Transforming CEV Into a Bessel-Type Process

From CEV Volatility to Unit Diffusion

In a constant-elasticity-of-variance model, the diffusion coefficient depends on a power of the state. When the volatility exponent lies strictly between zero and one, a suitable power transformation removes that state dependence. The price of this simplification is a drift term proportional to the reciprocal of the transformed state. That singular drift is the clue to how the process behaves near zero.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix constants \(\mu\in\mathbb{R}\), \(\sigma>0\), and \(0<\beta<1\). Consider a continuous solution, started at \(X_0=x_0>0\), of the CEV equation \(dX_t=\mu X_t\,dt+\sigma X_t^\beta\,dB_t\), up to its first approach to zero or explosion. Its stochastic integral is an Ito integral with progressively measurable integrand; on every interval where \(X\) is stopped inside a compact subset of \((0,\infty)\), that integrand is square-integrable in time and expectation. The coefficients are locally Lipschitz on \((0,\infty)\), so the standard local existence and uniqueness theorem for SDEs gives a unique strong solution up to the boundary of that domain.

The scale factor in this definition is chosen so that the diffusion coefficient of \(Y\) is one. Ito’s correction term then produces a singular drift, which we derive rather than infer from a formal power rule.

The Transformation and Its Bessel-Type Drift

Proof. Set \(f(x)=x^{1-\beta}/(\sigma(1-\beta))\) for \(x>0\). Its derivatives are \(f'(x)=x^{-\beta}/\sigma\) and \(f''(x)=-\beta x^{-\beta-1}/\sigma\). Ito’s formula has drift contribution \(f'(X_t)\mu X_t+\tfrac12 f''(X_t)\sigma^2X_t^{2\beta}\) and diffusion contribution \(f'(X_t)\sigma X_t^\beta\,dB_t\). Substituting the derivatives gives the diffusion coefficient \(1\), and the drift becomes \(\dfrac{\mu}{\sigma}X_t^{1-\beta}-\dfrac{\beta\sigma}{2}X_t^{\beta-1}\). Since \(X_t^{1-\beta}=\sigma(1-\beta)Y_t\) and \(X_t^{\beta-1}=1/(\sigma(1-\beta)Y_t)\), this is \(\mu(1-\beta)Y_t-\beta/[2(1-\beta)Y_t]\). Finally, \((\delta-1)/2=-\beta/[2(1-\beta)]\), proving the equivalent form. All calculations apply before the boundary because \(f\) is twice continuously differentiable on \((0,\infty)\). \(\square\)

The equation has unit diffusion, but it is not generally a Bessel process: the additional linear drift \(aY_t\) remains unless \(\mu=0\). Near zero, the reciprocal term is the singular part. Its coefficient can be written in the standard Bessel-type form \((\delta-1)/(2Y_t)\). For every \(0<\beta<1\), \(\delta<2\); it can be zero or negative, so “Bessel-type” does not mean a Bessel process of positive dimension in every case.

What the Singular Drift Says About Zero

A negative reciprocal drift points toward zero, but its sign alone is not a proof that zero can be reached. A precise boundary test uses the scale function and speed density. For a one-dimensional diffusion \(dY_t=b(Y_t)dt+dB_t\) on \((0,\infty)\), define a scale density \(s'(y)=\exp(-\int^y 2b(z)\,dz)\), a scale function \(s(y)=\int^y s'(z)\,dz\), and a speed density \(m'(y)=2/s'(y)\). The lower endpoint zero is accessible from the interior when, for some \(c>0\), the integral \(\int_0^c (s(y)-s(0))m'(y)\,dy\) is finite. This scale-speed accessibility test is a standard one-dimensional diffusion result; its general proof is beyond this tutorial.

For the transformed drift \(b(y)=ay+(\delta-1)/(2y)\), direct integration gives \(s'(y)=y^{1-\delta}e^{-ay^2}\), up to a positive constant factor. Because \(1-\delta=2\beta/(1-\beta)>0\), the scale function has a finite limit \(s(0)\). Near zero, \(e^{-ay^2}\) tends to \(1\), so \(s(y)-s(0)\) is of order \(y^{2-\delta}\), while \(m'(y)=2e^{ay^2}y^{\delta-1}\) is of order \(y^{\delta-1}\). Their product is of order \(y\), which is integrable at zero. The test therefore says that zero is accessible for every \(0<\beta<1\).

