Proving the Stratonovich Chain Rule

Why the Stratonovich Chain Rule Looks Ordinary

For a differentiable path $x_t$, the chain rule says $dF(x_t)=F'(x_t)\,dx_t$. An Ito process is not usually differentiable, so the corresponding Ito formula contains a second-derivative term. The Stratonovich integral is defined with a covariation correction to the Ito integral. For a function of the process, that correction exactly cancels Ito’s second-derivative term. The resulting identity resembles the ordinary chain rule, but it depends on using the Stratonovich integral and on the hypotheses that make the integrals and covariations meaningful.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. On a fixed finite interval $[0,T]$, let $X$ be the continuous Ito process $dX_t=\mu_t\,dt+\sigma_t\,dB_t$, where $\mu_t$ and $\sigma_t$ are progressively measurable. Assume $\int_0^T|\mu_t|\,dt<\infty$ and $\int_0^T\sigma_t^2\,dt<\infty$ almost surely. Integrals with respect to $B_t$ are Ito integrals; on any interval where the integrand is square-integrable, their integrands belong to $\mathcal{L}^2$, the class of progressively measurable $H$ satisfying $\mathbb{E}\int_0^T H_t^2\,dt<\infty$. In the general local setting, we apply the formulas after stopping so that these integrability conditions hold.

Definition: For continuous semimartingales $H$ and $X$ whose covariation $[H,X]$ exists, their Stratonovich integral is defined by $\int_0^t H_s\circ dX_s=\int_0^t H_s\,dX_s+\tfrac12[H,X]_t$. Here the first integral on the right is the Ito integral with respect to the semimartingale $X$. For $X_t=X_0+\int_0^t\mu_s\,ds+\int_0^t\sigma_s\,dB_s$, its quadratic variation is $[X]_t=\int_0^t\sigma_s^2\,ds$. Finite-variation terms contribute no quadratic variation.

The conversion identity is used for $H_t=F'(X_t)$. When only $F\in C^2$, the process $F'(X)$ need not itself be a semimartingale under a general choice of $F$. In this setting the Stratonovich integral in the theorem is understood through its conversion expression: the Ito integral plus half the covariation $[F'(X),X]$, which exists and has the formula established below. Thus the statement does not require an unjustified assumption that every $C^1$ function of an Ito process is a semimartingale.

The Chain Rule and Its Proof

Theorem: Let $X$ satisfy the conditions above and let $F:\mathbb{R}\to\mathbb{R}$ be twice continuously differentiable. Locally in time, and hence on every interval where the displayed integrals are defined, $$ dF(X_t)=F'(X_t)\circ dX_t. $$ More explicitly, both sides equal $$ F'(X_t)\mu_t\,dt+F'(X_t)\sigma_t\,dB_t +\tfrac12F''(X_t)\sigma_t^2\,dt. $$

Proof. First stop the process when its path leaves a compact interval and when the accumulated integrals of $|\mu|$ and $\sigma^2$ exceed a finite bound. On the stopped interval, $F'$ and $F''$ are bounded along the path, and the Ito integral with integrand $F'(X_t)\sigma_t$ is square-integrable. The localized identities below therefore apply; increasing the stopping bounds gives the local statement.

Ito’s formula applies because $F\in C^2$. Its time derivative is zero, its first spatial derivative is $F'$, and its second spatial derivative is $F''$. Using $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$, the formula is

dF(X_t) =F'(X_t)\,dX_t+\tfrac12F''(X_t)\,d[X]_t =F'(X_t)\mu_t\,dt+F'(X_t)\sigma_t\,dB_t +\tfrac12F''(X_t)\sigma_t^2\,dt.

Next compute the covariation in the conversion identity. The covariation chain rule for a continuous Ito process and a $C^1$ function $g$ is $[g(X),X]_t=\int_0^t g'(X_s)\,d[X]_s$. In this application $g=F'$, so $g'=F''$. This covariation rule follows by expanding increments: on a compact range, $\Delta g(X)=g'(X)\Delta X+r$, where $|r|\le\omega(\max|\Delta X|)|\Delta X|$ and $\omega(\delta)\to0$ as $\delta\to0$. Multiplying by $\Delta X$ and summing, the remainder tends to zero because the maximum increment tends to zero and the quadratic-variation sums remain bounded in probability. The main term converges to $\int g'(X)\,d[X]$, by the weighted quadratic-variation rule. Localization removes the compact-range restriction. Consequently,

[F'(X),X]_t =\int_0^t F''(X_s)\,d[X]_s =\int_0^t F''(X_s)\sigma_s^2\,ds.

By the definition of the Stratonovich integral,

\begin{aligned} \int_0^t F'(X_s)\circ dX_s &=\int_0^t F'(X_s)\,dX_s+\tfrac12[F'(X),X]_t\\ &=\int_0^t F'(X_s)\mu_s\,ds +\int_0^t F'(X_s)\sigma_s\,dB_s +\tfrac12\int_0^t F''(X_s)\sigma_s^2\,ds. \end{aligned}

This is exactly the integral form of Ito’s formula computed above. The two processes therefore agree, which proves the asserted differential identity. $\square$

Worked Examples: Seeing the Cancellation

Worked Example: The Square of Brownian Motion

Take $X_t=B_t$ and $F(x)=x^2$. Here $\mu_t=0$, $\sigma_t=1$, $F'(x)=2x$, and $F''(x)=2$. Ito’s formula gives

d(B_t^2)=2B_t\,dB_t+\tfrac12(2)\,dt =2B_t\,dB_t+dt.

