Time-Dependent Integrands and Conversion Corrections

Time Dependence Does Not Remove the Correction

In a Stratonovich integral, an integrand that depends on Brownian motion is evaluated using a symmetric, rather than purely left-endpoint, convention. The difference from the Itô integral records how the integrand changes along the Brownian path. If the integrand is $h(t,B_t)$, that change has two sources: explicit variation with time and variation with the Brownian state. Only the latter contributes to the Stratonovich-to-Itô correction. In particular, the correction involves the spatial derivative $h_x$, not the time derivative $h_t$.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to the filtration. Fix a finite horizon $T>0$. All stochastic integrals below are Itô integrals unless marked with a circle. An Itô integrand $H$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$ when $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; its integral is then a square-integrable martingale.

Definition: For a continuous semimartingale $H_t$ with a well-defined Itô integral against $B_t$, the Stratonovich integral is $\int_0^T H_t\circ dB_t=\int_0^T H_t\,dB_t+\tfrac12[H,B]_T$, where $[H,B]$ is quadratic covariation. The first term is an Itô integral; the second is an ordinary time-integral when the covariation has a density.

Here is the regularity setting used for the conversion result. Let $h:[0,T]\times\mathbb{R}\to\mathbb{R}$ be continuously differentiable in time and twice continuously differentiable in space. Assume its derivatives have sufficient growth that Itô’s formula applies to $h(t,B_t)$, and that

\mathbb{E}\int_0^T h(t,B_t)^2\,dt<\infty,\qquad \mathbb{E}\int_0^T |h_x(t,B_t)|\,dt<\infty,\qquad \mathbb{E}\int_0^T\bigl(|h_t(t,B_t)|+|h_{xx}(t,B_t)|\bigr)\,dt<\infty.

These assumptions make the Itô integral square-integrable, the correction integrable, and the finite-variation terms in Itô’s formula well-defined. Local versions follow by stopping when the relevant quantities become bounded; the displayed assumptions give the direct, nonlocalized statement on $[0,T]$.

The Conversion Formula

Theorem: Under the stated regularity and integrability assumptions, $h(t,B_t)$ is a continuous semimartingale and $$ \int_0^T h(t,B_t)\circ dB_t = \int_0^T h(t,B_t)\,dB_t +\frac12\int_0^T h_x(t,B_t)\,dt. $$ Explicit time dependence affects the value of the integrand and its Itô dynamics, but it does not create an additional correction involving $h_t$.

Proof. Apply Itô’s formula to the function $h(t,x)$ evaluated at $(t,B_t)$. Its partial derivatives are $h_t(t,x)$, $h_x(t,x)$, and $h_{xx}(t,x)$. Since $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$, Itô’s formula gives

dh(t,B_t)= \left(h_t(t,B_t)+\frac12h_{xx}(t,B_t)\right)dt +h_x(t,B_t)\,dB_t.

The first term is finite variation and therefore has zero quadratic covariation with $B$. The Itô covariation rule gives $\left[\int_0^\cdot h_x(s,B_s)\,dB_s,B\right]_T=\int_0^T h_x(s,B_s)\,ds$. This rule follows first for adapted step integrands from Brownian increments and then for square-integrable integrands by the Itô isometry; localization gives the corresponding locally integrable version. Thus $[h(\cdot,B),B]_T=\int_0^T h_x(t,B_t)\,dt$. Substituting this identity into the definition of the Stratonovich integral proves the formula. $\square$

The time derivative $h_t$ appears in the semimartingale decomposition of $h(t,B_t)$, but it multiplies $dt$. Since a finite-variation increment has zero covariation with Brownian motion, it cannot contribute to $[h(\cdot,B),B]$. This is the precise reason the conversion correction depends on $h_x$ alone.

Worked Conversions

Worked Example: A Linear Integrand with Time Dependence

Convert $\int_0^2 tB_t\circ dB_t$ to Itô form, and check its expectation by another calculation. Here $h(t,x)=tx$, so $h_x(t,x)=t$. The conversion theorem gives

\int_0^2 tB_t\circ dB_t = \int_0^2 tB_t\,dB_t+\frac12\int_0^2t\,dt = \int_0^2 tB_t\,dB_t+1.

The arithmetic in the correction is $\int_0^2t\,dt=[t^2/2]_0^2=2$, then $2/2=1$. The Itô integral is a martingale with mean zero: $tB_t$ is progressively measurable and $\mathbb{E}\int_0^2t^2B_t^2dt=\int_0^2t^3dt=4<\infty$, so its conditional expectation property follows from the square-integrable Itô integral theorem. Therefore the Stratonovich integral has expectation $1$.

For an independent check, take $F(t,x)=tx^2/2$, for which $F_x=tx$, $F_t=x^2/2$, and $F_{xx}=t$. Itô’s formula, including its second-derivative correction, gives

dF(t,B_t)=\frac{B_t^2}{2}\,dt+tB_t\,dB_t+\frac{t}{2}\,dt.

The Stratonovich chain rule, or equivalently the conversion just proved applied to $F_x$, yields $\int_0^2 tB_t\circ dB_t=F(2,B_2)-F(0,B_0)-\int_0^2 B_t^2/2\,dt$. Its expectation is $2-\frac12\int_0^2t\,dt=2-1=1$, agreeing with the conversion calculation. The integrability needed here holds because Brownian motion has finite second moments on this finite horizon.

