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Square-Root Diffusions and Other Models · Tutorial 258 of 600

CEV Dynamics Across Volatility Exponents

Learn how the CEV exponent controls local volatility regularity and what it suggests—and does not by itself determine—about reaching zero.

Advanced 11 min read

What You'll Learn

  • Write the CEV stochastic differential equation and identify its drift and diffusion coefficients.
  • Compare local regularity of the diffusion coefficient for exponents below, equal to, and above one.
  • Use the standard scale-and-speed boundary test in a driftless CEV example.
  • Verify the exponent-one solution directly using Itô’s formula.
  • Distinguish a candidate boundary classification from a conclusion that requires drift and boundary assumptions.

CEV Dynamics and the Role of the Exponent

The constant-elasticity-of-variance (CEV) model describes a positive state whose instantaneous volatility changes as a power of its current level. Its exponent determines how quickly the diffusion coefficient shrinks—or grows—near zero. Comparing exponents is useful, but the coefficient’s regularity and the actual boundary behavior are distinct questions: boundary conclusions also depend on the drift and on how the process is defined at zero.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For \(x\ge0\), a common CEV equation is

$$ dX_t=\kappa X_t\,dt+\sigma X_t^\beta\,dB_t,\qquad X_0=x_0>0, $$

where \(\kappa\in\mathbb{R}\), \(\sigma>0\), and, for this discussion, \(\beta\ge0\). The drift coefficient is \(\kappa x\), while the diffusion coefficient is \(\sigma x^\beta\). An SDE solution is understood to be adapted and continuous, and its stochastic integral is an Itô integral: on every finite time interval where it is used, the integrand is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For coefficients that may grow too quickly for this condition globally, the integral and equation can instead be considered up to exit times from bounded state intervals.

In applications, \(X_t\) is often restricted to nonnegative values. When discussing a process killed at zero, we mean that it follows the SDE until its first hit of zero and is then held at zero. This is one possible boundary convention, not an automatic consequence of writing down the equation. In particular, when a coefficient is not locally Lipschitz at zero, existence and uniqueness questions require care.

Definition: In the CEV equation, \(\beta\) is the volatility exponent: the diffusion coefficient at state \(x\) is \(\sigma x^\beta\). For \(x>0\), the elasticity of the diffusion coefficient with respect to \(x\), defined by \(x\,(\sigma x^\beta)'/(\sigma x^\beta)\), equals \(\beta\). The elasticity describes the relative response of volatility to a relative change in the state.

What Changes at Exponent One?

For \(0<\beta<1\), the coefficient tends to zero at zero, but less rapidly than a linear function. It is locally Hölder continuous with exponent \(\beta\) there, and is not locally Lipschitz at zero. At \(\beta=1\), the coefficient is linear and locally Lipschitz. For \(\beta>1\), it vanishes faster than linearly and is also locally Lipschitz at zero. These statements concern the coefficient as a function; they do not alone decide whether a sample path reaches the boundary.

Proposition: For \(\sigma>0\), the function \(a(x)=\sigma x^\beta\) on \([0,\infty)\) is locally Lipschitz at zero when \(\beta\ge1\), and is not locally Lipschitz at zero when \(0<\beta<1\). In the latter case it is Hölder continuous with exponent \(\beta\) on every bounded interval.

Proof. If \(\beta\ge1\), then \(a\) is continuous at zero and, for \(x>0\), \(a'(x)=\sigma\beta x^{\beta-1}\). On \([0,R]\) this derivative is bounded by \(\sigma\beta R^{\beta-1}\) when \(\beta\ge1\). The mean value theorem therefore gives a finite Lipschitz bound on \([0,R]\), including pairs with one endpoint zero. Now suppose \(0<\beta<1\). For \(x>0\), the quotient \(|a(x)-a(0)|/|x-0|=\sigma x^{\beta-1}\) tends to infinity as \(x\downarrow0\), so no finite local Lipschitz constant exists. To check the Hölder claim, for \(u,v\ge0\) use \(|u^\beta-v^\beta|\le |u-v|^\beta\), which follows from \((r+s)^\beta\le r^\beta+s^\beta\) for \(r,s\ge0\). Multiplication by \(\sigma\) proves the stated bound. \(\square\)

At \(\beta=0\), the coefficient is the positive constant \(\sigma\), so it is Lipschitz but does not vanish at zero. This endpoint is therefore different from the positive exponents below one: noise does not switch off at the boundary. A nonnegative-state model needs an explicit boundary convention in this case.

Worked Example: Comparing Coefficients Near Zero

Take \(\sigma=2\) and compare \(a(x)=2x^{1/2}\), \(a(x)=2x\), and \(a(x)=2x^2\) at \(x=0,\tfrac14,1\). Also test whether the first coefficient can be Lipschitz at zero.

