Almost-Sure Quadratic Variation Along Dyadic Partitions

From Mean-Square Convergence to Almost-Sure Convergence

For Brownian motion, the sum of squared increments over a fine partition approaches elapsed time in mean square. That statement controls an average squared error, but it does not by itself give convergence for almost every sample path. Along dyadic grids, a stronger conclusion follows because the variance errors are summable across grid levels. A Borel–Cantelli argument gives convergence at every dyadic time on one common event, and continuity then extends the result to all times.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with $B_0=0$, continuous paths, and independent increments $B_t-B_s\sim N(0,t-s)$ for $s<t$. For $n\ge0$, the dyadic grid has spacing $h_n=2^{-n}$. Define the sum up to time $t$ by including all complete grid intervals and the final, possibly partial, interval:

S_n(t)=\sum_{k=1}^{\lfloor 2^n t\rfloor} \left(B_{k2^{-n}}-B_{(k-1)2^{-n}}\right)^2 +\left(B_t-B_{\lfloor 2^n t\rfloor 2^{-n}}\right)^2.

The final term is zero when $t$ is a grid point. It matters for defining the sum on every time interval $[0,t]$, rather than only at grid endpoints. We will also use the sum over complete intervals alone, denoted by $R_n(t)$, the first term in this display.

Mean and Variance of a Dyadic Sum

Proposition: Fix $t\ge0$. If the lengths of the complete and partial intervals in $S_n(t)$ are $\ell_1,\ldots,\ell_m$, then $$ \mathbb{E}[S_n(t)]=t,\qquad \operatorname{Var}(S_n(t))=2\sum_{i=1}^{m}\ell_i^2\le 2t\,2^{-n}. $$ The same mean and variance formulas apply to $R_n(t)$ when $t$ is a grid point.

Proof. The increments over the listed intervals are independent, and the increment over an interval of length $\ell_i$ has distribution $N(0,\ell_i)$. If $Z\sim N(0,\ell_i)$, then $\mathbb{E}[Z^2]=\ell_i$ and $\operatorname{Var}(Z^2)=2\ell_i^2$, using $\mathbb{E}[Z^4]=3\ell_i^2$. Therefore, the mean of the sum of their squares is $\sum_i\ell_i=t$. Independence makes the variance of that sum the sum of the variances, namely $2\sum_i\ell_i^2$. Each $\ell_i\le2^{-n}$, so $\sum_i\ell_i^2\le2^{-n}\sum_i\ell_i=2^{-n}t$. At a grid point the final partial interval has length zero, so the same calculation applies to $R_n(t)$. $\square$

Worked Example: Mean and Variance on the Unit Interval

At $t=1$ and level $n=3$, there are $2^3=8$ intervals, each of length $1/8$. Thus

\mathbb{E}[R_3(1)]=8\cdot\frac18=1,\qquad \operatorname{Var}(R_3(1))=2\cdot8\left(\frac18\right)^2 =16\cdot\frac1{64}=\frac14.

For comparison, at level $n=4$ there are $16$ intervals of length $1/16$, so the variance is $2\cdot16\cdot(1/16)^2=32/256=1/8$, while the mean remains $16/16=1$.

Check: the bound $2t2^{-n}$ gives $2\cdot1\cdot1/8=1/4$ at level $3$, exactly the computed variance. At level $4$, it gives $2\cdot1\cdot1/16=1/8$, also exactly the computed variance.

At a fixed time, Chebyshev’s inequality gives convergence in probability from the variance bound. To obtain almost-sure convergence along the sequence of grid levels, we need the stronger fact that the error probabilities are summable over $n$.

Worked Example: A Summable Error Bound at Three-Quarters

Take the dyadic time $r=3/4$. For every $n\ge2$, it is a grid point, and its sum has $3\cdot2^{n-2}$ increments, each of length $2^{-n}$. Thus

\mathbb{E}[R_n(3/4)]=\frac34,\qquad \operatorname{Var}(R_n(3/4)) =2\left(3\cdot2^{n-2}\right)2^{-2n} =3\cdot2^{-(n+1)}.

