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Ito Processes and the General Ito Formula · Tutorial 170 of 600

Proving the Multidimensional Ito Formula

Follow the Taylor expansion across partitions to see exactly why the covariance matrix contracts with the Hessian in the multidimensional Ito formula.

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What You'll Learn

  • State the multidimensional Ito formula for a scalar function of a vector Ito process
  • Separate time, drift, and diffusion increments in a partition-based Taylor expansion
  • Control the Taylor remainders using continuity and quadratic variation bounds
  • Identify diagonal and mixed quadratic covariation limits
  • Apply the formula to correlated Brownian drivers and time-dependent functions
  • Recognize why the ordinary multivariable chain rule misses the Ito correction

From Taylor Expansion to the Hessian Correction

For a smooth path \(x(t)\), the ordinary chain rule accounts for first-order changes in time and space. An Ito process has an additional second-order contribution: its random increments are of order \(\sqrt{\Delta t}\), so products of two such increments can accumulate to a nonzero limit over a partition. In several dimensions, these accumulated products include both squared increments and mixed products. The mixed terms are why the correction involves the full covariance matrix, not just the variances of the individual coordinates.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a \(d\)-dimensional standard Brownian motion \(\mathbf{B}_t=(B_t^1,\ldots,B_t^d)\). Let \(X_t=(X_t^1,\ldots,X_t^n)\) satisfy \(dX_t^i=a_i(t)\,dt+\sum_{k=1}^d\sigma_{ik}(t)\,dB_t^k\). Assume the coefficients are progressively measurable, \(\mathbb{E}\int_0^T(|a(t)|^2+\|\sigma(t)\|^2)\,dt<\infty\), and the derivatives used below are bounded. Thus each Brownian integrand in the formula belongs to \(\mathcal{L}^2\), the class of progressively measurable \(H\) with \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). The boundedness assumptions are convenient sufficient conditions; the usual local formula follows by stopping when the coefficients or derivatives leave bounded sets.

Theorem: If \(f:[0,T]\times\mathbb{R}^n\to\mathbb{R}\) is \(C^{1,2}\), meaning that \(f_t\), its first spatial derivatives, and its second spatial derivatives are continuous, then $$ \begin{aligned} f(t,X_t)=f(0,X_0) &+\int_0^t\left(f_t(s,X_s)+\sum_{i=1}^n a_i(s)f_i(s,X_s) +\frac12\sum_{i,j=1}^n c_{ij}(s)f_{ij}(s,X_s)\right)ds\\ &+\sum_{k=1}^d\int_0^t\left(\sum_{i=1}^n f_i(s,X_s)\sigma_{ik}(s)\right)dB_s^k, \end{aligned} $$ where \(f_i=\partial f/\partial x_i\), \(f_{ij}=\partial^2 f/\partial x_i\partial x_j\), and \(c_{ij}(s)=\sum_{k=1}^d\sigma_{ik}(s)\sigma_{jk}(s)\). All stochastic integrals displayed here are Ito integrals.

The matrix \(c(s)=\sigma(s)\sigma(s)^\mathsf{T}\) is the instantaneous covariance matrix of the coordinates of \(X\). In particular, \(c_{ii}\) gives the variance rate of coordinate \(i\), while \(c_{ij}\) for \(i\ne j\) gives the covariation rate between coordinates \(i\) and \(j\). The proof below shows where each entry appears.

Partition Proof and Remainder Control

Fix \(t\le T\) and take partitions \(0=t_0<t_1<\cdots<t_m=t\) whose mesh tends to zero. Write \(\Delta t_r=t_r-t_{r-1}\), \(\Delta X_r=X_{t_r}-X_{t_{r-1}}\), and evaluate derivatives with subscript \(r-1\) at \((t_{r-1},X_{t_{r-1}})\). Split each change into a time change at the new spatial point and a spatial change at the old time:

$$ \begin{aligned} f(t_r,X_{t_r})-f(t_{r-1},X_{t_{r-1}}) ={}&f(t_r,X_{t_r})-f(t_{r-1},X_{t_r})\\ &+f(t_{r-1},X_{t_r})-f(t_{r-1},X_{t_{r-1}}). \end{aligned} $$

Continuity of \(f_t\) gives a first-order time increment \(f_t(t_{r-1},X_{t_{r-1}})\Delta t_r\), up to a remainder whose sum tends to zero in probability. To justify evaluating the time derivative at \(X_{t_{r-1}}\), rather than \(X_{t_r}\), use uniform continuity on the compact set reached by the stopped process: the accumulated difference is bounded by \(t\) times a modulus of continuity evaluated at \(\max_r|\Delta X_r|\), which tends to zero almost surely.

