Questions: Properties of Brownian Motion

Question 1

The total variation of Brownian motion on any interval [0,T] is:

• A. Equal to T, matching the quadratic variation
• B. Finite but random, depending on the particular sample path
• C. Almost surely infinite
• D. Zero, because the expected value of each increment is zero

Answer: (C)

Brownian paths are so rough that their total variation (sum of absolute increments) is infinite on every interval. Over a partition of [0,T] into n subintervals, the sum of |ΔW_i| is approximately Σ|N(0,T/n)| ~ n · √(T/n) · √(2/π) = √(2Tn/π) → ∞ as n → ∞. This infinite first variation but finite quadratic variation (equal to T) places Brownian motion in a regularity class between differentiable functions (finite first variation) and truly wild paths. It is the finite quadratic variation that makes stochastic calculus work.

Question 2

If W(t) is a standard Brownian motion, is the time-reversed process X(t) = tW(1/t) (with X(0) = 0) also a standard Brownian motion?

• T. True
• F. False

Answer: True

This is the time-inversion property. X(t) = tW(1/t) has Gaussian finite-dimensional distributions with E[X(t)] = 0 and Cov(X(s), X(t)) = st · Cov(W(1/s), W(1/t)) = st · min(1/s, 1/t) = st · (1/max(s,t)) = min(s,t). Since X has the correct mean, covariance, and Gaussian distributions, and its continuity at 0 follows from the law of the iterated logarithm, X is a standard Brownian motion. This is one of several non-obvious symmetries of the Wiener process.

Question 3

Why does the quadratic variation of Brownian motion being equal to t (rather than zero) fundamentally change the rules of calculus for stochastic integrals?

Model answer: In ordinary calculus, smooth functions have zero quadratic variation, and (df)² terms vanish in Taylor expansions. Brownian motion's quadratic variation [W,W]_t = t means (dW)² ≈ dt is a first-order term that cannot be discarded. When expanding f(W(t)) via Taylor series, the second-order term (1/2)f''(W)(dW)² contributes (1/2)f''(W)dt — a non-zero drift term. This produces the extra (1/2)σ²f'' correction in Itô's formula compared to the classical chain rule. The quadratic variation is finite and deterministic, which is what makes the resulting calculus consistent.

The heuristic dW² = dt captures the essential point: Brownian motion is rough enough that second-order terms in Taylor expansions survive. For any process with zero quadratic variation (all smooth deterministic functions), the classical chain rule suffices. The non-zero quadratic variation of Brownian motion is the single structural fact that necessitates Itô calculus.

Question 4

The scaling property states that for any c > 0, the process Y(t) = (1/√c)W(ct) is also a standard Brownian motion.

• T. True
• F. False

Answer: True

Check: Y(0) = 0, Y has Gaussian increments with Y(t) - Y(s) = (1/√c)(W(ct) - W(cs)) ~ N(0, (ct-cs)/c) = N(0, t-s), and Y inherits independence of increments and path continuity from W. The process Y is therefore a standard Brownian motion. This self-similarity means Brownian motion looks statistically the same at all scales — zooming in on a Brownian path yields another Brownian path, which is characteristic of fractal objects.