Differential Multiplication Rules for Ito Processes

Why Differential Multiplication Needs a Rule

For ordinary differentiable paths, products of small increments are often discarded when they are of higher order than the time increment. Brownian increments behave differently: their squared increments accumulate to quadratic variation. Ito differential multiplication is a compact way to keep precisely those second-order terms that survive. Its main practical use here is to calculate the quadratic-variation contribution that enters a function’s drift correction.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a \(d\)-dimensional Brownian motion \(\mathbf{B}_t=(B_t^1,\ldots,B_t^d)\) with respect to \((\mathcal{F}_t)\). Thus its components are independent standard Brownian motions. On a fixed finite interval \([0,T]\), every stochastic integral below is an Ito integral with progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T |H_s|^2\,ds<\infty\). Drift coefficients are assumed progressively measurable and integrable in time almost surely.

An Ito process \(X\) in this setting has a representation \(X_t=X_0+\int_0^t a_s\,ds+\sum_{k=1}^d\int_0^t \sigma_{k,s}\,dB_s^k\), where each \(\sigma_k\) satisfies the stated square-integrability condition. Its differential notation is \(dX_t=a_t\,dt+\sum_k\sigma_{k,t}\,dB_t^k\). The coefficient \(a_t\) is the drift; the \(\sigma_{k,t}\) are diffusion coefficients.

Definition: The Ito differential multiplication table is the bookkeeping convention \(dt\,dt=0\), \(dt\,dB_t^k=dB_t^k\,dt=0\), and \(dB_t^i\,dB_t^j=\delta_{ij}\,dt\), where \(\delta_{ij}=1\) if \(i=j\) and \(0\) otherwise. Products are expanded bilinearly, and terms are simplified using this table.

These rules are not ordinary identities between infinitesimal real numbers. In particular, \(dB_t\,dB_t=dt\) encodes the quadratic variation \([B]_t=t\), while \(dB_t\,dt=0\) and \(dt\,dt=0\) encode the disappearance of the corresponding mixed and finite-variation quadratic terms in limits of partition sums. The table keeps the surviving second-order terms; it does not replace the definition of an Ito integral.

Multiplying Ito Differentials

Proposition: Suppose \(X\) and \(Y\) are Ito processes driven by the same \(d\)-dimensional Brownian motion, with differentials \(dX_t=a_t\,dt+\sum_{k=1}^d\sigma_{k,t}\,dB_t^k\) and \(dY_t=b_t\,dt+\sum_{k=1}^d\tau_{k,t}\,dB_t^k\). Then their differential product is \(dX_t\,dY_t=\sum_{k=1}^d\sigma_{k,t}\tau_{k,t}\,dt\). In particular, the drift coefficients do not appear in this product.

Proof. Expand bilinearly. The drift-drift term is \(a_tb_t\,dt\,dt=0\). Each drift-diffusion term is a multiple of \(dt\,dB_t^k\) or \(dB_t^k\,dt\), hence is zero. The remaining terms are \(\sum_{i,j}\sigma_{i,t}\tau_{j,t}\,dB_t^i\,dB_t^j\). By the multiplication table, \(dB_t^i\,dB_t^j=\delta_{ij}dt\), so this sum is \(\sum_{i,j}\sigma_{i,t}\tau_{j,t}\delta_{ij}dt=\sum_k\sigma_{k,t}\tau_{k,t}dt\). This agrees with the quadratic-covariation formula \([X,Y]_t=\int_0^t\sum_k\sigma_{k,s}\tau_{k,s}\,ds\), the standard covariation result for Ito processes. \(\square\)

The proposition gives a short calculation rule: retain the diffusion coefficients, multiply matching Brownian components, and sum. The result is the instantaneous covariation density. For \(X=Y\), it gives \(dX_t\,dX_t=\sum_k\sigma_{k,t}^2\,dt\), the density of \([X]_t\). Cross terms matter when two processes share Brownian drivers; terms from distinct independent components vanish.

Worked Example: Two Processes Driven by One Brownian Motion

Let \(B\) be a standard Brownian motion on the filtered space above, and define \(X_t=\tfrac12t+2B_t\) and \(Y_t=-t+3B_t\). The constant integrands \(2\) and \(3\) are progressively measurable and satisfy \(\mathbb{E}\int_0^T 2^2\,ds=4T<\infty\) and \(\mathbb{E}\int_0^T 3^2\,ds=9T<\infty\). Compute the three differential products.