For the strict tail event $\{|R_n(3/4)-3/4|>1/10\}$, Chebyshev’s inequality gives

\mathbb{P}\left(\left|R_n(3/4)-\frac34\right|>\frac1{10}\right) \le 100\cdot 3\cdot2^{-(n+1)} =150\cdot2^{-n}.

The bounds are summable over $n\ge2$: their sum is $150\sum_{n=2}^{\infty}2^{-n}=150\cdot(1/2)=75<\infty$. The bound at the first level $n=2$ is $150/4=75/2$, which exceeds $1$ and is merely a valid, uninformative upper bound. Summability concerns the series, not whether each early bound is sharp.

Check: at $n=2$, the grid has three intervals of length $1/4$. Its variance is $2\cdot3\cdot(1/4)^2=6/16=3/8$, which agrees with $3\cdot2^{-3}=3/8$ from the general expression.

Convergence at Every Dyadic Time

Theorem: With probability one, $R_n(r)\to r$ for every nonnegative dyadic rational $r$, simultaneously.

Proof. Fix a dyadic rational $r>0$, and choose $m$ so that $r=k2^{-m}$ for some positive integer $k$. For $n\ge m$, $r$ is an endpoint of the level-$n$ grid. The proposition and Chebyshev’s inequality give, for each $\varepsilon>0$,

\mathbb{P}\bigl(|R_n(r)-r|>\varepsilon\bigr) \le \frac{\operatorname{Var}(R_n(r))}{\varepsilon^2} =\frac{2r\,2^{-n}}{\varepsilon^2}.

The sum of these upper bounds over $n\ge m$ is finite, since $\sum_{n=m}^{\infty}2^{-n}=2^{1-m}$. The first Borel–Cantelli implication says that if the probabilities of events $A_n$ have finite sum, then with probability one only finitely many $A_n$ occur. Indeed, for every $N$, $\mathbb{P}(\bigcup_{n\ge N}A_n)\le\sum_{n\ge N}\mathbb{P}(A_n)$, and the right-hand side tends to zero as $N\to\infty$. Applying this implication to the events $\{|R_n(r)-r|>1/j\}$, for each positive integer $j$, proves $R_n(r)\to r$ almost surely. There are countably many dyadic rationals and countably many such $j$; the intersection of their probability-one events still has probability one. At $r=0$, $R_n(0)=0$ for every $n$. Hence convergence holds simultaneously at all nonnegative dyadic rationals. $\square$

This simultaneous statement is stronger than proving convergence separately for each fixed $r$: it supplies one probability-one event on which the result holds at every dyadic endpoint. That common event is what permits a pathwise extension to arbitrary times.

Continuity Extends the Limit to All Times

Theorem: With probability one, for every finite $T\ge0$, $$ \sup_{0\le t\le T}|S_n(t)-t|\longrightarrow0 \qquad\text{as }n\to\infty. $$ In particular, the clipped dyadic squared-increment sums converge almost surely to $t$ at every time $t$.

Proof. Take an outcome in the probability-one event from the preceding theorem and in the event that the Brownian path is continuous. Fix a positive integer $M$ and a coarse dyadic level $q$. The coarse grid divides $[0,M]$ into finitely many intervals $[a,b]$ of length $2^{-q}$. For $n\ge q$, $R_n$ is nondecreasing in $t$: it adds nonnegative squared increments as the completed grid intervals accumulate. If $t\in[a,b]$, then $R_n(a)\le R_n(t)\le R_n(b)$. Consequently,

