-1)/2},
     \label{Hn_2nd_derivative<bis}
     \\
     H_n'''(m)
  &\le& c_9 \, \vartheta (n\vee M) \prod_{i=0}^{n-1} H_i(m)^{m-1},
     \label{Hn_3rd_derivative<bis}
 \end{eqnarray}

 \noindent They are immediate consequences of Lemma \ref{l:Pi_n<} and the monotonicity of $M\mapsto \vartheta(M)$.\qed

\end{remark}

\medskip

We also need an elementary inequality.

\medskip

\begin{lemma}
 \label{l:201810171111}

 Let $\ell \ge 4$ be an integer, and let
 \begin{equation}
     B_\ell
     :=
     \{ {\bf u} := (u_1, \ldots, u_m) \in ([0, \, \ell-1] \cap \z)^m: \, u_1+\cdots+u_m = \ell \} \, .
     \label{B_ell}
 \end{equation}

 \noindent There exists a constant $c_{18}>0$, depending only on $m$, such that for all $y\ge 3m$,\footnote{Strictly speaking, we should write $\prod_{i: \, 1\le i\le m, \, u_i\ge 3} \frac{1}{u_i(u_i-1)}$ for $\prod_{i: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}$. Notation: $\prod_\varnothing := 1$.}
 \begin{equation}
    \sum_{{\bf u} := (u_1, \ldots, u_m) \in B_\ell} y^{(\ell-\eta({\bf u})-2)^+} \prod_{i: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}
    \le
    c_{18}\, \frac{y^{\ell-4}}{\ell^2} \, ,
    \label{pf_lemma51_eq2}
 \end{equation}

 \noindent where $a^+ := \max\{ a, \, 0\}$ as before, and
 \begin{equation}
     \eta({\bf u})
     :=
     \sum_{i=1}^m {\bf 1}_{\{ u_i \ge 1\} }
     \ge 2 \, .
     \label{eta}
 \end{equation}

\end{lemma}

\medskip

\noindent {\it Proof.} The sum over ${\bf u} := (u_1, \ldots, u_m) \in B_\ell$ satisfying $u_{\max} := \max_{1\le i\le m} u_i \le 2$ is very simple: in this case, $\ell \le 2m$; since $\eta({\bf u}) \ge 2$, we have $(\ell-\eta({\bf u})-2)^+ \le \ell-4$. The number of such ${\bf u}$ being smaller than $3^m$, we get
$$
\sum_{{\bf u}\in B_\ell: \, u_{\max} \le 2} y^{(\ell-\eta({\bf u}) -2)^+} \prod_{i: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}
\le
3^m y^{\ell-4}
\le
c_{19} \, \frac{y^{\ell-4}}{\ell^2} \, ,
$$

\noindent with $c_{19} := 3^m (2m)^2$.  [Notation: $\sum_\varnothing := 0$.]

Let $\mathrm{LHS}_{\eqref{pf_lemma51_eq2}}$ denote the expression on the left-hand side of \eqref{pf_lemma51_eq2}. Then
$$
\mathrm{LHS}_{\eqref{pf_lemma51_eq2}}
\le
c_{19} \, \frac{y^{\ell-4}}{\ell^2}
+
\sum_{j=2}^{\ell \wedge m} \binom{m}{j} j!\, y^{(\ell-j-2)^+} \sum_{(u_1, \ldots, u_j)} \, \prod_{i: \, 1\le i\le j, \, u_i\ge 3} \frac{1}{u_i(u_i-1)} \, ,
$$

\noindent where, on the right-hand side, $\sum_{(u_1, \ldots, u_j)}$ sums over all $(u_1, \ldots, u_j) \in \z_+^j$ with $1\le u_1\le u_2\le \cdots\le u_j$ such that $u_1+\cdots+u_j=\ell$ and that $u_j \ge 3$. Note that $u_j\ge 3$ implies $j\le \ell-2$, thus $(\ell-j-2)^+ = \ell-j-2$. Moreover, we have $u_j \ge \frac{\ell}{j} \ge \frac{\ell}{m}$, thus $u_j(u_j-1) \ge \frac12 u_j^2 \ge \frac{\ell^2}{2m^2}$. Consequently, $\prod_{i\le j: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}$ is bounded by $\frac{2m^2}{\ell^2} \prod_{i\le j-1: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}$. This leads to (using $\binom{m}{j}j! \le m^j$):
\begin{eqnarray*}
    \mathrm{LHS}_{\eqref{pf_lemma51_eq2}}
 &\le& c_{19} \, \frac{y^{\ell-4}}{\ell^2}
    +
    \frac{2m^2}{\ell^2} \sum_{j=2}^{(\ell-2)\wedge m} m^j y^{\ell-j-2} \sum_{(u_1, \ldots, u_j)} \, \prod_{i\le j-1: \, u_i\ge 3} \frac{1}{u_i(u_i-1)}
    \\
 &\le& c_{19} \, \frac{y^{\ell-4}}{\ell^2}
    +
    \frac{2m^2}{\ell^2} \sum_{j=2}^{(\ell-2)\wedge m} m^j y^{\ell-j-2} \prod_{i=1}^{j-1} \Big( 1+\sum_{u=3}^\infty \frac{1}{u(u-1)} \Big) \, .
\end{eqnarray*}

\noindent Of course, $\sum_{u=3}^\infty \frac{1}{u(u-1)} = \frac12$; also, we bound $\sum_{j=2}^{(\ell-2)\wedge m}$ by $\sum_{j=2}^\infty$. This yields that
$$
\mathrm{LHS}_{\eqref{pf_lemma51_eq2}}
\le
c_{19} \, \frac{y^{\ell-4}}{\ell^2}
+
\frac{2m^2}{\ell^2} \sum_{j=2}^\infty m^j y^{\ell-j-2} (\frac32)^{j-1} \, .
$$

\noindent On the right-hand side, write $\sum_{j=2}^\infty m^j y^{\ell-j-2} (\frac32)^{j-1} = m^2 y^{\ell-4} \sum_{j=2}^\infty (\frac{m}{y})^{j-2} (\frac32)^{j-1}$; in view of our choice $y\ge 3m$, this is bounded by $m^2 y^{\ell-4} \sum_{j=2}^\infty (\frac13)^{j-2} (\frac32)^{j-1} = 3m^2 y^{\ell-4}$. As a consequence,
$$
\mathrm{LHS}_{\eqref{pf_lemma51_eq2}}
\le
c_{19} \, \frac{y^{\ell-4}}{\ell^2}
+
\frac{6m^4y^{\ell-4}}{\ell^2} \, ,
$$

\noindent yielding \eqref{pf_lemma51_eq2} with $c_{18} := c_{19}+6m^4$.\qed

\bigskip

We have all the ingredients for the proof of Theorem \ref{t:H_nkupper}.

\bigskip

\noindent {\it Proof of Theorem \ref{t:H_nkupper}.} Let $\gamma>0$ and let $(Z_n, \, n\ge 0)$ be a $\gamma$-dominable system. Write $H_n(u) := \E(u^{Z_n})$ as before. Recall $\vartheta(M) \ge 1$ (for all $M\ge M_0$) from \eqref{assump_H0k} and \eqref{assump_Qn}. Write, for brevity,
\begin{eqnarray*}
    Q_n
 &:=& \prod_{j=0}^{n-1} H_j(m)^{m-1}
    =
    \prod_{j=0}^{n-1} [\E(m^{Z_j})]^{m-1} ,
    \\
    M_n
 &:=& n \vee M ,
    \\
    \vartheta_n
 &:=& \vartheta(M_n)
    =
    \vartheta(n\vee M),
    \qquad
    n\ge 1, \; M\ge M_0.
\end{eqnarray*}

\noindent By assumption \eqref{assump_Qn},
\begin{equation}
    Q_n \le \vartheta_n Q_n \le \gamma \, M_n^2 ,
    \qquad
    n\ge 1 \, .
    \label{Qn}
\end{equation}

We claim that for all integer $k\ge 3$,
\begin{equation}
    H_n^{(k)}(m)
    \le
    c_4^{k-1}\, (k-2)! \, M_n^{k-3} \, \vartheta_n Q_n,
    \qquad
    n\ge 1,
    \label{e:momentupper1111}
\end{equation}

\noindent where $c_4 := \max\{ 4+\frac{c_{18}c_{20}}{m}, \, c_9^{1/2}, \, \gamma \}$, with $c_{20} := m^{m/(m-1)} (c_8^m \vee 1) (\gamma^{3m/2} \vee 1)$. Since $\vartheta_n Q_n \le \gamma \, M_n^2$ (see \eqref{Qn}), \eqref{e:momentupper1111} will imply Theorem \ref{t:H_nkupper}.

It remains to prove \eqref{e:momentupper1111}, which we do by induction in $k\ge 3$.

By \eqref{Hn_3rd_derivative<bis}, $H_n'''(m)\le c_9\, \vartheta_n Q_n$ for $n\ge 1$. So \eqref{e:momentupper1111} holds for $k=3$ since $c_9\le c_4^2$.

Let $\ell\ge 4$ be an integer. Suppose \eqref{e:momentupper1111} holds for all $k\in \{ 3, \, 4, \, \ldots, \, \ell-1\}$. We need to prove \eqref{e:momentupper1111} for $k=\ell$.

