Feller’s Criteria for Classifying Diffusion Boundaries

Why Boundary Classification Needs Two Tests

A one-dimensional diffusion can approach an endpoint in two different senses. It may reach that endpoint in finite time when started in the interior; and, separately, it may be possible to start at the endpoint and have the process enter the interior. These properties need not agree. A boundary reachable from inside may not permit entrance from outside, while an inaccessible boundary may still serve as an entrance point.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider a one-dimensional diffusion on an open interval $I=(l,r)$, where either endpoint may be infinite, described in the interior by $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$. Assume $\mu$ and $\sigma$ are continuous, $\sigma(x)>0$ on $I$, and the coefficients are locally Lipschitz there, so the interior equation has a unique solution up to its lifetime. The stochastic integral in this equation is the Ito integral: on each compact subinterval, its progressively measurable integrand $\sigma(X_t)$, stopped on leaving that subinterval, satisfies $\mathbb{E}\int_0^T \sigma(X_t)^2\,dt<\infty$. The drift is $\mu$, and the diffusion coefficient is $\sigma$.

Choose any reference point $x_0\in I$. The scale density measures the directional bias induced by the drift relative to the noise. The scale function is an antiderivative of that density, and the speed density weights how much time the diffusion spends near each location. Additive and positive multiplicative changes to the scale function are harmless; this will be verified below.

Definition: Define the scale density $s'(x)$, scale function $S(x)$, and speed density $m(x)$ by $$ s'(x)=\exp\!\left(-\int_{x_0}^{x}\frac{2\mu(y)}{\sigma^2(y)}\,dy\right),\qquad S(x)=\int_{x_0}^{x}s'(y)\,dy,\qquad m(x)=\frac{2}{\sigma^2(x)s'(x)}. $$ Thus $s'(x)>0$, so $S$ is strictly increasing, and $m(x)>0$. All endpoint integrals below are improper integrals of nonnegative functions.

The Endpoint Tests

Let $c\in(l,r)$. For the left endpoint, take the limits $S(l)=\lim_{x\downarrow l}S(x)$, allowing $S(l)=-\infty$. For the right endpoint, define $S(r)=\lim_{x\uparrow r}S(x)$, allowing $S(r)=+\infty$. If a scale difference in an integrand is infinite, the corresponding integral is infinite. This convention avoids ambiguous products of zero and infinity.

Theorem: The standard Feller boundary classification theorem gives the following tests for a regular one-dimensional diffusion. At the left endpoint $l$, define $$ \Sigma_l(c)=\int_l^c\big(S(x)-S(l)\big)m(x)\,dx,\qquad N_l(c)=\int_l^c\big(S(c)-S(x)\big)m(x)\,dx. $$ The endpoint is accessible from the interior exactly when $\Sigma_l(c)<\infty$, and is an admissible entrance point exactly when $N_l(c)<\infty$. At the right endpoint $r$, define $$ \Sigma_r(c)=\int_c^r\big(S(r)-S(x)\big)m(x)\,dx,\qquad N_r(c)=\int_c^r\big(S(x)-S(c)\big)m(x)\,dx. $$ The right endpoint is accessible exactly when $\Sigma_r(c)<\infty$, and is an admissible entrance point exactly when $N_r(c)<\infty$. The proof of this standard theorem, which connects these tests to the diffusion’s hitting and entrance behavior, is beyond this tutorial.

Here, accessible means that the endpoint can be reached from an interior starting point in finite time with positive probability. Entrance means that the process can be started at that endpoint and enter the interior, although the endpoint cannot be reached from the interior. These meanings give the four mutually exclusive classes:

Accessible from the interior?	Admissible as an entrance?	Boundary class
Yes	Yes	Regular
Yes	No	Exit
No	Yes	Entrance
No	No	Natural

The endpoint tests use a fixed interior point $c$, but their finiteness does not depend on its choice. Nor does the classification depend on the arbitrary normalization of scale.

Proposition: The finiteness or divergence of each endpoint test is unchanged if the scale function is replaced by $\widetilde S=aS+b$, where $a>0$ and $b\in\mathbb{R}$. The tests are also independent of the interior reference point $c$.