For \(\beta=\tfrac12\), the values are \(a(0)=0\), \(a(\tfrac14)=2\sqrt{\tfrac14}=1\), and \(a(1)=2\). For \(\beta=1\), they are \(0\), \(2\cdot\tfrac14=\tfrac12\), and \(2\). For \(\beta=2\), they are \(0\), \(2(\tfrac14)^2=\tfrac18\), and \(2\). All three coefficients vanish at zero, but the square-root coefficient decreases more slowly near zero than either of the other two.

For the square-root coefficient, the Lipschitz quotient between \(x\) and zero is \(2\sqrt{x}/x=2/\sqrt{x}\) for \(x>0\). At \(x=\tfrac14\) it is \(4\), and as \(x\downarrow0\) it is unbounded. Thus no finite Lipschitz constant works near zero. Check: the general proposition gives Hölder exponent \(\tfrac12\); indeed \(|2\sqrt{x}-2\sqrt{y}|\le2\sqrt{|x-y|}\) for \(x,y\ge0\).

The Exponent-One Case

When \(\beta=1\), the state and diffusion coefficients are linear. For positive initial state, the equation has the explicit geometric Brownian motion solution. Its pathwise expression gives a direct boundary conclusion for any finite time horizon: the process remains strictly positive. This special case is especially useful as a benchmark, but its conclusion should not simply be extended to other exponents.

Theorem: Let \(\kappa\in\mathbb{R}\), \(\sigma>0\), and \(x_0>0\). The solution of \(dX_t=\kappa X_t\,dt+\sigma X_t\,dB_t\) is \(X_t=x_0\exp((\kappa-\tfrac12\sigma^2)t+\sigma B_t)\). It is strictly positive at every finite time, almost surely.

Proof. Define \(Y_t=x_0\exp((\kappa-\tfrac12\sigma^2)t+\sigma B_t)\). Apply Itô’s formula to \(f(t,b)=x_0\exp((\kappa-\tfrac12\sigma^2)t+\sigma b)\). Its derivatives are \(f_t=(\kappa-\tfrac12\sigma^2)f\), \(f_b=\sigma f\), and \(f_{bb}=\sigma^2f\). The formula, including its second-derivative correction, gives \(dY_t=f_t\,dt+f_b\,dB_t+\tfrac12 f_{bb}\,dt=\kappa Y_t\,dt+\sigma Y_t\,dB_t\). Thus \(Y\) satisfies the equation. The coefficients are globally Lipschitz, so the standard existence-and-uniqueness result for Lipschitz SDEs identifies this solution with the unique solution. The exponential of a finite real number is positive, and Brownian paths are finite at finite times, so \(Y_t>0\). \(\square\)

Worked Example: Verifying a CEV Solution at Exponent One

Set \(x_0=2\), \(\kappa=\tfrac14\), and \(\sigma=\tfrac12\). Write the solution and compute its expectation at \(t=2\). Then verify the drift in the Itô calculation.

The formula gives \(X_t=2\exp((\tfrac14-\tfrac12(\tfrac12)^2)t+\tfrac12B_t)=2\exp(\tfrac18t+\tfrac12B_t)\). At \(t=2\), this is \(2\exp(\tfrac14+\tfrac12B_2)\). Since \(B_2\) is normal with mean zero and variance \(2\), the normal moment-generating function gives \(\mathbb{E}[\exp(\tfrac12B_2)]=\exp((\tfrac12)^2\cdot2/2)=\exp(\tfrac14)\). Hence \(\mathbb{E}[X_2]=2e^{1/4}e^{1/4}=2e^{1/2}\).

For the direct Itô check, \(f_t=\tfrac18 f\), \(f_b=\tfrac12 f\), and \(f_{bb}=\tfrac14 f\). Therefore the correction is \(\tfrac12f_{bb}\,dt=\tfrac18f\,dt\), and the total drift is \(\tfrac18f+\tfrac18f=\tfrac14f\), as required. The stochastic integrand is \(\tfrac12X_t\). It is square-integrable on every finite time interval since \(\mathbb{E}[X_t^2]=x_0^2e^{(2\kappa+\sigma^2)t}\) is finite and \(\int_0^T\mathbb{E}[X_t^2]\,dt<\infty\). Check: the solution’s mean is \(x_0e^{\kappa t}\); at \(t=2\), this is \(2e^{1/2}\), matching the computed value.

Candidate Zero-Boundary Behavior

A clean comparison uses the driftless CEV diffusion \(dX_t=\sigma X_t^\beta\,dB_t\) on the positive half-line, considered up to its first hit of zero or an upper level. The standard one-dimensional diffusion boundary test expresses accessibility through the scale and speed densities. For a diffusion with drift \(b(x)\) and diffusion coefficient \(a(x)>0\) on \((0,\infty)\), these densities are \(s'(x)=\exp(-\int^x 2b(y)/a(y)^2\,dy)\) and \(m(x)=2/(a(x)^2s'(x))\), up to a positive constant in the scale density. The Feller boundary classification, including its accessibility criterion, is a standard result of one-dimensional diffusion theory; its proof is beyond this tutorial.

