Multidimensional Itô's Lemma

The one-dimensional Itô formula (see Itô's Lemma) extends naturally to systems of SDEs driven by multiple Brownian motions. This page presents the multidimensional version using index notation, derives the quadratic covariation structure, and works through concrete examples.

1. Setting

Let \(W_t = (W_t^{1}, \ldots, W_t^{m})\) be an \(m\)-dimensional Brownian motion and \(X_t = (X_t^{1}, \ldots, X_t^{d})\) an \(\mathbb{R}^d\)-valued diffusion solving:

\[ dX_t^{i} = b^{i}(t, X_t)\,dt + \sigma^{i\alpha}(t, X_t)\,dW_t^{\alpha}, \qquad i = 1, \ldots, d, \quad \alpha = 1, \ldots, m \]

Notation:

• \(b^{i}: [0,T] \times \mathbb{R}^d \to \mathbb{R}\) is the \(i\)-th drift component
• \(\sigma^{i\alpha}: [0,T] \times \mathbb{R}^d \to \mathbb{R}\) is the \((i, \alpha)\) entry of the \(d \times m\) diffusion matrix
• Einstein summation convention: repeated indices are summed over their range. For example, in \(\sigma^{i\alpha} dW_t^{\alpha}\), the index \(\alpha\) is summed from \(1\) to \(m\).

Index notation convention

This page uses all superscripts for index notation, for consistency with the SDE structure above. This deviates from the standard tensor convention (which sums a lower index against an upper index). Here, any repeated index — regardless of height — denotes summation. Readers familiar with standard tensor notation should note this difference.

2. The Theorem

Let \(f: [0,T] \times \mathbb{R}^d \to \mathbb{R}\) be \(C^{1,2}\) (once differentiable in \(t\), twice in \(x\)). Write

\[ f_t = \frac{\partial f}{\partial t}, \qquad f_i = \frac{\partial f}{\partial x^i}, \qquad f_{ij} = \frac{\partial^2 f}{\partial x^i \partial x^j} \]

Theorem (Multidimensional Itô Formula). The process \(Y_t = f(t, X_t)\) satisfies:

\[ \boxed{ df(t, X_t) = f_t\,dt + f_i\,dX_t^{i} + \frac{1}{2} f_{ij}\,d\langle X^{i}, X^{j}\rangle_t } \]

where all partial derivatives are evaluated at \((t, X_t)\).

Quadratic covariation. Since \(dX_t^{i} = b^{i}\,dt + \sigma^{i\alpha}\,dW_t^{\alpha}\), only the Brownian parts contribute. The key rule for \(m\)-dimensional Brownian motion is

\[ d\langle W^{\alpha}, W^{\beta} \rangle_t = \delta^{\alpha\beta}\,dt \]

which reflects two facts: independent Brownian components (\(\alpha \neq \beta\)) have zero covariation, while the same component satisfies \((dW^\alpha)^2 = dt\) — see From Taylor to Itô for the 1D derivation. Applying this:

\[ \boxed{ d\langle X^{i}, X^{j} \rangle_t = \sigma^{i\alpha}(t, X_t)\,\sigma^{j\alpha}(t, X_t)\,dt =: a^{ij}(t, X_t)\,dt } \]

where \(\delta^{\alpha\beta}\) is the Kronecker delta and the repeated index \(\alpha\) is summed from \(1\) to \(m\).

The diffusion matrix \(a^{ij} := \sigma^{i\alpha}\sigma^{j\alpha} = (\sigma\sigma^T)^{ij}\) is a \(d \times d\) symmetric positive-semidefinite matrix. Non-negative definiteness follows from

\[ \sum_{i,j} \xi^i a^{ij} \xi^j = \sum_\alpha \left(\sum_i \sigma^{i\alpha}\xi^i\right)^2 = |\sigma^T\xi|^2 \geq 0 \]

for all \(\xi \in \mathbb{R}^d\). It is positive definite (elliptic) if and only if \(\sigma\) has full row rank.

3. Expanded and Standard Forms

Substituting \(dX_t^{i}\) and \(d\langle X^i, X^j \rangle_t\) explicitly:

\[ df(t, X_t) = f_t\,dt + f_i\!\left(b^{i}\,dt + \sigma^{i\alpha}\,dW_t^{\alpha}\right) + \frac{1}{2}f_{ij}\,\sigma^{i\alpha}\sigma^{j\alpha}\,dt \]

Collecting \(dt\) and \(dW\) terms gives the standard form:

\[ \boxed{ df(t, X_t) = \left(f_t + b^{i} f_i + \frac{1}{2}a^{ij} f_{ij}\right)dt + \sigma^{i\alpha} f_i\,dW_t^{\alpha} } \]

where all functions are evaluated at \((t, X_t)\).

• Drift of \(f\): time derivative + advection by drift + Itô correction from curvature
• Stochastic part of \(f\): \(f_i \sigma^{i\alpha} dW_t^{\alpha}\), i.e., the gradient of \(f\) contracted with the diffusion matrix \(\sigma\), acting on \(dW_t\)

The infinitesimal generator \(\mathcal{L}\) packages the spatial part of the drift (excluding \(f_t\)):

\[ (\mathcal{L}f)(t, x) := b^{i}(t, x)\,f_i(t,x) + \frac{1}{2}a^{ij}(t, x)\,f_{ij}(t,x) \]

so the standard form condenses to:

\[ df(t, X_t) = \bigl(f_t + \mathcal{L}f\bigr)(t, X_t)\,dt + f_i(t, X_t)\,\sigma^{i\alpha}(t, X_t)\,dW_t^{\alpha} \]

Note that \(\mathcal{L}\) acts on spatial variables only and does not include \(f_t\); the full drift is \(f_t + \mathcal{L}f\). The generator \(\mathcal{L}\) governs the expected infinitesimal evolution of \(f(t, X_t)\); it appears in the Kolmogorov forward and backward equations and the Feynman–Kac formula — see Kolmogorov Equations and Feynman–Kac Formula.

4. Examples

Example 1: Norm Squared

Let \(f(x) = |x|^2 = \sum_i (x^i)^2\). Then \(f_t = 0\), \(f_i = 2x^i\), \(f_{ij} = 2\delta^{ij}\) (where \(\delta^{ij} = 1\) if \(i = j\) and \(0\) otherwise). The standard form gives:

\[ d|X_t|^2 = \left(2\sum_i X_t^i b^i + \sum_i a^{ii}\right)dt + 2\sum_{i,\alpha} X_t^i \sigma^{i\alpha}\,dW_t^\alpha \]

where \(\sum_i a^{ii} = \operatorname{tr}(a) = \operatorname{tr}(\sigma\sigma^T)\) is the sum of all diffusion variances. The Itô correction inflates the drift by \(\operatorname{tr}(\sigma\sigma^T)\) relative to the classical chain rule.

Example 2: Inner Product of Two Diffusions

Let \(X_t\) and \(Y_t\) be two \(\mathbb{R}^d\)-valued diffusions with SDEs

\[ dX_t^i = b_X^i\,dt + \sigma_X^{i\alpha}\,dW_t^\alpha, \qquad dY_t^i = b_Y^i\,dt + \sigma_Y^{i\alpha}\,dW_t^\alpha \]

driven by the same \(m\)-dimensional Brownian motion. To apply Itô's formula, form the joint process \(Z_t = (X_t, Y_t) \in \mathbb{R}^{2d}\) and apply the multidimensional formula to \(f(x, y) = \sum_i x^i y^i\).

The gradient of \(f\) with respect to \((x,y)\) is the vector of partial derivatives \((\partial f/\partial x^i, \partial f/\partial y^i)_{i=1}^d = (y^i, x^i)_{i=1}^d\). That is, for each fixed \(i\): \(\partial f/\partial x^i = y^i\) and \(\partial f/\partial y^i = x^i\). The only non-zero second-order cross terms are \(\partial^2 f/\partial x^i \partial y^i = 1\) for each \(i\). Substituting into the formula yields the product rule:

\[ d\!\left(\sum_i X_t^i Y_t^i\right) = \sum_i Y_t^i\,dX_t^i + \sum_i X_t^i\,dY_t^i + \sum_i d\langle X^i, Y^i \rangle_t \]

Each covariation term evaluates to \(d\langle X^i, Y^i \rangle_t = \sum_\alpha \sigma_X^{i\alpha}\sigma_Y^{i\alpha}\,dt\), so summing over \(i\):

\[ \boxed{ d\!\left(\sum_i X_t^i Y_t^i\right) = \left(\sum_i Y_t^i b_X^i + \sum_i X_t^i b_Y^i + \sum_{i,\alpha}\sigma_X^{i\alpha}\sigma_Y^{i\alpha}\right)dt + \sum_{i,\alpha}\left(Y_t^i \sigma_X^{i\alpha} + X_t^i \sigma_Y^{i\alpha}\right)dW_t^\alpha } \]

This is the multidimensional Itô product rule. When \(X = Y\) it reduces to Example 1.

