A Route from an SDE to an Expectation
A useful way to organize diffusion calculations is to keep four objects distinct: the stochastic differential equation, its transition kernel, its infinitesimal generator, and the expectation of a test function. The kernel gives the distribution at a later time. The generator describes the instantaneous change of test-function expectations. Dynkin’s formula connects that instantaneous change to a finite-time expectation. Agreement between the kernel and generator calculations is a strong check on signs, coefficients, and time intervals.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). An Ito integral \(\int_0^t H_s\,dB_s\) below is taken for an adapted, progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^t H_s^2\,ds<\infty\); in that case it is a square-integrable martingale with conditional mean-zero increments. The examples use globally Lipschitz coefficients with linear growth, so each SDE has a unique strong solution. For a time-homogeneous diffusion, write \(P_t(x,A)=\mathbb{P}(X_t\in A\mid X_0=x)\) for its transition kernel and \(P_tf(x)=\mathbb{E}[f(X_t)\mid X_0=x]\) when the expectation exists.
The displayed formula is the fixed-time form of Dynkin’s formula, established in the earlier material on Dynkin’s formula. For the computations below, we also verify why the stochastic integral in Ito’s formula has expectation zero: its integrand is progressively measurable and square-integrable, so the integral is a true martingale, not merely a local martingale. The generator records the drift and diffusion contributions separately; the factor \(\tfrac12\) comes from Brownian quadratic variation \([B]_t=t\), not from an ordinary chain rule.
Worked Example: Brownian Motion and a Fourth-Moment Check
Worked Example: Brownian Kernel, Generator, and Dynkin’s Formula
Let \(X_t=x+B_t\), where \(x\in\mathbb{R}\), and take \(f(y)=y^4\). Find the transition kernel, compute \(P_tf(x)\), and verify the result using the generator.
Since \(B_t\) has distribution \(N(0,t)\), the transition kernel has density \(p_t(x,y)=\dfrac{1}{\sqrt{2\pi t}}\exp\!\left(-\dfrac{(y-x)^2}{2t}\right)\) for \(t>0\). Equivalently, \(X_t\) has the same distribution as \(x+\sqrt{t}Z\), where \(Z\sim N(0,1)\). Expanding the fourth power and using \(\mathbb{E}[Z]=\mathbb{E}[Z^3]=0\), \(\mathbb{E}[Z^2]=1\), and \(\mathbb{E}[Z^4]=3\),
The drift is \(\mu=0\), the diffusion coefficient is \(\sigma=1\), and \(f'(y)=4y^3\), \(f''(y)=12y^2\). Thus the generator is \(\mathcal{L}f(y)=\tfrac12(12y^2)=6y^2\). Since \(\mathbb{E}[X_s^2]=x^2+s\), Dynkin’s formula gives
For the martingale condition, Ito’s formula gives \(d(X_t^4)=4X_t^3\,dB_t+6X_t^2\,dt\). Here \(f'(X_t)=4X_t^3\), \(f''(X_t)=12X_t^2\), and the Ito correction is \(\tfrac12 f''(X_t)\,dt=6X_t^2\,dt\). On every finite interval, \(\mathbb{E}\int_0^t(4X_s^3)^2\,ds<\infty\), since \(X_s\) is Gaussian and has finite sixth moment. Therefore the stochastic integral has expectation zero. The kernel computation and the generator computation agree, as required.
Check: At \(x=0\), the formula gives \(P_tf(0)=3t^2\), the fourth moment of a centered normal random variable with variance \(t\). At \(t=0\), it returns \(x^4\), the initial test-function value.
Composing Gaussian Kernels
Semigroup composition is useful when a transition is described in several time steps. For a time-homogeneous Markov process, the Chapman–Kolmogorov relation is \(P_{s+t}(x,A)=\int P_t(y,A)\,P_s(x,dy)\). In calculations with Gaussian kernels, this amounts to taking the distribution after the first step and then averaging the second-step kernel over that distribution.
Worked Example: Composing Ornstein–Uhlenbeck Kernels
Consider the Ornstein–Uhlenbeck process \(dX_t=-X_t\,dt+\sqrt{2}\,dB_t\), with \(X_0=x\). Compute its transition law over two successive intervals of length \(1\), first by composing the one-step Gaussian kernels and then by using the direct two-unit kernel.
