Ito Formula for Vector-Valued Functions

From a Scalar Formula to a Vector Transformation

A vector-valued function can be transformed by applying the scalar multidimensional Ito formula to each of its output coordinates. The calculation is componentwise, but matrix notation makes the structure visible: the Jacobian transforms the diffusion, while each component's Hessian contributes a separate correction to its drift. The Hessian correction is essential even when the original process has no drift.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard $m$-dimensional Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^m)^\mathsf{T}$. Its coordinates are independent standard Brownian motions. Let $X_t$ take values in $\mathbb{R}^n$ and satisfy, for $0\le t\le T$,

dX_t=b_t\,dt+A_t\,d\mathbf{B}_t,

where $b_t\in\mathbb{R}^n$ is progressively measurable with $\mathbb{E}\int_0^T|b_t|\,dt<\infty$, and $A_t\in\mathbb{R}^{n\times m}$ is progressively measurable with $\mathbb{E}\int_0^T\|A_t\|^2\,dt<\infty$. Here $\|\cdot\|$ is the Frobenius matrix norm. Each stochastic integral below is an Ito integral: its integrand is progressively measurable and has finite expected integral of its squared norm on $[0,T]$. The notation $dB_t^k\,dB_t^\ell=\mathbf{1}_{\{k=\ell\}}dt$, $dB_t^k\,dt=0$, and $dt\,dt=0$ records the Ito multiplication rules.

Let $f=(f^1,\ldots,f^p)^\mathsf{T}:[0,T]\times\mathbb{R}^n\to\mathbb{R}^p$ have continuous first time derivatives and continuous first and second spatial derivatives. Assume the composed drift terms below are time-integrable and the composed stochastic integrands are in $\mathcal{L}^2$ on $[0,T]$. These conditions make the displayed integrals well-defined; local versions can also be obtained by stopping when the process or coefficients leave bounded sets.

Definition: The Jacobian $D_xf(t,x)$ is the $p\times n$ matrix with entry $\partial_i f^a(t,x)$ in row $a$, column $i$. For each output coordinate $a$, its Hessian $D_x^2f^a(t,x)$ is the $n\times n$ matrix with entry $\partial_{ij}f^a(t,x)$. Thus the vector-valued function has $p$ Hessian matrices, one for each output coordinate.

The Vector-Valued Ito Formula

Theorem: Under the stated assumptions, set $Y_t=f(t,X_t)$ and $C_t=A_tA_t^\mathsf{T}$. For each $a=1,\ldots,p$, $$ dY_t^a=\left(\partial_t f^a(t,X_t)+D_xf^a(t,X_t)b_t+\frac12\operatorname{tr}\!\left(C_tD_x^2f^a(t,X_t)\right)\right)dt +\sum_{k=1}^m\left(\sum_{i=1}^n\partial_i f^a(t,X_t)A_t^{ik}\right)dB_t^k. $$ Equivalently, the vector drift has coordinate $a$ equal to the parenthesized expression, and the $p\times m$ diffusion matrix is $D_xf(t,X_t)A_t$.

Proof. Fix an output coordinate $a$. The scalar multidimensional Ito formula applies to the scalar function $(t,x)\mapsto f^a(t,x)$ and the $n$-dimensional Ito process $X_t$. Its time derivative term is $\partial_t f^a\,dt$; its first-order spatial drift term is $\sum_i\partial_i f^a b_t^i\,dt$. The covariance matrix of the state diffusion is $C_t=A_tA_t^\mathsf{T}$, since $dX_t^i\,dX_t^j=(A_tA_t^\mathsf{T})_{ij}dt$. Therefore the second-order term is $\frac12\sum_{i,j}C_t^{ij}\partial_{ij}f^a\,dt=\frac12\operatorname{tr}(C_tD_x^2f^a)\,dt$. The coefficient of $dB_t^k$ is $\sum_i\partial_i f^a A_t^{ik}$. This proves the displayed equation for each $a$; stacking the $p$ equations gives the vector drift and the matrix product $D_xf A_t$. $\square$

The trace expression is a contraction of two $n\times n$ matrices, not a matrix product that produces another Hessian. Since $C_t$ is symmetric, it can also be written $\sum_{i,j}C_t^{ij}\partial_{ij}f^a$. In particular, off-diagonal covariances multiply mixed second derivatives. For a time-independent $f$, the $\partial_t f^a$ term is absent; for an affine $f$, every Hessian vanishes.