Proof. Under $\widetilde S=aS+b$, the scale density becomes $\widetilde s'=a s'$, and the corresponding speed density is $\widetilde m=m/a$. Each scale difference in an endpoint test is multiplied by $a$, while the speed density is divided by $a$. Therefore each integrand, and hence each integral, is unchanged.

For the left accessibility test, changing the upper limit $c$ adds or removes only an integral over a compact subinterval of $I$; there the scale differences and speed density are finite and continuous. If $S(l)=-\infty$, every left accessibility test is infinite by the stated convention. For the left entrance test, suppose $d>c$. On $x\in(l,c)$, the integrand for $d$ is the integrand for $c$ plus the nonnegative term $(S(d)-S(c))m(x)$; the additional integral over $(c,d)$ is finite. If the test at $c$ is finite, then $S(c)-S(x)$ is bounded below by a positive constant for $x$ sufficiently close to $l$, so $\int_l^c m(x)\,dx$ is finite there. Thus the added term is integrable. Conversely, finiteness at $d$ implies finiteness at $c$, since $S(d)-S(x)\ge S(c)-S(x)$ for $x<c$. The right endpoint arguments follow with the inequalities and integration directions reversed. Hence the test outcomes do not depend on $c$. $\square$

Worked Classifications

Worked Example: A Regular Left Endpoint

Classify the left endpoint $0$ for the diffusion on $(0,\infty)$ $dX_t=\tfrac12\,dt+\sqrt{2X_t}\,dB_t$. The equation is understood up to exit from compact subintervals of $(0,\infty)$; the stopped diffusion integrands are progressively measurable and square-integrable on finite time intervals.

Here $\mu(x)=\tfrac12$ and $\sigma^2(x)=2x$, so $2\mu(x)/\sigma^2(x)=1/(2x)$. Taking $x_0=1$ gives $s'(x)=\exp(-\tfrac12\log x)=x^{-1/2}$. Choose $S(x)=2\sqrt{x}$, which differs from the integral normalized at $1$ only by an additive constant. The speed density is $m(x)=2/(2x\cdot x^{-1/2})=x^{-1/2}$, and $S(0)=0$.

For any $c>0$, the accessibility integral is $$ \Sigma_0(c)=\int_0^c 2\sqrt{x}\,x^{-1/2}\,dx =\int_0^c 2\,dx=2c<\infty. $$ The entrance integral is $$ N_0(c)=\int_0^c(2\sqrt{c}-2\sqrt{x})x^{-1/2}\,dx =4\sqrt{c}\sqrt{c}-2c=2c<\infty. $$ Thus $0$ is regular. Check: the entrance integrand is nonnegative on $(0,c)$, and its direct evaluation agrees with its finite near-zero behavior, which is proportional to $x^{-1/2}$; that power is integrable at zero.

Worked Example: An Exit Left Endpoint

Consider $dX_t=-dt+\sqrt{X_t}\,dB_t$ on $(0,\infty)$, again interpreted up to exit from compact subintervals. Here $2\mu/\sigma^2=-2/x$. With $x_0=1$, $s'(x)=\exp(2\log x)=x^2$. Take $S(x)=x^3/3$, so $S(0)=0$; then $m(x)=2/(x\cdot x^2)=2x^{-3}$.

The accessibility integral is $$ \Sigma_0(c)=\int_0^c\frac{x^3}{3}\,2x^{-3}\,dx =\frac23\int_0^c1\,dx=\frac{2c}{3}<\infty. $$ For entrance, the exact integrand is $$ \big(S(c)-S(x)\big)m(x) =\frac{2}{3}(c^3-x^3)x^{-3} =\frac23\left(c^3x^{-3}-1\right). $$ As $x\downarrow0$, this is bounded below by a positive constant times $x^{-3}$, whose integral diverges. Hence $N_0(c)=\infty$, and $0$ is an exit boundary. Check: accessibility is finite because its integrand simplifies to the constant $2/3$; entrance fails because its integrand has a nonintegrable $x^{-3}$ term.

