Timeline for Robust estimation of $Ax=b$
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Dec 5, 2020 at 18:31 | comment | added | Daniel Shapero | To add to @FedericoPoloni's excellent answer, there are loads of other ways of solving L1-type problems besides IRLS. Proximal algorithms, such as FISTA, are often very effective and quite simple to implement. | |
Dec 5, 2020 at 17:48 | comment | added | Mark L. Stone | I have now edited my answer. | |
Dec 5, 2020 at 17:43 | comment | added | Mark L. Stone | Fair enough. But why no solve it as a Linear Programming Problem? Unlike a century ago, such solvers now exist. | |
Dec 5, 2020 at 17:41 | comment | added | Federico Poloni | @MarkL.Stone No, IRLS is an iterative method to solve a L1 loss problem in which a L2 problem is solved at each iteration. | |
Dec 5, 2020 at 17:35 | comment | added | Mark L. Stone | The problem statement says L1 loss, not least squares (L2) loss. | |
Dec 5, 2020 at 14:47 | comment | added | Federico Poloni | QR. Anyway, are you sure SVD is that expensive? It should be at most a factor 2 slower than A^TWA. | |
Dec 5, 2020 at 14:44 | comment | added | lalit | Thank you for your answer. I'll update on this. For m>>n, e.g., m=10000, n=100, SVD computation of W^(1/2) * A will be expensive both in terms of memory and time taken when compared to A^T * W * A. Can you suggest any method to reduce the computation time? | |
Dec 5, 2020 at 14:06 | history | edited | Federico Poloni | CC BY-SA 4.0 |
added 105 characters in body
|
Dec 5, 2020 at 12:30 | history | answered | Federico Poloni | CC BY-SA 4.0 |