Ito's Correction Versus the Ordinary Chain Rule

Why the Ordinary Chain Rule Needs a Correction

For a differentiable path $x(t)$, the ordinary chain rule says $d f(x(t))=f'(x(t))\,dx(t)$. Brownian paths are not differentiable, and their squared increments do not disappear when summed over a fine partition. Their quadratic variation is $[B]_t=t$. Consequently, the second-order term in Taylor’s formula survives, producing the correction $\tfrac12 f''(B_t)\,dt$. This tutorial derives that correction, then checks it on polynomials and an exponential.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Assume $B_t$ is a standard Brownian motion with respect to this filtration: it is adapted, has continuous paths, and for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. The independence from $\mathcal{F}_s$ is essential for the martingale properties used below.

An integrand $H_s$ for an Ito integral is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For such $H$, $\int_0^t H_s\,dB_s$ is square-integrable. In particular it is integrable, and for $0\le u\le t$,

\mathbb{E}\left[\int_0^t H_s\,dB_s\middle|\mathcal{F}_u\right] =\int_0^u H_s\,dB_s.

Why this martingale property holds. For adapted step integrands, it follows by conditioning each future increment on the past: Brownian increments after $u$ have conditional mean zero given $\mathcal{F}_u$. General progressively measurable $\mathcal{L}^2$ integrands are approximated in the norm $\mathbb{E}\int_0^T H_s^2\,ds$ by such step integrands. The Ito isometry gives convergence of their integrals in $L^2$; conditional expectation is a contraction in $L^2$, so the identity passes to the limit. The Brownian-with-respect-to-the-filtration assumption is what justifies the conditioning step.

Definition: In the Ito multiplication rules, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. These are shorthand for quadratic-variation and vanishing cross-sum limits, not ordinary algebraic identities between infinitesimal numbers. In particular, the Ito integral $\int H_s\,dB_s$ is distinct from a Stratonovich integral $\int H_s\circ dB_s$.

Deriving the Second-Derivative Term

Theorem: Suppose $f\in C^3(\mathbb{R})$, and $f'$, $f''$, and $f'''$ are bounded. For Brownian motion $B$ with respect to $(\mathcal{F}_t)$, and every finite $T$, the process $f'(B_s)$ is progressively measurable and belongs to $\mathcal{L}^2$. For $0\le t\le T$, $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$

Proof. Fix $t\le T$ and a partition $0=t_0<t_1<\cdots<t_n=t$. Write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and let $|\Pi|=\max_i\Delta t_i$. Taylor’s formula, with $M_3=\sup_x|f'''(x)|$, gives $$ f(B_{t_i})-f(B_{t_{i-1}}) =f'(B_{t_{i-1}})\Delta B_i+\frac12f''(B_{t_{i-1}})(\Delta B_i)^2+R_i, \qquad |R_i|\le\frac{M_3}{6}|\Delta B_i|^3. $$ Since $\Delta B_i\sim N(0,\Delta t_i)$, $\mathbb{E}|\Delta B_i|^3=2\sqrt{2/\pi}(\Delta t_i)^{3/2}$. Thus $$ \mathbb{E}\sum_i|R_i| \le \frac{M_3}{6}\,2\sqrt{\frac2\pi}\sum_i(\Delta t_i)^{3/2} \le \frac{M_3}{3}\sqrt{\frac2\pi}\,t\sqrt{|\Pi|}\longrightarrow0. $$ For the second-order sum, put $Y_i=f''(B_{t_{i-1}})((\Delta B_i)^2-\Delta t_i)$. Brownian increments are independent of $\mathcal{F}_{t_{i-1}}$, so $\mathbb{E}[Y_i\mid\mathcal{F}_{t_{i-1}}]=0$. The $Y_i$ are orthogonal in $L^2$: for $i<j$, $Y_i$ is $\mathcal{F}_{t_{j-1}}$-measurable, and conditioning $Y_j$ on that sigma-field gives zero. If $M_2=\sup_x|f''(x)|$, the normal fourth moment gives $$ \mathbb{E}[Y_i^2]\le 2M_2^2(\Delta t_i)^2, \qquad \mathbb{E}\left[\left(\sum_iY_i\right)^2\right] \le 2M_2^2\sum_i(\Delta t_i)^2 \le 2M_2^2t|\Pi|\longrightarrow0. $$ The remaining weighted time sum $\sum_i f''(B_{t_{i-1}})\Delta t_i$ converges pathwise to $\int_0^t f''(B_s)\,ds$, because $s\mapsto f''(B_s)$ is continuous. The first-order sum converges in $L^2$ to $\int_0^t f'(B_s)\,dB_s$ by the Ito isometry: if $M_2$ bounds $|f''|$, then $$ \mathbb{E}\int_0^t\left|f'(B_s)-f'(B_{t_{i-1}})\right|^2ds \le M_2^2\sum_i\int_{t_{i-1}}^{t_i}(s-t_{i-1})\,ds =\frac{M_2^2}{2}\sum_i(\Delta t_i)^2\longrightarrow0, $$ where the integrand in each interval uses its left endpoint. Summing the Taylor identities telescopes to $f(B_t)-f(B_0)$. Taking limits in probability in those identities gives the stated formula. Boundedness of $f'$ ensures $f'(B_s)\in\mathcal{L}^2$, since $\mathbb{E}\int_0^T f'(B_s)^2ds\le T\sup_x|f'(x)|^2$. $\square$

