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{\bf Stochastic analysis, spring  2013, Exercises-13, 02.05.2013 }


\begin{enumerate}
 \item 
(Bougerol' identity ) Consider 

$\sinh( W_t)$
where $W_t$ is Brownian motion and 
\begin{align*} &
\sinh(x)= \frac{ e^x - e^{-x} } 2 , \; \cosh(x)= \frac{ e^x + e^{-x} } 2, \; &\\ &
\frac d {dx } \sinh(x) = \cosh(x), \;  \frac d {dx } \cosh(x)=\sinh(x) , \; \cosh(x)^2 - \sinh(X)^2 = 1 &
\end{align*}

Let also $B_t$ and $\beta_t$ independent Brownian motions, and
\begin{align}
X_t = e^{B_t} \int_0^t e^{-B_s} d\beta_s 
\end{align}

Apply Ito's formula to
show that 

the processes $(X_t : t\ge 0)$   and $( \sinh( W_t) : t\ge 0)$ have the same distributions,

i.e. they satisfy the same stochastic differential equation in Ito sense.



\item We consider a $\C$-valued  continuous local martingale
\begin{align*}
 Z_t = X_t + i Y_t
\end{align*}
where $X_t$ and $Y_t$ are $\R$-valued  continuous local martingales in a filtration $\F$.

\begin{enumerate}

\item
Prove that there is an unique continuous $\C$-valued process of finite variation $\langle Z,Z \rangle_t$
such that 
\begin{align*}
  Z^2 - \langle Z,Z \rangle_t  
\end{align*}
is a complex local martingale.

\item Prove that  the following statements are equivalent:
\begin{enumerate}
\item $Z^2_t$ is a $\F$-local martingale.
\item $\langle Z,Z \rangle_t = 0$   $\forall t$
\item $\langle X,X \rangle=\langle Y,Y \rangle$  and $\langle X,Y \rangle=0$ 
\end{enumerate}
Such $\C$-valued local martingales are called {\it conformal  } local martingales.
For example, if $ B_t$ and $\beta_t$ are independent $\R$-valued Brownian motions in the 
filtration $\F$, the  $\C$-valued { \it planar Brownian motion  } 
\begin{align}
  W_t = B_t + i \beta_t
\end{align}
is a conformal martingale.


\item Use L\'evy characterization theorem to show that if $Z_t=X_t+ i Y_t$ is  a $C$-valued
continuous conformal local martingale in a filtration $F$, there exists
a $\C$-valued planar Brownian motion, such that
\begin{align*}
 Z_t =  W_{\langle X,X\rangle_t } 
\end{align*}
with
\begin{align*}
 W_u=  X_{\tau(u)} + i Y_{\tau(u) }  , \quad \tau(u)=\inf\bigl\{ t: \langle X,X \rangle_t \ge u \bigr\}
\end{align*}










\item Show that for a conformal local martingale $Z_t=X_t + i Y_t$
\begin{align}
 \langle \Re(Z), \Re(Z ) \rangle_t = \frac 1 2 \langle  Z, \overline Z \rangle_t
\end{align}



\item Let $H_t$ a $\C$-valued bounded progressive process, and $Z_t$ a conformal local martingale,
Then
\begin{align*}
 U_t =\int_0^t H_s dZ_s = \int_0^t \Re( H_s) dX_s -\int_0^t \Im(H_s) dY_s + i  \biggl( \int_0^t \Im(H_s) dX_t + \int_0^t \Re(H_s) dY_s \biggr)
\end{align*} 
is a conformal martingale with 
\begin{align*}
 \langle U, \overline U \rangle_t = \int_0^t | H_s| ^² d \langle Z,\overline Z \rangle_t
\end{align*}

\item Let $Z_t $ a continuous conformal local martingale and $F: \C \to \C$ twice differentiable as function of two real variables,
use Ito formula to show
\begin{align*} 
 F(Z_t) =  F(Z_0) + \int_0^t \frac{\partial }{\partial z} (Z_s) dZ_s + \int_0^t \frac{\partial }{\partial  \overline z} (Z_s) d\overline Z_s + 
\frac 1 4 \int_0^t \Delta F(Z_s) d\langle Z, \overline Z\rangle_s
\end{align*}
where  for $z=x+iy$
\begin{align*} & 
\Delta F(z) = \Delta(F(x,y) ) = \frac{\partial^2 }{\partial x^2} F(x,y)+ \frac{\partial^2 }{\partial y^2} F(x,y) & \\ &
 =  \biggl( \frac{\partial  }{\partial x} -i  \frac{\partial  }{\partial y} \biggr)  
\biggl( \frac{\partial  }{\partial x} +i  \frac{\partial  }{\partial y} \biggr)  F(x,y)=
 \frac{\partial }{\partial z}    \frac{\partial }{\partial \overline z}  F(x,y)&
\end{align*}


\item
We say that $F$ is harmonic if $\Delta F(z)=0$ $\forall z\in \C$.
Show that if $F$ is harmonic and $Z_t $ a continuous conformal local martingale, than
 $F(Z_t)$  is a local martingale.






\item Let 
\begin{align*}
 \frac{ \partial }{\partial z }= \frac 1 2 \biggl(
 \frac{ \partial }{\partial x } - i   \frac{ \partial }{\partial y } \biggr), \quad  \frac{ \partial }{\partial \overline 
z }= \frac 1 2 \biggl(
 \frac{ \partial }{\partial x } + i   \frac{ \partial }{\partial y } \biggr),
\end{align*}



 A function $F: \C \to \C$ is holomorphic if and only if 
\begin{align*}
\frac{\partial} {\partial \overline z} F (z) = 0\end{align*}
and in such  case we set 
\begin{align*}
F'(x)=  \frac{  \partial} {\partial  z} F (z) \; .
\end{align*} If $F(z)$ is holomorphic
it is necessarily harmonic.  Show that in such case, if $Z_t $ a continuous conformal local martingale,
then $F(Z_t)$ is  a continuous conformal local martingale with
\begin{align*}
 F(Z_t)=F(Z_0) +\int_0^t F'(Z_s) dZ_s
\end{align*}




\item Show that if $W_t$ is a $\C$-valued planar Brownian motion and $F: \C \to \C$ is 
holomorphic and non-constant
\begin{align*}
F( W_t)= F(W_0) + \widetilde W_{\langle X, X \rangle_t}
\end{align*}
where $X_t = \Re( F(W_t) )$ and
\begin{align*}
 \langle X, X \rangle_t =\int_0^t | F'(W_s) |^2 ds
\end{align*}
and $\widetilde W_t$ is a $\C$-valued planar Brownian motion.

In other words $F(W_t)$ is a  time changed Brownian motion.
We say that Brownian motion is {\it conformal invariant }.

{\bf Note} We can also show that $\langle X, X \rangle_{\infty}=\infty$.









































\end{enumerate}
\end{enumerate}


\end{document}