We first prove that the induction assumption yields that for $n\ge 1$ and ${\bf u} := (u_1, \ldots, u_m) \in B_\ell$ (defined in \eqref{B_ell}), we have, with $c_{20} := m^{m/(m-1)} (c_8^m \vee 1) (\gamma^{3m/2} \vee 1)$ as before,
\begin{equation}
    \prod_{i=1}^m H_n^{(u_i)}(m)
    \le
    c_{20} c_4^{\ell-2} M_n^{(\ell-\eta({\bf u}) -2)^+} \vartheta_n Q_n \prod_{i: \, u_i\ge 3} (u_i-2)! \, ,
    \label{e:diffmomentsupper}
\end{equation}

\noindent where $\eta({\bf u}):= \sum_{i=1}^m {\bf 1}_{\{ u_i \ge 1\} }$ is as in \eqref{eta}.

To check \eqref{e:diffmomentsupper}, let $n\ge 1$ and ${\bf u} \in B_\ell$. Since $H_n(m) \le m^{1/(m-1)} =: c_{21}$ (see \eqref{Gn(m)<}) and $H_n'(m) \le \frac{1}{m-1} \, \E(m^{Z_n}) \le \frac{c_{21}}{m-1} \le c_{21}$, we have
$$
\prod_{i=1}^m H_n^{(u_i)}(m)
\le
c_{21}^m H_n''(m)^{\lambda_2({\bf u})} \prod_{i: \, u_i\ge 3} H_n^{(u_i)}(m) \, ,
$$

\noindent where $\lambda_2({\bf u}) := \sum_{i=1}^m {\bf 1}_{\{ u_i =2\}}$. By \eqref{Hn_2nd_derivative<bis}, we have $H_n''(m) \le c_8 \, \vartheta_n^{1/2} \, Q_n^{1/2}$; thus with $c_{22} := c_{21}^m \max\{ c_8^m, \, 1\}$,
$$
\prod_{i=1}^m H_n^{(u_i)}(m)
\le
c_{22}\, (\vartheta_n Q_n)^{\lambda_2({\bf u})/2} \prod_{i: \, u_i\ge 3} H_n^{(u_i)}(m) \, .
$$

\noindent By the induction assumption in \eqref{e:momentupper1111}, $H_n^{(u_i)}(m) \le c_4^{u_i-1}(u_i-2)! \, M_n^{u_i-3}\, \vartheta_n Q_n$ if $u_i \ge 3$. As such, we have
\begin{eqnarray*}
    \prod_{i=1}^m H_n^{(u_i)}(m)
 &\le& c_{22}\, (\vartheta_n Q_n)^{\lambda_2({\bf u})/2} \prod_{i: \, u_i\ge 3} \Big( c_4^{u_i-1}(u_i-2)! \, M_n^{u_i-3} \vartheta_n Q_n\Big)
    \\
 &=& c_{22}\, (\vartheta_n Q_n)^{\lambda_2({\bf u})/2} \, (c_4 M_n)^{\sum_{i: \, u_i\ge 3} (u_i-1)} (M_n^{-2} \vartheta_n Q_n)^{\eta_3({\bf u})} \, \prod_{i: \, u_i\ge 3} (u_i-2)! \, ,
\end{eqnarray*}

\noindent where $\eta_3({\bf u}) := \sum_{i=1}^m {\bf 1}_{\{ u_i \ge 3\}}$. Note that $\sum_{i: \, u_i\ge 3} (u_i-1) = \sum_{i=1}^m (u_i-1)^+ - \lambda_2({\bf u}) = \ell - \eta({\bf u}) - \lambda_2({\bf u}) \le \ell-2$. So $c_4^{\sum_{i: \, u_i\ge 3} (u_i-1)} \le c_4^{\ell-2}$ (using $c_4>1$). This leads to:
\begin{eqnarray*}
    \prod_{i=1}^m H_n^{(u_i)}(m)
&\le& c_{22}\, c_4^{\ell-2}\, (\vartheta_n Q_n)^{\frac{\lambda_2({\bf u})}{2} + \eta_3({\bf u}) } \, M_n^{\ell - \eta({\bf u}) - \lambda_2({\bf u}) - 2\eta_3({\bf u})} \, \prod_{i: \, u_i\ge 3} (u_i-2)!
    \\
 &=& c_{22}\, c_4^{\ell-2}\, (\vartheta_n Q_n)^{\frac{\lambda_2({\bf u})}{2} + \eta_3({\bf u}) -1} \, M_n^{\ell - \eta({\bf u}) - \lambda_2({\bf u}) - 2\eta_3({\bf u})} \, \vartheta_n Q_n \prod_{i: \, u_i\ge 3} (u_i-2)! \; .
\end{eqnarray*}

Assume for the moment $\frac{\lambda_2({\bf u})}{2} + \eta_3({\bf u}) \ge 1$. By \eqref{Qn}, $\vartheta_n Q_n \le \gamma \, M_n^2$, so we have $(\vartheta_n Q_n)^{\frac{\lambda_2({\bf u})}{2} + \eta_3({\bf u}) -1} \le c_{23} \, M_n^{\lambda_2({\bf u}) + 2\eta_3({\bf u}) -2}$, with $c_{23} := \max\{ \gamma^{3m/2}, \, 1\}$. Since $c_{20} = c_{22}c_{23}$, we get
$$
\prod_{i=1}^m H_n^{(u_i)}(m)
\le
c_{20}\, c_4^{\ell-2}\, M_n^{\ell - \eta({\bf u}) - 2} \, \vartheta_n Q_n \prod_{i: \, u_i\ge 3} (u_i-2)! ,
$$

\noindent yielding \eqref{e:diffmomentsupper}. If, on the other hand, $\frac{\lambda_2({\bf u})}{2} + \eta_3({\bf u}) <1$, then $\lambda_2({\bf u}) \le 1$ and $\eta_3({\bf u}) =0$, i.e., $\max_{1\le i\le m} u_i\le 2$. This time, $\ell-\eta({\bf u}) = \sum_{i=1}^m (u_i-1)^+ \le 1$. The situation is very simple if we look at $\prod_{i=1}^m H_n^{(u_i)}(m)$ directly: at most one term among $H_n^{(u_i)}(m)$ is $H_n''(m)$ (which is bounded by $c_8 \, \vartheta_n^{1/2} \, Q_n^{1/2}$ as we have seen in \eqref{Hn_2nd_derivative<bis}), while all the rest is either $H_n(m)$ (which is bounded by $m^{1/(m-1)}=: c_{21}$) or $H_n'(m)$ (which is bounded by $1$ because by \eqref{Gn(m)<}, $H_n'(m) = \E(Z_n \, m^{Z_n-1}) \le \frac{m^{1/(m-1)}}{m(m-1)} \le 1$). Hence
$$
\prod_{i=1}^m H_n^{(u_i)}(m)
\le
c_{21}^m c_8 \, \vartheta_n^{1/2} \, Q_n^{1/2}
\le
c_{24} \, \vartheta_n Q_n \, ,
$$

\noindent with $c_{24} := c_{21}^m c_8$. [We have used $\vartheta_n Q_n \ge 1$.] This again gives \eqref{e:diffmomentsupper} because $c_{24} \le c_{20}$ and $c_4>1$.

Now that \eqref{e:diffmomentsupper} is proved, it is painless to complete the proof of Theorem \ref{t:H_nkupper}. Indeed, by \eqref{iteration_Gn},
$$
s H_{n+1}(s) = [H_n(s)]^m + (s-1) [H_n(0)]^m\, .
$$

\noindent On both sides, we differentiate $\ell$ times with respect to $s$ (recalling that $\ell \ge 4$, so the last term on the right-hand side, being affine in $s$, makes no contribution to the derivatives), and apply the general Leibniz rule to the first term on the right-hand side; this leads to:
\begin{eqnarray*}
    sH_{n+1}^{(\ell)}(s) + \ell H_{n+1}^{(\ell-1)}(s)
 &=& \sum_{(u_1, \ldots, u_m) \in \z_+^m: \, u_1+\cdots+u_m=\ell} \frac{\ell!}{u_1! \cdots u_m!} \prod_{i=1}^m H_n^{(u_i)}(s)
    \\
 &=& m H_n^{(\ell)}(s) H_n(s)^{m-1}
    +
    \sum_{{\bf u}\in B_\ell} \frac{\ell!}{u_1! \cdots u_m!} \prod_{i=1}^m H_n^{(u_i)}(s)\, .
\end{eqnarray*}

\noindent Note that the expression on the left-hand side is at least $sH_{n+1}^{(\ell)}(s)$. We take $s=m$ to see that
$$
H_{n+1}^{(\ell)}(m)
-
H_n^{(\ell)}(m) H_n(m)^{m-1}
\le
\frac1m \sum_{{\bf u}\in B_\ell} \frac{\ell!}{u_1! \cdots u_m!} \prod_{i=1}^m H_n^{(u_i)} (m).
$$

\noindent By \eqref{e:diffmomentsupper}, we have $\prod_{i=1}^m H_n^{(u_i)}(m) \le c_{20} c_4^{\ell-2} M_n^{(\ell-\eta({\bf u})-2)^+} \vartheta_n Q_n \prod_{i: \, u_i\ge 3} (u_i-2)!$, where $\eta ({\bf u}) := \sum_{i=1}^m {\bf 1}_{\{ u_i \ge 1\} }$ is as in \eqref{eta}. Hence
$$
H_{n+1}^{(\ell)}(m)
-
H_n^{(\ell)}(m) H_n(m)^{m-1}
\le
\frac{c_{20}}{m} c_4^{\ell-2} \ell! \, \vartheta_n Q_n \sum_{{\bf u}\in B_\ell} M_n^{(\ell-\eta({\bf u})-2)^+} \prod_{i: \, u_i\ge 3} \frac{1}{u_i(u_i-1)} \, ,
$$