Accessibility means that the boundary can be reached in finite time with positive probability from an interior state. It does not mean that zero is hit almost surely, nor does it prescribe how the model behaves after the hit. In the original CEV equation, an absorbing convention sets \(X_t=0\) after the first hit. The power transformation is only used before that time.

Why the Transformation Does Not Hide Explosion

The transformed equation is derived while the state remains positive. To use it up to the first hit of zero, one must also rule out explosion to \(+\infty\) before that hit. The estimate below handles the two boundaries together; it does not assume that a path which visits a small positive level cannot later recover.

Proof. Fix \(T>0\), \(n>x_0\), and \(0<\varepsilon<x_0\). Let \(\tau_{\varepsilon,n}\) be the first exit time from \((\varepsilon,n)\). Apply Ito’s formula to \(f(x)=x^2\), whose derivatives are \(f'(x)=2x\) and \(f''(x)=2\). For the stopped process, \(X_{t\wedge\tau_{\varepsilon,n}}^2=x_0^2+\int_0^{t\wedge\tau_{\varepsilon,n}}(2\mu X_s^2+\sigma^2X_s^{2\beta})\,ds+2\sigma\int_0^{t\wedge\tau_{\varepsilon,n}}X_s^{\beta+1}\,dB_s\). The stopped stochastic-integral integrand is bounded on \([\varepsilon,n]\), hence belongs to \(\mathcal{L}^2\) on \([0,T]\). By the Ito isometry and the martingale property of square-integrable Ito integrals, its expectation is zero.

For every \(x>0\), \(x^{2\beta}\le 1+x^2\): if \(x\le1\), then \(x^{2\beta}\le1\), and if \(x\ge1\), then \(x^{2\beta}\le x^2\). Put \(C=2|\mu|+\sigma^2\). Taking expectations and using the preceding bound gives \(\mathbb{E}[X_{t\wedge\tau_{\varepsilon,n}}^2]\le x_0^2+\sigma^2T+C\int_0^t\mathbb{E}[X_{s\wedge\tau_{\varepsilon,n}}^2]\,ds\) for \(0\le t\le T\). Gronwall’s inequality yields the bound \(K_T=(x_0^2+\sigma^2T)e^{CT}\), independent of \(\varepsilon\) and \(n\).

If the process exits through \(n\) before \(\varepsilon\) by time \(T\), then \(X_{T\wedge\tau_{\varepsilon,n}}^2=n^2\). Nonnegativity therefore gives \(\mathbb{P}(\tau_{\varepsilon,n}\le T,\ X_{\tau_{\varepsilon,n}}=n)\le K_T/n^2\). Now fix \(n\) and let \(\varepsilon\) decrease to zero. Any path that reaches \(n\) before hitting zero has a strictly positive minimum on the compact time interval up to its first visit to \(n\), by continuity. It is therefore counted as an upper exit before \(\varepsilon\) for every sufficiently small \(\varepsilon\). Thus the probability of reaching \(n\) before zero by time \(T\) is at most \(K_T/n^2\). An explosion before zero by time \(T\) would require reaching every sufficiently large \(n\) before zero by time \(T\); its probability is bounded by \(K_T/n^2\) for every such \(n\), and hence is zero as \(n\to\infty\). Taking integer \(T\) rules out finite-time explosion before zero. \(\square\)

Common Pitfalls

Dropping the Ito correction. A wrong calculation treats the power transform as an ordinary chain rule and writes only the transformed linear drift. For the square-root example, that would give \(dY_t=dB_t\), omitting \(-1/(2Y_t)\,dt\). The correction is not optional: \(f''(x)\) is nonzero and the quadratic variation rule is \(dB_t\,dB_t=dt\), while \(dB_t\,dt=0\) and \(dt\,dt=0\).

Confusing accessibility with certain hitting. The scale-speed test establishes that zero is accessible, not that every path reaches it. Also, the transformed equation has a singular drift at zero and is not continued there by simply substituting \(Y=0\). State the boundary convention for \(X\) separately.

Using an upper-exit estimate as if paths could not return from small values. The bound for exiting \((\varepsilon,n)\) through \(n\) applies to that stopped interval. It does not by itself rule out a later rise after a visit near zero. The uniform-in-\(\varepsilon\) bound above first controls reaching a fixed \(n\) before zero, then lets \(\varepsilon\downarrow0\), and only afterward lets \(n\to\infty\).