Long-Time Analysis Problems for One-Dimensional Diffusions

Turning Long-Time Questions into Calculations

For a one-dimensional diffusion, long-time analysis often asks three related questions: can moments of the process be controlled uniformly in time; does a stationary probability density exist; and what value does a time average of an observable approach? Lyapunov functions address the first question, while a normalized stationary density and an ergodic theorem address the latter two. The calculations are linked, but their conclusions are not interchangeable: a moment bound alone does not identify an invariant law, and an invariant law alone does not prove that a time average converges.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a Brownian motion $B_t$. Unless specified otherwise, the examples take $X_0=x$ deterministic and use a nonexplosive strong solution of $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$. A strong solution is adapted to the given filtration and driven by the given Brownian motion; a weak solution allows the probability space and Brownian motion to be chosen as part of the solution. Pathwise uniqueness means solutions driven by the same Brownian motion and with the same initial value agree almost surely; uniqueness in law means their distributions agree. These are distinct properties.

All stochastic integrals below are Itô integrals. An integrand $H_s$ is in $\mathcal{L}^2$ on $[0,T]$ when it is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; in particular, the resulting integral is a square-integrable martingale. When a calculation is localized, square-integrability of the stopped integrand is stated or assumed. The drift is $\mu$, the diffusion coefficient is $\sigma$, and the generator is $\mathcal{L}f=\mu f'+\tfrac12\sigma^2 f''$. The correction $\tfrac12\sigma^2 f''\,dt$ comes from $dB_t\,dB_t=dt$, whereas $dB_t\,dt=0$ and $dt\,dt=0$. No Stratonovich integral is used here.

A Drift Bound Controls Expected Moments

Theorem: Suppose $X_t$ is nonexplosive, $V\ge0$ is twice continuously differentiable, $\mathbb{E}[V(X_0)]<\infty$, and, for constants $C\ge0$ and $\lambda>0$, $\mathcal{L}V(x)\le C-\lambda V(x)$ for every state $x$. Assume Itô's formula applies to $e^{\lambda t}V(X_t)$, that $\mathbb{E}\int_0^T e^{\lambda s}|\mathcal{L}V(X_s)+\lambda V(X_s)|\,ds<\infty$, and that the stochastic integral with integrand $e^{\lambda s}V'(X_s)\sigma(X_s)$ is square-integrable on every finite interval. Then $$ \mathbb{E}[V(X_t)]\le e^{-\lambda t}\mathbb{E}[V(X_0)]+\frac{C}{\lambda}(1-e^{-\lambda t}). $$

Proof. Itô's formula, with the derivatives of $g(t,x)=e^{\lambda t}V(x)$ displayed, gives $g_t=\lambda e^{\lambda t}V(x)$, $g_x=e^{\lambda t}V'(x)$, and $g_{xx}=e^{\lambda t}V''(x)$. Thus $$ d\big(e^{\lambda t}V(X_t)\big) =e^{\lambda t}\big(\lambda V(X_t)+\mathcal{L}V(X_t)\big)\,dt +e^{\lambda t}V'(X_t)\sigma(X_t)\,dB_t, $$ where the second-order correction is $\tfrac12e^{\lambda t}V''(X_t)\sigma^2(X_t)\,dt$, included in $\mathcal{L}V$. The assumed integrability makes the drift integrable and the stochastic integral a square-integrable martingale with expectation zero. Taking expectations and using $\mathcal{L}V+\lambda V\le C$ yields $$ e^{\lambda t}\mathbb{E}[V(X_t)] \le \mathbb{E}[V(X_0)]+C\int_0^t e^{\lambda s}\,ds =\mathbb{E}[V(X_0)]+\frac{C}{\lambda}(e^{\lambda t}-1). $$ Dividing by $e^{\lambda t}$ proves the stated bound. $\square$

The expectation of $V(X_0)$ matters when the initial value is random. Replacing it by the random variable $V(X_0)$ on the right side of a deterministic expectation inequality is not valid. If $X_0=x$ is deterministic, the theorem's first term is $e^{-\lambda t}V(x)$. When $C=0$, it gives exponential decay of the expected Lyapunov function. For $C>0$, it gives a uniform long-time bound of $C/\lambda$ on the limit superior of that expectation. Neither conclusion by itself says that $X_t$ converges almost surely to a point.

