Partition Dependence and Quadratic Variation Pitfalls

Quadratic Variation Depends on How the Partition Is Chosen

For a continuous path, a quadratic-variation sum adds the squared increments over the intervals of a partition. Brownian motion has a robust quadratic-variation limit along deterministic partitions whose mesh tends to zero: on a fixed time interval \([0,T]\), the sums converge in probability to \(T\). This statement does not say that every fine partition, including one selected after looking at the path, gives the same answer. The choice of partition can depend on the very increments being squared.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(S=\mathbb{R}\), starting at \(B_0=0\). For a finite partition \(\Pi=\{0=t_0<t_1<\cdots<t_m=T\}\), define its mesh and its quadratic-variation sum by

A deterministic partition is fixed independently of the Brownian path. A path-dependent partition has at least one partition time chosen as a function of the observed path. For a sequence of random partitions, \(|\Pi_n|\to0\) in probability means that for every \(\varepsilon>0\), \(\mathbb{P}(|\Pi_n|>\varepsilon)\to0\). Our main path-dependent construction will have a mesh bounded by a deterministic quantity tending to zero, but its quadratic-variation limit will not be \(T\).

Deterministic Partitions: Convergence in Probability

Proof. Fix one partition \(\Pi=\{t_0,\ldots,t_m\}\), and write \(\Delta_r=t_{r+1}-t_r\) and \(Y_r=B_{t_{r+1}}-B_{t_r}\). By the defining increment properties of Brownian motion, the \(Y_r\) are independent, and \(Y_r\) has a centered normal distribution with variance \(\Delta_r\). For a centered normal variable of variance \(v\), its second moment is \(v\) and its fourth moment is \(3v^2\); consequently, the variance of its square is \(3v^2-v^2=2v^2\). Independence therefore gives

Thus \(\mathbb{E}[(Q(\Pi)-T)^2]=\operatorname{Var}(Q(\Pi))\le2T|\Pi|\). Apply this bound to each deterministic \(\Pi_n\). Since \(|\Pi_n|\to0\), the mean-square error tends to zero. For every \(\varepsilon>0\), Markov's inequality applied to \((Q(\Pi_n)-T)^2\) gives \(\mathbb{P}(|Q(\Pi_n)-T|>\varepsilon)\le 2T|\Pi_n|/\varepsilon^2\to0\). This proves both claims. \(\square\)

The hypotheses matter. The proof used that the partition was fixed independently of the path: it let us treat the increments on its intervals as independent normal variables with known interval lengths. The estimate is a convergence-in-probability result; it does not, by itself, assert almost-sure convergence along every deterministic sequence of partitions.

A Partition Chosen at Each Interval Maximum

We now build partitions that use the path itself. Divide \([0,T]\) into \(n\) deterministic intervals, each of length \(h=T/n\). In each interval, insert a time at which Brownian motion attains its maximum on that interval. If there is more than one maximizing time, choose the earliest. The continuity of Brownian paths ensures a maximum is attained, and the earliest maximizing time is defined for every continuous path. Remove any repeated endpoints, so the resulting set is a partition. Each new partition interval lies within one of the original intervals; hence the random partition \(\widehat{\Pi}_n\) satisfies \(|\widehat{\Pi}_n|\le T/n\).

To analyse its sum, define a standard Brownian motion \((W_s)_{0\le s\le1}\), its maximum \(M=\max_{0\le s\le1}W_s\), and the nonnegative random variable

The sum \(Z\) is the sum of the two squared increments when an interval is split at its maximum, after scaling that interval to length one. We use the standard Brownian Gaussian maximal-moment bound \(\mathbb{E}[\sup_{0\le s\le1}|W_s|^2]<\infty\); its proof is not needed here. Since \(0\le M\le\sup_s|W_s|\) and \(|W_1-M|\le |W_1|+M\), this bound implies \(\mathbb{E}Z<\infty\). Put \(c=\mathbb{E}Z\).

Proof. On the \(k\)th interval, Brownian scaling says that the path of increments, rescaled in time by \(h\) and in space by \(\sqrt h\), has the law of standard Brownian motion on \([0,1]\). Brownian increments on disjoint deterministic intervals are independent as processes. It follows that the contribution from the two pieces of each interval has the form \(hZ_{n,k}\), where, for each fixed \(n\), the \(n\) variables \(Z_{n,0},\ldots,Z_{n,n-1}\) are independent and each has the distribution of \(Z\). Therefore \(Q(\widehat{\Pi}_n)=h\sum_{k=0}^{n-1}Z_{n,k}\). The standard weak law of large numbers for independent, identically distributed integrable variables gives \(n^{-1}\sum_{k=0}^{n-1}Z_{n,k}\to c\) in probability. Since \(h=T/n\), this proves \(Q(\widehat{\Pi}_n)\to cT\) in probability.

It remains to check that \(c>1\). Since \(M\ge0\) and \(M\ge W_1\),

The inequality is strict with positive probability. In fact, the increments \(W_{1/2}\) and \(W_1-W_{1/2}\) are independent \(N(0,\tfrac12)\) variables, so the event \(W_{1/2}>1\) and \(W_1-W_{1/2}<-1\) has positive probability. On this event, \(M\ge W_{1/2}>1\) and \(M-W_1\ge W_{1/2}-W_1>1\), so \(Z-W_1^2>0\). As \(\mathbb{E}W_1^2=1\), taking expectations yields \(c>1\). Thus \(cT>T\) for \(T>0\), as claimed. \(\square\)

This proof establishes a strict difference without needing to calculate \(c\) exactly. The key distinction is how the split point is chosen: it is selected to be a maximum of the same path whose increments enter the sum. The deterministic-partition calculation cannot be applied to these selected increments as though the intervals had been fixed in advance.

The Independence Pitfall

A tempting but incorrect argument for a random partition is to write, for each random interval \([U,V]\), \(\mathbb{E}[(B_V-B_U)^2\mid U,V]=V-U\), and then sum to get \(T\). The formula for a Brownian increment over a fixed deterministic interval does not justify this conditional identity when the endpoints were selected using the path. In the maximum-based construction, the interval endpoints and the increments are linked by the choice of the maximum. Treating the selected pieces as ordinary independent increments discards precisely that dependence.

Common mistake and correction. On each base interval of length \(h\), an incorrect calculation replaces the maximum-split contribution by \(h\), on the grounds that its expected squared increments should add to the interval length. Summing over \(n\) intervals then gives \(nh=T\). But the actual contribution has expectation \(h\,\mathbb{E}Z=hc\), with \(c>1\), so the total expectation is \(nhc=cT\). The correction is not to alter the deterministic result: it is to recognize that the selected partition is path-dependent, so the fixed-interval increment calculation does not apply to its pieces as if they were chosen in advance.

Keep the scope of each conclusion precise. The deterministic theorem proves convergence in probability for every fixed sequence of deterministic partitions with mesh tending to zero. The maximum-based construction proves that one particular sequence of path-dependent partitions, also with mesh tending to zero, has a different limit. It does not imply that every random partition changes the limit, or that every method of choosing partition times has the same alternative limit. For a pathwise quadratic-variation claim, the class of allowed partitions and the mode of convergence must be specified.