Stochastic Differential Equations

Example

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<..<x_n=b and x_k≤x_k^*≤x_k+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=a_j)=f(a_j), j=1,2,… Let x₁ <..< x_n, and we can always choose the x_i’s such that x_i≠a_j for all i and j. Now if x_k and x_k+1 are such that none one the a_j’s are between them we have

F(x_k+1)=F(x_k)

and so

h(x_k )(F(x_k+1)-F(x_k)) = 0.

On the other hand say we have x_k<a_j <x_k+1. In that case let x_k^* = x_k. 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

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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

x_k = a+k(b-a)/2ⁿ, k=0,..,2ⁿ

and define

But we need to be careful. When approximating a function f in the interval [x_k ,x_k+1) what should we choose for e_k? 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:

Example

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(B_t) has two parts, one “usual” and one with dB_s. This motivates the following

Definition

A stochastic integral is stochastic process {X(t)} of the form

sometimes this is written as

dX_t = u dt +vdB_t

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(½B_t²) = ½ dt + B_t dB_t

In general we have

Theorem (Ito formula)

Let X_t be a stochastic integral of the form

dX_t = u dt +vdB_t

and let t → g(t,x) be some “reasonably” nice function. If Y_t = g(t,X_t) then Y_t is again a stochastic integral with

with the convention that

dt . dt = dt . dB_t = 0

dB_t . dB_t =dt

proof

omitted

Example

Let X_t = B_t and g(t,x) = ½x². Then dg/dt=0, dg/dx=x and d²g/dx²=1, and so

as before

Example

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

\[\int_0^t sdB_s= tB_t-\int_0^t B_s ds\]