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{\bf Stochastic analysis, spring  2013, Exercises-9, 28.03.2013 }


\begin{enumerate}
\item
Let $(B_t: t\ge 0)$ a  Brownian motion in the filtration $\F=({\mathcal F}_t:t \ge 0 )$
which means that $B_t$ and $(B_t^2 -t)$ are continuous $\F$-martingales.
\\

 Let $\tau(\omega)$ be a $\F$-stopping time with $E_P( \tau ) < \infty$.




\begin{enumerate}

\item
 Let also $\tau_0=0$ and  $\tau_n(\omega) = \tau(\omega) \wedge \inf \{ t: | B_t(\omega)| > n \}$, $\quad n \in \N$.
Show that these are $\F$-stopping times and $\tau_n(\omega) \uparrow  \tau(\omega)$ as $n\uparrow \infty$.



 \item  Show that for fixed $n\in \N$,  $\bigl( B_{t\wedge {\tau_n}}^2 -t\wedge \tau_n \bigr) : t\ge 0\bigr)$ is an uniformly
integrable $\F$-martingale.





 \item 
Show that $\bigl(B_{\tau_n}:n \in \N\bigr)$ and $\bigl(  (B_{\tau_n}^2 - \tau_n  : n \in \N\bigr)$ are martingales 
in the discrete time filtration  $({\mathcal F}_{\tau_n} : n \in \N)$.

\item 

Show that the telescopic sum $B_{\tau}= \sum_{n=1}^{\infty}  (  B_{\tau_n} - B_{\tau_{n-1} } )$
is convergent in $L^2(P)$.

Hint: by the discrete integration by parts formula
\begin{align*}
    \bigl( B_{\tau_n} -B_{\tau_{n-1} } \bigr)^2 =  B_{\tau_n}^2 - B_{\tau_{n-1}}^2  - 2 B_{\tau_{n-1} } ( B_{\tau_n} -B_{\tau_{n-1} } \bigr)
\end{align*}
then use the martingale properties.





\item
Show that $E_P( B^2_{\tau} ) = E_P( \tau )$.


\end{enumerate}

\item We do the same exercise for a continuous local martingale.

Let $(M_t: t\ge 0)$ a continuous local $\F$-martingale with $M_0=0$ and  predictable variation $\langle M \rangle_t$,
which means that $\langle M \rangle_t$ is $\F$-adapted continuous  and non-decreasing with 
\begin{align*} 
  M_t^2=  N_t - \langle M \rangle_t
\end{align*}
where $N_t$ is a local martingale with $N_0= \langle M\rangle_0=0$.


Let $\sigma_n (\omega) \uparrow \infty$ a localizing sequence of stopping times such that
for each $n$, the stopped processes $( M_{ t\wedge \sigma_n}: t\ge 0)$ and 
$(N_{t\wedge \sigma_n} : t\ge 0)$ are   true $\F$-martingales.

Let $\tau(\omega)$ an $\F$-stopping time, with $E_P\bigl( \langle M \rangle_{\tau} \bigr) < \infty$.

For $n\in \N$ let 
 $\tau_0=0$ and  
\begin{align*}
\tau_n(\omega) = \tau(\omega) \wedge \sigma_n(\omega) \wedge \inf \{ t: | M_t(\omega)| > n \}, \quad n \in \N.
\end{align*}

\begin{enumerate}\item
Check that $\tau_n(\omega) \uparrow \tau(\omega)$.




 \item  Show that for fixed $n\in \N$,  $\bigl( M_{t\wedge {\tau_n}}^2 - \langle M \rangle_{t\wedge \tau_n}  \bigr) : t\ge 0\bigr)$ is an uniformly
integrable $\F$-martingale.





 \item 
Show that $\bigl(M_{\tau_n}:n \in \N\bigr)$ and $\bigl(  (M_{\tau_n}^2 - \langle M \rangle_{\tau_n}  : n \in \N\bigr)$ are martingales 
in the discrete time filtration  $({\mathcal F}_{\tau_n} : n \in \N)$.

\item 

Show that the telescopic sum $M_{\tau}= \sum_{n=1}^{\infty}  (  M_{\tau_n} - M_{\tau_{n-1} } )$
is convergent in $L^2(P)$.
\begin{align*}
    \bigl( M_{\tau_n} -M_{\tau_{n-1} } \bigr)^2 =  M_{\tau_n}^2 - M_{\tau_{n-1}}^2  - 2 M_{\tau_{n-1} } ( M_{\tau_n} -M_{\tau_{n-1} } \bigr)
\end{align*}
Hint: by the discrete integration by parts formula and the martingale properties.



\item
Show that $E_P(  M ^2_{\tau} ) = E_P( \langle M \rangle_{\tau} )$.




\end{enumerate}


\item If $(M'_t: t\ge 0)$ is another continuous local martingale, with $M'_0=0$, and $\tau$ is a stopping time
with $E(\tau)<\infty$, show that 
\begin{align}
  E_P\bigl( M_{\tau} M'_{\tau}  \bigr) = E_P\bigl( \langle M, N\rangle_{\tau}\bigr)
\end{align}
where the predictable covariation $\langle M, N\rangle_{t}$ is the continuous $\F$-adapted
process of finite variation on compacts such that 
\begin{align*}
  M_t M'_t -M_0 M'_0 - \langle M, M'\rangle_t
\end{align*}
is a $\F$-local martingale. 



Hint: you can use polarization :  $M M' = \bigl( (M+M')^2 -(M-M')^2 \bigr)/4$.


 \item ( Martingale characterization)
 Let $M_t(\omega)$ adapted to the filtration $\F=({\mathcal F}_t:t\in \R^+)$,  with $E(|M_t|  ) < \infty $ $\forall t$.
Then $(X_t)$ is an $\F$-martingale if and only if $\forall $  $\F$-stopping times $\tau(\omega)$ which can take at most two  finite values
\begin{eqnarray*}
E( M_{\tau} ) = E(M_0 ) 
\end{eqnarray*}

Hint: the necessity is just Doob optional sampling theorem, for sufficency for $s\le t$ and $A \in {\mathcal F}_s$
define $\tau(\omega)=\bigl( t {\bf 1}_A  + s  {\bf 1}_{A^c } \bigr) $  and show that it is a stopping time.
Then show that $E\bigl(  (M_t-M_s) {\bf 1}_A \bigr) =0$.






\end{enumerate}

\end{document}
