Timeline for Analysis for checking if an Ensemble model is a better fit for a dataset than Primitive model
Current License: CC BY-SA 3.0
17 events
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| Oct 28, 2014 at 16:07 | history | edited | show_stopper | CC BY-SA 3.0 |
deleted 111 characters in body; edited tags; edited title
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| Oct 28, 2014 at 16:06 | comment | added | show_stopper | sorry for my lack of knowledge about terminologies. What I meant by linear model was : primitive models and what I meant by non-linear models was ensemble models. I have made the required edits in the question above | |
| Oct 28, 2014 at 15:46 | answer | added | charles | timeline score: 1 | |
| Oct 28, 2014 at 14:10 | comment | added | Robert Kubrick | @Scortchi Yes, my bad. I was confusing with $log(y) = \beta_0 + {\beta_1}log(x) + \epsilon$. | |
| Oct 28, 2014 at 13:45 | comment | added | Scortchi♦ | @RobertKubrick: The latter is the logarithm of the former. Nevertheless $\beta_0 + \beta_1\log x + \epsilon$ is not the logarithm of $\beta_0+x^{\beta_1} +\varepsilon$: the models are not equivalent. | |
| Oct 28, 2014 at 13:34 | comment | added | Robert Kubrick | @Scortchi How is $x^{\beta_1}$ different from $\beta_1 log(x)$? | |
| Oct 28, 2014 at 13:31 | comment | added | Scortchi♦ | @RobertKubrick: That's a quite different model, & one which is indeed non-linear in its parameters. $\beta_1$ enters the model as an exponent, not as a coefficient. | |
| Oct 28, 2014 at 13:27 | comment | added | Robert Kubrick | @Scortchi How is linear in the parameters if $y = \beta_0 + x^{\beta_1} + \epsilon$? | |
| Oct 28, 2014 at 13:23 | comment | added | Scortchi♦ | @RobertKubrick: $y =\beta_0 + \beta_1 \log{x} + \varepsilon$ is linear in $\beta_0$ & $\beta_1$. gung has good explanations here & here. The criterion is that the parameters to be estimated only enter the model as coefficients of additive terms, which can be any old functions of the predictors. | |
| Oct 28, 2014 at 13:04 | comment | added | Robert Kubrick | @Scortchi Fine, but then any $x$ log transformation implies non-linearity (because of the multiplicative relationship between the untransformed features). So the strict description would be 1 exponents and addictive relationship between the features. | |
| Oct 28, 2014 at 12:56 | comment | added | Scortchi♦ | @RobertKubrick: That's not quite the distinction - e.g. $y = \frac{\beta_0 x}{\beta_1 + x} + \varepsilon$ is non-linear in the parameters $\beta_0$ & $\beta_1$. But you're right it's got nothing to do with feature selection. | |
| Oct 28, 2014 at 12:15 | comment | added | Steve S |
Re: non-linear models should not be used unless absolutely necessary. That makes it sound as if you're saying that someone should only resort to non-linear models in extreme circumstances.
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| Oct 28, 2014 at 12:03 | answer | added | Steve S | timeline score: 1 | |
| Oct 28, 2014 at 11:39 | comment | added | Robert Kubrick | @nar Linear in the parameters means the parameters exponent are all set to 1. That has nothing to do with the number of features used in the model. | |
| Oct 28, 2014 at 10:05 | comment | added | show_stopper | My understanding. Linear model: a model which has a defined number of parameters, can be either polynomial, or log or an interaction term, but the term needs to be decided beforehand. Non-Linear: no defined number of parameters. You just give in the variables, and the parameters and the number of parameters are selected when the algorithm is fitting | |
| Oct 28, 2014 at 9:54 | comment | added | Scortchi♦ | GLMs are linear in the parameters. In empirical modelling a non-linear relationship between the expected response (transformed by the link function) & a predictor is often allowed for by representing that predictor with a polynomial or spline basis: the model is still linear. Are you using "linear" in the same sense? | |
| Oct 28, 2014 at 9:29 | history | asked | show_stopper | CC BY-SA 3.0 |