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{\bf Stochastic analysis, spring  2013, Exercises-7, 14.03.2013 }


\begin{enumerate}


\item
   Suppose we have an urn which contains at time $t=0$
 two balls, one black and one white. 
At each time $t\in N  $  we draw uniformly at random from the urn one 
ball, and we put it back together with a new ball of the
same colour.  \\

We introduce the random variables  

$X_t(\omega) = {\bf 1}\bigl\{
 \mbox { the ball drawn at time $t$ is black }\bigr\} $

and denote $S_t=(1+ X_1 + \dots + X_t)$,  

$M_t= S_t/(t+2) $, the proportion of black
 balls in the urn. 

We use the filtration $\{ {\mathcal F}_n \}$ with  ${\mathcal F}_n=\sigma\{  X_s : 
\; s\in \N, \; s\le t \}$.
\\


i) Compute the Doob decomposition of $(S_t)$, 
 $S_t= S_0 +N_t+ A_t$, where $(N_t)$
 is a martingale and $(A_t)$ is predictable.
\\

ii) Show that $(M_t)$ is a martingale and find the representation of
$(M_t)$ as a martingale transform $M_t= ( C\cdot N)_t$, where
  $(N_t)$ is the martingale part of $(S_t)$ and $(C_t)$ is predictable.
\\


iv) Note that the  martingale $(M_t)_{t\ge 0}$ is uniformly integrable (Why ?).
 Show that $P$ a.s. and in $L^1$ exists
$ M_{\infty}=\lim_{t\to\infty} M_t$ . Compute $E(M_{\infty})$.
\\

v) Show that $P( 0 < M_{\infty} < 1)>0$ .   \\

Since
 $M_{\infty}(\omega)\in [0,1]$, it is enough to show that $0<E( M_{\infty}^2 )<  E( M_{\infty})$ with
strict inequalities.
\\

Hint: 
compute the Doob decomposition of the submartingale $( M_{t}^2 )$,
and than take expectations before going to the limit to find the
value of $E( M_{\infty}^2 )$.




\item
Consider an i.i.d. random sequence
 $( U_t: t\in \N)$ with uniform distribution on $[0,1]$, $P( U_1 \in dx)= {\bf 1}_{[0,1]}(x) dx $.
Note that $E_P( U_t)= 1/2$.

Consider also the
random variable  $-\log(U_1(\omega))$ which is $1$-exponential w.r.t.  $P$.
\begin{eqnarray*}
 P(  -\log(U_1)  >  x )=\left\{
% use packages: array
\begin{array}{ll} \exp(-x)  
 & \mbox{ kun } x \ge 0   \\ 
 1 & \mbox{ kun  } x < 0 
                        \end{array} \right.
\end{eqnarray*}
$-\log(U_1)\in L^1(P)$ with $E_P(-\log( U_1))=1$.


\begin{enumerate}  

\item
Let $Z_0=1$, and
\begin{eqnarray*}
 Z_t(\omega)= 2^t \prod_{s=1}^t U_s(\omega)
\end{eqnarray*}
Show that  $(Z_t)$ is a  martingale in the filtration $\F=({\mathcal F}_t:t\in \N)$, 
with ${\mathcal F}_t= \sigma(Z_1,Z_2,\dots, Z_t)= \sigma(U_1,U_2, \dots, U_t)$.

\item Show that $E_P( Z_t)=1$.


\item 
Show that  the limit  $Z_{\infty}(\omega) =\lim_{t\to\infty} Z_t(\omega)$ exists $P$ almost surely.


\item Show that
\begin{eqnarray*}
 Z_{\infty}(\omega)=0 \quad \mbox{ $P$-a.s. }
\end{eqnarray*}
{\bf Hint} Compute first  the $P$-a.s. limit
\begin{eqnarray*}
 \lim_{t\to\infty} \frac 1 t \log( Z_t(\omega))
\end{eqnarray*}
(remember Kolmogorov's strong law of large numbers!).



\item Show that  the  martingale $(Z_t(\omega):t\in \N)$ is not uniformly integrable.

\item Show that  $\log(Z_t(\omega))$ is a supermartingale, does it satisfy the assumptions of   Doob's martingale
convergence theorem ?


\item At every time $t\in \N$, define the probability measure
\begin{eqnarray*}
 Q_t ( A ):= E_P( Z_t {\bf 1}_A) \quad \quad \forall A \in {\mathcal F}_t
\end{eqnarray*}
on the probability space $(\Omega,{\mathcal F})$.


Show that the random variables $(U_1,\dots,U_t)$ are i.i.d. also under $Q_t$, compute their
probability density under $Q_t$.

Hint: use Doob's maximal inequality.

\end{enumerate}







\item
( Paley's and Littlewood's  maximal function)
Consider a function in $f(x)\in L^1( \R^d ,{\mathcal B}(\R^d), dx )$.


Define  the $\sigma$-algebra
\begin{align*}
 {\mathcal F}_k = \sigma\bigl\{   Q_{k,z}=  \bigl(z 2^{-k} , (z+1) 2^{-k} \bigr] ,  \; z \in \Z^d \bigr\} \subseteq{\mathcal B}(\R^d), \quad k\in \Z
\end{align*}
and the two sided filtration $\F=( {\mathcal F}_k :k \in \Z)$
where the dyadic cubes $(Q_{k,z}: z\in \Z^d)$ form a partition of $\R^d$,
and the functions
\begin{eqnarray*}
   f_k( x)=  \sum_{z\in \Z^d } {\bf 1}(  x \in Q_{k,z} ) \frac 1 {  | Q_{k,z} | } \int_{Q_{k,z} } f(y) dy
\end{eqnarray*}
where for $k\in \Z$, $|Q_{k,z}| = 2^{-kd}$ is the Lebesgue measure of the $d$-dimensional dyadic cube

Show that $f_k(x)$ is an $\F$-martingale w.r.t. Lebesgue measure. 
Note that the definition of conditional expectation  martingales extends directly to
the case where we integrate with respect to $\sigma$-finite positive measures.
To work with a probability measure,
 we could take instead with  $f(x)\in L^1( [0,1]^d ,{\mathcal B}([0,1]^d), dx )$.

Show that $\lim_{k\to -\infty}  f_k(x) = 0$  both almost surely w.r.t. the Lebesgue measure but not $L^1(\R^d)$-sense.



Define the {\it  maximal function }
\begin{eqnarray*}
                   f^{\Box}(x):= \sup_{k\in \Z} f_k(x)
\end{eqnarray*}

Show that for $1<p<\infty$
\begin{eqnarray*}
   \parallel  f^{\Box}(x) \parallel_{L^p(\R^d) } \le \frac p {p-1} \sup_{k\in \Z}   \parallel  f_k \parallel_{L^p(\R^d) } 
\le \frac p {p-1}\parallel  f \parallel_{L^p(\R^d) } 
\end{eqnarray*}
and 
\begin{eqnarray*}
 c  P(  |f^{\Box}(x)| > c)  \le \sup_{k\in \Z}   \parallel  f_k\parallel_{L^1(\R^d) } \le \parallel f \parallel_{L^1(\R^d)} 
\end{eqnarray*}










\end{enumerate}

\end{document}
