Timeline for answer to Random matrix determinant problem by ofer zeitouni
Current License: CC BY-SA 3.0
Post Revisions
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2014 at 19:14 | comment | added | ofer zeitouni | Read the T-V text (I suspect it is close to your application) and then email me if you have further questions, and I'll do my best to answer. | |
| Jan 22, 2014 at 19:09 | comment | added | ofer zeitouni | Yes, but this discussion begins to be beyond MO comments... | |
| Jan 22, 2014 at 19:06 | comment | added | MLT | Thank you. Then we proceed. To find the free convolution we need the R-transform. The R-transform of the M-P distribution (say \mu_i)of the eigenvalues of $a_iv_iv_i^H$, with the dimension ratio $\beta$ is $R_i(x)=\frac{a_i}{1-\beta x}$. Then the R-transform of the distribution of eigenvalues $\lambda_i$ of $X$ is the sum of R-transforms $R_{\lambda_i}=\sum_{j \neq i}R_j=\sum_{j\neq i} \frac{a_j}{1-\beta x}$ ? | |
| Jan 22, 2014 at 18:55 | comment | added | ofer zeitouni | Indeed (the M-P distribution needs to be rescaled by the a_i, but that's a detail). | |
| Jan 22, 2014 at 18:31 | comment | added | MLT | Thanks. I would like to clarify if my understanding is correct please. The distribution of the eigenvalues of $a_iv_iv_i^H$ as dimension tends to $+\infty$, is given by the Marchenko–Pastur distribution. Then we need the distribution of sum of these Marchenko–Pastur distributions, which are independent. For some reason these distributions do not commute therefore we cannot do normal convolution and instead have to do something called free convolution? | |
| Jan 22, 2014 at 17:41 | comment | added | ofer zeitouni | In order to relate to the empirical measure of eigenvalues. Since you care about optimization, maximizing the quantity or the log of the quantity is the same. But even if not, the computation I sketched gives you the exponential rate of growth of the determinant. (Recall that the determinant is product of eigenvalues, hence taking log makes it into a sum). | |
| Jan 22, 2014 at 16:55 | comment | added | MLT | Could you please explain why you took the $\log$ of $B$? | |
| Jan 22, 2014 at 16:02 | history | answered | ofer zeitouni | CC BY-SA 3.0 |