The drift coefficient is \(\mu(y)=-y\), and the diffusion coefficient is \(\sigma(y)=\sqrt{2}\). The transition law over time \(t\) is Gaussian with mean \(e^{-t}x\) and variance \(1-e^{-2t}\). In particular, after one time unit the intermediate state \(Y\) has mean \(e^{-1}x\) and variance \(1-e^{-2}\). Conditional on \(Y=y\), the next unit-step state has mean \(e^{-1}y\) and variance \(1-e^{-2}\).
To compose, represent the two steps using independent standard normal variables \(Z_1,Z_2\). The final state has the same distribution as \(e^{-1}(e^{-1}x+\sqrt{1-e^{-2}}Z_1)+\sqrt{1-e^{-2}}Z_2\). Its mean is \(e^{-2}x\). Independence makes the variance the sum of the two scaled variances:
Thus the composed kernel is Gaussian with mean \(e^{-2}x\) and variance \(1-e^{-4}\). The direct kernel at time \(2\) has that same mean and variance, so the two calculations agree.
Check: Setting \(x=0\) gives mean \(0\) under both calculations. The composed variance is positive because \(1-e^{-4}>0\), and it is exactly the direct two-unit variance. More generally, the composition works because the second-step conditional mean is linear in the intermediate state and the new Brownian increment is independent of that state.
Proof. To verify the composition for any two nonnegative time intervals \(s,t\), write the first transition as \(Y=e^{-s}x+\sqrt{1-e^{-2s}}Z_1\), and the second as \(X=e^{-t}Y+\sqrt{1-e^{-2t}}Z_2\), with \(Z_1,Z_2\) independent standard normal variables. Substitution gives mean \(e^{-(s+t)}x\). The variance is \(e^{-2t}(1-e^{-2s})+(1-e^{-2t})=1-e^{-2(s+t)}\). These are the mean and variance of the stated Ornstein–Uhlenbeck kernel at time \(s+t\). If \(s=0\) or \(t=0\), the corresponding variance contribution is zero and the same equalities hold. Since a Gaussian distribution is determined by its mean and variance, the composed and direct kernels agree. \(\square\)
Worked Example: Ornstein–Uhlenbeck Expectations from the Generator
Worked Example: A Quadratic Test Function
For the Ornstein–Uhlenbeck process above, start at \(x=1\), take \(f(y)=y^2\), and evaluate \(\mathbb{E}[f(X_t)\mid X_0=1]\) at \(t=\log 2\). Check the result with Dynkin’s formula.
From the transition kernel, \(X_t\) has mean \(e^{-t}\) and variance \(1-e^{-2t}\). Therefore its second moment is \(e^{-2t}+1-e^{-2t}=1\). At \(t=\log 2\), this is still \(1\). For the generator calculation, \(f'(y)=2y\), \(f''(y)=2\), so \(\mathcal{L}f(y)=-y(2y)+\tfrac12(2)(2)=2-2y^2\).
Here Ito’s formula gives \(d(X_t^2)=2X_t\,dX_t+\tfrac12(2)(dX_t\,dX_t)\), where \(dX_t=-X_t\,dt+\sqrt{2}\,dB_t\) and \(dX_t\,dX_t=2\,dt\). Hence \(d(X_t^2)=(-2X_t^2+2)\,dt+2\sqrt{2}X_t\,dB_t\). The integrand \(2\sqrt{2}X_s\) is progressively measurable, and \(\mathbb{E}\int_0^t8X_s^2\,ds<\infty\) because the process has finite second moments on bounded intervals. Its Ito integral is therefore a true square-integrable martingale. Taking expectations gives Dynkin’s formula, whose generator term is zero at every time because \(\mathbb{E}[X_s^2]=1\) for \(X_0=1\). Thus the result is \(1+\int_0^t(2-2)\,ds=1\).
Check: At \(t=\log 2\), the kernel mean is \(e^{-\log 2}=\tfrac12\), and its variance is \(1-e^{-2\log 2}=1-\tfrac14=\tfrac34\). Their contribution to the second moment is \((\tfrac12)^2+\tfrac34=\tfrac14+\tfrac34=1\), confirming the generator calculation.