The differentials are \(dX_t=\tfrac12dt+2dB_t\) and \(dY_t=-dt+3dB_t\). For \(X\), expand every term:

(dX_t)^2=\left(\tfrac12dt\right)^2+2\left(\tfrac12dt\right)(2dB_t)+(2dB_t)^2 =0+0+4dt=4dt.

For \(Y\), the drift square and drift-diffusion terms vanish, leaving \((dY_t)^2=3^2dt=9dt\). For the cross product, the only surviving term is \((2dB_t)(3dB_t)=6dt\), so \(dX_t\,dY_t=6dt\).

Check: the diffusion coefficients are \(2\) and \(3\), so their squared values are \(4\) and \(9\), and their product is \(2\cdot3=6\). These reproduce all three results. The drift coefficients \(\tfrac12\) and \(-1\) contribute to none of these quadratic terms.

From Products to a Second-Order Drift Term

The multiplication table is especially useful inside a second-order Taylor expansion. For a twice continuously differentiable function \(f:\mathbb{R}^m\to\mathbb{R}\), the quadratic part of the local expansion at a state \(x\) is \(\tfrac12\sum_{i,j} f_{ij}(x)\,dX_t^i\,dX_t^j\), where \(f_{ij}\) denotes the second partial derivative with respect to coordinates \(i\) and \(j\). This tutorial computes that quadratic part only; the full time-dependent Ito formula is a separate result.

Proposition: If \(dX_t^i=a_t^i\,dt+\sum_{k=1}^d\sigma_{ik,t}\,dB_t^k\) for \(i=1,\ldots,m\), then the second-order Taylor contribution is \(\tfrac12\sum_{k=1}^d\left(\sum_{i,j=1}^m f_{ij}(X_t)\sigma_{ik,t}\sigma_{jk,t}\right)dt\). The stated formula assumes the displayed derivatives exist and the Ito coefficients satisfy the integrability conditions above.

Proof. Apply the product proposition to each pair \(X^i,X^j\): it gives \(dX_t^i\,dX_t^j=\sum_k\sigma_{ik,t}\sigma_{jk,t}\,dt\). Substitute this expression into the quadratic Taylor term and interchange the finite sums: \(\tfrac12\sum_{i,j}f_{ij}(X_t)\sum_k\sigma_{ik,t}\sigma_{jk,t}\,dt =\tfrac12\sum_k\sum_{i,j}f_{ij}(X_t)\sigma_{ik,t}\sigma_{jk,t}\,dt\). This is the claimed expression. \(\square\)

Worked Example: A Two-Variable Quadratic Correction

Use the processes in the first example and consider \(f(x,y)=x^2+3xy+2y^2\). Compute only the second-order Taylor contribution, not the full differential of \(f(X_t,Y_t)\).

The second partial derivatives are \(f_{xx}=2\), \(f_{xy}=f_{yx}=3\), and \(f_{yy}=4\). The products already computed are \(dX_t\,dX_t=4dt\), \(dX_t\,dY_t=dY_t\,dX_t=6dt\), and \(dY_t\,dY_t=9dt\). Therefore the quadratic contribution is

\tfrac12\left(2(4)+3(6)+3(6)+4(9)\right)dt =\tfrac12(8+18+18+36)dt=40dt.

Check: the common diffusion direction in \((X,Y)\) is \((2,3)\). The Hessian matrix is \(\begin{pmatrix}2&3\\3&4\end{pmatrix}\); multiplying it by \((2,3)^\mathsf{T}\) gives \((13,18)^\mathsf{T}\), and the dot product with \((2,3)\) is \(2(13)+3(18)=80\). Half of \(80\) is \(40\), matching the calculation.

Several Brownian Components and Cancellation

When the driving Brownian motion has several independent components, the cross product is a dot product of diffusion-coefficient vectors. In particular, two processes can each have nonzero quadratic variation while their mutual covariation is zero: their coefficient vectors may be orthogonal, or contributions from different shared components may cancel.

