Why the Product Rule Has an Extra Term
For differentiable functions of time, the product rule has two terms: \(d(XY)=X\,dY+Y\,dX\). For Ito processes, fluctuations in both factors can occur on the same time interval. The product of those fluctuations contributes a further term, their quadratic covariation \(d[X,Y]\). Its value depends on the processes’ diffusion coefficients: drift terms do not contribute to this covariation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard \(m\)-dimensional Brownian motion \(\mathbf{B}_t=(B_t^1,\ldots,B_t^m)\) with independent components. Consider continuous Ito processes of the form \(dX_t=a_t\,dt+\boldsymbol{\beta}_t\cdot d\mathbf{B}_t\) and \(dY_t=c_t\,dt+\boldsymbol{\gamma}_t\cdot d\mathbf{B}_t\). Here \(a_t,c_t\) are progressively measurable drift coefficients and \(\boldsymbol{\beta}_t,\boldsymbol{\gamma}_t\) are progressively measurable diffusion vectors. On each finite interval \([0,T]\), assume \(\mathbb{E}\int_0^T(|a_t|+|c_t|)\,dt<\infty\) and \(\mathbb{E}\int_0^T(\|\boldsymbol{\beta}_t\|^2+\|\boldsymbol{\gamma}_t\|^2)\,dt<\infty\), and take \(X_0,Y_0\) to be finite. These conditions define the drift integrals and place each Brownian integrand in \(\mathcal{L}^2\).
In particular, if \(X\) and \(Y\) share one Brownian driver, their covariation rate is the product of their diffusion coefficients. If they are driven by different independent components, that rate is zero. With correlated drivers, it is the covariance-weighted product of their coefficients; the multidimensional representation above encodes this by taking the corresponding vectors’ dot product.
The Ito Product Rule
Proof. Apply the multidimensional Ito formula to the function \(f(x,y)=xy\) and the two-dimensional Ito process \((X_t,Y_t)\). Its partial derivatives are \(f_x=y\), \(f_y=x\), \(f_{xx}=0\), \(f_{yy}=0\), and \(f_{xy}=f_{yx}=1\). The first-order terms are \(Y_t\,dX_t+X_t\,dY_t\). The second-order correction is \(\tfrac12(f_{xx}\,d[X]_t+2f_{xy}\,d[X,Y]_t+f_{yy}\,d[Y]_t)=d[X,Y]_t\). Substituting the drift and diffusion forms gives the stated drift, while the Brownian terms combine to \((X_t\boldsymbol{\gamma}_t+Y_t\boldsymbol{\beta}_t)\cdot d\mathbf{B}_t\). The formula holds locally if these product integrands are not globally square-integrable: stop the processes when their magnitudes or accumulated integrals reach finite bounds, apply Ito’s formula to the stopped processes, and then let the stopping levels increase. If the product diffusion vector is in \(\mathcal{L}^2\), its Ito integral is a true martingale: its expectation is finite by the Ito isometry, and its conditional expectation at a later time given \(\mathcal{F}_s\) equals its value at \(s\), because the Ito integral over \([s,t]\) has conditional expectation zero. Without that integrability, a local martingale need not be a true martingale. \(\square\)
The rule can also be checked using the differential multiplication table. Multiplying \(dX_t=a_tdt+\boldsymbol{\beta}_t\cdot d\mathbf{B}_t\) by \(dY_t=c_tdt+\boldsymbol{\gamma}_t\cdot d\mathbf{B}_t\), all terms containing \(dt\,dt\) or \(dt\,dB_t^i\) vanish, and the Brownian products give \(\boldsymbol{\beta}_t\cdot\boldsymbol{\gamma}_t\,dt\). Thus \(dX_t\,dY_t=d[X,Y]_t\). This multiplication is shorthand for the quadratic-covariation calculation; it does not assert that a single Brownian increment squared equals the time increment.
