Timeline for Square root of normal distribution
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 2, 2018 at 0:08 | comment | added | Iosif Pinelis | A complete answer to this question (and actually a more general one) is now available at arxiv.org/abs/1803.09838 | |
| Mar 13, 2015 at 16:58 | comment | added | Iosif Pinelis | helvio: $\frac{P(x)}{2\pi}$ would be the density of $\ln|X|$ -- provided that $P(x)$ is nonnegative for all real $x$. I think this nonnegativity is likely the case, but don't know how to prove that. The other approach that I suggested later, with the $\mu_j$'s and $\nu_j$'s, may be more promising. | |
| Mar 13, 2015 at 13:47 | vote | accept | FreeQuark | ||
| Mar 13, 2015 at 13:46 | comment | added | FreeQuark | Thank you very much @iosif-pinelis for your detailed answer, I learned a great deal with it. So if I understood correctly the probability density $P(x)$ is given by the Fourier transform of the square root of the characteristic function of the normal probability density, $$P(x) = \int_{-\infty}^{+\infty} \left( \frac{2^{it/2}}{\sqrt\pi} \Gamma\Big(\frac{1+it}{2}\Big) \right)^\frac{1}{2} e^{-itx} dt$$ right? I can't expect a closed form solution, and even if it existed it's probably useless in practice. But the whole argument is sound and solid, so I'll accept it. | |
| Mar 12, 2015 at 16:25 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
added to the previous version
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| Mar 12, 2015 at 3:19 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |