Brownian Increment Scaling and the Second-Order Term

Why Taylor Order and Brownian Scale Interact

Along a smooth path, an increment over a short time interval is typically proportional to the interval length. Along Brownian motion, an increment over an interval of length $h$ instead has standard deviation $\sqrt{h}$. That change of scale alters the effect of Taylor terms: a first-order term has the size of a Brownian increment, a second-order term has size $h$, and a third-order term has size $h^{3/2}$. When these terms are added over about $1/h$ intervals, the second-order terms can accumulate to order one, whereas the total absolute size of the third-order terms tends to zero.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$. For a fixed $T>0$, divide $[0,T]$ into $n$ equal intervals, set $h=T/n$, and write $t_i=ih$ and $\Delta_iB=B_{t_i}-B_{t_{i-1}}$. Brownian increments on disjoint intervals are independent, and $\Delta_iB$ has the normal distribution $N(0,h)$.

Definition: The mesh of a partition is its largest interval length. Here the mesh is $h=T/n$. A partition sum of order $k$ is a sum whose summands involve the $k$th power of an increment, such as $\sum_i(\Delta_iB)^k$; its absolute-power counterpart is $\sum_i|\Delta_iB|^k$.

The scaling can be read directly from a normal random variable. If $Z\sim N(0,1)$, then $\Delta_iB$ has the same distribution as $\sqrt{h}Z$. Thus, for every $p>0$,

\mathbb{E}\big[|\Delta_iB|^p\big] =h^{p/2}\mathbb{E}[|Z|^p].

One interval contributes a quantity of typical size $h^{p/2}$. A sum over $n=T/h$ intervals therefore has absolute-power expectation proportional to $h^{p/2-1}$. This predicts a distinct role for powers below, at, and above two. First-order increments are not estimated by summing their absolute values: their signed sum telescopes exactly to $B_T-B_0=B_T$. Second powers have total expected size independent of $h$. For powers $p>2$, the expected sum of absolute powers vanishes as $h$ tends to zero.

Partition-Sum Scaling

Theorem: For the equal partition of $[0,T]$, let $Q_n=\sum_{i=1}^n(\Delta_iB)^2$. Then $\mathbb{E}[Q_n]=T$, $\operatorname{Var}(Q_n)=2T^2/n$, and $Q_n\to T$ in $L^2(\mathbb{P})$. For every $p>2$, $\mathbb{E}\big[\sum_{i=1}^n|\Delta_iB|^p\big]=\mathbb{E}[|Z|^p]T^{p/2}n^{1-p/2}\to0$.

Proof. Each $\Delta_iB$ has mean zero, variance $h$, and fourth moment $3h^2$. The fourth moment follows from the standard normal moment $\mathbb{E}[Z^4]=3$, which can be obtained by integrating the normal density by parts. Therefore $\mathbb{E}[(\Delta_iB)^2]=h$, and summing gives $\mathbb{E}[Q_n]=nh=T$. Also, $\operatorname{Var}((\Delta_iB)^2)=3h^2-h^2=2h^2$. Independence of the Brownian increments gives independence of their squares, so $\operatorname{Var}(Q_n)=2nh^2=2T^2/n$. Consequently, $\mathbb{E}[(Q_n-T)^2]=2T^2/n\to0$, which is the asserted $L^2$ convergence. For the power sum, write each increment in distribution as $\sqrt{h}Z$. Its expected absolute-power sum is $n h^{p/2}\mathbb{E}[|Z|^p]=\mathbb{E}[|Z|^p]T^{p/2}n^{1-p/2}$. Since $p>2$, the exponent $1-p/2$ is negative, so this expression tends to zero. $\square$

This result describes accumulated scale, not the value of an individual increment. A single squared increment is random and is not equal to the time step. Rather, the sum of squared increments has mean $T$ and fluctuations that shrink as the partition is refined. This distinction is essential when interpreting a Taylor expansion term by term.

