Deriving the Ito Product Rule

Why Multiplying Ito Processes Produces an Extra Term

For differentiable functions of time, the product rule has two first-order terms. Ito processes have those terms too, but their increments also contain Brownian fluctuations of order $\sqrt{\Delta t}$. Multiplying two such fluctuations produces a term of order $\Delta t$, and summing those products over a partition can have a nonzero limit. The Ito product rule records that limit as quadratic covariation.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a $d$-dimensional Brownian motion $\mathbf{B}_t$. A vector integrand $H$ is in $\mathcal{L}^2$ on $[0,T]$ if it is progressively measurable and $\mathbb{E}\int_0^T\|H_s\|^2\,ds<\infty$; its Ito integral is defined in the square-integrable sense. In the result below, assume $X$ and $Y$ are continuous adapted processes, bounded by a deterministic constant on $[0,T]$, with representations

X_t=X_0+\int_0^t a_s\,ds+\int_0^t\boldsymbol{\sigma}_s\cdot d\mathbf{B}_s,\qquad Y_t=Y_0+\int_0^t b_s\,ds+\int_0^t\boldsymbol{\tau}_s\cdot d\mathbf{B}_s.

Here $a,b$ are progressively measurable real-valued processes and $\boldsymbol{\sigma},\boldsymbol{\tau}$ are progressively measurable $\mathbb{R}^d$-valued processes. Assume $\mathbb{E}\int_0^T(a_s^2+b_s^2+\|\boldsymbol{\sigma}_s\|^2+\|\boldsymbol{\tau}_s\|^2)\,ds<\infty$. Boundedness of $X,Y$ ensures that $X\boldsymbol{\tau}$ and $Y\boldsymbol{\sigma}$ are in $\mathcal{L}^2$, so the stochastic integrals in the product rule below are well-defined without an unstated fourth-moment assumption.

Definition: For these Ito processes, their quadratic covariation is $[X,Y]_t=\int_0^t\boldsymbol{\sigma}_s\cdot\boldsymbol{\tau}_s\,ds$. The dot product pairs the diffusion coefficients driven by the same Brownian components.

The Brownian multiplication conventions are $dB_t^i\,dB_t^j=\mathbf{1}_{\{i=j\}}dt$, $dB_t^i\,dt=0$, and $dt\,dt=0$. In particular, the ordinary chain rule omits a contribution that need not vanish: the product of the two diffusion increments.

The Second-Order Expansion

For real numbers $x,y,\Delta x,\Delta y$, direct multiplication gives the exact identity

(x+\Delta x)(y+\Delta y)-xy =y\,\Delta x+x\,\Delta y+\Delta x\,\Delta y.

This is also the second-order Taylor expansion of $f(x,y)=xy$, with no remainder: $f_x=y$, $f_y=x$, and its Hessian is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Thus the quadratic Taylor contribution is $\tfrac12(2\Delta x\Delta y)=\Delta x\Delta y$. For Ito processes, the first two terms lead to stochastic and ordinary integrals. The last term leads to $[X,Y]$, not generally to zero.

Standard covariation result: For continuous Ito processes with the representations above, sums of products of increments over deterministic partitions of $[0,t]$, as the mesh tends to zero, converge in probability to $\int_0^t\boldsymbol{\sigma}_s\cdot\boldsymbol{\tau}_s\,ds$. This is the quadratic-covariation result for Ito processes established in the earlier material on quadratic variation and covariation.

For intuition, on a small interval of length $\Delta t$, the diffusion parts of the increments are approximately $\boldsymbol{\sigma}\cdot\Delta\mathbf{B}$ and $\boldsymbol{\tau}\cdot\Delta\mathbf{B}$. Their product has conditional mean approximately $(\boldsymbol{\sigma}\cdot\boldsymbol{\tau})\Delta t$. The covariation result makes this accumulated second-order contribution precise; it does not assert that an individual increment product equals its mean.

