Why a Symmetric Integral Has a Correction
For an ordinary Riemann integral, changing a left-endpoint sum to a midpoint sum does not introduce a persistent correction as the mesh tends to zero. For Brownian-driven integrals, the difference between these sums contains products of increments of the integrand and Brownian motion. Those products can add up to a nonzero limit: their limit is governed by quadratic covariation. The resulting correction is the scalar Stratonovich-to-Itô conversion formula.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. For \(T>0\), an Itô integrand \(H\) is progressively measurable and belongs to \(\mathcal{L}^2\) on \([0,T]\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Its Itô integral \(\int_0^T H_s\,dB_s\) is defined by the \(L^2\) extension of integrals of adapted step processes.
The distinction between the two integrals is not a change of notation for the same limit. On each partition, writing \(\Delta H_i=H_{t_i}-H_{t_{i-1}}\) and \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\), the midpoint summand is the left-endpoint summand plus \(\tfrac12\Delta H_i\Delta B_i\). Thus the relevant quantity is \(\sum_i\Delta H_i\Delta B_i\), the covariation sum.
The Conversion Formula
Proof. Put \(A_t=\int_0^t a_s\,ds\) and \(M_t=\int_0^t\gamma_s\,dB_s\). First, the left-endpoint sums converge in probability to \(\int_0^T H_s\,dB_s\). Indeed, \(H\) is continuous and square integrable on this finite interval, and the Itô isometry shows that left-step approximations converge to \(H\) in the time-integrated \(L^2\) norm. The difference between a symmetric sum and its left-endpoint sum is exactly \(\tfrac12\sum_i\Delta H_i\Delta B_i\).
We show that \(\sum_i\Delta H_i\Delta B_i\) converges in probability to \(\int_0^T\gamma_s\,ds\). The drift part vanishes: if \(|a_s|\le C\), then \(|\Delta A_i|\le C\Delta t_i\), and Cauchy–Schwarz gives \(\big|\sum_i\Delta A_i\Delta B_i\big|\le C(\sum_i(\Delta t_i)^2)^{1/2}(\sum_i(\Delta B_i)^2)^{1/2}\). Here \(\sum_i(\Delta t_i)^2\le T|\Pi|\to0\), while \(\sum_i(\Delta B_i)^2\to T\) in probability by Brownian quadratic variation. Hence the product tends to zero in probability.
For the martingale part, Itô's product formula on each interval gives \(\Delta M_i\Delta B_i=\int_{t_{i-1}}^{t_i}(M_s-M_{t_{i-1}})\,dB_s+\int_{t_{i-1}}^{t_i}(B_s-B_{t_{i-1}})\,dM_s+\int_{t_{i-1}}^{t_i}\gamma_s\,ds\). The first two sums of stochastic integrals tend to zero in \(L^2\). To see the bounds, let \(|\gamma_s|\le G\). The Itô isometry and \(\mathbb{E}(M_s-M_{t_{i-1}})^2\le G^2(s-t_{i-1})\) bound the squared \(L^2\) norm of the first sum by \(G^2\sum_i(\Delta t_i)^2/2\). For the second sum, the isometry and \(\mathbb{E}(B_s-B_{t_{i-1}})^2=s-t_{i-1}\) give the same bound. Both bounds tend to zero. Consequently \(\sum_i\Delta M_i\Delta B_i\to\int_0^T\gamma_s\,ds\) in probability. Combining the drift and martingale parts with the exact difference of the sums proves the formula. The same computation identifies the covariation as \([H,B]_T=\int_0^T\gamma_s\,ds\). \(\square\)
The boundedness assumptions make the estimates direct. The formula also applies locally when the coefficients are only locally bounded: stop the processes when their coefficient bounds are exceeded, apply the result to the stopped processes, and then increase the stopping bounds. More generally, for a continuous semimartingale \(H\), the standard covariation form is \(\int H\circ dB=\int H\,dB+\tfrac12[H,B]\) whenever the integrals and covariation are defined. The theorem above proves the identity for the scalar Itô processes used here.
For an Itô process with the same Brownian motion as its noise, \(dH_t=a_t\,dt+\sigma(t,H_t)\,dB_t\), the coefficient of \(dB_t\) is the diffusion coefficient. The conversion correction is therefore \(\tfrac12\int_0^T\sigma(t,H_t)\,dt\). The drift coefficient \(a_t\) does not enter this correction. This is a statement about converting integrals; it is not the ordinary chain rule, which requires a separate argument.
Worked Example: The Integrand \(H_t=B_t\)
Convert \(\int_0^T B_s\circ dB_s\) to Itô form and evaluate it. Here \(H_0=0\), \(a_s=0\), and \(\gamma_s=1\). On every finite interval, \(B\) is in \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2\). The correction is \(\tfrac12\int_0^T1\,ds=T/2\).
To evaluate the Itô integral, apply Itô's formula to \(f(x)=x^2/2\). Its derivatives are \(f'(x)=x\) and \(f''(x)=1\), so the formula, including its correction term, is \(d(B_t^2/2)=B_t\,dB_t+\tfrac12\,dt\). Integrating gives \(\int_0^T B_s\,dB_s=\tfrac12B_T^2-\tfrac12T\). Therefore
Check: the Itô integral has mean zero: its integrand is in \(\mathcal{L}^2\), and the integral is a square-integrable martingale. For completeness, this martingale property follows first for adapted step integrands from the zero conditional mean of Brownian increments, and then for \(\mathcal{L}^2\) integrands by the Itô isometry and \(L^2\) approximation. The right-hand side of the conversion has expectation \(\tfrac12\mathbb{E}[B_T^2]=T/2\), matching the zero mean of the Itô integral plus the correction \(T/2\).
