Chapter 1 — From Random Walks to Brownian Motion

Guiding question

How can a process be continuous in time, random at every scale, and still mathematically tractable enough to support calculus, pricing, and risk management?

This chapter builds Brownian motion from the discrete random walk. The point is not to introduce a mysterious object by definition, but to discover why this object is almost inevitable once one asks for a non-degenerate continuous-time limit of many small independent shocks.

We will encounter Brownian motion in several equivalent languages:

1. as the scaling limit of random walks;
2. as a centered Gaussian process with covariance kernel $K(s,t)=\min(s,t)$;
3. as a continuous martingale with independent stationary increments;
4. as a random element of path space;
5. as the canonical source of quadratic variation.

These viewpoints are not decorative. Each one becomes essential later: the scaling limit explains volatility scaling, the Gaussian kernel gives geometry, the martingale property underlies no-arbitrage pricing, and quadratic variation forces Itô calculus.

---

1.1 Discrete randomness: the symmetric random walk

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. Let

$$\xi_1,\xi_2,\ldots$$

be independent identically distributed random variables satisfying

$$\mathbb P(\xi_k=1)=\mathbb P(\xi_k=-1)=\frac12.$$

Define the random walk

$$S_n = \sum_{k=1}^n \xi_k, \qquad S_0=0.$$

Then

$$\mathbb E[S_n]=0, \qquad \operatorname{Var}(S_n)=n.$$

Proof

Linearity gives

$$\mathbb E[S_n]=\sum_{k=1}^n \mathbb E[\xi_k]=0.$$

Since the $\xi_k$'s are independent and $\operatorname{Var}(\xi_k)=1$,

$$\operatorname{Var}(S_n) =\sum_{k=1}^n \operatorname{Var}(\xi_k) =n.$$

The variance grows linearly in the number of steps, so the standard deviation grows like $\sqrt n$. This square-root scaling is the first sign of Brownian geometry.

Geometric meaning of $\sqrt n$

After $n$ independent zero-mean shocks, positive and negative shocks partially cancel. The net displacement is not usually of order $n$, but of order $\sqrt n$. Randomness accumulates like Euclidean length: variances add, standard deviations scale as square roots.

---

1.2 Embedding the walk into continuous time

Fix a time horizon $T>0$. For each integer $n$, divide $[0,T]$ into $n$ intervals of length

$$\Delta t = \frac{T}{n}.$$

If each shock has size $\sqrt{\Delta t}$, define

$$X_t^{(n)} = \sqrt{\Delta t}\, S_{\lfloor t/\Delta t\rfloor}.$$

At time $t$, approximately $t/\Delta t$ shocks have occurred, so

$$\operatorname{Var}\bigl(X_t^{(n)}\bigr) = \Delta t\, \left\lfloor \frac{t}{\Delta t}\right\rfloor \approx t.$$

The scaling $\sqrt{\Delta t}$ is not arbitrary. It is the only natural scaling that keeps the limiting variance finite and nonzero.

Why not scale by $\Delta t$?

If the shock size were $\Delta t$, then variance over a fixed horizon would be approximately

$$\frac{t}{\Delta t}(\Delta t)^2 = t\Delta t \to 0.$$

The limit would become deterministic.

Why not use unscaled shocks?

If each shock had fixed size, variance after $t/\Delta t$ shocks would be approximately

$$\frac{t}{\Delta t}\to \infty.$$

The limiting object would explode.

Random walk scaling: discrete paths becoming diffusion-like

Number of steps: 200

The normalization S_floor(nt)/sqrt(n) keeps variance at time t of order t. Without this scaling, paths either collapse or explode.

Finance interpretation: square-root-of-time volatility

If independent return shocks accumulate through time, variances add. Therefore volatility scales as $\sqrt T$, not $T$. This is the primitive reason that daily volatility is annualized by multiplying by approximately $\sqrt{252}$, not by $252$.

---

1.3 Donsker's invariance principle

The classical central limit theorem says that for fixed time,

$$\frac{S_n}{\sqrt n}\Rightarrow N(0,1).$$

Donsker's theorem strengthens this from convergence of random variables to convergence of entire random paths.

Donsker's invariance principle, informal version

Let

$$X_t^{(n)} = \frac{1}{\sqrt n}S_{\lfloor nt\rfloor}, \qquad 0\le t\le 1,$$

with linear interpolation between grid points. Then

$$X^{(n)} \Rightarrow W$$

as random elements of the path space $C([0,1])$, where $W$ is standard Brownian motion.

The theorem says that Brownian motion is the universal scaling limit of many small independent shocks. The detailed distribution of the shocks is not essential; under suitable moment assumptions, the same limiting process appears.

