Scale and Speed as Boundary Tests
For a one-dimensional diffusion, the drift and diffusion coefficient together determine whether a boundary can be reached from the interior, or whether the process can be consistently started there and enter the state space. Scale and speed turn those questions into integrals. Here we use the Feller classification criteria and compute them for Brownian motion, the Ornstein–Uhlenbeck process, and Bessel processes.
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 \(\int_0^t H_s\,dB_s\) is understood for a progressively measurable integrand \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on each finite horizon where that condition holds. For coefficients that become unbounded near a boundary, the equation is interpreted up to the first exit from compact subintervals; its stochastic integrals are then defined by stopping so the integrands are bounded. In each SDE below, \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient.
To see why these densities have the stated form, the generator on twice differentiable functions is \(\mathcal{L}f=\mu f'+\tfrac12\sigma^2 f''\). A scale function satisfies \(\mathcal{L}s=0\). Substituting \(s'\) above gives \(s''/s'=-2\mu/\sigma^2\), and hence \(\mathcal{L}s=0\). The speed density is then \(2/(\sigma^2s')\). These are local formulas on the interior; the endpoint tests use their integrals.
“Accessible” concerns whether the diffusion can reach the endpoint from the interior. “Enterable” concerns whether the boundary can serve as a starting point from which the process enters the interior. The labels do not, by themselves, specify what boundary behavior to impose at a regular endpoint.
Brownian Motion and Ornstein–Uhlenbeck Endpoints
Worked Example: The Two Infinite Endpoints of Brownian Motion
For Brownian motion on \(\mathbb{R}\), the SDE is \(dX_t=dB_t\), so \(\mu=0\) and \(\sigma=1\). Choose \(s(x)=x\); then \(s'(x)=1\) and \(m(x)=2\). Take \(c=0\). At the right endpoint \(r=+\infty\), \(A_r=\int_0^\infty(+\infty-x)2\,dx=\infty\), and \(E_r=\int_0^\infty x\,2\,dx=\infty\). Thus \(+\infty\) is natural. By the same calculation with the integrals directed toward the left endpoint, both left-endpoint integrals are infinite, so \(-\infty\) is natural as well.
Check: with \(s(x)=x\), the scale length from \(0\) to \(R>0\) is \(R\), which tends to infinity as \(R\) increases. The infinite scale length agrees with neither endpoint being accessible.
For the Ornstein–Uhlenbeck process, take \(dX_t=-\theta X_t\,dt+\sigma\,dB_t\), with \(\theta>0\) and \(\sigma>0\). The drift pulls toward the origin, but the endpoint test depends on the full scale and speed integrals, not drift direction alone.
Worked Example: Both Ornstein–Uhlenbeck Endpoints
Set \(a=\theta/\sigma^2>0\). The scale density and speed density are \(s'(x)=e^{ax^2}\) and \(m(x)=\frac{2}{\sigma^2}e^{-ax^2}\). Both are symmetric about zero. At \(+\infty\), \(s(+\infty)=+\infty\), so \(A_r=\infty\). For the other integral, for \(x\ge2\) the interval \([x-1/x,x]\) lies to the right of \(1\), and \(y^2\ge (x-1/x)^2\ge x^2-2\) there. Consequently \(s(x)-s(0)=\int_0^x e^{ay^2}\,dy\ge e^{-2a}e^{ax^2}/x\). Multiplying by \(m(x)\) gives \((s(x)-s(0))m(x)\ge (2e^{-2a}/\sigma^2)(1/x)\). Its integral diverges, so \(E_r=\infty\). The right endpoint is natural. Symmetry gives the same result at \(-\infty\).
Check: \(A_r=\infty\) because \(\int_x^\infty e^{ay^2}\,dy=\infty\) for every finite \(x\), while the displayed lower bound proves divergence of \(E_r\) by comparison with \(\int_2^\infty dx/x\).
The speed density of the Ornstein–Uhlenbeck process is proportional to \(e^{-\theta x^2/\sigma^2}\), an integrable density on the real line. That fact is useful for its long-run behavior, but integrability of speed alone is not the classification test: both endpoint integrals must be checked.
Bessel Processes: The Dimension Threshold at Zero
A Bessel process of dimension \(\delta>0\), considered in the interior \((0,\infty)\), satisfies \(dR_t=dB_t+\frac{\delta-1}{2R_t}\,dt\). The coefficient \(1/R_t\) is unbounded near zero. To interpret the Ito integral on a finite horizon \(T\), one may stop at \(\tau_n=\inf\{t:R_t\notin(1/n,n)\}\); the stopped stochastic integrand is bounded on \([0,T]\), so it belongs to \(\mathcal{L}^2\). The boundary classification describes what happens as these stopping intervals approach the endpoints.
Here \(\sigma=1\) and \(\mu(x)=(\delta-1)/(2x)\). Therefore \(s'(x)=x^{1-\delta}\) and \(m(x)=2x^{\delta-1}\). At zero, the scale integral \(\int_0^c x^{1-\delta}\,dx\) is finite exactly when \(\delta<2\). This is the source of the dimension threshold.