\noindent which, in view of Lemma \ref{l:201810171111} (applied to $y:= M_n$), yields that, for $n\ge 1$,
$$
H_{n+1}^{(\ell)}(m)
\le
H_n^{(\ell)}(m)H_n(m)^{m-1}
+
c_{18} \frac{c_{20}}{m} c_4^{\ell-2}\, (\ell-2)! \, M_n^{\ell-4} \vartheta_n Q_n.
$$

\noindent Recall that $Q_n := \prod_{j=0}^{n-1} H_j(m)^{m-1}$. Iterating this inequality, and by means of the monotonicity of $n\mapsto M_n^{\ell-4} \vartheta_n Q_n$, we get
\begin{eqnarray*}
    H_n^{(\ell)}(m)
 &\le& H_0^{(\ell)}(m) Q_n
    +
    \sum_{j=0}^{n-1} \frac{c_{18}c_{20}}{m} c_4^{\ell-2} (\ell-2)! \, M_n^{\ell-4} \, \vartheta_n Q_n
    \\
 &=& \Big( \frac{H_0^{(\ell)}(m)}{\vartheta_n} +n\frac{c_{18}c_{20}}{m} c_4^{\ell-2} (\ell-2)! \, M_n^{\ell-4} \Big) \vartheta_n Q_n \, .
\end{eqnarray*}

\noindent We use $n \le M_n$ so that $n\, M_n^{\ell-4} \le M_n^{\ell-3}$. On the other hand, $H_0^{(\ell)}(m) \le M^{\ell-3} (\vartheta(M) + \ell !)$ (by assumption \eqref{assump_H0k}), which is bounded by $M_n^{\ell-3} \vartheta_n\, (1+ \ell !)$. Thus
\begin{eqnarray*}
    H_n^{(\ell)}(m)
 &\le& \Big( 1+ \ell! + \frac{c_{18}c_{20}}{m} c_4^{\ell-2} (\ell-2)! \Big) M_n^{\ell-3} \, \vartheta_n Q_n
    \\
 &\le& (1 + \ell(\ell-1) +\frac{c_{18}c_{20}}{m} c_4^{\ell-2}) \, (\ell-2)! \, M_n^{\ell-3} \, \vartheta_n Q_n \, .
\end{eqnarray*}

\noindent Since $\ell \ge 4$, we have $1 + \ell(\ell-1) \le \ell^2 \le 2^\ell \le 4 \, c_4^{\ell-2}$ (because $c_4\ge 2$), so $1 + \ell(\ell-1) +\frac{c_{18}c_{20}}{m} c_4^{\ell-2} \le 4 \, c_4^{\ell-2} + \frac{c_{18}c_{20}}{m} c_4^{\ell-2}\le c_4^{\ell-1}$ by means of the fact that $c_4 := \max\{ 4+\frac{c_{18}c_{20}}{m}, \, c_9^{1/2}, \, \gamma \}$. Consequently,
$$
H_n^{(\ell)}(m)
\le
c_4^{\ell-1} (\ell-2)! \, M_n^{\ell-3} \, \vartheta_n Q_n \, ,
$$

\noindent implying \eqref{e:momentupper1111} for $k=\ell$, and completing the proof of Theorem \ref{t:H_nkupper}.\qed

\bigskip

\noindent {\it Proof of Corollary \ref{p:sXsub}.} Only \eqref{cor1} needs proving because \eqref{e:922745} will follow immediately from \eqref{cor1} and assumption \eqref{assump_Qn}.

Let $n\ge 1$ and $s\in [m, \, m + \frac{1}{2c_4 M_n}]$, where we keep using the notation
$$
M_n := n \vee M\, .
$$

\noindent Write $H_n(u) := \E(u^{Z_n})$ as before. Then
$$
\E(Z_n^2 s^{Z_n})
=
s H_n'(s) + s^2H_n''(s) \, .
$$

\noindent Since $u\mapsto H_n''(u)$ is non-decreasing, we have $H_n'(s) \le H_n'(m) + (s-m) H_n''(s)$; hence
$$
\E(Z_n^2 s^{Z_n})
\le
sH_n'(m) + s(s-m)H_n''(s) + s^2H_n''(s)
=
sH_n'(m) + s(2s-m)H_n''(s)\, .
$$

\noindent On the right-hand side, we use $H_n'(m) \le 1$ (which has already been observed as a consequence of \eqref{Gn(m)<}), and $m\le s\le m+1$ (so $s(2s-m) \le (m+1)(m+2)$), to see that
$$
\E(Z_n^2 s^{Z_n})
\le
m+1 + (m+1)(m+2) H_n''(m+\frac{1}{2c_4 M_n}) \, .
$$

\noindent To bound $H_n''(m+\frac{1}{2c_4 M_n})$, we recall that $Z_n$ is bounded for each $n$ (which is a consequence of the boundedness of $Z_0$), so by Taylor expansion,
$$
H_n''(m+\frac{1}{2c_4 M_n})
-
H_n''(m)
=
\sum_{k=3}^\infty \frac{(2c_4 M_n)^{-(k-2)}}{(k-2)!} H_n^{(k)}(m) \, .
$$

\noindent By \eqref{e:momentupper1111}, we have $H_n^{(k)}(m)\le c_{4}^{k-1}\, (k-2)! \, M_n^{k-3} \, \vartheta_n Q_n$ (for $k\ge 3$). Hence
$$
H_n''(m+\frac{1}{2c_4 M_n})
-
H_n''(m)
\le
\sum_{k=3}^\infty \frac{(2c_4 M_n)^{-(k-2)}}{(k-2)!} c_{4}^{k-1}\, (k-2)! \, M_n^{k-3} \, \vartheta_n Q_n
=
c_4\, \frac{\vartheta_n Q_n}{M_n} \, .
$$

\noindent As such, we arrive at:
$$
\E(Z_n^2 s^{Z_n})
\le
m+1 + (m+1)(m+2) \Big( H_n''(m) + c_4\, \frac{\vartheta_n Q_n}{M_n} \Big) \, .
$$

\noindent By \eqref{Hn_2nd_derivative<bis}, $H_n''(m)\le c_8 \, \vartheta_n^{1/2} Q_n^{1/2}$, whereas according to assumption \eqref{assump_Qn}, $M_n \ge \frac{1}{\gamma^{1/2}} \, \vartheta_n^{1/2} Q_n^{1/2}$, this readily yields \eqref{cor1} (recalling that $\vartheta_n Q_n \ge 1$). Corollary \ref{p:sXsub} is proved.\qed

\section{Proof of Theorem \ref{t:moment_ub}}
\label{s:moment_s>m}

Let $(Y_n, \, n\ge 0)$ be a critical system such that $\E(s^{Y_0})<\infty$ for some $s>m$. We claim that with $c_{25} := (\sup_{x>0} x\, (\frac{s}{m})^{-x/\ee}) \vee 1 \in [1, \, \infty)$, we have, for all integer $k\ge 1$,
\begin{equation}
    \E(Y_0^k m^{Y_0}) \le c_{25}^k \, \E(s^{Y_0}) \, k!\, .
    \label{case_big_moments}
\end{equation}

Indeed, by definition,
$$
x\le c_{25} \, (\frac{s}{m})^{x/\ee} \, ,
$$

\noindent for all $x>0$. Taking to the power $k$ on both sides and with $x:= \frac{\ee\, \ell}{k}$, we see that for all integers $k\ge 1$ and $\ell \ge 1$,
$$
(\frac{\ee\, \ell}{k})^k \le c_{25}^k \, (\frac{s}{m})^\ell \, .
$$

\noindent Since $k! \ge (\frac{k}{\ee})^k$ by Stirling's formula, this yields $\ell^k m^\ell\le c_{25}^k \, s^\ell \, k!$, from which \eqref{case_big_moments} follows.

Let $L\ge 1$ be an integer, and let $Z_0 = Z_0(M, \, L) := Y_0 \, {\bf 1}_{\{ Y_0 \le L\} }$. Then $Z_0$ is bounded, and does not depend on $M$, though we still treat it as indexed by $M$. Let us check that assumptions \eqref{assump_H0k} and \eqref{assump_Qn} in Definition \ref{d:dominable} are satisfied.

Since $Z_0\le Y_0$, it follows from \eqref{case_big_moments} that $\E(Z_0^k m^{Z_0}) \le c_{26} c_{25}^k \, k!$ for $k\ge 1$, where $c_{26} =c_{26}(s) := \E(s^{Y_0}) \ge 1$. Let $M\ge c_{26} c_{25}^4$. Then $M^{k-3} \ge (c_{26} c_{25}^3)^{k-3} c_{25}^{k-3} \ge c_{26} c_{25}^3 c_{25}^{k-3} = c_{26} c_{25}^k$ for all integer $k\ge 4$, so $c_{26} c_{25}^k \, k! \le M^{k-3} k!$ for $k\ge 4$. For $k=3$, we have $c_{26} c_{25}^3 \, 3! \le c_{27} + 3!$ with $c_{27} := 6(c_{26} c_{25}^3-1) \vee 1 \in [1, \, \infty)$. As such, we see that for all integers $k\ge 3$ and $M\ge c_{26} c_{25}^4$,
$$
\E(Z_0^k m^{Z_0}) \le M^{k-3} (c_{27} + k!) ;
$$

\noindent in words, assumption \eqref{assump_H0k} is satisfied with $\vartheta (M) := c_{27}$ for all $M \ge M_0 := \lceil c_{26} c_{25}^4\rceil$.

