Showing the equivalence between the regularized regression and their constraint formulas using KKTThe proof of equivalent formulas of ridge regressionRidge regression formulation as constrained versus penalized: How are they equivalent?Equivalence between Elastic Net formulationsCalculating $R^2$ for Elastic NetEquivalence between Elastic Net formulationsBridge penalty vs. Elastic Net regularizationRegularized linear regression fails to predict my dataLogistic regression coefficients are wildlyHow to explain differences in formulas of ridge regression, lasso, and elastic netIntuition Behind the Elastic Net PenaltyRegularized Logistic Regression: Lasso vs. Ridge vs. Elastic NetCan you predict the residuals from a regularized regression using the same data?Elastic Net and collinearity
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Showing the equivalence between the regularized regression and their constraint formulas using KKT
The proof of equivalent formulas of ridge regressionRidge regression formulation as constrained versus penalized: How are they equivalent?Equivalence between Elastic Net formulationsCalculating $R^2$ for Elastic NetEquivalence between Elastic Net formulationsBridge penalty vs. Elastic Net regularizationRegularized linear regression fails to predict my dataLogistic regression coefficients are wildlyHow to explain differences in formulas of ridge regression, lasso, and elastic netIntuition Behind the Elastic Net PenaltyRegularized Logistic Regression: Lasso vs. Ridge vs. Elastic NetCan you predict the residuals from a regularized regression using the same data?Elastic Net and collinearity
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
According to the references Book 1, Book 2 and paper.
It has been mentioned that there is an equivalence between the regularized regression (Ridge, LASSO and Elastic Net) and their constraint formulas.
I have also looked at Cross Validated 1, and Cross Validated 2, but I can not see a clear answer show that equivalence or logic.
My question is
How to show that equivalence using Karush–Kuhn–Tucker (KKT)?
The following formulas are for Ridge regression.
The following formulas are for LASSO regression.
The following formulas are for Elastic Net regression.
NOTE
This question is not homework. It is only to increase my comprehension of this topic.
regression optimization lasso ridge-regression elastic-net
asked Apr 4 at 16:05
jezajeza
425420
$endgroup$
This question has an open bounty worth +50
reputation from jeza ending ending at 2019-04-13 16:45:13Z">in 6 days.
This question has not received enough attention.
Required detailed answer step by step with a practical example.
add a comment |
$begingroup$
According to the references Book 1, Book 2 and paper.
It has been mentioned that there is an equivalence between the regularized regression (Ridge, LASSO and Elastic Net) and their constraint formulas.
I have also looked at Cross Validated 1, and Cross Validated 2, but I can not see a clear answer show that equivalence or logic.
My question is
How to show that equivalence using Karush–Kuhn–Tucker (KKT)?
The following formulas are for Ridge regression.
The following formulas are for LASSO regression.
The following formulas are for Elastic Net regression.
NOTE
This question is not homework. It is only to increase my comprehension of this topic.
regression optimization lasso ridge-regression elastic-net
asked Apr 4 at 16:05
jezajeza
425420
$endgroup$
This question has an open bounty worth +50
reputation from jeza ending ending at 2019-04-13 16:45:13Z">in 6 days.
This question has not received enough attention.
Required detailed answer step by step with a practical example.
add a comment |
$begingroup$
According to the references Book 1, Book 2 and paper.
It has been mentioned that there is an equivalence between the regularized regression (Ridge, LASSO and Elastic Net) and their constraint formulas.
I have also looked at Cross Validated 1, and Cross Validated 2, but I can not see a clear answer show that equivalence or logic.
My question is
How to show that equivalence using Karush–Kuhn–Tucker (KKT)?
The following formulas are for Ridge regression.
The following formulas are for LASSO regression.
The following formulas are for Elastic Net regression.
NOTE
This question is not homework. It is only to increase my comprehension of this topic.
regression optimization lasso ridge-regression elastic-net
asked Apr 4 at 16:05
jezajeza
425420
$endgroup$
According to the references Book 1, Book 2 and paper.
It has been mentioned that there is an equivalence between the regularized regression (Ridge, LASSO and Elastic Net) and their constraint formulas.
I have also looked at Cross Validated 1, and Cross Validated 2, but I can not see a clear answer show that equivalence or logic.
My question is
How to show that equivalence using Karush–Kuhn–Tucker (KKT)?
The following formulas are for Ridge regression.
The following formulas are for LASSO regression.
The following formulas are for Elastic Net regression.
NOTE
This question is not homework. It is only to increase my comprehension of this topic.
regression optimization lasso ridge-regression elastic-net
regression optimization lasso ridge-regression elastic-net
asked Apr 4 at 16:05
jezajeza
425420
asked Apr 4 at 16:05
jezajeza
425420
edited 9 hours ago
jeza
asked Apr 4 at 16:05
jezajeza
425420
asked Apr 4 at 16:05
jezajeza
425420
asked Apr 4 at 16:05
jezajeza
425420
425420
This question has an open bounty worth +50
reputation from jeza ending ending at 2019-04-13 16:45:13Z">in 6 days.
This question has not received enough attention.
