From the Generator to the Semigroup
A diffusion’s generator describes its instantaneous action on test functions, while its transition semigroup describes expectations after a positive amount of time. The semigroup-generator identity connects these two views: the change in an expectation over a time interval is the time integral of the expected generator. Its proof is short once the integrability conditions are explicit, but those conditions matter: Itô’s formula produces a stochastic integral, and taking expectations of that term requires more than calling it a local martingale.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For each initial state \(x\in\mathbb{R}\), consider the Ornstein-Uhlenbeck process \(dX_t=-\lambda X_t\,dt+\sigma\,dB_t,\ X_0=x\), where \(\lambda>0\) and \(\sigma\ge0\). Its drift is \(-\lambda x\) and its diffusion coefficient is \(\sigma\). The coefficients are globally Lipschitz and have linear growth, so the SDE has a unique strong solution. Its transition semigroup is \(P_t f(x)=\mathbb{E}[f(X_t^x)]\) whenever this expectation exists. The semigroup property is \(P_{t+s}f=P_t(P_sf)\).
For every stochastic integral below, the integrand is understood to be adapted and progressively measurable. The class \(\mathcal{L}^2\) on \([0,T]\) consists of such processes \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). An integral \(\int_0^T H_s\,dB_s\) with \(H\in\mathcal{L}^2\) is a square-integrable martingale at its upper limit, so it has mean zero.
The factor \(\tfrac12\sigma^2 f''\) is the Itô correction: in Itô’s multiplication rules, \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). It is not part of the ordinary chain rule.
The Identity and Its Generator Consequence
Proof. The process \(X^x\) is continuous and satisfies the stated SDE. Itô’s formula, with the function \(f\) and derivatives \(f'\) and \(f''\), gives $$ f(X_t^x)=f(x)+\int_0^t f'(X_s^x)(-\lambda X_s^x)\,ds +\int_0^t \sigma f'(X_s^x)\,dB_s +\frac12\int_0^t \sigma^2 f''(X_s^x)\,ds. $$ Combining the two ordinary time integrals gives \(\int_0^t\mathcal{L}f(X_s^x)\,ds\). The assumed square-integrability makes \(H_s=\sigma f'(X_s^x)\) an element of \(\mathcal{L}^2\) on \([0,T]\). Consequently \(\int_0^t H_s\,dB_s\) is a true martingale with conditional mean-zero increments, and in particular has expectation zero. Taking expectations is justified for the other terms by the stated integrability assumptions. The definition of the semigroup gives \(\mathbb{E}[f(X_t^x)]=P_tf(x)\) and \(\mathbb{E}[\mathcal{L}f(X_s^x)]=P_s\mathcal{L}f(x)\). Thus taking expectations in the displayed Itô formula proves the identity. \(\square\)
In the usual pointwise sense, the infinitesimal generator \(A\) of the semigroup is defined where the limit exists by \(Af(x)=\lim_{h\downarrow0}(P_hf(x)-f(x))/h\). The identity identifies this derivative with \(\mathcal{L}f(x)\) under an additional, convenient condition.
Proof. Divide the identity by \(h>0\): $$ \frac{P_hf(x)-f(x)}h=\frac1h\int_0^h\mathbb{E}[\mathcal{L}f(X_s^x)]\,ds. $$ The explicit solution of the Ornstein-Uhlenbeck SDE is \(X_s^x=e^{-\lambda s}x+\sigma\int_0^s e^{-\lambda(s-r)}\,dB_r\). The deterministic integrand \(e^{-\lambda(s-r)}\) is progressive and square-integrable on each finite time interval, so this Itô integral is well-defined. It has mean zero and variance \(\int_0^s e^{-2\lambda(s-r)}\,dr\), which tends to zero as \(s\downarrow0\). Hence \(X_s^x\to x\) in probability. Boundedness and continuity of \(\mathcal{L}f\) imply \(\mathbb{E}[\mathcal{L}f(X_s^x)]\to\mathcal{L}f(x)\): boundedness controls the contribution where \(X_s^x\) is far from \(x\), and continuity controls the contribution near \(x\). The average over \(0\le s\le h\) therefore tends to \(\mathcal{L}f(x)\) as \(h\downarrow0\), proving the claim. \(\square\)
Worked Ornstein-Uhlenbeck Expectations
The explicit solution also gives \(X_t^x\) a normal distribution with mean \(m_t=e^{-\lambda t}x\) and variance \(v_t=\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})\). Indeed, the stochastic-integral term has a centered normal distribution and variance \(\sigma^2\int_0^t e^{-2\lambda(t-r)}\,dr=v_t\). We use this distribution to calculate three semigroup expectations and check them against the identity.