Worked Example: An Exponential Time Factor and a Quadratic State Term

Convert $\int_0^1 e^t(1+B_t^2)\circ dB_t$. Here $h(t,x)=e^t(1+x^2)$, so $h_x(t,x)=2e^t x$. Hence

\int_0^1 e^t(1+B_t^2)\circ dB_t = \int_0^1 e^t(1+B_t^2)\,dB_t+\int_0^1e^tB_t\,dt.

The correction is one half of $\int_0^1 2e^tB_t\,dt$, which is exactly $\int_0^1e^tB_t\,dt$. The Itô integral is a square-integrable martingale: its integrand is progressively measurable and its expected squared norm is $\int_0^1 e^{2t}\mathbb{E}(1+B_t^2)^2dt$, finite since $\mathbb{E}(1+B_t^2)^2=1+2t+3t^2$. Its expectation is zero. The correction also has expectation zero because $\mathbb{E}B_t=0$, so the converted integral has mean zero.

Check by setting $F(t,x)=e^t(x+x^3/3)$, so $F_x=e^t(1+x^2)$, $F_t=e^t(x+x^3/3)$, and $F_{xx}=2e^t x$. Itô’s formula gives

dF(t,B_t)= \left(e^t(B_t+B_t^3/3)+e^tB_t\right)dt +e^t(1+B_t^2)\,dB_t.

Subtracting the $F_t\,dt$ contribution in the Stratonovich chain rule leaves the correction $e^tB_t\,dt$, exactly as above. Also, $F(1,B_1)$ and $F(0,B_0)$ have mean zero because each is an odd function of a centered Gaussian variable; the remaining expectation from $-\int_0^1F_t(t,B_t)dt$ is zero for the same reason. This independently confirms the mean check.

Worked Example: A Constant Spatial Derivative

Convert $\int_0^1(t^2+B_t)\circ dB_t$. For $h(t,x)=t^2+x$, $h_x(t,x)=1$, so the correction equals $\frac12\int_0^1 1\,dt=\frac12$. Therefore

\int_0^1(t^2+B_t)\circ dB_t = \int_0^1(t^2+B_t)\,dB_t+\frac12.

The Itô integrand is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^1(t^2+B_t)^2dt=\int_0^1(t^4+t)dt=1/5+1/2=7/10$; the cross term has expectation zero. Thus its Itô integral is a square-integrable martingale of mean zero, and the Stratonovich integral has mean $1/2$.

For a second check, let $F(t,x)=t^2x+x^2/2$. Then $F_t=2tx$, $F_x=t^2+x$, and $F_{xx}=1$. Applying Itô’s formula with the correction explicitly included gives

dF(t,B_t)=2tB_t\,dt+(t^2+B_t)\,dB_t+\frac12\,dt.

The Stratonovich chain rule gives $\int_0^1(t^2+B_t)\circ dB_t=F(1,B_1)-F(0,B_0)-\int_0^1 2tB_t\,dt$. Its expectation is $0+\mathbb{E}(B_1^2)/2-0=1/2$. The last time integral has mean zero, since $\mathbb{E}B_t=0$, so this agrees with the converted form.

A Reusable Check and a Common Error

When an antiderivative $F$ with $F_x=h$ is available, the Stratonovich chain rule gives a useful independent check: $\int_0^T h(t,B_t)\circ dB_t=F(T,B_T)-F(0,B_0)-\int_0^T F_t(t,B_t)\,dt$. To verify the relation rather than use it as a formal rule, apply Itô’s formula to $F(t,B_t)$: its drift is $F_t+\frac12F_{xx}$, its diffusion term is $F_x\,dB_t$, and $F_{xx}=h_x$. The conversion theorem adds $\frac12\int h_xdt$ to the Itô integral, cancelling exactly the second-derivative drift in the rearranged identity.

Common mistake: omitting the correction because the integrand depends explicitly on time. For the first example, the incorrect calculation would assert $\int_0^2tB_t\circ dB_t=\int_0^2tB_t\,dB_t$, and hence claim expectation zero. But $h_x=t$, so the omitted term is $\frac12\int_0^2t\,dt=1$. The corrected expectation is $1$, as the antiderivative calculation confirms. A different error is to use $h_t$ in the correction: here $h_t=B_t$, which would produce a random term rather than the required deterministic $1$. The covariation calculation shows why that is wrong: $h_tdt$ is finite variation, while the coefficient of $dB_t$ in $dh(t,B_t)$ is $h_x$.

The formula requires the spatial dependence to be differentiated with respect to the state variable driven by Brownian motion. If $h(t,x)=g(t)$ has no state dependence, then $h_x=0$, and its Stratonovich and Itô integrals agree whenever the integral is defined. For a state-dependent $h$, even an explicit time factor such as $t$ or $e^t$ does not change which derivative supplies the correction. In every case, confirm that the Itô integrand is progressively measurable and satisfies the stated square-integrability condition before treating its integral as a true martingale; without that integrability, a local martingale need not be a true martingale.

Check Your Understanding

For each conversion, identify the Itô integrand and compute the correction separately.

Convert $\int_0^T t^2B_t\circ dB_t$ to Itô form. Which derivative determines the correction?
For $h(t,x)=e^{-t}\sin x$, write the correction on $[0,T]$, and state the regularity facts that allow the conversion.
Use $F(t,x)=tx+x^2/2$ and Itô’s formula to check the conversion for $h(t,x)=t+x$.
If $h(t,x)=t^3$, explain why no correction appears even though $h$ depends explicitly on time.
For $h(t,x)=tx$ on $[0,2]$, compute the correction and the expectation of the Stratonovich integral, and explain why the Itô integral has mean zero.

Time-Dependent Integrands and Conversion Corrections

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