Stochastic Processes and Financial Mathematics
(part two)
11.4 Properties of Brownian motion
We now examine some of the more detailed properties of Brownian motion. Recall that \(B_t\) denotes a standard Brownian motion.
Symmetry
The normal distribution \(Z\sim N(0,\sigma ^2)\) is symmetric about \(0\), in the sense that \(-Z\) also has the distribution \(N(0,\sigma ^2)\). This symmetry about \(0\) is also present in Brownian motion.
Proof: We must check that \(W_t=-B_t\) satisfies the three defining properties of Brownian motion. By the first property, \(B_t\) is almost surely continuous, hence \(-B_t\) is also almost surely continuous. Also, for \(0\leq u\leq t\) we have
\[W_t-W_u=-(B_t-B_u).\]
Since, by the second property, \(B_t-B_u\) is independent of \(\mc {F}_u\), so is \(W_t-W_u\), and we have
\[\sigma (W_v\-v\leq u)=\sigma (-W_v\-v\leq u)=\sigma (B_v\-v\leq u)=\mc {F}_u.\]
Thus \(W_t-W_u\) is independent of \(\sigma (W_v\-v\leq u)\).
Lastly, if \(Z\sim N(0,t)\) then, using the symmetry of normal random variables, \(-Z\sim N(0,t)\), so we have
\[W_t-W_u=-(B_t-B_u)\sim N(0,t-u)\]
by the third property. Hence, all three properties also hold for \((W_t)\), so \(W_t\) is a Brownian motion. Since \(W_0=-B_0=0\), we have that \((W_t)\) is a standard Brownian motion. ∎
Lemma 11.4.1 is often referred to as an example of a ‘self-symmetry’ of Brownian motion, meaning a transformation of a Brownian motion that results in another Brownian motion. It turns out that Brownian motion has many self-symmetries, and they are very important to the theory of Brownian motion.
Non-differentiability
As we’ve seen in Section 11.1, the paths of Brownian motion look very jagged and erratic. We can express this idea formally: the sample paths of Brownian motion are not differentiable!
Proof: Using the second property of Brownian motion, and the scaling properties of normal random variables,
\[\frac {B_{t+h}-B_t}{h}\;\sim \;\frac {B_h}{h}\;\sim \; \frac {X}{\sqrt {h}}.\]
where \(X\sim N(0,1)\). Note that \(X\) is positive half the time and negative half the time (and \(\P [X=0]=0\)). Hence, as \(h\to 0\), we obtain that
\(\seteqnumber{0}{11.}{5}\)\begin{equation} \label {eq:limXsqrth} \frac {X}{\sqrt {h}}\stackrel {a.s.}{\to } X_\infty = \begin{cases} \infty & \text { with probability }1/2, \\ -\infty & \text { with probability }1/2. \end {cases} \end{equation}
Form Lemma 6.1.2, almost sure convergence implies convergence in distribution, so this limit also holds in distribution. Since \(\frac {B_{t+h}-B_t}{h}\) has the same distribution as \(\frac {X}{\sqrt {h}}\), we obtain that \(\frac {B_{t+h}-B_t}{h}\) converges in distribution to \(X_\infty .\)
Consider the event \(E=\{B_t\text { is differentiable at }t\}\). When the event \(E\) occurs, \(\frac {B_{t+h}-B_t}{h}\) converges to a finite quantity as \(h\to 0\). However, we saw above that \(\frac {B_{t+h}-B_t}{h}\) had the limit \(X_\infty \), with \(\P [X_\infty \in \{\infty ,-\infty \}]=1\), so the probability that this limit is a finite quantity is zero. Therefore, \(\P [E]=0\). ∎
Pure mathematicians discovered functions that were nowhere differentiable at around the start of the \(20^{th}\) century. At first, they were widely thought to be mathematical curiosities, with little or no importance in the ‘real’ world. A few decades later, the discovery that Brownian motion played a key role in physics, biology and mathematical finance had reversed this viewpoint.
Relationship to martingales
It turns out that there are many martingales associated to Brownian motion. Here’s two, with two more to come in exercise 11.6, and others in later sections of the course.
Proof: It is enough to look at the case \((B_t)\) of standard Brownian motion, since adding and subtracting a deterministic constant does not change if a process is a martingale.
Since \(B_t\sim N(0,t)\) we have \(\var (B_t)<\infty \), which implies that \(B_t\in L^1\). Since the filtration \((\mc {F}_t)\) is the generated filtration of \(B_t\), is immediate that \(B_t\) is adapted. Lastly, for any \(0\leq u\leq t\) we have
\(\seteqnumber{0}{11.}{6}\)\begin{align*} \E [B_t\|\mc {F}_u]&=\E [B_t-B_u\|\mc {F}_u]+\E [B_u\|\mc {F}_u]\\ &=\E [B_t-B_u]+B_u\\ &=B_u. \end{align*} Here, we use the properties of Brownian motion: \(B_t-B_s\) is independent of \(\mc {F}_u\) and \(\E [B_t]=\E [B_u]=0\). ∎
Proof: Since \(B_t\sim N(0,t)\) we have \(\var (B_t)<\infty \), which implies \(B^2_t\in L^1\). Hence also \(B_t^2-t\in L^1\). Since \(B_t^2-t\) is a deterministic function of \(B_t\), we have that \(B_t^2-t\) is adapted to the generated filtration of \(B_t\). Lastly, for \(0\leq u\leq t\),
\(\seteqnumber{0}{11.}{6}\)\begin{align*} \E [B_t^2-t\|\mc {F}_u] &=\E [(B_t-B_u)^2+2B_tB_u-B_u^2\|\mc {F}_u]-t\\ &=\E [(B_t-B_u)^2\|\mc {F}_u]+2B_u\E [B_t\|\mc {F}_u]-B_u^2-t\\ &=\E [(B_t-B_u)^2]+2B_u^2-B_u^2-t\\ &=(t-u)+B_u^2-t\\ &=B_u^2-u \end{align*} as required. Here we use that \(B_t-B_u\) is independent of \(\mc {F}_u\), along with both (11.1) and Lemma 11.4.3. ∎