From Ito’s Formula to the Generator
The generator describes the first-order change in expected test-function values over a short time. For a diffusion, that change has two sources: the drift moves the state systematically, while the Brownian fluctuations contribute through the second derivatives of the test function. Ito’s formula makes both effects explicit. Taking expectations removes its stochastic-integral term, and a short-time limit yields the generator.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying an \(m\)-dimensional Brownian motion \(\mathbf{B}_t\). Let \(X_t\) take values in \(\mathbb{R}^d\) and satisfy the time-homogeneous Ito SDE \(dX_t=b(X_t)\,dt+\sigma(X_t)\,d\mathbf{B}_t\), with initial state \(X_0=x\). Here \(b:\mathbb{R}^d\to\mathbb{R}^d\) is the drift and \(\sigma:\mathbb{R}^d\to\mathbb{R}^{d\times m}\) is the diffusion coefficient. The stochastic integral is an Ito integral: its matrix integrand is progressively measurable and has finite expected integrated squared Frobenius norm on every finite time interval.
Write \(\nabla f\) for the column vector of first derivatives and \(\nabla^2 f\) for the Hessian matrix. The covariance matrix per unit time is \(a(x)=\sigma(x)\sigma(x)^{\mathsf T}\). Ito’s formula uses the quadratic-variation rules \(dB_t^i\,dB_t^j=\delta_{ij}\,dt\), \(dB_t^i\,dt=0\), and \(dt\,dt=0\). In particular, the second-derivative contribution has a factor of one-half. This is an Ito formula, not the ordinary chain rule or the Stratonovich chain rule.
The Expectation Identity
Proof. Set \(a(z)=\sigma(z)\sigma(z)^{\mathsf T}\). Ito’s formula, with the function and derivative terms displayed, gives \(df(X_t)=\sum_{i=1}^d\partial_i f(X_t)\,dX_t^i+\tfrac12\sum_{i,j=1}^d\partial_{ij}f(X_t)\,d[X^i,X^j]_t\). Since \(dX_t^i=b_i(X_t)\,dt+\sum_{k=1}^m\sigma_{ik}(X_t)\,dB_t^k\) and \(d[X^i,X^j]_t=a_{ij}(X_t)\,dt\), this becomes \(df(X_t)=\big(\sum_i b_i(X_t)\partial_i f(X_t)+\tfrac12\sum_{i,j}a_{ij}(X_t)\partial_{ij}f(X_t)\big)dt+\sum_{i,k}\partial_i f(X_t)\sigma_{ik}(X_t)\,dB_t^k\). Thus the drift is exactly \((\mathcal{L}f)(X_t)\), and the remaining term is the Ito integral with integrand \(H_{k,s}=\sum_i\partial_i f(X_s)\sigma_{ik}(X_s)\). The derivatives of \(f\) are bounded, so linear growth of \(\sigma\) and the finite-time second-moment bound for a Lipschitz SDE imply \(\mathbb{E}\int_0^t\sum_k H_{k,s}^2\,ds<\infty\). Hence \(H\) belongs to the Ito-integrand class \(\mathcal{L}^2\). Its Ito integral is a square-integrable martingale: it is integrable by the Ito isometry, and its conditional expectation of an increment given \(\mathcal{F}_s\) is zero, first for adapted step integrands by independent Brownian increments and then for \(\mathcal{L}^2\) integrands by the isometry and \(L^2\) approximation. Taking conditional expectations in the integrated Ito formula therefore removes that martingale term and gives the asserted identity. \(\square\)
The Short-Time Limit
Proof. Fix \(x\), and let \(X_0=x\). First, \(X_s\) approaches \(x\) in \(L^2\) as \(s\downarrow0\). To see the needed bounds, linear growth gives \(|b(z)|^2+\|\sigma(z)\|_F^2\le C_1(1+|z|^2)\) for a finite constant \(C_1\). The integral form of the SDE, the inequality \(|u+v+w|^2\le3(|u|^2+|v|^2+|w|^2)\), Cauchy–Schwarz for the time integral, and the Ito isometry give, for \(s\le1\), \(\mathbb{E}|X_s|^2\le3|x|^2+3s\int_0^s\mathbb{E}|b(X_u)|^2\,du+3\int_0^s\mathbb{E}\|\sigma(X_u)\|_F^2\,du\). The growth bound and Gronwall’s inequality imply \(\sup_{0\le s\le1}\mathbb{E}|X_s|^2<\infty\). Applying the same integral estimates to \(X_s-x\) now gives \(\mathbb{E}|X_s-x|^2\le C_x(s^2+s)\) for a finite constant \(C_x\); thus \(X_s\to x\) in \(L^2\), and in probability.
The function \(\mathcal{L}f\) is continuous. It is also bounded: its derivative factors vanish outside the compact support of \(f\), and \(b,\sigma\) are bounded on that support. Write \(M=\sup_z|\mathcal{L}f(z)|<\infty\). For any \(\varepsilon>0\), continuity at \(x\) gives \(\delta>0\) such that \(|\mathcal{L}f(z)-\mathcal{L}f(x)|<\varepsilon\) whenever \(|z-x|<\delta\). Consequently, \(\mathbb{E}|(\mathcal{L}f)(X_s)-(\mathcal{L}f)(x)|\le\varepsilon+2M\,\mathbb{P}(|X_s-x|\ge\delta)\to\varepsilon\) as \(s\downarrow0\). Since \(\varepsilon\) was arbitrary, \(\mathbb{E}[(\mathcal{L}f)(X_s)]\to(\mathcal{L}f)(x)\).
