Let’s consider the following simple growth model: if there are N individuals at time t a certain percentage of them have offspring, so at the next time period the number of individuals will be
N(t+1) = a(t)N(t)
if we allow for a continuous change in the population we get the differential equation
dN/dt = a(t)N(t)
Now in real live this will never be exact, there will always be some randomness in the number of offspring, so a more realistic model would be
dN/dt = (a(t)+noise)N(t)
and of course “noise” is a random variable, so now we have a stochastic differential equation.
In order to be more precise we need to go back to Calculus for bit. There integrals are usually defined via approximating sums called Riemann sums:
where a=x1<..<xn=b and xk≤xk*≤xk+1
Another way of making sense of this definition is as follows: let’s consider a function f and say we want to find the integral above. That is we want to find the area under the curve f(x). One way to this (for example numerically) is to approximate f by a step function
We will return to this idea in a bit, but first we need to study an extension of the Riemann integral called the Riemann-Stiltjies integral, defined as follows:
Here f is a function as before, and g is a function for which we have some restrictions. The most common one is to require g to be a function of bounded variation. For a discussion of this concept see http://en.wikipedia.org/wiki/Bounded_variation, for our purpose the following is enough:
Lemma
say g=f-h where both f and h are monotone functions, then g is of bounded variation
First of it is obvious that the Riemann-Stiltjies integral is a generalization of the Riemann integral, just take g(x)=x. Moreover we have
Lemma:
say g has a derivative, then
As with Riemann integration we have some formulas to help carry out the work. Especially nice is
Theorem (Integration by parts)
One reason Riemann-Stiltjies integrals are useful in Probability theory is the following
Theorem
Let the rv X have distribution F, and h some function. Then
E[h(X)] = ∫h(x)dF(x)
proof
if X is a continuous rv, F has derivative f and so
∫h(x)dF(x) =∫h(x)f(x)dx
Say X is a discrete rv with P(X=aj)=f(aj), j=1,2,… Let x1 <..< xn, and we can always choose the xi’s such that xi≠aj for all i and j. Now if xk and xk+1 are such that none one the aj’s are between them we have
F(xk+1)=F(xk)
and so
h(xk )(F(xk+1)-F(xk)) = 0.
On the other hand say we have xk<aj <xk+1. In that case let xk* = xk. Then
So one advantage of the Riemann-Stiltjies integral in Probability theory is that often we do not need different formulas for discrete and continuous random variables.
What is the reason for introducing the Riemann-Stiltjies integral? In effect it allows us to generalize the idea of the length of an interval. Instead of the “classic” b-a for the interval (a,b) we now have g(b)-g(a), and this allows us to do integration in much more general circumstances. Of course in Real Analysis this is not the end. An even more general notion of integral comes from allowing an even more general idea of the “size” of an interval, or more generally a set. This leads to the most widely used integral in Real Analysis, the Lebesgue integral defined by
∫fdμ
where μ is the so-called Lebesgue measure
Now back to stochastic differential equations. In essence we can consider introducing a random variable in two ways into an integral , either as
∫X(t,ω)dt
or
∫f(t,ω)dX(t,ω)
Now the first of these is in principle easy because this is just the regular Riemann integral. But unfortunately this alone does not lead to a useful theory. We need to try and define the second one.
What type of stochastic process {X(t)} can we use here? There are several possible choices, but not surprisingly the most useful one is based on Brownian motion.
How should we define the integral? Here we return to the idea of the Riemann integral to approximate the function f(.,ω) by a step function, so we need to decide how to define the integral for a step function. So let f be a step function. Because ultimately we can use any step function for the approximation let’s divide the interval [a,b] into 2n subintervals of equal lengths, so
xk = a+k(b-a)/2n, k=0,..,2n
and define
But we need to be careful. When approximating a function f in the interval [xk ,xk+1) what should we choose for ek? Let’s consider two possibilities:
so unlike in ordinary integration here the choice of point in the subintervals matters!
One can now play around with different choices (left endpoint? right endpoint? midpoint? …) One of them leads to the
Definition (Ito integral)
let g(x) be a continuous function and let {B(t),t≥ 0} be a standard Brownian motion. For each fixed t>0, there exists a random variable
which is the limit of the approximating sums
as n→∞.
Another common choice is the midpoint which is called the Stratonovich Integral. In what follows, though, we will only consider the Ito integral, named after Kiyoshi Ito
Theorem
say f and g are functions, a is a constant, then
Ψ(f+g) = Ψ(f)+Ψ(g)
Ψ(af) = aΨ(f)
proof
all of these follow directly from the definition of a sum of rv’s
Theorem
The random variable Ψ(g) is normally distributed with mean 0 and variance
proof
That the approximating sums are normal is easy as sums of normal rv’s are normal. That this is also true in the limit is a consequence of the CLT.
Now
Theorem
Ψ(f) is a martingale
proof omitted
All the usual ways to find integrals have their equivalence here. For example, there is a version of the integration by parts formula:
Theorem
Let g be some differentiable function, then
proof
similar to the one for Riemann-Stiltjies integrals
Using this formula we can find a number of useful stochastic processes:
Let’s find
in analogy to the Riemann-Stiltjies integral we might guess the following:
but this turns out to be wrong! The problem is that now the “integrand” is also a random variable. Instead we will show that
First let
and so
∫B(s)dB(s) = ∫e(n)dB(s)
as n→∞
Next
and so we see that Ito integrals don’t behave like ordinary integrals. In the example above we eventually found the integral directly from the definition as a sum, but just like in ordinary integration this is very tedious. In order to make use of Ito integrals we need some formulas to help. We already have an equivalent to the integration by parts formula. We will now derive an equivalent of the change of variables formula.
Let’s again consider the above example, where we found
so we see that a function like g(Bt) has two parts, one “usual” and one with dBs. This motivates the following
Definition
A stochastic integral is stochastic process {X(t)} of the form
sometimes this is written as
dXt = u dt +vdBt
but note that this is just a shorthand for the equation above, despite the fact that this field is called stochastic differential equations there is actually only stochastic integration, not stochastic differentiation!
In the case of our example then we find
d(½Bt2) = ½ dt + Bt dBt
In general we have
Theorem (Ito formula)
Let Xt be a stochastic integral of the form
dXt = u dt +vdBt
and let t → g(t,x) be some “reasonably” nice function. If Yt = g(t,Xt) then Yt is again a stochastic integral with
with the convention that
dt . dt = dt . dBt = 0
dBt . dBt =dt
proof
omitted
Let Xt = Bt and g(t,x) = ½x2. Then dg/dt=0, dg/dx=x and d2g/dx2=1, and so
as before
What is
Of course we can do this using the integration by parts formula:
but we can also use the Ito formula:
so
\[d(tB_t)=B_t \dot dt +tdB_t\]
\[tB_t=\int_0^t B_s ds+\int_0^t sdB_s\]
and so
\[\int_0^t sdB_s= tB_t-\int_0^t B_s ds\]