Worked Example: An Affine Transformation of a Two-Dimensional Process

Let $B^1,B^2$ be independent Brownian motions, and suppose $dX_t^1=dt+dB_t^1+2dB_t^2$ and $dX_t^2=-dt+2dB_t^1-dB_t^2$. Transform the state by $f(x_1,x_2)=(x_1+x_2,x_1-x_2)^\mathsf{T}$.

The drift is $b=(1,-1)^\mathsf{T}$, the diffusion matrix is $A=\begin{pmatrix}1&2\\2&-1\end{pmatrix}$, and the Jacobian and Hessians are $D_xf=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, $D_x^2f^1=D_x^2f^2=0$. The time derivatives are also zero. Thus the transformed drift is $D_xf\,b=(1-1,1+1)^\mathsf{T}=(0,2)^\mathsf{T}$, and the transformed diffusion matrix is

D_xf\,A= \begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}1&2\\2&-1\end{pmatrix} = \begin{pmatrix}3&1\\-1&3\end{pmatrix}.

Therefore $dY_t^1=3dB_t^1+dB_t^2$ and $dY_t^2=2-dB_t^1+3dB_t^2$. Check: directly adding the two state equations gives $d(X_t^1+X_t^2)=3dB_t^1+dB_t^2$; subtracting the second from the first gives $d(X_t^1-X_t^2)=2dt-dB_t^1+3dB_t^2$, in agreement.

Hessian Corrections and Output Covariation

Proposition: Under the theorem's assumptions, the quadratic covariation of output coordinates $Y^a=f^a(t,X_t)$ and $Y^c=f^c(t,X_t)$ is $$ [Y^a,Y^c]_t=\int_0^t\left(D_xf^a(s,X_s)A_sA_s^\mathsf{T}D_xf^c(s,X_s)^\mathsf{T}\right)ds. $$

Proof. The finite-variation drift terms in the Ito formula have zero quadratic covariation with continuous Ito processes. The row diffusion coefficient of $Y^a$ is $D_xf^a A$, and that of $Y^c$ is $D_xf^c A$. The quadratic covariation rule for Ito integrals gives their covariation density as $(D_xf^a A)(D_xf^c A)^\mathsf{T}=D_xf^aAA^\mathsf{T}(D_xf^c)^\mathsf{T}$. Integrating this density from $0$ to $t$ proves the identity. $\square$

Worked Example: A Product with Correlated State Noise

Let $X_0=(1,2)^\mathsf{T}$ and $dX_t^1=dt+dB_t^1$, $dX_t^2=-dt+dB_t^1+dB_t^2$, where $B^1,B^2$ are independent. For $f(x_1,x_2)=x_1x_2$, calculate the transformed equation and verify its mean.

Here $b=(1,-1)^\mathsf{T}$, $A=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, and $C=AA^\mathsf{T}=\begin{pmatrix}1&1\\1&2\end{pmatrix}$. The derivatives are $\partial_1f=x_2$, $\partial_2f=x_1$, and $\partial_{11}f=\partial_{22}f=0$, $\partial_{12}f=\partial_{21}f=1$. The drift from the first derivatives is $x_2-x_1$. The Hessian correction is $\frac12(C_{12}+C_{21})=1$. The diffusion coefficient row is $(x_2,x_1)A=(x_1+x_2,x_1)$. Hence

$$ d(X_t^1X_t^2)=(X_t^2-X_t^1+1)dt+(X_t^1+X_t^2)dB_t^1+X_t^1dB_t^2. $$

In fact, $X_t^1=1+t+B_t^1$ and $X_t^2=2-t+B_t^1+B_t^2$. Thus $\mathbb{E}[X_t^1]=1+t$, $\mathbb{E}[X_t^2]=2-t$, and $\operatorname{Cov}(X_t^1,X_t^2)=t$. Consequently $\mathbb{E}[X_t^1X_t^2]=(1+t)(2-t)+t=2+2t-t^2$. The stochastic integrands in the transformed equation are linear in Brownian motion and have finite expected squared integrals on every finite time interval, so their Ito integrals are square-integrable martingales with mean zero. Taking expectations in the equation therefore gives initial mean $2$ plus $\int_0^t((2-s)-(1+s)+1)\,ds=2+\int_0^t(2-2s)\,ds=2+2t-t^2$, confirming the independent calculation.

