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How to show 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 formulationsWhy is “relaxed lasso” different from standard lasso?Bridge penalty vs. Elastic Net regularizationLogistic 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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$begingroup$
According to the following 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)?
These formulas are for Ridge regression.
These formulas are for LASSO regression.
These 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 7 hours ago
jezajeza
470420
$endgroup$
add a comment |
$begingroup$
According to the following 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)?
These formulas are for Ridge regression.
These formulas are for LASSO regression.
These 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 7 hours ago
jezajeza
470420
$endgroup$
add a comment |
$begingroup$
According to the following 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)?
These formulas are for Ridge regression.
These formulas are for LASSO regression.
These 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 7 hours ago
jezajeza
470420
$endgroup$
According to the following 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)?
These formulas are for Ridge regression.
These formulas are for LASSO regression.
These 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 7 hours ago
jezajeza
470420
asked 7 hours ago
jezajeza
470420
edited 2 hours ago
jeza
asked 7 hours ago
jezajeza
470420
asked 7 hours ago
jezajeza
470420
asked 7 hours ago
jezajeza
470420
470420
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
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left(1-alpha) sum_j=1^p $$
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 7 hours ago
stats_modelstats_model
20216
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
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active
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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
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left(1-alpha) sum_j=1^p $$
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 7 hours ago
stats_modelstats_model
20216
$endgroup$
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
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left(1-alpha) sum_j=1^p $$
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 7 hours ago
stats_modelstats_model
20216
$endgroup$
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
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left(1-alpha) sum_j=1^p $$
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 7 hours ago
stats_modelstats_model
20216
$endgroup$
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
$$mathcal L(beta) = undersetbetamathrmargmin,leftsum_i=1^N left(y_i - sum_j=1^p x_ij beta_jright)^2right + mu left(1-alpha) sum_j=1^p $$
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 7 hours ago
stats_modelstats_model
20216
edited 5 hours ago
answered 7 hours ago
stats_modelstats_model
20216
answered 7 hours ago
stats_modelstats_model
20216
answered 7 hours ago
stats_modelstats_model
20216
20216
add a comment |
add a comment |
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