Brownian motion §

The four axioms each pin down one aspect of the process, and together they force the entire law of $W$.

Step 1 — Start at zero and move continuously. Axioms 1 and 2 say the path begins at the origin and never jumps: $t \mapsto W_t$ is a continuous curve. This is what separates Brownian motion from a jump process such as the Poisson process. Continuity alone, however, does not fix the distribution — a deterministic drift $W_t = \mu t$ is also continuous.

Step 2 — Independent, stationary increments. Axioms 3 and 4 say the “steps” of the path are memoryless and time-homogeneous: the increment over any interval of length $h$ has the same law, $\mathcal{N}(0, h)$, and increments over disjoint intervals are independent. This is the continuous-time analogue of the i.i.d. increments of a random walk. It makes $W$ a Markov process (the future depends only on the present) and a martingale (the best prediction of $W_t$ given the past is $W_s$).

Step 3 — The Gaussian law is forced. Because the increments are independent and stationary, $W_t$ is a sum of $t$ worth of independent Gaussian increments; the only law consistent with all four axioms is $W_t \sim \mathcal{N}(0, t)$. The covariance follows by conditioning on the overlap of the two intervals:

Example. Take $s = 2$, $t = 5$. Then $W_2 \sim \mathcal{N}(0,2)$, $W_5 \sim \mathcal{N}(0,5)$, and $\mathrm{Cov}(W_2, W_5) = 2$. The correlation is $\mathrm{Corr}(W_2, W_5) = 2/\sqrt{2 \cdot 5} \approx 0.63$: the two values are positively correlated because the path from $0$ to $2$ is shared.

Authoritative sources

• Wiener process (Wikipedia)

Step 1 — The CLT is a statement about one time. The central limit theorem says $S_n/(\sigma\sqrt{n}) \xrightarrow{d} \mathcal{N}(0,1)$: at a single time $t = 1$, the rescaled walk looks Gaussian. Donsker’s theorem upgrades this to a statement about the whole path: the entire rescaled trajectory converges, not just its value at one point.

Step 2 — Why the $\sqrt{n}$ scaling. The walk $S_n$ has standard deviation $\sigma\sqrt{n}$, so dividing by $\sigma\sqrt{n}$ keeps the vertical scale fixed. Horizontally, $n$ steps are compressed into the unit interval $[0,1]$, so each step spans $1/n$ in time. The vertical step size is $\sigma/\sqrt{n}$ and the horizontal step size is $1/n$; their ratio is $\sigma/\sqrt{n} : 1/n = \sigma\sqrt{n} : 1$, which grows like $\sqrt{n}$. This is why the limit path is continuous yet wildly rough — the steps shrink in both directions, but the vertical steps shrink more slowly.

Step 3 — The limit inherits the axioms. The rescaled process has (approximately) independent, stationary increments and a Gaussian marginal at each time; Donsker’s theorem makes this rigorous and shows the limit is continuous, hence a Brownian motion.

Example. A simple symmetric random walk ($X_i = \pm 1$ with probability $1/2$) has $\sigma^2 = 1$. After $n = 10^4$ steps the walk is typically at height $\pm 100$; rescaled by $1/\sqrt{n} = 1/100$ and plotted over $[0,1]$, it looks like a Brownian path.

Authoritative sources

• Invariance principle (Wikipedia)

Step 1 — Continuity does not imply smoothness. A continuous function can still be nowhere differentiable (think of the Weierstrass function). Brownian paths are even rougher: the derivative $dW_t/dt$ does not exist anywhere. Intuitively, zooming into any interval reveals the same jagged structure at every scale — the path is self-similar.

Step 2 — Self-similarity. For any $c > 0$, the process $c^{-1/2} W_{ct}$ is again a standard Brownian motion. This scaling symmetry is the continuous-time echo of the $\sqrt{n}$ scaling in the random-walk limit, and it is why the roughness is the same at every magnification.

Step 3 — Quadratic variation. For a smooth function $f$, the sum $\sum (f_{t_{i+1}} - f_{t_i})^2$ over a fine partition tends to $0$ (the squared increments are of order $(\Delta t)^2$). For Brownian motion the increments are of order $\sqrt{\Delta t}$, so the squared increments are of order $\Delta t$, and the sum converges to the length of the interval:

This is the single fact that makes Itô calculus different from ordinary calculus: the “infinitesimal” $dW_t$ satisfies $(dW_t)^2 = dt$, not $0$.

Example. Over $[0,1]$, refine the partition from $10$ to $1000$ intervals. The sum of squared increments stays near $1$ (the value of $t$), not near $0$ as it would for a smooth curve.

Authoritative sources

• Quadratic variation (Wikipedia)

Step 1 — Why the ordinary chain rule fails. The chain rule $df(W_t) = f'(W_t)\,dW_t$ is wrong for Brownian motion because $W$ has nonzero quadratic variation. The missing term is $\tfrac{1}{2} f''(W_t)\,dt$, which comes from the rule $(dW_t)^2 = dt$.

Step 2 — Itô’s lemma. For $f \in C^2$, a Taylor expansion $f(W_{t+\Delta t}) - f(W_t) \approx f'(W_t)\Delta W + \tfrac{1}{2} f''(W_t)(\Delta W)^2$ plus the fact that $(\Delta W)^2 \approx \Delta t$ (in the quadratic-variation sense) gives

Example. Take $f(x) = x^2$. Then $d(W_t^2) = 2 W_t\,dW_t + dt$. Integrating: $W_t^2 = 2\int_0^t W_s\,dW_s + t$, so $\int_0^t W_s\,dW_s = \tfrac{1}{2}(W_t^2 - t)$. The $-t$ term is the Itô correction that has no classical analogue.

Step 3 — SDEs and geometric Brownian motion. An SDE $dX_t = \mu\,dt + \sigma\,dW_t$ is solved by $X_t = X_0 + \mu t + \sigma W_t$. The geometric SDE $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ keeps $S_t > 0$ (multiplicative noise) and is the Black–Scholes price model. Applying Itô’s lemma to $f(x) = \ln x$ gives $d\ln S_t = (\mu - \tfrac{1}{2}\sigma^2)\,dt + \sigma\,dW_t$, so

The $-\tfrac{1}{2}\sigma^2$ in the exponent is again the Itô correction: it is why the expected price $\mathbb{E}[S_t] = S_0 e^{\mu t}$ grows at rate $\mu$ even though the log-drift is $\mu - \tfrac{1}{2}\sigma^2$.

Authoritative sources

• Itô’s lemma (Wikipedia)
• Geometric Brownian motion (Wikipedia)

A stock price follows geometric Brownian motion $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ with $S_0 = 100$, $\mu = 0.08$, and $\sigma = 0.2$.

(a) Find $\mathbb{E}[S_1]$.

(b) Find the drift of $\ln S_t$.