A Problem-Solving Workshop on Coupled Ito Systems

A Reliable Workflow for Coupled Ito Systems

For a coupled system, a transformation can mix the two processes in ways that are easy to miss by inspection. The dependable approach is to record the drift and diffusion coefficients first, compute the covariance of the process increments, differentiate the chosen function, and then collect the resulting drift and Brownian terms. In particular, correlated Brownian drivers contribute through the mixed second derivative; treating them as independent when they are not changes the answer.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying two Brownian motions $B^1_t,B^2_t$, adapted to the filtration, with constant correlation $\rho\in[-1,1]$. Thus their quadratic variations are $t$ and their covariation is $[B^1,B^2]_t=\rho t$. The endpoints $\rho=\pm1$ are allowed; the two drivers need not be independent. All stochastic integrals below are Ito integrals. On a fixed finite interval $[0,T]$, the integrands are progressively measurable and belong to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For locally square-integrable coefficients, the formula is applied up to stopping times that keep the coefficients bounded.

Consider continuous Ito processes satisfying

\begin{aligned} dX_t&=a_t\,dt+b_t\,dB^1_t,\\ dY_t&=c_t\,dt+e_t\,dB^2_t. \end{aligned}

The drift coefficients are $a_t,c_t$; the diffusion coefficients are $b_t,e_t$. The differential multiplication rules are $dB^1_t\,dB^1_t=dt$, $dB^2_t\,dB^2_t=dt$, $dB^1_t\,dB^2_t=\rho\,dt$, $dB^i_t\,dt=0$, and $dt\,dt=0$. In particular, $dX_t\,dY_t=\rho b_te_t\,dt$. This covariance is the entry that produces the mixed derivative correction.

Definition: For a twice continuously differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, the transformed process is $Z_t=f(X_t,Y_t)$. Its drift is the coefficient of $dt$ in its Ito differential; its diffusion terms are the coefficients multiplying the Brownian differentials.

The Coefficient-and-Covariance Formula

Result: If $f\in C^2(\mathbb{R}^2)$, then $$ df(X_t,Y_t)=\left(f_xa_t+f_yc_t+\tfrac12 f_{xx}b_t^2+\rho f_{xy}b_te_t+\tfrac12 f_{yy}e_t^2\right)dt+f_xb_t\,dB^1_t+f_ye_t\,dB^2_t, $$ where all derivatives are evaluated at $(X_t,Y_t)$. The formula holds locally for continuous Ito processes with locally square-integrable coefficients; if the displayed stochastic integrands are in $\mathcal{L}^2$ on $[0,T]$, the corresponding Ito integrals are defined on that interval.

Proof. The two-dimensional Ito formula gives the first-order terms $f_x\,dX_t+f_y\,dY_t$, together with one half of the Hessian applied to the quadratic covariations. Here $d[X]_t=b_t^2dt$, $d[Y]_t=e_t^2dt$, and $d[X,Y]_t=\rho b_te_tdt$. Thus the second-order contribution is $\tfrac12 f_{xx}b_t^2dt+f_{xy}\rho b_te_tdt+\tfrac12 f_{yy}e_t^2dt$; the factor of two in the off-diagonal Hessian terms cancels the overall one half. Substituting the two equations for $dX_t,dY_t$ into the first-order terms and collecting coefficients gives the displayed formula. This is the standard multidimensional Ito formula, established in the earlier material on that formula; the calculation here identifies each coefficient for the coupled system. $\square$

The workflow follows directly from this result. First write the two equations and their driver covariance. Next compute $f_x,f_y,f_{xx},f_{xy},f_{yy}$. Then place the first derivatives in the drift and diffusion terms, and place the second derivatives beside the corresponding covariances. The factor on the mixed term is $\rho b_te_t$, not $\tfrac12\rho b_te_t$: the two equal off-diagonal contributions in the Hessian have already combined.

Worked Transformations

Worked Example: Transforming a Product with Correlated Drivers

Suppose $dX_t=2\,dt+dB^1_t$ and $dY_t=-dt+2\,dB^2_t$, with $\rho=\tfrac12$, $X_0=1$, and $Y_0=3$. Find the Ito equation for $Z_t=X_tY_t$, including its drift and both diffusion terms.

Take $f(x,y)=xy$. Its derivatives are $f_x=y$, $f_y=x$, $f_{xx}=0$, $f_{xy}=1$, and $f_{yy}=0$. The coefficients are $a=2$, $b=1$, $c=-1$, and $e=2$. The drift is therefore $f_xa+f_yc+\rho f_{xy}be=2y-x+\tfrac12(1)(1)(2)=2y-x+1$. The diffusion terms are $y\,dB^1_t$ and $2x\,dB^2_t$, so

d(X_tY_t)=(2Y_t-X_t+1)\,dt+Y_t\,dB^1_t+2X_t\,dB^2_t.

At the initial point, the drift is $2(3)-1+1=6$, and the two diffusion coefficients are $3$ and $2$. As an independent check, the Ito product rule gives $d(XY)=X\,dY+Y\,dX+d[X,Y]$. Here $X\,dY$ contributes drift $-X$, $Y\,dX$ contributes drift $2Y$, and $d[X,Y]=1\cdot2\cdot\tfrac12\,dt=dt$. The resulting drift is $2Y-X+1$, as above.

Worked Example: Squared Radius with Constant Drifts

Let $dX_t=dt+dB^1_t$ and $dY_t=-dt+dB^2_t$, with $X_0=1$, $Y_0=0$, and any $\rho\in[-1,1]$. Find the equation for $R_t=X_t^2+Y_t^2$, then verify its expected value at time $t$.

