Timeline for Average value of $\frac{x'A^2x}{x'A^3x}$ over surface of $n$-dimensional sphere
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1, 2021 at 8:03 | vote | accept | Yaroslav Bulatov | ||
| Aug 31, 2021 at 23:36 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Aug 31, 2021 at 20:54 | comment | added | Yaroslav Bulatov | was using $x\sim \mathcal{N}\left(0, I_n\right)$ to denote sampling from standard Gaussian in $n$ dimension -- ie, 0 mean, identity covariance matrix, both real valued | |
| Aug 31, 2021 at 15:21 | comment | added | Pietro Majer | Is $x$ in $\mathbb {R}^n$ or in $\mathbb{C}^n$? And what is $I_n$? Thx | |
| Aug 31, 2021 at 13:23 | comment | added | Iosif Pinelis | @YaroslavBulatov : Those asymptotic expansions are valid only for large $s$, whereas the integral is over all $s>0$. So, there is no reason to substitute those asymptotic expressions for the actual ones. | |
| Aug 31, 2021 at 13:13 | answer | added | Iosif Pinelis | timeline score: 0 | |
| Aug 31, 2021 at 12:58 | comment | added | Carlo Beenakker | this follows if your replace the sum over $k$ in the expressions for $G_n$ and $\log F_n$ by an integral $\int_0^\infty dk$; I don't think that is a controlled approximation from which you can make definite conclusions. | |
| Aug 31, 2021 at 11:48 | comment | added | Yaroslav Bulatov | Substituting infinite $s$ expansions from that answer, the integral can be solved exactly, and it evaluates to 0.5513. I'm confused whether this provides evidence that the limit in question is lower than 2 | |
| Aug 31, 2021 at 11:22 | comment | added | Carlo Beenakker | I doubt very much one can do better than math.stackexchange.com/a/4228443 | |
| Aug 31, 2021 at 8:57 | history | asked | Yaroslav Bulatov | CC BY-SA 4.0 |