From the Generator to an Expected Change
For a diffusion, the generator records the instantaneous expected change of a smooth test function. Dynkin’s formula turns that local description into an identity over a fixed time interval: the expected change in the test function equals the expected accumulated generator. The key qualification is integrability. Itô’s formula produces a stochastic integral, and its expectation is zero only when that integral is a true martingale with the required integrability—not merely a local martingale.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a solution, started at \(X_0=x\), of the time-homogeneous SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\). Here \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. One standard sufficient condition for a unique, non-explosive strong solution is that both coefficients are globally Lipschitz and satisfy linear growth; the standard existence theorem for Lipschitz SDEs supplies that conclusion. The formula below assumes the solution exists and states the additional conditions needed for the expectation identity.
Write \(\mathbb{E}_x\) for expectation under the law of this solution with initial state \(x\). For a twice continuously differentiable function \(f:\mathbb{R}\to\mathbb{R}\), its diffusion generator is \(\mathcal{L}f(y)=\mu(y)f'(y)+\tfrac12\sigma(y)^2f''(y)\). The first term comes from the drift; the second is the correction from Brownian quadratic variation.
In the results and examples below, every Itô integral \(\int_0^t H_s\,dB_s\) is understood with a progressively measurable integrand \(H\) in \(\mathcal{L}^2\): for each finite horizon \(T\), \(\mathbb{E}_x\int_0^T H_s^2\,ds<\infty\). This condition makes the integral a square-integrable martingale. The drift integral is an ordinary time integral, and we require its absolute integrability when taking expectations.
The Martingale Step and Dynkin’s Formula
Proof. Itô’s formula, with the function \(f\), its derivatives \(f'\) and \(f''\), and the quadratic-variation correction displayed, gives \(df(X_t)=f'(X_t)dX_t+\tfrac12f''(X_t)d[X]_t\). For this diffusion, \(d[X]_t=\sigma(X_t)^2dt\), so substitution of the SDE yields \(df(X_t)=f'(X_t)\mu(X_t)dt+f'(X_t)\sigma(X_t)dB_t+\tfrac12f''(X_t)\sigma(X_t)^2dt\). Thus \(M_t=\int_0^t f'(X_s)\sigma(X_s)\,dB_s\). The integrand is progressively measurable because \(X\) is adapted and continuous and the coefficients and \(f'\) are measurable. The first integrability assumption says precisely that it belongs to \(\mathcal{L}^2\) on each finite horizon. By the Itô isometry, this stochastic integral is square-integrable; its martingale property gives \(\mathbb{E}_x[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). Hence \(M\) is a martingale. \(\square\)
Proof. The Proposition gives the martingale identity \(\mathbb{E}_x[M_t]=\mathbb{E}_x[M_0]\). Since \(M_0=0\), its expectation is zero. By the second integrability assumption, \(\int_0^t\mathcal{L}f(X_s)\,ds\) is integrable, so we may rearrange the definition of \(M_t\) and take expectations. This gives the stated formula. \(\square\)
The square-integrability condition on the stochastic integrand is a useful sufficient condition, not a claim that it is necessary in every application. If it fails, Itô’s formula may still identify a local martingale, but a local martingale need not have constant expectation. In that case, the formula cannot be justified by simply deleting its stochastic-integral term; further integrability or localization arguments would be needed.
Worked Checks for Brownian Motion
Worked Example: A Quadratic Test Function
Let \(X_t=x+B_t\), so \(\mu=0\), \(\sigma=1\), and take \(f(y)=y^2\). Compute both sides of Dynkin’s formula for fixed \(t\ge0\).
The derivatives are \(f'(y)=2y\) and \(f''(y)=2\), hence \(\mathcal{L}f(y)=0\cdot 2y+\tfrac12\cdot1^2\cdot2=1\). The stochastic-integral condition holds because \(\mathbb{E}_x\int_0^t(2X_s)^2ds=4\int_0^t(x^2+s)ds=4x^2t+2t^2<\infty\). The generator integral is \(\int_0^t1\,ds=t\), so Dynkin’s formula gives \(\mathbb{E}_x[X_t^2]=x^2+t\).
Check: \(X_t=x+B_t\), and \(B_t\) has mean zero and variance \(t\). Directly, \(\mathbb{E}_x[(x+B_t)^2]=x^2+2x\mathbb{E}[B_t]+\mathbb{E}[B_t^2]=x^2+0+t=x^2+t\). For \(x=2,t=3\), both calculations give \(4+3=7\).
Worked Example: An Exponential Test Function
For \(X_t=B_t\) started at \(0\), take \(f(y)=e^y\) and a fixed \(t\ge0\). Find the generator integral and verify the resulting expectation.
Here \(f'(y)=e^y\), \(f''(y)=e^y\), and \(\mathcal{L}f(y)=\tfrac12e^y\). For every finite \(T\), \(\mathbb{E}\int_0^T e^{2B_s}\,ds=\int_0^T e^{2s}\,ds<\infty\), using the normal moment-generating function. Thus the stochastic integrand \(e^{B_s}\) belongs to \(\mathcal{L}^2\). The expected generator integral is \(\tfrac12\int_0^t\mathbb{E}[e^{B_s}]\,ds=\tfrac12\int_0^t e^{s/2}ds=e^{t/2}-1\). Since \(f(0)=1\), Dynkin’s formula gives \(\mathbb{E}[e^{B_t}]=e^{t/2}\).
Check: \(B_t\) has distribution \(N(0,t)\), whose moment-generating function at \(1\) is \(\mathbb{E}[e^{B_t}]=e^{t/2}\). This agrees with \(1+(e^{t/2}-1)=e^{t/2}\), including at \(t=0\), when the generator integral is zero.
