Covariation with Continuous Finite-Variation Processes

Why Finite Variation Removes Covariation

For Brownian motion, squared increments accumulate to a nonzero limit: its quadratic variation on $[0,T]$ is $T$. A continuous path of finite variation behaves differently when paired with any continuous path. Its increments have a finite total absolute sum, while the largest increment of the other path tends to zero as the partition is refined. Their products therefore add up to zero in the limit. This is why a continuous finite-variation term contributes no covariation, even when the other process has nonzero quadratic variation.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. The pathwise result below applies to continuous processes $X$ and $A$; it does not require Brownian motion or adaptedness. When the processes are random, we will specify the almost-sure path assumptions. No stochastic integral is used here.

Fix $T>0$. A partition of $[0,T]$ is a finite set $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, and its mesh is $|\Pi|=\max_i(t_i-t_{i-1})$. Write $\Delta_iX=X_{t_i}-X_{t_{i-1}}$ and $\Delta_iA=A_{t_i}-A_{t_{i-1}}$. The covariation sum along $\Pi$ is $\sum_i\Delta_iX\,\Delta_iA$. For a path $a$, its total variation on $[0,T]$ is the supremum, over all partitions, of the sums of absolute increments; finite variation means this supremum is finite.

Definition: A continuous path $a$ has finite variation on $[0,T]$ if $$ \operatorname{Var}_{[0,T]}(a)=\sup_{\Pi}\sum_i|a(t_i)-a(t_{i-1})|<\infty. $$ The partition covariation sums of paths $x,a$ are $C_\Pi(x,a)=\sum_i\Delta_ix\,\Delta_i a$. Their limit, when it exists as the mesh tends to zero, is their covariation on $[0,T]$.

The Pathwise Estimate

Lemma: For any two paths $x,a$ on $[0,T]$, if $a$ has finite variation, then every partition $\Pi$ satisfies $$ |C_\Pi(x,a)|\le \max_i|\Delta_i x|\,\operatorname{Var}_{[0,T]}(a). $$

Proof. The triangle inequality gives $|\sum_i\Delta_i x\,\Delta_i a|\le\sum_i|\Delta_i x|\,|\Delta_i a|$. Each $|\Delta_i x|$ is at most $\max_j|\Delta_jx|$, so the right-hand side is at most $\max_j|\Delta_jx|\sum_i|\Delta_i a|$. The sum of absolute increments along this particular partition is at most the supremum defining $\operatorname{Var}_{[0,T]}(a)$. Combining the inequalities proves the bound. $\square$

Theorem: Let $x$ be continuous on $[0,T]$, and let $a$ be continuous and of finite variation there. For every sequence of partitions $\Pi_m$ with $|\Pi_m|\to0$, $$ C_{\Pi_m}(x,a)\longrightarrow0. $$

Proof. A continuous function on the compact interval $[0,T]$ is uniformly continuous. Thus, for every $\varepsilon>0$, there is a $\delta>0$ such that $|x(t)-x(s)|<\varepsilon$ whenever $|t-s|<\delta$. Once $|\Pi_m|<\delta$, every partition interval has length less than $\delta$, so $\max_i|\Delta_i x|<\varepsilon$. The lemma now gives \[ |C_{\Pi_m}(x,a)|\le\varepsilon\,\operatorname{Var}_{[0,T]}(a). \] The variation is a fixed finite number. If it is zero, every increment of $a$ is zero and every covariation sum is already zero. If it is positive, taking $\varepsilon$ arbitrarily small shows that the covariation sums tend to zero. $\square$

This is a pathwise result: there is no expectation or independence argument in its proof. Consequently, if $X$ and $A$ are random processes whose paths are continuous and whose $A$-paths have finite variation almost surely on $[0,T]$, the conclusion holds almost surely along every fixed deterministic sequence of partitions with mesh tending to zero. Indeed, on the probability-one event where both path assumptions hold, apply the deterministic theorem to each pair of paths. The same argument works simultaneously for all such partition sequences on that event, since the deterministic conclusion holds for every sequence.

Worked Example: A Smooth Path Paired with Time

Take $x(t)=t^2$ and $a(t)=t$ on $[0,1]$. Use the partition $t_i=i/n$, where $i=0,\ldots,n$, and compute the covariation sum exactly.

For each $i=1,\ldots,n$, $\Delta_i x=(i^2-(i-1)^2)/n^2=(2i-1)/n^2$, and $\Delta_i a=i/n-(i-1)/n=1/n$. Therefore \[ C_{\Pi}(x,a)=\sum_{i=1}^n\frac{2i-1}{n^3} =\frac{1}{n^3}\left(2\sum_{i=1}^n i-n\right) =\frac{1}{n^3}\big(n(n+1)-n\big) =\frac1n. \] This tends to zero. Check: when $n=4$, the products are $1/64,3/64,5/64,7/64$; their sum is $16/64=1/4$, agreeing with $1/n$. The lemma also bounds the sum by $\max_i|\Delta_i x|\operatorname{Var}(a)=(2n-1)/n^2$, which tends to zero.

