A Reliable Audit for Brownian Chain-Rule Calculations
Most errors in a Brownian chain-rule calculation are local: a missing factor of one-half, the wrong derivative in the correction, or an omitted time derivative. A separate conceptual error is to read \(dB_t\,dB_t=dt\) as an equality between individual random increments. A short audit can catch all four problems: write down the relevant derivatives, place the time term explicitly, apply the quadratic-variation rules, and check the result against an independent calculation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For a progressively measurable process \(H_s\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), the Itô integral \(\int_0^t H_s\,dB_s\) is a square-integrable martingale on \([0,T]\). We use the Brownian Itô formula established in the earlier material on Itô's formula: if \(f\) is twice continuously differentiable on an interval containing the Brownian values and the displayed integrals are defined, then, for a time-independent \(f\),
For a time-dependent function \(f(t,x)\) with one continuous time derivative and two continuous space derivatives, the formula is
Whenever we claim the stochastic-integral term is a true martingale, we will check the square-integrability condition above. The symbol \(d\) in these formulas is shorthand for the corresponding increment identity; it does not turn Brownian paths into differentiable paths.
What the Half Factor and Second Derivative Actually Say
The factor one-half comes from the second-order Taylor term, \(\tfrac12 f''(x)(\Delta x)^2\), together with Brownian quadratic variation. In the differential bookkeeping, \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). These rules retain the second-order Brownian contribution and discard the mixed and time-squared terms. In particular, \(dB_t\,dB_t=dt\) records the limiting quadratic variation \([B]_t=t\); it does not assert that \((B_{t+h}-B_t)^2=h\) for any particular increment.
Proof. Substitute the process \(B_t\) into the time-dependent Itô formula with \(f(t,x)=f(x)\). Its time derivative is zero, its first space derivative is \(f'(x)\), and its second space derivative is \(f''(x)\). The resulting identity is the time-independent formula displayed above, including the coefficient \(\tfrac12\). The integral form also shows that its correction over \([0,t]\) is exactly \(\tfrac12\int_0^t f''(B_s)\,ds\); the symbolic term \(\tfrac12 f''(B_t)\,dt\) is its differential notation. \(\square\)
Worked Example: The Square Function and the Half Factor
Apply Itô's formula to \(f(x)=x^2\). Compute its derivatives first: \(f'(x)=2x\) and \(f''(x)=2\). Therefore
Integrated from zero, using \(B_0=0\), this gives \(B_t^2=2\int_0^t B_s\,dB_s+t\). Equivalently, \(\int_0^t B_s\,dB_s=\tfrac12(B_t^2-t)\).
Check: For finite \(t\), \(\mathbb{E}\int_0^t B_s^2\,ds=\int_0^t s\,ds=t^2/2<\infty\), so the stochastic integral is a square-integrable martingale and has expectation zero. Taking expectations in the identity gives \(\mathbb{E}[B_t^2]=t\), as required for a centered normal variable with variance \(t\). At \(t=0\), both sides of the integrated identity are zero.
Check the Derivative Before the Correction
A frequent slip is to write a correction involving \(f'(B_t)\), or to differentiate twice but forget the factor one-half. The safest order is to compute \(f'\) and \(f''\) on a separate line, then substitute them into the formula without simplifying away the source of the coefficient.
Worked Example: A Cubic Function and the Second Derivative
Let \(f(x)=x^3\). The derivatives are \(f'(x)=3x^2\) and \(f''(x)=6x\). Substitution gives
The corrected calculation does not have \(3B_t^2\,dt\): that expression uses the first derivative in the correction. Nor is the correction \(6B_t\,dt\), which drops the one-half factor. Integrating from zero yields \(B_t^3=3\int_0^t B_s^2\,dB_s+3\int_0^t B_s\,ds\).
Check: The stochastic integrand is square-integrable because \(\mathbb{E}\int_0^t 9B_s^4\,ds=\int_0^t 27s^2\,ds=9t^3<\infty\), using \(\mathbb{E}[B_s^4]=3s^2\). Its integral is therefore a martingale of mean zero. Also, \(\mathbb{E}\int_0^t B_s\,ds=\int_0^t\mathbb{E}[B_s]\,ds=0\). The formula consequently gives \(\mathbb{E}[B_t^3]=0\), which agrees with the symmetry of the centered normal distribution.
These expectation checks can reveal an error, but they do not prove a pathwise formula by themselves: distinct random variables can have the same expectation. The derivative substitution is the calculation; the check is an independent consistency test.
Time Dependence Requires Its Own Term
When a function depends explicitly on time, the ordinary time derivative contributes \(f_t\,dt\). It is separate from the Brownian correction \(\tfrac12f_{xx}\,dt\). A useful error check is to set aside the Brownian term briefly and ask whether changing \(t\) while holding \(x\) fixed changes the function. If it does, \(f_t\) must appear.