Path-space viewpoint

For each $\omega\in\Omega$, the map

$$t\mapsto X_t^{(n)}(\omega)$$

is a curve. Donsker's theorem says the probability distribution on these curves converges to a probability distribution on continuous curves. Brownian motion is not one curve; it is a probability law on path space.

Brownian motion as a distribution over paths

One outcome omega gives one continuous curve t → W_t(omega). Brownian motion is the full distribution on such curves, not a single curve.

---

1.4 Definition of Brownian motion

A stochastic process $W=(W_t)_{t\ge0}$ is called a standard Brownian motion if:

1. $W_0=0$ almost surely.
2. For $0\le t_0<t_1<\cdots<t_m$, the increments

$$W_{t_1}-W_{t_0},\; W_{t_2}-W_{t_1},\ldots, W_{t_m}-W_{t_{m-1}}$$

are independent. 3. For $0\le s<t$,

$$W_t-W_s\sim N(0,t-s).$$

4. The sample paths $t\mapsto W_t(\omega)$ are continuous almost surely.

Continuity is a path property, not a marginal property

Knowing only that each $W_t$ is normal does not imply that paths are continuous. Brownian motion is a process-level object. Its defining properties constrain the joint distribution of all times, not merely each isolated time.

---

1.5 Brownian motion as a Gaussian process

A process $(X_t)_{t\in I}$ is a Gaussian process if every finite vector

$$(X_{t_1},\ldots,X_{t_m})$$

has a multivariate normal distribution.

Brownian motion is a centered Gaussian process with covariance function

$$K(s,t)=\operatorname{Cov}(W_s,W_t)=\min(s,t).$$

Covariance derivation

Assume $s\le t$. Then

$$W_t = W_s + (W_t-W_s).$$

The increment $W_t-W_s$ is independent of $W_s$, and has mean zero. Therefore

$$\begin{aligned} \operatorname{Cov}(W_s,W_t) &=\operatorname{Cov}\bigl(W_s, W_s+(W_t-W_s)\bigr)\\ &=\operatorname{Var}(W_s)+\operatorname{Cov}(W_s,W_t-W_s)\\ &=s+0\\ &=s. \end{aligned}$$

By symmetry,

$$\operatorname{Cov}(W_s,W_t)=\min(s,t).$$

Covariance geometry: K(s,t) = min(s,t)

Darker means larger covariance. Early time is embedded in later time: W_t = W_s + independent future increment.

Covariance as geometry

For a centered Gaussian vector, the covariance matrix determines the shape of its probability ellipsoid. For Brownian motion, the kernel $K(s,t)=\min(s,t)$ says that later values contain the accumulated history of earlier values. The process is not a collection of independent normal variables; it is a highly structured random curve.

---

1.6 Martingale property

Let $(\mathcal F_t)_{t\ge0}$ be the natural filtration of Brownian motion:

$$\mathcal F_t = \sigma(W_s:0\le s\le t).$$

This represents the information generated by observing the path up to time $t$.

Brownian motion is a martingale

For $0\le s\le t$,

$$\mathbb E[W_t\mid \mathcal F_s]=W_s.$$

Proof

Write

$$W_t = W_s + (W_t-W_s).$$

Since $W_s$ is $\mathcal F_s$-measurable,

$$\mathbb E[W_s\mid\mathcal F_s]=W_s.$$

Since $W_t-W_s$ is independent of $\mathcal F_s$ and has mean zero,

$$\mathbb E[W_t-W_s\mid\mathcal F_s]=0.$$

Therefore

$$\mathbb E[W_t\mid\mathcal F_s]=W_s.$$

No predictable instantaneous gain

The martingale property says Brownian motion has no drift that can be predicted from its own past. In arbitrage-free pricing, discounted asset prices are martingales under a risk-neutral measure. Brownian martingales are therefore the primitive noise sources for continuous-time pricing models.

---

1.7 Markov property

Brownian motion also satisfies a Markov property. Conditional on the present value $W_s$, the future evolution is independent of the earlier path.

More precisely, for bounded measurable $f$,

$$\mathbb E[f(W_t)\mid \mathcal F_s] = \mathbb E[f(W_t)\mid W_s], \qquad s\le t.$$

Since

$$W_t = W_s + (W_t-W_s),$$

and $W_t-W_s\sim N(0,t-s)$, the conditional distribution of $W_t$ given $W_s=x$ is

$$W_t\mid W_s=x \sim N(x,t-s).$$

Markov geometry

The past path matters only through its endpoint. From the current point $(s,W_s)$, the future is a fresh Brownian cloud translated by $W_s$. This is why Brownian transition densities satisfy heat equations.

