random
math
Brownian Motion and SDE
Nguyen Minh Hieu
05/19/2023

This is a brief techical note on Brownian Motion and SDE. For more exhaustive and rigorous material, checkout the references.

Brownian Motion

To define Brownian motion, we first need to know that such disribution exists.


Theorem 1 (existence) \exist a probability distribution over the set of all continuous function B:R0RB: \mathbb{R}_{\geq 0}\mapsto \mathbb{R} s.t.

  1. P(B(0)=0)=1P(B(0) = 0) = 1
  2. (Stationary) 0st,B(t)B(s)N(0,ts)\forall 0\leq s \leq t, B(t) - B(s)\sim\mathcal{N}(0, t-s)
  3. (Independent Increment) If intervals {[si,ti]}\{[s_i, t_i]\} are non-overlapping, then B(ti)B(si)B(t_i) - B(s_i) are independent.

Now we define the Brownian motion:


Definition The distribution given by Theorem 1 is called Brownian motion.


One important property of Brownian motion is that it’s NOT differentiable! We will not prove this. Since Brownian motion can’t be properly computed using normal calculus, we will use Ito’s Calculus to derive all the equations. To see how we can derive Ito’s Lemma, we first check an important theorem:


Theorem 2 (Quadractic Variation) T>0\forall T>0, we have:

limnt=1n(B(tnT))(B(t1nT))2=T\lim_{n\rightarrow \infty} \sum_{t=1}^n \left(B\left(\frac{t}{n}T\right)\right) - \left(B\left(\frac{t-1}{n}T\right)\right) ^ 2 = T

This is a surprising fact about Brownian motion because for any continuous and differentiable function ff, we have that

i=1n(f(ti+1)f(ti))2i=1nf(si)(ti+1ti) Mean Value Thm. =(max0sTf(s)2)i=1n(ti+1ti)2=(max0sTf(s)2)T2n\begin{align*} \sum_{i=1}^n (f(t_{i+1}) - f(t_i))^2 &\leq \sum_{i=1}^n f'(s_i)(t_{i+1} - t_i) \quad \because \text{ Mean Value Thm. } \\ &= \left( \max_{0\leq s \leq T} f'(s)^2 \right) \sum_{i=1}^n (t_{i+1} - t_i) ^2 \\ &= \left( \max_{0\leq s \leq T} f'(s)^2 \right) \frac{T^2}{n} \end{align*}

and as n0n\rightarrow 0, i=1n(f(ti+1)f(ti))20\sum_{i=1}^n (f(t_{i+1}) - f(t_i))^2 \rightarrow 0. Thus, for normal functions, the quadratic variation is zero. We can denote the quadractic variation of Brownian motion as (dB)2=dt(dB)^2 = dt. We say that square of difference in Brownian motion goes to the difference in time with probability 1.

Intuitively, the quadratic variation simply denote the variance of the Brownian motion between an infinitesimal step, which, by definition, equals to the time step. For deterministic function, the variance is simply zero. This accumulation of the second order term makes the Linear appoximation of Brownian motion much different from ordinary functions.


Ito’s Calculus

Suppose we want to compute f(Bt)f(B_t) for some smooth function ff. The Taylor Expansion of the ff is:

df=fdBt+f2(dBt)2+f3!(dBt)3+...\begin{align*} df = f'dB_t + \frac{f''}{2} (dB_t)^2 + \frac{f'''}{3!} (dB_t)^3 + ... \end{align*}

Normally, we would ignore terms after the first order. However, from Theorem 2, we know that (dBt)2=dt(dB_t)^2 = dt, and thus we can no longer ignore the second order term. Finally, we can write the (simple) Ito’s Lemma:


Lemma (Simple Ito)

df=fdBt+12fdtdf = f'dB_t + \frac{1}{2} f''dt

For a stronger Ito’s Lemma, consider the Taylor Expansion of the function f(t,Bt)f(t, B_t):

df=ftdt+fBtdBt+12(2ft2dt2+22ftBt+2fBt2(dBt)2)+...df = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial B_t} dB_t + \frac{1}{2} \left( \frac{\partial^2 f}{\partial t^2} dt^2 + 2 \frac{\partial^2 f} {\partial t \partial B_t} + \frac{\partial^2 f}{\partial B_t^2} (dB_t)^2\right) + ...

