Worked Transformations of Two-Dimensional Ito Processes

Transforming a Pair of Correlated Itô Processes

A transformation of a two-dimensional diffusion depends not only on the variance of each coordinate but also on their covariation. If the two coordinates share correlated noise, the second-order correction for a smooth function includes a mixed-derivative term. Omitting it can change the drift even when both first-order noise coefficients have been calculated correctly.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a two-dimensional Brownian motion $(B^1_t,B^2_t)$ with constant correlation $\rho\in[-1,1]$. Thus $[B^1]_t=[B^2]_t=t$ and $[B^1,B^2]_t=\rho t$. The endpoints $\rho=\pm1$ are allowed; the covariance matrix is positive semidefinite throughout this range. All stochastic integrals below are Itô integrals. On each finite interval $[0,T]$, an integrand $H$ is in $\mathcal{L}^2$ when it is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. When coefficients or transformed integrands satisfy only local versions of these conditions, the formulas are understood after stopping on intervals where the required integrals are defined.

Suppose $X,Y$ satisfy $dX_t=a_t\,dt+u_{1,t}\,dB^1_t+u_{2,t}\,dB^2_t$ and $dY_t=b_t\,dt+v_{1,t}\,dB^1_t+v_{2,t}\,dB^2_t$. The drift coefficients $a,b$ are progressively measurable and locally integrable in time, and the diffusion coefficients $u_1,u_2,v_1,v_2$ are progressively measurable and locally square-integrable in time. In particular, local integrability of a diffusion coefficient alone is not enough: its square must be time-integrable locally for the Itô integral to be defined. The three instantaneous covariance rates are

c_{XX}=u_1^2+u_2^2+2\rho u_1u_2,\qquad c_{YY}=v_1^2+v_2^2+2\rho v_1v_2,\qquad c_{XY}=u_1v_1+u_2v_2+\rho(u_1v_2+u_2v_1).

Here and below, coefficients on the right-hand side are evaluated at the current time. These rates mean $d[X]_t=c_{XX,t}\,dt$, $d[Y]_t=c_{YY,t}\,dt$, and $d[X,Y]_t=c_{XY,t}\,dt$. In the differential multiplication rules, $dB^i_t\,dB^i_t=dt$, $dB^1_t\,dB^2_t=\rho\,dt$, $dB^i_t\,dt=0$, and $dt\,dt=0$. These are Itô rules, not ordinary differential chain rules. Stratonovich integrals use a different convention and require a correction when converted to Itô form; no Stratonovich integral is used here.

The Transformation Formula

Theorem: Let $f:\mathbb{R}^2\to\mathbb{R}$ be twice continuously differentiable, and let $X,Y$ be the Itô processes above. On any interval where the terms are defined (or after localization), $Z_t=f(X_t,Y_t)$ satisfies $$ dZ_t=\left(f_xa+f_yb+\tfrac12 f_{xx}c_{XX}+f_{xy}c_{XY}+\tfrac12 f_{yy}c_{YY}\right)dt +(f_xu_1+f_yv_1)dB^1_t+(f_xu_2+f_yv_2)dB^2_t. $$ For a time-dependent $f(t,x,y)$, add $f_t$ to the drift coefficient. Each derivative is evaluated at $(X_t,Y_t)$.

Proof. Apply the multidimensional Itô formula to $f(X_t,Y_t)$. Its first-order terms are $f_x\,dX_t+f_y\,dY_t$, giving drift $f_xa+f_yb$ and the two displayed Brownian coefficients. The second-order term is $\tfrac12 f_{xx}\,d[X]_t+f_{xy}\,d[X,Y]_t+\tfrac12 f_{yy}\,d[Y]_t$. Substituting the three covariance rates gives exactly the stated drift. For a time-dependent function, the formula adds $f_t\,dt$. The formula applies locally as well: stop where the drift integrals are finite and the diffusion integrals are square-integrable, apply the formula there, and read the resulting identity up to those stopping times. Local integrability of the drift does not replace square-integrability of the diffusion. $\square$

The formula can also be applied component by component to a vector-valued transformation $F=(f,g)$, provided both components are twice continuously differentiable. Each output coordinate has its own first and second derivatives, but both use the same covariance rates. This is a direct application of the scalar formula, not a new independence assumption about the transformed coordinates.

Quadratic Forms: A Useful General Calculation

For a column vector $z=(x,y)^\mathsf{T}$ and a symmetric matrix $A$, define the quadratic form $q(z)=z^\mathsf{T}Az$. Write the drift vector as $m=(a,b)^\mathsf{T}$, the noise vectors as $\gamma_1=(u_1,v_1)^\mathsf{T}$ and $\gamma_2=(u_2,v_2)^\mathsf{T}$, and let $R=\begin{pmatrix}1&\rho\\ \rho&1\end{pmatrix}$. Then the covariance-rate matrix of $(X,Y)^\mathsf{T}$ is $C=\sum_{i,j=1}^2 R_{ij}\gamma_i\gamma_j^\mathsf{T}$.

