Brownian motion and Gaussian increments

Definition Let $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$ be a filtered probability space. A process $W = (W_t)_{t \ge 0}$ is a standard Brownian motion with respect to $(\mathcal{F}_t)$ when $W_0 = 0$, the process is adapted, almost every path $t \mapsto W_t$ is continuous, and for $0 \le s < t$

where the second line says that the increment is independent of every event in $\mathcal{F}_s$. That such a process exists is Wiener’s theorem, which the course states without constructing the process. The path regularity theorems are also stated without proof, for example that almost every path is nowhere differentiable. With the natural filtration of $W$, the independence condition says that increments over disjoint intervals are independent. With a larger filtration it says more, because the increment must be independent of everything in $\mathcal{F}_s$, and enlarging the filtration with information about the future can destroy the property. A price model built on $W$, such as $S_t = S_0 + \mu t + \sigma W_t$, is an assumption about prices and not a consequence of the definition.

Definition For times $0 = t_0 < t_1 < \dots < t_n$, the increments $W_{t_1} - W_{t_0}, \dots, W_{t_n} - W_{t_{n-1}}$ of a standard Brownian motion are independent, and the $k$th one is $N(0, t_k - t_{k-1})$. Over an interval of length $h > 0$ the increment has variance $h$ and standard deviation $\sqrt{h}$, and it has the same law as $\sqrt{h}\, Z$ with $Z \sim N(0, 1)$. The levels $(W_{t_1}, \dots, W_{t_n})$ are partial sums of the increments, so this vector is jointly Gaussian with mean zero. With $t$ in years, $W_t$ is measured in square-root years, and a coefficient $\sigma$ multiplying it carries the units of the modeled quantity per square-root year. Over a step of length $h$ a drift $\mu$ contributes $\mu h$, while the shock $\sigma (W_{t+h} - W_t)$ has standard deviation $\lvert \sigma \rvert \sqrt{h}$, so when $\sigma \ne 0$ and $h$ is small the typical shock is much larger than the drift, because $\sqrt{h}/h = 1/\sqrt{h}$ grows without bound.

Definition For $h > 0$ the Brownian transition density is

the density in $y$ of the normal law with mean $x$ and variance $h$. For $s \ge 0$, $h > 0$ and every bounded measurable function $f$,

almost surely, because $W_{s+h} = W_s + (W_{s+h} - W_s)$, where $W_s$ is known at time $s$ and the increment is independent of $\mathcal{F}_s$. This is the Markov property, which says that the conditional law of a future level depends on the past only through the current level. The densities compose by the Chapman-Kolmogorov equation

for $h, k > 0$,

and the joint density of $(W_{t_1}, \dots, W_{t_n})$ is the product of the transition densities over the successive intervals, starting from $x_0 = 0$. There is no density at $h = 0$, where the law is the point mass at $x$.

Definition For all $s, t \ge 0$,

For $s \le t$ write $W_t = W_s + (W_t - W_s)$. The increment is independent of $W_s$ and has mean zero, so

For $0 < s \le t$ the correlation of the two levels is $\sqrt{s/t}$. Conversely, a process with $W_0 = 0$, continuous paths and centered jointly Gaussian finite-dimensional laws with covariance $\min(s, t)$ is a standard Brownian motion for its natural filtration, because its increments are jointly Gaussian with variance $t - s$ and are uncorrelated over disjoint intervals, and uncorrelated jointly Gaussian variables are independent.

Definition For a fixed $c > 0$ let $W^{(c)}_t = c^{-1/2} W_{ct}$. Then $W^{(c)}$ is a standard Brownian motion with respect to the filtration $\mathcal{G}_t = \mathcal{F}_{ct}$, so

an equality of the laws of the whole processes. The proof checks the definition. The new process starts at $0$, has continuous paths and is adapted to $(\mathcal{G}_t)$, and for $s < t$ its increment $c^{-1/2}(W_{ct} - W_{cs})$ is independent of $\mathcal{F}_{cs} = \mathcal{G}_s$ and normal with variance $(ct - cs)/c = t - s$. The statement compares distributions, not values on one path. It concerns the driftless standard Brownian motion. For $X_t = \mu t + W_t$ the rescaled process $c^{-1/2} X_{ct} = \mu \sqrt{c}\, t + W^{(c)}_t$ has drift $\mu \sqrt{c}$, so a drifted Brownian motion does not keep its law under this rescaling unless $\mu = 0$ or $c = 1$.

Definition For $s \ge 0$, $h > 0$ and a threshold $b$, the transition density with $f = \mathbf{1}_{(b, \infty)}$ gives

almost surely, where $\Phi$ is the standard normal distribution function. Written with the regular conditional distribution given $W_s = x$, the probability is $1 - \Phi((b - x)/\sqrt{h})$. For a continuous variable this is not a ratio of probabilities, since the event $\{W_s = x\}$ has probability zero. The answer is the same for $W_{s+h} \ge b$, because the normal law has no atoms. As $h \to 0$ it tends to $1$ when $x > b$ and to $0$ when $x < b$, and it equals $\tfrac12$ for every $h$ when $x = b$. The event concerns the level at time $s + h$ only. The probability that the path reaches $b$ at some time in $[s, s + h]$ is larger, and for $x \le b$ it equals $2\bigl(1 - \Phi((b - x)/\sqrt{h})\bigr)$ by the reflection principle, a result not derived here.