By the proposition, the generator difference quotient is the average \(\frac1h\int_0^h\mathbb{E}[(\mathcal{L}f)(X_s)]\,ds\). The integrand tends to \((\mathcal{L}f)(x)\) as \(s\downarrow0\), so its average over \([0,h]\) has the same limit: for any \(\varepsilon>0\), once the integrand differs from \((\mathcal{L}f)(x)\) by less than \(\varepsilon\) for \(0\le s\le h\), the average differs by at most \(\varepsilon\). Taking \(h\downarrow0\) proves the formula. \(\square\)
Worked Generator Calculations
Worked Example: A Scalar Diffusion at the Origin
Let \(dX_t=dt+2\,dB_t\) and evaluate the generator at \(x=0\) for the compactly supported test function \(f(y)=(1-y^2)^3\) when \(|y|<1\), and \(f(y)=0\) when \(|y|\ge1\). The function is \(C^2\), since its value and first two derivatives match zero at the endpoints.
Inside \((-1,1)\), \(f'(y)=-6y(1-y^2)^2\) and \(f''(y)=-6(1-y^2)^2+24y^2(1-y^2)\). Therefore \(f'(0)=0\) and \(f''(0)=-6\). Here \(b(0)=1\) and \(\sigma(0)=2\), so \(\mathcal{L}f(0)=1\cdot0+\tfrac12(2^2)(-6)=-12\).
Check: In one dimension the matrix trace is simply \(\sigma(0)^2f''(0)\). Substitution gives \(\tfrac12\cdot4\cdot(-6)=-12\), agreeing with the generator calculation.
Worked Example: A Two-Dimensional Covariance Matrix
At \(z=(0,0)\), take constant drift \(b=(1,-1)^{\mathsf T}\), diffusion matrix \(\sigma=\begin{pmatrix}1&2\\0&1\end{pmatrix}\), and \(f(z)=(1-z_1^2-z_2^2)^3\) for \(|z|<1\), extended by zero outside the unit disk. This extension is \(C^2\). Find \(\mathcal{L}f(0,0)\).
Writing \(r^2=z_1^2+z_2^2\), differentiation gives \(\partial_i f=-6z_i(1-r^2)^2\), so \(\nabla f(0,0)=0\). The Hessian at the origin is \(-6I\), since each diagonal second derivative is \(-6\) and the mixed derivative is zero. Direct matrix multiplication gives \(\sigma\sigma^{\mathsf T}=\begin{pmatrix}5&2\\2&1\end{pmatrix}\). Thus the drift term is zero, and \(\tfrac12\operatorname{tr}((\sigma\sigma^{\mathsf T})(-6I))=\tfrac12[-6(5)-6(1)]=-18\).
Check: Expanding the trace entry by entry, the diagonal Hessian terms contribute \(5(-6)+1(-6)=-36\), while both mixed-Hessian terms vanish. Multiplication by one-half gives \(-18\).
Worked Example: A Nonzero First-Derivative Contribution
Consider \(dX_t=(2-X_t)\,dt+(1+X_t)\,dB_t\), and at \(x=0\) use \(f(y)=y(1-y^2)^3\) for \(|y|<1\), extended by zero for \(|y|\ge1\). Both coefficients are globally Lipschitz and of linear growth. Expand the test function near zero: \(f(y)=y-3y^3+3y^5-y^7\). It follows by differentiation that \(f'(0)=1\) and \(f''(0)=0\). Since \(b(0)=2\) and \(\sigma(0)=1\), \(\mathcal{L}f(0)=2\cdot1+\tfrac12(1^2)\cdot0=2\).
Check: The drift contribution is \(2\), while the Ito correction is zero because the second derivative at the evaluation point is zero. Their sum is \(2\).
Common Mistake: Dropping the One-Half
For the first example, a common incorrect calculation is to use \(\sigma^2 f''\) instead of \(\tfrac12\sigma^2 f''\). It gives \(2^2(-6)=-24\), twice the correct answer. The correction comes from the second-order term in Ito’s formula, which is \(\tfrac12 f''(X_t)\,d[X]_t\); here \(d[X]_t=\sigma^2dt=4dt\), so the contribution is \(\tfrac12(-6)(4)=-12\). The drift and diffusion terms also have different roles: the drift multiplies first derivatives, and the covariance multiplies second derivatives.
The matrix expression is equivalent to summing over coordinates: \(\operatorname{tr}(\sigma\sigma^{\mathsf T}\nabla^2 f)=\sum_{i,j=1}^d(\sigma\sigma^{\mathsf T})_{ij}\partial_{ij}f\). Using only diagonal entries of \(\sigma\), or replacing \(\sigma\sigma^{\mathsf T}\) with \(\sigma\), generally gives a different and incorrect result. The formula applies here to Ito SDEs under the stated assumptions; a Stratonovich SDE must first be converted to Ito form if this generator formula is to be used.
Check Your Understanding
Use the definition by the short-time expectation quotient and the proved generator formula.
- For \(dX_t=3\,dt+dB_t\), what is \(\mathcal{L}f(x)\) for \(f\in C_c^2(\mathbb{R})\)?
- In the multidimensional formula, which matrix plays the role of covariance per unit time?
- For constant \(\sigma\) and a test function with \(\nabla f(x)=0\), which part of the generator remains at \(x\)?
- Why does the stochastic integral in Ito’s formula have zero conditional expectation in the proof?
- For the first worked example, recompute the generator if the diffusion coefficient is \(1\) instead of \(2\), keeping the same drift, test function, and evaluation point.