Worked Example: The Linear Test Function
Take \(f(x)=x\). Then \(f'(x)=1\), \(f''(x)=0\), and \(\mathcal{L}f(x)=-\lambda x\). The relevant stochastic-integral integrand is the constant \(\sigma\), which belongs to \(\mathcal{L}^2\) on every finite interval. The linear function and its generator have finite expectations under the normal transition law.
The mean of \(X_t^x\) is \(e^{-\lambda t}x\), so \(P_tf(x)=e^{-\lambda t}x\). The right-hand side of the identity is $$ \int_0^t P_s\mathcal{L}f(x)\,ds =\int_0^t-\lambda e^{-\lambda s}x\,ds =x(e^{-\lambda t}-1). $$ This equals \(P_tf(x)-f(x)\).
Check: set \(\lambda=1\), \(x=2\), and \(t=\ln 2\). Then \(P_tf(x)=e^{-\ln2}\cdot2=1\), so the change is \(1-2=-1\). The integral gives \(2(e^{-\ln2}-1)=2(\tfrac12-1)=-1\), as required.
Worked Example: The Quadratic Test Function
Take \(f(x)=x^2\). Its derivatives are \(f'(x)=2x\) and \(f''(x)=2\), so $$ \mathcal{L}f(x)=-2\lambda x^2+\sigma^2. $$ All required expectations are finite because a normal random variable has finite moments of every order, including the second moment needed for the stochastic-integral condition.
For a normal random variable with mean \(m_t\) and variance \(v_t\), the second moment is \(m_t^2+v_t\). Thus $$ P_tf(x)=e^{-2\lambda t}x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t}). $$ At time \(s\), applying the semigroup to the generator gives $$ P_s\mathcal{L}f(x)=\sigma^2-2\lambda\left(e^{-2\lambda s}x^2+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda s})\right) =e^{-2\lambda s}(\sigma^2-2\lambda x^2). $$ Integrating yields \(\frac{1-e^{-2\lambda t}}{2\lambda}(\sigma^2-2\lambda x^2)\). Subtracting \(f(x)\) from the displayed expression for \(P_tf(x)\) gives exactly that same quantity.
Check: take \(\lambda=1\), \(\sigma=2\), \(x=1\), and \(t=\ln2\). Since \(e^{-2t}=e^{-2\ln2}=\tfrac14\), the moment formula gives \(P_tf(x)=\tfrac14+\frac42(1-\tfrac14)=\tfrac14+\tfrac32=\tfrac74\). The change is \(\tfrac74-1=\tfrac34\). The integral gives \(\frac{1-\tfrac14}{2}(4-2)=\frac38\cdot2=\tfrac34\).
Worked Example: An Exponential Test Function
Let \(f(x)=e^{ux}\), where \(u\in\mathbb{R}\). Here \(f'(x)=ue^{ux}\), \(f''(x)=u^2e^{ux}\), and $$ \mathcal{L}f(x)=\left(-\lambda ux+\frac12\sigma^2u^2\right)e^{ux}. $$ The exponential moments of a normal random variable are finite for every real \(u\). The stochastic-integral condition also holds on finite time intervals: \(\mathbb{E}[e^{2uX_s^x}]\) and \(\mathbb{E}[(X_s^x)^2e^{2uX_s^x}]\) are finite and bounded as \(s\) ranges over such an interval.
