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{\bf Stochastic analysis, spring  2013, Exercises-12, 25.04.2013 }




 It is not true that
all uniformly integrable local martingales are
true martingales.
Even local martingales bounded in $L^2(P)$ need not to be
true martingales !
Here we study such counterexample.
\\


Let $B_t=(B^{(1)}_t, B^{(2)}_t, B^{(3)}_t)$ a 3-dimensional
brownian motion starting from $0$ at time $0$ ,
 with 
independent components, so that $\langle B^{(i)}, B^{(j)} \rangle_t= \delta_{ij}$.

The process
 \begin{eqnarray*}
R_t= | B_t| =\sqrt{ \sum_{i=1}^3  \bigl( B_s^{(i)} \bigr)^2 }
\end{eqnarray*} 
is called  the $3$-dimensional
Bessel process. 



\begin{enumerate}

\item
Use Ito formula to compute the semimartingale decomposition of $R_t$
into a continuous local martingale part $W_t$ and a  continuous process of
finite variation.





\item 
Compute $\langle R \rangle_t=\langle W \rangle_t$ and
use  Paul L\'evy's characterization theorem for Brownian motion  to show that the local martingale
part of $R_t$ which satisifies
\begin{eqnarray*}
 W_t =           R_t- \int_0^s \frac 1 {R_s} ds 
\end{eqnarray*} 
is  a Brownian motion in the filtration $\F$ generated by $(B_t)$.


\item
Show that $R_t$ is a $\F$-submartingale.
\\



Let $M_t=R_t^{-1}$ for $t\ge 1$. We start the process
at time $1$ since $R_0=0$.
\\

\item
 Use Ito formula to show that $(M_t)_{t\ge 1 }$ is a local martingale,
 and write its Ito integral representation.


\item Compute $\langle M \rangle_t$.



\item
Show that 
$(M_t)$ is a supermartingale. Hint: it is non-negative.

\item 
Let $\tau_a:=\inf \{ t \ge 1 : R_t = a \}$,  $a>0$,
with the convention  $\inf\{\emptyset \}=\infty$.

 Show that the stopped process
 $(M^{\tau_a}_t)_{t\ge 1 } $ is a martingale and consequently
$( \tau_{1 /n } :n\in \N )$  

is a localizing sequence for the local martingale $(M_t :t\ge 1 )$



\item \label{ex:iii}
Let $0<r'<y<r''$. Use the martingale property
 of $(M_{t\wedge \tau_r}:t \ge 1 ) $ to compute $P(\tau_{r'}<\tau_{r''} | R_1 = y)$.
By the conditioning we mean that we start $R_t$ at time $t=1$ in position $y$.


\item For $0<r<y$ compute also  $P(\tau _r<\infty | R_1=y)$.



\item  Using  (g), show that the $3$-dimensional Brownian motion  
   is transient,  
$|B_t|\to \infty$ $P$ a.s., meaning that
 it  leaves eventually any ball centered
around the origin withou coming back, and therefore 
 $M_{\infty}=\lim\limits_{t\to\infty} M_t=0$.





\item Using the multivariate gaussian density in polar coordinates, compute
the probability densities of $R_t$  and $M_t$, and
show that the local martingale $( M_t : t \ge 1 )$ is 
 bounded in $L^2$ , so that in particular it is 
uniformly integrable.
( We start the martingale at time
 $1$ since there is the square norm explodes near
the origin ).
\\


 Note however that 
 $(M_t)_{t\ge 1 } $ 
 is not a martingale. Otherwise  $(X_t)$ would be 
   an   uniformly integrable martingale so that
    $M_t=E(M_{\infty} | {\mathcal F}_t)$, $t\ge 1 $.
 But in dimension 3 the Brownian motion is
 transient, which means that $M_{\infty}=0$.



\item Compute also the probability distribution of 
$R_t^2$. 

Show first that  for $t=1$,
\begin{eqnarray*}
P( R_1^2 \in dx  ) = {\bf 1}( x \ge 0 )  \frac  1 { \Gamma(3/2 ) 2^{3/2} } \exp(- x/2  ) x^{\frac 3 2 -1 }  dx
\end{eqnarray*}
which is the distribtion of a Gamma random variable with shape parameter  $3/2$ and scale parameter $2$, 
(also called chi-square with $3$ degrees of freedom
and use the
scaling property of Brownian motion.

\item  Show that $E( \langle M \rangle_t ) = \infty $  $\forall t\ge 1$.

{\bf Remark } This is not in contradiction with $E( M_{\infty}^2) <  \infty$,
since 
\begin{align*}
 E\bigl( (M_t -M_1)^ 2 \bigr ) = E  \bigl(  \langle M \rangle_t - \langle M \rangle_1\bigr )
\end{align*}
holds for martingales but does not need to hold for local martingales.
Even if the local martingale $M_t$ is bounded in $L^2$,  it means
that $M_t^2$ is bounded in $L^1$ which does not give the 
uniform integrability condition which is necessary to take the
limit of a localizing sequence under the expectation.



\end{enumerate}



{\bf Remark} In general, when $M_t$ is a continuous local martingale
with localizing sequence $\tau_n \uparrow\infty$,
to show that it is a true martingale, you need to show that
for $s\le t $ and $A\in {\mathcal F}_s$

\begin{eqnarray*} &&
E_P(  M_{t\wedge\tau_n } {\bf 1}_A ) \stackrel{?}{\to} E_P(  M_{t } {\bf 1}_A )  \\ &&
E_P(  M_{s\wedge\tau_n } {\bf 1}_A ) \stackrel{?}{\to} E_P(  M_{s } {\bf 1}_A ) 
\end{eqnarray*}
where the left sides are equal by since $(M_{t\wedge \tau_n}:t \ge 0 )$ is a martingale.
When the local martingale $( M_t : t \ge 0 )$ is uniformly integrable or bounded in $L^2(P)$,
 for fixed $t$, the sequence 
  $( M_{t\wedge\tau_n} : n\in \N )$ does not need to be  uniformly integrable,
which is the condition that we need to take the limit inside the expectation.

Of course if the local martingale is bounded on finite intervals, $|M_t(\omega)|\le c(t)<\infty $  $P$ a.s.,
then it is a true martingale.

Also, when $M_t \in L^2(P)$  is a martingale and   $\forall t$ and $Y_t$ is a progressive integrand with
\begin{eqnarray}  \label{integrability_square}
 E\biggl( \int_0^t Y_s^2 d\langle M \rangle_s \biggr)< \infty \quad \forall t
\end{eqnarray}
then the Ito integral $( Y\cdot M )_t \in L^2(P)$ is a true martingale.

If $M_t$ is not a square integrable martingale,  or  $Y$ does not satisfy the condition  
 \ref{integrability_square}, the stochastic integral $( Y\cdot M )_t$ is just local
martingale.



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