Assumption \eqref{assump_Qn} is easily seen to be satisfied: by \eqref{prod_Gi(m)<}, $\prod_{i=0}^{n-1} [\E(m^{Y_i})]^{m-1} \le c_7 n^2$ for all $n\ge 1$, so \eqref{assump_Qn} holds with $\gamma := c_7 c_{27}$.

So we are entitled to apply Theorem \ref{t:H_nkupper} to see that for $k\ge 3$, $n\ge 1$ and $M\ge M_0:= \lceil c_{26} c_{25}^4\rceil$,
$$
\E[Z_n(Z_n-1) \cdots (Z_n-k+1)\, m^{Z_n-k}] \le k! \, c_4^k \, (n\vee M)^{k-1} \, .
$$

\noindent We take $M := M_0$. Recall that $Z_0 = Z_0(M_0, \, L) := Y_0 \, {\bf 1}_{\{ Y_0 \le L\} }$. Letting $L\to \infty$, and applying the monotone convergence theorem, we get, for $k\ge 3$ and $n\ge 1$,
$$
\E[Y_n(Y_n-1) \cdots (Y_n-k+1)\, m^{Y_n-k}] \le k! \, c_4^k \, (n\vee M_0)^{k-1} \, .
$$

\noindent This implies the desired inequality for $k\ge 3$. The case $k=1$ has already been implicitly treated in the proof of Corollary \ref{p:sXsub} in Section \ref{s:domination}: $\E(Y_n \, m^{Y_n-1}) = \frac{\E(m^{Y_n})}{m(m-1)} \le \frac{m^{1/(m-1)}}{m(m-1)}$ (see \eqref{Gn(m)<}). The case $k=2$ follows from the cases $k=1$ and $k=3$ by the Cauchy--Schwarz inequality.\qed

\section{Truncating the critical system}
\label{s:truncating}

Let $(Y_n, \, n\ge 0)$ be a Derrida--Retaux system satisfying $\E(m^{Y_0}) = (m-1) \E(Y_0\, m^{Y_0})<\infty$ (so the system is critical). Recall that the system $(Y_n, \, n\ge 0)$ is of ``finite variance" if $\E(Y_0^3 \, m^{Y_0})<\infty$, and is stable if $\P(Y_0=j) \sim c_0 \, m^{-j}j^{-\alpha}$, $j\to \infty$, for some $c_0>0$ and $2<\alpha<4$ as in \eqref{alpha}. The following theorem tells that in either case, we can define
$$
Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le a(M)\}},
$$

\noindent for some appropriate $a(M) \in [1, \, \infty]$ such that $(Z_n, \, n\ge 0)$ is dominable in the sense of Definition \ref{d:dominable}; the values of $a(M)$ and $\vartheta(M)$ (as defined in Definition \ref{d:dominable}) are also given as they are often useful in the applications.

\medskip

\begin{theorem}
\label{t:dominable}

 Let $(Y_n, \, n\ge 0)$ be a Derrida--Retaux system satisfying $\E(m^{Y_0}) = (m-1) \E(Y_0\, m^{Y_0})<\infty$. If it is either of ``finite variance" or stable, then there is a dominable system $(Z_n, \, n\ge 0)$ such that $Z_0 \le Y_0$ \hbox{\rm a.s.} More precisely,

 {\rm (i)} if the system is of ``finite variance", we can choose $Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le M \zeta(M)\} }$ where $\zeta(M) := - \log \E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{ Y_0 >M\} }) \le \infty$,\footnote{So $\lim_{M\to \infty} \zeta(M) =\infty$ by the ``finite variance" assumption.} with $\vartheta (M) := \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$;

 {\rm (ii)} if the system is stable, we can choose $Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le M\} }$, with $\vartheta(M) := c_{30}\, M^{4-\alpha}$, where $c_{30}$ is the constant in \eqref{theta(M)_stable} below.

\end{theorem}

\medskip

For the sake of clarity, the two situations (``finite variance", stable) are discussed in distinct parts.

\subsection{Proof of Theorem \ref{t:dominable}: the ``finite variance" case}

Assume $\E(m^{Y_0}) = (m-1) \E(Y_0\, m^{Y_0})<\infty$ and $\E(Y_0^3 \, m^{Y_0}) <\infty$. Let
$$
\zeta(M)
:=
- \log \E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{ Y_0 >M\} })
\le
\infty \, .
$$

\noindent Since $\lim_{M\to \infty} \zeta(M) \to \infty$, we can choose $M$ sufficiently large so that $M \, \zeta(M) \ge 2$. Let $Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le M \zeta(M)\} }$.

We claim that assumption \eqref{assump_H0k} is satisfied with $\vartheta (M) := \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$ and that $Z_0$ is bounded. For any integer $k\ge 3$, we write
\begin{eqnarray*}
   \E(Z_0^km^{Z_0})
 &=& \E(Y_0^k m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}})
   +
   \E(Y_0^k m^{Y_0} \, {\bf 1}_{\{M< Y_0\le M\zeta(M)\}})
   \\
 &\le& M^{k-3}\E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}})
   +
   \E(Y_0^k m^{Y_0} \, {\bf 1}_{\{M< Y_0\le M\zeta(M)\}}) \, .
\end{eqnarray*}

\noindent The first term on the right-hand side is easy to handle: we have $\E(Y_0^k m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}}) \le M^{k-3}\E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}}) \le \E(Y_0^3 m^{Y_0}) \, M^{k-3}$. In case $\zeta(M) =\infty$, we have $Y_0\le M$ a.s., so $Z_0$ is bounded and the second term on the right-hand side vanishes, which yields \eqref{assump_H0k} with $\vartheta (M) := \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$.

To treat the case $\zeta(M)<\infty$ (in which case $Z_0$ is obviously bounded), let us look at the second term on the right-hand side: since $\E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{ Y_0>M\}}) = \ee^{-\zeta(M)}$, we have
$$
\E(Y_0^k m^{Y_0} \, {\bf 1}_{\{M< Y_0\le M\zeta(M)\}})
\le
M^{k-3}\zeta(M)^{k-3}\E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{ Y_0>M\}})
=
M^{k-3}\zeta(M)^{k-3}\, \ee^{-\zeta(M)}\, .
$$

\noindent Applying the inequality $\ee^x \ge \frac{x^{k-3}}{(k-3)!}$ (for $x\ge 0$ and $k\ge 3$) to $x:= \zeta(M)$ yields \eqref{assump_H0k} again with $\vartheta (M) := \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$.

Consequently, regardless of whether $\zeta(M)$ is finite or infinite, $Z_0$ is bounded, and assumption \eqref{assump_H0k} is satisfied with $\vartheta (M) := \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$. Note that $\vartheta(M)$ does not depend on $M$.

Assumption \eqref{assump_Qn} is also satisfied: by \eqref{prod_Gi(m)<}, $\prod_{i=0}^{n-1} [\E(m^{Y_i})]^{m-1} \le c_7 n^2$ for all $n\ge 1$; since $\E(m^{Z_i}) \le \E(m^{Y_i})$ for all $i\ge 0$, \eqref{assump_Qn} is satisfied with $\gamma := c_7 \max\{ \E(Y_0^3 m^{Y_0}), \, 1\}$.\qed

\subsection{Proof of Theorem \ref{t:dominable}: the stable case}

We start with a simple inequality.

\medskip

\begin{lemma}
\label{l:prod_subcritical}

 Let $(X_n, \, n\ge 0)$ be a Derrida--Retaux system satisfying $(m-1)\E(X_0 \, m^{X_0}) <\E(m^{X_0})  < \infty$. Then
 $$
 \prod_{i=0}^\infty [\E(m^{X_i})]^{m-1}
 \le
 \frac{1}{\E(m^{X_0})- (m-1) \E(X_0\, m^{X_0})} \, .
 $$

\end{lemma}

\medskip

\noindent {\it Proof.} For the moment, let $(X_n, \, n\ge 0)$ be an arbitrary Derrida--Retaux system satisfying $\E(X_0 \, m^{X_0})<\infty$, and such that $\E(m^{X_0}) \not= (m-1) \E(X_0\, m^{X_0})$. By \eqref{iteration_Collet},
\begin{equation}
   \prod_{i=0}^{n-1} [\E(m^{X_i})]^{m-1}
    =
   \frac{\E(m^{X_n})- (m-1)\E(X_n\, m^{X_n})}{\E(m^{X_0})- (m-1) \E(X_0\, m^{X_0})} \, .
   \label{recursion_Coll}
\end{equation}

\noindent For the nominator in \eqref{recursion_Coll}, we observe that $\E(m^{X_n})- (m-1)\E(X_n\, m^{X_n}) = \E[(1-(m-1)X_n) m^{X_n}] \le \P(X_n=0) \le 1$. If we assume $(m-1)\E(X_0 \, m^{X_0}) <\E(m^{X_0})$, then the denominator in \eqref{recursion_Coll} is positive; the lemma follows immediately from the monotone convergence theorem by letting $n\to \infty$.\qed

