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\section{Introduction to Stochastic Calculus}
This discussion is adapted from {\it Financial Calculus: An
introduction to derivative pricing} by M Baxter, and
A. Rennie, Cambridge University Press, 1996, pages 52-62.
\subsection{ A brief review of calculus and differential equations}
Differentiable functions, no matter how difficult or how strange their
global behavior, are at basis composed of nearly straight line
segments. Differential calculus is the formal acknowledgment and
application of this simple idea.
With differential calculus, we could decide to build up a family of
nice functions by specifying how they are built up locally out of our
basis function, the straight line. We would write the local change
in value of the function over a time interval of infinitesimal length
$dt$ informally as
$$
df = \mu \, dt
$$
or more typically and formally, as
$$
\frac{df}{dt} = \mu(t, f(t)), f(t_0) = x_0
$$
This says that the initial point is specified, and the slope at the point
is determined (through $\mu$) by the time and the function value.
This is the usual expression of an Ordinary Differential Equation (ODE).
Then we can infinitesimally extend the function with that information
(along the tangent straight line) and repeat. Of course, this is
just an expression in words of the Euler method for numerically
solving the differential expression above.
{\it Simple Example 1} The simplest differential equation is $df = \mu
\, dt$ where
$\mu$ is a constant is just the expression of a straight line, since
it says that the slope is everywhere constant.
{\it Simple Example 2} The next simplest differential equation is $df
= t \, dt$. This differential equation says that the slope is
increasing, in fact increasing exactly as the time. The solution may
be easily guessed (using the Fundamental Theorem of Calculus!) as $f =
t^2/2 + C$, where $C$ would be specified by the initial value.
{\it Simple Example 3} The next simplest and first non-trivial
differential equation is $df = f \, dt$. Here the differential
equation says that the slope is the same as the function value. The
solution may be guessed to be $f(t) = C e^t$ where the constant $C$ is
determined from the initial condition.
\subsection{ Stochastic differentials}
We know that zooming in on Brownian motion does not produce a straight
line, it only produces another scaled imaged of Brownian motion
again. This self-similarity is the result of the scaling property.
This self-similarity is ideal for an infinitesimal building block, we
could build global Brownian motion out of lots of local ``chunks'' of
small Brownian motions. And then that suggests we could build other
stochastic processes out of suitably scaled Brownian motion. If we in
addition include straight line segments we can include differentiable
functions as well.
A stochastic process called a diffusion will have both a Newtonian or
straight line differential term based on $dt$, and a stochastic term
based on the infinitesimal component of standard Brownian motion $X$,
which we will call $dX$. Each infinitesimal can be scaled, the
scaling can depend on the time $t$, the function value $Y$ and even
the history up to $t$ of the stochastic process:
$$
dY = \mu \, dt + \sigma \, dX
$$
We call $\mu$ the drift of $Y$ at time $t$ and $\sigma$ the volatility
of the process. A special case when the drift $\mu$ and the
volatility $\sigma$ depend only on the time $t$ and the process value
$Y(t)$ is called a stochastic differential equation (SDE)
$$
dY = \mu(t, Y(t) ) \, dt + \sigma(t, Y(t)) \, dX
$$
We can guess some solutions of SDEs by using the intuition we have
already gained about previous stochastic processes.
{\it Simple Example 1} Suppose the drift an the volatility are both
constant:
$$
dY = \mu \, dt + \sigma \, dX
$$
Then we might reasonably guess that the stochastic process with this
behavior is scaled Brownian motion with drift $\mu$.
{\it Simple Example 2} Suppose we have the slightly more complicated
example where the volatility and the drift are proportional to the
current value of the process:
$$
dY = \mu Y \, dt + \sigma Y \, dX
$$
Then, unless we are a very good guesser, we would not be able to name
what process has precisely this kind of differential behavior.
\subsection{Ito's Formula and Ito calculus}
The important thing to know about normal differential calculus is that
it is the chain rule, the Fundamental Theorem of Calculus and Taylor's
formula that enable us to calculate with functions. we need some
operational rules to allow us to manipulate functions with calculus.
Likewise we similar rules and formulas for stochastic differentials,
we can't rely on simple guessing. Ito's formula will perform that
function for us.
The next example will show us that we will need some new rules for
stochastic differentials, the old rules from calculus will no longer
make sense.
{\it Example} Consider the process which is the square of Brownian
motion:
$$
Y(t) = X(t)^2
$$
We wonder what is the stochastic differential of $Y(t)$ as discussed
above? Using naive calculus, we might conjecture:
$$
dY = 2 X(t) dX(t)
$$
If that were true then
$$
Y(t) = \int_0^t dY = \int_0^t 2 X(t) dX(t)
$$
should also be true. But consider $\int_0^t 2 X(t) dX(t)$. It ought
to correspond to a summation of some sort (for instance a
Riemann-Stieltjes left sum):
$$
\int_0^t 2 X(t) dX(t) \approx \sum_{i=1}^{n} X( (i-1)t/n) [ X(it/n) -
X((i-)t/n) ]
$$
But look at this carefully: $X((i-1)t/n) = X((i-1)t/n) - X(0)$ is independent
of $ [ X(it/n) - X((i-)t/n) ]$ by property 2 of the definition of
Brownian motion. Therefore, the summation term will have zero mean.
But the mean of $Y(t) = X(t)^2$ is the variance of Brownian motion,
namely $t$ by property 1 of the definition of Brownian motion,
which is definitely not zero! Even at the first moment, the two
processes don't agree, something is not working!
\begin{thm} (Ito's formula)
If $Y$ is a stochastic process, satisfying $dY = \mu \, dt + \sigma
\, dX$ and $f$ is a twice continuously differentiable function, then
$Z(t) = f(Y(t))$ is also a stochastic process and is given by
$$
dZ = (\mu f'(Y) + (\sigma^2/2) f''(Y)) \, dt + (\sigma f'(Y)) \, dX
$$
\end{thm}
In words, Ito's formula in this form tells us how to compute (or
expand, in analogy with Taylor's formula) a process which is defined
as a function of scaled Brownian motion with drift.
{\it Example} Here the function $f$ is the squaring function, $f(x) =
x^2$, $f'(x) = 2x$ and $f''(x) = 2$, and $Y =
X$ so $dY = dX$, i.e. $\mu = 0$ and $\sigma = 1$. Then according to
Ito's formula:
$$
d(X^2) = (1/2)(2) dt + 2 X(t) dX = dt + 2 X(t) dX
$$
Notice the additional $dt$ term!
In fact, the case where $dY = dX$, that is the scaled Brownian motion
with drift is just standard Brownian motion with zero drift, so $Z =
f(X)$,
directly occurs
commonly enough that we record Ito's formula for this special case:
$$
dZ = df(X) = (1/2)f''(X) \, dt + f'(X) \, dX.
$$
\emph{ Example} Consider geometric Brownian motion
$$
\exp(\sigma X(t) + \mu t).
$$
What SDE does geometric Brownian motion follow?
Take $Y(t) = \sigma X(t) + \mu t$, so that $dY = \mu dt + \sigma dX$.
Then geometric Brownian motion can be written as $Z(t) = \exp(Y(t))$,
so $f$ is the exponential function. One application of Ito's formula
gives
$$
dZ = (\mu f'(Y(t)) + (1/2)\sigma^2 f'' Y(t)) + \sigma f'(Y) dX
$$
Computing the derivative of the exponential function, $f'(Y(t)) = \exp(Y(t))
= Z(t)$ and likewise for the second derivative. Hence
$$
dZ = (\mu + (1/2)\sigma^2) Z(t) dt + \sigma Z(t) dX
$$
\subsection{Guessing Processes from SDEs with Ito's Formula}
One of the key needs we will have is to go in the direction and convert
SDEs to processes, in other words to solve SDEs.
A few rare examples can be solved and just like ODEs in that they depend on
inspired guesses (based on a thorough understanding of the formulas of
calculus!) and then a proof that the proposed solution is an actual solution.
Such a solution to an SDE is called a diffusion.
\emph{Example}
Suppose we are asked to solve the SDE
$$
dY(t) = \sigma Y(t) dX
$$
We need an inspired guess, so we notice that the stochastic term (or
Brownian motion differential term) is the same as the SDE we generated
for geometric Brownian motion above. Moreover, if we choose $\mu$ to be
$(1/2)\sigma^2$ then the drift term in the geometric differential equation
would match the SDE we have to solve as well. We therefore guess
$$
Y(t) = \exp( \sigma X(t)- (1/2)\sigma^2 t).
$$
We can double check by applying Ito's formula, but it is clear that it will
work.
Soluble SDEs are scarce, and this one is special enough to give a name.
It is the \emph{Dol\`{e}an's exponential of Brownian motion}.
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