Worked Example: An Entrance Left Endpoint

Now use $dX_t=\tfrac12\,dt+\sqrt{X_t}\,dB_t$ on $(0,\infty)$, with the same local interpretation of the stochastic equation. The ratio $2\mu/\sigma^2$ is $1/x$. Therefore $s'(x)=x^{-1}$; choose $S(x)=\log x$, and compute $m(x)=2/(x\cdot x^{-1})=2$.

Since $S(0)=-\infty$, the left accessibility integral is infinite. The entrance integral, for $c>0$, is $$ N_0(c)=2\int_0^c\log(c/x)\,dx=2c<\infty. $$ For the evaluation, substitute $u=x/c$ to obtain $2c\int_0^1-\log u\,du=2c$; integration by parts gives $\int_0^1-\log u\,du=1$, since $u\log u\to0$ at zero. Thus $0$ is an entrance boundary. Check: the scale tends to $-\infty$, which makes the accessibility test fail, while the logarithmic divergence in the entrance integrand is integrable.

Worked Example: A Natural Left Endpoint

For geometric Brownian motion on $(0,\infty)$, $dX_t=X_t\,dB_t$, the drift is zero and $\sigma^2(x)=x^2$. The scale density is $s'(x)=1$; take $S(x)=x$. The speed density is $m(x)=2/x^2$.

At zero, $S(0)=0$. For $c>0$, $$ \Sigma_0(c)=2\int_0^c\frac{dx}{x}=\infty,\qquad N_0(c)=2\int_0^c\frac{c-x}{x^2}\,dx=\infty. $$ The first integral diverges logarithmically. In the second, the integrand is $2c/x^2-2/x$, whose positive leading term has a divergent integral at zero. Neither test is finite, so zero is natural. Check: the conclusion uses both tests: failure of accessibility alone would not distinguish an entrance endpoint from a natural one.

Using the Classification Carefully

The tests answer two separate questions, so checking only one integral is not enough to name the boundary class. In particular, “inaccessible” does not mean “natural”: an inaccessible endpoint with finite entrance integral is an entrance boundary. Likewise, “reachable” does not imply that the process can be started there and enter the state space; that is why exit and regular endpoints are distinguished.

There is also a useful way to read the integrands. The scale difference vanishes at the endpoint when the scale has a finite endpoint value, while the speed density describes the local weight in the test. A vanishing scale difference can compensate for a large speed density, as in the regular example. Conversely, a finite speed density does not guarantee accessibility if the scale difference is infinite, as in the entrance example.

Common mistake. A tempting but incorrect shortcut is to inspect only whether $S(l)$ is finite and declare the endpoint accessible. For the entrance example, $S(0)=-\infty$, so this shortcut happens to say “inaccessible,” but it gives no answer to whether entrance is possible. The actual second test is finite, $N_0(c)=2c$, and the boundary is entrance, not natural. The correction is to compute both scale-speed integrals and then use the two-question classification table.

These criteria turn qualitative boundary behavior into endpoint integrability calculations. Their practical value is that they separate hitting from entrance and allow coefficients that look complicated to be classified through their scale and speed asymptotics. For any given diffusion, first verify the interior assumptions, then compute $s'$, $S$, and $m$, and finally test the relevant endpoint integrals. The local behavior near the endpoint decides convergence; contributions over compact interior intervals are finite.

Check Your Understanding

For each question, use the endpoint tests with a fixed interior point $c\in(l,r)$, and distinguish strict finite-integral conditions from divergence.

State the left-endpoint accessibility and entrance integrals in terms of $S$ and $m$.
If both left-endpoint integrals are finite, what is the boundary class? What if only the entrance integral is finite?
For $dX_t=X_t\,dB_t$, use $S(x)=x$ and $m(x)=2/x^2$ to explain why the left endpoint is not entrance.
Explain why replacing $S$ by $aS+b$, with $a>0$, does not change either endpoint classification.
For $dX_t=\tfrac12\,dt+\sqrt{X_t}\,dB_t$, identify which endpoint test at zero diverges and which is finite, and state the resulting class.