Worked Example: Comparing Second, Third, and Fourth Powers

On $[0,1]$, use four equal intervals. Compute the expected sum of squared increments, its variance, the expected sum of absolute cubed increments, and the expected sum of fourth powers.

Here $n=4$, $T=1$, and $h=1/4$. The second-power sum has mean $nh=4\cdot\tfrac14=1$. Its variance is $2nh^2=2\cdot4\cdot(\tfrac14)^2=\tfrac12$.

For a standard normal variable, $\mathbb{E}[|Z|^3]=2\sqrt{2/\pi}$. Hence the expected sum of absolute cubed increments is

4\left(2\sqrt{\frac{2}{\pi}}\right)\left(\frac14\right)^{3/2} =4\left(2\sqrt{\frac{2}{\pi}}\right)\frac18 =\sqrt{\frac{2}{\pi}}.

The standard normal fourth moment is $3$, so the expected sum of fourth powers is $4\cdot3\cdot(\tfrac14)^2=\tfrac34$.

Check: for $n$ equal intervals on $[0,1]$, these three expectations are $1$, $2\sqrt{2/\pi}/\sqrt{n}$, and $3/n$, respectively. Substituting $n=4$ gives $1$, $\sqrt{2/\pi}$, and $\tfrac34$, matching the calculations. As $n$ grows, the second-power expectation stays at $1$, while the absolute third- and fourth-power sums tend to zero in expectation.

How the Taylor Terms Accumulate

For a twice differentiable function $f$, a one-step Taylor expansion around the left endpoint has a first-order term $f'(B_{t_{i-1}})\Delta_iB$, a second-order term $\tfrac12f''(B_{t_{i-1}})(\Delta_iB)^2$, and a remainder. The scale of the increment suggests that the first-order term is of size $\sqrt{h}$ on one interval, while the second-order term is of size $h$. The first-order terms are signed and can combine through cancellation and telescoping; the second-order terms have a nonzero accumulated scale because there are $T/h$ of them, each of order $h$.

The following result makes the smallness of the Taylor remainder precise under a convenient global smoothness assumption. It does not assert a limit for the weighted second-order sum; it isolates why the third-order error disappears when summed.

Proposition: Suppose $f:\mathbb{R}\to\mathbb{R}$ is three times continuously differentiable and $|f'''(x)|\le M$ for every $x\in\mathbb{R}$, for some finite constant $M$. For the equal partitions of $[0,T]$, write $$ f(B_{t_i})-f(B_{t_{i-1}}) =f'(B_{t_{i-1}})\Delta_iB+\frac12f''(B_{t_{i-1}})(\Delta_iB)^2+R_i. $$ Then $\mathbb{E}[\sum_{i=1}^n|R_i|]\to0$ as $n\to\infty$.

Proof. Taylor's theorem with remainder, applied to $f$ between $B_{t_{i-1}}$ and $B_{t_i}$, gives $|R_i|\le (M/6)|\Delta_iB|^3$. Taking expectations and summing yields $\mathbb{E}[\sum_i|R_i|]\le (M/6)n h^{3/2}\mathbb{E}[|Z|^3]$. Since $nh=T$, the right side is $(M/6)\mathbb{E}[|Z|^3]T\sqrt{h}$, which tends to zero as $h=T/n\to0$. This proves the assertion. $\square$

Worked Example: Taylor Expansion of the Square

Let $f(x)=x^2$ and partition $[0,2]$ into four equal intervals. Display the first- and second-order Taylor terms and check their sum.

Here $h=2/4=1/2$, and $f'(x)=2x$, $f''(x)=2$. The Taylor formula is exact because $f$ is quadratic:

B_{t_i}^2-B_{t_{i-1}}^2 =2B_{t_{i-1}}\Delta_iB+(\Delta_iB)^2.

Summing over all four intervals, the left side telescopes to $B_2^2-B_0^2=B_2^2$. Thus the accumulated first-order term is $B_2^2-Q_4$, where $Q_4=\sum_{i=1}^4(\Delta_iB)^2$, and the accumulated second-order term is $Q_4$. Their sum is exactly $B_2^2$.