|R_n(t)-t| \le \max_{v\in G_q}|R_n(v)-v|+2^{-q},

where $G_q=\{k2^{-q}:0\le k2^{-q}\le M\}$ is the finite set of coarse grid points. To verify the bound, $R_n(t)-t\le R_n(b)-t\le |R_n(b)-b|+(b-t)$, and $b-t\le2^{-q}$; also $t-R_n(t)\le t-R_n(a)\le(t-a)+|a-R_n(a)|$, with $t-a\le2^{-q}$. Because $G_q$ is finite and convergence holds at every dyadic point, $\max_{v\in G_q}|R_n(v)-v|\to0$ as $n\to\infty$. Given any $\varepsilon>0$, choose $q$ with $2^{-q}<\varepsilon/2$, then choose $n$ sufficiently large that this finite maximum is less than $\varepsilon/2$. This proves uniform convergence of $R_n(t)$ to $t$ on $[0,M]$.

It remains to include the partial final increment. Define the path modulus of continuity on $[0,M]$ by $\omega_M(\delta)=\sup\{|B_s-B_u|:s,u\in[0,M],\,|s-u|\le\delta\}$. Continuity on the compact interval implies $\omega_M(\delta)\to0$ as $\delta\to0$. For every $t\in[0,M]$, the partial interval has length at most $2^{-n}$, so $0\le S_n(t)-R_n(t)\le\omega_M(2^{-n})^2$. Thus this difference tends uniformly to zero, and $S_n(t)\to t$ uniformly on $[0,M]$. Taking the intersection over positive integer $M$ gives one probability-one event on which the conclusion holds on every compact time interval, and hence for every finite $T$. $\square$

Worked Example: A Clipped Sum at a Non-Grid Time

At level $n=2$, the grid spacing is $1/4$. Take $t=11/16$. The complete intervals end at $1/4$ and $1/2$; the last partial interval has length $11/16-1/2=3/16$. The three independent increments have lengths $1/4,1/4,3/16$. Therefore

\mathbb{E}[S_2(11/16)] =\frac14+\frac14+\frac3{16} =\frac{11}{16}, $$ $$ \operatorname{Var}(S_2(11/16)) =2\left[\left(\frac14\right)^2+\left(\frac14\right)^2+ \left(\frac3{16}\right)^2\right] =2\left(\frac{16+16+9}{256}\right) =\frac{41}{128}.

Check: the general variance bound is $2t2^{-n}=2\cdot(11/16)\cdot(1/4)=11/32=44/128$, which is at least the exact variance $41/128$. The partial increment is included in the sum, so the mean is the full elapsed time $11/16$, not just the last grid endpoint $1/2$.

Why Summability Matters

A common mistake is to infer almost-sure convergence directly from a variance tending to zero. For example, $\operatorname{Var}(R_n(r))=2r2^{-n}\to0$ proves convergence in mean square, and hence in probability, but that implication alone does not establish almost-sure convergence. The valid argument fixes an error threshold $\varepsilon$, sums the Chebyshev bounds $2r2^{-n}/\varepsilon^2$, and then applies Borel–Cantelli. The geometric decay makes that sum finite. Finally, using countably many thresholds and dyadic times produces a single probability-one event; continuity and monotonicity then fill in the times between grid points.

The result is specific to the dyadic sequence used here. It does not by itself establish almost-sure convergence along every deterministic sequence of partitions. The important mechanism in this argument is that the errors across dyadic levels admit summable probability bounds, not merely that the error at each level tends to zero.

Check Your Understanding

Use the variance calculation and the almost-sure convergence argument to answer these questions.

For a dyadic endpoint $r>0$, derive the Chebyshev bound for $\mathbb{P}(|R_n(r)-r|>\varepsilon)$ and explain why its sum over $n$ is finite.
Why is it necessary to take a countable intersection over error thresholds when proving convergence at a fixed dyadic endpoint?
Explain how monotonicity of $R_n$ bounds its error at a time between two coarse dyadic endpoints.
Why does continuity of the Brownian path make the partial final squared increment negligible uniformly on a compact interval?
Does convergence in probability at each fixed time alone imply simultaneous almost-sure convergence at every time? Explain the role of the summability argument.