5. Summary

\[ \boxed{ df(t, X_t) = \left(f_t + b^i f_i + \tfrac{1}{2}a^{ij}f_{ij}\right)dt + \sigma^{i\alpha} f_i\,dW_t^\alpha } \]

	1D	\(d\)-dimensional
SDE	\(dX = \mu\,dt + \sigma\,dW\)	\(dX^i = b^i\,dt + \sigma^{i\alpha}\,dW^\alpha\)
Itô correction	\(\tfrac{1}{2}\sigma^2 f_{xx}\)	\(\tfrac{1}{2}a^{ij}f_{ij}\), \(\;a^{ij} = \sigma^{i\alpha}\sigma^{j\alpha}\)
Generator \(\mathcal{L}\) (spatial only)	\(\mu f_x + \tfrac{1}{2}\sigma^2 f_{xx}\)	\(b^i f_i + \tfrac{1}{2}a^{ij}f_{ij}\)
Full drift of \(f\)	\(f_t + \mu f_x + \tfrac{1}{2}\sigma^2 f_{xx}\)	\(f_t + b^i f_i + \tfrac{1}{2}a^{ij}f_{ij}\)

The 1D formula is recovered by setting \(d = m = 1\), \(b^1 = \mu\), \(\sigma^{11} = \sigma\), so \(a^{11} = \sigma^2\).

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Exercises

Exercise 1. Let \(d = 2\), \(m = 2\), with \(dX_t^1 = \sigma_1\,dW_t^1\) and \(dX_t^2 = \sigma_2\,dW_t^2\) (two independent Brownian motions, no drift). Compute the diffusion matrix \(a^{ij} = \sigma^{i\alpha}\sigma^{j\alpha}\) and verify that it is diagonal.

Solution to Exercise 1

The diffusion matrix is \(\sigma = \begin{pmatrix} \sigma_1 & 0 \\ 0 & \sigma_2 \end{pmatrix}\) (each component driven by its own independent Brownian motion). The diffusion matrix \(a^{ij} = \sigma^{i\alpha}\sigma^{j\alpha}\) is computed by summing over \(\alpha = 1, 2\):

• \(a^{11} = \sigma^{11}\sigma^{11} + \sigma^{12}\sigma^{12} = \sigma_1^2 + 0 = \sigma_1^2\)
• \(a^{12} = \sigma^{11}\sigma^{21} + \sigma^{12}\sigma^{22} = 0 + 0 = 0\)
• \(a^{21} = \sigma^{21}\sigma^{11} + \sigma^{22}\sigma^{12} = 0 + 0 = 0\)
• \(a^{22} = \sigma^{21}\sigma^{21} + \sigma^{22}\sigma^{22} = 0 + \sigma_2^2 = \sigma_2^2\)

So \(a = \begin{pmatrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \end{pmatrix}\), which is diagonal. This confirms that independent Brownian drivers produce a diagonal diffusion matrix — the two components have no quadratic covariation.

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Exercise 2. Let \(f(x^1, x^2) = x^1 x^2\) with the same setup as Exercise 1. Apply the multidimensional Itô formula to compute \(d(X_t^1 X_t^2)\). Show that the Itô correction term vanishes because the two processes are driven by independent Brownian motions.

Solution to Exercise 2

For \(f(x^1, x^2) = x^1 x^2\): \(f_1 = x^2\), \(f_2 = x^1\), \(f_{11} = 0\), \(f_{22} = 0\), \(f_{12} = f_{21} = 1\). From Exercise 1, \(a^{ij}\) is diagonal with \(a^{12} = 0\). No drift (\(b^i = 0\)). The Itô correction is:

\[ \frac{1}{2}a^{ij}f_{ij} = \frac{1}{2}(a^{11} \cdot 0 + a^{12} \cdot 1 + a^{21} \cdot 1 + a^{22} \cdot 0) = \frac{1}{2}(0 + 0 + 0 + 0) = 0 \]

The multidimensional Itô formula gives:

\[ d(X_t^1 X_t^2) = X_t^2\,dX_t^1 + X_t^1\,dX_t^2 + 0 = \sigma_1 X_t^2\,dW_t^1 + \sigma_2 X_t^1\,dW_t^2 \]

The Itô correction vanishes because \(d\langle X^1, X^2\rangle_t = 0\) — the two processes are driven by independent Brownian motions.

