Deriving the Multidimensional Conversion Formula

Why the Correction Is Componentwise

A Stratonovich integral against one Brownian component is corrected by the covariation of its integrand with that same component. If the integrand depends on several coordinates of an Itô process, each coordinate can contribute: what matters is how each coordinate covaries with the chosen Brownian integrator. The correction is therefore a sum of spatial derivatives multiplied by the corresponding diffusion coefficients, not just a derivative in one preferred coordinate.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying an $m$-dimensional standard Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^m)$, whose components have covariations $[B^j,B^k]_t=\delta_{jk}t$. Let $X_t=(X_t^1,\ldots,X_t^d)$ satisfy, on $[0,T]$,

dX_t^i=b_i(t)\,dt+\sum_{j=1}^m\sigma_{ij}(t)\,dB_t^j,\qquad 1\le i\le d.

Assume $X_0$ is deterministic and the progressively measurable coefficients $b_i,\sigma_{ij}$ are bounded on $[0,T]$. For a concrete sufficient regularity condition below, take $f_k(t,x)$ to be $C^{1,2}$, with $f_k$, its first time derivative, and its first and second spatial derivatives continuous, and with those derivatives bounded on $[0,T]\times\mathbb{R}^d$. It is enough instead to impose direct integrability conditions ensuring the displayed integrals exist.

Definition: A process $H$ is in $\mathcal{L}^2$ on $[0,T]$ if it is progressively measurable and $\mathbb{E}\int_0^T H_t^2\,dt<\infty$. An Itô integral below is taken with an integrand in $\mathcal{L}^2$. For such $H$, the Stratonovich integral against $B^k$ is defined by $\int_0^T H_t\circ dB_t^k=\int_0^T H_t\,dB_t^k+\tfrac12[H,B^k]_T$, when the covariation exists.

Under the stated bounded-coefficient assumptions, $\mathbb{E}|X_t|^2$ is finite and bounded over $[0,T]$: this follows by writing each coordinate as its initial value plus a bounded drift integral and bounded-coefficient Itô integrals, and applying Cauchy–Schwarz and the Itô isometry. Bounded first spatial derivatives and bounded $f_k(t,0)$ imply $|f_k(t,x)|\le C(1+|x|)$. Thus $H_t^k=f_k(t,X_t)$ belongs to $\mathcal{L}^2$. The Itô drift and diffusion terms obtained by applying Itô’s formula are integrable as well. In particular, the Stratonovich integral in the definition is well defined.

Cross-Variation and the Conversion Formula

Theorem: Under the assumptions above, for $H_t^k=f_k(t,X_t)$ and each $1\le k\le m$, $$ \int_0^T f_k(t,X_t)\circ dB_t^k = \int_0^T f_k(t,X_t)\,dB_t^k +\frac12\int_0^T\sum_{i=1}^d \frac{\partial f_k}{\partial x_i}(t,X_t)\,\sigma_{ik}(t)\,dt. $$ Consequently, the correction for the sum of component integrals is the sum of these componentwise corrections.

Proof. Put $H_t^k=f_k(t,X_t)$. Itô’s formula, with all derivatives evaluated at $(t,X_t)$, gives

dH_t^k= \left( \partial_t f_k+\sum_{i=1}^d b_i\,\partial_i f_k +\frac12\sum_{i=1}^d\sum_{\ell=1}^d\sum_{j=1}^m \sigma_{ij}\sigma_{\ell j}\,\partial_{i\ell}f_k \right)dt +\sum_{j=1}^m\left(\sum_{i=1}^d\partial_i f_k\,\sigma_{ij}\right)dB_t^j.

The quadratic covariation of a finite-variation integral with Brownian motion is zero. For Itô integrals, $dB^j\,dB^k=\delta_{jk}\,dt$; hence only the $j=k$ Brownian coefficient in this differential contributes to $[H^k,B^k]$. Therefore

[H^k,B^k]_T =\int_0^T\sum_{i=1}^d\partial_i f_k(t,X_t)\,\sigma_{ik}(t)\,dt.

Substitute this covariation into the definition of the Stratonovich integral. This proves the formula. Applying the same identity to each component and adding the resulting equalities proves the assertion for the sum. $\square$

The term $b_i\,dt$ does not appear in the correction: its paths have finite variation and thus zero covariation with Brownian motion. The time derivative and second spatial derivatives do appear in the full Itô differential of $H^k$, but they are finite-variation terms and likewise do not enter $[H^k,B^k]$.

Worked Applications

Worked Example: Two Coordinates Both Covary with the Integrator

Let $B^1,B^2$ be independent Brownian components, $X_t^1=B_t^1$, $X_t^2=B_t^1+B_t^2$, and $H_t=X_t^1X_t^2$. Convert $\int_0^T H_t\circ dB_t^1$ to an Itô integral.

Here $f(x_1,x_2)=x_1x_2$, so $\partial_1f=x_2$, $\partial_2f=x_1$, and both coordinates have coefficient $1$ in front of $dB^1$. The correction density is therefore $X_t^2+X_t^1=2B_t^1+B_t^2$, giving

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

Check: Directly, $H=(B^1)^2+B^1B^2$. The product rule gives $dH=(2B^1+B^2)dB^1+B^1dB^2+dt$, since $[B^1,B^2]=0$ and $d[B^1]_t=dt$. Its coefficient against $dB^1$ is exactly $2B^1+B^2$, confirming the covariation and the correction. In particular, the $B^1dB^2$ term is present in $dH$, even though it does not contribute to $[H,B^1]$.