Worked Example: Cross Terms from Two Independent Drivers

Let \((B^1,B^2)\) be a two-dimensional Brownian motion and set \(X_t=\tfrac14t+2B_t^1+B_t^2\), \(Y_t=-\tfrac12t+B_t^1-2B_t^2\). The constant integrands \(2,1,1,-2\) are progressive and square-integrable on every finite interval, since the integral of each squared constant over \([0,T]\) is finite. Find the differential products.

The differentials have coefficient vectors \((2,1)\) and \((1,-2)\). Thus \(dX_t\,dX_t=(2^2+1^2)dt=5dt\), and \(dY_t\,dY_t=(1^2+(-2)^2)dt=5dt\). For the cross term, match components rather than multiplying the coefficient vectors in every possible pairing:

dX_t\,dY_t=(2\cdot1+1\cdot(-2))dt=(2-2)dt=0.

Check: expanding the Brownian terms gives \(2dB^1dB^1-4dB^1dB^2+dB^2dB^1-2dB^2dB^2\). The two mixed products are zero, and the remaining terms are \(2dt-2dt=0\). The drift terms vanish as well, confirming the cross product.

Time-Varying Coefficients and the Role of Drift

Worked Example: Squaring a Time-Varying Ito Differential

Define \(X_t=\int_0^t(1+s)\,ds+\int_0^t s\,dB_s\) on \([0,2]\). The drift \(1+s\) is integrable and the progressive integrand \(s\) satisfies \(\mathbb{E}\int_0^2s^2\,ds=8/3<\infty\). Compute the differential square and the quadratic variation over this interval.

At time \(t\), \(dX_t=(1+t)dt+t\,dB_t\). Bilinear expansion gives \((dX_t)^2=(1+t)^2dt\,dt+2t(1+t)dt\,dB_t+t^2dB_t\,dB_t=t^2dt\). Integrating the quadratic-variation density from \(0\) to \(2\) gives \([X]_2=\int_0^2t^2\,dt=[t^3/3]_0^2=8/3\).

Check: the only diffusion coefficient is \(t\), so the quadratic-variation formula gives \([X]_2=\int_0^2t^2\,dt=8/3\) directly. The drift coefficient \(1+t\) does not enter. This second calculation confirms both the differential square and its integrated value.

Common Pitfall: Squaring the Drift as if It Survived

A tempting but incorrect calculation is to treat all terms in an Ito differential as ordinary algebraic increments. For \(dX_t=2dt+3dB_t\), one might write \((dX_t)^2=4dt+12dt\,dB_t+9dt\), interpreting \((2dt)^2\) as \(4dt\). This is wrong: \((2dt)^2=4dt\,dt=0\), and the mixed term is \(12dt\,dB_t=0\). The correct result is \((dX_t)^2=9(dB_t)^2=9dt\). The drift still affects the first-order change in \(X\); it is only absent from this quadratic product.

Another error is to set \((dB_t)^2=0\) because a single Brownian increment tends to zero as the time step shrinks. The relevant quantity is the sum of squared increments: it converges to elapsed time, not zero. Thus \(dB_t\,dB_t=dt\) records quadratic variation, whereas \(dB_t\,dt=0\) and \(dt\,dt=0\) record vanishing quadratic terms.

Product	Rule	What survives
Drift with drift	\(dt\,dt=0\)	Nothing
Drift with Brownian increment	\(dt\,dB_t^k=0\)	Nothing
Matching Brownian components	\(dB_t^k\,dB_t^k=dt\)	Coefficient product times \(dt\)
Distinct Brownian components	\(dB_t^i\,dB_t^j=0\) for \(i\ne j\)	Nothing

Check Your Understanding

For each question, distinguish drift terms from diffusion terms and apply the table component by component.

If \(dU_t=3dt-2dB_t\), what is \(dU_t\,dU_t\), and which terms vanish?
For \(dU_t=dt+2dB_t^1+dB_t^2\) and \(dV_t=4dt-dB_t^1+3dB_t^2\), calculate \(dU_t\,dV_t\).
For a scalar process with diffusion coefficients \(\sigma_{1,t},\ldots,\sigma_{d,t}\), write its differential square.
Explain why two processes can each have nonzero quadratic variation but zero mutual covariation.
For \(f(x,y)=x^2+y^2\), identify the second-order Taylor contribution when \(dX_t=dB_t^1\) and \(dY_t=dB_t^2\).