Brownian Paths Have Infinite Total Variation

From Quadratic Variation to Infinite Variation

Brownian motion has finite quadratic variation on every compact time interval, but that does not mean its paths have finite total variation. In fact, the positive quadratic variation forces the opposite conclusion. The key comparison is simple: if a continuous path has finite total variation, then the sum of its squared increments along increasingly fine partitions must tend to zero. Brownian motion’s corresponding sums instead converge to the length of the interval.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Thus $B_0=0$, its paths are continuous almost surely, and for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. This tutorial uses no stochastic integrals. All pathwise variation statements below concern continuous sample paths.

Definition: For a real-valued function $f$ on $[a,b]$, its total variation is $$ \operatorname{Var}_{[a,b]}(f)=\sup_{\{a=t_0<\cdots<t_m=b\}}\sum_{j=1}^{m}|f(t_j)-f(t_{j-1})|, $$ where the supremum is over all finite partitions of $[a,b]$. The function has finite variation on $[a,b]$ when this supremum is finite. For a partition $\Pi$, its quadratic-variation sum is $Q_\Pi(f)=\sum_j(f(t_j)-f(t_{j-1}))^2$.

Why Finite Variation Makes the Square Sum Vanish

The finite-variation criterion needed here is a pathwise estimate. It applies to any continuous function, not just a stochastic process. Write $|\Pi|=\max_j(t_j-t_{j-1})$ for the mesh of a partition, and let $m_\Pi=\max_j|f(t_j)-f(t_{j-1})|$ be its largest increment.

Proposition: If $f$ is continuous and has finite variation $V=\operatorname{Var}_{[a,b]}(f)$, then for every partition $\Pi$, $$ Q_\Pi(f)\le m_\Pi V. $$ Consequently, for every sequence of partitions with mesh tending to zero, $Q_\Pi(f)\to0$.

Proof. For each increment, $(f(t_j)-f(t_{j-1}))^2=|f(t_j)-f(t_{j-1})|\,|f(t_j)-f(t_{j-1})|\le m_\Pi|f(t_j)-f(t_{j-1})|$. Summing these inequalities gives $Q_\Pi(f)\le m_\Pi\sum_j|f(t_j)-f(t_{j-1})|\le m_\Pi V$. For a sequence of partitions whose mesh tends to zero, continuity on the compact interval implies uniform continuity. Therefore the largest increment $m_\Pi$ tends to zero: given $\varepsilon>0$, uniform continuity gives $\delta>0$ such that $|f(t)-f(s)|<\varepsilon$ whenever $|t-s|<\delta$, and eventually every partition interval has length less than $\delta$. Since $V$ is finite, $0\le Q_\Pi(f)\le m_\Pi V\to0$. This also covers $V=0$, for which every increment is zero. $\square$

Worked Example: A Finite-Variation Path

Take $f(t)=t$ on $[2,5]$, and divide the interval into $n$ equal subintervals. Each time increment and each path increment is $3/n$. The quadratic-variation sum is therefore $$ Q_n=\sum_{j=1}^{n}\left(\frac{3}{n}\right)^2 =n\frac{9}{n^2}=\frac{9}{n}. $$

The total variation is $3$, since every increment is nonnegative and the sum of the increments is $5-2=3$. The largest increment is $3/n$, so the proposition gives $Q_n\le(3/n)\cdot3=9/n$; equality holds here. In particular, $Q_n\to0$.

Check: for $n=3$, the three increments are all $1$, so the sum of their squares is $1+1+1=3$, which agrees with $9/3=3$. The bound also gives $(3/3)\cdot3=3$.

Positive Square-Sum Limits on Dyadic Intervals

A dyadic rational is a number of the form $k/2^q$, where $k$ is an integer and $q$ is a nonnegative integer. Consider any fixed dyadic rationals $a<b$. For each positive integer $n$, divide $[a,b]$ into $2^n$ equal subintervals and denote the resulting quadratic-variation sum of Brownian motion by $Q_n(a,b)$. The increments are independent, each has distribution $N(0,(b-a)2^{-n})$, and the normal moments give an exact mean and variance.

Proposition: For each fixed pair of dyadic rationals $a<b$, $$ Q_n(a,b)\longrightarrow b-a\quad\text{almost surely as }n\to\infty. $$

Proof. Set $L=b-a>0$ and $h_n=L2^{-n}$. The $2^n$ increments in the sum are independent $N(0,h_n)$ random variables. A centered normal variable with variance $h_n$ has second moment $h_n$ and fourth moment $3h_n^2$. Hence the square of each increment has expectation $h_n$ and variance $3h_n^2-h_n^2=2h_n^2$. Adding expectations and, by independence, variances gives $$ \mathbb{E}[Q_n(a,b)]=2^nh_n=L,\qquad \operatorname{Var}(Q_n(a,b))=2^n(2h_n^2)=2L^2\,2^{-n}. $$ For every $\varepsilon>0$, Chebyshev’s inequality therefore gives $$ \mathbb{P}(|Q_n(a,b)-L|>\varepsilon) \le \frac{2L^2\,2^{-n}}{\varepsilon^2}. $$ For fixed $\varepsilon$, the right-hand side is summable in $n$. The first Borel–Cantelli lemma implies that the event $|Q_n(a,b)-L|>\varepsilon$ occurs only finitely often, almost surely. Applying this conclusion to the countable choices $\varepsilon=1/k$, $k=1,2,\ldots$, yields $Q_n(a,b)\to L$ almost surely. $\square$