The proof shows exactly where the correction comes from: the centered part of the squared-increment sum vanishes, while its conditional mean is $\Delta t_i$. In differential notation, the second-order Taylor term therefore contributes $\tfrac12 f''(B_t)dt$. For a differentiable finite-variation path, the sum of squared increments vanishes instead, which is why the ordinary chain rule has no such term.

Worked Polynomial Examples

Worked Example: The Square Function

Apply the formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$, $f''(x)=2$, and $f'''(x)=0$. The theorem’s global boundedness assumptions do not hold for this polynomial, but the formula applies by stopping Brownian motion on $[-n,n]$, applying the bounded-derivative result to a smooth function agreeing with $x^2$ there, and then letting $n$ increase. Brownian paths are bounded on every compact time interval, so these stopping times eventually exceed any fixed $t$, almost surely. The resulting identity is $$ B_t^2=2\int_0^t B_s\,dB_s+t. $$ The integrand $2B_s$ is progressively measurable and belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^t4B_s^2\,ds=4\int_0^t s\,ds=2t^2<\infty$.

Check: Taking expectations, the Ito integral has mean zero by the martingale property established above. Hence the right side has expectation $t$, matching $\mathbb{E}[B_t^2]=t$. Omitting the correction would give only $2\int_0^t B_s\,dB_s$, whose expectation is zero, so it could not equal $B_t^2$.

Worked Example: The Cube Function

For $f(x)=x^3$, the derivatives are $f'(x)=3x^2$, $f''(x)=6x$, and $f'''(x)=6$. As for the square, localize to bounded intervals and then let the stopping levels increase. Ito’s formula yields $$ B_t^3=3\int_0^t B_s^2\,dB_s+3\int_0^t B_s\,ds. $$ The stochastic integrand $3B_s^2$ is progressively measurable and lies in $\mathcal{L}^2$, because $\mathbb{E}\int_0^t9B_s^4\,ds=9\int_0^t3s^2\,ds=9t^3<\infty$. The ordinary chain rule would have no $3\int_0^tB_s\,ds$ term.

Check: Both sides have expectation zero. The left side has expectation zero by the symmetry of $N(0,t)$. On the right, the Ito integral has mean zero by its square-integrable martingale property, and $\mathbb{E}\int_0^tB_s\,ds=\int_0^t\mathbb{E}[B_s]\,ds=0$. The correction is not generally zero path by path, but its expectation is zero in this example.