The normal moment-generating formula gives $$ P_tf(x)=\exp\left(ue^{-\lambda t}x+\frac{u^2\sigma^2}{4\lambda}(1-e^{-2\lambda t})\right). $$ To check the integral identity directly, set \(F(s)=\exp(um_s+\tfrac12u^2v_s)\), with \(m_s=e^{-\lambda s}x\) and \(v_s=\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda s})\). The normal moment formula, differentiated with respect to \(u\), gives \(\mathbb{E}[X_s^xe^{uX_s^x}]=(m_s+uv_s)F(s)\). Therefore $$ P_s\mathcal{L}f(x)=\left(-\lambda u(m_s+uv_s)+\frac12\sigma^2u^2\right)F(s). $$ Since \(m_s'=-\lambda m_s\) and \(v_s'=\sigma^2-2\lambda v_s\), differentiation gives \(F'(s)=P_s\mathcal{L}f(x)\). Integrating from \(0\) to \(t\) gives \(F(t)-F(0)=P_tf(x)-f(x)\).
Check: choose \(\lambda=1\), \(\sigma=\sqrt{2}\), \(x=0\), \(u=1\), and \(t=\ln2\). The mean is zero and the variance is \(\frac22(1-\tfrac14)=\tfrac34\), so \(P_tf(0)=e^{(1/2)(3/4)}=e^{3/8}\). Also \(F(0)=1\), and the direct derivative calculation gives \(\int_0^{\ln2}P_s\mathcal{L}f(0)\,ds=F(\ln2)-F(0)=e^{3/8}-1\), matching the change in the expectation.
Why the Identity Is Useful—and a Common Error
The identity turns a time-evolved expectation into an integral involving the generator of the original test function. This is useful when \(P_tf\) is difficult to obtain directly but \(P_s\mathcal{L}f\) can be calculated or bounded. It also explains why the generator is the first-order, small-time description of a diffusion semigroup. For the Ornstein-Uhlenbeck process, the linear and quadratic examples show how the identity yields evolution equations for moments without confusing the drift contribution with the Itô correction.
Common mistake: replacing the integral by \(t\mathcal{L}f(x)\) treats the generator as though it stayed at its initial value throughout the time interval. For \(f(x)=x\), that incorrect calculation gives \(P_tf(x)-f(x)=-\lambda tx\). In fact, the generator’s expectation at time \(s\) is \(P_s\mathcal{L}f(x)=-\lambda e^{-\lambda s}x\), which changes with \(s\), and its integral is \(x(e^{-\lambda t}-1)\). For \(\lambda=1\), \(x=2\), and \(t=\ln2\), the incorrect expression is \(-2\ln2\), whereas the correct change is \(-1\).
The theorem’s assumptions prevent a second common error: taking expectations of a stochastic integral that is only known to be a local martingale. Here the square-integrability condition puts the Itô integrand in \(\mathcal{L}^2\), making the integral a true martingale with mean zero. If that condition is unavailable, localization and additional estimates may be needed before the same expectation argument is justified.
Check Your Understanding
For each question, state any parameter assumptions you use and distinguish the semigroup expectation from the generator value.
- For \(f(x)=x^3\), calculate \(\mathcal{L}f(x)\) and state which moments are needed to apply the identity.
- Use the Ornstein-Uhlenbeck mean and variance to calculate \(P_t(x^2)(x)\) when \(\lambda=2\), \(\sigma=2\), and \(x=0\).
- Explain why square-integrability of \(\sigma f'(X_s^x)\) allows the stochastic integral in Itô’s formula to have expectation zero.
- For \(f(x)=x\), compare \(\int_0^tP_s\mathcal{L}f(x)\,ds\) with \(t\mathcal{L}f(x)\). For which initial states do these expressions agree for every \(t\ge0\)?
- State the extra condition in the corollary that lets the semigroup-generator identity identify the pointwise generator with \(\mathcal{L}f\).