Cross-Variation in Ito Product-Rule Calculations

Why a Product Rule Has a Cross Term

For differentiable processes, the product rule is $d(XY)=X\,dY+Y\,dX$. For Ito processes driven by the same Brownian motion, this misses a term: the product of their Brownian increments contributes at first order in time. Its sign depends on the signs of both diffusion coefficients. Keeping track of that term is essential when deriving a product identity or checking a stochastic calculation.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$. On a fixed finite interval $[0,T]$, an Ito integral $\int_0^t H_s\,dB_s$ is defined for a progressively measurable integrand $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, all integrals below are Ito integrals, not Stratonovich integrals.

Definition: Let $X$ and $Y$ be continuous Ito processes driven by this same $B$: $$ dX_t=a_t\,dt+b_t\,dB_t,\qquad dY_t=c_t\,dt+d_t\,dB_t. $$ Here $a,b,c,d$ are progressively measurable, the drift coefficients $a,c$ and diffusion coefficients $b,d$ satisfy $\mathbb{E}\int_0^T(a_t^2+b_t^2+c_t^2+d_t^2)\,dt<\infty$, and $X_0,Y_0$ are square-integrable. The quadratic covariation $[X,Y]_t$ is the second-order term in the Ito product formula; equivalently, it is the probability limit of sums of products of increments over partitions whose mesh tends to zero.

For the product formula stated below, also assume $\mathbb{E}\int_0^T(X_t^2d_t^2+Y_t^2b_t^2)\,dt<\infty$. This ensures that the stochastic integrands in the formula belong to $\mathcal{L}^2$. These assumptions are sufficient rather than necessary; local versions can be obtained by stopping the processes.

The Product Rule and Quadratic Covariation

Theorem: Under the assumptions above, for $0\le t\le T$, $$ [X,Y]_t=\int_0^t b_s d_s\,ds $$ and $$ X_tY_t=X_0Y_0+\int_0^t(X_sc_s+Y_sa_s+b_sd_s)\,ds +\int_0^t X_s d_s\,dB_s+\int_0^t Y_s b_s\,dB_s. $$ Equivalently, $d(X_tY_t)=X_t\,dY_t+Y_t\,dX_t+d[X,Y]_t$.

Proof. Apply the two-variable Ito formula to $f(x,y)=xy$ and the process $(X_t,Y_t)$. Its derivatives are $f_x=y$, $f_y=x$, $f_{xx}=0$, $f_{yy}=0$, and $f_{xy}=1$. The second-order Ito term is therefore $f_{xy}\,d[X,Y]_t$. The first-order terms are $Y_t\,dX_t+X_t\,dY_t$. Since the Brownian parts of $X$ and $Y$ are $b_t\,dB_t$ and $d_t\,dB_t$, their cross-variation is $b_td_t\,dt$: the multiplication rules are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. Thus $[X,Y]_t=\int_0^t b_sd_s\,ds$. Substituting the two SDEs into the first-order terms gives the displayed formula. The drift integrals are integrable by Cauchy–Schwarz and the stated square-integrability assumptions; $X_sd_s$ and $Y_sb_s$ are in $\mathcal{L}^2$ by the additional assumption. Hence both Ito integrals are well-defined. $\square$

The term $b_td_t\,dt$ is the signed cross-variation. It is positive when the diffusion coefficients have the same sign, negative when they have opposite signs, and zero if either coefficient is zero. In contrast, the drift coefficients do not enter $[X,Y]$. This is a statement about quadratic covariation, not ordinary multiplication of differential symbols: it comes from the shared Brownian increment and the rule $dB_t\,dB_t=dt$.

Worked Example: Two Processes with Positive Cross-Variation

Let $X_t=2+B_t$ and $Y_t=3+2B_t$. Compute their product using the Ito product rule, then check it against direct multiplication.

The drift coefficients are $a_t=c_t=0$, and the diffusion coefficients are $b_t=1$, $d_t=2$. Thus $[X,Y]_t=\int_0^t 1\cdot2\,ds=2t$. The product rule gives $$ d(X_tY_t)=2X_t\,dB_t+Y_t\,dB_t+2\,dt=(2X_t+Y_t)\,dB_t+2\,dt. $$ Since $2X_t+Y_t=2(2+B_t)+(3+2B_t)=7+4B_t$, integration yields $$ X_tY_t=6+\int_0^t(7+4B_s)\,dB_s+2t. $$

To evaluate the Ito integral, apply Ito's formula to $f(x)=x^2$: $f'(x)=2x$, $f''(x)=2$, so $d(B_t^2)=2B_t\,dB_t+\tfrac12(2)\,dt=2B_t\,dB_t+dt$. Consequently $\int_0^t B_s\,dB_s=\tfrac12(B_t^2-t)$, and $$ \int_0^t(7+4B_s)\,dB_s=7B_t+4\cdot\tfrac12(B_t^2-t)=7B_t+2B_t^2-2t. $$ Substitution gives $X_tY_t=6+7B_t+2B_t^2$. Direct multiplication independently gives $(2+B_t)(3+2B_t)=6+7B_t+2B_t^2$, as required.

Check: the positive correction $2t$ in the product-rule expression cancels the $-2t$ generated when evaluating the Ito integral. Omitting the cross term would leave an erroneous $-2t$ in the final product.

Worked Example: Negative Cross-Variation and a Direct Ito Check

Set $X_t=t+B_t$ and $Y_t=2t-3B_t$. Find the differential of $X_tY_t$ by the product rule and verify it by applying Ito's formula directly.