Theorem: Under the boundedness and integrability assumptions stated above, for every $t\in[0,T]$, almost surely, $$ X_tY_t=X_0Y_0+\int_0^t X_s b_s\,ds+\int_0^t Y_s a_s\,ds +\int_0^t X_s\boldsymbol{\tau}_s\cdot d\mathbf{B}_s +\int_0^t Y_s\boldsymbol{\sigma}_s\cdot d\mathbf{B}_s +[X,Y]_t. $$

Proof. Fix $t$, and take a deterministic partition $0=t_0<\cdots<t_n=t$. Write $\Delta X_i=X_{t_i}-X_{t_{i-1}}$ and similarly for $Y$. The exact product expansion gives

X_tY_t-X_0Y_0 =\sum_{i=1}^n X_{t_{i-1}}\Delta Y_i +\sum_{i=1}^n Y_{t_{i-1}}\Delta X_i +\sum_{i=1}^n\Delta X_i\Delta Y_i.

As the mesh tends to zero, the first sum converges to $\int_0^t X_s b_s\,ds+\int_0^t X_s\boldsymbol{\tau}_s\cdot d\mathbf{B}_s$, and the second to $\int_0^t Y_s a_s\,ds+\int_0^t Y_s\boldsymbol{\sigma}_s\cdot d\mathbf{B}_s$. To justify these limits, the left-endpoint step processes converge pointwise to $X_s,Y_s$ by continuity and are bounded by the deterministic bound on the processes. Dominated convergence gives convergence of the drift integrals in $L^1$, since $\mathbb{E}\int_0^T(|a_s|+|b_s|)\,ds<\infty$. For the stochastic integrals, the Ito isometry and dominated convergence give $L^2$ convergence: for example, the squared error for the integrand $X\boldsymbol{\tau}$ is bounded by a constant times $\|\boldsymbol{\tau}_s\|^2$, whose expectation integrated over time is finite.

These stochastic integrals are square-integrable martingales: their integrands are in $\mathcal{L}^2$, so the Ito isometry gives finite second moments, and the martingale property follows first for adapted step integrands from the independent, mean-zero Brownian increments and then for $\mathcal{L}^2$ integrands by approximation and conditional-expectation convergence. Finally, the third sum converges in probability to $[X,Y]_t$ by the stated quadratic-covariation result. Taking limits in the exact identity proves the formula: the left side is fixed, and limits in probability are unique up to almost-sure equality. $\square$

The proof identifies precisely where the correction enters: it is the limit of $\sum_i\Delta X_i\Delta Y_i$. If the processes have no shared Brownian noise, the dot product of their diffusion coefficients is zero and this covariation term vanishes. If they share noise, it may be nonzero, positive, or negative.

Worked Applications and Independent Checks

Worked Example: Two Bounded Functions of the Same Brownian Motion

Let $B_t$ be a one-dimensional Brownian motion and set $X_t=2+\sin B_t$, $Y_t=3+\cos B_t$. Both processes are bounded. Apply Ito's formula to each: for $g(x)=\sin x$, $g'(x)=\cos x$ and $g''(x)=-\sin x$; for $h(x)=\cos x$, $h'(x)=-\sin x$ and $h''(x)=-\cos x$. Hence

dX_t=-\tfrac12\sin B_t\,dt+\cos B_t\,dB_t,\qquad dY_t=-\tfrac12\cos B_t\,dt-\sin B_t\,dB_t.

The diffusion coefficients are $\cos B_t$ and $-\sin B_t$, so $d[X,Y]_t=-\sin B_t\cos B_t\,dt$. Substitution in the product rule gives diffusion coefficient $-(2+\sin B_t)\sin B_t+(3+\cos B_t)\cos B_t$, which equals $-2\sin B_t-\sin^2B_t+3\cos B_t+\cos^2B_t$. Its drift is

-\tfrac12(2+\sin B_t)\cos B_t -\tfrac12(3+\cos B_t)\sin B_t -\sin B_t\cos B_t =-\cos B_t-\tfrac32\sin B_t-2\sin B_t\cos B_t.

Check: Directly, $X_tY_t=6+3\sin B_t+2\cos B_t+\sin B_t\cos B_t$. For $q(x)=6+3\sin x+2\cos x+\sin x\cos x$, $q'(x)=3\cos x-2\sin x+\cos^2x-\sin^2x$ and $q''(x)=-3\sin x-2\cos x-4\sin x\cos x$. Ito's formula gives diffusion $q'(B_t)$, equal to the coefficient above, and drift $\tfrac12q''(B_t)=-\tfrac32\sin B_t-\cos B_t-2\sin B_t\cos B_t$, also equal to the product-rule drift.

