Brownian motion in continuous time

The main component of the asset price in this context is a Brownian motion in continuous time. This can be intuitively thought of as the limit of the random walks

$\displaystyle \Delta _{\delta }B\left( t\right) =B\left( t+\delta \right) -B\le...
...ght) =\sqrt{\delta }\varepsilon \left( t+\delta \right) +o\left( \delta
\right)$   ,

where the innovations $ \varepsilon \left( t\right) $ satisfy the usual conditions, and the limit is taken as $ \delta \rightarrow
0$. For example one can visualize how the convergence is achieved when the number of steps go from $ 40$ to $ 2000$ in figure 7.1. The mean is taken to be equal to one, and the volatility equal to 2.

Figure 7.1: Brownian and Geometric Brownian Motion paths. One can also observe the expected value and the $ \pm 40 \%$ intervals. Samples of 2000 and 40 points are generated to show the convergence from discrete to continuous time.
\begin{figure}\centering %%
\epsfig{file=c:/miktex/mymiktex/derivatives/v2003/NE...
...=c:/miktex/mymiktex/derivatives/v2003/NET/Brown04.eps,height=5cm}%%
\end{figure}

Now the process $ B\left( t\right) $, thought of as a function of time has two very important properties:

The Brownian motion $ B\left( t\right) $ is generating a filtration $ %%
\left\{ \mathscr{F}\left( t\right) \right\} _{t\geq 0}$ in a similar way that a discrete time random walk generated the filtration $ \left\{ \mathscr{F}_{t}\right\} _{t=0}^{n}$. It includes the information that one obtains by observing the Brownian motion up to time $ t$. It is easy to show that $ B\left( t\right) $ forms a martingale, $ \mathbf{E}\left[ B\left(
t+s\right) \vert\mathscr{F}\left( t\right) \right] =B\left( t\right)
$. In addition,

$\displaystyle B\left( t\right) -B\left( s\right) \sim \mathcal{N}\left( 0,\left\vert
t-s\right\vert \right)$   .

Using the Brownian motion one can define other simple processes, called stochastic differential equations [SDEs], of the form

$\displaystyle dX\left( t\right) =\alpha dt+\beta dB\left( t\right)$   .

The parameter $ \alpha $ is the drift while the parameter $ \beta $ is the volatility of the process.

If one wants to solve the above process assuming $ X\left( 0\right) =0$, and to find the value of $ X\left( t\right) $, elementary calculus would suggest a solution of the form

$\displaystyle X\left( t\right) =\int_{0}^{t}\alpha ds+\int_{0}^{t}\beta dB\left( s\right)$   .

At this point the problems start to begin: since the function $ B\left( t\right) $ is discontinuous everywhere, one cannot approach the second integral of the above expression in the Riemann sense7.1. Such integrals called Itô integrals; their general form is

$\displaystyle I\left( t\right) =\int_{0}^{t}\eta \left( s\right) dB\left( s\right)$   ,

where $ \eta \left( t\right) $ is $ \mathscr{F}\left( t\right) $-measurable [or adapted] and square integrable $ \int_{0}^{t}\eta
^{2}\left( s\right) ds<\infty $ [just a technical condition]. Itô integrals have the following properties:

Adapted-ness
$ I\left( t\right) $ is $ \mathscr{F}\left( t\right) $ adapted;

Linearity
For every $ \alpha _{1},\alpha _{2}\in \mathbbm{R}$, and $ %%
\eta _{1}\left( t\right) ,\eta _{2}\left( t\right) $ $ \mathscr{F}\left( t\right) $-measurable functions $ \int_{0}^{t}\left[ \alpha _{1}\eta _{1}\left( s\right) +a_{2}\eta
_{2}\left( s...
...t( s\right) +\alpha
_{2}\int_{0}^{t}\eta _{2}\left( s\right) dB\left( s\right) $;

Martingale
$ I\left( t\right) $ is a martingale [w.r.t. its filtration $ \mathscr{F}\left( t\right) $]

Continuity
$ I\left( t\right) $ is a continuous function of the upper limit of the integration, $ t$.

Itô isometry
$ \mathbf{E}I^{2}\left( t\right) =\mathbf{E}%%
\int_{0}^{t}\eta ^{2}\left( s\right) ds$ [hence the technical condition].

Example 41   The process $ X\left( t\right) $ above will have
$\displaystyle \mathbf{E}X\left( t\right)$ $\displaystyle =$ $\displaystyle \mathbf{E}\left[ \int_{0}^{t}\alpha
ds+\int_{0}^{t}\beta dB\left( s\right) \right]$  
  $\displaystyle =$ $\displaystyle \alpha t$, and  


$\displaystyle \mathbf{E}\left[ X\left( t\right) -\alpha t\right] ^{2}$ $\displaystyle =$ $\displaystyle \mathbf{E}%%
\int_{0}^{t}\beta ^{2}ds$  
  $\displaystyle =$ $\displaystyle \beta ^{2}t$.  

We have already said that if $ t\geq s$ we have $ \mathbf{E}\left[
B\left( t\right) -B\left( s\right) \right] ^{2}=t-s$ [which is just the variance]. We can also show that $ \mathbf{Var}\left[ B\left( t\right)
-B\left( s\right) \right] ^{2}=2\left[ t-s\right] ^{2}$. This can be rewritten as

$\displaystyle \mathbf{E}\left[ \Delta _{\delta }B\left( t\right) \right] ^{2}$ $\displaystyle =$ $\displaystyle \delta$   , while  
$\displaystyle \mathbf{Var}\left[ \Delta _{\delta }B\left( t\right) \right] ^{2}$ $\displaystyle =$ $\displaystyle o\left(
\delta \right)$   .  

This can also be expressed as $ \left[ \Delta _{\delta }B\left( t\right) %%
\right] ^{2}=\delta +o\left( \delta \right) $. Now if we take the limit $ %%
s\rightarrow t$ [or $ \delta \rightarrow
0$] we will have
$\displaystyle \mathbf{E}\left[ dB\left( t\right) \right] ^{2}$ $\displaystyle =$ $\displaystyle dt$, while  
$\displaystyle \mathbf{Var}\left[ dB\left( t\right) \right] ^{2}$ $\displaystyle =$ 0.  

Therefore although $ dB\left( t\right) $ represents an instantaneous random movement, $ \left[ dB\left( t\right) \right] ^{2}$ turns out to be deterministic, and in fact equal to $ dt$. We write this relationship informally as

$\displaystyle dB\left( t\right) \times dB\left( t\right) =dt$.

Kyriakos 2003-03-17