Worked Example: A Quadratic-Variation Sum on a Fixed Interval

Take $[a,b]=[1,3/2]$ and divide it into four equal intervals, so $n=2$. The interval length is $L=3/2-1=1/2$, and each time step is $h=L/4=1/8$. There are four independent Brownian increments, each distributed as $N(0,1/8)$. Thus $$ \mathbb{E}[Q_2(1,3/2)]=4\left(\frac18\right)=\frac12 $$ and $$ \operatorname{Var}(Q_2(1,3/2)) =4\left(2\left(\frac18\right)^2\right) =4\left(\frac{2}{64}\right)=\frac18. $$

Check: the general formulas give $L=1/2$ for the mean and $2L^2 2^{-2}=2(1/4)(1/4)=1/8$ for the variance. The exact limit as the number of subdivisions tends to infinity is the interval length $1/2$, not zero.

One Event That Works on Every Interval

The fixed-interval proposition gives an almost-sure assertion for each chosen dyadic interval. To conclude that the property holds simultaneously on every interval, the exceptional events must be handled together. There are only countably many pairs of dyadic rationals $a<b$. For each pair, let $A_{a,b}$ be the probability-one event on which $Q_n(a,b)\to b-a$. The intersection of these events over all such pairs still has probability one, because a countable union of probability-zero complements has probability zero. Intersect it also with the probability-one event on which the Brownian paths are continuous. Call the resulting probability-one event $A$.

Theorem: With probability one, every Brownian path has infinite total variation on every compact interval $[u,v]$ with $0\le u<v<\infty$. Consequently, its variation is infinite on every time interval containing a nontrivial compact subinterval.

Proof. Fix an outcome in $A$, and suppose, for contradiction, that its Brownian path has finite variation on some $[u,v]$ with $u<v$. There are dyadic rationals $a,b$ with $u\le a<b\le v$: the dyadic rationals are dense, so one can choose two distinct ones strictly between $u$ and $v$. The restriction of a finite-variation path on $[u,v]$ to $[a,b]$ also has finite variation, since every partition sum on $[a,b]$ is bounded by the total variation on $[u,v]$. The finite-variation proposition applied to the equal-subdivision partitions of $[a,b]$ implies $Q_n(a,b)\to0$. But the choice of outcome in $A$ gives $Q_n(a,b)\to b-a>0$. A sequence cannot have both limits, a contradiction. Thus the path has infinite variation on every such $[u,v]$. If a time interval contains any compact $[u,v]$ of positive length, its variation is at least the variation on $[u,v]$, and is therefore infinite as well. $\square$

Worked Example: Applying the Simultaneous Result

Consider the interval $[1/5,11/10]$. The dyadic interval $[1/4,1]$ is contained in it because $1/5<1/4$ and $1<11/10$. Its length is $1-1/4=3/4$. Along its equal subdivisions, the Brownian quadratic-variation sums converge almost surely to $3/4$. On the event $A$, the finite-variation proposition would force those same sums to converge to zero if the path had finite variation on $[1/5,11/10]$. Therefore the path’s variation on $[1/5,11/10]$ is infinite.

Check: for $n=1$, the two subintervals of $[1/4,1]$ have length $3/8$. The sum has expectation $2(3/8)=3/4$ and variance $2\cdot2(3/8)^2=9/16$. The expectation already matches the limiting interval length; the proposition proves almost-sure convergence as $n$ increases.

Common Mistake: Treating Brownian Increments as Smooth

A tempting but incorrect calculation is to treat every Brownian increment over a step of length $h$ as though it were of size $h$. That would suggest a squared increment of order $h^2$, and hence a sum of order $2^n h^2$, which tends to zero on a fixed interval as $h$ decreases. Brownian increments instead have variance $h$, so their squared increments have expectation $h$. With $2^n$ steps of length $L2^{-n}$, the expected sum is $2^nL2^{-n}=L$, not zero.

For example, on an interval of length $1$, using four equal steps gives expected square-sum $4(1/4)=1$. The mistaken smooth-increment calculation would use $(1/4)^2$ for each square and predict $4(1/16)=1/4$. The latter is not the expectation of the Brownian quadratic-variation sum: each increment has variance $1/4$, so its squared value has expectation $1/4$, not $1/16$.

The conclusion is pathwise: on one probability-one event, no nontrivial interval can support finite variation. The argument does not say that every individual increment is large, nor that every partition sum is bounded away from zero. It uses the limiting behavior along a specified sequence of increasingly fine partitions and the fact that finite variation would force that sequence’s square-sums to vanish.

Check Your Understanding

Use the finite-variation bound and the dyadic-interval limit to answer these questions.

For a continuous finite-variation function with total variation $V$, why does $Q_\Pi(f)\le m_\Pi V$?
For a fixed interval of length $L$, what are the expectation and variance of the Brownian square-sum with $2^n$ equal subintervals?
Why does convergence of $Q_n(a,b)$ for each fixed dyadic interval not, by itself, give a single probability-one event for all intervals?
Explain why finding one dyadic interval inside a larger interval is enough to rule out finite variation on the larger interval.
What is the precise error in predicting that the square-sum on an interval of length $1$, divided into four steps, has expectation $1/4$?