Stochastic ODEs

Introduction

In classical probability, a random variable assigns a real number to each outcome $ω \in Ω$ .
A stochastic process (or, loosely speaking, a curve-valued random variable) assigns an entire function (a curve $t \mapsto x (t)$ ) to each $ω$ . Formally:

X : Ω \to Functions ([0, T], R)

X (t) = f (t, W (t))

where $W (t)$ is Brownian motion, the result is still a random function, because $W (t)$ itself is a random process.

We can visualize $X$ : we can generate several samples of $W (t)$ for a particular time increment $d t$ , and then we compute $f (t, X (t))$ .

SDEs generalize ordinary differential equations by incorporating a random noise term, often modeled by Brownian motion.
General form:

d x (t) = F (x (t), t) d t + G (x (t), t) d W (t)

Think of $\frac{d W}{d t}$ as if it were a random number from a normal $N (0, \infty)$ distribution. Then,

d W (t) = \frac{d W}{d t} \cdot d t

is a random number from a $N (0, d t)$ distribution.

The solution to an SDE is not a single function but a stochastic process — i.e., a curve-valued random variable.
The simplest example: $d x (t) = a d t + b d W (t) \Rightarrow x (t) = x_{0} + a t + b W (t)$

d S (t) = μ S (t) d t + σ S (t) d W (t)

This models:
- $μ$ : expected growth rate (drift)
- $σ$ : market volatility (randomness)
- $W (t)$ : Brownian motion (random shocks)
The solution is:

S (t) = S_{0} \exp ((μ - \frac{σ^{2}}{2}) t + σ W (t))

This is a random curve over time — each realization gives one possible stock price trajectory.

S (t) = S_{0} \exp ((μ - \frac{σ^{2}}{2}) t + σ W (t)),

you might expect the chain rule to apply like in standard calculus, and wonder why this form solves the SDE:

d S (t) = μ S (t) d t + σ S (t) d W (t) .

But you're forgetting that Itô lemma (the stochastic analog of the chain rule) introduces extra terms due to the randomness of $W (t)$ .

Let:

f (t, W (t)) = \exp (A t + σ W (t)), where A = μ - \frac{σ^{2}}{2} .

Then:

Now plug into Ito’s formula:

d S = (A f + \frac{1}{2} σ^{2} f) d t + σ f d W = f ((A + \frac{1}{2} σ^{2}) d t + σ d W)

Recall $A = μ - \frac{σ^{2}}{2}$ , so:

A + \frac{1}{2} σ^{2} = μ

Therefore:

d S = f (μ d t + σ d W) = S (t) (μ d t + σ d W)

From Stratonovich integral#Conversion rule we can show the following. Suppose you have an Itô SDE

d X_{t} = a (t, X_{t}) d t + b (t, X_{t}) d W_{t},

with $X_{t} \in R$ , $W_{t}$ one-dimensional Brownian motion.
The Stratonovich version is

d X_{t} = (a (t, X_{t}) - \frac{1}{2} b (t, X_{t}) \partial_{x} b (t, X_{t})) d t + b (t, X_{t}) \circ d W_{t} .

That is:

\int_{0}^{t} b (s, X_{s}) \circ d W_{s} = \int_{0}^{t} b (s, X_{s}) d W_{s} + \frac{1}{2} \int_{0}^{t} \partial_{x} b (s, X_{s}) b (s, X_{s}) d s .