Timeline for Marchenko-Pastur Law under general covariance structure
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2020 at 14:07 | comment | added | ofer zeitouni | The covariance matrix (what the OP called $\Sigma$) | |
| Sep 10, 2020 at 13:53 | comment | added | R Salimi | @oferzeitouni, I'm not an expert, please demystify what do you mean with the expression R. | |
| Mar 12, 2020 at 15:52 | history | edited | ofer zeitouni | CC BY-SA 4.0 |
edited body
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| Mar 12, 2020 at 15:52 | comment | added | ofer zeitouni | The matrix $\sqrt{R} YY^* \sqrt{R}$ has the same spectrum as $Y^*RY$, so no difference between the two as far as spectrum is concerned. More to the point, in your notation $Y$ is $p\times n$ dimensional, so $YRY^*$ makes no sense at all. My answer of course should have $Y^* RY$. Corrected now. | |
| Mar 11, 2020 at 22:01 | comment | added | Learning math | @oferzeitouni Sorry I meant: if we take $X:= \sqrt {R} Y,$ of dimension $p \times n,$ then the sample covariance is $1/p XX^{*}= 1/p \sqrt R YY^{*} \sqrt R, $ (dim = $p \times p$) and the Gram matrix (dual covriance) is: $1/n X^{*}X = 1/n Y^{*}R Y$ (dim = $n \times n$). Here, $p=$ no if features, and $n =$ no of samples. None of these two are proportional to $YRY^{*}$, so I'm just checking with you regarding the symbols. I know it's a minor difference, but as a newbie to RMT, it'd still be helpful for me. Thanks in advance! | |
| Mar 11, 2020 at 18:42 | comment | added | ofer zeitouni | I can't decipher your message, sorry. Y^*Y=1/n Y^*RY?? | |
| Mar 11, 2020 at 16:06 | comment | added | Learning math | Sorry I wrote $Y*$, but I really meant $Y^{*}$, the adjoint of $Y$. | |
| Mar 11, 2020 at 15:50 | comment | added | Learning math | Hello and thanks for your answer! I've a very rudilentary follow-up question if it's ok. Since $x_i = \sqrt{R}y_i, R \mathbb{R}^{p \times p} $, then the data matrix is: $X:=[x_1 \dots x_n] \mathbb{R}^{p \times n}$. So if I take the sample cov $1/p XX* = 1/p \sqrt R YY* \sqrt R.$, an dif I take the Gram (or dual covariance) matrix, then $Y*Y = 1/n Y*RY$. Since I'm very new to this area, I'm just checking with you to see if I got this right. I'll read the paper you mentioned. | |
| Jan 4, 2020 at 15:58 | comment | added | neverevernever | Thanks for pointing out the reference! | |
| Jan 4, 2020 at 15:58 | vote | accept | neverevernever | ||
| Jan 3, 2020 at 23:45 | history | answered | ofer zeitouni | CC BY-SA 4.0 |