Quadratic Variation as the Source of the Ito Correction

Why Squared Increments Produce a Correction

For a twice differentiable function, a small change $x\mapsto x+\Delta x$ contributes a second-order Taylor term $\tfrac12 f''(x)(\Delta x)^2$. Along a smooth path, the sum of these squared increments vanishes as the partition is refined. For Brownian motion, the squared increments instead accumulate to elapsed time. Thus the second-order Taylor term need not disappear: its coefficient $\tfrac12$ remains, and its squared-increment factor becomes a time integral. This is the source of the one-half second-derivative correction in Ito calculus.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Throughout, $B_t$ is Brownian motion with respect to this filtration: it is adapted, has continuous paths and independent Gaussian increments, and, for every $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. This independence from the filtration’s past is important; independence of the increments from each other alone would not justify the conditional-expectation arguments below.

Fix $T>0$ and a deterministic partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$. Write $h_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and $|\Pi|=\max_i h_i$. The quadratic variation sum is $Q_\Pi=\sum_{i=1}^n(\Delta B_i)^2$. For stochastic integrals, the integrand class used here is $\mathcal{L}^2$: progressively measurable $H$ such that $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The integral $\int_0^T H_s\,dB_s$ is the Ito integral.

Quadratic Variation Converges to Elapsed Time

Theorem: For each deterministic partition of $[0,T]$, $\mathbb{E}[Q_\Pi]=T$ and $\operatorname{Var}(Q_\Pi)=2\sum_i h_i^2\le 2T|\Pi|$. Consequently, as $|\Pi|\to0$, $Q_\Pi\to T$ in $L^2(\mathbb{P})$.

Proof. Since $\Delta B_i\sim N(0,h_i)$, its second moment is $h_i$, and its fourth moment is $3h_i^2$. Therefore $\mathbb{E}[Q_\Pi]=\sum_i h_i=T$, while $\operatorname{Var}((\Delta B_i)^2)=3h_i^2-h_i^2=2h_i^2$. The Brownian increments over disjoint intervals are independent, so their squares are independent and the variance of their sum is the sum of their variances. Finally, $h_i^2\le|\Pi|h_i$, whence $\sum_i h_i^2\le|\Pi|T$. Since the mean of $Q_\Pi$ is $T$, we obtain $\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)\le2T|\Pi|$, which tends to zero with the mesh. $\square$

Worked Example: A Two-Interval Quadratic Variation Sum

For Brownian motion on $[0,1]$ with partition $0<\tfrac12<1$, compute the mean and variance of $Q_\Pi$.

Both time increments have length $h_1=h_2=\tfrac12$, so $Q_\Pi=(\Delta B_1)^2+(\Delta B_2)^2$. Its mean is $\tfrac12+\tfrac12=1$. Each squared increment has variance $2(\tfrac12)^2=\tfrac12$; independence gives $\operatorname{Var}(Q_\Pi)=\tfrac12+\tfrac12=1$.

Check: the theorem’s bound is $2T|\Pi|=2\cdot1\cdot\tfrac12=1$, agreeing with the exact variance. This particular partition has a random quadratic variation sum, not a sum that equals $1$ on every Brownian path.

Weighted Squared Increments

The second-order Taylor term is weighted by the second derivative evaluated at the left endpoint. The next result explains why such weights do not change the basic quadratic-variation mechanism. The filtration assumption in the Brownian definition ensures that the next increment is independent of the information available at its left endpoint.

