



\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{\G}{\mathbb{G}} 
 \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-8, 21.03.2013 }


\begin{enumerate}
\item
Let $\tau(\omega) \in [0,+\infty]$ be a random time, $F(t)=P(\tau \le t)$ for $t\in [0, \infty)$.


Consider the single jump {\it  counting process } ${\bf 1}(\tau(\omega) \le t)$ which generates the filtration $\F=({\mathcal F}_t)$
with ${\mathcal F}_t^N = \sigma( N_s: s\le t)$.

\begin{enumerate}

\item Show that $\tau$ is a stopping time in the filtration $\F$.



\item
Show that first that   for every Borel function $f(x)$,
the random variable
  \begin{align*}
   f(\tau(\omega)) {\bf 1}( \tau(\omega) \le s )
  \end{align*}
 is $ {\mathcal F}_s$-measurable.


\item


Define the  {\it cumulative hazard function }
\begin{align*}
 \Lambda(t) = \int_0^t \frac 1 { 1-F(s-) } F(ds) 
\end{align*}
where $F(s-)=P( \tau < s)$ denotes the limit from the left.



Show that
 \begin{align*}
  M_t=  N_t - \Lambda_{t\wedge \tau}
 \end{align*}
is a  an $\F$-martingale. 

Hint:  use the definition, and show that for $s\le t$  and  every $A \in {\mathcal F}_s$
\begin{align*}
 E_P \biggl(   (N_t - N_s ) {\bf 1}_A \biggr) = E_P \biggl(   (\Lambda_{t\wedge \tau}  - \Lambda_{s\wedge\tau}  ) {\bf 1}_A \biggr) 
\end{align*}

It turns out that it is enough to do the computation for $A= \{ \omega: \tau(\omega)>s  \}$    (why ?).
Fubini's theorem may be also useful.



\item   Assume that $t\mapsto F(t)$ and therefore also $t\mapsto \Lambda(t)$ are continuous, which means
$P( \tau= t)=0$ $\forall t \in \R^+$.

Show that $\Lambda_{\tau}$ has 1-exponential distribution:

\begin{align*}
   P\bigl(   \Lambda_{\tau } > x ) = \exp( -x),  \quad   x \ge 0
\end{align*}
 
Hint: one line of proof compute the Laplace transform 
\begin{align*}
{\mathcal L}(\theta):= E_P\biggl( \exp( - \theta \Lambda_{\tau} \bigr) \biggr)   \quad \theta > 0
\end{align*}
and compare it with the Laplace transform of the 1-exponential distribution.

\item Show that the martingale $M_t$ is uniformly integrable, what is $M_{\infty}?$.

\end{enumerate}


\item Let $(M_t:t\in \R^+)$ a $\F$-martingale, and $\G$ a filtration with ${\mathcal G}_t \subseteq {\mathcal F}_t$
We assume that $(M_t)$ is  also $\G$-adapted. Show that $(M_t)$ is  a martingale in the smaller filtration $\G$.

\item Let $(M_t:t\in \R)$ a $F$-martingale under $P$, and ${\mathcal G}_t$ a filtration such that $\forall t \ge 0$,
 the $\sigma$-algebrae ${\mathcal G}_{t} $ and $\sigma( M_s:s\le t)$
are $P$-independent.


Show that under $P$, $(M_t:t\in \R^+)$ is  a martingale in the enlarged filtration $( {\mathcal F}_t \vee {\mathcal G}_t :  t\ge 0 )$.
\\

 Let $(B_t:t\ge 0)$ a Brownian motion in the filtration $\F$,
which means
\begin{itemize}
 \item $B_0(\omega) = 0$
\item $ t\mapsto B_t(\omega)$ is  continuous  
 \item $ \forall 0 \le s \le t $ , $(B_t-B_s) $ is $P$-independent from ${\mathcal F}_s$, conditionally gaussian
with conditional mean $E( B_t-B_s| {\mathcal F}_s)=0$ and conditional variance  $E( (B_t-B_s)^2| {\mathcal F}_s)=t-s$
\end{itemize}

\item
Show that for $a>0$ the process
 $ (  a^{-1/2}  B_{at}:t \in \R^+)$ is also a Brownian motion.

\item The process $W_0=0$, $W_t= t B_{1/t}$ is also a Brownian motion.


\item
Let $\theta \in \R$,  and $i= \sqrt{-1}$ the imaginary unit

Show that
\begin{align*}
  E_P\bigl( \exp( i \theta B_t ) \bigr ) = \exp\bigl( -\frac 1 2 \theta^2 t  \bigr)
\end{align*}

Hint: Use complex integration over the rectangular contour   delimited by in the complex plane
by the points  $R , (R+ i \theta), (-R+ i \theta), -R$  with $R\in \R$ and let $R\to \infty$.




\item  For $\theta\in \R$,
consider now  
\begin{align*}
M_t = \exp\bigl( i \theta B_t + \frac 1 2 \theta^2 t \bigr) = \biggl\{
  \exp\bigl( \frac 1 2 \theta^2 t \bigr) \cos( \theta B_t) 
+ \sqrt{-1}  \exp\bigl( \frac 1 2 \theta^2 t \bigr) \sin( \theta B_t) \biggr\}\;
 \in \C
\end{align*}
where $i= \sqrt{-1}$ is the imaginary unit.

Recall that $E( \exp(  i \theta G) )= \exp( - \theta^2  \sigma^2/2 )$  when $G (\omega) \sim{\mathcal N}(0,\sigma^2)$.
\begin{itemize}

\item
Show that $M_t$ is complex valued $\F$-martingale, which means that real and imaginary
parts are $\F$-martingales.

\item 
Show that $\lim_{t\to\infty} |M_t(\omega)|= \infty$
\end{itemize}









\end{enumerate}

\end{document}