The conversion correction for the right side of the Stratonovich chain rule is $\tfrac12[2B,B]_t=\tfrac12(2t)=t$. Therefore $\int_0^t2B_s\circ dB_s=2\int_0^tB_s\,dB_s+t=B_t^2-B_0^2=B_t^2$. The Ito integrand is square-integrable on $[0,T]$, since $\mathbb{E}\int_0^T(2B_s)^2ds=4\int_0^T s\,ds=2T^2<\infty$.

Check: at $t=0$, both sides are $B_0^2=0$. Also, Ito’s formula rearranges to $2\int_0^tB_s\,dB_s=B_t^2-t$; adding the conversion correction $t$ gives $B_t^2$, as required.

Worked Example: An Exponential of a Drifted Brownian Motion

Let $X_t=t+B_t$ and $F(x)=e^x$. Then $F'(x)=e^x$, $F''(x)=e^x$, $\mu_t=1$, and $\sigma_t=1$. Ito’s formula, including its correction, gives

d(e^{X_t})=e^{X_t}\,dt+e^{X_t}\,dB_t+\tfrac12e^{X_t}\,dt =\tfrac32e^{X_t}\,dt+e^{X_t}\,dB_t.

For the Stratonovich side, $[e^X,X]_t=\int_0^t e^{X_s}\,ds$. Thus $\int_0^t e^{X_s}\circ dX_s =\int_0^t e^{X_s}(ds+dB_s)+\tfrac12\int_0^t e^{X_s}ds$, which is the same expression as the Ito formula. The Brownian integrand is square-integrable because $\mathbb{E}\int_0^T e^{2X_s}ds=\int_0^T e^{2s}\mathbb{E}[e^{2B_s}]ds=\int_0^T e^{4s}ds<\infty$.

Check: the normal moment formula gives $\mathbb{E}[e^{X_t}]=e^t e^{t/2}=e^{3t/2}$. The drift in the computed differential is $\tfrac32e^{X_t}$, consistent with the derivative of this expectation; the Ito stochastic integral has mean zero by square-integrability.

Worked Example: A Sine Function with Nonzero Drift and Diffusion

Take $X_t=2t+\tfrac12B_t$ and $F(x)=\sin x$. The derivatives are $F'(x)=\cos x$ and $F''(x)=-\sin x$. Here $\mu_t=2$ and $\sigma_t=\tfrac12$, so Ito’s formula gives

d(\sin X_t) =2\cos(X_t)\,dt+\tfrac12\cos(X_t)\,dB_t -\tfrac12\sin(X_t)\left(\tfrac12\right)^2dt =\left(2\cos(X_t)-\tfrac18\sin(X_t)\right)dt +\tfrac12\cos(X_t)\,dB_t.

The covariation is $[\cos(X),X]_t=\int_0^t-\sin(X_s)\cdot\tfrac14\,ds$. Its half is $-\tfrac18\int_0^t\sin(X_s)ds$. Adding that correction to $\int_0^t\cos(X_s)(2ds+\tfrac12dB_s)$ reproduces the displayed Ito formula. The stochastic integrand is square-integrable since $|\cos(X_s)/2|\le\tfrac12$, and hence $\mathbb{E}\int_0^T\cos^2(X_s)/4\,ds\le T/4$.

Check: the normal characteristic function gives $\mathbb{E}[\sin(X_t)]=e^{-t/8}\sin(2t)$. Differentiating this expression gives $e^{-t/8}(2\cos(2t)-\tfrac18\sin(2t))$, which equals $\mathbb{E}[2\cos(X_t)-\tfrac18\sin(X_t)]$. The agreement checks both the drift and the sign of the correction.

The Correction Is the Point

The ordinary-looking Stratonovich chain rule does not mean that the Ito correction has disappeared from stochastic calculus. It has moved into the definition of the Stratonovich integral. For $X$ as above, the Ito differential for $F(X)$ contains $\tfrac12F''(X)\sigma^2dt$; converting $F'(X)\,dX$ to $F'(X)\circ dX$ adds precisely the same term.

Common mistake: omitting the conversion correction. For $X_t=B_t$ and $F(x)=x^2$, an incorrect calculation might write $d(B_t^2)=2B_t\,dB_t$ and conclude that the chain rule fails. Ito’s formula instead gives $d(B_t^2)=2B_t\,dB_t+dt$. The Stratonovich integral satisfies $2B_t\circ dB_t=2B_t\,dB_t+dt$, so the chain rule holds. The error is treating an Ito integral as though it were a Stratonovich integral.

In general, the drift $\mu_t$ contributes $F'(X_t)\mu_tdt$, while the diffusion coefficient $\sigma_t$ contributes both $F'(X_t)\sigma_t dB_t$ and the correction $\tfrac12F''(X_t)\sigma_t^2dt$. The latter is not an additional drift in the Stratonovich chain rule; it is the covariation term required to convert the integral.

Check Your Understanding

For each question, keep the Ito and Stratonovich integrals distinct until the conversion correction has been included.

For $X_t=3t+B_t$ and $F(x)=x^2$, write Ito’s formula and compute $[F'(X),X]_t$.
For $X_t=\sigma B_t$ and $F(x)=\cos x$, identify the second-derivative term in Ito’s formula and the conversion correction.
State the covariation chain rule used to compute $[F'(X),X]$, including the regularity required of $F$.
Why does the proof use localization when $\mu$, $\sigma$, or the derivatives of $F$ are not globally bounded?
In the Brownian-square example, what term is missing if $2B_t\,dB_t$ is incorrectly used in place of $2B_t\circ dB_t$?

Proving the Stratonovich Chain Rule

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