Proof. Fix \(c>0\). First suppose \(0<\delta<2\). Then \(s(0)\) is finite and \(s(x)-s(0)=x^{2-\delta}/(2-\delta)\) for the normalization \(s'(x)=x^{1-\delta}\). Thus \((s(x)-s(0))m(x)=2x/(2-\delta)\), whose integral over \((0,c)\) is finite. Also \(s(c)-s(x)\) remains bounded as \(x\downarrow0\), while \(m(x)=2x^{\delta-1}\) is integrable there because \(\delta>0\). Both left integrals are finite, so zero is regular.
Now suppose \(\delta\ge2\). Then \(s(0)=-\infty\), so \(A_0=\infty\). For \(0<x<c\), \(s(c)-s(x)=\int_x^c y^{1-\delta}\,dy\). If \(\delta=2\), this is \(\log(c/x)\), and its product with \(m(x)=2x\) is integrable near zero. If \(\delta>2\), the integral is \((x^{2-\delta}-c^{2-\delta})/(\delta-2)\). Its product with \(2x^{\delta-1}\) is bounded above by \(2x/(\delta-2)\), which is integrable near zero. Hence \(E_0<\infty\), and zero is entrance.
Finally consider \(+\infty\). If \(0<\delta\le2\), then \(s(+\infty)=+\infty\), so \(A_{+\infty}=\infty\). If \(\delta>2\), then \(s(+\infty)\) is finite and \(s(+\infty)-s(x)=x^{2-\delta}/(\delta-2)\). Multiplying by \(m(x)=2x^{\delta-1}\) gives \(2x/(\delta-2)\), whose integral to infinity diverges. Thus \(A_{+\infty}=\infty\) in this case too. For \(E_{+\infty}\), if \(\delta<2\), \(s(x)-s(c)\) grows as a positive constant times \(x^{2-\delta}\); its product with \(m(x)\) grows as a positive constant times \(x\). If \(\delta=2\), the product is \(2x\log(x/c)\). Both have divergent integrals. If \(\delta>2\), \(s(x)-s(c)\) is bounded below by the positive constant \(s(2c)-s(c)\) for \(x\ge2c\), while \(m(x)=2x^{\delta-1}\) has divergent integral there. So \(E_{+\infty}=\infty\) for all \(\delta>0\). Both integrals are infinite, making \(+\infty\) natural. \(\square\)
Worked Example: Exact Tests for Dimensions One and Two
Take \(c=1\). For dimension \(\delta=1\), \(s'(x)=1\), \(s(x)=x\), and \(m(x)=2\). At zero, \(A_0=\int_0^1 2x\,dx=1\) and \(E_0=\int_0^1 2(1-x)\,dx=1\). Both are finite, so zero is regular.
For dimension \(\delta=2\), \(s'(x)=1/x\), \(s(x)=\log x\), and \(m(x)=2x\). At zero, \(A_0=\infty\) because \(s(0)=-\infty\). The other integral is \(E_0=\int_0^1 (-\log x)2x\,dx=\tfrac12\), so zero is entrance.
Check: integration by parts gives \(\int_0^1 x(-\log x)\,dx=\tfrac14\), hence \(E_0=2\cdot\tfrac14=\tfrac12\). The two dimensions lie on opposite sides of the classification threshold: dimension one is regular, while dimension two is entrance.
When \(0<\delta<2\), a Bessel process in the interior can reach zero; the regular classification means boundary behavior must be specified if the process is to continue there. When \(\delta\ge2\), zero is not accessible from the interior, even though the entrance classification describes it as a possible boundary from which an appropriately defined process may start. That distinction prevents a common inference: being able to define a process from a boundary does not imply that an interior-started process reaches it.
A Common Classification Pitfall
Wrong calculation: “Dimension two has \(s(x)=\log x\), which is finite at every \(x>0\), so zero is accessible.” This uses only interior values of the scale function and ignores the improper limit \(s(0)=-\infty\). Correction: the left accessibility integral contains \(s(x)-s(0)\); since \(s(0)=-\infty\), that integral is infinite. The entrance integral is finite, as the exact calculation above shows. Zero is entrance, not regular. Always evaluate endpoint limits and both Feller integrals rather than classifying from the drift or from the scale function at interior points alone.
Check Your Understanding
For each question, use improper integrals at the endpoint and distinguish accessibility from enterability.
- For Brownian motion on \(\mathbb{R}\), what are the scale density and speed density, and how do the two right-endpoint integrals behave?
- Why does integrability of the Ornstein–Uhlenbeck speed density not replace checking both Feller integrals?
- For Bessel dimension \(\delta=3\), determine the classification of zero and identify which integral is finite there.
- For Bessel dimension \(\delta=1\), compute the two zero-endpoint integrals with \(c=2\).
- Explain why an entrance boundary need not be reachable from an interior starting point.