The Multidimensional Ito Formula for Scalar Functions

From Scalar Ito Calculus to Vector-Valued States

For a one-dimensional Ito process, the second-order correction in Ito's formula depends on the squared diffusion coefficient and the second derivative of the function. For a vector-valued process, each pair of coordinates can contribute: the relevant quantity is the rate of their covariation. The scalar function's Hessian records its second derivatives, while the instantaneous covariance matrix records the covariations of the process. Their contraction gives the full correction, including terms from correlated Brownian drivers.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying Brownian drivers $B^1,\ldots,B^m$. Suppose these drivers have constant instantaneous correlations $R_{ab}$, so $d[B^a,B^b]_t=R_{ab}\,dt$, where the symmetric matrix $R$ is positive semidefinite and $R_{aa}=1$. Independent standard Brownian drivers correspond to $R$ equal to the identity matrix. The vector process $X_t=(X_t^1,\ldots,X_t^n)^\mathsf{T}$ has drift $b_t=(b_t^1,\ldots,b_t^n)^\mathsf{T}$ and diffusion matrix $\sigma_t=(\sigma_{ta})$, with dynamics $dX_t=b_t\,dt+\sigma_t\,d\mathbf{B}_t$. Thus the $i$th coordinate has diffusion term $\sum_{a=1}^m\sigma_{ia,t}\,dB_t^a$.

Definition: The instantaneous covariance matrix of $X$ is $A_t=\sigma_t R\sigma_t^\mathsf{T}$. Its entries satisfy $d[X^i,X^j]_t=(A_t)_{ij}\,dt$. For a scalar function $f(t,x)$, its gradient is $\nabla f=(\partial_{x_i}f)_{i=1}^n$, and its Hessian is $D^2f=(\partial_{x_i x_j}f)_{i,j=1}^n$.

For the formula below, take coefficients progressively measurable, with $\int_0^T|b_t|\,dt<\infty$ and $\int_0^T\|\sigma_t\|^2\,dt<\infty$ almost surely on each finite horizon. The stochastic integrals are Ito integrals: their integrands are progressively measurable and, whenever an expectation statement is made, belong to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_t^2\,dt<\infty$. For correlated drivers it suffices to impose this condition on each component of the integrand; the matrix $R$ is bounded. Let $f$ be once continuously differentiable in time and twice continuously differentiable in space, $f\in C^{1,2}$. The formula is understood up to localization if the derivatives or coefficients are not bounded. The standard multidimensional Ito formula is stated here; its proof is given in the later course material on proving the multidimensional Ito formula.

Theorem: Under these conditions, $$ df(t,X_t)=\left(\partial_t f(t,X_t)+\nabla f(t,X_t)^\mathsf{T}b_t+\frac12\operatorname{tr}\!\left(D^2f(t,X_t)A_t\right)\right)dt +\nabla f(t,X_t)^\mathsf{T}\sigma_t\,d\mathbf{B}_t. $$ Here $\operatorname{tr}$ denotes the sum of the diagonal entries, and the stochastic term means $\sum_{a=1}^m\left(\sum_{i=1}^n\partial_{x_i}f(t,X_t)\sigma_{ia,t}\right)dB_t^a$. Equivalently, the correction is $\frac12\sum_{i,j=1}^n\partial_{x_i x_j}f(t,X_t)(A_t)_{ij}$.

Both expressions for the correction are the same because $D^2f$ and $A$ are symmetric: $\operatorname{tr}(D^2f\,A)=\sum_{i,j}\partial_{x_i x_j}f\,A_{ij}$. In particular, the formula uses $dX^i_t\,dX^j_t=(A_t)_{ij}\,dt$: products involving a time increment vanish, $dt\,dB_t^a=0$ and $dt\,dt=0$, while $dB_t^a\,dB_t^b=R_{ab}\,dt$. These are Ito multiplication rules, not ordinary chain-rule identities.

Why the Hessian and Covariance Must Be Paired

The second derivative measures curvature in each pair of coordinate directions. Covariation determines how much each pair of directions contributes at order $dt$. The resulting correction is not generally a sum of only the pure second derivatives: if two state coordinates covary, their mixed derivative contributes too.

