Quadratic Variation Along Partitions

Why Sum Squared Increments?

A Brownian path is continuous, but its small increments do not behave like those of a smooth curve. To quantify this difference, divide a time interval into short pieces, square each Brownian increment, and add the squares. The resulting sum is called a quadratic variation sum. As the pieces become shorter, the sum converges in probability to the length of the time interval.

This result is a basic reason Brownian motion is a useful noise model in stochastic differential equations: accumulated squared increments remain substantial even as the time steps shrink. The quantity that survives is elapsed time, not zero. The focus here is the precise partition result and its calculations, rather than later rules for manipulating stochastic integrals.

The result concerns any sequence of deterministic partitions \(\pi_k\) for which \(|\pi_k|\to 0\). The partitions need not be equally spaced or nested. Standard Brownian motion has independent increments, and each increment over an interval of length \(h\) has distribution \(N(0,h)\). Those are the facts needed for the derivation.

Deriving the Mean and Variance

Write \(\Delta_i=t_i-t_{i-1}\) and \(X_i=W(t_i)-W(t_{i-1})\). The increments \(X_i\) are independent, with \(X_i\sim N(0,\Delta_i)\). For a centered normal random variable with variance \(v\), its second moment is \(v\) and its fourth moment is \(3v^2\). Thus \(E[X_i^2]=\Delta_i\), and $$ \operatorname{Var}(X_i^2)=E[X_i^4]-(E[X_i^2])^2=3\Delta_i^2-\Delta_i^2=2\Delta_i^2. $$ The independence of the increments also gives independence of their squares. Therefore the sum has expectation equal to the sum of the interval lengths, while its variance is the sum of the individual variances.

The expectation is exactly the elapsed time for every partition, not merely in the limit. The variability of the sum is what shrinks with the mesh. Since each \(\Delta_i\leq|\pi|\), $$ \sum_{i=1}^{m}\Delta_i^2\leq |\pi|\sum_{i=1}^{m}\Delta_i=|\pi|(b-a). $$ This bound gives a direct proof of the theorem.

For completeness, mean-square convergence implies convergence in probability here by Markov's inequality applied to the nonnegative squared error. For any \(\varepsilon>0\), $$ P\bigl(|Q_{\pi_k}-(b-a)|>\varepsilon\bigr) \leq \frac{E[(Q_{\pi_k}-(b-a))^2]}{\varepsilon^2} \leq \frac{2|\pi_k|(b-a)}{\varepsilon^2}\longrightarrow 0. $$ This is precisely convergence in probability. The variance bound also provides a usable finite-partition error bound. \(\square\)

Equal-Spaced Partitions

For an equal-spaced partition of \([0,T]\) into \(n\) intervals, each interval has length \(T/n\). The increment on interval \(i\) can be written \(X_i=\sqrt{T/n}\,Z_i\), where the \(Z_i\) are independent \(N(0,1)\) random variables. Consequently, $$ Q_n=\sum_{i=1}^{n}X_i^2=\frac{T}{n}\sum_{i=1}^{n}Z_i^2. $$ The sum of squares of \(n\) independent standard normal variables has a chi-square distribution with \(n\) degrees of freedom. Hence \(Q_n\) has the distribution \((T/n)\chi_n^2\), with mean \(T\) and variance \(2T^2/n\).

What the Result Does—and Does Not—Say

The convergence statement is probabilistic: for every fixed positive tolerance, the probability of an error larger than that tolerance tends to zero as the mesh shrinks. It does not say that every finite partition gives exactly the elapsed time. Even though \(E[Q_\pi]=b-a\) at every resolution, individual sums vary; their variance quantifies that variation.

It is also important to distinguish the squared-increment sum from the square of the total increment. With two increments \(x_1\) and \(x_2\), the endpoint square is $$ (x_1+x_2)^2=x_1^2+x_2^2+2x_1x_2. $$ It includes both squared contributions as well as a cross-product, but combines them rather than retaining the separate squared increments. For example, if \(x_1=1\) and \(x_2=2\), the endpoint square is \(9\), while the squared-increment sum is \(1^2+2^2=5\). Thus the endpoint square is not a substitute for the partition sum. For Brownian increments, the endpoint square and the squared-increment sum can even have the same expectation, without being the same quantity.

The result assumes deterministic partitions. The proof uses the fact that the interval lengths are fixed and the corresponding Brownian increments are independent normal variables with specified variances. A partition chosen using observed path values is not covered by this argument. The result also does not, by itself, make a claim about convergence along every path. Those are different questions from the convergence-in-probability statement proved here.

Diagnostics and Practical Interpretation

For an equal-spaced calculation, the variance \(2T^2/n\) is a quick diagnostic: if a reported uncertainty does not decrease as \(n\) grows, check the interval length, the squared-increment calculation, and whether the increments have actually been treated as Brownian increments. For an irregular partition, calculate \(2\sum_i\Delta_i^2\); using \(2|\pi|(b-a)\) is a safe upper bound, not generally the exact variance.

In a pharmacokinetic model, this result gives mathematical context for Brownian-driven process variation between observation times. It does not imply that observed concentration differences themselves are Brownian increments: the state dynamics, sampling schedule, and measurement error all matter. Nor is quadratic variation an ordinary slope or endpoint displacement. For a smooth deterministic trajectory with bounded derivative, increments are typically proportional to interval length, so the sum of their squares tends to zero; Brownian increments have variance proportional to interval length, producing a nonzero limiting sum instead.