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Can the alpha, lambda values of a glmnet object output determine whether ridge or Lasso?

Why do Lars and Glmnet give different solutions for the Lasso problem?What does the varImp function in the caret package actually compute for a glmnet (elastic net) objectWhat are the differences between Ridge regression using R's glmnet and Python's scikit-learn?Using lasso regression in Matlab with constraints on lambda valuesHow to interpret the results when both ridge and lasso separately perform well but produce different coefficientsUsing different alpha values in glmnet when comparing two feature sets?r: coefficients from glmnet and caret are different for the same lambdaWhat is the difference between the lambda function in ridge regression vs in lasso regression?In LASSO, does it make sense to choose lambda based on the mean error associated with different lambda values, over multiple cross-validations?glmnet: Nested cross validation, tuning alpha and lambda

2

$begingroup$

Given a glmnet object using train() where trControl method is "cv" and number of iterations is 5, I obtained that the bestTune alpha and lambda values are alpha=0.1 and lambda= 0.007688342. On running the glmnet object, I notice that the alpha values start from 0.1.
Can the inference here be that the method used is Lasso and not ridge because of the non-negative alpha value?

In general, can the values of alpha, lambda indicate which model is being used?

regression generalized-linear-model cross-validation caret

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asked 2 hours ago

red4life93red4life93

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2

$begingroup$

Given a glmnet object using train() where trControl method is "cv" and number of iterations is 5, I obtained that the bestTune alpha and lambda values are alpha=0.1 and lambda= 0.007688342. On running the glmnet object, I notice that the alpha values start from 0.1.
Can the inference here be that the method used is Lasso and not ridge because of the non-negative alpha value?

In general, can the values of alpha, lambda indicate which model is being used?

regression generalized-linear-model cross-validation caret

share|cite|improve this question

asked 2 hours ago

red4life93red4life93

111

New contributor

red4life93 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Given a glmnet object using train() where trControl method is "cv" and number of iterations is 5, I obtained that the bestTune alpha and lambda values are alpha=0.1 and lambda= 0.007688342. On running the glmnet object, I notice that the alpha values start from 0.1.
Can the inference here be that the method used is Lasso and not ridge because of the non-negative alpha value?

In general, can the values of alpha, lambda indicate which model is being used?

regression generalized-linear-model cross-validation caret

share|cite|improve this question

asked 2 hours ago

red4life93red4life93

111

New contributor

red4life93 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 2 hours ago

red4life93red4life93

111

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red4life93 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

asked 2 hours ago

red4life93red4life93

111

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red4life93 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

red4life93red4life93

111

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Check out our Code of Conduct.

asked 2 hours ago

red4life93red4life93

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2 Answers
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Absolutely! The $alpha$ parameter can be adjusted to either fit a Lasso or a Ridge regression (or something in between). Recall that the loss function which Elastic Net minimizes is $$frac12Nsum^N_i=1(y_i-beta_0-x_i^tbeta)^2+lambdasum_j=1^p(frac12(1-alpha)beta_j^2+alpha|beta_j|).$$
Focus on the second big sum (the one multiplied by $lambda$). If you let $alpha=1$, the first term inside this sum becomes $0$, and the whole function becomes exactly the function that Lasso minimizes (or the Lasso loss function). If you let $alpha=0$, the second term becomes $0$ and you are left with Ridge.

You can check the loss for Ridge and Lasso in this book and for elastic net in this paper.

share|cite|improve this answer

answered 2 hours ago

BananinBananin

1795

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like a good answer but can you edit to include citations for the hyperlinks? Over time, links die.
  $endgroup$
  – Sycorax
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

As far as I understand glmnet, $alpha=0$ would actually be a ridge penalty, and $alpha=1$ would be a Lasso penalty (rather than the other way around) and as far as glmnet is concerned you can fit those end cases.

The penalty with $alpha=0.1$ would be fairly similar to the ridge penalty but it is not the ridge penalty; if it's not considering $alpha$ below $0.1$ you can't necessarily infer much more than that just from the fact that you had that endpoint. If you know that an $alpha$ value that was only slightly larger was worse then it would be likely that a larger range might have chosen a smaller $alpha$, but it doesn't suggest it would have been $0$; I expect it would not. If the grid of values is coarse it may well have been that a larger value than $0.1$ would be better.

[You may want to check whether there was some other reason that $alpha$ might have been at an endpoint; e.g. I seem to recall $lambda$ got set to an endpoint in forecasting if coefficients for lambdaOpt were not saved.]

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Glen_b♦Glen_b

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1

$begingroup$

Absolutely! The $alpha$ parameter can be adjusted to either fit a Lasso or a Ridge regression (or something in between). Recall that the loss function which Elastic Net minimizes is $$frac12Nsum^N_i=1(y_i-beta_0-x_i^tbeta)^2+lambdasum_j=1^p(frac12(1-alpha)beta_j^2+alpha|beta_j|).$$
Focus on the second big sum (the one multiplied by $lambda$). If you let $alpha=1$, the first term inside this sum becomes $0$, and the whole function becomes exactly the function that Lasso minimizes (or the Lasso loss function). If you let $alpha=0$, the second term becomes $0$ and you are left with Ridge.

