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{\bf Stochastic analysis, spring  2013, Exercises-4, 14.02.2013 }
\begin{enumerate}






\item Let $\tau_1(\omega)$ and $\tau_2(\omega)$ stopping times with respect to the filtration $\F= ( {\mathcal F}_t:t\in T)$ taking values in $T$  .
Here $T$ could be either  $\R^+$ or $\N$.
\\

Use the definition of stopping time to 
show that $\sigma(\omega)= \min( \tau_1(\omega), \tau_2(\omega) )$ is a $\F$-stopping time. 

\item Let $(M_t:t\in \R^+)$ a $\F$-martingale, and $\tau$ a $\F$-stopping time. 

Show that the stopped process $( M_{t\wedge \tau}: t\in \R^+)$
\begin{eqnarray*}
  M^{\tau}_t(\omega)= M_{t\wedge \tau}(\omega) =  
M_t(\omega) {\bf 1}( t\le \tau(\omega) ) + M_{\tau(\omega)}(\omega) {\bf 1}( t > \tau(\omega))
\end{eqnarray*}
is a $\F$-martingale.

We have shown this when $T=\N$ is discrete by using the martingale transform. 
In continuous time we have not yet defined such martingale transforms. Prove the
statement directly  by using the definitions.




\item 
 Let $(M_t(\omega) )_{t\in T}$ a martingale
with respect to the filtration $\F= ( {\mathcal F}_t)$
 with $M_0(\omega)=0$.  Here $T$ could be either $\R^+$ or $\N$.

Define the family of  random times $\tau_x : x\in \R$
\begin{eqnarray*}
 \tau_x (\omega) =\left\{ \begin{matrix}  \inf \{  s: M_s \ge x \} & \mbox{ for } x \ge 0  \\
                           \inf \{  s: M_s \le x \} & \mbox{ for } x <  0
                          \end{matrix} \right.
\end{eqnarray*}
  Show that $\tau_x$ is a stopping time.
 




\item Let
\begin{eqnarray*}
 M_t(\omega)= \sum_{s=1}^t X_s(\omega)
\end{eqnarray*}
is a binary random walk where $t\in \N$ and
$(X_s: s\in \N) $ are i.i.d. random variables with
\begin{eqnarray*}
   P( X_s=\pm 1 ) = P( X_s= \pm 1 | {\mathcal F}_{s-1}   )=   1/2
\end{eqnarray*}
$X_s$ is ${\mathcal F}_s$ measurable and $P$-independent from ${\mathcal F}_{s-1}$.

\begin{itemize}
\item
Show that $(M_t)_{t\in \N}$ and  $(M_t^2-t)_{t\in \N}$ are   $\F$-martingales.

\item
Consider the stopping time $\sigma(\omega)=\min( \tau_a, \tau_b )$ where $a < 0 < b  \in \N $,
and the stopped martingales $( M_{t\wedge \sigma})_{ t\in \N}$ and  $( M_{t\wedge \sigma}^2 - t\wedge \sigma )_{ t\in \N}$.

Show that Doob's martingale convergence theorem  applies and 
 \begin{eqnarray*}
\lim_{t\to \infty}  M_{t\wedge \sigma }(\omega)  = M_{\sigma}(\omega)
  \end{eqnarray*}
exists $P$-almost surely. 

\item 
Consider now $(M_{t\wedge \sigma}^2-t\wedge\sigma)$. Use the martingale property together
with the reverse Fatou lemma
to show that $E( \sigma )< \infty$ which implies
 $P(\sigma< \infty)=1$.


\item
For $a < 0 < b  \in \N $, compute $P( \tau_a < \tau_b)$.
\end{itemize}

Hint: a martingale has constant expectation $E_P( M_t)= E_P( M_0)$.
This holds also for the stopped martingale $M^{\tau}_t = M_{t\wedge \tau}$.




\item
 Let  $M_t(\omega)= B_t(\omega)$,  $t\in \R^+$, a Brownian motion which
is assumed to be  $\F$-adapted, and such that for all $0<s<t$
the increment $(B_t-B_s)$ is $P$-independent from the $\sigma$-algebra ${\mathcal F}_s$.

Note this since by assumption the Brownian motion is $\F$-adapted,
it follows that ${\mathcal F}^B_t = \sigma( B_s : 0 \le s \le t) \subseteq {\mathcal F}_t$,
which could be strictly bigger.

\begin{itemize}
\item



Show that $B_t$, $M_t=(B_t^2-t)$  and  $Z_t^a = \exp( a B_t - a^2 t /2)$  are $\F$-martingales.

\item

Let $\sigma(\omega)=\min( \tau_a(\omega), \tau_b(\omega))$, for $a < 0 < b  \in \R$.
We will see in the lectures  that the
Doob martingale convergence theorem applies also to continuous
martingales in continuous time. By following the same line of proof as in the
random walk case check that $P( \sigma < \infty)=1$.
\item
Let $a < 0 < b  \in \R$. Compute $P( \tau_a < \tau_b)$,
\end{itemize}
{\bf Hints} 

When $M$ is either a Brownian motion or a random walk, the 
stopped process $M_{t\wedge \sigma}(\omega)$  is a uniformly bounded martingale.
To compute $P( \tau_a < \tau_b)$,
use first  the martingale property
\begin{eqnarray*}
E(M_{t\wedge \sigma} ) = E( M_0)=0 ,
\end{eqnarray*}
then for $t\to\infty$  use the bounded convergence theorem.


\end{enumerate}

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