Justification of the contraction. If $H=D^2f$ and $A=\sigma R\sigma^\mathsf{T}$, then matrix multiplication gives $\operatorname{tr}(HA)=\sum_i\sum_j H_{ij}A_{ji}$. Since $A_{ji}=A_{ij}$, this is $\sum_{i,j}H_{ij}A_{ij}$. Also, for any vector $v$, $v^\mathsf{T}Av=(\sigma^\mathsf{T}v)^\mathsf{T}R(\sigma^\mathsf{T}v)\ge0$, since $R$ is positive semidefinite. Thus $A$ is itself a symmetric positive semidefinite covariance matrix. $\square$

Worked Example: The Product of Correlated Brownian Coordinates

Let $(B^1,B^2)$ be standard Brownian drivers with correlation $\rho=\tfrac12$, and set $X_t=(B_t^1,B_t^2)^\mathsf{T}$. Apply the formula to $f(x_1,x_2)=x_1x_2$, then check the drift using the expectation of $B_t^1B_t^2$.

The drift is zero and the diffusion matrix is the identity, so $A=R=\begin{pmatrix}1&\tfrac12\\ \tfrac12&1\end{pmatrix}$. The derivatives are $\partial_{x_1}f=x_2$, $\partial_{x_2}f=x_1$, and $D^2f=\begin{pmatrix}0&1\\1&0\end{pmatrix}$; also $\partial_tf=0$. The correction is $\frac12\operatorname{tr}(D^2f A)=\frac12(\frac12+\frac12)=\frac12$. The stochastic term is $B_t^2\,dB_t^1+B_t^1\,dB_t^2$. Therefore

d(B_t^1B_t^2)=B_t^2\,dB_t^1+B_t^1\,dB_t^2+\frac12\,dt.

Check: the expectation of each stochastic integral is zero: on $[0,T]$, $\mathbb{E}\int_0^T(B_s^2)^2\,ds=T^2/2<\infty$, and the same holds for $B^1$. The covariance of the two Brownian coordinates is $\mathbb{E}[B_t^1B_t^2]=\rho t=t/2$, agreeing with the integrated drift $t/2$. This also checks the sign and size of the mixed-derivative correction.

Worked Example: A Quadratic Function with Drift

Let $B^1,B^2$ be independent Brownian motions and define $X_t^1=t+2B_t^1$, $X_t^2=-t+B_t^2$, with $X_0=(0,0)$. Find the Ito differential of $f(x_1,x_2)=x_1^2+3x_2^2$, then check its expectation.

Here $b=(1,-1)^\mathsf{T}$, $\sigma=\begin{pmatrix}2&0\\0&1\end{pmatrix}$, and $A=\sigma\sigma^\mathsf{T}=\begin{pmatrix}4&0\\0&1\end{pmatrix}$. The derivatives are $\nabla f=(2x_1,6x_2)^\mathsf{T}$ and $D^2f=\begin{pmatrix}2&0\\0&6\end{pmatrix}$. The time derivative is zero. The drift contribution from $\nabla f^\mathsf{T}b$ is $2x_1-6x_2$, and the Hessian correction is $\frac12(2\cdot4+6\cdot1)=7$. The diffusion contribution is $4x_1\,dB_t^1+6x_2\,dB_t^2$, so

df(X_t)=(2X_t^1-6X_t^2+7)\,dt+4X_t^1\,dB_t^1+6X_t^2\,dB_t^2.

Check: directly, $\mathbb{E}[(X_t^1)^2]=t^2+4t$ and $\mathbb{E}[(X_t^2)^2]=t^2+t$, since the Brownian terms have mean zero and variances $t$. Hence $\mathbb{E}[f(X_t)]=(t^2+4t)+3(t^2+t)=4t^2+7t$. In the differential, the stochastic integrands are square-integrable on every finite horizon because $X^1,X^2$ have finite second moments integrated over time. The stochastic integrals are therefore martingales with expectation zero. The expected drift at time $s$ is $2s-6(-s)+7=8s+7$; its integral from $0$ to $t$ is $4t^2+7t$, matching the direct calculation.

A Stopped Example on a Punctured Domain

The formula applies locally when the function's derivatives are only defined away from a boundary, provided the process is stopped before reaching that boundary. The next calculation illustrates how a nonzero Hessian can still give zero correction when its contraction with the covariance matrix vanishes.

Worked Example: The Logarithm of Squared Radius

Let $X_t=(1+B_t^1,B_t^2)$, where the Brownian coordinates are independent. Fix $0<\varepsilon<1<R$, and let $\tau$ be the first time $X_t$ leaves the annulus $\{x:\varepsilon<|x|<R\}$. Apply Ito's formula to $f(x_1,x_2)=\log(x_1^2+x_2^2)$ up to $t\wedge\tau$.