Required detailed answer step by step with a practical example.
This question has an open bounty worth +50
reputation from jeza ending ending at 2019-04-13 16:45:13Z">in 6 days.
This question has not received enough attention.
Required detailed answer step by step with a practical example.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The more technical answer is because the constrained optimization problem can be written in terms of Lagrange multipliers. In particular, the Lagrangian associated with the constrained optimization problem is given by
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left + alpha sum_j=1^p beta_j^2right$$
where $mu$ is a multiplier chosen to satisfy the constraints of the problem. The first order conditions (which are sufficient since you are working with nice proper convex functions) for this optimization problem can thus be obtained by differentiating the Lagrangian with respect to $beta$ and setting the derivatives equal to 0 (it's a bit more nuanced since the LASSO part has undifferentiable points, but there are methods from convex analysis to generalize the derivative to make the first order condition still work). It is clear that these first order conditions are identical to the first order conditions of the unconstrained problem you wrote down.
However, I think it's useful to see why in general, with these optimization problems, it is often possible to think about the problem either through the lens of a constrained optimization problem or through the lens of an unconstrained problem. More concretely, suppose we have an unconstrained optimization problem of the following form:
$$max_x f(x) + lambda g(x)$$
We can always try to solve this optimization directly, but sometimes, it might make sense to break this problem into subcomponents. In particular, it is not hard to see that
$$max_x f(x) + lambda g(x) = max_t left(max_x f(x) mathrm s.t g(x) = tright) + lambda t$$
So for a fixed value of $lambda$ (and assuming the functions to be optimized actually achieve their optima), we can associate with it a value $t^*$ that solves the outer optimization problem. This gives us a sort of mapping from unconstrained optimization problems to constrained problems. In your particular setting, since everything is nicely behaved for elastic net regression, this mapping should in fact be one to one, so it will be useful to be able to switch between these two contexts depending on which is more useful to a particular application. In general, this relationship between constrained and unconstrained problems may be less well behaved, but it may still be useful to think about to what extent you can move between the constrained and unconstrained problem.
answered Apr 4 at 16:34
stats_modelstats_model
21216
$endgroup$
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The more technical answer is because the constrained optimization problem can be written in terms of Lagrange multipliers. In particular, the Lagrangian associated with the constrained optimization problem is given by
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left + alpha sum_j=1^p beta_j^2right$$
where $mu$ is a multiplier chosen to satisfy the constraints of the problem. The first order conditions (which are sufficient since you are working with nice proper convex functions) for this optimization problem can thus be obtained by differentiating the Lagrangian with respect to $beta$ and setting the derivatives equal to 0 (it's a bit more nuanced since the LASSO part has undifferentiable points, but there are methods from convex analysis to generalize the derivative to make the first order condition still work). It is clear that these first order conditions are identical to the first order conditions of the unconstrained problem you wrote down.
However, I think it's useful to see why in general, with these optimization problems, it is often possible to think about the problem either through the lens of a constrained optimization problem or through the lens of an unconstrained problem. More concretely, suppose we have an unconstrained optimization problem of the following form:
$$max_x f(x) + lambda g(x)$$
We can always try to solve this optimization directly, but sometimes, it might make sense to break this problem into subcomponents. In particular, it is not hard to see that
$$max_x f(x) + lambda g(x) = max_t left(max_x f(x) mathrm s.t g(x) = tright) + lambda t$$
So for a fixed value of $lambda$ (and assuming the functions to be optimized actually achieve their optima), we can associate with it a value $t^*$ that solves the outer optimization problem. This gives us a sort of mapping from unconstrained optimization problems to constrained problems. In your particular setting, since everything is nicely behaved for elastic net regression, this mapping should in fact be one to one, so it will be useful to be able to switch between these two contexts depending on which is more useful to a particular application. In general, this relationship between constrained and unconstrained problems may be less well behaved, but it may still be useful to think about to what extent you can move between the constrained and unconstrained problem.
answered Apr 4 at 16:34
stats_modelstats_model
21216
$endgroup$
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
add a comment |
$begingroup$
The more technical answer is because the constrained optimization problem can be written in terms of Lagrange multipliers. In particular, the Lagrangian associated with the constrained optimization problem is given by
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left + alpha sum_j=1^p beta_j^2right$$
where $mu$ is a multiplier chosen to satisfy the constraints of the problem. The first order conditions (which are sufficient since you are working with nice proper convex functions) for this optimization problem can thus be obtained by differentiating the Lagrangian with respect to $beta$ and setting the derivatives equal to 0 (it's a bit more nuanced since the LASSO part has undifferentiable points, but there are methods from convex analysis to generalize the derivative to make the first order condition still work). It is clear that these first order conditions are identical to the first order conditions of the unconstrained problem you wrote down.
