Timeline for p.d.f. of $\left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2$, where $\textbf{x}$ and $\textbf{y}$ are complex Gaussians?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Aug 3, 2019 at 18:16 | comment | added | Felipe Augusto de Figueiredo | OK, many thanks for your reply! If you have any idea or hint, please, let me know. | |
| Aug 3, 2019 at 15:14 | comment | added | Carlo Beenakker | Not immediately, I’m afraid. | |
| Aug 3, 2019 at 10:46 | comment | added | Felipe Augusto de Figueiredo | Dear Carlo, do you think you could help me with this problem? mathoverflow.net/questions/337341/… thanks! | |
| Feb 16, 2019 at 18:51 | comment | added | Carlo Beenakker | for the moments of $Z$, see mathoverflow.net/a/323393/11260 | |
| Feb 16, 2019 at 18:38 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Feb 16, 2019 at 18:26 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Feb 16, 2019 at 18:20 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Feb 16, 2019 at 18:10 | comment | added | Carlo Beenakker | for large $M$ the sum of the square of $2M$ normally distributed independent variables self-averages to $2M$, so $\xi_{2M}$ in the numerator and $\xi_2+\xi_{2M-2}$ in the denominator can both be replaced by $2M$. | |
| Feb 16, 2019 at 17:39 | comment | added | Felipe Augusto de Figueiredo | Please, could you explain how you arrived at the limit when $M \gg 1$? | |
| Feb 16, 2019 at 17:09 | vote | accept | Felipe Augusto de Figueiredo | ||
| Feb 16, 2019 at 14:23 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Feb 16, 2019 at 14:03 | comment | added | Felipe Augusto de Figueiredo | Thanks for your answer. Do you think it is possible to find a closed-form expression for the moments of $Z$? | |
| Feb 16, 2019 at 13:38 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Feb 16, 2019 at 13:01 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |