Partition [0,T] into n equal pieces and let Q_n = sum_i (W_{t_{i+1}} - W_{t_i})^2 for standard Brownian motion. Compute Var(Q_n) exactly and say what it tells you about the quadratic variation of the path.

Partition [0,T] into n equal pieces and let Q_n = sum_i (W_{t_{i+1}} - W_{t_i})^2 for standard Brownian motion. Compute Var(Q_n) exactly and say what it tells you about the quadratic variation of the path.

Approach: Each squared increment is (T/n) times a chi-squared variable with one degree of freedom. Use independence across the partition and the variance of a chi-squared.

2T^2/n. Each increment is N(0, T/n), so (W_{t_{i+1}} - W_{t_i})^2 = (T/n) X with X chi-squared on one degree of freedom, giving mean T/n and variance 2(T/n)^2 since Var(X) = 2. The increments are independent, so the variances add over the n cells and Var(Q_n) = n * 2(T/n)^2 = 2T^2/n, while E[Q_n] = T for every n. Therefore Q_n converges in L^2 to the constant T as n grows, which is the statement that the quadratic variation of a Brownian path on [0,T] is T. Note that the limit is deterministic even though every Q_n is random, and that the rate is n^{-1/2} in standard deviation, so a path simulated on 252 steps estimates T with about 8.9% relative error.

Follow-up: What is the limit of sum_i |W_{t_{i+1}} - W_{t_i}|^3, and what does it imply about third order terms in a Taylor expansion?

Key concepts: quadratic variation, chi-squared, L^2 convergence, independent increments.