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Exercise 3. Now suppose \(dX_t^1 = \sigma_1\,dW_t^1\) and \(dX_t^2 = \sigma_2\,dW_t^1\) (both driven by the same Brownian motion). Compute the diffusion matrix \(a^{ij}\) and the quadratic covariation \(d\langle X^1, X^2 \rangle_t\). Apply the multidimensional Itô formula to \(f(x^1, x^2) = x^1 x^2\) and identify the Itô correction.

Solution to Exercise 3

Now \(\sigma = \begin{pmatrix} \sigma_1 \\ \sigma_2 \end{pmatrix}\) (a \(2 \times 1\) matrix, since \(m = 1\)). The diffusion matrix is:

\[ a^{ij} = \sigma^{i1}\sigma^{j1} = \begin{pmatrix} \sigma_1^2 & \sigma_1\sigma_2 \\ \sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix} \]

The quadratic covariation is \(d\langle X^1, X^2\rangle_t = \sigma_1\sigma_2\,dt\).

For \(f(x^1, x^2) = x^1 x^2\): \(f_{12} = 1\), all other second derivatives are zero. The Itô correction is:

\[ \frac{1}{2}a^{ij}f_{ij} = \frac{1}{2}(0 + \sigma_1\sigma_2 \cdot 1 + \sigma_1\sigma_2 \cdot 1 + 0) = \sigma_1\sigma_2 \]

The full formula gives:

\[ d(X_t^1 X_t^2) = X_t^2 \sigma_1\,dW_t + X_t^1 \sigma_2\,dW_t + \sigma_1\sigma_2\,dt \]

\[ = \sigma_1\sigma_2\,dt + (\sigma_1 X_t^2 + \sigma_2 X_t^1)\,dW_t \]

The Itô correction \(\sigma_1\sigma_2\,dt\) is non-zero because both processes share the same Brownian motion.

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Exercise 4. Let \(X_t = (X_t^1, X_t^2)\) be a 2D process with

\[ dX_t^1 = dW_t^1, \qquad dX_t^2 = dW_t^1 + dW_t^2 \]

where \(W^1\) and \(W^2\) are independent. Compute the \(2 \times 2\) diffusion matrix \(a^{ij} = (\sigma\sigma^T)^{ij}\) and verify that it is positive definite.

Solution to Exercise 4

The diffusion matrix is \(\sigma = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) (rows correspond to \(X^1, X^2\); columns to \(W^1, W^2\)). The diffusion matrix is:

\[ a = \sigma\sigma^T = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \]

To verify positive definiteness, check that both eigenvalues are positive. The determinant is \(1 \cdot 2 - 1 \cdot 1 = 1 > 0\) and the trace is \(3 > 0\), so both eigenvalues are positive. Alternatively, \(\det(a) = 1 > 0\) and \(a^{11} = 1 > 0\), confirming positive definiteness by Sylvester's criterion.

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Exercise 5. For the setting in Example 1 (norm squared), let \(d = 3\), \(m = 3\), with \(dX_t^i = \mu^i\,dt + dW_t^i\) for \(i = 1, 2, 3\) (independent standard Brownian motions with constant drifts). Compute \(d|X_t|^2\) explicitly. What is the Itô correction term \(\operatorname{tr}(\sigma\sigma^T)\) in this case?

Solution to Exercise 5

With \(dX_t^i = \mu^i\,dt + dW_t^i\) for \(i = 1, 2, 3\), the diffusion matrix is \(\sigma = I_{3\times 3}\) (the identity), so \(a = \sigma\sigma^T = I\). For \(f(x) = |x|^2 = (x^1)^2 + (x^2)^2 + (x^3)^2\):

• \(f_i = 2x^i\), \(f_{ij} = 2\delta^{ij}\)

The Itô correction is \(\frac{1}{2}a^{ij}f_{ij} = \frac{1}{2}\delta^{ij} \cdot 2\delta^{ij} = \sum_i 1 = 3\). That is, \(\operatorname{tr}(\sigma\sigma^T) = \operatorname{tr}(I) = 3\).

The full formula:

\[ d|X_t|^2 = \left(2\sum_{i=1}^3 X_t^i \mu^i + 3\right)dt + 2\sum_{i=1}^3 X_t^i\,dW_t^i \]

The Itô correction contributes \(+3\,dt\) to the drift, reflecting the fact that three independent Brownian motions each contribute \(+1\,dt\) through their quadratic variation.