Worked Example: A Moment Bound for a Nonlinear Diffusion

Consider $dX_t=(-X_t-X_t^3)\,dt+\sqrt{2}\,dB_t$, with deterministic $X_0=x$, and take $V(x)=x^2$. The coefficients are locally Lipschitz. The standard Lyapunov nonexplosion criterion applies here: the computation below gives $\mathcal{L}V\le2$, so the stopped second-moment estimate rules out explosion. The noise integrand is the constant $\sqrt{2}$, which is progressively measurable and satisfies $\mathbb{E}\int_0^T2\,ds=2T<\infty$.

Here $V'(x)=2x$, $V''(x)=2$, and the generator calculation, including the Itô correction, is $$ \mathcal{L}V(x)=2x(-x-x^3)+\tfrac12(2)(2)=-2x^2-2x^4+2\le 2-2V(x). $$ Thus $C=2$ and $\lambda=2$. The theorem gives $$ \mathbb{E}[X_t^2]\le e^{-2t}x^2+(1-e^{-2t}). $$ For example, if $x=2$ and $t=\log 2$, then $e^{-2t}=1/4$, so the bound is $4/4+1-1/4=3/2$.

Check: at $t=0$ the bound is $x^2$, and as $t$ grows it approaches $1$. The generator also satisfies $\mathcal{L}V\le2$, the weaker estimate used in the nonexplosion criterion.

From Zero Flux to a Normalized Invariant Density

For a diffusion on the real line with sufficiently regular coefficients and $\sigma(x)>0$, the stationary Fokker–Planck equation has probability current $J(x)=\mu(x)p(x)-\tfrac12(\sigma^2(x)p(x))'$. The zero-flux case sets $J(x)=0$. The following derivation identifies a candidate density; it is an invariant probability density only when it satisfies the applicable boundary conditions and can be normalized. This is the zero-current stationary-density formula from the earlier material on the Fokker–Planck equation, not a general assertion that every diffusion has a stationary density.

Formula: If $J=0$, then a candidate density is proportional to $$ \frac{1}{\sigma^2(x)} \exp\left(\int_{x_*}^{x}\frac{2\mu(y)}{\sigma^2(y)}\,dy\right). $$ It gives a probability density only if its integral over the state space is finite and positive, and the resulting stationary solution satisfies the boundary conditions.

Derivation. The equation $J=0$ says $(\sigma^2p)'=2\mu p$. Set $q=\sigma^2p$. Since $p=q/\sigma^2$, the equation becomes $q'/q=2\mu/\sigma^2$ wherever $q>0$. Integrating from $x_*$ to $x$ gives $q(x)=q(x_*)\exp(\int_{x_*}^x 2\mu(y)/\sigma^2(y)\,dy)$. Dividing by $\sigma^2(x)$ gives the displayed form up to a constant. If that function has finite, positive integral $Z$, the constant $1/Z$ normalizes it. This calculation derives the zero-flux candidate; stationarity still requires the stated boundary conditions. $\square$

Worked Example: Normalizing an Ornstein–Uhlenbeck Density

For $dX_t=-2X_t\,dt+2\,dB_t$, the drift is $\mu(x)=-2x$ and the diffusion coefficient is $\sigma(x)=2$. The Brownian integrand is the constant $2$, in $\mathcal{L}^2$ on every finite time interval. The exponent in the zero-flux formula is $$ \int_0^x\frac{2(-2y)}{4}\,dy =\int_0^x-y\,dy=-\frac{x^2}{2}. $$ The factor $1/\sigma^2$ is $1/4$, a constant absorbed by normalization. Since $\int_{\mathbb{R}}e^{-x^2/2}\,dx=\sqrt{2\pi}$, the normalized candidate is $p(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. It is the standard normal density.

Its mean is zero by symmetry. For the second moment, integration by parts gives $$ \int_{\mathbb{R}}x^2p(x)\,dx =-\int_{\mathbb{R}}x p'(x)\,dx =\int_{\mathbb{R}}p(x)\,dx=1, $$ because $p'(x)=-xp(x)$ and the boundary term $xp(x)$ vanishes at both infinities. Thus this invariant law has variance $1$. The boundary term in the zero-flux equation also vanishes, since $\mu p-\tfrac12(4p)'=-2xp+2xp=0$.

Check: the density integrates to one by the Gaussian integral, and the zero-flux expression is identically zero. These checks establish normalization and the stationary equation; the boundary conditions are compatible with the Gaussian tails.