\bigskip

We now proceed to the proof of Theorem \ref{t:dominable} for stable systems. Assume $\E(m^{Y_0}) = (m-1) \E(Y_0\, m^{Y_0})<\infty$ and $\P(Y_0=j) \sim c_0 \, m^{-j}j^{-\alpha}$, $j\to \infty$, for some $c_0>0$ and $2<\alpha<4$ as in \eqref{alpha}. This yields the existence of constants $c_{28} \ge c_{29} >0$ and an integer $j_0\ge 1$, all depending on $m$ and on the law of $Y_0$, such that
\begin{eqnarray}
    \P(Y_0=j)
 &\le& c_{28} \, m^{-j}j^{-\alpha} ,
    \qquad
    j\ge 1,
    \label{stable_tail_ub}
    \\
    \P(Y_0=j)
 &\ge& c_{29} \, m^{-j}j^{-\alpha} .
    \qquad
    j\ge j_0,
    \label{stable_tail_lb}
\end{eqnarray}

\noindent Let $M\ge j_0$ be an integer and let
\begin{equation}
    Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le M\} } ,
    \label{Z0}
\end{equation}

\noindent which is a bounded random variable. For integer $k\ge 3$, we write
$$
\E(Z_0^k m^{Z_0})
=
\E(Y_0^k m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}})
\le
M^{k-3} \, \E(Y_0^3 m^{Y_0} \, {\bf 1}_{\{Y_0\le M\}}) ,
$$

\noindent so by \eqref{stable_tail_ub}, we have
\begin{eqnarray*}
    \E(Z_0^k m^{Z_0})
 &\le& c_{28} M^{k-3} \sum_{j=1}^M j^{3-\alpha}
    \le
    c_{28} M^{k-3} \int_0^{M+1} x^{3-\alpha} \d x
    \\
 &=& c_{28} M^{k-3} \, \frac{(M+1)^{4-\alpha}}{4-\alpha}
    \le
    c_{28} M^{k-3} \, \frac{2^{4-\alpha} \, M^{4-\alpha}}{4-\alpha}  \, .
\end{eqnarray*}

\noindent As such, assumption \eqref{assump_H0k} is satisfied with
\begin{equation}
    \vartheta (M) := c_{30}\, M^{4-\alpha},
    \label{theta(M)_stable}
\end{equation}

\noindent where $c_{30} := \max\{ c_{28} \frac{2^{4-\alpha}}{4-\alpha}, \, 1\}$. In particular, $M\mapsto \vartheta (M)$ is non-decreasing.

It remains to check assumption \eqref{assump_Qn}, which in this case states that for some constant $c_{31}>0$,
\begin{equation}
    \prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1}
    \le
    c_{31} \, (n\vee M)^{\alpha-2},
    \qquad
    M\ge j_0, \; n\ge 1.
    \label{check_assump2_stable}
\end{equation}

Since $\E(m^{Z_i}) \le \E(m^{Y_i})$, there is nothing to prove if $n \le j_0$: it suffices to take $c_{31}$ such that $c_{31} \ge \prod_{i=0}^{j_0-1} [\E(m^{Y_i})]^{m-1}$. Let us assume $n>j_0$. We write $\prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1} \le \prod_{i=0}^\infty [\E(m^{Z_i})]^{m-1}$, so by Lemma \ref{l:prod_subcritical},
$$
\prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1}
\le
\frac{1}{\E(m^{Z_0})- (m-1) \E(Z_0\, m^{Z_0})} \, .
$$

\noindent By definition, $Z_0 = Y_0 \, {\bf 1}_{\{ Y_0 \le M\} }$, and by assumption, $\E(m^{Y_0}) = (m-1) \E(Y_0\, m^{Y_0})$. So
\begin{eqnarray}
 &&\E(m^{Z_0})- (m-1) \E(Z_0\, m^{Z_0})
    \nonumber
    \\
 &=& \E[((m-1)Y_0 -1)\, m^{Y_0}\, {\bf 1}_{\{ Y_0 > M\} }) + \P(Y_0 > M)
    \label{E(mZ0)}
    \\
 &\ge& \E[((m-1)Y_0 -1)\, m^{Y_0}\, {\bf 1}_{\{ Y_0 > M\} }) \, ,
    \nonumber
\end{eqnarray}

\noindent which, by \eqref{stable_tail_lb}, is $\ge c_{29} \sum_{j=M+1}^\infty ((m-1)j -1) j^{-\alpha} \ge \frac{c_{32}}{M^{\alpha-2}}$ for some constant $c_{32}>0$ and all $M\ge j_0$. Consequently,
\begin{equation}
    \prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1}
    \le
    \frac{M^{\alpha-2}}{c_{32}} \, ,
    \label{e:Q_nupper_alpha}
\end{equation}

\noindent which is bounded by $\frac{(n\vee M)^{\alpha-2}}{c_{32}}$ as $\alpha>2$. This yields \eqref{check_assump2_stable}.\qed

\medskip

\begin{remark}
\label{r:Z2upperQ}

 For further use, let us note that in the stable case, $\vartheta (M) := c_{30}\, M^{4-\alpha}$ for all $M\ge M_0$ (by \eqref{theta(M)_stable}), whereas $\prod_{i=0}^{n-1} [\E(m^{Z_i})]^{m-1} \le \frac{M^{\alpha-2}}{c_{32}}$ (by \eqref{e:Q_nupper_alpha}), so inequality \eqref{cor1} in Corollary \ref{p:sXsub} implies that for $M\ge M_0$, $n\ge 1$ and $v := m + \frac{1}{2c_4 (n\vee M)}$ ($c_4 \ge 1$ being as before the constant in Theorem $\ref{t:H_nkupper}$),
\begin{equation}
    \E(Z_n^2 v^{Z_n})
    \le
    c_{33} (n\vee M)^{(4-\alpha)/2} M^{(\alpha-2)/2} \, ,
    \label{cor1_variant_stable}
\end{equation}

\noindent where $c_{33} := \frac{c_5 \, c_{30}^{1/2}}{c_{32}^{1/2}}$.\qed

\end{remark}

\section{Proof of Theorem \ref{t:main}}
\label{s:pf_thm}

Let $(Y_n, \, n\ge 0)$ be a Derrida--Retaux system such that $\E(m^{Y_0}) =(m-1) \,\E(Y_0 \, m^{Y_0})<\infty$. We assume that $\P(Y_0=j) \sim c_0 \, m^{-j}j^{-\alpha}$, $j\to \infty$, for some $c_0>0$ and $2<\alpha<4$ as in \eqref{alpha}.

\subsection{Upper bound}

We start with a lemma.

\medskip

\begin{lemma}
\label{l:general<}

 Let $(X_n, \, n\ge 0)$ be a Derrida--Retaux system satisfying $\E(X_0 \, m^{X_0}) < \infty$. There exists a constant $c_{34} >0$, depending only on $m$, such that $(\frac{n^2}{0} := \infty)$
 $$
 \prod_{i=0}^{n-1} [\E(m^{X_i})]^{m-1}
 \le
 c_{34} \, \frac{n^2}{\E(X_0^3 m^{X_0} {\bf 1}_{\{ 2\le X_0 \le 3n\} })} ,
 \qquad
 n\ge 1 \, .
 $$

\end{lemma}

\medskip

\noindent {\it Proof.} The lemma was known in various forms. Recall from \cite{bmxyz_questions} that for $s\in (\frac{m}{2}, \, m)$,
\begin{equation}
    \prod_{i=0}^{n-1} [\E(m^{X_i})]^{m-1}
    \le
    \Big(\frac{m}{2s-m}\Big)^n \frac{1}{\Delta_0(s)},
    \label{tech1}
\end{equation}

\noindent where
$$
\Delta_0(s)
:=
\sum_{k=1}^\infty m^k ((m-1)k-1) (1-(k+1)x^k + kx^{k+1}) \P(X_0=k),
$$

\noindent with $x:= \frac{s}{m}\in (\frac12, \, 1)$. We now reproduce some elementary computations from \cite{bmvxyz_conjecture_DR}. For $k\ge 1$, we have $x^k \le \ee^{-(1-x)k}$, so $1-(1+k)x^k+ kx^{1+k} \ge 1-(1+u)\, \ee^{-u}$, where $u:= (1-x)k>0$. Since $1 - (1+v)\,\ee^{-v} \ge 1 -\frac{2}{\ee}$ for $v\ge 1$ (because $v\mapsto 1 - (1+v)\,\ee^{-v}$ is increasing on $(0, \, \infty)$) and $1 - (1+v)\, \ee^{-v} \ge \frac{v^2}{2 \ee}$ for $v\in (0,\, 1]$ (because $v \mapsto 1 - (1+v)\, \ee^{-v} - \frac{v^2}{2\ee}$ is increasing on $(0, \, 1]$), we get, for $k\ge 1$,
$$
1-(k+1)x^k + kx^{k+1}
\ge
c_{35} \, \min\{ (1-x)^2 k^2, \, 1\} \, ,
$$

