Scale Functions and Speed Measures for One-Dimensional Diffusions

Why Scale and Speed Are Useful

A one-dimensional diffusion moves through its state interval under the combined effects of drift and random fluctuation. Two quantities reorganize those effects in a way that is especially useful for studying the process: scale describes the spatial coordinate in which the generator has no drift, while speed records the weight associated with time spent across the state space. Here we define these objects and compute them for three basic diffusions. Boundary tests and formulas for exit probabilities are separate questions; the definitions here provide tools for addressing them.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Let $I$ be an open interval and consider a diffusion, up to its lifetime in $I$, satisfying $dX_t=b(X_t)\,dt+\sigma(X_t)\,dB_t$. Assume $b$ and $\sigma$ are continuous on $I$, $\sigma(x)>0$ on $I$, and that the SDE has a solution up to the first exit from each compact subinterval of $I$. Each Ito integral is understood locally: after stopping $X$ on a compact subinterval, its progressively measurable integrand is square-integrable in expectation on finite time intervals, so it belongs to $\mathcal{L}^2$. The generator on twice continuously differentiable functions is $\mathcal{L}f(x)=b(x)f'(x)+\tfrac12\sigma^2(x)f''(x)$.

Definition: Fix a reference point $x_0\in I$. The scale density and the corresponding scale function, with this normalization, are $$ p(x)=\exp\!\left(-\int_{x_0}^{x}\frac{2b(y)}{\sigma^2(y)}\,dy\right), \qquad s(x)=\int_{x_0}^{x}p(y)\,dy. $$ The associated speed measure is $m(dx)=m'(x)\,dx$, where its density with respect to Lebesgue measure is $$ m'(x)=\frac{2}{\sigma^2(x)p(x)}. $$ These definitions apply on the interior interval where $\sigma>0$. Other normalizations multiply $p$ and $s'$ by a positive constant, add a constant to $s$, and change the speed density by the reciprocal constant.

The factor $2$ in the speed density is part of the convention used here. In particular, speed measure is a measure, not necessarily a probability measure. It need not have finite total mass, and its definition alone does not assert that it is a stationary distribution.

Why the Scale Function Removes the Drift

Proposition: The scale function satisfies $\mathcal{L}s=0$ on $I$. Consequently, $s(X_t)$, stopped before $X$ leaves any compact subinterval of $I$, is a local martingale.

Proof. The integrand $2b/\sigma^2$ is continuous, so $p$ is continuously differentiable and $s$ is twice continuously differentiable. Differentiating their definitions gives $s'(x)=p(x)$ and $p'(x)=-\frac{2b(x)}{\sigma^2(x)}p(x)$. Thus $\mathcal{L}s(x)=b(x)p(x)+\tfrac12\sigma^2(x)p'(x)=b(x)p(x)-b(x)p(x)=0$. To identify the process, apply Ito’s formula to $s(X_t)$ up to exit from a compact subinterval. For $f(x)=s(x)$, the time derivative is zero, $f'(x)=p(x)$, and $f''(x)=p'(x)$; the second-order correction is $\tfrac12p'(X_t)\sigma^2(X_t)\,dt$. Therefore $$ ds(X_t)=p(X_t)b(X_t)\,dt +\tfrac12p'(X_t)\sigma^2(X_t)\,dt +p(X_t)\sigma(X_t)\,dB_t =p(X_t)\sigma(X_t)\,dB_t. $$ On the stopped compact interval, $p$ and $\sigma$ are bounded. The progressively measurable integrand $p(X_t)\sigma(X_t)$, with the stopping included, is therefore in $\mathcal{L}^2$ on every finite time interval. Its Ito integral is a true martingale there by the Ito isometry and the conditional-mean-zero property of square-integrable Ito integrals. Without such a bound or another integrability condition, the un-stopped stochastic integral is only asserted to be a local martingale; it need not be a true martingale. This proves the claim. $\square$

The cancellation depends on the minus sign in the definition of $p$. The drift contributes $b p$, and differentiating the scale density contributes exactly $-b p$ through the Ito correction. The scale function is therefore a coordinate adapted to the generator, not simply an ordinary change of variable chosen for convenience.

Speed and the Divergence Form of the Generator

Proposition: For every twice continuously differentiable $f$ on $I$, $$ \mathcal{L}f(x)=\frac{1}{m'(x)}\frac{d}{dx}\!\left(\frac{f'(x)}{p(x)}\right). $$

Proof. From $p'=-2bp/\sigma^2$, differentiation of the expression inside the derivative gives $\frac{d}{dx}(f'/p)=f''/p-f'p'/p^2=f''/p+2bf'/(\sigma^2p)$. The reciprocal speed density is $1/m'=\sigma^2p/2$. Multiplying the differentiated expression by this reciprocal yields $\frac{\sigma^2}{2}f''+bf'=\mathcal{L}f$, as required. All quantities are defined because $\sigma$ and $p$ are strictly positive on $I$. $\square$

This identity displays how scale and speed fit together: the derivative is taken after dividing by the scale density, and the result is weighted by the reciprocal speed density. It also gives a direct check on calculations. If either proposed density has the wrong sign or factor, the divergence expression will not recover the original drift and diffusion terms.

Three Standard Diffusions

Worked Example: Brownian Motion

For Brownian motion on $I=\mathbb{R}$, take $b(x)=0$, $\sigma(x)=1$, and reference point $x_0=0$. The scale density is $p(x)=\exp(-\int_0^x 0\,dy)=1$. Integrating from the reference point gives $s(x)=\int_0^x1\,dy=x$. The speed density is $m'(x)=2/(1^2\cdot1)=2$, so $m(dx)=2\,dx$.