Worked Example: A Diffusion Matrix with Several Nonzero Entries

Let $X^1=2B^1+B^2$, $X^2=B^1+3B^2$, and $H=(X^1)^2+X^2$. Convert $\int_0^T H_t\circ dB_t^2$.

The derivatives are $\partial_1f=2X^1$, $\partial_2f=1$, and the coefficients of $dB^2$ in $dX^1,dX^2$ are $1,3$, respectively. Thus the correction density is $(2X^1)(1)+(1)(3)=2X^1+3$, and the correction is $\int_0^T(X_t^1+\tfrac32)\,dt$.

For an independent check, $dX^1=2dB^1+dB^2$ and $dX^2=dB^1+3dB^2$. Since $d[X^1]_t=(2^2+1^2)dt=5dt$, Itô’s formula for $(X^1)^2+X^2$ gives

$$ dH_t=(4X_t^1+1)dB_t^1+(2X_t^1+3)dB_t^2+5dt. $$

The $dB^2$ coefficient again gives $[H,B^2]_T=\int_0^T(2X_t^1+3)dt$, exactly the correction density computed from the derivatives. As a special-value check, when $X^1=0$, the density is $3$, so the half-correction density is $3/2$, not zero: the $X^2$ coordinate still covaries with $B^2$.

Worked Example: A Time-Dependent Integrand

Let $X_t=t+2B_t^1+B_t^2$ and $H_t=tX_t$. Convert $\int_0^2 H_t\circ dB_t^2$.

Here $f(t,x)=tx$, so $\partial_t f=x$, $\partial_x f=t$, and $\partial_{xx}f=0$. The coefficient of $dB^2$ in $dX$ is $1$. Itô’s formula yields $dH_t=(X_t+t)dt+2t\,dB_t^1+t\,dB_t^2$, and hence $[H,B^2]_2=\int_0^2t\,dt=2$. The half-correction is $1$, so

\int_0^2 tX_t\circ dB_t^2 = \int_0^2 tX_t\,dB_t^2+1.

Check: the derivative-and-diffusion formula gives $\tfrac12\int_0^2(\partial_x f)(1)\,dt=\tfrac12\int_0^2t\,dt=\tfrac12(2)=1$, agreeing with the covariation computed from the full differential. The drift $1$ in $dX$ contributes to the $dt$ term of $dH$, but not to the correction.

What the Formula Requires

The Itô integrals in these examples are square-integrable. More generally, the conversion applies when $H^k$ is progressively measurable and in $\mathcal{L}^2$, and its covariation with $B^k$ exists with an integrable correction. In the examples, polynomial functions of Brownian motion have finite second moments on every finite time interval, so these conditions hold. An Itô integral with an $\mathcal{L}^2$ integrand is a square-integrable martingale: its conditional expectation at a later time, given the current filtration, equals its current value. This follows first for adapted step integrands from the zero conditional mean of future Brownian increments, and then for $\mathcal{L}^2$ integrands by the Itô isometry and $L^2$ approximation.

If the driving Brownian components are correlated, with $[B^j,B^k]_t=\rho_{jk}t$, the coefficient of $dB^k$ in the covariation calculation is replaced by the covariance-weighted sum. Specifically, for $dX^i=b_i\,dt+\sum_j\sigma_{ij}\,dB^j$, the correction density against $B^k$ is $\sum_{i,j}\partial_i f_k\,\sigma_{ij}\rho_{jk}$. The standard independent-component formula above is the case $\rho_{jk}=\delta_{jk}$.

Common Mistake: Keeping Only One State Derivative

For $H=X^1X^2$ in the first example, it would be wrong to use only $\partial_1 f\,\sigma_{11}=X^2$ as the correction density against $B^1$. That calculation omits the second coordinate’s contribution, $\partial_2f\,\sigma_{21}=X^1$. The correct density is $X^2+X^1=2B^1+B^2$, because both state coordinates contain the same Brownian component $B^1$. Conversely, a coordinate whose diffusion coefficient against the chosen integrator is zero contributes nothing, regardless of the size of its derivative.

The practical procedure is to identify the Brownian component being integrated, list the diffusion coefficient of every state coordinate against that component, multiply each by the corresponding spatial derivative, and sum. The drift and the remaining Itô terms do not enter this covariation correction.

Check Your Understanding

For each question, identify the Brownian component used as integrator before computing the correction.

If $X^1=B^1$, $X^2=B^1+B^2$, and $f(x_1,x_2)=x_1x_2$, what is the correction density against $B^2$?
For $dX^1=b_1dt+2dB^1+dB^2$, $dX^2=b_2dt+dB^1+3dB^2$, and $f(x_1,x_2)=x_1^2+x_2$, compute the half-correction density against $B^1$.
Why does $\partial_t f$ contribute to Itô’s formula for $f(t,X_t)$ but not directly to its covariation with $B^k$?
Suppose the coefficient of $dB^k$ in every coordinate of $X$ is zero. What is the Stratonovich correction for $f(t,X_t)$ against $B^k$, and why?
For correlated Brownian components, which covariance factor replaces the Kronecker delta in the componentwise correction?