Worked Example: A Geometric Brownian Transition
Worked Example: The Second Moment of a Multiplicative Diffusion
Let \(X_0=x>0\) and \(dX_t=X_t\,dB_t\). Find the transition distribution and use the generator to calculate the second moment.
Applying Ito’s formula to \(\log X_t\), with \(g(y)=\log y\), \(g'(y)=1/y\), and \(g''(y)=-1/y^2\), gives \(d(\log X_t)=dB_t-\tfrac12dt\); the correction is \(\tfrac12g''(X_t)X_t^2dt=-\tfrac12dt\). Thus \(X_t\) has the same distribution as \(x\exp(\sqrt{t}Z-\tfrac12t)\), where \(Z\sim N(0,1)\). Its transition density for \(y>0\) is \(\dfrac{1}{y\sqrt{2\pi t}}\exp\!\left(-\dfrac{(\log(y/x)+t/2)^2}{2t}\right)\).
For \(f(y)=y^2\), the derivatives are \(f'(y)=2y\) and \(f''(y)=2\); the drift is zero and the diffusion coefficient is \(y\). Consequently, \(\mathcal{L}f(y)=\tfrac12y^2(2)=y^2\). The kernel gives \(\mathbb{E}[X_t^2]=x^2e^t\), since \(\mathbb{E}[e^{2\sqrt{t}Z-t}]=e^{2t-t}=e^t\). Dynkin’s formula gives the same value: \(x^2+\int_0^t\mathbb{E}[X_s^2]\,ds=x^2+\int_0^t x^2e^s\,ds=x^2e^t\).
For the martingale check, Ito’s formula for \(X_t^2\) has derivatives \(f'(X_t)=2X_t\), \(f''(X_t)=2\), and correction \(\tfrac12f''(X_t)X_t^2dt=X_t^2dt\), yielding \(d(X_t^2)=2X_t^2\,dB_t+X_t^2dt\). The integrand \(2X_s^2\) is progressively measurable and square-integrable on every bounded time interval because the lognormal distribution has finite fourth moment there. Hence the Ito integral is a true martingale. The resulting expectation equation agrees with the kernel calculation.
Check: At \(t=0\), both calculations give \(x^2\). The second moment increases by the factor \(e^t\), consistent with the positive generator value \(\mathcal{L}f(y)=y^2\) for \(y>0\).
Common Pitfall: Mixing Up the Generator and the Kernel
The generator is not itself a transition density, and it does not give a finite-time expectation by simply evaluating \(\mathcal{L}f\) at the starting point. For example, in the geometric Brownian calculation, \(\mathcal{L}(y^2)=y^2\). Replacing the Dynkin integral by \(t\mathcal{L}f(x)=tx^2\) would produce \(x^2+tx^2\), not \(x^2e^t\). The generator must be evaluated along the evolving process and averaged: the correct integral is \(\int_0^t\mathbb{E}[X_s^2]\,ds=x^2\int_0^te^s\,ds\).
Likewise, when composing kernels, the second transition is conditional on the intermediate state. In the Ornstein–Uhlenbeck example its mean is \(e^{-1}y\), not \(e^{-1}x\) for every intermediate state \(y\). Integrating over the intermediate kernel, or equivalently substituting the random intermediate value, is essential. These distinctions make kernel and generator calculations mutually useful checks rather than interchangeable shortcuts.
Check Your Understanding
For each question, state the assumptions needed for the expectation or kernel calculation.
- For \(X_t=x+B_t\), use the Gaussian kernel to compute \(P_t f(x)\) for \(f(y)=y^2\), and verify the answer using \(\mathcal{L}f\).
- For the Ornstein–Uhlenbeck process in this tutorial, compose kernels over intervals \(s\) and \(t\), and show that the resulting variance is \(1-e^{-2(s+t)}\).
- For \(dX_t=X_t\,dB_t\), calculate \(\mathcal{L}(y^3)\) and identify the moments needed to justify the Ito integral in the corresponding Ito formula.
- Explain why the generator value \(\mathcal{L}f(x)\) alone does not generally determine \(P_tf(x)\) for positive \(t\).