Using an Invariant Law to Evaluate a Time Average

To pass from a stationary density to a pathwise long-run average, one needs an ergodic result, not just a density calculation. We use the standard continuous-time ergodic theorem for positive recurrent, ergodic diffusions: if $\pi$ is the invariant probability law and $f$ is integrable under $\pi$, then, under the theorem's recurrence and irreducibility hypotheses, $T^{-1}\int_0^T f(X_s)\,ds\to\int f\,d\pi$ almost surely. This is a standard result from the earlier material on ergodic averages; its proof is not repeated here. The conclusion is about a time average, not necessarily convergence of $X_t$ itself.

Worked Example: The Long-Run Average of a Shifted Diffusion

Let $dX_t=-3(X_t-2)\,dt+\sqrt{6}\,dB_t$ with deterministic $X_0=x$. The Brownian integrand $\sqrt{6}$ is progressively measurable and has squared integral $6T$ on $[0,T]$. Applying the zero-flux formula gives an exponent $$ \int_2^x\frac{2[-3(y-2)]}{6}\,dy =-\int_2^x(y-2)\,dy=-\frac{(x-2)^2}{2}. $$ After normalization, the invariant density is $p(x)=\frac{1}{\sqrt{2\pi}}e^{-(x-2)^2/2}$, a normal density with mean $2$ and variance $1$. Therefore $\int x^2p(x)\,dx=2^2+1=5$. The observable $f(x)=x^2$ is integrable under this law. The ergodic theorem gives $T^{-1}\int_0^T X_s^2\,ds\to5$ almost surely.

An independent transient check comes from solving the linear SDE with an integrating factor: $$ X_t=2+(x-2)e^{-3t}+\sqrt{6}\int_0^t e^{-3(t-s)}\,dB_s. $$ The integral is square-integrable because $6\int_0^t e^{-6(t-s)}\,ds=1-e^{-6t}<\infty$, and its mean is zero and variance is $1-e^{-6t}$ by the Itô isometry. Hence $$ \mathbb{E}[X_t^2]=\big(2+(x-2)e^{-3t}\big)^2+1-e^{-6t}\longrightarrow5. $$ The limiting expected observable agrees with the invariant-density calculation. Consequently the expected time average also tends to $5$: integrating this expectation over $[0,T]$ and dividing by $T$, the decaying exponential terms vanish as $T\to\infty$. This expectation calculation is a check; the almost-sure time-average conclusion uses the ergodic theorem.

Check: if $x=2$, the formula reduces to $\mathbb{E}[X_t^2]=4+1-e^{-6t}=5-e^{-6t}$, which starts at $4$ and tends to $5$, as predicted by the invariant law.

Common Pitfalls in Long-Time Problems

Do not confuse moment control with almost-sure stability. The bound in the first example controls $\mathbb{E}[X_t^2]$; it does not show that every sample path converges to zero. Noise can keep a diffusion fluctuating even when its expected Lyapunov function is bounded.

Do not omit the expectation for random initial data. A tempting but incorrect statement is $\mathbb{E}[V(X_t)]\le e^{-\lambda t}V(X_0)+C(1-e^{-\lambda t})/\lambda$. For $dX_t=-X_t\,dt$, $V(x)=x^2$, $C=0$, $\lambda=2$, and $X_0$ equal to $0$ or $1$ with probability $1/2$, the left side is $\tfrac12e^{-2t}$. On the event $X_0=0$, the proposed right side is zero. The correct right side is $e^{-2t}\mathbb{E}[V(X_0)]=\tfrac12e^{-2t}$, as the proved theorem states.

Do not treat a formal density as an invariant probability law automatically. The zero-flux expression must be integrable, normalizable, and compatible with the boundary conditions. And even after obtaining an invariant law, a claim about almost-sure time averages requires the recurrence and ergodicity assumptions of an ergodic theorem.

Check Your Understanding

For each question, state the assumptions needed for the conclusion as well as carrying out the calculation.

If $\mathcal{L}V\le 3-2V$ and $\mathbb{E}[V(X_0)]=4$, what upper bound does the Lyapunov theorem give for $\mathbb{E}[V(X_t)]$?
For $dX_t=-aX_t\,dt+\sigma\,dB_t$, with $a>0$ and $\sigma>0$, derive the exponent in the zero-flux density formula and state its normalized variance.
Why does a finite, normalized zero-flux density not by itself prove that every time average converges almost surely?
For the shifted diffusion in the worked example, compute $\int (x-2)^2p(x)\,dx$, and hence the long-run average of $(X_t-2)^2$ under the ergodic theorem.
In the Lyapunov proof, why is the expectation of the stochastic integral zero, and which integrability condition ensures this?