Brownian Motion

In the previous blog, we have discussed the mathematical definition of Brownian motion, it’s distribution and alternative characterization. Now we are going to see the filtration of the Brownian motion. We previously have discussed little bit about filtration in the random walk. Let us go little bit more into the filtration.

Filtration for Brownian motion: Let \((\Omega, \mathbb{F}, \mathbb{P})\) be a probability space on which the Brownian motion \(W(t), \ t \geq 0\). A Filterartion for the Brownian motion is the collection of \(\sigma-\text{algebras}\) \(\mathbb{F}(t), \text{for} \ t\geq 0\) satisfying:

• Information accumulates: For \(0 \leq s < t\) every set in \(\mathbb{F}(s)\) is also in \(\mathbb{F}(t)\). In other words, there is at least as much information available at later time \(\mathbb{F}(t)\) as there is at the earlier time \(\mathbb{F}(s)\).
• Adaptivity: For \(t \geq 0\), the Brownian motion W(t) at time t is \(\mathbb{F(t)}\)-measurable. In other words, the information available at time t is enough to evaluate Brownian motion at time t, i.e. \(W(t)\).
• Independence of future increments: For \(0 \leq t < u\), the increment \(W(t)-W(u)\) is independent of \(\mathbb{F(t)}\). In other words, an increment of the Brownian motion after t is independent of the information available at time t.

Let \(\Delta(t), \ t \geq 0\) be a stochastic process. Then \(\Delta(t)\) is adapted to the filtration \(\mathbb{F(t)}\) if- for each \(t\geq 0\) the random variable \(\Delta(t)\) is \(\mathbb{F(t)}\) measurable.

Martingale Property of Brownian motion:

In the random walk, we proved it’s martingale property. Similarly here for the Brownian motion we will try to prove the same.

Theorem: Brownian motion is a martingale.

Proof: Let \(0 \leq s \leq t\) is given. Then- \[\begin{aligned} E[W(t)|\mathbb{F(s)}] &= E[\{W(s)+W(t)-W(s)\}|\mathbb{F(s)}] \\ &= E[W(s)|\mathbb{F(s)}]+E[W(t)-W(s)|\mathbb{F(s)}] \\ &= W(s)+ 0 = W(s) \end{aligned}\] Given \(\mathbb{F(s)}\) we know the value of \(W(s)\) by property (1). So, \(E[W(s)|\mathbb{F(s)}] = W(s)\).

By property (2) we have- \([W(t)-W(s)|\mathbb{F(s)}] = W(t)-W(s)\).

Which gives us- \(E[W(t)-W(s)|\mathbb{F(s)}] =E[W(t)-W(s)]\).

Now if you remember, for \(0 \leq s \leq t\) we have \(W(t)-W(s) \sim \text{N(0, t-s)}\), hence, \(E[W(t)-W(s)] = 0\). Q.E.D

In the next blog we are going to see the quadratic variation of the Brownian motion. Happy reading!!