Partition Sums as Definitions of Quadratic Variation

From Increments to Partition Sums

Quadratic variation is defined through sums of squared increments over increasingly fine partitions. For a Brownian path, increments over short time intervals are typically of size comparable to the square root of the interval length. Squaring those increments therefore produces contributions of the same order as the time intervals themselves. This makes the accumulated sum a candidate to have a nonzero limit, unlike the corresponding sum for a smooth path.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $T>0$. A deterministic partition of $[0,T]$ is a finite set $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$. Write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and define the mesh by $|\Pi|=\max_{1\le i\le n}\Delta t_i$.

Definition: The quadratic-variation sum of $B$ along $\Pi$ is the random variable $Q_\Pi=\sum_{i=1}^n(\Delta B_i)^2$. For a sequence of partitions $(\Pi_m)$, its mesh tends to zero if $|\Pi_m|\to0$. The quadratic variation along that sequence is the limit of $Q_{\Pi_m}$, if the sums converge in the specified sense.

The qualification “in the specified sense” matters. For random variables $X_m$ and $X$, convergence in probability means $\mathbb{P}(|X_m-X|>\varepsilon)\to0$ for every $\varepsilon>0$. Convergence in mean square, or in $L^2(\mathbb{P})$, means $\mathbb{E}[(X_m-X)^2]\to0$. Almost-sure convergence means that $X_m(\omega)\to X(\omega)$ for all $\omega$ outside a set of probability zero. These are statements about a sequence of random variables, not merely about their means.

What a Brownian Partition Sum Does

For each interval of the partition, Brownian independent increments give $\Delta B_i\sim N(0,\Delta t_i)$, and the variables $\Delta B_i$ are independent. The elementary normal moments used below are $\mathbb{E}[Z^2]=1$ and $\mathbb{E}[Z^4]=3$ for $Z\sim N(0,1)$. In particular, if $Y\sim N(0,h)$, scaling $Y=\sqrt{h}Z$ gives $\mathbb{E}[Y^2]=h$ and $\mathbb{E}[Y^4]=3h^2$.

Proposition: For any deterministic partition $\Pi$ of $[0,T]$, $\mathbb{E}[Q_\Pi]=T$ and $\operatorname{Var}(Q_\Pi)=2\sum_{i=1}^n(\Delta t_i)^2$. Consequently, for every $\varepsilon>0$, $$ \mathbb{P}(|Q_\Pi-T|>\varepsilon)\le \frac{2T|\Pi|}{\varepsilon^2}, \qquad \mathbb{E}[(Q_\Pi-T)^2]\le 2T|\Pi|. $$

Proof. For each $i$, the second-moment formula gives $\mathbb{E}[(\Delta B_i)^2]=\Delta t_i$. Summing and using $\sum_i\Delta t_i=T$ yields $\mathbb{E}[Q_\Pi]=T$. The fourth-moment formula gives $$ \operatorname{Var}((\Delta B_i)^2) =\mathbb{E}[(\Delta B_i)^4]-\big(\mathbb{E}[(\Delta B_i)^2]\big)^2 =3(\Delta t_i)^2-(\Delta t_i)^2 =2(\Delta t_i)^2. $$ Independence of the increments implies independence of their squares, so the variance of their sum is the sum of their variances. This proves the stated variance. Since $\Delta t_i\le|\Pi|$, we have $\sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i=T|\Pi|$. Also, $Q_\Pi-T$ has mean zero, so its second moment equals its variance. This gives the mean-square bound. Finally, applying Markov's inequality to the nonnegative random variable $(Q_\Pi-T)^2$ gives the probability bound. $\square$

Worked Example: A Nonuniform Partition of a Unit Interval

Take $T=1$ and $\Pi=\{0,\tfrac14,\tfrac12,1\}$. The interval lengths are $\tfrac14,\tfrac14,\tfrac12$, so the mesh is $\tfrac12$. Compute the mean, variance, and the resulting probability bound for $\varepsilon=\tfrac12$.

The mean is $T=1$. The variance is $$ 2\left(\left(\tfrac14\right)^2+\left(\tfrac14\right)^2+\left(\tfrac12\right)^2\right) =2\left(\tfrac1{16}+\tfrac1{16}+\tfrac14\right) =2\cdot\tfrac38 =\tfrac34. $$ The general bound gives $$ \mathbb{P}(|Q_\Pi-1|>\tfrac12) \le \frac{2\cdot1\cdot\tfrac12}{(\tfrac12)^2} =4. $$ Because a probability is at most $1$, this bound is valid but uninformative. Check: the variance calculation also equals $2\sum_i(\Delta t_i)^2=\tfrac34$, and the inequality $\tfrac34\le 2T|\Pi|=1$ holds.

Three Modes of Convergence

The proposition gives a direct mean-square conclusion for any sequence of deterministic partitions whose mesh tends to zero: $\mathbb{E}[(Q_{\Pi_m}-T)^2]\to0$. This also gives convergence in probability. In general, mean-square convergence implies convergence in probability because for every $\varepsilon>0$, $$ \mathbb{P}(|X_m-X|>\varepsilon) \le \frac{\mathbb{E}[(X_m-X)^2]}{\varepsilon^2}. $$ Almost-sure convergence also implies convergence in probability, but its proof uses a different argument: if $X_m\to X$ almost surely, then the indicators $\mathbf{1}_{\{|X_m-X|>\varepsilon\}}$ converge to zero almost surely and are bounded by $1$. Dominated convergence gives that their expectations, which are the probabilities of these events, tend to zero.