Worked Checks for the Ornstein–Uhlenbeck Process
For the remaining examples, let \(X\) solve the Ornstein–Uhlenbeck SDE \(dX_t=-\kappa X_t\,dt+\eta\,dB_t\), where \(\kappa>0\), with \(X_0=x\). Its drift is \(\mu(y)=-\kappa y\), its diffusion coefficient is \(\sigma(y)=\eta\), and \(\mathcal{L}f(y)=-\kappa y f'(y)+\tfrac12\eta^2 f''(y)\). These linear coefficients satisfy the usual Lipschitz and linear growth conditions, and the explicit solution has finite moments of every order at finite times. Consequently the integrability requirements for the linear and quadratic tests below hold.
Worked Example: The Mean of an Ornstein–Uhlenbeck Process
Use \(f(y)=y\) to find \(\mathbb{E}_x[X_t]\), and check the result from the explicit solution.
The derivatives are \(f'(y)=1\) and \(f''(y)=0\), so \(\mathcal{L}f(y)=-\kappa y\). The stochastic integrand is the constant \(\eta\), and \(\mathbb{E}_x\int_0^t\eta^2ds=\eta^2t<\infty\). The explicit solution is \(X_s=xe^{-\kappa s}+\eta\int_0^s e^{-\kappa(s-u)}\,dB_u\). The stochastic integral here has a deterministic square-integrable integrand, so its expectation is zero; therefore \(\mathbb{E}_x[X_s]=xe^{-\kappa s}\). Dynkin’s formula now gives \(\mathbb{E}_x[X_t]=x-\kappa\int_0^t xe^{-\kappa s}ds=x-x(1-e^{-\kappa t})=xe^{-\kappa t}\). The formula also covers \(x=0\): both the initial term and expected generator integral are zero.
Check: Taking \(x=3\), \(\kappa=1\), and \(t=\ln 2\), the formula yields \(3e^{-\ln 2}=3/2\). The explicit solution has mean \(3/2\) because its stochastic-integral term has mean zero.
Worked Example: The Second Moment of an Ornstein–Uhlenbeck Process
Take \(x=1\), \(\kappa=1\), \(\eta=2\), and \(t=\ln 2\). Use \(f(y)=y^2\) to compute \(\mathbb{E}_1[X_t^2]\), then verify the answer independently.
The derivatives are \(f'(y)=2y\) and \(f''(y)=2\). Thus \(\mathcal{L}f(y)=-2y^2+4\). The stochastic integrand is \(2\eta X_s=4X_s\), which is square-integrable on finite intervals because the explicit solution has a finite second moment. From the explicit solution and the Itô isometry, \(\mathbb{E}_1[X_s^2]=e^{-2s}+4\int_0^s e^{-2(s-u)}du=e^{-2s}+2(1-e^{-2s})=2-e^{-2s}\). Therefore \(\mathbb{E}_1[\mathcal{L}f(X_s)]=-2(2-e^{-2s})+4=2e^{-2s}\). Its integral from \(0\) to \(\ln 2\) is \(\int_0^{\ln 2}2e^{-2s}ds=1-e^{-2\ln 2}=1-\tfrac14=\tfrac34\). Dynkin’s formula gives \(\mathbb{E}_1[X_{\ln 2}^2]=1+\tfrac34=\tfrac74\).
Check: The explicit second-moment expression at \(s=\ln 2\) is \(2-e^{-2\ln 2}=2-\tfrac14=\tfrac74\). This agrees with the generator calculation. At \(t=0\), the integral vanishes and the formula returns the initial second moment \(1\).
Why the Conditions Matter
The formula is a fixed-time identity, not a claim that \(f(X_t)\) changes at the ordinary rate \(\mathcal{L}f(X_t)\) along each path. The stochastic term in Itô’s formula remains present pathwise; the expectation identity follows because, under the stated conditions, that term is a square-integrable martingale with mean zero. The drift term and the quadratic-variation correction together form the generator.
A common mistake is to apply the ordinary chain rule to \(f(B_t)=B_t^2\) and write \(d(B_t^2)=2B_t\,dB_t\). That calculation omits the Itô correction. The correct derivatives are \(f'(y)=2y\) and \(f''(y)=2\), so Itô’s formula gives \(d(B_t^2)=2B_t\,dB_t+\tfrac12(2)\,dt=2B_t\,dB_t+dt\). Taking expectations yields \(\mathbb{E}[B_t^2]=t\), not \(0\). The correction is precisely what makes the generator of Brownian motion send \(y^2\) to \(1\).
In applications, check the stochastic-integral and drift-integral conditions separately. A finite expected generator integral does not by itself ensure that the stochastic integral is a true martingale. Conversely, when the integrand is in \(\mathcal{L}^2\), the Itô isometry supplies square integrability and the conditional-expectation property needed in the proof.
Check Your Understanding
For each question, identify the generator and check the integrability needed before applying the fixed-time identity.
- For \(X_t=x+B_t\) and \(f(y)=y^3\), calculate \(\mathcal{L}f(y)\) and state the resulting Dynkin identity.
- For Brownian motion started at \(x\), apply \(f(y)=\cos y\) and compute both the generator and the expected generator integral.
- For the Ornstein–Uhlenbeck process with parameters \(\kappa,\eta\), calculate \(\mathcal{L}f\) when \(f(y)=y^3\).
- Explain why knowing only that the stochastic term in Itô’s formula is a local martingale is insufficient to conclude that its expectation is zero.
- In the quadratic Brownian example, identify the term omitted by the ordinary chain rule and explain how it changes the expected value.