Worked Example: A Time-Dependent Function with Cancellation
Take \(f(t,x)=e^{t/2}\cos x\). Its partial derivatives are \(f_t(t,x)=\tfrac12e^{t/2}\cos x\), \(f_x(t,x)=-e^{t/2}\sin x\), and \(f_{xx}(t,x)=-e^{t/2}\cos x\). Thus the time and correction terms are
The full result is \(d(e^{t/2}\cos B_t)=-e^{t/2}\sin(B_t)\,dB_t\). The cancellation is not a reason to omit either term: each is nonzero before they are added.
Check: The stochastic integrand is progressively measurable, and \(\mathbb{E}\int_0^T e^s\sin^2(B_s)\,ds\le\int_0^T e^s\,ds=e^T-1<\infty\). Its integral is a square-integrable martingale, so the expectation of \(e^{t/2}\cos(B_t)\) equals its initial value \(1\). Independently, \(B_t\) has distribution \(N(0,t)\), hence \(\mathbb{E}[\cos(B_t)]=e^{-t/2}\). Therefore \(\mathbb{E}[e^{t/2}\cos(B_t)]=e^{t/2}e^{-t/2}=1\), agreeing with the formula.
A Four-Point Error Audit
For \(f(t,x)\), list \(f_t\), \(f_x\), and \(f_{xx}\). Do not infer the correction from \(f_x\).
Write \(f_t\,dt+f_x\,dB_t+\tfrac12 f_{xx}\,dt\), evaluated at \((t,B_t)\). For a time-independent function, set \(f_t=0\).
Add \(f_t\) and \(\tfrac12f_{xx}\) as coefficients of \(dt\). Keep \(f_x\,dB_t\) as the stochastic-integral term.
Check initial values and, when integrability permits, expectations or a known distributional moment. Such checks complement, but do not replace, the derivative calculation.
For example, the incorrect line \(d(B_t^2)=2B_t\,dB_t+2\,dt\) comes from using \(f''=2\) but forgetting the one-half. The correction is \(\tfrac12(2)\,dt=dt\), not \(2\,dt\). This wrong expression would imply \(\mathbb{E}[B_t^2]=2t\), contradicting the Brownian variance \(t\). Likewise, writing \(d(tB_t)=t\,dB_t\) omits explicit time dependence: for \(f(t,x)=tx\), the derivatives are \(f_t=x\), \(f_x=t\), and \(f_{xx}=0\), so the correct result is \(d(tB_t)=B_t\,dt+t\,dB_t\).
What the Differential Multiplication Rules Do—and Do Not—Mean
The multiplication rules are a compact way to retain the terms that survive in Itô's formula. They are not ordinary algebraic identities between finite increments. In particular, an increment \(\Delta B\) over a time interval of length \(h\) is random; its square is not fixed at \(h\). Instead, sums of squared Brownian increments converge to elapsed time as partitions become fine. That quadratic variation is why a second derivative contributes even though the ordinary chain rule has no second-derivative term.
| Calculation item | Correct Brownian rule | Common error |
|---|---|---|
| Second-order correction | \(\tfrac12 f_{xx}\,dt\) | Omit \(\tfrac12\) or use \(f_x\) |
| Explicit time dependence | Add \(f_t\,dt\) | Apply the time-independent formula unchanged |
| Squared Brownian differential | Use \(dB_t\,dB_t=dt\) as quadratic-variation bookkeeping | Claim each squared increment equals its time step |
| Mixed and time-squared terms | \(dB_t\,dt=0\), \(dt\,dt=0\) | Retain them as second-order contributions |
A practical habit is to keep the unsimplified expression until every derivative and coefficient has been checked. Then use a check suited to the function: a moment, the value at time zero, a cancellation, or the formula's integrated form. This makes the calculation auditable and helps distinguish a genuine Itô correction from an algebraic slip.
Check Your Understanding
For each question, identify the relevant derivatives before writing the Itô differential.
- For \(f(x)=x^4\), compute \(f'\), \(f''\), and the drift correction in \(d f(B_t)\).
- For \(f(t,x)=t^2+x\), identify \(f_t\), \(f_x\), and \(f_{xx}\), then write the differential.
- Explain why \(dB_t\,dB_t=dt\) does not imply \((B_{t+h}-B_t)^2=h\) for each \(h>0\).
- For \(f(x)=\sin x\), write the stochastic and correction terms, keeping the one-half factor visible.
- What integrability condition on an integrand \(H_s\) ensures that \(\int_0^t H_s\,dB_s\) is a square-integrable martingale on a fixed interval \([0,T]\)?