1.1 Brownian Motion

Theorem 1.1.1. Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space and $\bracs{B_t|t \ge 0}$ be a $\real^{d}$-valued, $\bracs{\cf_t}$-progressively measurable process such that $B_{0} = 0$ almost surely, then the following are equivalent:

$\bracs{B_t|t \ge 0}$ is a Brownian motion.
For each $\xi \in \real^{d}$, the process
\[X^{\theta}_{t} = \exp\braks{i \dpn{B_t, \xi}{\real^d} + \frac{1}{2}\norm{\xi}_{\real^d}^2t}\]
is a $\bracs{\mathcal{F}_t}$-martingale.
For each $\phi \in BC^{2}(\real^{d})$, the process
\[Y^{\phi}_{t} = \phi(B_{t}) - \frac{1}{2}\int_{0}^{t} \Delta f(B_{s})ds\]
is a $\bracs{\mathcal{F}_t}$-martingale.

Proof [Theorem 4.1.1, SV97]. (1) $\Rightarrow$ (2): For each $1 \le s \le t$ and $\xi \in \real^{d}$,

\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&= \ev\braks{e^{i \dpn{B_s, \xi}{\real^d}}e^{i \dpn{B_t - B_s, \xi}{\real^d}} \big| \cf_s}\\&= e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}

Therefore

\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d} + t\norm{\xi}^2/2} \big| \cf_s}= e^{i \dpn{B_s, \xi}{\real^d} + s\norm{\xi}_{\real^d}^2/2}\]

(2) $\Rightarrow$ (3): For each $0 \le s < t$ and $\xi \in \real^{d}$,

\begin{align*}\frac{d}{dt}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}} \big| \cf_s}&= -\frac{\norm{\xi}_{\real^d}^{2}}{2}e^{i \dpn{B_s, \xi}{\real^d}}e^{-(t - s)\norm{\xi}_{\real^d}^2/2}\end{align*}

Therefore for any $0 \le s < t$,

\begin{align*}\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}&= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} e^{i \dpn{B_s, \xi}{\real^d}}e^{-(r - s)\norm{\xi}_{\real^d}^2/2}dr \\&= -\frac{\norm{\xi}_{\real^d}^{2}}{2}\int_{s}^{t} \ev\braks{e^{i \dpn{B_r, \xi}{\real^d}} \big | \cf_{s}}dr\end{align*}

If $\phi_{\xi}(x) = e^{-i \dpn{x, \xi}{\real^d}}$, then

\[\ev\braks{e^{i \dpn{B_t, \xi}{\real^d}}-e^{i \dpn{B_s, \xi}{\real^d}} \big| \cf_s}= \ev\braks{\int_s^t \frac{1}{2}(\Delta \phi_\xi)(B_r)dd \bigg | \cf_s}\]

so the process

\[Z^{\xi}_{t} = e^{i \dpn{B_t, \xi}{\real^d}}+ \int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r})dr\]

is a $\bracs{\mathcal{F}_t}$-martingale.

Let $\phi \in \mathcal{S}(\real^{d})$, then by the Fourier Inversion Theorem, there exists $\psi \in \mathcal{S}(\real^{d})$ such that for any $x \in \real^{d}$,

\[\phi(x) = \int e^{i \dpn{\xi, x}{\real^d}}\psi(\xi)d\xi\]

In which case, for any $t \ge 0$,

\[\phi(B_{t}) + \int_{0}^{t} \frac{1}{2}(\Delta \phi)(B_{r})dr = \int_{\real^d}e^{i \dpn{\xi, B_t}{\real^d}}\psi(\xi)d\xi + \int_{\real^d}\int_{0}^{t} \frac{1}{2}(\Delta \phi_{\xi})(B_{r}) \psi(\xi) dr d\xi\]

By the Dominated Convergence Theorem, the above also holds for any $\phi \in BC^{2}(\real^{d})$. Therefore the process $\bracsn{Y^\phi_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale as well.$\square$

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1.1 Brownian Motion

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