Brownian motion: definition and properties

I choose to open this series of notes with a discussion of Brownian motion, arguably the most popular stochastic process in the theory of SDE. Prerequisites for this part are a basic knowledge of probability and measure theory.

Stochastic processes

A stochastic process is a family of random variables

\[ X=\{X_t\}_{t\in T} \]

defined on the same probability space

\[ (\Omega,\mathcal{F},\mathbb{P}), \]

with values in some state space \(E\).

Usually,

\[ T=[0,\infty) \]

and

\[ E=\mathbb{R}^d. \]

The process can equivalently be viewed as a function of two variables,

\[ X:T\times\Omega\to E, \]

with

\[ X(t,\omega)=X_t(\omega). \]

There are two complementary ways to look at it.

For fixed \(t\),

\[ \omega\mapsto X_t(\omega) \]

is a random variable.

For fixed \(\omega\),

\[ t\mapsto X_t(\omega) \]

is a function of time, called a sample path, trajectory, or realization of the process.

Stationary processes

A process is called stationary when its statistical behaviour does not change under shifts in time.

There are two important notions.

Strict stationarity

A stochastic process \(\{X_t\}\) is strictly stationary if, for every \(n\), every choice of times \(t_1,\ldots,t_n\), and every time shift \(h\),

\[ (X_{t_1},\ldots,X_{t_n}) \overset{d}{=} (X_{t_1+h},\ldots,X_{t_n+h}), \]

whenever the shifted times remain in the time domain.

Thus all finite-dimensional distributions are invariant under time translation.

Strict stationarity is therefore a statement about the entire law of the process.


Weak stationarity

Suppose \(X_t\in L^2\).

The process is weakly stationary, or second-order stationary, if

\[ \mathbb{E}[X_t] = \mu \]

is independent of \(t\), and the covariance depends only on the time difference:

\[ \operatorname{Cov}(X_t,X_s) = C(t-s). \]

For a centered process,

\[ \mathbb{E}[X_t]=0, \]

this becomes

\[ C(t-s) = \mathbb{E}[X_tX_s]. \]

The function \(C\) is called the autocovariance or correlation function.

Thus weak stationarity says that the first two moments are invariant under time translation.

Strict stationarity is stronger in general:

\[ \text{strict stationarity} + X_t\in L^2 \quad\Longrightarrow\quad \text{weak stationarity}. \]

The converse is generally false.

Independent increments

\(X_t\) has independent increments if, for all

\[ t_0 < t_1 < \cdots < t_n, \]

the random variables

\[ X_{t_1}-X_{t_0}, \quad X_{t_2}-X_{t_1}, \quad \ldots \]

are independent.

If, for all \(t_1,t_2,s\in T\) and \(B\subset\mathbb R\),

\[ \mathbb P(X_{t_1+s}-X_{t_1}\in B) = \mathbb P(X_{t_2+s}-X_{t_2}\in B), \]

then \(X_t\) has stationary independent increments.

Brownian motion

A Brownian motion

\[ W(t):\mathbb R^+\to\mathbb R \]

is a real-valued stochastic process with a.s. continuous paths such that

\[ W(0)=0, \]

it has independent increments, and, for all \(t>s\geq 0\),

\[ W(t)-W(s) \]

has a Gaussian distribution with mean \(0\) and variance \(t-s\):

\[ W(t)-W(s)\sim\mathcal N(0,t-s). \]

The density of the random variable \(W(t)-W(s)\) is

\[ g(x;t,s) = \frac{1}{\sqrt{2\pi(t-s)}} \exp\left( -\frac{x^2}{2(t-s)} \right). \]

Although Brownian motion has stationary increments, it is not a stationary process. The probability density of the one-dimensional Brownian motion is

\[ g(x,t) = \frac{1}{\sqrt{2\pi t}} \exp\left( -\frac{x^2}{2t} \right). \]

Similarly, a \(d\)-dimensional Brownian motion

\[ W(t):\mathbb R^+\to\mathbb R^d \]

is a vector of \(d\) independent Brownian motions,

\[ W(t) = \left( W_1(t),\ldots,W_d(t) \right). \]

The density of the Gaussian random variable \(W(t)-W(s)\) is

\[ g(x;t,s) = \frac{1}{[2\pi(t-s)]^{d/2}} \exp\left( -\frac{\|x\|^2}{2(t-s)} \right). \]

For \(d\)-dimensional Brownian motion,

\[ \mathbb E[W(t)] = 0, \qquad \forall t\geq 0, \]

and

\[ \mathbb E \left[ (W(t)-W(s)) \otimes (W(t)-W(s)) \right] = (t-s)I. \]

Moreover,

\[ \mathbb E \left[ W(t)\otimes W(s) \right] = \min(t,s)I. \]

Properties of Brownian motion

The following properties will be central to the development of stochastic calculus. I choose to present them in a non-technical way, using heuristic arguments rather than proofs.