For the spatial change, Taylor's theorem at fixed time gives $$ f(t_{r-1},X_{t_r})-f(t_{r-1},X_{t_{r-1}}) =\sum_i f_{i,r-1}\Delta X_r^i+ \frac12\sum_{i,j}f_{ij,r-1}\Delta X_r^i\Delta X_r^j+R_r. $$ On a compact set, continuity of the Hessian bounds the remainder by \(|R_r|\le\omega(\max_q|\Delta X_q|)|\Delta X_r|^2\), where \(\omega(u)\to0\) as \(u\to0\). The stopped process has continuous paths, so \(\max_r|\Delta X_r|\to0\) almost surely. Also \(\sum_r|\Delta X_r|^2\) is bounded in probability: its coordinate sums converge to finite quadratic variations, and the cross terms are bounded by the corresponding sums of squares. Consequently \(\sum_r|R_r|\to0\) in probability. These arguments can be applied on successively larger stopping regions, giving the local result without assuming the process stays in a fixed compact set.

Summing the Taylor expansions leaves first-order sums and weighted quadratic sums. The drift parts of the first-order sums converge to ordinary time integrals; their Brownian parts converge to Ito integrals by the Ito isometry and approximation of continuous adapted integrands by their left-endpoint values. For the quadratic sums, use the standard quadratic-covariation theorem for Ito processes: for each \(i,j\), \(\sum_r\Delta X_r^i\Delta X_r^j\) converges in probability to \(\int_0^t c_{ij}(s)\,ds\). This is the quadratic-variation and covariation result for Ito processes; its partition-limit proof belongs to that earlier material. The weighted version applies to the continuous adapted weights \(f_{ij}(s,X_s)\), by approximating the weights with step processes and localizing. Thus the quadratic Taylor term converges to \(\frac12\int_0^t\sum_{i,j}f_{ij}(s,X_s)c_{ij}(s)\,ds\). The time, first-order, and quadratic limits are exactly the terms in the theorem, proving the formula. \(\square\)

Corollary: The Hessian correction can also be written as \(\frac12\operatorname{tr}(c(s)D^2f(s,X_s))\,ds\). This includes mixed covariations; the sum over \(i,j\) counts both \(i,j\) and \(j,i\).

Proof. The covariance matrix \(c\) and Hessian \(D^2f\) are symmetric. By the definition of matrix trace, \(\operatorname{tr}(cD^2f)=\sum_i\sum_j c_{ij}f_{ji}\). Equality of mixed partial derivatives gives \(f_{ji}=f_{ij}\), so this is precisely the double sum in the theorem. \(\square\)

Worked Applications

Worked Example: A Product with Correlated Coordinates

Let \(B^1,B^2\) be independent Brownian coordinates, \(X_t^1=B_t^1\), and \(X_t^2=B_t^1+B_t^2\). Apply the formula to \(f(x_1,x_2)=x_1x_2\).

The derivatives are \(f_1=x_2\), \(f_2=x_1\), \(f_{11}=0\), \(f_{22}=0\), and \(f_{12}=f_{21}=1\). The diffusion rows are \((1,0)\) and \((1,1)\), so \(c_{11}=1\), \(c_{22}=2\), and \(c_{12}=c_{21}=1\). The drift is zero. The correction is \(\frac12(0+1+1+0)\,dt=dt\). The Brownian integrands are \(f_1+f_2=X_t^1+X_t^2\) for \(B^1\), and \(f_2=X_t^1\) for \(B^2\). Therefore $$ d(X_t^1X_t^2)=(X_t^1+X_t^2)\,dB_t^1+X_t^1\,dB_t^2+dt. $$ Check: the expectation of each Ito integral is zero: its integrands are square-integrable on every bounded time interval, since Brownian coordinates have finite second moments. Hence \(\mathbb{E}[X_t^1X_t^2]=t\), which also follows directly from \(\mathbb{E}[B_t^1(B_t^1+B_t^2)]=t+0=t\). The two mixed Hessian terms together are essential to this check.

Worked Example: An Exponential of a Linear Combination

Use the same independent Brownian coordinates, and set \(X_t^1=B_t^1\), \(X_t^2=2B_t^1+B_t^2\). For \(f(x_1,x_2)=e^{x_1-x_2}\), calculate the differential.

