Timeline for Can the wavelet bispectrum be normalised so that its integral "gives the right answer"?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| S Mar 7, 2018 at 1:32 | history | bounty ended | CommunityBot | ||
| S Mar 7, 2018 at 1:32 | history | notice removed | CommunityBot | ||
| Mar 4, 2018 at 22:45 | history | edited | Julian Newman | CC BY-SA 3.0 |
added update
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| Mar 2, 2018 at 22:06 | comment | added | Julian Newman | Unfortunately, I think I can prove that my conjectured answer just integrates to give $0$. | |
| Feb 27, 2018 at 14:26 | comment | added | Julian Newman | Since the Morlet reconstruction formula can be used to obtain the first moment and the more standard reconstruction formula is used to derive the second moment (Kaiser, p69), perhaps the general reconstruction formula in Note 2 of the top answer at math.stackexchange.com/questions/579199/… can be used to obtain the answer to my question. | |
| Feb 27, 2018 at 14:17 | comment | added | Julian Newman | Thanks, this is a very insightful first step. By the way, where you write $\xi_1^{-3/2}\xi_2^{-3/2}\xi_3^{-3/2}$, I think it should just be $\xi_1\xi_2\xi_3$; for the sake of analogy with the Fourier transform, I have not defined the wavelet transform in quite the standard way. | |
| Feb 27, 2018 at 13:07 | comment | added | Carlo Beenakker | This the best I can come up with (using Morlet's reconstruction formula), four integrals instead of three: $$\int_{-\infty}^\infty x(t)^3\,dt=\frac{1}{D^3_\psi}\int_{-\infty}^\infty dt\int_0^\infty d\xi_1\int_0^\infty d\xi_2\int_0^\infty d\xi_3$$ $$\xi_1^{-3/2}\xi_2^{-3/2}\xi_3^{-3/2}\mathring{x}(\xi_1,t)\mathring{x}(\xi_2,t)\mathring{x}(\xi_3,t)$$ | |
| Feb 26, 2018 at 23:51 | history | edited | Julian Newman | CC BY-SA 3.0 |
edited body
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| S Feb 26, 2018 at 23:50 | history | bounty started | Julian Newman | ||
| S Feb 26, 2018 at 23:50 | history | notice added | Julian Newman | Draw attention | |
| Feb 23, 2018 at 16:26 | comment | added | Julian Newman | There was an important typo that I've now corrected: $\xi$ is not squared in the integrand of the rightmost-hand side of $(1)$. | |
| Feb 23, 2018 at 16:24 | history | edited | Julian Newman | CC BY-SA 3.0 |
brackets around remark
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| Feb 23, 2018 at 3:08 | history | asked | Julian Newman | CC BY-SA 3.0 |