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{\bf Stochastic analysis, spring  2013, Exercises-10, 11.04.2013 }


\begin{enumerate}
\item
 Consider a Brownian motion $(B_t:t \ge 0)$ in the filtration $\F =( {\mathcal F}_t)_{t\ge 0}$,
which means that $B_t$ is $\F$-adapted, time continuous, and for each $0\le s \le t$,

the conditional distribution of the increment $(B_t -B_s)$ given ${\mathcal F}_s$  is  a Gaussian with zero mean and variance $(t-s)$.


\begin{enumerate} \item
Show that the Brownian motion has the Markov property: $\forall$  $s\le t$ and bounded Borel function $f(x)$,
\begin{align*}&
 E_P( f( B_t ) | {\mathcal F}_s) (\omega) = E_P( f( B_t ) | \sigma(B_s) ) (\omega) = E_P\biggl(  f (   x + B_t -B_s ) \biggr) \bigg\vert_{x= B_s(\omega) }
= & \\ &
 E_P\biggl(  f (   x + B_{t-s} ) \biggr) \bigg\vert_{x= B_s(\omega) }   = \varphi( B_s(\omega )  ) 
&\end{align*}
for some bounded Borel-measurable function $\varphi(x)$.
\item
Show  also that for $0 \le t_0 \le t_1 \le \dots \le t_d$ and $f( x_1,\dots,x_d)$ bounded and Borel measurable,
\begin{align*}
 & E_P( f( B_{t_1},\dots, B_{t_d} ) |  {\mathcal F}_{t_0} )(\omega) = E_P( f( B_{t_1},\dots, B_{t_d} ) |  \sigma( B_{t_0} ) )(\omega) = 
& \\ &E_P\biggl( f( x + B_{t_1-t_0} , \dots ,x + B_{t_d-t_0} )  \biggr) \bigg\vert_{x= B_{t_0}(\omega) } =
 \psi(B_{t_0}(\omega)) &
\end{align*}
for some Borel measurable function $\psi(x)$.  Hint: use induction.





\item  Let $\tau(\omega)$ an $\F$-stopping time taking finitely many values.
Show first the {\bf strong Markov property } of Brownian motion:
for $f(x)$ bounded measurable function,
\begin{eqnarray*}&&
  E_P\bigl( f( B_{\tau +t} ) \big\vert {\mathcal F}_{\tau}  \bigr) (\omega) =   E_P\bigl( f( B_{\tau+t} ) \big\vert  \sigma(B_{\tau})  
\bigr ) (\omega) = \\ && 
  E_P\bigl(  f( x + B_t ) \bigr ) \bigg\vert_{x = B_{\tau}(\omega) } = \varphi( B_{\tau}(\omega ) 
\end{eqnarray*}

Hint: remember the definition of stopped $\sigma$-algebra ${\mathcal F}_{\tau}$.

\item Show that
 $( B_{\tau+ t} -B_{\tau} ) \stackrel{ P}{ \indep} {\mathcal F}_{\tau}$, and the 
conditional distribution of $( B_{\tau+ t} -B_{\tau} ) $ given $ {\mathcal F}_{\tau}$ is Gaussian with zero mean and variance $t$.

This means that at every stopping time the Brownian motion restarts from the position $B_{\tau}$ independently of the past.

\item
Show that the strong Markov property for a general $\F$-stopping time $\tau$. Assume that the filtration $\F$ is right
continuous. We have shown that there is a sequence of $\F$-stopping times $\tau_n(\omega) \downarrow \tau$
approximating $\tau$ from above, with each $\tau_n$ taking only finitely many values. Note also that ${\mathcal F}_{\tau_n} \supseteq {\mathcal F}_{\tau}$.

\item Show that if $\tau$ is an $\F$-stoppping time
$0 \le t_1 \le \dots \le t_d$ and $f( x_1,\dots,x_d)$ bounded and Borel measurable,
\begin{eqnarray*} &&
 E_P\bigl( f( B_{\tau+t_1},\dots, B_{\tau + t_d} ) \big\vert  {\mathcal F}_{\tau} \bigr )(\omega) = 
E_P\bigl( f( B_{\tau +t_1},\dots, B_{\tau+t_d} ) \big\vert \sigma( B_{\tau+t_0} )\bigr )(\omega) =   \\ &&
E_P\bigl( f( x + B_{t_1-t_0} , \dots ,x + B_{t_d-t_0} ) \bigr ) \bigg\vert_{x= B_{\tau}(\omega)} = \psi(B_{\tau}(\omega))
\end{eqnarray*}
for some Borel measurable function $\psi(x)$.  Hint: use induction.


\end{enumerate}


\item (Reflection priociple )
Let 
\begin{align*}
B_t^* = \max_{0 \le s \le t } B_s
\end{align*}
the running maximum of Brownian motion.

We show that for $x>0$, $P( B^*_t  >  x   ) = 2 P( B_t > x )$.

Consider the stopping time $\tau_x= \inf\{  s : B_s > x \}$ and note that $\{ B^*_t > x \} = \{ \tau_x < t \}$.

Consider the process 
\begin{align*} \widetilde B_t = \left\{ \begin{matrix}   B_t  &   t \le \tau_x \\
                                            2x-B_t  &  t  > \tau_x    
                                        \end{matrix} \right.
\end{align*}
which is Brownian motion reflected at level $t$.

\begin{enumerate}
\item
Use the strong Markov property to  show that $\widetilde B_t$ is a Brownian motion in the filtration $\F$.

\item
Note that 
\begin{align} \label{eq:1}
\{ B^*_t > x \} =  \{  B_t \ge x  \}\cup \{  \widetilde B_t > x \}  
\end{align}
with $\{ B_t \ge x \} \cap \{ \widetilde B_t > x \} = \emptyset$, and $P( B_t=x)=0$.

Compute the probability density function of $B_t^*$.

\item Compute 
\begin{align*}
  P(  B^*_t >  x  , B_t >  Y )
\end{align*}
Hint: use ( \ref{eq:1} ).


\item Compute the joint probability density of $(B_t^*, B_t )$.


\item The running maximum  $( B_t^* : t \ge 0 )$ is not a  Markov process. Show that the pair $(  B^*_t ,B_t) $ is a strong Markov process,
Hint: use the strong Markov property of Brownian motion.
\end{enumerate}


\item The same reflection principle argument holds for a symmetric random walk on $\Z$.
Consider a filtration $\F=( {\mathcal F}_n :n \in \N)$ in discrete time.
Let $(X_n : n \in N)$  an $\F$-adapted  process with  
\begin{align*}
  P( X_n = 1  | {\mathcal F}_{n-1} ) = P( X_n = -1 | {\mathcal F}_{n-1} ) = 1/2
\end{align*}
 (which means $X_n$ is indepependent from the past)
and
\begin{align*}
  S_n = X_1 + \dots + X_n
\end{align*}
Note that the proabability law of $S_n$ is the binomial distribution
\begin{align*}
  P( S_n = k ) = \binom{n}{k} 2^{-n}
\end{align*}
Let
\begin{align*}
S^*_n = \max_{ 1\le k \le n } S_k 
\end{align*}
the running maximum of the random walk.

\begin{enumerate}
\item Show that $(S_n: n \in \N)$ is a strong Markov process in the filtration $\F$.

\item Compute the joint probabiliy
$P( S^*_n = \ell , S_n = k )$.

\item Show that $( S^*_n, S_n )_{n\in \N}$ is  a strong Markov process in the filtration $\F$.
\end{enumerate}




\end{enumerate}




\end{document}
