From an SDE to Its Generator
For a diffusion, the drift and diffusion coefficients describe its local change, but the generator packages their effect on smooth test functions. Applying it to a carefully chosen nonnegative function can reveal whether the process tends to move toward regions where that function is smaller. Such a function is called a Lyapunov function. The resulting drift inequality is a useful route to moment bounds; by itself, it does not establish every form of stability or a stationary distribution.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a continuous, adapted, nonexplosive solution of the one-dimensional Itô SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with initial value \(X_0\). Here \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. Whenever an Itô integral \(\int_0^t H_s\,dB_s\) is used, its integrand \(H\) is progressively measurable and belongs to \(\mathcal{L}^2\) on the time interval in question: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). We assume the solution and coefficients are regular enough for Itô's formula to apply.
To see the source of the formula, Itô's formula gives, for \(f\in C^2\), \(df(X_t)=f'(X_t)\,dX_t+\tfrac12 f''(X_t)\sigma(X_t)^2\,dt\). Substituting the SDE and using \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) yields the generator term \(Lf(X_t)\,dt\). These multiplication rules are Itô rules; they are not ordinary differential identities.
The semigroup interpretation can be justified directly in a convenient bounded setting. Suppose the coefficients are bounded and continuous, \(f\in C^2\) has bounded derivatives, and \(Lf\) is bounded and continuous. Start the solution at a deterministic \(x\). Then the stochastic-integral integrand \(f'(X_s)\sigma(X_s)\) is bounded and progressively measurable, so it is in \(\mathcal{L}^2\) on every finite interval. Its Itô integral is a square-integrable martingale and has mean zero. Taking expectations in Itô's formula gives \(P_hf(x)-f(x)=\int_0^h\mathbb{E}_x[Lf(X_s)]\,ds\). Continuity of \(X\) at zero and continuity and boundedness of \(Lf\) imply \(\mathbb{E}_x[Lf(X_s)]\to Lf(x)\) as \(s\downarrow0\). Dividing by \(h\) and taking the limit proves the asserted semigroup-generator formula in this setting.
The Drift Inequality and an Expected-Value Bound
The drift inequality has a precise finite-time consequence, provided all terms are integrable. This qualification matters: a stochastic integral with a square-integrable integrand is a martingale, but adding a nonintegrable initial value does not produce an integrable martingale.
Proof. Itô's formula, with \(f=V\), \(f'=V'\), and \(f''=V''\), states \(V(X_t)=V(X_0)+\int_0^t LV(X_s)\,ds+\int_0^t V'(X_s)\sigma(X_s)\,dB_s\), where \(LV=\mu V'+\tfrac12\sigma^2V''\). The last integrand is progressively measurable and satisfies the stated \(\mathcal{L}^2\) condition. By the Itô isometry, its integral is square-integrable; its conditional expectation at a later time given \(\mathcal{F}_s\) equals its value at \(s\), so it is a true martingale. Also \(V(X_0)\) is integrable, and the assumed integrability of \(LV\) permits taking expectations in the identity. Thus \(m(t)=\mathbb{E}V(X_t)\) satisfies \(m(t)=m(0)+\int_0^t\mathbb{E}[LV(X_s)]\,ds\). The integrability assumptions make \(m\) absolutely continuous on finite intervals. The pointwise drift condition gives \(m'(t)\le c-\lambda m(t)\) for almost every \(t\). Multiplying by \(e^{\lambda t}\) gives \(\frac{d}{dt}(e^{\lambda t}m(t))\le ce^{\lambda t}\) almost everywhere. Integrating from \(0\) to \(t\), and using \(\int_0^t ce^{\lambda s}\,ds=\frac{c}{\lambda}(e^{\lambda t}-1)\), proves the displayed bound. Taking the limit superior proves the final assertion. \(\square\)
The estimate controls the expected value of \(V(X_t)\), not necessarily the whole distribution of \(X_t\). If \(V\) is coercive, large values of the state force large values of \(V\), so the bound limits the probability of being far away: for \(R>0\) with \(a_R=\inf_{|x|\ge R}V(x)>0\), Markov's inequality gives \(\mathbb{P}(|X_t|\ge R)\le \mathbb{E}V(X_t)/a_R\). This does not by itself prove convergence to an invariant distribution, uniqueness of such a distribution, or ergodicity.
Worked Generator Calculations
Worked Example: A Quadratic Lyapunov Function for an Ornstein–Uhlenbeck Diffusion
Consider \(dX_t=-2X_t\,dt+3\,dB_t\), with deterministic \(X_0=x_0\), and \(V(x)=1+x^2\). Compute \(LV\) and identify a drift inequality.
Here \(\mu(x)=-2x\), \(\sigma(x)=3\), \(V'(x)=2x\), and \(V''(x)=2\). Therefore \(LV(x)=(-2x)(2x)+\tfrac12(3^2)(2)=-4x^2+9\). Since \(V(x)=1+x^2\), this is \(13-4V(x)\). The drift condition holds with \(c=13\) and \(\lambda=4\), both of which meet the required signs.
The bound is \(\mathbb{E}V(X_t)\le e^{-4t}(1+x_0^2)+\tfrac{13}{4}(1-e^{-4t})\). As an independent check, the explicit linear-diffusion solution is \(X_t=e^{-2t}x_0+3\int_0^t e^{-2(t-s)}\,dB_s\). The integral has mean zero and variance \(\int_0^t9e^{-4(t-s)}\,ds=\tfrac94(1-e^{-4t})\), so \(\mathbb{E}V(X_t)=1+x_0^2e^{-4t}+\tfrac94(1-e^{-4t})\). Expanding the bound gives \(e^{-4t}(1+x_0^2)+\tfrac{13}{4}(1-e^{-4t})=1+x_0^2e^{-4t}+\tfrac94(1-e^{-4t})\), exactly the same value. The theorem's integrability conditions hold: this explicit Gaussian solution has finite second moments on every finite interval, and \(V'\sigma=6X\).