Worked Calculations
Worked Example: Opposite Brownian Diffusions
Let \(X_t=1+B_t\) and \(Y_t=2-B_t\). Find the drift and diffusion of \(X_tY_t\), then verify them by applying Ito’s formula to the product as a function of \(B_t\).
The coefficients are \(a_t=c_t=0\), \(\beta_t=1\), and \(\gamma_t=-1\). Therefore \(d[X,Y]_t=(1)(-1)dt=-dt\). The product-rule drift is \(X_tc_t+Y_ta_t+\beta_t\gamma_t=-1\), and its diffusion coefficient is \(X_t\gamma_t+Y_t\beta_t=-X_t+Y_t\). Substitution gives \(-(1+B_t)+(2-B_t)=1-2B_t\). Hence \(d(X_tY_t)=-dt+(1-2B_t)dB_t\).
For an independent check, write \(g(x)=(1+x)(2-x)=2+x-x^2\). Then \(g'(x)=1-2x\) and \(g''(x)=-2\). Ito’s formula gives \(dg(B_t)=g'(B_t)dB_t+\tfrac12g''(B_t)dt=(1-2B_t)dB_t-dt\), matching the product-rule calculation. The Ito integral is a true martingale on \([0,T]\): \(\mathbb{E}\int_0^T(1-2B_t)^2dt=\int_0^T(1+4t)dt=T+2T^2<\infty\). Consequently \(\mathbb{E}[X_tY_t]=2-t\), also obtained directly from \(2+B_t-B_t^2\) and \(\mathbb{E}[B_t]=0\), \(\mathbb{E}[B_t^2]=t\).
Worked Example: Drift and Diffusion from Two Ito Processes
Let \(X_t=t+B_t\) and \(Y_t=2t-B_t\). Compute the product’s drift and diffusion, and check the result by expanding the product and applying Ito’s formula to each random term.
Here \(a_t=1\), \(\beta_t=1\), \(c_t=2\), and \(\gamma_t=-1\). The covariation rate is \(\beta_t\gamma_t=-1\). The drift is \(2X_t+Y_t-1=2(t+B_t)+(2t-B_t)-1=4t+B_t-1\). The diffusion coefficient is \(-X_t+Y_t=-(t+B_t)+(2t-B_t)=t-2B_t\). Thus \(d(X_tY_t)=(4t+B_t-1)dt+(t-2B_t)dB_t\).
For the check, expand \(X_tY_t=(t+B_t)(2t-B_t)=2t^2+tB_t-B_t^2\). The deterministic derivative of \(2t^2\) is \(4t\,dt\). For \(tB_t\), use \(f(t,x)=tx\): \(f_t=x\), \(f_x=t\), \(f_{tt}=f_{xx}=0\), and \(f_{tx}=1\). The time-Brownian cross term vanishes since \(dt\,dB_t=0\), so \(d(tB_t)=B_tdt+t\,dB_t\). For \(B_t^2\), use \(h(x)=x^2\): \(h'(x)=2x\), \(h''(x)=2\), and the correction is \(\tfrac12h''(B_t)dt=dt\). Hence \(d(B_t^2)=2B_t\,dB_t+dt\). Combining these three differentials gives \((4t+B_t-1)dt+(t-2B_t)dB_t\), as required.
The diffusion integrand \(t-2B_t\) is in \(\mathcal{L}^2\) on every finite interval because \(\mathbb{E}\int_0^T(t-2B_t)^2dt=\int_0^T(t^2+4t)dt=T^3/3+2T^2<\infty\). Thus its integral has mean zero, and the product’s expectation is determined by its initial value and drift.
Worked Example: Correlated Brownian Drivers
Let \(B_t^1,B_t^2\) be independent Brownian components and set \(X_t=B_t^1\), \(Y_t=\tfrac35B_t^1+\tfrac45B_t^2\). Compute the covariation and the differential of the product.
