Antiderivatives for Brownian Stratonovich Integrals
The Stratonovich chain rule makes some stochastic integrals look like ordinary calculus: an integral of \(f(B)\) against \(B\) is evaluated by finding an antiderivative of \(f\). The important qualification is that the integral is Stratonovich, not Ito. The Ito integral has a quadratic-variation correction, and computing that correction gives a direct way to check the answer.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), with \(B_0=0\). Fix a finite horizon \(T>0\). An Ito integrand \(H\) is in \(\mathcal{L}^2\) on \([0,T]\) if it is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For such an integrand, \(\int_0^t H_s\,dB_s\) is the Ito integral. It is a square-integrable martingale: its increments have conditional expectation zero, and the Ito isometry gives \(\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds\).
For a continuous semimartingale integrand \(H\), the Stratonovich integral is defined, when its terms exist, by conversion to Ito form. In particular, if \(H\) is in \(\mathcal{L}^2\) and its covariation with \(B\) is defined, then \(\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t\). Here \([H,B]\) denotes quadratic covariation. Brownian quadratic variation is \([B]_t=t\), encoded by \(dB_t\,dB_t=dt\); the multiplication rules also include \(dB_t\,dt=0\) and \(dt\,dt=0\). These rules are not the ordinary chain rule: the nonzero quadratic variation is exactly why Ito and Stratonovich integrals differ.
Proof. The process \(f(B_s)\) is adapted and continuous, hence progressively measurable. The first integrability assumption puts it in \(\mathcal{L}^2\), so its Ito integral exists. Applying Ito's formula to the \(C^2\) function \(f\) shows that \(f(B)\) is a continuous semimartingale. Its martingale part is \(\int_0^t f'(B_s)\,dB_s\), and its finite-variation part is \(\tfrac12\int_0^t f''(B_s)\,ds\). Consequently its covariation with \(B\) is \([f(B),B]_t=\int_0^t f'(B_s)\,ds\). This correction is integrable under the second assumption.
Now apply Ito's formula to \(F(B_t)\). Its derivatives are \(F'=f\) and \(F''=f'\), so the formula, including its quadratic-variation term, is \(F(B_t)-F(B_0)=\int_0^t f(B_s)\,dB_s+\tfrac12\int_0^t f'(B_s)\,ds\). By the definition of the Stratonovich integral and the covariation just computed, the right-hand side is \(\int_0^t f(B_s)\circ dB_s\). The continuous versions of these processes give the identity for all \(t\in[0,T]\) on a single event of probability one. \(\square\)
The regularity assumption matters. The Ito integral alone being well defined does not ensure that \(f(B)\) is a semimartingale, which is required for the Stratonovich integral defined here. The \(C^2\) hypothesis is a convenient sufficient condition: it lets Ito's formula establish the needed semimartingale decomposition. The two integrability assumptions separately ensure that the Ito integral and the covariation correction are integrable on the chosen horizon.
Three Evaluations and Their Ito Checks
Worked Example: The Integral of Brownian Motion
Evaluate \(\int_0^t B_s\circ dB_s\), for \(0\le t\le T\), and verify the answer by conversion. Here \(f(x)=x\), \(F(x)=x^2/2\), and \(f(B)\) is a continuous semimartingale. Also, \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty\), so the Ito integrand is in \(\mathcal{L}^2\).
The antiderivative gives \(\int_0^t B_s\circ dB_s=B_t^2/2\), since \(B_0=0\). For the conversion check, \([B,B]_t=t\), so \(\int_0^t B_s\circ dB_s=\int_0^t B_s\,dB_s+t/2\). Applying Ito's formula to \(G(x)=x^2/2\), with \(G'(x)=x\), \(G''(x)=1\), and the explicit correction \(\tfrac12G''(B_s)\,ds=\tfrac12ds\), gives \(B_t^2/2=\int_0^t B_s\,dB_s+t/2\). Thus the converted expression agrees with the antiderivative answer.
Check: The Ito integral has mean zero by square integrability. Since \(\mathbb{E}[B_t^2]=t\), its converted expression has mean \(\mathbb{E}[B_t^2/2-t/2]=0\), as required. The Stratonovich integral itself has mean \(\mathbb{E}[B_t^2/2]=t/2\); it is not the same martingale as the Ito integral.
Worked Example: The Integral of the Square of Brownian Motion
Evaluate \(\int_0^t B_s^2\circ dB_s\) and verify the correction. For \(f(x)=x^2\), take \(F(x)=x^3/3\). The integrand is in \(\mathcal{L}^2\), because \(\mathbb{E}\int_0^T B_s^4\,ds=\int_0^T 3s^2\,ds=T^3<\infty\). Also, \(\mathbb{E}\int_0^T 2|B_s|\,ds<\infty\), so the covariation correction is integrable.