Hölder continuity

Brownian paths are almost surely locally Hölder continuous with any exponent

\[ \alpha \in \left(0,\frac12\right). \]

That is, for every \(T>0\) and \(\alpha<1/2\), there exists an a.s. finite random constant \(C\) such that

\[ |W_t-W_s| \leq C|t-s|^\alpha, \qquad s,t\in[0,T]. \]

Brownian motion has unbounded variation

For a partition

\[ \Pi=\{0=t_0<t_1<\cdots<t_n=t\}, \]

define

\[ \Delta W_k=W_{t_{k+1}}-W_{t_k}, \qquad \Delta t_k=t_{k+1}-t_k, \]

Since (remember that \(\Delta W_k\sim\mathcal N(0,\Delta t_k)\))

\[ \mathbb E[(\Delta W_k)^2]=\Delta t_k, \qquad \mathbb E[(\Delta W_k)^4]=3(\Delta t_k)^2, \]

we have

\[ \mathbb E[\sum_{t_k<t}(\Delta W_k)^2] = \sum_{t_k<t}\Delta t_k = t. \]

Moreover, by independence of the increments,

\[ \begin{aligned} \mathbb E[(\sum_{t_k<t}(\Delta W_k)^2-t)^2] &= \sum_{t_k<t} \mathbb E \left[ \left((\Delta W_k)^2-\Delta t_k\right)^2 \right] \\ &= \sum_{t_k<t} \left( 3(\Delta t_k)^2-(\Delta t_k)^2 \right) \\ &= 2\sum_{t_k<t}(\Delta t_k)^2 \longrightarrow 0. \end{aligned} \]

Hence

\[ \sum_{t_k<t}(\Delta W_k)^2. = \sum_{t_k<t}(\Delta W_k)^2 \longrightarrow t \qquad\text{in }L^2. \]

Now suppose, by contradiction, that a Brownian path has bounded variation on \([0,t]\). Then

\[ V_t(W) = \sup_{\Pi} \sum_{t_k<t}|\Delta W_k| <\infty, \]

so, for every partition \(\Pi\),

\[ \sum_{t_k<t}|\Delta W_k| \leq V_t(W). \]

Moreover,

\[ \begin{aligned} \sum_{t_k<t}(\Delta W_k)^2 &= \sum_{t_k<t}|\Delta W_k|\,|\Delta W_k| \\ &\leq \max_{t_k<t}|\Delta W_k| \sum_{t_k<t}|\Delta W_k| \\ &\leq \max_{t_k<t}|\Delta W_k|\,V_t(W). \end{aligned} \]

As \(|\Pi|\to0\), continuity of the Brownian path on the compact interval \([0,t]\) implies uniform continuity, and therefore

\[ \max_{t_k<t}|\Delta W_k| \longrightarrow0. \]

Since \(V_t(W)<\infty\) by assumption,

\[ 0 \leq \sum_{t_k<t}(\Delta W_k)^2 \leq \max_{t_k<t}|\Delta W_k|\,V_t(W) \longrightarrow0. \]

Hence a continuous path of bounded variation must have zero quadratic variation:

\[ \sum_{t_k<t}(\Delta W_k)^2\longrightarrow0. \]

But for Brownian motion we have just shown that

\[ \sum_{t_k<t}(\Delta W_k)^2\longrightarrow t. \]

For \(t>0\) this is a contradiction. Therefore

\[ V_t(W) = \sup_{\Pi} \sum_{t_k<t}|W_{t_{k+1}}-W_{t_k}| = \infty \qquad\text{a.s.} \]

Non-differentiability

Brownian paths are almost surely nowhere differentiable.

This should not be surprising: Brownian motion is continuous, but its paths have infinite variation on every non-trivial time interval.

Quadratic variation

We have just shown that, in \(L^2\),

\[ \sum_{t_k<t}(\Delta W_k)^2 \longrightarrow t. \]

This is conventionally written

\[ (dW_t)^2=dt \]

or, informally over a small increment,

\[ (\Delta W)^2\sim\Delta t. \]

Since the second-order term contributes at the same order as an ordinary time increment it cannot be discarded when Taylor-expanding functions of Brownian motion.

Two other increment rules

For a Brownian increment,

\[ \Delta W = W_{t+\Delta t}-W_t \sim \mathcal N(0,\Delta t). \]

Equivalently,

\[ \Delta W=\sqrt{\Delta t}\,Z, \qquad Z\sim\mathcal N(0,1). \]

Thus a Brownian increment is of order \(\sqrt{\Delta t}\). Consequently,

\[ (\Delta W)^2 = \Delta t\,Z^2, \]

is of order \(\Delta t\), while

\[ \Delta W\,\Delta t = (\Delta t)^{3/2}Z \]

and

\[ (\Delta t)^2 \]

are of higher order.

As before we could show that, in \(L^2\),

\[ \sum_k\Delta W_k\,\Delta t_k \longrightarrow 0, \qquad \sum_k(\Delta t_k)^2 \longrightarrow 0. \]

which gives

\[ dW_t\,dt=0, \qquad (dt)^2=0. \]

Brownian scaling

Let \(W_t\) be a one-dimensional Brownian motion and let \(c>0\). Define

\[ \frac{1}{c}W_{c^2t}. \]

is again a Brownian motion.

Equivalently, setting \(a=c^2\),

\[ \frac{1}{\sqrt a}W_{at} \overset{d}{=}W_t. \]

This is the Brownian scaling property.

Shift invariance

For every \(c>0\), the process

\[ X_t=W_{c+t}-W_c \]

is a Brownian motion independent of

\[ \{W_u:u\in[0,c]\}. \]

Time reversal

Let

\[ X_t=W_{1-t}-W_1, \qquad t\in[0,1]. \]

Then

\[ X_t\overset{d}{=}W_t. \]

Time inversion

Define

\[ X_0=0, \qquad X_t=tW_{1/t}, \quad t>0. \]

Then

\[ X_t\overset{d}{=}W_t. \]