Timeline for "Square root" of Beta(a,b) distribution
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 28, 2010 at 13:56 | comment | added | Jon Peterson | Ivan, are you referring to a one-dimensional or a two-dimensional Brownian motion. I know that the fraction of time a one-dimensional Brownian motion spends to the right of the origin is $\beta(1/2,1/2)$. However, I am not aware of any such statement for the amount of time a two-dimensional Brownian motion spends in the first quadrant, and in fact a quick computation shows that it cannot be $\beta(1/2,1/2)$. | |
| Sep 27, 2010 at 20:19 | comment | added | Ivan Dornic | Hmm, $\beta(1/2,1/2)$ is also the distribution of the fraction of time spent on the positive quadrant by Brownian Motion started at the origin. Have you pursued ideas of this sort taking two independent copies of this random variable ? | |
| Sep 7, 2010 at 7:38 | comment | added | Steve Kroon | It turns out that I'm now looking for the square root of the Beta(0.5, 0.5) distribution, so non-integral b has suddenly become more important to me. | |
| Aug 17, 2010 at 16:04 | comment | added | Did | Yes. Call $B$ the set of parameters $b$ such that $f_{a,b}$ exists for every positive $a$. We know that $B$ is closed by addition and that it contains $1$, hence that $B$ contains every positive integer. And $B$ might also be a closed subset of the real line. But it is not clear (to me) that $B$ should contain every positive $b$. In particular the positive definiteness condition which Jon alluded to, might fail for small positive values of $b$. Who knows... :-) | |
| Aug 16, 2010 at 14:05 | vote | accept | Steve Kroon | ||
| Aug 16, 2010 at 14:05 | |||||
| Aug 16, 2010 at 13:55 | comment | added | Steve Kroon | Thank you for the feedback. I am also interested in existence, but I would love to be able to find the square root distribution if it exists, given a and b. Since you indicate that it is likely they exist, and Didier's work indicates how to obtain it for integral b, the best way forward for me might be to attempt to extend that characterization to non-integral b. | |
| Jul 28, 2010 at 17:47 | comment | added | Jon Peterson | Didier informed me of a small mistake in my above explanation. I said that $\psi(u)$ decays like $|u|^-b$. However, what is true is that $\phi(u)$ decays roughly at that rate, and so for $\psi$ to be in $L^1$ we would need $b>2$. Due to Didier's above proof in the cases where $b$ is a positive integer I believe that the "square root" distribution probably exists for any $a,b>0$. I still don't know what to do about proving the "positive definite" condition, but the integrability assumption on $\psi$ can probably be removed by someone who knows inverse Fourier transforms better than me. | |
| Jul 26, 2010 at 19:26 | history | answered | Jon Peterson | CC BY-SA 2.5 |