Brownian Motion Paired with a Finite-Variation Path

The theorem does not say that Brownian increments are small at a deterministic rate. Instead, continuity guarantees that their largest increment tends to zero along partitions whose mesh tends to zero. The finite variation of the other path supplies the summable factor in the estimate.

Worked Example: Brownian Motion Paired with Time

Let $X_t=B_t$ and $A_t=t$ on $[0,2]$. On the equal partition with $n$ intervals, each time increment is $2/n$, so \[ C_\Pi(B,A)=\sum_{i=1}^n(B_{2i/n}-B_{2(i-1)/n})\frac2n =\frac2n B_2. \] The last equality follows by telescoping the Brownian increments and using $B_0=0$. For every continuous Brownian path this tends to zero as $n\to\infty$, since $B_2$ is a finite real number on that path.

The mean is $0$, and the variance is $(2/n)^2\operatorname{Var}(B_2)=4\cdot2/n^2=8/n^2$, so the sum also tends to zero in mean square. Check: at $n=4$ the sum is $B_2/2$, with variance $\operatorname{Var}(B_2)/4=2/4=1/2$; the formula $8/n^2$ gives $8/16=1/2$. This independent moment calculation agrees with the exact telescoping expression.

Here $B$ supplies the continuous path and $t$ has variation $2$. The theorem applies even though Brownian motion itself does not have finite variation. The essential requirement is that at least one factor in the products has finite variation, while the other path is continuous.

A Second Finite-Variation Example

Worked Example: A Path with a Corner

On $[0,1]$, let $x(t)=|t-\tfrac12|$ and $a(t)=t^2$. For an even number $n$ of equal intervals, compute $C_\Pi(x,a)$. Put $m=n/2$. On the first $m$ intervals, $\Delta_i x=-1/n$; on the last $m$, $\Delta_i x=1/n$. On every interval, $\Delta_i a=(2i-1)/n^2$. Hence \[ C_\Pi(x,a)=\frac{1}{n^3} \left(\sum_{i=m+1}^{n}(2i-1)-\sum_{i=1}^{m}(2i-1)\right). \] The first $m$ positive odd integers sum to $m^2$, and the first $n$ sum to $n^2$. The parenthesized difference is therefore $(n^2-m^2)-m^2=n^2-2m^2=n^2/2$. Thus $C_\Pi(x,a)=1/(2n)$, which tends to zero.

Check: at $n=4$, the increments of $x$ are $-1/4,-1/4,1/4,1/4$, and those of $a$ are $1/16,3/16,5/16,7/16$. The products sum to $(-1-3+5+7)/64=8/64=1/8$, matching $1/(2n)=1/8$. The corner in $x$ causes no difficulty: continuity, not differentiability, is the hypothesis.

What the Result Means for Covariation

Apply the theorem with $x=a$ when a continuous path $a$ has finite variation. It follows that its quadratic-variation sums $\sum_i(\Delta_i a)^2$ tend to zero. Applying it instead to a continuous path $x$ and a continuous finite-variation path $a$ gives the corresponding zero cross-covariation. Thus, when $X$ and $A$ are continuous semimartingales and $A$ has finite-variation paths, their bracket satisfies $[X,A]_t=0$. This agrees with the informal multiplication rule $dX_t\,dA_t=0$ in this setting. The rule is shorthand for a limit of partition sums, not a claim that each product of increments is exactly zero.

For example, the bookkeeping rules $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ describe limiting quadratic and covariation behavior. They do not replace the pathwise argument: the theorem explains why a continuous finite-variation increment contributes zero when paired with a continuous increment.

Common Pitfall: Treating a Limiting Rule as an Exact Identity

A mistaken calculation might say that $dt\,dt=0$, and therefore $\Delta t_i\Delta t_i=0$ on every partition. The finite-partition calculation shows why this is wrong. For $a(t)=t$ on $[0,1]$ with $n$ equal intervals, every product is $(1/n)(1/n)=1/n^2$, and the sum is $n/n^2=1/n$, not zero. The sum becomes zero only in the limit $n\to\infty$. The infinitesimal multiplication table summarizes that limit; it is not an exact arithmetic rule for finite increments.

A different mistake is to assert a fixed bound such as $|\Delta_i x|\le C(t_i-t_{i-1})$ for every continuous $x$. Continuity alone does not imply such a Lipschitz bound. The proof needs only $\max_i|\Delta_i x|\to0$, which follows from uniform continuity, and the finite total variation of the other path. No differentiability or deterministic rate of decay is needed.

Check Your Understanding

Use the pathwise estimate and distinguish finite-partition identities from limiting statements.

State the bound for $|C_\Pi(x,a)|$ when $a$ has finite variation, and identify which factor tends to zero as the mesh decreases.
For $x(t)=t^2$ and $a(t)=t$ on $[0,1]$, compute the covariation sum on $n$ equal intervals.
Explain why the result applies pathwise to Brownian motion paired with time, even though Brownian paths do not have finite variation.
For $x(t)=|t-\tfrac12|$ and $a(t)=t^2$, what is the sum on an even equal partition with $n$ intervals?
Why does $dB_t\,dt=0$ not imply that each product $\Delta_iB\,\Delta_i t$ is exactly zero?