\noindent where $c_{35} := \min\{ 1 -\frac{2}{\ee}, \, \frac{1}{2\ee}\} > 0$. Let $n\ge 1$. We take $s=s_n:= (1-\frac{1}{3n})m$ (so $x= 1-\frac{1}{3n}$), to see that
\begin{eqnarray*}
    \Delta_0(s_n)
 &\ge& c_{35} \sum_{k=1}^\infty m^k ((m-1)k-1) \min\{ \frac{k^2}{(3n)^2}, \, 1\} \P(X_0=k)
    \\
 &\ge& \frac{c_{35}}{(3n)^2} \sum_{k=1}^{3n} k^2 m^k ((m-1)k-1) \P(X_0=k)\, .
\end{eqnarray*}

\noindent We use $(m-1)k-1 \ge \frac{m-1}{2} \, k$ for $k\ge 2$, so that with $c_{36} := c_{35} \, \frac{m-1}{18} >0$
$$
\Delta_0(s_n)
\ge
\frac{c_{36}}{n^2} \sum_{k=2}^{3n} k^3 m^k \P(X_0=k)
=
\frac{c_{36}}{n^2} \, \E(X_0^3 m^{X_0} \, {\bf 1}_{\{ 2\le X_0 \le 3n\} } ).
$$

\noindent This, in view of \eqref{tech1} (applied to $s=s_n$), yields the lemma.\qed

\bigskip

\noindent {\it Proof of the upper bound in Theorem \ref{t:main}.} By Lemma \ref{l:general<}, for all $n\ge 1$,
$$
\prod_{i=0}^{n-1} [\E(m^{Y_i})]^{m-1}
\le
c_{34} \, \frac{n^2}{\E(Y_0^3 m^{X_0} {\bf 1}_{\{ 2\le Y_0 \le 3n\} })} .
$$

\noindent Since $\P(Y_0=j) \sim c_0 \, m^{-j}j^{-\alpha}$, $j\to \infty$, and $2<\alpha<4$, we have $\E(Y_0^3 m^{Y_0} {\bf 1}_{\{ 2\le Y_0 \le 3n\} }) \ge c_{37}\, n^{4-\alpha}$ for some constant $c_{37}>0$ and all sufficiently large $n$; this implies the upper bound in Theorem \ref{t:main}.\qed

\subsection{Lower bound in Theorem \ref{t:main}}

The proof of the lower bound in Theorem \ref{t:main} needs some preparation.

\medskip

\begin{lemma}
\label{l:pf_thm_1}

 There exists a constant $c_{38}>0$ such that for all sufficiently large integer $n$, say $n\ge n_0$, we have
 $$
 \hbox{either } \prod_{i\in (\frac{n}{2}, \, n] \cap \z} [\E(m^{Y_i})]^{m-1} \ge 8,
 \quad
 \hbox{or}
 \quad
 \prod_{i=0}^n [\E(m^{Y_i})]^{m-1} \ge c_{38} \, n^{\alpha-2}\, .
 $$

\end{lemma}

\medskip

\noindent {\it Proof.} Let $M_0\ge 1$ and $c_{33}\ge 1$ be the constants in \eqref{cor1_variant_stable} in Remark \ref{r:Z2upperQ}. Let $c_{39}:= [120(m-1)c_{33}]^{2/(\alpha-2)} \ge 120$. Let $n\ge \lceil 3 c_{39} \, M_0 \rceil =: n_0$ be an integer, and let $M = M(n):= \lfloor \frac{n}{c_{39}} \rfloor$. So $\frac{n}{2c_{39}} \le M \le \frac{n}{120}$. Let $u_n := m- \frac{c_{40}}{n}$, where $c_{40}:= 30m$. We can enlarge the value of $M_0$ if necessary to ensure that $u_n > \frac{m}{2}$.

We discuss on two possible situations, each leading to one of the inequalities stated in the lemma.

{\it First situation: $\E[(1-(m-1)Y_i)u_n^{Y_i}] < \frac12$ for all $i\in (\frac{n}{2}, \, n] \cap \z$.} In this situation, we have, for all $i\in (\frac{n}{2}, \, n] \cap \z$, $(m-1) \E (Y_i \, u_n^{Y_i}) \ge \E (u_n^{Y_i}) - \frac12 \ge \frac12$, thus
$$
\E (Y_i \, u_n^{Y_i}) \ge \frac{1}{2(m-1)} \, .
$$

\noindent Consider the function $s\mapsto f_i(s) := \E(s^{Y_i})$, $s\in [0, \, m]$. We have
$$
\E(m^{Y_i})
=
f_i(m)
\ge
f_i(u_n) + (m-u_n)f_i'(u_n)
\ge
1+(m-u_n)f_i'(u_n) .
$$

\noindent Since $m-u_n= \frac{30m}{n}$ and $f_i'(u_n) = \frac{1}{u_n}\, \E (Y_i \, u_n^{Y_i}) \ge \frac1m \, \frac{1}{2(m-1)} = \frac{1}{2m(m-1)}$, we get, for all $i\in (\frac{n}{2}, \, n] \cap \z$,
$$
\E(m^{Y_i}) \ge 1+ \frac{15}{(m-1)n} \, .
$$

\noindent Consequently,
$$
\prod_{i\in (\frac{n}{2}, \, n] \cap \z} [\E(m^{Y_i})]^{m-1}
\ge
\Big( 1+ \frac{15}{(m-1)n} \Big)^{\! (m-1)n/2}
\ge 8 \, .
$$

{\it Second (and last) situation: $\E[(1-(m-1)Y_\ell)u_n^{Y_\ell}] \ge \frac12$ for some $\ell = \ell(n) \in (\frac{n}{2}, \, n] \cap \z$.} We will be working with this particular $\ell$ in the rest of the proof. Let $Z_0 := Y_0 \, {\bf 1}_{\{ Y_0 \le M\} }$ as in \eqref{Z0}. Let $(Z_n, \, n\ge 0)$ be a Derrida--Retaux system whose initial distribution is given by $Z_0$.

Since $u_n \in [1, \, m]$, the function $x\mapsto (1-(m-1)x)u_n^x$ is decreasing on $[0, \, \infty)$, so
\begin{equation}
    \E[(1-(m-1)Z_\ell)u_n^{Z_\ell}]
    \ge
    \E[(1-(m-1)Y_\ell)u_n^{Y_\ell}]
    \ge
    \frac12\, .
    \label{phi(un)}
\end{equation}

\noindent Consider the function
$$
\varphi(s) := \E[(1-(m-1)Z_\ell)s^{Z_\ell}] ,
\qquad
s\ge 0 ,
$$

\noindent which is well-defined because $Z_\ell$ is bounded. We have just proved that $\varphi(u_n) \ge \frac12$. Let $v_n := m+ \frac{1}{2c_4 n}$, where $c_4 \ge 1$ is the constant in Theorem $\ref{t:H_nkupper}$. Since $v_n > m$, we have, by concavity of $\varphi(\cdot)$,
$$
\varphi(v_n) \ge \varphi(m) + (v_n-m) \varphi'(v_n) \, .
$$

\noindent By assumption, the system $(Y_n, \, n\ge 0)$ is critical, so $(Z_n, \, n\ge 0)$ is subcritical (or critical in case $Y_0 \le M$ a.s.), which implies that $(m-1) \E(Z_\ell \, m^{Z_\ell}) \le \E(m^{Z_\ell})$. This means that $\varphi(m) \ge 0$. On the other hand, $v_n-m = \frac{1}{2c_4 n}$, whereas
$$
\varphi'(v_n)
=
\E[(1-(m-1)Z_\ell)Z_\ell v_n^{Z_\ell-1}]
\ge
-(m-1) \E(Z_\ell^2 v_n^{Z_\ell-1}),
$$

\noindent which is $= - \frac{m-1}{v_n} \E(Z_\ell^2 v_n^{Z_\ell}) \ge - \frac{m-1}{m} \E(Z_\ell^2 v_n^{Z_\ell})$. Assembling these pieces together yields that
$$
\varphi(v_n) \ge - \frac{1}{2c_4 n} \, \frac{m-1}{m} \, \E(Z_\ell^2 v_n^{Z_\ell})\, .
$$

\noindent Since $v_n = m+ \frac{1}{2c_4 n} \le m+ \frac{1}{2c_4 \ell} = m+ \frac{1}{2c_4 (\ell\vee M)}$ (to obtain the last equality, we have used the fact that $\ell\ge \frac{n}{2} \ge M$), we are entitled to apply inequality \eqref{cor1_variant_stable} in Remark \ref{r:Z2upperQ} to see that
$$
\E(Z_\ell^2 v_n^{Z_\ell})
\le
c_{33} (\ell\vee M)^{(4-\alpha)/2} M^{(\alpha-2)/2} \, .
$$

\noindent Since $(\ell \vee M) = \ell \le n$ and $M\le \frac{n}{c_{39}}$, this yields
$$
\E(Z_\ell^2 v_n^{Z_\ell})
\le
c_{33} n^{(4-\alpha)/2} (\frac{n}{c_{39}})^{(\alpha-2)/2}
=
\frac{c_{33}}{c_{39}^{(\alpha-2)/2}} \, n\, .
$$

\noindent Accordingly,
$$
\varphi(v_n)
\ge
- \frac{1}{2c_4 n} \, \frac{m-1}{m} \, \frac{c_{33}}{c_{39}^{(\alpha-2)/2}} \, n
=
- \frac{c_{41}}{c_{39}^{(\alpha-2)/2}} \, ,
$$

