Solution: Covariance Structure of Brownian Motion

Exercise: Covariance Structure of Brownian Motion

Proof

Assume $0 \le s \le t$ . Decompose $W_t$ as:

W_t = W_s + (W_t - W_s)

Since $\mathbb{E}[W_s] = 0$ and $\mathbb{E}[W_t] = 0$ (by (BM1) + (BM3)):

\operatorname{Cov}(W_s, W_t) = \mathbb{E}[W_s W_t] = \mathbb{E}[W_s(W_s + (W_t - W_s))] = \mathbb{E}[W_s^2] + \mathbb{E}[W_s(W_t - W_s)]

The first term is $\operatorname{Var}(W_s) = s$ by (BM3). For the second term, (BM2) says $W_s$ (which is $\mathcal{F}_s$ -measurable) is independent of the increment $W_t - W_s$ , so:

\mathbb{E}[W_s(W_t - W_s)] = \mathbb{E}[W_s] \cdot \mathbb{E}[W_t - W_s] = 0 \cdot 0 = 0

Hence $\operatorname{Cov}(W_s, W_t) = s = \min(s, t)$ . By symmetry, the identity holds for all $s, t \ge 0$ . $\square$

Part 1 — Simulation

# Python
import numpy as np

rng = np.random.default_rng(42)
N = 100_000
dt = 1e-3
n_steps = 1000           # t ∈ [0, 1] with Δt = 1e-3
s_idx, t_idx = 300, 700  # times 0.3 and 0.7

increments = rng.normal(0, np.sqrt(dt), size=(N, n_steps))
W = np.cumsum(increments, axis=1)
W_s = W[:, s_idx - 1]
W_t = W[:, t_idx - 1]

cov_hat = np.cov(W_s, W_t, ddof=0)[0, 1]
print(f"Sample Cov(W_0.3, W_0.7) = {cov_hat:.4f}")
print(f"Theoretical min(0.3, 0.7) = 0.3")
# Sample Cov(W_0.3, W_0.7) = 0.3005
# Theoretical min(0.3, 0.7) = 0.3

The sample covariance matches the theoretical value to three decimal places — Monte Carlo error at $N = 10^5$ is on the order of $1/\sqrt{N} \approx 3 \times 10^{-3}$ , consistent with the observed gap of $5 \times 10^{-4}$ .

Part 2 — Correlation

Using $\operatorname{Var}(W_s) = s$ , $\operatorname{Var}(W_t) = t$ , and $\operatorname{Cov}(W_s, W_t) = s$ for $s \le t$ :

\operatorname{Corr}(W_s, W_t) = \frac{\operatorname{Cov}(W_s, W_t)}{\sqrt{\operatorname{Var}(W_s)\operatorname{Var}(W_t)}} = \frac{s}{\sqrt{st}} = \sqrt{\frac{s}{t}}

The correlation depends only on the ratio

s/t \in (0, 1]

, not on the individual values. So

(W_1, W_2)

and

(W_{100}, W_{200})

have the same correlation

\sqrt{1/2} \approx 0.707

Part 3 — Long-range dependence

If $s$ and $t$ grow proportionally, say $t = \lambda s$ for a fixed $\lambda > 1$ , then:

\operatorname{Corr}(W_s, W_{\lambda s}) = \sqrt{\frac{s}{\lambda s}} = \frac{1}{\sqrt{\lambda}}

This does not depend on

s

. So even as the absolute gap

t - s = (\lambda - 1)s

grows without bound, the correlation stays fixed at

1/\sqrt{\lambda}

. Compare this with a stationary AR(1) or OU process, where correlation decays exponentially with the gap; Brownian motion is non-stationary and its correlation is scale-invariant.

Financially: if you sample a stock's Brownian-driven price at two times proportionally spaced (say, every doubling of horizon), the correlation between levels is constant. This is what makes volatility forecasts stable across different holding-period horizons but makes price-level forecasts unstable — the level of

W_t

is persistent, even though the increments are independent.

Takeaways

$\operatorname{Cov}(W_s, W_t) = \min(s, t)$ is the single identity that characterises Brownian motion among Gaussian processes, once you fix mean zero.
Correlation depends only on the ratio $s/t$ . This is a restatement of Brownian motion's self-similarity — the covariance structure is scale-invariant.
Long-range dependence in levels, independence in increments. These are not contradictory: the level $W_t$ is the cumulative sum of many independent past increments, so it remembers them; the next increment $W_{t+h} - W_t$ does not.
Simulation check is cheap and worth doing once. When an identity falls out of the definition, running a 10-second Monte Carlo verification builds intuition that algebra alone cannot.