Here $a_t=1$, $b_t=1$, $c_t=2$, and $d_t=-3$, so $[X,Y]_t=\int_0^t(-3)\,ds=-3t$. The product rule gives drift $X_tc_t+Y_ta_t+b_td_t=2(t+B_t)+(2t-3B_t)-3=4t-B_t-3$, and diffusion coefficient $X_td_t+Y_tb_t=-3(t+B_t)+(2t-3B_t)=-t-6B_t$. Therefore $$ d(X_tY_t)=(4t-B_t-3)\,dt+(-t-6B_t)\,dB_t. $$

For an independent check, write $X_tY_t=F(t,B_t)$, where $F(t,x)=(t+x)(2t-3x)=2t^2-tx-3x^2$. Its derivatives are $F_t=4t-x$, $F_x=-t-6x$, and $F_{xx}=-6$. Ito's formula, including its correction, gives $$ dF(t,B_t)=F_t\,dt+F_x\,dB_t+\tfrac12F_{xx}\,dt =(4t-B_t-3)\,dt+(-t-6B_t)\,dB_t, $$ which matches the product-rule calculation. Direct expansion also gives $X_tY_t=2t^2-tB_t-3B_t^2$; at $t=0$, both the product and the integrated formula start at zero.

Check: the diffusion coefficients have opposite signs, so the cross-variation is negative. Its contribution is $-3\,dt$, not $+3\,dt$; changing that sign would change the drift from $4t-B_t-3$ to $4t-B_t+3$, contradicting the direct Ito calculation.

When the Cross Term Vanishes—and When It Does Not

If $X$ has no Brownian component, then $b_t=0$, and its quadratic covariation with $Y$ is zero. More generally, the cross term vanishes when $b_td_t=0$ for almost every time, almost surely. The fact that the drift of one process is zero does not make the cross term vanish: it is the diffusion coefficients, not the drift coefficients, that determine it.

Worked Example: Detecting an Omitted Cross Term

Take $X_t=B_t$ and $Y_t=-B_t$. The product rule gives $d(X_tY_t)=-2B_t\,dB_t-dt$, because $b_t=1$ and $d_t=-1$, so $d[X,Y]_t=-dt$. Check the result directly.

For $g(x)=-x^2$, $g'(x)=-2x$ and $g''(x)=-2$. Ito's formula gives $$ d(-B_t^2)=(-2B_t)\,dB_t+\tfrac12(-2)\,dt=-2B_t\,dB_t-dt, $$ as the product rule requires. Integrating and using the earlier identity $\int_0^tB_s\,dB_s=\tfrac12(B_t^2-t)$ gives $$ -2\int_0^tB_s\,dB_s-t=-(B_t^2-t)-t=-B_t^2=X_tY_t. $$

Check: the Ito integral $\int_0^tB_s\,dB_s$ is square-integrable on $[0,T]$, since $\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty$. Ito integrals of $\mathcal{L}^2$ integrands are true martingales: their increments have conditional expectation zero, first for adapted simple integrands by independence and the zero mean of Brownian increments, and then for $\mathcal{L}^2$ integrands by the Ito isometry and $L^2$ approximation. Thus $\mathbb{E}\int_0^tB_s\,dB_s=0$, and the integrated identity gives $\mathbb{E}[X_tY_t]=-t$. Directly, $\mathbb{E}[-B_t^2]=-t$, confirming the correction and its sign.

Common Errors in Product-Rule Calculations

Omitting the cross term. A common incorrect calculation is $d(B_t^2)=2B_t\,dB_t$, obtained by applying the ordinary chain rule to $x^2$. The correction is $\tfrac12 f''(B_t)\,dt=\tfrac12(2)\,dt=dt$, so the Ito identity is $d(B_t^2)=2B_t\,dB_t+dt$. This is the special case $X=Y=B$, for which $[B,B]_t=t$.

Using the wrong sign. The cross term is the product $b_td_t\,dt$, not $|b_td_t|\,dt$. For $X=B$, $Y=-B$, the product is $-dt$. Replacing it with $+dt$ conflicts with $XY=-B^2$ and its direct Ito differential.

Confusing Ito and Stratonovich products. The formulas here use Ito integrals. For continuous semimartingales, the Stratonovich product rule has the ordinary form $d(XY)=X\circ dY+Y\circ dX$; the correction is incorporated when converting between Stratonovich and Ito integrals. It must not be dropped from the Ito product rule. Under only local integrability, the stochastic-integral terms may be local martingales rather than true martingales, so their expectations cannot automatically be set to zero. The $\mathcal{L}^2$ assumptions above ensure true martingale terms when that property is used.

The practical check is to write both processes with their drift and diffusion coefficients before multiplying. The drift of the product contains $Xc+Ya+bd$, while its diffusion coefficient is $Xd+Yb$. Checking those two coefficients separately catches many omissions and sign errors.

Check Your Understanding

In each question, assume the processes are defined on a finite interval and satisfy sufficient integrability for the stated Ito integrals.

If $dX_t=2\,dt+3\,dB_t$ and $dY_t=-dt+4\,dB_t$, what is $d[X,Y]_t$?
Write the drift and diffusion coefficients of $d(X_tY_t)$ when $dX_t=a_t\,dt+b_t\,dB_t$ and $dY_t=c_t\,dt+d_t\,dB_t$.
For $X_t=B_t$ and $Y_t=5-B_t$, determine the sign and value of the cross-variation over $[0,t]$.
Explain why a zero drift coefficient for $X$ does not imply $[X,Y]=0$.
For $X_t=t+B_t$ and $Y_t=2t-3B_t$, identify the term in the product differential that would be wrong if the cross-variation were omitted.

Cross-Variation in Ito Product-Rule Calculations

What You'll Learn