91Ó°ÊÓ

For a standard Brownian motion \(W_t\), at any time t, the value \(W_t\) only depends on the information up to time t, namely \(\mathcal{F}_t\). Therefore, the process is adapted to the filtration \(\mathcal{F}_t\).

Recall that for a standard Brownian motion \(W_t\), \(W_t \sim \mathcal{N}(0, t)\). Thus, the expectation value of \(W_t\) is given by: \(E[|W_t|] = \int_{-\infty}^{\infty}|x|\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx\) Splitting the integral into the two intervals, where x is positive and negative: \(E[|W_t|] = 2\int_{0}^{\infty}x\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx\) Using integration by substitution, let \(u = \frac{x^2}{2t}\), we have: \(E[|W_t|] = 2\int_{0}^{\infty}\sqrt{2tu}\frac{1}{\sqrt{2\pi t}}e^{-u}d(\sqrt{2tu}) = \sqrt{\frac{2t}{\pi}}\) Since t is finite, \(E[|W_t|]\) must also be finite.

Now let's compute the conditional expectation \(E[W_t|\mathcal{F}_s]\), where \(0 \le s \le t\): \(E[W_t|\mathcal{F}_s] = E[(W_t - W_s) + W_s|\mathcal{F}_s] = E[W_t - W_s|\mathcal{F}_s] + E[W_s|\mathcal{F}_s]\) Since \(W_t - W_s \sim \mathcal{N}(0, t-s) \) is independent of \(\mathcal{F}_s\), the term \(E[W_t - W_s|\mathcal{F}_s]\) becomes: \(E[W_t - W_s|\mathcal{F}_s] = E[W_t - W_s] = 0\) For the term \(E[W_s|\mathcal{F}_s]\), it is equal to \(W_s\) as the process is adapted and the value of \(W_s\) depends only on \(\mathcal{F}_s\). Thus, we have: \(E[W_t|\mathcal{F}_s] = 0 + W_s = W_s\) Since the three conditions of a Martingale process are satisfied, we can conclude that the standard Brownian motion is a Martingale.