On the annulus, put $q=x_1^2+x_2^2$, which is strictly positive. The first derivatives are $\partial_{x_1}f=2x_1/q$ and $\partial_{x_2}f=2x_2/q$. Differentiating again gives $\partial_{x_1x_1}f=2/q-4x_1^2/q^2$, $\partial_{x_2x_2}f=2/q-4x_2^2/q^2$, and $\partial_{x_1x_2}f=-4x_1x_2/q^2$. The time derivative is zero. Here $b=0$ and $A$ is the identity matrix, so the correction is

\frac12\left(\partial_{x_1x_1}f+\partial_{x_2x_2}f\right) =\frac12\left(\frac4q-\frac{4(x_1^2+x_2^2)}{q^2}\right) =\frac12\left(\frac4q-\frac{4q}{q^2}\right)=0.

Consequently, up to and including the stopping time,

d f(X_{t\wedge\tau}) =\mathbf{1}_{\{t<\tau\}}\left(\frac{2X_t^1}{(X_t^1)^2+(X_t^2)^2}\,dB_t^1+ \frac{2X_t^2}{(X_t^1)^2+(X_t^2)^2}\,dB_t^2\right).

The indicator expresses that the integrands are zero after stopping; the value at the single time $t=\tau$ does not affect the Ito integral. Check: the two diagonal second derivatives sum to $4/q-4q/q^2=0$ at every point with $q>0$, not just at the starting point. On the stopped annulus the displayed integrands are bounded, so they are in $\mathcal{L}^2$ on every finite horizon. The stochastic integral is a true martingale there, and its expectation is zero. This check is confined to the stopped process; it does not assert that the unstopped logarithm is integrable at or near the origin.

Expectation Identity and a Common Error

Proposition: Suppose the terms in the multidimensional Ito formula are integrable on $[0,T]$, and the stochastic-integral integrand $G_t=\nabla f(t,X_t)^\mathsf{T}\sigma_t$ satisfies $\mathbb{E}\int_0^T G_t R G_t^\mathsf{T}\,dt<\infty$. Then $$ \mathbb{E}[f(T,X_T)]=\mathbb{E}[f(0,X_0)]+\mathbb{E}\int_0^T\left(\partial_t f+\nabla f^\mathsf{T}b+\frac12\operatorname{tr}(D^2f\,A)\right)(t,X_t)\,dt. $$

Proof. Integrate the Ito formula from $0$ to $T$. The stochastic term is an Ito integral with a progressively measurable integrand whose squared norm in the Brownian covariance metric has finite expected integral by assumption. It is therefore a square-integrable martingale starting at zero; in particular, its expectation is zero. Taking expectations of the integrated formula gives the stated identity. The assumed integrability of the other terms ensures both sides and the time integral are well-defined. $\square$

Common mistake: ignoring off-diagonal covariance. In the correlated Brownian example, using only the diagonal entries of $A$ would give $\frac12(0\cdot1+0\cdot1)=0$ for the product's correction. That calculation discards $A_{12}=A_{21}=\tfrac12$, even though $f_{x_1x_2}=f_{x_2x_1}=1$. The correct contraction is $\frac12(1\cdot\tfrac12+1\cdot\tfrac12)=\tfrac12$. The omission would incorrectly predict zero expectation for $B_t^1B_t^2$, while its covariance is $t/2$.

For independent drivers, $R$ is the identity, but the coordinates of $X$ can still covary when they share Brownian drivers. In either case, calculate $A=\sigma R\sigma^\mathsf{T}$ first, then contract it with the Hessian. The drift $b$ contributes through the first derivatives; the diffusion contributes both to the stochastic integral and to the second-order correction. These roles are distinct.

Check Your Understanding

For each question, state any covariance or stopping assumptions you use.

For two Brownian drivers with correlation $\rho$, compute the Ito correction for $f(x_1,x_2)=x_1^2+x_1x_2$.
If $A$ is diagonal, which entries of the Hessian contribute to $\operatorname{tr}(D^2f\,A)$?
For $f(x_1,x_2)=x_1^2+3x_2^2$ and $A=\begin{pmatrix}4&0\\0&1\end{pmatrix}$, calculate the correction independently from the trace and the index-sum expressions.
Why is the stopped logarithmic example's stochastic integral square-integrable on a finite time interval?
What condition on a stochastic-integral integrand allows the expectation identity to use expectation zero for its Ito integral?

The Multidimensional Ito Formula for Scalar Functions

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