Quadratic Variation Along Deterministic Partitions

Quadratic Sums Along a Partition

Brownian paths have infinite total variation, so summing the absolute sizes of their increments does not produce a finite measure of their accumulated movement. Squaring the increments changes the behavior: over a fixed time interval, the sum of squared increments becomes close to the length of that interval when the observation times become dense. The conclusion is a limit in probability, not an identity for each path at each finite partition.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying a standard Brownian motion \((B_t)_{t\ge0}\), with \(B_0=0\), continuous paths, and independent increments such that \(B_v-B_u\) is normal with mean \(0\) and variance \(v-u\) whenever \(0\le u<v\). We may take the filtration to be the natural filtration \(\mathcal{F}_t=\sigma(B_s:0\le s\le t)\); no conditional expectation or martingale argument is needed below.

The target is \(T\), the elapsed time. Each increment has variance equal to the length of its time interval, so the expected quadratic sum is \(T\). To see why the random sum concentrates around this target, we also need its variance. Brownian increments on disjoint deterministic intervals are independent, and the variance of the square of a centered normal variable with variance \(h\) is \(2h^2\). These facts yield a direct convergence proof.

The Convergence Result

Proof. Fix a deterministic partition \(\Pi=\{0=t_0<\cdots<t_m=T\}\), and put \(h_i=t_i-t_{i-1}\) and \(Y_i=B_{t_i}-B_{t_{i-1}}\). Each \(Y_i\) is normal with mean \(0\) and variance \(h_i\), and the variables \(Y_1,\ldots,Y_m\) are independent. The second and fourth moments of a centered normal variable of variance \(h_i\) are \(\mathbb{E}[Y_i^2]=h_i\) and \(\mathbb{E}[Y_i^4]=3h_i^2\). For the fourth moment, write \(Y_i=\sqrt{h_i}Z\) for a standard normal \(Z\); integration by parts in the normal density gives \(\mathbb{E}[Z^4]=3\mathbb{E}[Z^2]=3\). Therefore \[ \mathbb{E}[Y_i^2]=h_i,\qquad \operatorname{Var}(Y_i^2)=\mathbb{E}[Y_i^4]-(\mathbb{E}[Y_i^2])^2 =3h_i^2-h_i^2=2h_i^2. \] Linearity of expectation and \(\sum_i h_i=T\) give \(\mathbb{E}[Q_\Pi(T)]=T\). Since the \(Y_i^2\) are independent, their covariances vanish, and hence \[ \operatorname{Var}(Q_\Pi(T))=\sum_{i=1}^m\operatorname{Var}(Y_i^2) =2\sum_{i=1}^m h_i^2 \le 2|\Pi|\sum_{i=1}^m h_i =2T|\Pi|. \] The inequality uses \(h_i\le|\Pi|\), so \(h_i^2\le|\Pi|h_i\) for each \(i\). Chebyshev's inequality, applied to the centered variable \(Q_\Pi(T)-T\), now gives \[ \mathbb{P}\bigl(\lvert Q_\Pi(T)-T\rvert>\varepsilon\bigr) \le\frac{\operatorname{Var}(Q_\Pi(T))}{\varepsilon^2} \le\frac{2T|\Pi|}{\varepsilon^2}. \] For the sequence \(\Pi_r\), the right-hand side tends to zero because \(T\) and \(\varepsilon>0\) are fixed and \(|\Pi_r|\to0\). This proves the stated convergence in probability. \(\square\)

The argument does not require equal interval lengths. It only uses that the intervals are deterministic, their Brownian increments are independent, and the largest interval length tends to zero. At a given partition the quadratic sum is random; the theorem says that the probability of a fixed-size error becomes small as the mesh shrinks.

What the Limit Does—and Does Not—Say

Convergence in probability to \(T\) means that for every fixed \(\varepsilon>0\), the probability of the strict-tail event \(\lvert Q_{\Pi_r}(T)-T\rvert>\varepsilon\) tends to zero. It does not say that every quadratic sum equals \(T\), nor does it assert that the sums converge for every sample path. In particular, the finite-mesh variance is generally positive.

The mesh condition matters. If the same partition of \([0,1]\) consisting only of the endpoints \(0\) and \(1\) is used at every index \(r\), then \(Q_{\Pi_r}(1)=B_1^2\) for every \(r\). This sequence does not converge in probability to \(1\): for example, the event \(\lvert B_1^2-1\rvert>1\) contains the event \(B_1^2>2\), which has positive probability because \(B_1\) has a normal density that is positive on every real interval. Thus the probability of the error event does not tend to zero. Here the mesh remains \(1\), rather than tending to zero.

Common mistake: treating the sum as exactly its expectation. On the two-interval partition \(0,\tfrac12,1\), the sum is \[ Q_\Pi(1)=B_{1/2}^2+(B_1-B_{1/2})^2. \] Its expectation is \(\tfrac12+\tfrac12=1\). A mistaken calculation might replace the random sum by \(1\) path by path, reasoning that the two increment variances add to \(1\). The correction is that variances determine the expectation of each square, not its realized value. In fact, the variance of this sum is \[ 2(\tfrac12)^2+2(\tfrac12)^2 =2\cdot\tfrac14+2\cdot\tfrac14=1, \] so the sum is not constant. The theorem concerns concentration as the mesh tends to zero, not exact equality at a fixed partition.

Partition-based quadratic variation is useful because it records a characteristic feature of Brownian motion that ordinary path length misses: squared increments accumulate at a rate governed by elapsed time. The proof is also robust to irregular deterministic observation times. What matters for the conclusion is not a regular grid but a vanishing largest interval.