Proposition: For the quadratic form above, the Itô drift rate is $2z^\mathsf{T}Am+\operatorname{tr}(AC)$, and the coefficient of $dB^i_t$ is $2z^\mathsf{T}A\gamma_i$, for $i=1,2$.

Proof. Since $A$ is symmetric, differentiating $q(z)=z^\mathsf{T}Az$ gives gradient $2Az$ and Hessian $2A$. The first-order drift is therefore $(2Az)^\mathsf{T}m=2z^\mathsf{T}Am$, and the coefficient multiplying $dB^i_t$ is $(2Az)^\mathsf{T}\gamma_i=2z^\mathsf{T}A\gamma_i$. The second-order correction is one half of the sum of each Hessian entry times its corresponding covariance rate; in matrix notation this is $\tfrac12\operatorname{tr}((2A)C)=\operatorname{tr}(AC)$. This proves the drift and both noise coefficients, including the case of a singular covariance matrix such as $|\rho|=1$. $\square$

Worked Example: A Quadratic Form with Correlated Noise

Let $X_0=1,\ Y_0=0$ and $dX_t=dt+dB^1_t,\quad dY_t=-dt+2\,dB^2_t$, with $\rho=\tfrac12$. Find the Itô differential of $Q_t=X_t^2+2X_tY_t+3Y_t^2$, then check its expected value independently.

Set $q(x,y)=x^2+2xy+3y^2$. Its derivatives are $q_x=2x+2y,\ q_y=2x+6y,\ q_{xx}=2,\ q_{xy}=2,\ q_{yy}=6$. The diffusion coefficients are $u_1=1,u_2=0,v_1=0,v_2=2$, so

c_{XX}=1,\qquad c_{YY}=4,\qquad c_{XY}=1\cdot2\cdot\tfrac12=1.

The drift is $q_x(1)+q_y(-1)+\tfrac12q_{xx}(1)+q_{xy}(1)+\tfrac12q_{yy}(4)$. The first two terms combine to $(2x+2y)-(2x+6y)=-4y$; the correction is $\tfrac12(2)+2+\tfrac12(24)=1+2+12=15$. The $dB^1_t$ coefficient is $q_x$, and the $dB^2_t$ coefficient is $2q_y$. Hence

dQ_t=(15-4Y_t)\,dt+(2X_t+2Y_t)\,dB^1_t+(4X_t+12Y_t)\,dB^2_t.

For an independent expectation check, $\mathbb{E}X_t=1+t,\ \mathbb{E}Y_t=-t,\ \operatorname{Var}(X_t)=t,\ \operatorname{Var}(Y_t)=4t$, and $\operatorname{Cov}(X_t,Y_t)=t$. Thus $\mathbb{E}[X_t^2]=(1+t)^2+t=1+3t+t^2$, $\mathbb{E}[X_tY_t]=(1+t)(-t)+t=-t^2$, and $\mathbb{E}[Y_t^2]=t^2+4t$. Therefore $\mathbb{E}Q_t=1+3t+t^2-2t^2+3(t^2+4t)=1+15t+2t^2$. Taking expectations in the differential gives drift expectation $15-4\mathbb{E}Y_t=15+4t$, whose integral from $0$ to $t$ is $15t+2t^2$, agreeing with the direct calculation. The expectations are finite because these coordinates are Gaussian with finite second moments on finite time intervals.

A Smooth Nonlinear Transformation

For a general smooth function, the mixed term $f_{xy}c_{XY}$ is just as essential as the two pure second-derivative terms. It is present when both the function and the diffusion generate covariation; it vanishes only when the mixed derivative or the covariance rate is zero.

Worked Example: An Exponential-Trigonometric Transform

Take $dX_t=dt+dB^1_t,\ dY_t=2dt+2dB^2_t$, again with $\rho=\tfrac12$, and set $F(x,y)=e^x\sin y$. Compute $dF(X_t,Y_t)$.

The derivatives are $F_x=e^x\sin y=F$, $F_y=e^x\cos y=:G$, $F_{xx}=F$, $F_{xy}=G$, and $F_{yy}=-F$. The covariance rates are $c_{XX}=1,\ c_{YY}=4,\ c_{XY}=1$. The drift is $F+2G+\tfrac12F+G-2F=-\tfrac12F+3G$. The $dB^1_t$ coefficient is $F_x=F$, while the $dB^2_t$ coefficient is $2F_y=2G$. Thus

dF(X_t,Y_t)=\left(-\tfrac12 e^{X_t}\sin Y_t+3e^{X_t}\cos Y_t\right)dt +e^{X_t}\sin Y_t\,dB^1_t+2e^{X_t}\cos Y_t\,dB^2_t.

