Timeline for Between arithmetic and geometric Brownian motions: when are negative values possible?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2015 at 23:13 | comment | added | Stephan Sturm | For SDEs Bernt Oksendal's "Stochastic Differential Equations: An Introduction with Applications" might be what you are looking for (about SPDEs I do not know). | |
| Oct 30, 2015 at 13:36 | comment | added | 8one6 | Thanks again for your time and consideration. I'd like to learn more about this topic, what's a good first text? Specifically I'd like to learn about the techniques for answering questions like the one above...when will a process specified by a particular SDE/SPDE have various properties, etc. I'm less concerned with the theoretical foundations (though I don't mind them) than I am with understanding how to answer specific questions for concrete examples. | |
| Oct 30, 2015 at 1:11 | vote | accept | 8one6 | ||
| Oct 29, 2015 at 22:45 | comment | added | Stephan Sturm | homepage.alice.de/murusov/papers/mu-mart.pdf | |
| Oct 29, 2015 at 22:45 | comment | added | Stephan Sturm | 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: | |
| Oct 29, 2015 at 15:17 | comment | added | 8one6 | 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? | |
| Oct 29, 2015 at 15:02 | comment | added | Stephan Sturm | @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 $0<\beta<1$. | |
| Oct 29, 2015 at 11:12 | comment | added | 8one6 | I misstated the problem in my original post: the drift term should have the same scaling that the stochastic term does. Will fix above. | |
| Oct 28, 2015 at 19:31 | history | answered | Stephan Sturm | CC BY-SA 3.0 |