Theorem: Suppose $G_t$ is adapted, has continuous paths, and satisfies $|G_t|\le C$ for all $t\in[0,T]$ almost surely, for a deterministic finite constant $C$. Then, along deterministic partitions with mesh tending to zero, $$ \sum_{i=1}^n G_{t_{i-1}}(\Delta B_i)^2 \longrightarrow \int_0^T G_t\,dt \quad\text{in }L^2(\mathbb{P}). $$

Proof. Subtract the ordinary time sum and put $Y_i=G_{t_{i-1}}\big((\Delta B_i)^2-h_i\big)$. The variable $G_{t_{i-1}}$ is $\mathcal{F}_{t_{i-1}}$-measurable. Brownian motion with respect to the filtration has $\Delta B_i$ independent of $\mathcal{F}_{t_{i-1}}$; hence $\mathbb{E}[Y_i\mid\mathcal{F}_{t_{i-1}}]=0$. Also, the conditional second moment of $(\Delta B_i)^2-h_i$ is $2h_i^2$, so $\mathbb{E}[Y_i^2]\le2C^2h_i^2$. If $i<j$, then $Y_i$ is $\mathcal{F}_{t_{j-1}}$-measurable. Conditioning on $\mathcal{F}_{t_{j-1}}$ gives $\mathbb{E}[Y_iY_j]=\mathbb{E}[Y_i\mathbb{E}[Y_j\mid\mathcal{F}_{t_{j-1}}]]=0$. Thus $$ \mathbb{E}\left[\left(\sum_iY_i\right)^2\right] =\sum_i\mathbb{E}[Y_i^2] \le2C^2\sum_i h_i^2 \le2C^2T|\Pi|\longrightarrow0. $$ It remains to pass from $\sum_iG_{t_{i-1}}h_i$ to $\int_0^T G_t\,dt$. For every continuous path of $G$, these are left-endpoint Riemann sums and converge to the integral. Their absolute values and the integral are bounded by $CT$; consequently dominated convergence gives convergence in $L^2$ as well. Combining the two convergences proves the claim. $\square$

In particular, if $f$ is twice continuously differentiable and $f''$ is bounded, then $G_t=f''(B_t)$ meets the hypotheses: it is adapted and continuous, and is bounded by the bound on $f''$. Therefore the weighted squared-increment sum converges to $\int_0^T f''(B_t)\,dt$. Taylor’s second-order term carries an additional factor $\tfrac12$, so its quadratic-variation contribution converges to $\tfrac12\int_0^T f''(B_t)\,dt$. This identifies the correction’s source. It does not, by itself, prove that all Taylor remainders vanish or establish the full Ito formula; those require further estimates.

Worked Consequences of the Second-Order Term

Worked Example: The Quadratic Function

Take $f(x)=x^2$ on $[0,2]$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$, so the Taylor second-order term on interval $i$ is exactly $\tfrac12(2)(\Delta B_i)^2=(\Delta B_i)^2$. Summing these terms gives $Q_\Pi$, which converges in $L^2$ to $2$. Equivalently, the limiting correction is $\tfrac12\int_0^2 2\,dt=2$.

The finite-partition identity also makes the role of this term exact. Expanding each square gives $B_{t_i}^2-B_{t_{i-1}}^2=2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2$. Summing and using $B_0=0$ yields $$ B_2^2=2\sum_iB_{t_{i-1}}\Delta B_i+Q_\Pi. $$ The left-endpoint sums converge to $\int_0^2 B_s\,dB_s$. Indeed, $B\in\mathcal{L}^2$ on this interval because $\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2$; the Ito isometry gives the squared error as $\sum_i\int_{t_{i-1}}^{t_i}(s-t_{i-1})\,ds=\tfrac12\sum_i h_i^2$, which tends to zero. Taking the limit in the identity gives $B_2^2=2\int_0^2B_s\,dB_s+2$.

Check: the Ito integral here has mean zero: its defining left-endpoint sums have mean zero because $B_{t_{i-1}}$ is measurable at the left endpoint and the next Brownian increment is independent of that past and has mean zero; convergence in $L^2$ preserves the mean. Also $\mathbb{E}[B_2^2]=2$, so the mean of the right side is $2\cdot0+2=2$, as required.

Worked Example: The Cosine Function

Take $f(x)=\cos x$. Then $f'(x)=-\sin x$ and $f''(x)=-\cos x$, with bounded second derivative. The weighted quadratic-variation result shows that the sum of the second-order Taylor terms converges to $-\tfrac12\int_0^T\cos(B_t)\,dt$.

