



\documentclass[a4paper,10pt]{article}

\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage[geometry]{ifsym}
%\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
%\usepackage[finnish]{babel}
\usepackage[english]{babel}
\usepackage{graphicx}
\usepackage{verbatim}
\begin{document}

\newtheorem{lemma}{Lemma}
\newtheorem{thm}{Theorem}
\newtheorem{coro}{Corollary}
\newtheorem{example}{Example}
\newtheorem{ex}{Exercise}
\newtheorem{definition}{Definition}
\newcommand{\indep}{\perp\!\!\perp}
 \newcommand{\R}{\mathbb{R}} 
 \newcommand{\Q}{\mathbb{Q}} 
 \newcommand{\N}{\mathbb{N}} 
\newcommand{\F}{\mathbb{F}} 
 \newcommand{\C}{\mathbb{C}} 
\newcommand{\Z}{\mathbb{Z}} 
 \newcommand{\PP}{{\mathbf {P}}} 
\newcommand{\Tau}{{\mathcal {T}}} 

\date{\today}











%opening
%\title{}
%\author{}

{\bf Stochastic analysis, spring  2013, Exercises-6, 28.02.2013 }


A branching process  $(Z_t)_{t\in\N }$ with integer values,
represents the size of 
a population evolving randomly in discrete time.

We start with  $Z_0(\omega)=1$ individual  at time $t=0$.

Inductively each of the $Z_{t-1}(\omega)$ individuals in the $(t-1)$ generation
has a random number of offspring  $Y_{i,t}$. 
These  offspring numbers are independent and identically distributed  
with law $\pi=(\pi(n): n=0,1, \dots )$,

$\pi(n)=P\bigl( Y=n  \bigr)$,   $Y=Y_{1,1}$.

The size of the new generation at time $t$ is then
\begin{eqnarray*} 
Z_t(\omega) = \sum_{i=1}^{Z_{t-1}(\omega)}  Y_{i,t}(\omega)
\end{eqnarray*}

We assume 
that the mean offspring number is finite
\begin{eqnarray*} 
\mu=E_\pi(Y)= \sum_{n=0}^{\infty} n \pi(n) < \infty 
\end{eqnarray*}






Note that if $Z_{t} (\omega)= 0$, then $Z_{u}(\omega)= 0 $ $\forall  u \ge t$.
In this case we say that the process is extinct.
Clearly $P( Z_t = 0) \le P( Z_u  =0)$ for $t\le u$.
\\

Note also that $P( Y=0) > 0$ implies  $P( Z_t= 0 ) > 0  $, $\forall t \ge 1 $.
\\












 Consider the filtration $\F=( {\mathcal F}_t : t\in \N)$
with ${\mathcal F}_t= \sigma(Z_s : 0 \le s \le t)$.

Actually we could consider the larger filtration
 $\F'=( {\mathcal F}_t' : t\in \N)$
with 
\begin{align*}
{\mathcal F}_t'= \sigma\bigl(Z_0, Y_{s,i} {\bf 1}( Z_{s-1} \ge i ) : 0 \le s \le t, \; i\in \N\bigr) .
\end{align*}
or  $\F''=( {\mathcal F}_t'' : t\in \N)$
with 
\begin{align*}
{\mathcal F}_t''= \sigma\bigl(Z_0, Y_{s,i}  : 0 \le s \le t, \; i\in \N\bigr) .
\end{align*}
Although ${\mathcal F}_t \subset {\mathcal F}_t' \subset {\mathcal F}_t''$, 
the martingale properties we use in this exercise  for all these
filtrations.

\begin{enumerate}


 \item 
Show that $ Z_t(\omega)$ is a $\F$-martingale,
(respectively supermartingale, submartingale )
when $\mu=1$ (respectively  $0\le \mu< 1$, $1<\mu <\infty$,
in the filtration generated by the process $Z$ itself.

\item For $\mu\ne 1 $, write the Doob decomposition
of  the supermartingale  (respectively martingale) $Z_t$  
as sum of a martingale  and a non-increasing (respectively non-decreasing ) $\F$-predictable process,
and compute the mean $E(Z_t)$ for $t\in \N$.







\item Assume that $\mu \le 1 $, and that the offspring distribution is non-trivial,
meaning that $0 \le \pi( Y = 1) < 1 $.  The case $P( Y=1 )=1$ is trivial, nothing  happens, the size of the population
is constant.

 
Show that when $\mu \le 1 $ (subcritical and critical cases)
 \begin{eqnarray*} 
 \lim_{t\to\infty} Z_t(\omega) = 0 \quad \mbox{ $P$ a.s. } 
\end{eqnarray*}

Hint: first show that a finite limit $Z_{\infty}(\omega)$ exists $P$ a.s.  with $E( Z_{\infty} ) <\infty$.  
Use the indepdence  of $Y_{1,1}$ from $( Y_{t,i} : t\ge 2, i \in \N)$ to prove
\begin{eqnarray*}
P( Z_{\infty}= 0 | Z_1 = n ) = P( Z_{\infty}= 0)^n   
\end{eqnarray*} 
where $P(Z_{\infty} =0)$ is the probability that the descendance of a  single
individual becomes extinct.
\\


By computing  first the conditional probability $P( Z_{\infty}=0 | \sigma(Z_1) )(\omega)$ and 
taking  expectation, show that the unknown $q= P( Z_{\infty}=0)$ 
satisfies the equation
\begin{eqnarray*}
 q= E_P( q^Y ), \quad  q \in  [\pi(0) ,1]
\end{eqnarray*}
 where $P( Y= n) = \pi(n)$ is the offspring distribution.
\\

Note that since $\mu =E(Y)\le  1 $   and $\pi(1)=P( Y = 1 )< 1 $, necessarily
$\pi(0)=P(Y=0)>0$, and $P(Z_{\infty}=0) \ge P(Y=0)>0$.
Therefore  $q=0$ is  not a solution.
\\


$q=1$ is also a solution. We show that there are no other solutions.