Check: Here $\mathcal{L}f=\tfrac12 f''$. The divergence expression gives $\frac{1}{2}\frac{d}{dx}(f'/1)=\tfrac12 f''$, agreeing with the generator. Also $\mathcal{L}s=\tfrac12(x)''=0$.

Worked Example: Geometric Brownian Motion

Consider $dX_t=X_t\,dt+\sqrt{2}X_t\,dB_t$ on $I=(0,\infty)$, with $x_0=1$. Thus $b(x)=x$ and $\sigma^2(x)=2x^2$. The ratio in the scale definition is $2b(x)/\sigma^2(x)=2x/(2x^2)=1/x$. It follows that $$ p(x)=\exp\!\left(-\int_1^x\frac{1}{y}\,dy\right) =\exp(-\log x)=\frac{1}{x}, \qquad s(x)=\int_1^x\frac{1}{y}\,dy=\log x. $$ The speed density is $m'(x)=2/(2x^2\cdot(1/x))=1/x$, hence $m(dx)=dx/x$.

Check: The generator is $\mathcal{L}f=xf'+x^2f''$. For $s(x)=\log x$, the derivatives are $s'=1/x$ and $s''=-1/x^2$, so $\mathcal{L}s=x(1/x)+x^2(-1/x^2)=1-1=0$. For a separate divergence-form check, take $f(x)=x^2$. Directly, $f'=2x$, $f''=2$, and $\mathcal{L}f=2x^2+2x^2=4x^2$. The divergence expression gives $(1/m')\frac{d}{dx}(f'/p)=x\frac{d}{dx}(2x^2)=x(4x)=4x^2$, the same value.

Worked Example: Ornstein–Uhlenbeck Process

Let $dX_t=2(1-X_t)\,dt+2\,dB_t$ on $I=\mathbb{R}$, and choose $x_0=1$. Here $b(x)=2(1-x)$ and $\sigma^2(x)=4$, so $2b(x)/\sigma^2(x)=1-x$. Integrating from 1 to $x$, $\int_1^x(1-y)\,dy=-\tfrac12(x-1)^2$. Therefore $$ p(x)=\exp\!\left(\frac{(x-1)^2}{2}\right), \qquad s(x)=\int_1^x\exp\!\left(\frac{(y-1)^2}{2}\right)\,dy, \qquad m'(x)=\frac12\exp\!\left(-\frac{(x-1)^2}{2}\right). $$ The scale-function integral is an exact representation; it is not generally an elementary function.

Check: Differentiating gives $p'(x)=(x-1)\exp((x-1)^2/2)$. The generator is $\mathcal{L}f=2(1-x)f'+2f''$, so $\mathcal{L}s=2(1-x)p+2p'=2(1-x)p+2(x-1)p=0$. For $f(x)=x$, the generator gives $\mathcal{L}f=2(1-x)$. In divergence form, $f'/p=\exp(-(x-1)^2/2)$, whose derivative is $-(x-1)\exp(-(x-1)^2/2)$. Multiplying by $1/m'=2\exp((x-1)^2/2)$ gives $-2(x-1)=2(1-x)$, as required.

Normalization and a Common Sign Error

Changing the reference point changes the numerical normalization but not the underlying scale coordinate up to a positive affine transformation. For example, multiplying $p$ by $c>0$ multiplies $s'$ by $c$, and the speed density $2/(\sigma^2p)$ by $1/c$. The divergence-form identity remains unchanged because the two factors cancel. Thus a calculation should state its reference point and normalization, but equivalent choices describe the same diffusion structure.

Common pitfall: reversing the sign in the exponential. In the geometric Brownian example, using $p(x)=\exp(+\int_1^xdy/y)=x$ would give the candidate $\tilde{s}(x)=\int_1^x y\,dy=(x^2-1)/2$. But $\tilde{s}'=x$, $\tilde{s}''=1$, and the actual generator gives $\mathcal{L}\tilde{s}=x(x)+x^2(1)=2x^2$, which is not zero on $(0,\infty)$. The correct density $p(x)=1/x$ instead gives $\mathcal{L}(\log x)=0$. Checking $\mathcal{L}s=0$ is an efficient way to catch this sign error.

Scale and speed are useful because they encode the coefficients in complementary forms: scale makes the generator’s drift cancellation explicit, while speed supplies its natural weight in divergence form. They do not, by themselves, say whether a boundary can be reached, whether a process explodes, or whether a speed measure is finite. Those conclusions require additional criteria and hypotheses. In particular, an infinite speed measure cannot be treated as a stationary probability distribution merely by calling it “speed.”

Check Your Understanding

Use the definitions with a fixed reference point in each calculation.

For $dX_t=3\,dt+2\,dB_t$ on $\mathbb{R}$, find the scale density and speed density, normalized at $x_0=0$.
For $dX_t=\mu X_t\,dt+\sigma X_t\,dB_t$ on $(0,\infty)$, with $\sigma>0$, derive the scale density in terms of $\mu$, $\sigma$, and $x/x_0$.
Verify the divergence-form identity for Brownian motion by substituting an arbitrary $f\in C^2(\mathbb{R})$.
Explain why a locally defined scale-transformed process is asserted to be a local martingale without further integrability assumptions.
For the Ornstein–Uhlenbeck example, compute $\mathcal{L}f$ and the divergence expression when $f(x)=x^2$, and compare them.