For $T=2$, compute the expectation of this correction. Since $B_t\sim N(0,t)$, the characteristic function of a centered normal variable gives $\mathbb{E}[\cos(B_t)]=e^{-t/2}$. The correction is bounded in absolute value by $T/2=1$, so expectation can be taken under the time integral. Therefore $$ \mathbb{E}\left[-\frac12\int_0^2\cos(B_t)\,dt\right] =-\frac12\int_0^2e^{-t/2}\,dt =-\frac12\left[-2e^{-t/2}\right]_0^2 =-(1-e^{-1}). $$ The same value is obtained as the limit of the expectations of the partition sums: for a left endpoint $t_{i-1}$, independence gives $\mathbb{E}[\cos(B_{t_{i-1}})(\Delta B_i)^2]=h_i\mathbb{E}[\cos(B_{t_{i-1}})]=h_ie^{-t_{i-1}/2}$. Multiplying by $-\tfrac12$ and summing gives Riemann sums for the displayed integral.

Worked Example: Why a Smooth Path Has No Such Accumulation

Let $x(t)=t$ on $[0,1]$, partitioned into $n$ equal intervals, and use $f(x)=x^2$. Each increment of the path is $\Delta x_i=1/n$. The sum of the second-order terms is $$ \sum_{i=1}^n\frac12 f''(x(t_{i-1}))(\Delta x_i)^2 =\sum_{i=1}^n\frac12\cdot2\cdot\frac1{n^2} =\frac1n\longrightarrow0. $$ The exact endpoint difference is $f(x(1))-f(x(0))=1-0=1$; the first-order Taylor terms sum to $\sum_i2(i-1)n^{-1}\cdot n^{-1}=(n-1)/n$, and adding the second-order sum $1/n$ gives $1$ for every $n$. Thus the second-order contribution disappears in the smooth limit, unlike the Brownian case.

Check: at $n=2$, the two first-order terms are $0$ and $1/2$; the two second-order terms are $1/4$ and $1/4$. Their total is $0+\tfrac12+\tfrac14+\tfrac14=1$, matching the endpoint difference.

Common Pitfall: Replacing Every Square by Its Mean

A tempting but incorrect calculation is to replace every $(\Delta B_i)^2$ by $h_i$ as though the two quantities were equal on each interval. For the two-interval partition of $[0,1]$, this would claim $(\Delta B_1)^2=(\Delta B_2)^2=\tfrac12$ on every outcome. In fact, each square has mean $\tfrac12$ and variance $2(\tfrac12)^2=\tfrac12$, so it is not a constant. What is true is that the sum of squared increments converges to elapsed time in $L^2$ as the mesh tends to zero.

Likewise, applying the ordinary chain rule to $B_t^2$ and writing $B_2^2=2\int_0^2B_s\,dB_s$ omits the accumulated second-order term. The exact partition identity above shows what is missing: $Q_\Pi$, whose limit on $[0,2]$ is $2$. In differential shorthand, this mechanism is recorded as $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$. These rules summarize limits of increment products; they do not assert that an individual random squared increment equals its time increment.

Check Your Understanding

Use the squared-increment and weighted-sum arguments to answer the following.

For a deterministic partition of $[0,3]$, give the mean and variance of $Q_\Pi$ in terms of its interval lengths, and state a bound that tends to zero with the mesh.
Why does the weighted-sum proof require Brownian increments to be independent of $\mathcal{F}_{t_{i-1}}$, not merely independent of earlier Brownian increments?
For $f(x)=\sin x$, write $f'$, $f''$, and the limiting second-order Taylor contribution on $[0,T]$.
For $f(x)=x^2$, identify the finite-partition sum that converges to the correction and its limit on $[0,T]$.
In the smooth-path example $x(t)=t$ with $f(x)=x^2$, what is the second-order sum for $n$ equal intervals, and why does its limit differ from the Brownian case?