Check: At the test point $(x,y)=(0,0)$, the derivatives give $F=0,G=1$, so the drift is $3$ and the noise coefficients are $0$ and $2$. Substitution in the displayed answer gives the same values. At $(0,\tfrac{\pi}{2})$, $F=1,G=0$; the drift is $-\tfrac12$, with noise coefficients $1$ and $0$. In particular, the mixed correction $F_{xy}c_{XY}=G$ changes the drift at the first test point from what would result if the correlation contribution were dropped.

Domain Restrictions and Stopping

Some transformations are defined only on part of the plane. The formula is then applied only while the process remains in that domain. Stopping at the boundary, or first stopping on compact subsets within the domain, gives a precise version of the calculation. The next example uses a logarithm, whose argument must remain strictly positive.

Worked Example: The Logarithm of the Squared Radius

Let $X_t=2+B^1_t,\ Y_t=1+B^2_t$, with $\rho=\tfrac12$, and let $F(x,y)=\log(x^2+y^2)$ on $\mathbb{R}^2\setminus\{(0,0)\}$. Define $\tau$ as the first time $X_t^2+Y_t^2\le\tfrac14$ or $X_t^2+Y_t^2\ge16$. Compute the differential up to $\tau$.

Write $s=x^2+y^2>0$. The derivatives are $F_x=2x/s,\ F_y=2y/s,\ F_{xx}=2/s-4x^2/s^2,\ F_{yy}=2/s-4y^2/s^2,$ and $F_{xy}=-4xy/s^2$. Here $a=b=0$ and $c_{XX}=c_{YY}=1,\ c_{XY}=\tfrac12$. The drift correction is

\tfrac12F_{xx}+F_{xy}c_{XY}+\tfrac12F_{yy} =\tfrac12\left(\frac4s-\frac{4(x^2+y^2)}{s^2}\right)-\frac{2xy}{s^2} =-\frac{2xy}{s^2}.

Since each coordinate has coefficient $1$ against its own Brownian driver and no other driver, the noise coefficients are $2x/s$ and $2y/s$. Therefore, for times before $\tau$,

dF(X_t,Y_t)=-\frac{2X_tY_t}{(X_t^2+Y_t^2)^2}\,dt +\frac{2X_t}{X_t^2+Y_t^2}\,dB^1_t +\frac{2Y_t}{X_t^2+Y_t^2}\,dB^2_t.

At the initial point $(2,1)$, $s=2^2+1^2=5$. Direct differentiation gives $F_{xx}=2/5-16/25=-6/25,\ F_{yy}=2/5-4/25=6/25,\ F_{xy}=-8/25$. The correction is $\tfrac12(-6/25)+\tfrac12(-8/25)+\tfrac12(6/25)=-4/25$; the displayed drift gives $-2(2)(1)/25=-4/25$. The noise coefficients are $4/5$ and $2/5$ by either derivative evaluation or direct substitution. Stopping keeps the argument of the logarithm at least $1/4$ and at most $16$, so the derivatives are bounded on the stopped region and the displayed stochastic integrands are square-integrable on each finite stopped interval.

Common Pitfalls

Dropping the mixed derivative. In the exponential-trigonometric example, at $(0,0)$ the correct drift is $3$. If one omits $F_{xy}c_{XY}$, the calculation becomes $F+2G+\tfrac12F+\tfrac12(-F)=2$ at that point. The missing unit is precisely $F_{xy}c_{XY}=1\cdot1$. The correction does not disappear merely because the Brownian drivers are written separately; their correlation determines their covariation.

Using only the ordinary chain rule. The first-order expression $f_x\,dX+f_y\,dY$ misses all second-order terms. For quadratic forms, those terms contribute $\operatorname{tr}(AC)$ to the drift and can be nonzero even when the original drift vector is zero.

Confusing drift integrability with diffusion integrability. A drift needs local time-integrability for its ordinary time integral. A diffusion coefficient needs local square-integrability for its Itô integral. For example, $t^{-3/4}$ is integrable near zero, but its square $t^{-3/2}$ is not; it cannot serve as a diffusion coefficient for an Itô integral started at time zero. Stopping does not repair a failure of square-integrability at the initial time.

The transformation formula is useful for deriving the dynamics of energies, distances, and other functions of coupled state variables. Its drift has three sources: the original drifts, the curvature of the function in each coordinate, and the curvature shared across coordinates through covariation. A reliable calculation records all three before simplifying.

Check Your Understanding

For each question, assume the stated derivatives and stochastic integrals exist; when a function has a restricted domain, work only up to a stopping time that keeps the process inside it.

For $dX_t=a\,dt+dB^1_t,\ dY_t=b\,dt+dB^2_t$ with correlation $\rho$, compute the drift rate of $X_t^2+Y_t^2$.
For $f(x,y)=xy$, identify the mixed-derivative correction when $c_{XY}\ne0$.
In the quadratic-form proposition, explain why the second-order correction is $\operatorname{tr}(AC)$, rather than $\tfrac12\operatorname{tr}(AC)$.
For the logarithmic-radius example, which part of the proof requires stopping away from the origin?
State the separate local integrability requirements on drift and diffusion coefficients in a two-dimensional Itô process.