The off-diagonal entries of $C$ encode the covariation of the state coordinates. In this example $dX_t^1\,dX_t^2=dt$, so the product has a $+dt$ correction. Treating the two state noises as independent would incorrectly discard this term.

Worked Example: A Two-Coordinate Nonlinear Transformation

Let $X_t=(1+B_t^1,1+B_t^2)^\mathsf{T}$, with independent Brownian coordinates, and define $f(x_1,x_2)=(x_1^2-x_2^2,2x_1x_2)^\mathsf{T}$. Find both transformed equations and check their means.

The state drift is zero and $A=I_2$, so $C=I_2$. For the first output, the time derivative is zero, the gradient is $(2x_1,-2x_2)$, and the Hessian is $\begin{pmatrix}2&0\\0&-2\end{pmatrix}$. Its correction is $\frac12(2-2)=0$, and its diffusion row is $(2x_1,-2x_2)$. For the second output, the time derivative is zero, the gradient is $(2x_2,2x_1)$, and the Hessian is $\begin{pmatrix}0&2\\2&0\end{pmatrix}$. Its correction is $\frac12(0+0)=0$, because $C$ has zero off-diagonal entries; its diffusion row is $(2x_2,2x_1)$. Thus

dY_t^1=2X_t^1dB_t^1-2X_t^2dB_t^2,\qquad dY_t^2=2X_t^2dB_t^1+2X_t^1dB_t^2.

For each finite $T$, the displayed integrands are progressively measurable and square-integrable because $\mathbb{E}[(X_t^i)^2]=1+t$ and $\int_0^T(1+t)\,dt<\infty$. The resulting Ito integrals are square-integrable martingales. Independently, $\mathbb{E}[Y_t^1]=\mathbb{E}[(X_t^1)^2]-\mathbb{E}[(X_t^2)^2]=(1+t)-(1+t)=0$. Independence gives $\mathbb{E}[X_t^1X_t^2]=1$, so $\mathbb{E}[Y_t^2]=2$. Both expectations match the initial outputs $(0,2)$ and the zero drift terms.

A Common Error: Omitting the Correction

A tempting but incorrect calculation is to transform only the drift by the Jacobian and set the transformed drift equal to $D_xf\,b$. In the product example, this gives $x_2-x_1$, which at the initial state $(1,2)$ equals $1$. The correct drift there is $x_2-x_1+1=2$. The missing $1$ is $\frac12\operatorname{tr}(AA^\mathsf{T}D_x^2f)$, arising from the common $B^1$ noise in the two coordinates.

Keep the two roles separate: $D_xf\,A$ is the output diffusion matrix, while the Hessian-covariance contraction supplies one scalar correction for each output coordinate. The Ito formula here is not the ordinary chain rule, and it is not a Stratonovich chain rule: the displayed second-order drift is specific to Ito integration.

Check Your Understanding

For each question, use the component Hessian belonging to the indicated output coordinate and include any nonzero state covariations.

For $n=2$, $m=1$, and $f(x_1,x_2)=(x_1+x_2,x_1x_2)^\mathsf{T}$, write the Jacobian and both Hessian matrices.
If $A=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, compute $AA^\mathsf{T}$ and the Ito correction for $f(x_1,x_2)=x_1x_2$.
State the dimensions of $D_xf$, $A$, $D_xf A$, and each component Hessian when $f:\mathbb{R}^n\to\mathbb{R}^p$ and the Brownian motion has $m$ coordinates.
For two transformed coordinates, explain why their quadratic covariation depends on their diffusion rows but not on their drift terms.
For $f(x_1,x_2)=(x_1^2-x_2^2,2x_1x_2)^\mathsf{T}$ with identity state diffusion, identify the Hessian correction for each output and explain why both vanish.

Ito Formula for Vector-Valued Functions

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