\noindent where $c_{41} :=\frac{c_{33}}{2c_4} \, \frac{m-1}{m}$. On the other hand, we have $\varphi(u_n) \ge \frac12$ (see \eqref{phi(un)}). Since $m= \beta u_n + (1-\beta) v_n$ with $\beta := \frac{1}{1+2c_4c_{40}} \in (0, \, 1)$, it follows from concavity of $\varphi(\cdot)$ that
$$
\varphi(m)
\ge
\beta \, \varphi(u_n) + (1-\beta) \, \varphi(v_n)
\ge
\frac{\beta}{2} - (1-\beta) \frac{c_{41}}{c_{39}^{(\alpha-2)/2}}
=
\frac{1- \frac{4c_4c_{40}c_{41}}{c_{39}^{(\alpha-2)/2}}}{2(1+2c_4c_{40})} \, .
$$

\noindent Our choice of the constant $c_{39}$ ensures $\frac{4c_4c_{40}c_{41}}{c_{39}^{(\alpha-2)/2}} = \frac12$; hence $\varphi(m) \ge \frac{1}{4(1+2c_4c_{40})} =: c_{42}$, i.e.,
$$
\E(m^{Z_\ell})- (m-1)\E(Z_\ell\, m^{Z_\ell})
\ge
c_{42} \, .
$$

\noindent Recall from \eqref{recursion_Coll} that
$$
\prod_{i=0}^{\ell-1} [\E(m^{Z_i})]^{m-1}
=
\frac{\E(m^{Z_\ell})- (m-1)\E(Z_\ell\, m^{Z_\ell})}{\E(m^{Z_0})- (m-1) \E(Z_0\, m^{Z_0})} \, .
$$

\noindent On the right-hand side, the numerator is at least $c_{42}$, whereas the denominator has already appeared in \eqref{E(mZ0)}:
\begin{eqnarray*}
    \E(m^{Z_0})- (m-1) \E(Z_0\, m^{Z_0})
 &=& \E[((m-1)Y_0 -1)\, m^{Y_0}\, {\bf 1}_{\{ Y_0 > M\} }) + \P(Y_0 > M)
    \\
 &\le& (m-1) \E(Y_0 \, m^{Y_0}\, {\bf 1}_{\{ Y_0 > M\} }) \, .
\end{eqnarray*}

\noindent Our assumption $\P(Y_0=j) \sim c_0 \, m^{-j}j^{-\alpha}$ (for $j\to \infty$) yields that $\E(Y_0 \, m^{Y_0}\, {\bf 1}_{\{ Y_0 > M\} }) \le \frac{c_{43}}{M^{\alpha-2}}$ for some constant $c_{43}>0$ depending on the law of $Y_0$. As a consequence,
$$
\prod_{i=0}^{\ell-1} [\E(m^{Z_i})]^{m-1}
\ge
\frac{c_{42}}{c_{43}\, M^{-(\alpha-2)}}
=
c_{44} \, M^{\alpha-2} \, ,
$$

\noindent where $c_{44}:= \frac{c_{42}}{c_{43}}$. This implies that
$$
\prod_{i=0}^{n-1} [\E(m^{Y_i})]^{m-1}
\ge
\prod_{i=0}^{\ell-1} [\E(m^{Y_i})]^{m-1}
\ge
\prod_{i=0}^{\ell-1} [\E(m^{Z_i})]^{m-1}
\ge
c_{44}\, M^{\alpha-2}\, .
$$

\noindent Since $M\ge \frac{n}{2c_{39}}$, this completes the proof of the lemma.\qed

\bigskip

We have all the ingredients for the proof of the lower bound in Theorem \ref{t:main}.

\bigskip

\noindent {\it Proof of the lower bound in Theorem \ref{t:main}.} It follows quite easily from Lemma \ref{l:pf_thm_1}. Indeed, let $n_0$ be the integer in Lemma \ref{l:pf_thm_1}, and let $n\ge 2n_0$. According to Lemma \ref{l:pf_thm_1}, there are two possibilities:

$\bullet$ either we have $\prod_{i\in (n/2^{j+1}, \, n/2^j] \cap \z} [\E(m^{Y_i})]^{m-1} \ge 8$ for all $0\le j \le \lfloor \frac{\log (n/n_0)}{\log 2}\rfloor$, in which case we have
$$
\prod_{i=0}^n [\E(m^{Y_i})]^{m-1}
\ge
8^{\lfloor \frac{\log (n/n_0)}{\log 2}\rfloor}
\ge
8^{\frac{\log (n/n_0)}{\log 2}-1}
=
\frac{n^3}{8n_0^3}\, ,
$$

\noindent which yields the lower bound in Theorem \ref{t:main} because $3>\alpha-2$.

$\bullet$ or there exists an integer $0\le j^* = j^*(n) \le \lfloor \frac{\log (n/n_0)}{\log 2}\rfloor$ such that $\prod_{i=0}^{n/2^{j^*}} [\E(m^{Y_i})]^{m-1} \ge c_{38} \, (\frac{n}{2^{j^*}})^{\alpha-2}$ and that $\prod_{i\in (n/2^{j+1}, \, n/2^j] \cap \z} [\E(m^{Y_i})]^{m-1} \ge 8$ for all non-negative integer $j<j^*$ (if there is any); in this case, we have (recalling notation: $\prod_\varnothing := 1$)
\begin{eqnarray*}
    \prod_{i=0}^n [\E(m^{Y_i})]^{m-1}
 &\ge& \prod_{i=0}^{n/2^{j^*}} [\E(m^{Y_i})]^{m-1} \times \prod_{j=0}^{j^*-1} \prod_{i\in (n/2^{j+1}, \, n/2^j] \cap \z} [\E(m^{Y_i})]^{m-1}
    \\
 &\ge& c_{38} \, (\frac{n}{2^{j^*}})^{\alpha-2} \times 8^{j^*} ,
\end{eqnarray*}

\noindent which implies that
$$
\prod_{i=0}^n [\E(m^{Y_i})]^{m-1}
\ge
c_{38} \, n^{\alpha-2}\, 2^{(5-\alpha)j^*}
\ge
c_{38} \, n^{\alpha-2}\, .
$$

In both situations, the lower bound in Theorem \ref{t:main} is valid.\qed

\bigskip
\bigskip

\noindent {\bf Acknowledgements.} We are grateful to Victor Dagard, Bernard Derrida, Yueyun Hu and Mikhail Lifshits for stimulating discussions on various aspects of the Derrida--Retaux model throughout the last few years. We also wish to thank two anonymous referees whose comments have led to improvements in the presentation of the paper.

\bigskip
\bigskip


TITLE: Jump Diffusion Infinitesimal generator
QUESTION [3 upvotes]: I have this difussion process
$dX(t)=\mu X(t)dt+\sigma X(t)dW(t)+u X(t)  dN(t),\qquad X(0)=x > 0$
where $W(t)$ is a Brownian Motion and $N(t)$ is a Poisson process.
And I need to know the infinitesimal generator but I can't . Can someone help me?
$\mu,\sigma,u$ are constants.
Thank you so much