The second-order term has expected value $4\cdot\tfrac12=2$, and the theorem gives $\operatorname{Var}(Q_4)=2\cdot4\cdot(\tfrac12)^2=2$. As an independent check, $B_2\sim N(0,2)$, so $\mathbb{E}[B_2^2]=2$. The expectation of the first-order sum is therefore $\mathbb{E}[B_2^2-Q_4]=2-2=0$, consistent with the zero conditional mean of each $B_{t_{i-1}}\Delta_iB$.

Worked Example: A Cubic Taylor Remainder

For $f(x)=x^3$ on $[0,1]$, show how the exact one-step expansion separates the first-, second-, and third-order terms, and calculate the expected total absolute remainder for four equal intervals.

Since $f'(x)=3x^2$ and $f''(x)=6x$, direct expansion gives, for every interval,

B_{t_i}^3-B_{t_{i-1}}^3 =3B_{t_{i-1}}^2\Delta_iB+3B_{t_{i-1}}(\Delta_iB)^2+(\Delta_iB)^3.

The first term is the first-order Taylor term. The second term is $\tfrac12f''(B_{t_{i-1}})(\Delta_iB)^2$. The third term is the exact remainder; its absolute value is $|\Delta_iB|^3$. For $n=4$, $h=1/4$, so

\mathbb{E}\left[\sum_{i=1}^4|\Delta_iB|^3\right] =4\left(2\sqrt{\frac{2}{\pi}}\right)\left(\frac14\right)^{3/2} =\sqrt{\frac{2}{\pi}}.

Check: for general $n$ on $[0,1]$, the same calculation gives $2\sqrt{2/\pi}/\sqrt{n}$, which tends to zero. The finite-partition identity also checks directly: summing its four terms on the right gives $B_1^3-B_0^3=B_1^3$, exactly the telescoping sum on the left.

Why the Second-Order Term Cannot Be Dropped

The ordinary chain rule along a differentiable path retains the first-order Taylor contribution and discards squares of increments because those squares vanish when summed. Brownian increments have a different scale. Over $n$ equal intervals, each squared increment has mean $T/n$, and there are $n$ such terms. Their accumulated mean is $T$, not zero. Meanwhile, the total absolute cubic remainder has expectation proportional to $n^{-1/2}$, so it vanishes as the mesh shrinks. These comparisons explain why a second-order Taylor term survives in Brownian calculations even though third- and higher-order terms are negligible under the stated smoothness conditions.

Common mistake. Applying the ordinary chain rule to $f(x)=x^2$ and writing $B_T^2-B_0^2=2\sum_iB_{t_{i-1}}\Delta_iB$ drops the term $\sum_i(\Delta_iB)^2$. The exact algebra instead gives $B_T^2-B_0^2=2\sum_iB_{t_{i-1}}\Delta_iB+Q_n$. The omitted term has expectation $T$, and $Q_n$ converges to $T$ in $L^2$, so it cannot be dismissed as a negligible remainder.

This scaling analysis is the preparation for the more precise treatment of quadratic variation and the correction that appears in Ito's formula. The calculation here identifies the relevant orders; it does not replace the convergence arguments needed to handle general, state-dependent second-derivative weights.

Check Your Understanding

Use the equal-partition scaling relations above to answer the following.

For a partition of $[0,3]$ into $n$ equal intervals, what are the mean and variance of the sum of squared Brownian increments?
For $p=6$, how does the expected sum of absolute $p$th powers depend on $n$ when $T$ is fixed?
Why can the signed first-order increments sum to a quantity of order one even though each increment has mean zero?
For $f(x)=x^2$, identify the term lost by the ordinary chain-rule calculation and give its expectation on $[0,T]$.
Under the bounded-third-derivative assumption, which bound shows that the total Taylor remainder vanishes as the partition is refined?

Brownian Increment Scaling and the Second-Order Term

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