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Exercise 6. Consider a 2D geometric Brownian motion:

\[ dS_t^1 = \mu_1 S_t^1\,dt + \sigma_1 S_t^1\,dW_t^1, \qquad dS_t^2 = \mu_2 S_t^2\,dt + \sigma_2 S_t^2\,dW_t^2 \]

with independent Brownian motions. Apply the multidimensional Itô formula to \(f(x^1, x^2) = \log(x^1) + \log(x^2) = \log(x^1 x^2)\) to derive the SDE for \(\log(S_t^1 S_t^2)\).

Solution to Exercise 6

For \(f(x^1, x^2) = \log(x^1) + \log(x^2)\):

• \(f_1 = 1/x^1\), \(f_2 = 1/x^2\)
• \(f_{11} = -1/(x^1)^2\), \(f_{22} = -1/(x^2)^2\), \(f_{12} = 0\)

Since \(W^1\) and \(W^2\) are independent, \(a^{12} = 0\), \(a^{11} = \sigma_1^2(S_t^1)^2\), \(a^{22} = \sigma_2^2(S_t^2)^2\). The drift of \(S^i\) is \(\mu_i S_t^i\). Applying the formula:

\[ d(\log(S_t^1 S_t^2)) = \left(\frac{\mu_1 S_t^1}{S_t^1} + \frac{\mu_2 S_t^2}{S_t^2} + \frac{1}{2}\left(-\frac{\sigma_1^2(S_t^1)^2}{(S_t^1)^2} - \frac{\sigma_2^2(S_t^2)^2}{(S_t^2)^2}\right)\right)dt + \frac{\sigma_1 S_t^1}{S_t^1}\,dW_t^1 + \frac{\sigma_2 S_t^2}{S_t^2}\,dW_t^2 \]

Simplifying:

\[ d(\log(S_t^1 S_t^2)) = \left(\mu_1 + \mu_2 - \frac{\sigma_1^2}{2} - \frac{\sigma_2^2}{2}\right)dt + \sigma_1\,dW_t^1 + \sigma_2\,dW_t^2 \]

This is the sum of the individual log-dynamics: \(d\log(S_t^1) + d\log(S_t^2)\), each with its own convexity adjustment \(-\sigma_i^2/2\).

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Exercise 7. The infinitesimal generator is defined as \((\mathcal{L}f)(t, x) = b^i f_i + \frac{1}{2}a^{ij}f_{ij}\). For \(d = 2\), \(m = 1\), with \(dX_t^1 = \sigma_1\,dW_t\) and \(dX_t^2 = \sigma_2\,dW_t\) (same Brownian motion, no drift), compute \(\mathcal{L}f\) for \(f(x^1, x^2) = (x^1)^2 + (x^2)^2\). Verify your answer by applying the multidimensional Itô formula directly and reading off the \(dt\) coefficient.

Solution to Exercise 7

With \(d = 2\), \(m = 1\), \(b^i = 0\), \(\sigma^{11} = \sigma_1\), \(\sigma^{21} = \sigma_2\):

\[ a^{ij} = \sigma^{i1}\sigma^{j1} = \begin{pmatrix} \sigma_1^2 & \sigma_1\sigma_2 \\ \sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix} \]

For \(f(x^1, x^2) = (x^1)^2 + (x^2)^2\): \(f_1 = 2x^1\), \(f_2 = 2x^2\), \(f_{11} = 2\), \(f_{22} = 2\), \(f_{12} = 0\). The generator is:

\[ \mathcal{L}f = b^i f_i + \frac{1}{2}a^{ij}f_{ij} = 0 + \frac{1}{2}(\sigma_1^2 \cdot 2 + \sigma_1\sigma_2 \cdot 0 + \sigma_1\sigma_2 \cdot 0 + \sigma_2^2 \cdot 2) = \sigma_1^2 + \sigma_2^2 \]

Verification via the multidimensional Itô formula. Since \(f_t = 0\), the \(dt\) coefficient is \(f_t + \mathcal{L}f = \sigma_1^2 + \sigma_2^2\), and the \(dW_t\) coefficient is:

\[ \sigma^{i1}f_i = \sigma_1 \cdot 2X_t^1 + \sigma_2 \cdot 2X_t^2 \]

So:

\[ d((X_t^1)^2 + (X_t^2)^2) = (\sigma_1^2 + \sigma_2^2)\,dt + 2(\sigma_1 X_t^1 + \sigma_2 X_t^2)\,dW_t \]

The \(dt\) coefficient \(\sigma_1^2 + \sigma_2^2\) matches \(\mathcal{L}f\), confirming the computation.