Timeline for Solving Linear Regression with Fused Lasso Regularization by MLE
Current License: CC BY-SA 4.0
21 events
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| Aug 29, 2018 at 7:40 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 29, 2018 at 7:28 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 29, 2018 at 7:21 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 29, 2018 at 6:20 | comment | added | AlexConfused | Let us continue this discussion in chat. | |
| Aug 28, 2018 at 18:23 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 17:18 | comment | added | Sextus Empiricus | @AlexConfused I was adding the regressors not the entries. See the final part with the example how it works out (the $x_i$ are the vectors there with the index $_i$ referring to the $i$-th regressor). So in that example there are three regressors but there can be many more measurements. Could you explain why you think this is wrong. | |
| Aug 28, 2018 at 16:25 | comment | added | AlexConfused | I think it should refer to the i-th entry. Only then is the transformation correct. | |
| Aug 28, 2018 at 16:23 | review | Suggested edits | |||
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| Aug 28, 2018 at 15:34 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 14:57 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 14:56 | comment | added | Sextus Empiricus | It refers to the i-th vector | |
| Aug 28, 2018 at 14:44 | comment | added | AlexConfused | In your transformation $z_j = \sum_{i=j}^N x_i$ , is $x_i$ the $i$-th vector from the training set (as introduced in question) or the $i$-th entry of some vector $x$? | |
| Aug 28, 2018 at 14:20 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 14:06 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 14:01 | comment | added | Sextus Empiricus | I changed the the transformation (which was indeed in error). The $\beta^\star$ will indeed not be necessarily close to each other. Instead the $\beta^\star$ will be as small as possible such that the neighbors in $\beta_j=\sum_{i=1}^j \beta_i^\star$ will be as close as possible. | |
| Aug 28, 2018 at 13:59 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 13:58 | comment | added | AlexConfused | I don't think this is correct. Firstly, I am not sure if your transformation works and even if it works you will solve a regression problem with ridge loss in the end, i.e. your parameters $\beta$ will not have a form such that neighbor parameters are as close as $S$ to each other. | |
| S Aug 28, 2018 at 13:50 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 13:50 | review | Suggested edits | |||
| S Aug 28, 2018 at 13:50 | |||||
| Aug 28, 2018 at 13:43 | history | edited | Sextus Empiricus | CC BY-SA 4.0 |
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| Aug 28, 2018 at 13:33 | history | answered | Sextus Empiricus | CC BY-SA 4.0 |