Timeline for Restricted Isometry Property (Non Sparse Gaussian)
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2016 at 21:41 | comment | added | Samrat Mukhopadhyay | Is it the $\delta$ dependent on sparsity, or the probability $p$, as far as I know, it should be the probability $p$ that depends on sparsity as well as $\delta$, though that does not prevent the matrix to satisfy the RIP property, just the no. of rows of the matrix has to satisfy some bound of that depends on the sparsity of the unknown vector. So, in that sense I think the answer is yes, most of the random linear mappings with i.i.d. entries will satisfy the RIP property, of course with conditions on $N-M$ and $K$ | |
| Jun 26, 2013 at 12:48 | comment | added | rodms | To the best of my knowledge, no. The probabilistic guarantees in the RIP property concern the distribution of the sensing matrix $A$, and not of the vector $x$. However, Fourier, Gaussian, and Bernoulli matrices are known to satisfy RIP with high probability, but again, if the support $x$ is large, $\delta$ will be too. | |
| Jun 25, 2013 at 17:53 | comment | added | Mykie | I only require that RIP be satisfied with a high probability and not for every $x$. So the $\delta$ may not be too large. Are there any formal results too show this? | |
| Jun 25, 2013 at 13:08 | history | answered | rodms | CC BY-SA 3.0 |