The diffusion vectors are \(\boldsymbol{\beta}=(1,0)\) and \(\boldsymbol{\gamma}=(\tfrac35,\tfrac45)\), with zero drifts. Their dot product is \(1\cdot\tfrac35+0\cdot\tfrac45=\tfrac35\), so \(d[X,Y]_t=\tfrac35dt\). The product’s drift is \(\tfrac35\); its diffusion vector is \(X_t\boldsymbol{\gamma}+Y_t\boldsymbol{\beta}=(Y_t+\tfrac35X_t,\tfrac45X_t)\). Therefore \(d(X_tY_t)=\tfrac35dt+(Y_t+\tfrac35X_t)dB_t^1+\tfrac45X_t\,dB_t^2\).
Check the covariation independently: \([B^1,B^1]_t=t\), \([B^1,B^2]_t=0\), so \([X,Y]_t=\tfrac35[B^1,B^1]_t+\tfrac45[B^1,B^2]_t=\tfrac35t\). The stochastic integrands are square-integrable on finite intervals: \(\mathbb{E}\int_0^T X_t^2dt=\int_0^T t\,dt=T^2/2\), and \(\mathbb{E}\int_0^T Y_t^2dt=\int_0^T((\tfrac35)^2+(\tfrac45)^2)t\,dt=T^2/2\). The diffusion coefficients in the displayed product are linear combinations of these processes, so they too are square-integrable on \([0,T]\). The stochastic integrals are therefore true martingales. In particular, \(\mathbb{E}[X_tY_t]=\tfrac35t\), consistent with \(\mathbb{E}[B_t^1B_t^1]=t\) and \(\mathbb{E}[B_t^1B_t^2]=0\).
What the Correction Does—and Does Not—Say
The correction can be positive, negative, or zero. It is positive when the instantaneous diffusion directions reinforce each other and negative when they oppose each other. If one process has finite variation, its diffusion vector is zero, so its covariation with the other process is zero. In the special case \(X=Y\), the rule gives \(d(X_t^2)=2X_t\,dX_t+d[X]_t\), including the familiar positive quadratic-variation correction for the square of a Brownian-driven process.
The Ito product rule is not the ordinary product rule with a negligible error. The term \(d[X,Y]_t\) records the non-vanishing limit of products of simultaneous Brownian increments. Nor should it be confused with the Stratonovich product rule: for continuous semimartingales, Stratonovich integration incorporates half the relevant covariation correction, and the product rule for Stratonovich differentials has the ordinary two-term form. Converting an Ito expression to Stratonovich form requires accounting for that correction; simply deleting \(d[X,Y]\) from an Ito calculation changes the result.
Common mistake. For \(X_t=1+B_t\) and \(Y_t=2-B_t\), omitting covariation gives the incorrect calculation \(d(X_tY_t)\overset{\text{wrong}}{=}X_t\,dY_t+Y_t\,dX_t=(1-2B_t)dB_t\). The two stochastic terms do combine to \((1-2B_t)dB_t\), but this leaves out \(d[X,Y]_t=-dt\). The corrected differential is \((1-2B_t)dB_t-dt\). The direct calculation \(X_tY_t=2+B_t-B_t^2\), with \(d(B_t^2)=2B_t\,dB_t+dt\), confirms the missing drift is \(-dt\).
Check Your Understanding
For each calculation, identify the drift, diffusion, and covariation terms separately.
- If \(dX_t=a_tdt+\beta_tdB_t\) and \(dY_t=c_tdt+\gamma_tdB_t\), write the drift and diffusion coefficients of \(X_tY_t\).
- For \(X_t=B_t\) and \(Y_t=3B_t\), compute \([X,Y]_t\) and \(\mathbb{E}[X_tY_t]\).
- For \(X_t=B_t^1\) and \(Y_t=B_t^2\) with independent Brownian components, determine the product drift and explain why the covariation is zero.
- State an integrability condition that makes the stochastic integral in the product formula a true martingale rather than merely a local martingale.
- Explain why \(dt\,dB_t=0\) does not justify dropping \(d[X,Y]_t\) when both processes have Brownian diffusion terms.