However, I think it's useful to see why in general, with these optimization problems, it is often possible to think about the problem either through the lens of a constrained optimization problem or through the lens of an unconstrained problem. More concretely, suppose we have an unconstrained optimization problem of the following form:
$$max_x f(x) + lambda g(x)$$
We can always try to solve this optimization directly, but sometimes, it might make sense to break this problem into subcomponents. In particular, it is not hard to see that
$$max_x f(x) + lambda g(x) = max_t left(max_x f(x) mathrm s.t g(x) = tright) + lambda t$$
So for a fixed value of $lambda$ (and assuming the functions to be optimized actually achieve their optima), we can associate with it a value $t^*$ that solves the outer optimization problem. This gives us a sort of mapping from unconstrained optimization problems to constrained problems. In your particular setting, since everything is nicely behaved for elastic net regression, this mapping should in fact be one to one, so it will be useful to be able to switch between these two contexts depending on which is more useful to a particular application. In general, this relationship between constrained and unconstrained problems may be less well behaved, but it may still be useful to think about to what extent you can move between the constrained and unconstrained problem.
answered Apr 4 at 16:34
stats_modelstats_model
21216
$endgroup$
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
add a comment |
$begingroup$
The more technical answer is because the constrained optimization problem can be written in terms of Lagrange multipliers. In particular, the Lagrangian associated with the constrained optimization problem is given by
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left + alpha sum_j=1^p beta_j^2right$$
where $mu$ is a multiplier chosen to satisfy the constraints of the problem. The first order conditions (which are sufficient since you are working with nice proper convex functions) for this optimization problem can thus be obtained by differentiating the Lagrangian with respect to $beta$ and setting the derivatives equal to 0 (it's a bit more nuanced since the LASSO part has undifferentiable points, but there are methods from convex analysis to generalize the derivative to make the first order condition still work). It is clear that these first order conditions are identical to the first order conditions of the unconstrained problem you wrote down.
However, I think it's useful to see why in general, with these optimization problems, it is often possible to think about the problem either through the lens of a constrained optimization problem or through the lens of an unconstrained problem. More concretely, suppose we have an unconstrained optimization problem of the following form:
$$max_x f(x) + lambda g(x)$$
We can always try to solve this optimization directly, but sometimes, it might make sense to break this problem into subcomponents. In particular, it is not hard to see that
$$max_x f(x) + lambda g(x) = max_t left(max_x f(x) mathrm s.t g(x) = tright) + lambda t$$
So for a fixed value of $lambda$ (and assuming the functions to be optimized actually achieve their optima), we can associate with it a value $t^*$ that solves the outer optimization problem. This gives us a sort of mapping from unconstrained optimization problems to constrained problems. In your particular setting, since everything is nicely behaved for elastic net regression, this mapping should in fact be one to one, so it will be useful to be able to switch between these two contexts depending on which is more useful to a particular application. In general, this relationship between constrained and unconstrained problems may be less well behaved, but it may still be useful to think about to what extent you can move between the constrained and unconstrained problem.
answered Apr 4 at 16:34
stats_modelstats_model
21216
$endgroup$
The more technical answer is because the constrained optimization problem can be written in terms of Lagrange multipliers. In particular, the Lagrangian associated with the constrained optimization problem is given by
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left + alpha sum_j=1^p beta_j^2right$$
where $mu$ is a multiplier chosen to satisfy the constraints of the problem. The first order conditions (which are sufficient since you are working with nice proper convex functions) for this optimization problem can thus be obtained by differentiating the Lagrangian with respect to $beta$ and setting the derivatives equal to 0 (it's a bit more nuanced since the LASSO part has undifferentiable points, but there are methods from convex analysis to generalize the derivative to make the first order condition still work). It is clear that these first order conditions are identical to the first order conditions of the unconstrained problem you wrote down.
However, I think it's useful to see why in general, with these optimization problems, it is often possible to think about the problem either through the lens of a constrained optimization problem or through the lens of an unconstrained problem. More concretely, suppose we have an unconstrained optimization problem of the following form:
$$max_x f(x) + lambda g(x)$$
We can always try to solve this optimization directly, but sometimes, it might make sense to break this problem into subcomponents. In particular, it is not hard to see that
$$max_x f(x) + lambda g(x) = max_t left(max_x f(x) mathrm s.t g(x) = tright) + lambda t$$
So for a fixed value of $lambda$ (and assuming the functions to be optimized actually achieve their optima), we can associate with it a value $t^*$ that solves the outer optimization problem. This gives us a sort of mapping from unconstrained optimization problems to constrained problems. In your particular setting, since everything is nicely behaved for elastic net regression, this mapping should in fact be one to one, so it will be useful to be able to switch between these two contexts depending on which is more useful to a particular application. In general, this relationship between constrained and unconstrained problems may be less well behaved, but it may still be useful to think about to what extent you can move between the constrained and unconstrained problem.
answered Apr 4 at 16:34
stats_modelstats_model
21216
edited Apr 4 at 18:42
answered Apr 4 at 16:34
stats_modelstats_model
21216
answered Apr 4 at 16:34
stats_modelstats_model
21216
answered Apr 4 at 16:34
stats_modelstats_model
21216
21216
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
add a comment |
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
$begingroup$
could you please provide us with a detailed answer step by step with a practical example if that possible.
$endgroup$
– jeza
40 mins ago
add a comment |
Thanks for contributing an answer to Cross Validated!
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