Here \(f_1=f\), \(f_2=-f\), \(f_{11}=f\), \(f_{22}=f\), and \(f_{12}=f_{21}=-f\). The diffusion rows are \((1,0)\) and \((2,1)\), giving \(c_{11}=1\), \(c_{22}=5\), and \(c_{12}=c_{21}=2\). The Hessian correction is \(\frac12(f-2f-2f+5f)\,dt=f\,dt\). The coefficient of \(dB^1\) is \(f_1+2f_2=-f\), and that of \(dB^2\) is \(f_2=-f\). Thus $$ df(t,X_t)=f(X_t)(-dB_t^1-dB_t^2+dt). $$ Check: \(X_t^1-X_t^2=-B_t^1-B_t^2\), so the transformed process is \(e^{-B_t^1-B_t^2}\). Applying the one-dimensional Ito formula to this single Brownian sum, whose quadratic variation rate is \(1+1=2\), gives diffusion \(-e^{-B_t^1-B_t^2}(dB_t^1+dB_t^2)\) and correction \(\frac12(2)e^{-B_t^1-B_t^2}dt\), agreeing with the result.

Worked Example: A Time-Dependent Function with a Mixed Correction

Let \(X_t^1=t+B_t^1\) and \(X_t^2=2t+B_t^1+2B_t^2\), and take \(f(t,x_1,x_2)=t x_1x_2\). Find its Ito differential.

The drift vector is \((1,2)\); the diffusion rows are \((1,0)\) and \((1,2)\). Therefore \(c_{11}=1\), \(c_{22}=5\), and \(c_{12}=c_{21}=1\). The derivatives are \(f_t=x_1x_2\), \(f_1=tx_2\), \(f_2=tx_1\), \(f_{11}=f_{22}=0\), and \(f_{12}=f_{21}=t\). The time and drift terms sum to \(x_1x_2+tx_2+2tx_1\). The Hessian correction is \(\frac12(t+t)\,dt=t\,dt\). The Brownian coefficients are \(t(x_1+x_2)\) for \(B^1\) and \(2tx_1\) for \(B^2\). Hence $$ df(t,X_t)=\big(X_t^1X_t^2+tX_t^2+2tX_t^1+t\big)\,dt +t(X_t^1+X_t^2)\,dB_t^1+2tX_t^1\,dB_t^2. $$ Check: at \(t=0\), the drift coefficient is \(X_0^1X_0^2=0\), and both diffusion coefficients vanish because they contain \(t\). This agrees with \(f(0,X_0)=0\) and confirms that the time derivative term has not been omitted.

Why the Mixed Terms Matter

The differential multiplication rules \(dB_t^k\,dB_t^\ell=\mathbf{1}_{\{k=\ell\}}dt\), \(dB_t^k\,dt=0\), and \(dt\,dt=0\) summarize the relevant quadratic-covariation limits. In a vector process, the same Brownian coordinate can enter several coordinates of \(X\); their increments then have nonzero mixed covariation. The Hessian contraction records exactly these effects.

A common mistake is to retain only the diagonal terms. For the product example, that incorrect calculation would use \(f_{11}c_{11}=0\) and \(f_{22}c_{22}=0\), concluding that there is no correction. But \(f_{12}=f_{21}=1\) and \(c_{12}=c_{21}=1\), so the actual correction is \(\frac12(1+1)dt=dt\). The error amounts to pretending the coordinates have zero covariation even though both contain \(B^1\).

The Ito formula is also not the ordinary multivariable chain rule: its Hessian term comes from quadratic variation, with \([B]_t=t\), rather than from differentiability of Brownian paths. For Stratonovich integrals, a correction is needed when converting to Ito form; the formula proved here is for Ito integrals and uses the Ito correction directly. In particular, one must not remove the Hessian term merely by applying the ordinary chain rule.

Check Your Understanding

For each question, identify the derivatives and covariance entries before writing the correction.

  1. If \(X^1=B^1\) and \(X^2=B^1-B^2\), what are \(c_{11},c_{22},c_{12}\)?
  2. For \(f(x_1,x_2)=x_1^2+x_2^2\), which Hessian entries contribute to the correction when \(c_{12}\ne0\)?
  3. Why does continuity of the Hessian make the sum of spatial Taylor remainders vanish in probability?
  4. For \(f(t,x)=t x_1\), list the time derivative, spatial derivative, and Hessian terms in the formula.
  5. Explain why the formula's Brownian integrands must satisfy the stated square-integrability condition for the Ito integrals to be defined in \(\mathcal{L}^2\).