Hint take the
the  derivative
\begin{eqnarray*}
 \frac{ d }{ dq } E_P( q^Y )
\end{eqnarray*}
 checking that it is allowed to  take a derivative inside the expectation, and show that
$q<1$  is in contradiction with $E_P( q^Y)=q$.


\item
In the critical case $\mu=1$, show that the martingale
 $(Z_t: t\in \N)$ is not uniformly integrable



\item Next we work with the supercritical case, with $\mu=E_P( Y) \in (1,\infty) $.

\item
Show that
\begin{eqnarray*}
 W_t = Z_t (\omega) \mu^{-t} 
\end{eqnarray*}
is a martingale.


\item Show that $P$ almost surely $\lim_{t\to  \infty} W_t \to W_{\infty}$ with $W_{\infty}\in L^1(P)$.


\item The next result is a theorem from  Kesten and Stigum (1966)
which states that $W_t $ is an uniformly integrable martingale if and only if the offspring distribution
satisfies
\begin{align*}
  E_P( Y \log(Y) ) = 0
\end{align*}
where it is understood that $0 \log(0)=\lim_{x\downarrow 0} x\log(x)=0$.


Write the increments:

\begin{align*}
W_t -W_{t-1} =   \frac 1 { \mu^t } \sum_{i=1}^{Z_{t-i} } (Y_{t,i} -\mu)
\end{align*}

and truncate them in the following way: for
\begin{align*} &
\widetilde W_t =   \frac 1 { \mu^t } \sum_{i=1}^{Z_{t-i} }  Y_{t,i}{ \bf 1 }( Y_{t,i} \le \mu^t ), & \\ &
 R_t = \frac 1 { \mu^t } \sum_{i=1}^{Z_{t-i} } E\biggl(   Y{ \bf 1 }( Y > \mu^t ) \biggr)
\end{align*}
We decompose
\begin{align*} &
 W_t -W_{t-1} = \underbrace{ \biggl(  W_t -  \widetilde W_t - R_t \biggr) }_{\text{I}}+
\underbrace{\biggl( \widetilde W_t  + R_t - W_{t-1} \biggr) }_{\text{II}} =
& \\ &
 \frac 1 {\mu^t} 
\sum_{i=1}^{Z_{t-1} } \biggl\{
  Y_{t,i} {\bf 1} ( Y_{t,i} >  \mu^t )  - E\bigl ( Y {\bf 1}(Y > \mu_t) \bigr) \biggr\}
+ \frac 1 {\mu^t} \sum_{i=1}^{Z_{t-1 }}  
\biggl\{ Y_{t,i} {\bf 1}( Y_{t,i} \le \mu^t) - E\bigl ( Y {\bf 1}(Y \le \mu_t) \bigr)
 \biggr\}&\end{align*}
where (I) and (II) are martingale differences.

\item
Show that 
\begin{align*} &
E\biggl( \biggl\{  \sum_{t=1}^{\infty} 
\frac 1 {\mu^t} \sum_{i=1}^{Z_{t-1 }} 
 \biggl( Y_{t,i} {\bf 1}( Y_{t,i} \le \mu^t) - E\bigl ( Y {\bf 1}(Y \le \mu^t) \bigr) \biggr) \biggr\}^2 \biggr)
< \infty
\end{align*}
Therefore  by summing the increments (I), we obtain a martingale bounded in $L^2(P)$ which is
also uniformly integrable.

\item
Show also that, when $1<E(Y)<\infty$, without any additional assumptions
\begin{align*}
  \sum_{t=1}^{\infty } P( \widetilde W_t \ne W_t )< \infty
\end{align*}
and by the Borel Cantelli lemma, with probability one $\widetilde W_t \ne W_t$  only for  finitely many $t$.

\item
Show that  the series
\begin{align*}
  \sum_{t=1}^{\infty } \mu^{-t} \sum_{i=1}^{Z_{t-1} }  \biggl\{ 
Y_{t,i} {\bf 1}( Y_{t,i} > \mu^t ) - E_P\bigl( Y {\bf 1}( Y > \mu^t) \bigr) \biggr\}
\end{align*}
converges in $L^1(P)$ if and only if $E_P( Y \log Y)< \infty$.

Hint: it is enough to show that 
\begin{align*}
  \sum_{t=1}^{\infty } \mu^{-t}  E_P\biggl( \sum_{i=1}^{Z_{t-1} }  \biggl\{ 
Y_{t,i} {\bf 1}( Y_{t,i} > \mu^t )  \biggr) < \infty
\end{align*}



\item Show that when $1<E(Y)<\infty$,  $W_t$ is uniformly integrable if and only if
$E_P( Y \log Y) < \infty$.


\end{enumerate}

\end{document}
