> [!theorem] > > Let $C$ be a semipositive matrix on $\real^d$, and $B_t$ be a [[Lévy Process]] associated with $\pi_{0, I, 0} = \gamma_{0, I}$, then $X_t = C^{1/2}B_t$ is a Lévy Process associated with $\pi_{0, C, 0}$. > > *Proof*. For all $s$, > $ > \ev\braks{e^{i\angles{\xi, X_t}}} = \ev\braks{e^{i\angles{C^{1/2}\xi, B_t}}} = \exp\paren{-\frac{\xi C \xi}{2}} > $ > [!theorem] > > Let $B_t$ be a Lévy process associated with $\gamma_{0, I}$, then $B_t$ is the **standard Brownian motion**. The [[Probability|probability]] measure $\mathcal{W}$ on the product $\sigma$-algebra associated with $B$, is known as the Wiener measure. > > 1. For each $0 \le s \le t$, $\mu_{B_t} = \gamma_{0, tI}$, $\mu_{B_t - B_s} = \gamma_{0, (t - s)I}$, with $B_t - B_s$ independent from $B_r$ for all $r \le s$. > 2. If $B_t = (B_t^{1}, \cdots, B_t^{(d)})$, then each $B_t^{(j)}$ is a standard one-dimensional Brownian motion, with the coordinates being [[Probabilistic Independence|mutually independent]]. > 3. For each $1 \le j \le d$ and $s, t \ge 0$, $\ev\braks{B_s^{(j)} \cdot B_t^{(j)}} = \min(t, s)$. > 4. For each $s, t \ge 0$, $\ev\braks{\angles{B_s, B_t}} = d\min(t, s)$. > 5. $\bracs{\angles{B_t, \xi}: t \ge 0, \xi \in \real^d}$ is a centred Gaussian family. That is, for any $(t_1, \cdots, t_k)$ and $(\xi_1, \cdots, \xi_k)$, the family $\bracs{\angles{B_j, \xi_j}: j \le k}$ has a multivariate Gaussian distribution. Moreover, $\ev\braks{\angles{B_s, \xi} \cdot \angles{B_t, \zeta}} = \min(s, t)\angles{\xi, \zeta}$. > [!theorem] > > Let $B_t$ be a standard one-dimensional Brownian motion. Then the following processes are [[Martingale|martingales]] with respect to the natural [[Filtration|filtration]]: > 1. $B_t$ itself. > 2. $B_t^2 - t$. > 3. For any $\xi \in \real$, $\exp(\xi B_t \pm t\xi^2/2)$. > > Moreover, if the process is arbitrary, any part of $(3)$ holds and the process has RCLL sample paths, then the process is the Brownian motion. In addition, $(1) + (2)$ combined with continuous sample paths implies that the process is a Brownian motion. > [!theorem] > > For any $T > 0$ and $\alpha \in (0, 1/2)$, there exists a modification $B'$ of $B$ such that the mapping $t \mapsto B_t'$ is [[Almost Everywhere|a.s.]] locally [[Hölder Space|Hölder continuous]] with parameter $\alpha$. > > *Proof*. Let $n \ge 1$, then for any $T > 0$ and $s, t \le T$, > $ > \ev\braks{\abs{X_t - X_s}^{2n}}^{1/2n} \le C_{n, d}\abs{t - s}^{1/2} > $ > where $p = 2n$ and $r = 1/2 - 1/2n$. By choosing $n$ to be arbitrarily large, the result can hold for anything below $1/2$. By Kolmogorov's continuity criterion, $B$ admits a modification with the desired properties. > [!theorem] > > Let $B$ be the standard Brownian motion, then for any $\alpha > 1/2$, > $ > \bp\bracs{\exists s \ge 0: \limsup_{t \downto s}\frac{\abs{B_t - B_s}}{\abs{t - s}^\alpha} < \infty} = 0 > $ > In other words, > $ > \limsup_{t \downto s}\frac{\abs{B_t - B_s}}{\abs{t - s}^\alpha} = \infty > $ > and Brownian motion is almost surely nowhere Hölder-$\alpha$ continuous for all $\alpha > 1/2$, and thus almost surely nowhere differentiable, and does not have bounded variation. > > *Proof*. Let $E$ be the event where $B$ is Hölder-$\alpha$ continuous at some point in $[0, 1]$. Let $L \in \nat$. Suppose that $\omega \in E$, then there exists $s \in [0, 1]$, $A > 0$, and $\delta > 0$ such that for all $t \in B(s, \delta)$, $\abs{B_t(\omega) - B_s(\omega)} \le A \abs{t - s}^\alpha$. Moreover, if $t, t' \in B(s, \delta)$, then > $ > \abs{B_t(\omega) - B_{t'}(\omega)} \le A(\abs{t - s}^\alpha + \abs{t' - s}^\alpha) > $ > Divide the interval into $n$ equal pieces, then there exists $m \le n$ such that $s \in [(m-1)/n, m/n]$. Choose $n$ large enough such that $\bracs{(m+j)/n: 0 \le j \le L}$ is contained in $B(s, \delta)$. Applying the above inequality on the segments yield that > $ > \begin{align*} > \abs{B_{(m+k+1)/n}(\omega) - B_{(m+k)/n}} &\le A\braks{\abs{\frac{m+k+1}{n} - s}^\alpha + \abs{\frac{m+k}{n} - s}^\alpha} > \end{align*} > $ > Therefore > $ > E \subset \bigcup_{A = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{n \ge N} \bigcup_{m \le n} \bigcap_{k < L} \bracs{\abs{B_{(m+k+1)/n} - B_{(m+k)/n}}\le 2A\paren{\frac{L+1}{n}}^\alpha} > $ > where factoring through independence gives > $ > \begin{align*} > &\bigcap_{n \ge N} \bigcup_{m \le n} \bigcap_{k < L} \bracs{\abs{B_{(m+k+1)/n} - B_{(m+k)/n}}\le 2A\paren{\frac{L+1}{n}}^\alpha} \\ > &\le \liminf_{n \to \infty}n\braks{\gamma_{0, I/n}(B_{0, C_{A, L, \alpha}n^{-\alpha}})}^2 \\ > &\le \liminf_{n \to \infty}n\gamma_{0, I}(B_{0, Cn^{1/2 - \alpha}})^2 \\ > &\le C\liminf_{n \to \infty}n^{(1/2 - \alpha)dL+1} > \end{align*} > $ > By choosing a sufficiently large $L$, the resulting limit goes to $0$. > [!theorem] > > Since Brownian motion is almost surely continuous, $\mathcal{W}$ is a measure on $C([0, \infty), \real^d)$ with respect to the product $\sigma$-algebra, known as the **classical Wiener measure**. Let $\mathcal W_t = (\pi_t)_* \mathcal W$ be the pushforward by the projection, then > - Under $\mathcal W$, $\pi_t - \pi_s$ is independent from $\pi_r$ whenever $r \le s, t$. > - $\mathcal W$ assigns probability $1$ to the space of continuous functions. > - $\mathcal W$ assigns probability $0$ to the space of differentiable functions. > [!theorem] > > $\mathcal W$ is invariant under the following transformations: > 1. **Rescaling**: for any $\lambda > 0$, the map $f \mapsto \sqrt{\lambda} \cdot f \circ \lambda$. > 2. **Time Inversion**: $f \mapsto tf(1/t)$, filling in $0$ at $t = 0$. > 3. **Symmetry**: $f \mapsto -f$. > 4. **Time Reversal**: Given a finite time interval $[0, T]$, $f \mapsto f(T) - f(T - t)$. > [!theorem] > > Let $B_t$ be the standard Brownian motion and $0 \in (a, b) \subset \real$. Define > $ > \tau_1 = \inf\bracs{t \ge 0: B_t \ge b} \quad \tau_2 = \inf\bracs{t \ge 0: B_t \le a} > $ > and $\tau = \tau_1 \wedge \tau_2$ as the **exiting time** of $(a, b)$. > > So $B_{t \wedge \tau}$ has sample paths contained in $[a, b]$ for all $t$. By [[Hunt's Theorem]], > $ > \ev\braks{B_{\tau \wedge \tau}} = \ev\braks{B_{0 \wedge \tau}} = 0 > $ > thus > $ > b\bp\bracs{\tau = \tau_b} + (-a)\bp\bracs{\tau = \tau_a} > $ > which allows solving > $ > \bp\bracs{\tau = \tau_a} = \frac{b}{a+b} > $