Timeline for Upper bounding a inner product between gaussian Wigner matrix and a rank 2 matrix
Current License: CC BY-SA 4.0
15 events
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| Jun 8 at 17:26 | comment | added | tony | Thank you so much! Since $n\sqrt{n}$ doesn't suit my purpose, maybe adding certain constraint on $Q$ can achieve a smaller upper bound | |
| Jun 8 at 17:12 | comment | added | tony | I still have one question out of curiosity: (1) If I understood correctly, since $Q$ is rank 2, doing svd on $Q$ and use triangular inequality gives us $\mathbb E\sup_{q\in\{-1,=1\}^n}\langle Z,QQ^T \rangle\leq 2 \mathbb E\sup_{q\in[-1,+1]}\langle Z,qq^T \rangle$. Is there another way to argue it? In some cases, $QQ^T$ is replaced by a certain $f(QQ^T)$, which is not rank two. | |
| Jun 8 at 17:07 | vote | accept | tony | ||
| Jun 8 at 15:21 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 13:57 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jun 8 at 13:11 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 12:53 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 12:46 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 12:38 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 12:28 | comment | added | Jarosław Błasiok | Fixed a few mistakes in the proof, but the answer is correct: $n^{3/2}$ is the right grow of the supremum. We can reduce to the case where $q$ is rank one, but it is a matrix of form $Q = q q^T$ where all the entries of $q$ are between $-1$ and $1$ (for the lower bound it is enough to consider entries being either $-1$ or $1$), sup $\mathbb{E} \sup \langle Z, Q Q^T \rangle$ is actually non-trivial supremum. | |
| Jun 8 at 12:27 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 10:13 | comment | added | tony | I never thought about we could directly consider rank 1 matrix, which simplify the problem a lot! And about this, I have an addition information on $Q$: if $Q$ is rank 1, then $Q=1_n1_n^T$, where $1_n$ is all one n-dimensional vector. In this case, according to your answer $\mathbb E\sup \langle W,QQ^T \rangle \approx \mathbb E\sup \langle Z,QQ^T \rangle$ which concentrates on $n$. Is this correct? | |
| Jun 8 at 9:10 | comment | added | tony | Thank you! what is the value of $\gamma_2(\{\pm 1\}^{n/4},d_2)$ in the lower bound $\sqrt{n} \gamma_2(\{\pm 1\}^{n/4},d_2)$? is it $\gamma_2(\{\pm 1 \}^{n/4},d_2)\approx \mathbb E\sup_{q\in\{\pm 1\}^{n/4}}\langle q,G\rangle$ where $G$ is a gaussian vector in $\mathbb R^{n/4}$? this gives $\gamma_2(\{\pm \}^{n/4},d_2)\approx \mathbb E\|G\|_1\approx n/4$? Thus the lower bound is $\sqrt{n}n^{1/4}=n^{3/4}$ | |
| Jun 8 at 2:15 | history | edited | Jarosław Błasiok | CC BY-SA 4.0 |
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| Jun 8 at 1:57 | history | answered | Jarosław Błasiok | CC BY-SA 4.0 |