---

1.8 Scaling invariance

For $c>0$, define

$$\widetilde W_t = \frac{1}{\sqrt c}W_{ct}.$$

Brownian scaling

The process $\widetilde W$ is again a standard Brownian motion.

Proof sketch

We check the defining properties.

First,

$$\widetilde W_0=0.$$

For $s<t$,

$$\widetilde W_t-\widetilde W_s =\frac{1}{\sqrt c}(W_{ct}-W_{cs}).$$

Since

$$W_{ct}-W_{cs}\sim N(0,c(t-s)),$$

we get

$$\widetilde W_t-\widetilde W_s\sim N(0,t-s).$$

Independent increments and continuity are inherited from $W$.

This scaling property is not just an algebraic curiosity. It says that Brownian motion has no intrinsic time scale. Zooming in produces statistically similar roughness.

---

1.9 Path roughness and non-differentiability

For a deterministic differentiable function $f$, small increments satisfy

$$f(t+h)-f(t)\approx f'(t)h.$$

The increment is order $h$. For Brownian motion,

$$W_{t+h}-W_t\sim N(0,h),$$

so typical size is order $\sqrt h$. Therefore

$$\frac{W_{t+h}-W_t}{h}$$

has typical magnitude of order

$$\frac{\sqrt h}{h}=\frac1{\sqrt h}\to\infty.$$

This heuristic explains why Brownian paths are almost surely nowhere differentiable.

Heuristic versus theorem

The above argument explains the scaling obstruction to differentiability. The rigorous theorem is stronger: Brownian paths are almost surely nowhere differentiable. Proving this requires controlling increments over countably many rational intervals and then extending to all times.

---

1.10 Quadratic variation

For a partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, define

$$Q_\Pi(W)=\sum_{k=0}^{n-1}(W_{t_{k+1}}-W_{t_k})^2.$$

If the mesh $|\Pi|=\max_k(t_{k+1}-t_k)$ tends to zero, then

$$Q_\Pi(W)\to T$$

in probability, and under suitable nested partitions, almost surely.

Quadratic variation: smooth paths versus Brownian paths

Partition size: 50

Smooth path

sum (Delta x)^2 ≈ 0.3943

Brownian path

sum (Delta W)^2 ≈ 0.7491

As the partition refines, smooth quadratic variation tends to 0, while Brownian quadratic variation stabilizes near the horizon length T=1. This is the geometric reason (dW_t)^2 = dt.

Mean and variance computation for uniform partitions

Let $t_k=kT/n$. Then

$$\Delta W_k = W_{t_{k+1}}-W_{t_k}\sim N(0,T/n),$$

and increments are independent. Hence

$$\mathbb E[(\Delta W_k)^2]=T/n.$$

Therefore

$$\mathbb E\left[\sum_{k=0}^{n-1}(\Delta W_k)^2\right] =\sum_{k=0}^{n-1}T/n=T.$$

For a normal random variable $Z\sim N(0,\sigma^2)$,

$$\operatorname{Var}(Z^2)=2\sigma^4.$$

Thus

$$\operatorname{Var}\left(\sum_{k=0}^{n-1}(\Delta W_k)^2\right) =n\cdot 2(T/n)^2 =\frac{2T^2}{n}\to0.$$

So the quadratic variation converges to $T$ in $L^2$, hence in probability.

Why quadratic variation is the birth of Itô calculus

Smooth paths have increments of order $\Delta t$, so squared increments are order $(\Delta t)^2$, and their sum vanishes. Brownian increments are order $\sqrt{\Delta t}$, so squared increments are order $\Delta t$, and their sum survives. This surviving second-order term is exactly why stochastic calculus differs from classical calculus.

---

1.11 The symbolic rule $(dW_t)^2=dt$

The expression

$$(dW_t)^2=dt$$

is not ordinary algebra. It is shorthand for the quadratic variation statement

$$\sum (\Delta W_k)^2 \to T.$$

Similarly, because Brownian increments are order $\sqrt{dt}$, one informally keeps the products

$$(dW_t)^2\sim dt,$$

but discards

$$dt\,dW_t \sim dt^{3/2}, \qquad (dt)^2\sim dt^2.$$

This bookkeeping becomes precise in Itô's formula.

---

1.12 Finance-facing interpretation

Consider the model

$$dX_t = \mu\,dt + \sigma\,dW_t.$$

Over a short interval $\Delta t$, this means approximately

$$\Delta X\approx \mu\Delta t+\sigma\sqrt{\Delta t}Z, \qquad Z\sim N(0,1).$$

The deterministic drift is order $\Delta t$, but the random shock is order $\sqrt{\Delta t}$. Therefore at very short time scales,

$$\sqrt{\Delta t}\gg \Delta t.$$

Local randomness dominates drift

At high frequency, drift is statistically difficult to estimate because it is smaller than noise at small horizons. Volatility is locally visible; drift is not. This is one reason volatility modeling is often more statistically stable than expected-return modeling.