Worked Example: A Stable Geometric Diffusion
Let \(dX_t=-X_t\,dt+X_t\,dB_t\), \(X_0=x_0\) deterministic, and again take \(V(x)=1+x^2\). Find the drift condition and check the expected-value bound.
The coefficients are \(\mu(x)=-x\) and \(\sigma(x)=x\), while \(V'(x)=2x\) and \(V''(x)=2\). Hence \(LV(x)=(-x)(2x)+\tfrac12x^2(2)=-x^2=1-V(x)\). Thus \(c=1\) and \(\lambda=1\). The theorem gives \(\mathbb{E}V(X_t)\le e^{-t}(1+x_0^2)+(1-e^{-t})=1+x_0^2e^{-t}\).
For a direct check, set \(g(t,b)=x_0e^{b-3t/2}\). Its derivatives are \(g_t=-\tfrac32g\), \(g_b=g\), and \(g_{bb}=g\). Itô's formula, including the correction term, gives \(dg(t,B_t)=g_t\,dt+g_b\,dB_t+\tfrac12g_{bb}\,dt=-g\,dt+g\,dB_t\). So \(X_t=x_0e^{B_t-3t/2}\) solves the SDE. Since \(B_t\) is normal with variance \(t\), \(\mathbb{E}X_t^2=x_0^2e^{-3t}\mathbb{E}e^{2B_t}=x_0^2e^{-3t}e^{2t}=x_0^2e^{-t}\). Therefore \(\mathbb{E}V(X_t)=1+x_0^2e^{-t}\), attaining the bound. The martingale-integrand condition also holds: \(V'(X_t)\sigma(X_t)=2X_t^2\), and \(\mathbb{E}\int_0^T4X_s^4\,ds=4x_0^4\int_0^T e^{2s}\,ds=2x_0^4(e^{2T}-1)<\infty\). This check uses a deterministic initial value; a merely finite second moment for a random \(X_0\) would not guarantee this fourth-moment integrability.
Worked Example: Brownian Motion Does Not Have This Quadratic Drift
For \(dX_t=dB_t\), \(X_0=x_0\) deterministic, test \(V(x)=1+x^2\) against a condition \(LV\le c-\lambda V\) with \(c\ge0\) and \(\lambda>0\).
Now \(\mu=0\), \(\sigma=1\), \(V'=2x\), and \(V''=2\). Thus \(LV=0+\tfrac12(1)(2)=1\) at every state. No constants \(c\ge0,\lambda>0\) can satisfy the proposed inequality for all \(x\): its right-hand side \(c-\lambda(1+x^2)\) tends to \(-\infty\) as \(|x|\to\infty\), while its left-hand side remains \(1\).
The direct moment calculation confirms the absence of a uniform bound: \(X_t=x_0+B_t\), so \(\mathbb{E}V(X_t)=1+x_0^2+t\). Here \(V'(X_t)\sigma(X_t)=2X_t\), whose squared expectation is \(4(x_0^2+t)\), integrable over every finite interval. The failure is the drift inequality, not a lack of finite-time integrability.
What the Condition Does—and Does Not—Say
A negative term \(-\lambda V\) in the generator offsets the constant input \(c\). The theorem quantifies this balance: the initial contribution is multiplied by \(e^{-\lambda t}\), while the long-time upper bound is \(c/\lambda\). A larger \(\lambda\) means stronger negative drift relative to the same \(c\); it does not automatically give a sharper bound if \(c\) changes too.
Common pitfall: omitting the Itô correction. In the Ornstein–Uhlenbeck example, an incorrect ordinary-chain-rule calculation might write \(LV=V'\mu=-4x^2\), forgetting the second derivative term. The correct calculation is \(V'\mu+\tfrac12V''\sigma^2=(-4x^2)+\tfrac12(2)(9)=-4x^2+9\). That missing \(9\) changes the drift inequality and the resulting bound. The diffusion coefficient contributes through its square, even when it is constant.
Another pitfall is overstating the conclusion. The inequality controls \(\mathbb{E}V(X_t)\) under explicit integrability assumptions. It is not, without additional arguments, a proof of pathwise convergence, an invariant-measure theorem, or ergodicity. Conversely, failure of this particular inequality for a chosen \(V\) does not prove instability: it only says that this test function and this drift condition do not furnish the stated bound.
Check Your Understanding
Use the generator definition and keep the integrability hypotheses in view.
- For \(dX_t=-3X_t\,dt+2\,dB_t\) and \(V(x)=1+x^2\), compute \(LV\) and find constants \(c\ge0\) and \(\lambda>0\) for a drift inequality.
- For \(dX_t=\alpha X_t\,dt+\beta X_t\,dB_t\), compute \(L(x^2)\). Which combination of \(\alpha\) and \(\beta\) determines the sign of this expression?
- Why does \(\mathbb{E}\int_0^T(V'(X_s)\sigma(X_s))^2\,ds<\infty\) make the Itô integral a true martingale, and what separate condition is needed on \(V(X_0)\)?
- For Brownian motion with \(V(x)=1+x^2\), explain why no positive \(\lambda\) can give \(LV\le c-\lambda V\) for all real \(x\).
- State one conclusion that the Lyapunov expectation bound gives, and one long-time conclusion it does not establish by itself.