REPLY [0 votes]: I decided to post my answer because I always wanted to derive those things from scratch rather than using results from books I can hardly have access to.
The exponential Levy jump diffusion model is defined as follows. $X_t = x \cdot \exp(L_t) $ where the process $L_t$ reads:
\begin{equation}
L_t = \mu \cdot t + \sigma \cdot B_t  + \sum\limits_{i=1}^{N_t} Y_i
\end{equation}
Here $\mu \in {\mathbb R} $ and $\sigma > 0 $ are fixed numbers. Then $B_t = \sqrt{t} \cdot B_1 $ is a Brownian motion (in particular $B_1 = N(0,1)$), $N_t$ is a Poisson process and $(Y_i)_{i\ge 1}$ are iid random variables which are independent from $B_1$ and which have a common pdf $\omega_Y(y)$ such that  the expectation $E[\exp(m Y)]$ exists for every $m=0,1,2,\cdots$.
The process $X_t$ is used as a model for the stock prices see 1.
In order to find the infinitesimal generator we will firstly compute the characteristic function (the Fourier transform of the pdf) of the process in question. We have:
\begin{eqnarray}
&&\chi_{X_t}(k) := E\left[ e^{\imath k \cdot X_t}\right] = \\
&&\sum\limits_{n=0}^\infty
E\left[\exp\left(\imath k x \cdot e^{\mu t} \cdot e^{\sigma \sqrt{t} \cdot B_1} \cdot e^{\sum\limits_{i=1}^n Y_i}\right)\right] \cdot
\frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t}
\end{eqnarray}
where in the last line above we have conditioned on the number of jumps. Note that the expectation is taken with respect to both the Brownian motion part and the jumps.
Now let us compute the $m$th moment (where $m \in {\mathbb N}$) of the process in question. We have:
\begin{eqnarray}
&&E\left[ (X_t)^m \right] = \left. \frac{d^m}{d (\imath k)^m} \chi_{X_t}(k) \right|_{k=0} \\
&& x^m e^{m \mu t} \sum\limits_{n=0}^\infty
E \left[ e^{m \sigma \sqrt{t} \cdot B_1} \cdot e^{m \sum\limits_{i=1}^n Y_i} \right]\cdot \frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t} \\
&&x^m e^{m \mu t} \sum\limits_{n=0}^\infty
e^{m^2 \sigma^2 t/2} \cdot (E\left[ e^{m Y}\right])^n \cdot \frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t} \\
&& x^m e^{m \mu t} e^{m^2 \sigma^2 t/2} \sum\limits_{n=0}^\infty
\frac{(\lambda t\cdot E\left[ e^{m Y}\right])^n}{n!} \cdot e^{-\lambda \cdot t} \\
&& x^m e^{m \mu t} e^{m^2 \sigma^2 t/2} \cdot e^{\lambda t \left[ E\left[ e^{m Y}\right] - 1\right]}
\end{eqnarray}
As we can see all the integer moments exist.
Now, we are ready to compute the infinitesimal generator ${\mathbb L}_X $. We take a smooth function $F(x)$ and we have:
\begin{eqnarray}
&&{\mathbb L}_X F(x) := \lim\limits_{t \rightarrow 0} \frac{E\left[ F(X_t)\right] - F(x)}{t} = \\
&& \sum\limits_{m=0}^\infty  \frac{F^{(m)}(0)}{m!} \cdot \lim\limits_{t \rightarrow 0} \frac{E\left[ (X_t)^m \right]-x^m}{t}  = \\
&& \sum\limits_{m=0}^\infty  \frac{F^{(m)}(0)\cdot x^m}{m!} \cdot \lim\limits_{t \rightarrow 0} \frac{e^{m \mu t} e^{m^2 \sigma^2 t/2} \cdot e^{\lambda t \left[ E\left[ e^{m Y}\right] - 1\right]}-1}{t} \\
&& \sum\limits_{m=0}^\infty  \frac{F^{(m)}(0)\cdot x^m}{m!} \cdot
\left(m \mu  + m^2 \frac{\sigma^2 }{2} + \lambda \left[ E\left[ e^{m Y}\right] - 1\right] \right) = \\
&&\left(\mu x \frac{d}{d x} + \frac{\sigma^2 }{2} x \frac{d}{d x} x \frac{d}{d x} \right) F(x) +
\lambda \int\limits_{{\mathbb R}} \left[ F(x e^{\xi}) - F(x)\right] \cdot \omega_Y(\xi) d \xi \\
&& \left( (\mu + \frac{\sigma^2}{2}) x \frac{d}{d x} +  \frac{\sigma^2 x^2}{2} \frac{d^2}{d x^2}\right) F(x) +
\lambda \int\limits_{{\mathbb R}} \left[ F(x e^{\xi}) - F(x)\right] \cdot \omega_Y(\xi) d \xi
\end{eqnarray}
Now if we go to the body of the question and take $\omega_Y(y) = \delta(y- u) $ then we get ${\mathbb L}_X F(x) = \left( (\mu + \frac{\sigma^2}{2}) x \frac{d}{d x} +  \frac{\sigma^2 x^2}{2} \frac{d^2}{d x^2}\right) F(x) + \lambda \left[ F(x e^{u}) - F(x)\right]$.
Unfortunately this is different from the answer given by Sean Meyn above. There are two differences, firstly the Ito correction, i.e. the coefficient at the first derivative and secondly the term in square brackets on the very right, which has $e^u$ instead of $1+u$.
1: Ernst Eberlein, Jump-type Levy processes


TITLE: Reccurent sequence limit
QUESTION [0 upvotes]: $c>=0$
$a_1 = c$
$a_{n+1}=1+\frac{a_n}{a_n + 1}$
How can I calculate $\lim_{n\to\infty}a_n$ in dependency on $c$?

REPLY [2 votes]: Note that $a_{n+1}=2-\frac{1}{a_n+1}$. Thus for every integer $n \ge 2$, $1<a_n<2$.
Also $a_{n+1} \ge a_n$ if and only if $\frac{1-\sqrt{5}}{2}\le a_n\le \frac{1+\sqrt{5}}{2}$.
If $c\le\frac{1+\sqrt{5}}{2}$, $a_{n+1}=1+\frac{a_n}{a_n+1}\ge a_n$ is true, so by the Monotone Sequence Theorem the limit of the sequence exists.
If $c>\frac{1+\sqrt{5}}{2}$, it means that $a_{n+1}<a_n$, so the given sequence decreases. By the Monotone Sequence Theorem, the limit of the sequence exists.
Now let the limit $L$, then $L=1+\frac{L}{L+1}$ gives $L=\frac{1+\sqrt{5}}{2}$ regardless of the choice of $c$.


TITLE: Show that $\mathbb{R}^{\infty}$ can be written as a direct sum of some subgroups
QUESTION [0 upvotes]: This is Example 1 in Munkres' section 67. Consider the cartesian product $\mathbb{R}^{\omega}$ being an abelian group under the operation of coordinate-wise addition.
He constructed subgroups of $\mathbb{R}^{\omega}$ whose direct sum is $\mathbb{R}^{\infty}$, in the following way:

Consider the set $G_{n}$ consisting of those tuples $(x_{i})$ such that $x_{i}=0$ for all $i\neq n$. It is really quick to see that $G_{n}$ is a group, and it is isomorphic to $\mathbb{R}$, and thus subgroups of $\mathbb{R}^{\omega}$. Then $G_{n}$ generate the subgroup $\mathbb{R}^{\infty}$ of $\mathbb{R}^{\omega}$; indeed, $\mathbb{R}^{\infty}$ is their direct sum.

I understand $G_{n}\cong\mathbb{R}$ and thus subgroups of $\mathbb{R}^{\omega}$.
However, how could $G_{n}$ generate $\mathbb{R}^{\infty}$? Note that for each $x\in\mathbb{R}^{\infty}$, it is an $\infty$ tuples $$x=(x_{1}, x_{2}, x_{3},\cdots),$$ which is the sum of $$(x_{1},0,0,\cdots)+(0,x_{2},0,\cdots)+\cdots$$ and this sum is the sum from the element of each $G_{n}$. By then definition given by Munkres, groups $G_{\alpha}$ generates $G$ if every element $x\in G$ can be written as a finite sum of elements of the groups $G_{\alpha}$, but the sum above is not even finite sum.
Is there other way to express $x\in\mathbb{R}^{\omega}$ as a finite sum from the element of $G_{n}$? Why does the alternative way give a unique summation?
Thank you!

REPLY [1 votes]: $\mathbb{R}^{\infty}$ is defined in Exercise 7 in Section 19, as the set of sequences with only finitely many nonzero terms. So if
$$
x = (x_1, x_2, x_3, \dots)
$$
is in $\mathbb{R}^{\infty}$, then it equals
$$
x = (x_1, x_2, x_3, \dots, x_n, 0, 0, 0, \dots)
$$
for some $n$, and so is a finite sum of elements from the subgroups $G_i$.


TITLE: Use Discrete Markov Chain to predict n steps ahead
QUESTION [1 upvotes]: I modeled a Markov Chain with M states. Assuming that the process is homogeneous in time. But, each state has a differente resident time. Moreover, each state has a self-loop transition and a transition to other state. I assume that transition happens in one second. Then, I have built the transition matriz by count processes. In the transition matrix, the probability to keep in the same state is  98%. Using the steady state probabilities formula $\pi^{n} = \pi P^{n}$, I try to find what will be the probability to be in the next state for N steps ahead.

REPLY [1 votes]: I suspect the confusion may be in what you mean by "step" in the phrase "$N$ steps ahead".
If you mean the one-second time step of your discrete time Markov process, then what you're calculating is exactly correct — with a $0.98$ self-loop transition probability, the process has a less than $0.5$ probability of moving anywhere within $N < 35$ steps.
If, however, by "step" you mean a transition in which the process does not stay in the same state, then you need to eliminate the self-loop transitions from your matrix (since you don't count them as steps) by setting their probability to $0$ and normalizing the remaining transition probabilities for each state so that they again sum to $1$.
Of course, what you lose by doing so is the information about the residence time in each state.  If you wanted to know where the process will be after $N$ non-self-loop steps and how long it'll take to get there, that would require a rather more complicated calculation, and the resulting discrete phase-type distribution would, in general, be unlikely to have a simple form (although it might be possible to approximate it with some nicer distributions).

In the comments, you write that your transition matrix is (approximately) $$P=\pmatrix{0.99&0.0208&0&0\\0.0019&0.9792&0.0104&0\\0&0&0.9896&0.0108\\0.0028&0&0&0.9907}$$ and that "second option it is what I'm looking for", i.e. that you want to calculate the probability distribution after $N$ steps, a step denoting a transition between two distinct states.  If so, what you need to do is to calculate another stochastic matrix $P'$ which describes the same process, but which does not include the self-loop transitions (i.e. whose diagonal entries are zero).
To obtain $P'$ from $P$, simply zero out the diagonal entries and re-normalize each row so that the matrix stays stochastic (i.e. so that each row sums to $1$).  Since most of the rows in $P$ only have one non-zero off-diagonal element to begin with, that element becomes $1$, leaving row 2 as the only non-trivial case: $$P' = \pmatrix{0&1&0&0\\0.1545&0&0.8455&0\\0&0&0&1\\1&0&0&0}$$  From this, you can calculate the state distribution $\pi_n = \pi P'^n$ after $n$ steps, given a starting distribution $\pi$.
However, note that, whereas the original transition matrix $P$ was ergodic, $P'$ isn't — in particular, a process in an even-numbered state always ends up in an odd-numbered state after one step and vice versa, so all states have an even period.  This lack of ergodicity doesn't really affect the calculation of $\pi_n$ as such, but it does mean e.g. that $\pi_n$ won't generally converge to a stationary distribution as $n \to \infty$.