You can check the loss for Ridge and Lasso in this book and for elastic net in this paper.

share|cite|improve this answer

answered 2 hours ago

BananinBananin

1795

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like a good answer but can you edit to include citations for the hyperlinks? Over time, links die.
  $endgroup$
  – Sycorax
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Absolutely! The $alpha$ parameter can be adjusted to either fit a Lasso or a Ridge regression (or something in between). Recall that the loss function which Elastic Net minimizes is $$frac12Nsum^N_i=1(y_i-beta_0-x_i^tbeta)^2+lambdasum_j=1^p(frac12(1-alpha)beta_j^2+alpha|beta_j|).$$
Focus on the second big sum (the one multiplied by $lambda$). If you let $alpha=1$, the first term inside this sum becomes $0$, and the whole function becomes exactly the function that Lasso minimizes (or the Lasso loss function). If you let $alpha=0$, the second term becomes $0$ and you are left with Ridge.

You can check the loss for Ridge and Lasso in this book and for elastic net in this paper.

share|cite|improve this answer

answered 2 hours ago

BananinBananin

1795

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This looks like a good answer but can you edit to include citations for the hyperlinks? Over time, links die.
  $endgroup$
  – Sycorax
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Absolutely! The $alpha$ parameter can be adjusted to either fit a Lasso or a Ridge regression (or something in between). Recall that the loss function which Elastic Net minimizes is $$frac12Nsum^N_i=1(y_i-beta_0-x_i^tbeta)^2+lambdasum_j=1^p(frac12(1-alpha)beta_j^2+alpha|beta_j|).$$
Focus on the second big sum (the one multiplied by $lambda$). If you let $alpha=1$, the first term inside this sum becomes $0$, and the whole function becomes exactly the function that Lasso minimizes (or the Lasso loss function). If you let $alpha=0$, the second term becomes $0$ and you are left with Ridge.

You can check the loss for Ridge and Lasso in this book and for elastic net in this paper.

share|cite|improve this answer

answered 2 hours ago

BananinBananin

1795

$endgroup$

share|cite|improve this answer

answered 2 hours ago

BananinBananin

1795

answered 2 hours ago

BananinBananin

1795

answered 2 hours ago

BananinBananin

1795

1

$begingroup$

As far as I understand glmnet, $alpha=0$ would actually be a ridge penalty, and $alpha=1$ would be a Lasso penalty (rather than the other way around) and as far as glmnet is concerned you can fit those end cases.

The penalty with $alpha=0.1$ would be fairly similar to the ridge penalty but it is not the ridge penalty; if it's not considering $alpha$ below $0.1$ you can't necessarily infer much more than that just from the fact that you had that endpoint. If you know that an $alpha$ value that was only slightly larger was worse then it would be likely that a larger range might have chosen a smaller $alpha$, but it doesn't suggest it would have been $0$; I expect it would not. If the grid of values is coarse it may well have been that a larger value than $0.1$ would be better.

[You may want to check whether there was some other reason that $alpha$ might have been at an endpoint; e.g. I seem to recall $lambda$ got set to an endpoint in forecasting if coefficients for lambdaOpt were not saved.]

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Glen_b♦Glen_b

213k22412762

$endgroup$

add a comment | 

1

$begingroup$

As far as I understand glmnet, $alpha=0$ would actually be a ridge penalty, and $alpha=1$ would be a Lasso penalty (rather than the other way around) and as far as glmnet is concerned you can fit those end cases.

The penalty with $alpha=0.1$ would be fairly similar to the ridge penalty but it is not the ridge penalty; if it's not considering $alpha$ below $0.1$ you can't necessarily infer much more than that just from the fact that you had that endpoint. If you know that an $alpha$ value that was only slightly larger was worse then it would be likely that a larger range might have chosen a smaller $alpha$, but it doesn't suggest it would have been $0$; I expect it would not. If the grid of values is coarse it may well have been that a larger value than $0.1$ would be better.

[You may want to check whether there was some other reason that $alpha$ might have been at an endpoint; e.g. I seem to recall $lambda$ got set to an endpoint in forecasting if coefficients for lambdaOpt were not saved.]

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Glen_b♦Glen_b

213k22412762

$endgroup$

add a comment | 

$begingroup$

As far as I understand glmnet, $alpha=0$ would actually be a ridge penalty, and $alpha=1$ would be a Lasso penalty (rather than the other way around) and as far as glmnet is concerned you can fit those end cases.

The penalty with $alpha=0.1$ would be fairly similar to the ridge penalty but it is not the ridge penalty; if it's not considering $alpha$ below $0.1$ you can't necessarily infer much more than that just from the fact that you had that endpoint. If you know that an $alpha$ value that was only slightly larger was worse then it would be likely that a larger range might have chosen a smaller $alpha$, but it doesn't suggest it would have been $0$; I expect it would not. If the grid of values is coarse it may well have been that a larger value than $0.1$ would be better.

[You may want to check whether there was some other reason that $alpha$ might have been at an endpoint; e.g. I seem to recall $lambda$ got set to an endpoint in forecasting if coefficients for lambdaOpt were not saved.]

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Glen_b♦Glen_b

213k22412762

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

Glen_b♦Glen_b

213k22412762

answered 2 hours ago

Glen_b♦Glen_b

213k22412762

answered 2 hours ago

Glen_b♦Glen_b

213k22412762