Worked Example: A Time-Weighted Brownian Integrand
Convert \(\int_0^T sB_s\circ dB_s\) and give an explicit Itô-form expression. Take \(H_s=sB_s\), so its diffusion coefficient is \(\gamma_s=s\). Its integrability follows from \(\mathbb{E}\int_0^T s^2B_s^2\,ds=\int_0^T s^3\,ds=T^4/4\). The correction is \(\tfrac12\int_0^T s\,ds=T^2/4\).
For the Itô integral, use \(f(t,x)=tx^2/2\). The derivatives are \(f_t(t,x)=x^2/2\), \(f_x(t,x)=tx\), and \(f_{xx}(t,x)=t\). Itô's formula gives \(d(tB_t^2/2)=(B_t^2/2)\,dt+tB_t\,dB_t+(t/2)\,dt\); the last term is the explicit \(\tfrac12 f_{xx}\,dt\) correction. Rearranging and integrating, using \(f(0,B_0)=0\), yields
Adding the correction gives \(\int_0^T sB_s\circ dB_s=\tfrac{T B_T^2}{2}-\tfrac12\int_0^T B_s^2\,ds\). Check: the Itô integral has expectation zero by the square-integrable martingale argument above. In the displayed Itô expression, the expectation of the first term is \(T^2/2\), since \(\mathbb{E}[B_T^2]=T\); the expectation of \(\tfrac12\int_0^T B_s^2\,ds\) is \(\tfrac12\int_0^T s\,ds=T^2/4\), and the final constant is \(T^2/4\). These cancel, as required. The Stratonovich expression consequently has expectation \(T^2/4\), exactly the correction.
Worked Example: An Affine Itô Integrand
Let \(H_t=\alpha+at+\sigma B_t\), where \(\alpha,a,\sigma\) are constants. Its drift is \(a\), its diffusion coefficient is \(\sigma\), and the correction is \(\sigma T/2\). The integrand is in \(\mathcal{L}^2\) on \([0,T]\), because \(\mathbb{E}\int_0^T(\alpha+as+\sigma B_s)^2\,ds\le 3\alpha^2T+3a^2T^3/3+3\sigma^2T^2/2\), which is finite.
Compute the Itô integral term by term. The constant term gives \(\int_0^T\alpha\,dB_s=\alpha B_T\). For \(\int_0^T s\,dB_s\), apply Itô's formula to \(f(t,x)=tx\), with \(f_t=x\), \(f_x=t\), and \(f_{xx}=0\). Thus \(d(tB_t)=B_t\,dt+t\,dB_t\), and \(\int_0^T s\,dB_s=TB_T-\int_0^T B_s\,ds\). The Brownian term is \(\int_0^T B_s\,dB_s=\tfrac12(B_T^2-T)\), as computed in the first example. Therefore
The conversion formula now gives \(\int_0^T H_s\circ dB_s=\alpha B_T+a(TB_T-\int_0^T B_s\,ds)+\tfrac{\sigma}{2}B_T^2\). Check by symmetric sums: the constant part telescopes to \(\alpha B_T\). The time-weighted part tends to \(a(TB_T-\int_0^T B_s\,ds)\), by the discrete summation-by-parts identity and convergence of Riemann sums. For the Brownian part, each symmetric summand satisfies \(\tfrac12(B_{t_{i-1}}+B_{t_i})\Delta B_i=\tfrac12(B_{t_i}^2-B_{t_{i-1}}^2)\). Its sum is exactly \(\tfrac12 B_T^2\), so the direct symmetric-sum calculation agrees with the conversion.
What the Correction Measures
The correction is half the covariation between the integrand and the Brownian motion, not half the drift and not half the integrand itself. If \(dH_t=a_t\,dt+\gamma_t\,dB_t\), then the drift part contributes no covariation with \(B\), while the shared Brownian noise contributes \([H,B]_T=\int_0^T\gamma_s\,ds\). This explains why an adapted process with no Brownian component can have zero conversion correction, even if it changes substantially over time.
The symbolic multiplication rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) summarize the limiting increment products behind this calculation. They do not assert that a single Brownian increment squared equals its time increment. In the conversion proof, the term that survives is the accumulated product \(\sum_i\Delta H_i\Delta B_i\); the drift cross-products vanish as the mesh shrinks.
Common Mistake: Using the Drift Instead of the Diffusion
For \(H_t=3B_t\), the drift is zero and the diffusion coefficient is \(3\). A wrong calculation that uses the drift as the correction coefficient gives \(\int_0^T3B_s\circ dB_s\overset{\text{wrong}}{=}\int_0^T3B_s\,dB_s\). The correct covariation is \([3B,B]_T=3T\), so the correction is \(3T/2\). Since \(\int_0^T3B_s\,dB_s=\tfrac32(B_T^2-T)\), the correct result is \(\tfrac32B_T^2\). Substitution verifies the conversion: \(\tfrac32(B_T^2-T)+\tfrac32T=\tfrac32B_T^2\). The diffusion coefficient, not the drift, determines the correction.
Check Your Understanding
In each question, distinguish the Itô integral from its covariation correction.
- If \(H_t=c+\lambda B_t\) for constants \(c,\lambda\), what is \([H,B]_T\), and what is the Stratonovich correction?
- For \(H_t=t^2B_t\), identify its diffusion coefficient and compute the conversion correction on \([0,T]\).
- Explain why a bounded drift term \(a_t\,dt\) in \(dH_t=a_t\,dt+\gamma_t\,dB_t\) makes no contribution to \([H,B]\).
- For \(H_t=B_t\), use the Itô integral formula to check that the Stratonovich integral has expectation \(T/2\).
- In the affine example, which part of the final expression changes if the drift constant \(a\) changes while \(\sigma\) stays fixed?