For $f(x,y)=x^2+y^2$, the derivatives are $f_x=2x$, $f_y=2y$, $f_{xx}=2$, $f_{xy}=0$, and $f_{yy}=2$. The drift coefficients are $a=1$, $c=-1$, and both diffusion coefficients equal $1$. Hence the drift is $2x-2y+\tfrac12(2)+0+\tfrac12(2)=2x-2y+2$, while the diffusion terms are $2x\,dB^1_t$ and $2y\,dB^2_t$. Thus

dR_t=(2X_t-2Y_t+2)\,dt+2X_t\,dB^1_t+2Y_t\,dB^2_t.

Here $X_t=1+t+B^1_t$ and $Y_t=-t+B^2_t$, so $\mathbb{E}[X_t]=1+t$, $\mathbb{E}[Y_t]=-t$, and each Brownian marginal has variance $t$. Directly, $\mathbb{E}[R_t]=(1+t)^2+t+(-t)^2+t=1+4t+2t^2$. From the differential, the expected drift at time $s$ is $2(1+s)-2(-s)+2=4+4s$. The stochastic integrals have mean zero: their integrands are progressively measurable and $\mathbb{E}\int_0^t(2X_s)^2ds=4\int_0^t((1+s)^2+s)ds<\infty$, with the corresponding integral for $2Y_s$ also finite. The Ito integrals are therefore square-integrable martingales, and their expectations vanish. Integrating the expected drift gives $1+\int_0^t(4+4s)\,ds=1+4t+2t^2$, agreeing with the direct calculation.

The correlation does not enter this particular transformation because $f_{xy}=0$. This is a property of the chosen function, not a claim that the drivers are independent.

Worked Example: A Mixed Derivative Correction in a Cubic Transform

Take $dX_t=dt+dB^1_t$ and $dY_t=-dt+2\,dB^2_t$, with $\rho=-\tfrac12$, $X_0=1$, and $Y_0=2$. Find the equation for $Z_t=X_t^2Y_t$, and evaluate its coefficients at the initial point.

For $f(x,y)=x^2y$, the derivatives are $f_x=2xy$, $f_y=x^2$, $f_{xx}=2y$, $f_{xy}=2x$, and $f_{yy}=0$. The coefficients are $a=1$, $b=1$, $c=-1$, and $e=2$. The drift is $2xy-x^2+\tfrac12(2y)(1)^2+(-\tfrac12)(2x)(1)(2)+0=2xy-x^2+y-2x$. The diffusion terms are $2xy\,dB^1_t$ and $2x^2\,dB^2_t$. Therefore

d(X_t^2Y_t)=(2X_tY_t-X_t^2+Y_t-2X_t)\,dt+2X_tY_t\,dB^1_t+2X_t^2\,dB^2_t.

At $(X_0,Y_0)=(1,2)$, the drift is $2(1)(2)-(1)^2+2-2(1)=3$; the diffusion coefficients are $2(1)(2)=4$ and $2(1)^2=2$. To check the mixed correction separately, $d(X^2)=2X\,dX+dt$, and the product rule gives $d(X^2Y)=Y\,d(X^2)+X^2\,dY+d[X^2,Y]$. Since $d[X^2,Y]=2X\,d[X,Y]=2X(-1)dt=-2Xdt$, the drift is $2XY+Y-X^2-2X$, matching the formula. In particular, omitting the mixed term would give the wrong initial drift, $5$ instead of $3$.

Why the Covariance Calculation Matters

The three examples illustrate how to organize a calculation without relying on a memorized transformation rule. The covariance matrix of the noise in $(X,Y)$ is $\begin{pmatrix}b_t^2&\rho b_te_t\\ \rho b_te_t&e_t^2\end{pmatrix}$. The diagonal entries multiply pure second derivatives, while the off-diagonal entries multiply the mixed derivative. If $\rho=0$, the mixed correction vanishes; if $f_{xy}=0$, it vanishes even when the drivers are correlated. If either diffusion coefficient is zero, the mixed correction also vanishes.

Common mistake. Setting $dB^1_t\,dB^2_t=0$ just because the drivers have different labels silently assumes independence. For the third example, that mistake removes the term $-2X_t\,dt$. At the initial point it changes the drift from $3$ to $5$. The correction is determined by the actual covariation, $[B^1,B^2]_t=-\tfrac12t$, rather than by the names of the Brownian motions.

A second useful check is to compare a transformed equation with an applicable product rule. Such a check catches sign errors in the mixed term and errors in the first-order drift. For a proposed expectation calculation, also check that each stochastic integrand is square-integrable before using the zero-mean property of its Ito integral. A local martingale alone would not justify that step.

Check Your Understanding

For each question, use the stated correlation and keep drift, diffusion, and covariance terms separate until the final line.

If $dX_t=a\,dt+b\,dB^1_t$ and $dY_t=c\,dt+e\,dB^2_t$, what is $d[X,Y]_t$ when the driver correlation is $\rho$?
For $f(x,y)=x^2+y^2$, explain why the mixed derivative correction is zero even if $\rho\ne0$.
Apply the formula to $f(x,y)=xy$ when $b=3$, $e=2$, and $\rho=-\tfrac13$. What is the mixed contribution to the drift?
In the cubic example, recompute the initial drift after incorrectly setting the mixed covariance to zero. Which term accounts for the discrepancy?
What integrability condition lets you conclude that an Ito integral in the transformed equation is a square-integrable martingale with mean-zero increments?

A Problem-Solving Workshop on Coupled Ito Systems

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