Worked Example: A Deterministic Factor Has No Covariation

On a fixed finite interval, let $X_t=t$ and $Y_t=\sin B_t$. Here $a_t=1$, $\boldsymbol{\sigma}_t=0$, $b_t=-\tfrac12\sin B_t$, and $\boldsymbol{\tau}_t=\cos B_t$. The covariation is zero because one diffusion coefficient is zero. The product rule gives

d(t\sin B_t)=\left(\sin B_t-\tfrac12t\sin B_t\right)dt+t\cos B_t\,dB_t.

Check: apply the time-dependent Ito formula to $f(t,x)=t\sin x$. Its derivatives are $f_t=\sin x$, $f_x=t\cos x$, and $f_{xx}=-t\sin x$; the correction is $\tfrac12 f_{xx}dt=-\tfrac12t\sin x\,dt$. The resulting drift and diffusion match the product-rule calculation. On $[0,T]$, $X$ is bounded by $T$, and $Y$ is bounded by $1$, so the stated theorem's boundedness condition holds.

Worked Example: Independent Brownian Drivers

Let $\mathbf{B}_t=(B_t^1,B_t^2)$ have independent standard Brownian components, and take $X_t=\sin B_t^1$, $Y_t=\cos B_t^2$. Ito's formula gives $dX_t=-\tfrac12\sin B_t^1\,dt+\cos B_t^1\,dB_t^1$ and $dY_t=-\tfrac12\cos B_t^2\,dt-\sin B_t^2\,dB_t^2$. Their diffusion vectors are $(\cos B_t^1,0)$ and $(0,-\sin B_t^2)$; their dot product is zero. Therefore

d(X_tY_t)=-\sin B_t^1\cos B_t^2\,dt +\cos B_t^1\cos B_t^2\,dB_t^1 -\sin B_t^1\sin B_t^2\,dB_t^2.

Check: for $f(x,y)=\sin x\cos y$, $f_x=\cos x\cos y$, $f_y=-\sin x\sin y$, $f_{xx}=-\sin x\cos y$, and $f_{yy}=-\sin x\cos y$. The two-dimensional Ito correction is $\tfrac12(f_{xx}+f_{yy})dt=-\sin x\cos y\,dt$, confirming the drift and both diffusion coefficients. The two paths are individually rough, but their shared-noise covariation is zero because they use independent Brownian components.

Common Mistake: Dropping the Increment Product

A tempting but incorrect calculation is to write $\Delta(XY)\approx X\Delta Y+Y\Delta X$ and discard $\Delta X\Delta Y$ because each increment tends to zero. Smallness of each term does not imply smallness of their sum. For $X=Y=B$, the exact identity is $\Delta(B^2)=2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2$. Summing the omitted terms gives $\sum_i(\Delta B_i)^2$, which converges in probability to $t$, not zero. The correct rule is therefore $d(B_t^2)=2B_t\,dB_t+dt$. Independently, Ito's formula for $f(x)=x^2$, with $f'(x)=2x$ and $f''(x)=2$, gives $2B_t\,dB_t+\tfrac12(2)dt$, the same result.

The ordinary chain and product rules remain valid for finite-variation terms, but Brownian diffusion terms have nonzero quadratic variation: $dB_t\,dB_t=dt$. Cross terms $dB_t\,dt$ and $dt\,dt$ vanish. For two processes, the relevant second-order term is the dot product of their diffusion coefficients, not a generic correction independent of their drivers.

Check Your Understanding

Use the increment expansion and the diffusion coefficients to justify each answer.

For $X_t=B_t$ and $Y_t=3B_t$, identify the covariation term in $d(X_tY_t)$.
For scalar Brownian-driven processes with diffusion coefficients $\sigma_t$ and $\tau_t$, what is the density of their quadratic covariation with respect to time?
Why does boundedness of $X,Y$, together with square-integrable diffusion coefficients, ensure that the stochastic integrands $X\boldsymbol{\tau}$ and $Y\boldsymbol{\sigma}$ are square-integrable?
For $X_t=\sin B_t$ and $Y_t=\cos B_t$, calculate the sign of their covariation term.
Explain why the fact that $\Delta X_i\Delta Y_i\to0$ for each shrinking interval does not by itself imply that their sum tends to zero.

Deriving the Ito Product Rule

What You'll Learn

Why Multiplying Ito Processes Produces an Extra Term

The Second-Order Expansion