An Exponential and a Moment Check

Worked Example: The Exponential Function

Take $f(x)=e^x$, with $f'(x)=e^x$, $f''(x)=e^x$, and $f'''(x)=e^x$. These derivatives are not bounded on all of $\mathbb{R}$, so apply the formula first after stopping at $\tau_n=\inf\{s\ge0:|B_s|\ge n\}$, then let $n$ increase. This gives $$ e^{B_t}=1+\int_0^t e^{B_s}\,dB_s+\frac12\int_0^t e^{B_s}\,ds. $$ The unstopped integrand is progressively measurable and in $\mathcal{L}^2$: the normal moment-generating function gives $\mathbb{E}[e^{2B_s}]=e^{2s}$, so $\mathbb{E}\int_0^T e^{2B_s}\,ds=\int_0^T e^{2s}ds=(e^{2T}-1)/2<\infty$.

Independent check: Take expectations. The stochastic integral has mean zero by its square-integrable martingale property. Also $\mathbb{E}[e^{B_s}]=e^{s/2}$, so the expected right side is $$ 1+\frac12\int_0^t e^{s/2}\,ds =1+(e^{t/2}-1) =e^{t/2}. $$ This agrees with $\mathbb{E}[e^{B_t}]=e^{t/2}$, calculated directly from $B_t\sim N(0,t)$. The drift correction is necessary for this check to work.

Ordinary, Ito, and Midpoint Calculations

A common error is to apply the ordinary chain rule directly to Brownian motion. For example, the incorrect calculation $d(B_t^2)=2B_t\,dB_t$ omits the nonzero quadratic variation. The correct Ito calculation is $d(B_t^2)=2B_t\,dB_t+dt$. This is not a choice of algebraic convention: it records the limit of the squared Brownian increments.

There is a related distinction between Ito and Stratonovich integration. For the particular integrand $B_s$, the midpoint sums satisfy the exact identity $$ \sum_i\frac{B_{t_i}+B_{t_{i-1}}}{2}\Delta B_i =\sum_i B_{t_{i-1}}\Delta B_i+\frac12\sum_i(\Delta B_i)^2. $$ As the mesh tends to zero, the first sum converges to the Ito integral and the second to $t$. Thus $\int_0^t B_s\circ dB_s=\int_0^t B_s\,dB_s+\tfrac12t=\tfrac12B_t^2$. The midpoint expression has the ordinary-looking chain rule in this example; the Ito integral instead carries the quadratic-variation correction.

Calculation	Rule for $f(B_t)$	Why
Ordinary differentiable path	$df(x(t))=f'(x(t))\,dx(t)$	Squared increments vanish in the limit.
Brownian motion, Ito integral	$df(B_t)=f'(B_t)\,dB_t+\tfrac12f''(B_t)\,dt$	$[B]_t=t$.
Brownian midpoint sum for $B$	$\int B_s\circ dB_s=\tfrac12B_t^2$	The midpoint sum includes half the quadratic-variation sum.

The correction is often the key to a correct derivation: it changes drift terms even when the diffusion term looks unchanged. Always identify which integral is being used, check the regularity needed for the formula, and verify integrability before taking expectations. The martingale property applies to the Ito integral here because the integrand is in $\mathcal{L}^2$ and $B$ is Brownian with respect to the stated filtration; adaptedness alone would not suffice.

Check Your Understanding

For each question, distinguish the quadratic-variation correction from an ordinary chain-rule term.

For $f(x)=x^4$, write $f'(x)$ and $f''(x)$, then state the Ito differential for $f(B_t)$.
Why does the proof of the correction use independence of $B_{t_i}-B_{t_{i-1}}$ from $\mathcal{F}_{t_{i-1}}$?
For $f(x)=\cos x$, state the first- and second-derivative terms in Ito’s formula.
What integrability condition makes $\int_0^t H_s\,dB_s$ square-integrable, and what conditional-expectation identity verifies its martingale property?
Using midpoint sums, explain why $\int_0^t B_s\circ dB_s$ differs from $\int_0^t B_s\,dB_s$.

Calculation	Rule for \(f(B_t)\)	Why
Ordinary differentiable path	\(df(x(t))=f'(x(t))\,dx(t)\)	Squared increments vanish in the limit.
Brownian motion, Ito integral	\(df(B_t)=f'(B_t)\,dB_t+\tfrac12f''(B_t)\,dt\)	\([B]_t=t\).
Brownian midpoint sum for \(B\)	\(\int B_s\circ dB_s=\tfrac12B_t^2\)	The midpoint sum includes half the quadratic-variation sum.