Why Brownian Quadratic Variation Equals Elapsed Time

Quadratic variation measures the limiting sum of squared changes along a partition. For standard Brownian motion on

[0,T]

, each increment over length

h

has variance

h

. The squared increments add to

T

on average, and the fluctuations of that sum disappear as the mesh shrinks. Consequently

[W]_T=T

Why the random error vanishes under refinement

Let

0=t_0<\cdots<t_n=T

be a deterministic partition and write

h_i=t_{i+1}-t_i

. Independent Brownian increments have the distribution

\Delta W_i=\sqrt{h_i}Z_i

, where the

Z_i

are independent standard normals. Thus

\mathbb E\!\left[\sum_i(\Delta W_i)^2\right]=\sum_i h_i=T.

Since $\mathbb E[Z_i^4]=3$ and $\mathbb E[Z_i^2]=1$ , we have $\operatorname{Var}(Z_i^2)=2$ . Independence gives

\operatorname{Var}\!\left(\sum_i(\Delta W_i)^2\right) =2\sum_i h_i^2\le 2T\max_i h_i.

The right-hand side tends to zero, proving convergence in mean square and therefore in probability at fixed $T$ . Along dyadic partitions the error probabilities are summable by Chebyshev's inequality, giving almost sure convergence for that sequence. For $X_t=\sigma W_t$ with constant $\sigma$ , the squared sum is multiplied by $\sigma^2$ , so $[X]_T=\sigma^2T$ .

For an uneven grid, let $|\pi|$ denote the longest interval. Each $h_i^2$ is at most $|\pi|h_i$ , so the variance is bounded by $2T|\pi|$ . This tells you exactly which part of the sampling design matters. Adding a lot of points to the first half does little to control error if you leave one long interval in the second half. It also separates two different random quantities. The endpoint displacement is a single draw with variance $T$ , while the squared-step sum estimates $T$ from the whole path. On nested dyadic grids you can see the estimate improve overall without expecting each refinement to move closer to $T$ .

The grid experiment has another useful comparison. On a uniform grid with $n$ intervals, the variance of the estimate is $2T^2/n$ , so doubling $n$ reduces its standard deviation by a factor of $1/\sqrt 2$ . To cut that standard deviation in half you need four times as many intervals. Move the slider and watch the measured sum fluctuate. The path and the theoretical value stay fixed. A coarser grid can land nearer $T$ by chance, so judge convergence by the shrinking error scale rather than by monotonicity.

How noisy is a finite-grid estimate?

With uniform sampling, the squared sum has distribution

(T/n)\chi_n^2

. For one year and 25 equal steps its standard deviation is

\sqrt{2/25}\approx0.283

. With 400 equal steps it falls to approximately

0.071

. Increasing the sample frequency makes each path's realized variation a tighter estimate of one. The Brownian endpoint

W_T

still has variance

T

, regardless of the observation grid.

Compare the same path across observation grids

import numpy as np

rng = np.random.default_rng(42)
horizon = 1.0
for steps in (25, 100, 400):
    increments = rng.normal(0, np.sqrt(horizon / steps), (4_000, steps))
    realized = np.sum(increments**2, axis=1)
    print(steps, realized.mean(), realized.std(ddof=1))

Each row describes 4,000 independent paths on the same time horizon. Compare the sample standard deviation with

T\sqrt{2/n}

Why Brownian Quadratic Variation Equals Elapsed Time

Why the random error vanishes under refinement

How noisy is a finite-grid estimate?

Compare the same path across observation grids

Explore a Brownian path at different resolutions

References and next steps