The antiderivative gives \(\int_0^t B_s^2\circ dB_s=B_t^3/3\). To check it, Ito's formula for \(x\mapsto x^2\) gives \(d(B_t^2)=2B_t\,dB_t+dt\). Therefore the martingale part of \(B^2\) is \(\int_0^t 2B_s\,dB_s\), and \([B^2,B]_t=\int_0^t 2B_s\,ds\). Conversion now yields \(\int_0^t B_s^2\circ dB_s=\int_0^t B_s^2\,dB_s+\int_0^t B_s\,ds\).
For an independent check, apply Ito's formula to \(G(x)=x^3/3\). Its derivatives are \(G'(x)=x^2\) and \(G''(x)=2x\); the correction is \(\tfrac12G''(B_s)\,ds=B_s\,ds\). Hence \(B_t^3/3=\int_0^t B_s^2\,dB_s+\int_0^t B_s\,ds\), exactly the converted expression.
Check: Brownian symmetry gives \(\mathbb{E}[B_t^3/3]=0\). The Ito integral has mean zero, and \(\mathbb{E}\int_0^t B_s\,ds=0\), so the converted expression also has mean zero.
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^t \cos(B_s)\circ dB_s\). Here \(f(x)=\cos x\), \(F(x)=\sin x\), and \(f(B)\) is bounded, so \(\mathbb{E}\int_0^T \cos^2(B_s)\,ds\le T\). Its derivative is bounded as well, giving the required correction integrability.
The antiderivative gives \(\int_0^t \cos(B_s)\circ dB_s=\sin(B_t)\). For conversion, \(f'(x)=-\sin x\), so \(\int_0^t \cos(B_s)\circ dB_s=\int_0^t\cos(B_s)\,dB_s-\tfrac12\int_0^t\sin(B_s)\,ds\). Apply Ito's formula to \(G(x)=\sin x\): \(G'(x)=\cos x\), \(G''(x)=-\sin x\), and the correction is \(-\tfrac12\sin(B_s)\,ds\). Thus \(\sin(B_t)=\int_0^t\cos(B_s)\,dB_s-\tfrac12\int_0^t\sin(B_s)\,ds\), which verifies the result and the sign of the conversion term.
Check: The Ito integral has mean zero. Brownian symmetry implies \(\mathbb{E}[\sin(B_t)]=0\) and \(\mathbb{E}[\sin(B_s)]=0\) for each \(s\), so the converted expression also has mean zero, in agreement with the antiderivative.
What the Conversion Check Does—and Does Not—Say
For these examples, the Ito integral is a true martingale because the integrand belongs to \(\mathcal{L}^2\): its second moment is finite by the Ito isometry, and its conditional expectation at a later time equals its present value. In general, a locally square-integrable Ito integral is first known to be a local martingale; without an integrability condition that makes it a true martingale, its expectation need not be preserved. A Stratonovich integral is an Ito integral plus a covariation term, usually of finite variation, and therefore need not be a martingale. The first example demonstrates this distinction: its Stratonovich integral has mean \(t/2\), not zero.
A useful way to organize an evaluation is to keep the calculus and the conversion separate:
For \(f(B)\circ dB\), choose \(F\) with \(F'=f\), after checking that the integrand is a continuous semimartingale and the integrals are defined.
Use \(\int f(B)\circ dB=\int f(B)\,dB+\tfrac12\int f'(B)\,ds\). The sign follows from \(f'\), not from a choice of antiderivative.
For \(F(B)\), display \(F'\), \(F''\), and the term \(\tfrac12F''(B_s)\,ds\); its differential must reproduce the converted integral.
Common mistake: applying the ordinary chain rule to the Ito integral. The incorrect calculation \(\int_0^t B_s\,dB_s=B_t^2/2\) omits the quadratic-variation term. Ito's formula instead gives \(d(B_t^2/2)=B_t\,dB_t+\tfrac12dt\), and integrating shows \(\int_0^t B_s\,dB_s=B_t^2/2-t/2\). The Stratonovich answer is \(B_t^2/2\) because conversion adds \(t/2\). The ordinary-looking chain rule belongs to the Stratonovich integral; it does not remove the correction from the Ito integral.
The antiderivative method is useful whenever the integrand depends on Brownian motion alone and the required regularity and integrability conditions hold. Time-dependent integrands require additional terms, so the simple antiderivative identity should not be applied to them without checking the relevant formula.
Check Your Understanding
For each question, distinguish the Stratonovich integral from its Ito form and state which assumptions justify the integrals.
- Convert \(\int_0^t B_s^3\circ dB_s\) to Ito form and identify its antiderivative expression.
- For \(f(x)=e^x\), write the Ito correction for \(\int_0^t f(B_s)\circ dB_s\). On a fixed finite horizon, which integrability checks are needed?
- Apply Ito's formula to \(G(x)=x^4/4\), displaying both derivatives and the correction term, to verify the conversion for the first question.
- Why does \(\int_0^t B_s\,dB_s\) have mean zero while \(\int_0^t B_s\circ dB_s\) has mean \(t/2\)?
- Explain why the existence of the Ito integral alone does not establish that the corresponding Stratonovich integral is defined.