Worked Example: Equal-Length Partitions

Partition $[0,2]$ into $n$ equal intervals. Find the mean, variance, and a probability bound for the deviation from $2$, using an arbitrary $\varepsilon>0$.

Each interval has length $2/n$, and there are $n$ such intervals. Thus $\mathbb{E}[Q_{\Pi_n}]=2$, while $$ \operatorname{Var}(Q_{\Pi_n}) =2n\left(\frac{2}{n}\right)^2 =\frac{8}{n}. $$ The probability bound is $$ \mathbb{P}(|Q_{\Pi_n}-2|>\varepsilon) \le\frac{8}{n\varepsilon^2}. $$ For fixed $\varepsilon>0$, the right-hand side tends to zero as $n\to\infty$, so $Q_{\Pi_n}\to2$ in probability. The mean-square error is exactly $\mathbb{E}[(Q_{\Pi_n}-2)^2]=8/n$, so convergence also holds in mean square. Check: at $n=4$, each length is $1/2$, the variance is $2\cdot4\cdot(1/2)^2=2$, and the formula $8/n$ gives $8/4=2$.

For a particular sequence of partitions, mean-square convergence is stronger than convergence in probability. Neither statement, by itself, is the definition of almost-sure convergence. The distinction can be seen in a Brownian example not involving partition sums.

Worked Example: Convergence in Probability Without Almost-Sure Convergence

Let $Z_n=B_n-B_{n-1}$, for integers $n\ge1$. For each $n\ge2$, choose $a_n$ so that a standard normal variable $Z$ satisfies $\mathbb{P}(Z>a_n)=1/n$, and set $X_n=\sqrt{n}\,\mathbf{1}_{\{Z_n>a_n\}}$. The threshold exists because the standard normal tail is continuous and takes every value strictly between $0$ and $1$.

Since $Z_n\sim N(0,1)$, $\mathbb{P}(X_n\ne0)=1/n$. Therefore, for every $\varepsilon>0$, $\mathbb{P}(|X_n|>\varepsilon)\le1/n\to0$, so $X_n\to0$ in probability. But $\mathbb{E}[X_n^2]=n\mathbb{P}(Z_n>a_n)=n(1/n)=1$, so $X_n$ does not converge to zero in mean square.

Moreover, the increments $Z_n$ are independent, so the events $\{Z_n>a_n\}$ are independent. Their probabilities sum to $\sum_{n=2}^{\infty}1/n=\infty$. The second Borel–Cantelli lemma therefore implies that these events occur infinitely often with probability one. On that event $X_n=\sqrt n$ infinitely often, and hence $X_n$ does not converge to zero almost surely. This example separates convergence in probability from both mean-square and almost-sure convergence; it is not a claim about the quadratic-variation sums.

Worked Example: One Brownian Path at Times Approaching Zero

Set $Y_n=B_{1/n}$. Brownian path continuity at time zero implies $Y_n\to B_0=0$ almost surely. Also, $Y_n\sim N(0,1/n)$, so $\mathbb{E}[Y_n^2]=1/n\to0$. Thus $Y_n\to0$ in mean square, and the mean-square implication above gives convergence in probability as well. Check: the three conclusions agree here, but they follow from distinct facts: path continuity gives almost-sure convergence, and the Brownian variance gives mean-square convergence.

Reading the Definition Correctly

For Brownian motion, the proposition proves that along every deterministic sequence of partitions with mesh tending to zero, the quadratic-variation sums converge to $T$ in mean square and hence in probability. It does not assert that each squared increment equals its time increment. At a fixed interval, $(\Delta B_i)^2$ is random; the result concerns the accumulated sum as the partition is refined.

Nor should convergence along a specified sequence be silently replaced by almost-sure convergence. The present estimates establish mean-square and probability convergence, but they do not by themselves show that the sums converge for almost every path along every possible sequence of partitions. Almost-sure claims require their own argument and must specify the sequence under consideration.

A common mistake is to infer almost-sure convergence just because the variance tends to zero. For equal partitions of $[0,2]$, the variance $8/n$ tends to zero, which proves mean-square convergence and therefore convergence in probability. It is not, by itself, a proof of almost-sure convergence: the two properties have different definitions, and convergence in probability need not imply almost-sure convergence, as the Brownian-increment example shows.

Takeaway: A partition specifies which increments are squared and summed; its mesh specifies the scale of refinement; and the mode of convergence specifies what limit the sums are claimed to have. For Brownian motion the estimates here give the limit $T$ in mean square and in probability. An almost-sure conclusion requires a separate justification.

Check Your Understanding

Use the definitions and estimates above to answer the following.

For a partition with interval lengths $\tfrac13,\tfrac23$ on $[0,1]$, compute the mean and variance of $Q_\Pi$.
Why does convergence in mean square imply convergence in probability? State the inequality used.
For equal partitions of $[0,T]$ into $n$ intervals, derive the variance of $Q_{\Pi_n}$.
What additional conclusion, if any, follows from the partition-sum estimates here about almost-sure convergence?
In the Brownian-increment example, why does $X_n\to0$ in probability even though the events $\{X_n\ne0\}$ occur infinitely often almost surely?