---

1.13 Equivalent characterizations to remember

Brownian motion may be understood as:

Viewpoint	Mathematical object	Why it matters
Scaling limit	$S_{\lfloor nt\rfloor}/\sqrt n\Rightarrow W_t$	universality, volatility scaling
Gaussian process	covariance $\min(s,t)$	finite-dimensional laws, geometry
Martingale	$\mathbb E[W_t\mid\mathcal F_s]=W_s$	no-arbitrage pricing
Markov process	future depends on present	PDE and transition densities
Rough path	quadratic variation $[W]_t=t$	Itô calculus

---

1.14 Problems

Problem 1: covariance and correlation

Let $0<s<t$. Prove

$$\operatorname{Cov}(W_s,W_t)=s, \qquad \operatorname{Corr}(W_s,W_t)=\sqrt{\frac{s}{t}}.$$

Problem 2: Brownian bridge preview

Compute the conditional distribution of $W_s$ given $W_t=x$, where $0<s<t$. Interpret the conditional mean geometrically.

Problem 3: drifted Brownian motion

Let

$$X_t=x_0+\mu t+\sigma W_t.$$

Find $\mathbb E[X_t]$, $\operatorname{Var}(X_t)$, and $\operatorname{Cov}(X_s,X_t)$.

Problem 4: quadratic variation of drifted Brownian motion

For

$$X_t=\mu t+\sigma W_t,$$

compute the quadratic variation $[X]_T$.

Problem 5: why ordinary chain rule fails

Assume informally that $dW_t^2=2W_t dW_t$, as in ordinary calculus. Compare this with what Itô's formula will later imply. What term is missing?

---

1.15 Selected solutions

Solution 1

For $s<t$, write

$$W_t=W_s+(W_t-W_s).$$

The second term is independent of $W_s$, so

$$\operatorname{Cov}(W_s,W_t) =\operatorname{Var}(W_s)=s.$$

Since $\operatorname{Var}(W_s)=s$ and $\operatorname{Var}(W_t)=t$,

$$\operatorname{Corr}(W_s,W_t) =\frac{s}{\sqrt{s}\sqrt t} =\sqrt{\frac{s}{t}}.$$

Solution 2

The vector $(W_s,W_t)$ is jointly Gaussian with mean zero and covariance matrix

$$\begin{pmatrix} s & s\\ s & t \end{pmatrix}.$$

For jointly Gaussian variables,

$$\mathbb E[W_s\mid W_t=x] =\frac{\operatorname{Cov}(W_s,W_t)}{\operatorname{Var}(W_t)}x =\frac{s}{t}x.$$

The conditional variance is

$$\operatorname{Var}(W_s\mid W_t) =s-\frac{s^2}{t} =\frac{s(t-s)}{t}.$$

Thus

$$W_s\mid W_t=x\sim N\left(\frac{s}{t}x,\frac{s(t-s)}{t}\right).$$

Geometrically, the conditional mean is the straight line interpolating from $(0,0)$ to $(t,x)$. The remaining variance describes random fluctuation around that line.

Solution 3

We have

$$\mathbb E[X_t]=x_0+\mu t, \qquad \operatorname{Var}(X_t)=\sigma^2 t.$$

For $s\le t$, constants and deterministic drift do not affect covariance, so

$$\operatorname{Cov}(X_s,X_t)=\sigma^2\min(s,t)=\sigma^2 s.$$

Solution 4

The finite-variation drift $\mu t$ contributes zero quadratic variation. The Brownian part contributes $\sigma^2[W]_T$. Therefore

$$[X]_T=\sigma^2T.$$

Solution 5

Ordinary calculus would suggest no second-order correction. Itô calculus gives

$$d(W_t^2)=2W_t\,dW_t+dt.$$

The missing term is $dt$, arising from $(dW_t)^2=dt$. This term is the analytic signature of Brownian quadratic variation.

---

1.16 What this chapter gives you

You should now understand Brownian motion as a single object with several mutually reinforcing representations:

• a limit of rescaled random walks;
• a centered Gaussian process with kernel $\min(s,t)$;
• a martingale and Markov process;
• a probability law on continuous paths;
• a rough path with nonzero quadratic variation.

The next chapter constructs the Itô integral

$$\int_0^T H_t\,dW_t,$$

which is the first step toward solving stochastic differential equations.