Why Continuous Finite-Variation Paths Have Zero Quadratic Variation

The Key Estimate

Quadratic variation is detected by adding the squares of path increments over increasingly fine partitions. For Brownian motion these sums converge to a nonzero limit, but a continuous path whose total variation is finite behaves differently: its squared-increment sums tend to zero. The point is not merely that each increment becomes small. The number of increments can grow, so we need a bound for the entire sum that accounts for their number and sizes together.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). An Ito integral \(\int_0^T H_s\,dB_s\), when used, has a progressively measurable integrand \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). No stochastic integral is needed in the argument below: it is a deterministic statement about continuous paths, and then a pathwise consequence for processes.

Fix \(T>0\). A partition of \([0,T]\) is a finite set \( \Pi=\{0=t_0<t_1<\cdots<t_n=T\}\), its mesh is \(|\Pi|=\max_i(t_i-t_{i-1})\), and its increments for a function \(x:[0,T]\to\mathbb{R}\) are \(\Delta_i x=x(t_i)-x(t_{i-1})\). Define the squared-increment sum by \(Q_\Pi(x)=\sum_{i=1}^n(\Delta_i x)^2\). The total variation of \(x\) on \([0,T]\) is the extended nonnegative number

where the supremum is taken over all finite partitions of the interval. The path has finite total variation if this supremum is finite. For a fixed partition, let \(m_\Pi(x)=\max_i|\Delta_i x|\), the largest absolute increment.

Why Continuity Makes the Bound Vanish

Proof. For every index \(i\), nonnegativity gives \((\Delta_i x)^2=|\Delta_i x|\,|\Delta_i x|\le m_\Pi(x)|\Delta_i x|\). Summing these inequalities yields

Now suppose \(x\) is continuous. Since \([0,T]\) is compact, \(x\) is uniformly continuous. In particular, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(|x(t)-x(s)|<\varepsilon\) whenever \(|t-s|<\delta\). If \(|\Pi|<\delta\), every adjacent pair \(t_{i-1},t_i\) is less than \(\delta\) apart, so \(m_\Pi(x)<\varepsilon\). Thus \(m_\Pi(x)\to0\) as \(|\Pi|\to0\). The variation is a fixed finite number, independent of the partition. Consequently \(0\le Q_\Pi(x)\le m_\Pi(x)\operatorname{Var}_{[0,T]}(x)\to0\). This proves the estimate and the claimed convergence for every sequence of partitions whose mesh tends to zero. \(\square\)

Proof. For each adjacent pair of partition points, \(t_i-t_{i-1}\le|\Pi|\), so \(|\Delta_i x|\le\omega_x(|\Pi|)\). Therefore \(m_\Pi(x)\le\omega_x(|\Pi|)\). Substitution into the key estimate proves the stated bound. Uniform continuity gives \(\omega_x(\delta)\to0\) as \(\delta\downarrow0\); finite variation then makes the right-hand side tend to zero. \(\square\)

Worked Examples

Processes with Finite-Variation Paths

Proof. By assumption, there is an event \(A\in\mathcal{F}\) with \(\mathbb{P}(A)=1\) on which the sample path \(t\mapsto X_t(\omega)\) is continuous and has finite total variation on \([0,T]\). Fix any \(\omega\in A\). Apply the deterministic theorem to the function \(x(t)=X_t(\omega)\) and to the specified sequence of partitions. It gives \(Q_{\Pi_k}(X(\omega))\to0\). This holds for every \(\omega\in A\), so the convergence is almost sure. The argument does not require adaptedness; it is a pathwise conclusion. \(\square\)

This corollary explains the contrast with Brownian quadratic variation: the earlier result for Brownian motion gives a nonzero limiting squared-increment sum, whereas continuous finite-variation paths have limit zero. In particular, the result does not say that every continuous process has zero quadratic variation; the finite-variation hypothesis is essential to this bound. No assertion about covariation or stochastic integration is needed here.

Common Pitfall: Small Increments Do Not Suffice

It is not enough to observe that each increment becomes small and then conclude that the sum of their squares vanishes. The number of terms changes with the partition, and small terms can accumulate. The proof instead bounds the whole sum by the largest increment multiplied by the sum of absolute increments. The latter is controlled uniformly by total variation.

For a concrete error, consider the path \(x(t)=2t\) on \([0,1]\) with partition \(0<\tfrac12<1\). Both path increments equal \(1\), so \(Q_\Pi(x)=1^2+1^2=2\), while the largest absolute increment is \(1\). Thus the proposed inequality \(Q_\Pi(x)\le m_\Pi(x)\) would assert \(2\le1\), which is false. The missing factor is the total variation: here \(\operatorname{Var}_{[0,1]}(x)=2\), and the correct estimate gives \(Q_\Pi(x)\le1\cdot2=2\), attained. As the partition is refined into \(n\) equal intervals, each increment is \(2/n\), the sum of absolute increments remains \(2\), and the estimate gives \(Q_\Pi(x)\le(2/n)\cdot2=4/n\). The exact sum is \(n(2/n)^2=4/n\), which tends to zero.