Please note edits after original post changing the specific form of the setup

Let's say we have a stochastic differential equation: $$ \mathrm{d}S_t = |S^\beta| {(\mu \mathrm{d}t + \sigma\mathrm{d}W_t)} $$ where $W_t$ is a standard brownian motion.

When $\beta=0$, the whole thing continues to be a brownian motion (albeit with drift and with non-unit variance). In particular, there is a non-zero probability that $S_t$ will take on negative values. Conversely, when $\beta=1$ this describes a particular instance of geometric brownian motion and as a result, there process almost certainly avoids taking on negative values.

My question: for $0 < \beta < 1$ (note: strict inequalties), what statements can be made about the probability of the process taking on negative values? In particular, are negative values almost never achieved for whenever $\beta < 1$, or is there some critical $0 < \beta_c < 1$ such that processes with $\beta < \beta_c$ are able to take on negative values with non-zero probability while those with $\beta > \beta_c$ are not? Does the answer depend on the value of $\mu$? I'm most interested in the case where $\mu=0$ but left the drift term there to see if someone could offer insight about the more general setup.


Assuming that $\mu$ and $S_0$ are positive, the process stays almost surely non-negative. This is easily seen as when $S$ hits zero, it has a deterministic drift upwards. However, the process does not necessarily stay strictly positive, the hitting probability of zero can be positive.

This model is well studied in the financial mathematics literature and goes there under the name CEV model. The generic reference on it and the close relationship to Bessel processes is the following paper by Delbaen and Shirakawa: A Note of Option Pricing for Constant Elasticity of Variance Model, https://people.math.ethz.ch/~delbaen/ftp/preprints/CEV.pdf

  • $\begingroup$ I misstated the problem in my original post: the drift term should have the same scaling that the stochastic term does. Will fix above. $\endgroup$ – 8one6 Oct 29 '15 at 11:12
  • $\begingroup$ @8one6 The new formulation of the question changes nothing on the details, except that now the process is absorbed at zero as soon as it hits zero. This cannot happen a.s. if $\beta=1$, but can happen with positive probability for $0<\beta<1$. $\endgroup$ – Stephan Sturm Oct 29 '15 at 15:02
  • $\begingroup$ Thanks for taking the time to explain. So let's talk only about the $\mu=0$ case: why is it that the probability of hitting 0 (in finite time) is 0 for $\beta=1$ but non-zero for any $\beta<1$? What's the logic for/proof for the fact that the cases split at $\beta=1$ rather than, say $\beta=1/2$ or $\beta=2/3$ or other? $\endgroup$ – 8one6 Oct 29 '15 at 15:17
  • $\begingroup$ You can write $S_t$, using stochastic calculus, as exponential $S_t = \mathcal{E}\bigl(\int_0^{\cdot} \frac{1}{S_s^{1-\beta}} \, ds\bigr)_t$. If $\beta = 1$, the exponent is just a Brownian motion which takes almost surely finite values. However, for $\beta < 1$ you have as integrand something what explodes when $S$ approaches zero, so you lose integrability. For $\beta>1$, clearly the exponent goes to zero when $S$ approaches zero, this is not a problem for the integrability. For more details you might have a look on Example 3.2 of the following paper by Mijatovic and Urusov: $\endgroup$ – Stephan Sturm Oct 29 '15 at 22:45
  • $\begingroup$ homepage.alice.de/murusov/papers/mu-mart.pdf $\endgroup$ – Stephan Sturm Oct 29 '15 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.