Again, we apply the Theorem 2 to get Ito’s Lemma:


Ito Lemma

df=ftdt+fBtdBt+122fBt2(dBt)2=(ft+122fBt2)dt+fBtdBt\begin{align*} df &= \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial B_t} dB_t + \frac{1}{2}\frac{\partial^2 f}{\partial B_t^2} (dB_t)^2 \\ &= \left(\frac{\partial f}{\partial t} + \frac{1}{2} \frac{\partial^2 f}{\partial B_t^2}\right) dt + \frac{\partial f}{\partial B_t} dB_t \end{align*}

Now, let’s consider a more general stochastic process XtX_t defined as

dXt=μdt+σdBtdX_t = \mu dt + \sigma dB_t

Where μdt\mu dt is called the drift term. We can use Ito Lemma to write the df(t,Xt)df(t, X_t) as follow:

df=(ft+122fXt2)dt+fXtdXt\begin{align*} df &= \left(\frac{\partial f}{\partial t} + \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2}\right) dt + \frac{\partial f}{\partial X_t} dX_t \end{align*}

Stochastic Differential Equation (SDE)

Definition

A SDE is a differential equation of the form

dXt=μ(t,Xt)dt+σ(t,Xt)dBt\begin{align*} d\mathbf{X}_t = \mu(t,\mathbf{X}_t)dt + \sigma(t, \mathbf{X}_t) d\mathbf{B}_t \end{align*}

Then, we have the following theorem:

Theorem

SDE has a solution and if given X0\mathbf{X}_0 then the solution is unique as long as μ,σ\mu, \sigma satisfy certain condition\dagger.

Let’s consider when μ\mu and σ\sigma are proportional to Xt\mathbf{X}_t:

dXt=μXtdt+σXtdBt,σ,μR,X0=x0\begin{align*} d\mathbf{X}_t = \mu\mathbf{X}_tdt + \sigma\mathbf{X}_t d\mathbf{B}_t, \quad \sigma, \mu\in \mathbb{R}, \mathbf{X}_0 = \mathbf{x}_0 \end{align*}

Using Ito Lemma, we can show that the solution for the given SDE is Xt=X0eσXt+(μ12σ2)tX_t = X_0 e^{\sigma \mathbf{X}_t + (\mu - \frac{1}{2}\sigma^2)t}. Now, let’s consider when the stochastic term σ(t,Xt)\sigma(t, \mathbf{X}_t) is independent of the current state Xt\mathbf{X}_t:

dXt=αXtdt+σdBt,α>0,μR,X0=x0\begin{align*} d\mathbf{X}_t = -\alpha\mathbf{X}_tdt + \sigma d\mathbf{B}_t, \quad \alpha > 0, \mu\in \mathbb{R}, \mathbf{X}_0 = \mathbf{x}_0 \end{align*}

This is called the Ornstein-Uhlenbeck (OU) Process. And we have the following solution for the OU process:

Xt=eαtX0+0tσeα(st)dBs\mathbf{X}_t = e^{-\alpha t}\mathbf{X}_0 + \int_0^t \sigma e^{\alpha(s-t)}d\mathbf{B}_s

For more detailed derivation of the solution, check out the awesome lecture note [2] on SDE.


Connection to the Heat Equation

For fun, let’s see the connection between Brownian Motion and the heat equation. This profound connection was discovered by none other than Albert Einstein in 1905 when he was 26 and working on Special Relativity and the Photoelectric Effect :). Let’s first define a Heat Equation:

ut=2ux2\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}

with a given boundary condition u(0,x)u(0, x) and u(t,x)u(t,x) is the probability density function at time tt and position xx. Due to the linearity property of the heat equation (if u1,u2u_1, u_2 are solutions then u1+u2u_1+u_2 is also a solution), the intergral usds\int u_s ds is a solution if all usu_s are solutions to the heat equation. So, for all initial condition, it’s sufficient to solve for u(0,x)=δ(x)u(0,x) = \delta (x). Let denote the solution to this initial condition as uδ(t,x)u_\delta(t,x). We can easily see this if we have some complex initial condition u0u_0, we can decompose u0u_0 into delta functions:

u0(x)=δ(xs)u0(s)ds=δ(x)u0(x)u_0(x) = \int_{-\infty}^\infty \delta (x-s)u_0(s) ds = \delta(x) * u_0(x)

Then, the solution given u0u_0 can also be decomposed into solutions of the u0=δu_0 = \delta:

u(t,x)=uδ(t,xs)u0(s)ds=uδ(t,x)u0(x)u(t, x) = \int_{-\infty}^\infty u_\delta (t,x-s)u_0(s) ds = u_\delta(t,x) * u_0(x)

Finally, the solution uδ(t,x)u_\delta(t,x) is also known as the heat kernel or gaussian kernel and is given as

uδ(t,x)=12πtex24tu_\delta (t,x) = \frac{1}{2\sqrt{\pi t}} e^{-\frac{x^2}{4t}}

To connect this to Brownian Motion, consider the following SDE:

dXt=2kdBtX0=0d\mathbf{X}_t = \sqrt{2k} d\mathbf{B}_t \quad \mathbf{X}_0 = 0

then, the probability density function of XtX_t is given as

12πktex24kt\frac{1}{2\sqrt{\pi k t}} e^{-\frac{x^2}{4kt}}

Which is also the solution to the heat equation

ut=k2ux2u(0,x)=δ(x)\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2} \quad u(0,x) = \delta(x)

References

[0] MIT OCW, Ito Calculus
[1] MIT OCW, Stochastic Differential Equations
[2] Gautam Iyer, Stochastic Calculus for Finance Brief Lecture Notes