For zero drift, choose \(s'(x)=1\), so \(s(x)=x\), and \(m(x)=2/(\sigma^2x^{2\beta})\). The accessibility integral at zero is finite precisely when \(\int_0^c (s(x)-s(0))m(x)\,dx\) is finite, for any fixed \(c>0\). Here its integrand is a constant multiple of \(x^{1-2\beta}\). The integral is finite for \(\beta<1\), and infinite for \(\beta\ge1\). Under this driftless comparison, zero is accessible for exponents below one and is not accessible from the interior for exponents at least one. Accessibility means the boundary can be hit; the behavior after hitting it still requires a boundary convention.

The coefficient comparison and this boundary test tell a consistent local story, but they are not interchangeable. Local Lipschitz regularity is a property of the coefficient, while accessibility is a property of the diffusion and its boundary setup. For \(\beta>1\), a nonzero linear drift changes the scale density near zero, so the driftless test cannot be quoted as a classification for every \(\kappa\). For \(\beta<1\), the scale factor from a linear drift has a finite, nonzero limit at zero because the relevant integral has integrand proportional to \(x^{1-2\beta}\), which is integrable there; the same accessibility threshold follows in this setting. At \(\beta=1\), the explicit solution already settles the positive-initial-state case for any \(\kappa\).

Worked Example: Applying the Driftless Boundary Test

Take \(\sigma=2\), and compare \(\beta=\tfrac12\) with \(\beta=\tfrac32\). Compute the accessibility integral from zero to one and interpret the result.

For either exponent, \(s(x)=x\) and \(m(x)=2/(4x^{2\beta})=\tfrac12x^{-2\beta}\). When \(\beta=\tfrac12\), the integral is \(\int_0^1 x\cdot\tfrac12x^{-1}\,dx=\int_0^1\tfrac12\,dx=\tfrac12\), which is finite. Thus zero is accessible in this driftless comparison. When \(\beta=\tfrac32\), the integral is \(\int_0^1 x\cdot\tfrac12x^{-3}\,dx=\tfrac12\int_0^1x^{-2}\,dx=\infty\), so zero is not accessible from the interior under the same test.

Check: the general integrand is proportional to \(x^{1-2\beta}\). Its exponent is \(0\) for \(\beta=\tfrac12\), yielding a finite integral, and \(-2\) for \(\beta=\tfrac32\), yielding a divergent integral. The two outcomes agree with the threshold \(\beta=1\).

Interpretation and Common Pitfalls

The exponent indicates how strongly noise acts near zero. Below one, the diffusion coefficient is not locally Lipschitz when \(0<\beta<1\), and the driftless boundary test predicts accessibility. At one, the geometric model has a positive explicit solution. Above one, the coefficient is locally Lipschitz and vanishes faster than linearly, while the driftless boundary test predicts inaccessibility. These are useful reference cases, not a substitute for stating the drift, initial condition, state space, and boundary convention.

A common error is to infer “zero cannot be reached” merely from the fact that the coefficient vanishes at zero. For example, in the driftless model with \(\beta=\tfrac12\), the coefficient \(a(x)=\sigma\sqrt{x}\) is zero at zero, yet the accessibility integral is finite. Conversely, the coefficient’s lack of local Lipschitz regularity does not by itself prove that every solution hits zero or determine what happens after a hit. Those conclusions require the diffusion boundary test and a specified solution and boundary setup.

ExponentCoefficient near zeroRegularity at zeroDriftless boundary comparison
\(0<\beta<1\)Vanishing, slower than linearHölder; not locally LipschitzZero accessible
\(\beta=1\)LinearLocally LipschitzZero inaccessible from a positive start
\(\beta>1\)Vanishing, faster than linearLocally LipschitzZero inaccessible for zero drift

Check Your Understanding

Use the coefficient comparison and the stated boundary assumptions when answering.

  1. For \(\beta=\tfrac34\), is \(x^\beta\) locally Lipschitz at zero? What Hölder exponent does the coefficient satisfy?
  2. For \(\beta=2\) and zero drift, what power of \(x\) appears in the integrand of the scale-and-speed accessibility test?
  3. Why does the explicit exponent-one solution remain positive for every finite time?
  4. Does a diffusion coefficient that vanishes at zero, by itself, imply that the process cannot hit zero? Explain using the driftless boundary comparison.
  5. What additional model information is needed before making a boundary claim for an exponent above one with nonzero drift?