Equivalent formulation of LASSO?
$begingroup$
I am currently trying to tell wheter or not those two problems are equivalent :
$$min_x |x|_1 text { s.t. } |Ax-y|^2_2 le varepsilon.$$
And
$$min_x |Ax-y|^2_2 text { s.t. } |x|_1le t.$$
With $x$ and $y$ vectors and $A$ a compatible matrix.
I know that the second problem is equivalent to :
$$min_x |Ax-y|^2_2 + lambda |x|_1.$$
(Lagrangian form)
The same process for the first problem would give :
$$min_x |x|_1 + mu |Ax-y|^2_2$$
Does this lead somewhere ?
Thank for your help.
optimization linear-regression
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
$endgroup$
add a comment |
$begingroup$
I am currently trying to tell wheter or not those two problems are equivalent :
$$min_x |x|_1 text { s.t. } |Ax-y|^2_2 le varepsilon.$$
And
$$min_x |Ax-y|^2_2 text { s.t. } |x|_1le t.$$
With $x$ and $y$ vectors and $A$ a compatible matrix.
I know that the second problem is equivalent to :
$$min_x |Ax-y|^2_2 + lambda |x|_1.$$
(Lagrangian form)
The same process for the first problem would give :
$$min_x |x|_1 + mu |Ax-y|^2_2$$
Does this lead somewhere ?
Thank for your help.
optimization linear-regression
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
$endgroup$
add a comment |
$begingroup$
I am currently trying to tell wheter or not those two problems are equivalent :
$$min_x |x|_1 text { s.t. } |Ax-y|^2_2 le varepsilon.$$
And
$$min_x |Ax-y|^2_2 text { s.t. } |x|_1le t.$$
With $x$ and $y$ vectors and $A$ a compatible matrix.
I know that the second problem is equivalent to :
$$min_x |Ax-y|^2_2 + lambda |x|_1.$$
(Lagrangian form)
The same process for the first problem would give :
$$min_x |x|_1 + mu |Ax-y|^2_2$$
Does this lead somewhere ?
Thank for your help.
optimization linear-regression
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
$endgroup$
I am currently trying to tell wheter or not those two problems are equivalent :
$$min_x |x|_1 text { s.t. } |Ax-y|^2_2 le varepsilon.$$
And
$$min_x |Ax-y|^2_2 text { s.t. } |x|_1le t.$$
With $x$ and $y$ vectors and $A$ a compatible matrix.
I know that the second problem is equivalent to :
$$min_x |Ax-y|^2_2 + lambda |x|_1.$$
(Lagrangian form)
The same process for the first problem would give :
$$min_x |x|_1 + mu |Ax-y|^2_2$$
Does this lead somewhere ?
Thank for your help.
optimization linear-regression
optimization linear-regression
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
edited Dec 20 '18 at 10:05
nicomezi
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
asked Dec 20 '18 at 9:44
nicomezinicomezi
4,1591920
4,1591920
add a comment |
add a comment |
1 Answer
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$begingroup$
The first problem
$$min_x |x|_1 text { subject to } |Ax-y|^2_2 le epsilon$$
can be described by the equivalent saddle point problem:
$$
min_xmax_{lambdage 0}left[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right],
$$ by Lagrange's multiplier theorem. Let $lambda_0ge 0$ be the solution of the above problem such that
$$
max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=min_xleft[ |x|_1 +lambda_0(|Ax-y|_2^2-epsilon)right].
$$If we have $lambda_0 = 0$, then the minimizer should be $x=0$ and $|y|^2_2leqepsilon.$ Let us exclude this trivial case and assume that $lambda_0>0$. By complementary slackness, we should have $|Ax_0-y|_2^2 =epsilon$ where $x_0$ is the solution of the first problem.
In the same manner, the second problem can be equivalently described as
$$
min_{x'}max_{muge 0}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]=max_{muge 0}min_{x'}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]quadcdots(*),
$$ where $t = |x_0|_1$. If we look at the formulation of the second problem carefully, we can see that the solution $(mu_0, x_0')$ should be $(frac{1}{lambda_0},x_0)$. To see this, one can plug $mu=lambda_0^{-1}$ and $x_0'=x_0$ into the $(*)$ to check if equality holds. In fact, equality holds since given $mu=lambda_0^{-1}$, $x=x_0$ is the minimizer and conversely, given $x=x_0$, $mu =lambda_0^{-1}$ is the maximizer. This establishes that Lagrangian dual problems in this case (in fact, generally) are equivalent.
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
add a comment |
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1 Answer
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active
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1 Answer
1
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$begingroup$
The first problem
$$min_x |x|_1 text { subject to } |Ax-y|^2_2 le epsilon$$
can be described by the equivalent saddle point problem:
$$
min_xmax_{lambdage 0}left[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right],
$$ by Lagrange's multiplier theorem. Let $lambda_0ge 0$ be the solution of the above problem such that
$$
max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=min_xleft[ |x|_1 +lambda_0(|Ax-y|_2^2-epsilon)right].
$$If we have $lambda_0 = 0$, then the minimizer should be $x=0$ and $|y|^2_2leqepsilon.$ Let us exclude this trivial case and assume that $lambda_0>0$. By complementary slackness, we should have $|Ax_0-y|_2^2 =epsilon$ where $x_0$ is the solution of the first problem.
In the same manner, the second problem can be equivalently described as
$$
min_{x'}max_{muge 0}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]=max_{muge 0}min_{x'}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]quadcdots(*),
$$ where $t = |x_0|_1$. If we look at the formulation of the second problem carefully, we can see that the solution $(mu_0, x_0')$ should be $(frac{1}{lambda_0},x_0)$. To see this, one can plug $mu=lambda_0^{-1}$ and $x_0'=x_0$ into the $(*)$ to check if equality holds. In fact, equality holds since given $mu=lambda_0^{-1}$, $x=x_0$ is the minimizer and conversely, given $x=x_0$, $mu =lambda_0^{-1}$ is the maximizer. This establishes that Lagrangian dual problems in this case (in fact, generally) are equivalent.
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
add a comment |
$begingroup$
The first problem
$$min_x |x|_1 text { subject to } |Ax-y|^2_2 le epsilon$$
can be described by the equivalent saddle point problem:
$$
min_xmax_{lambdage 0}left[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right],
$$ by Lagrange's multiplier theorem. Let $lambda_0ge 0$ be the solution of the above problem such that
$$
max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=min_xleft[ |x|_1 +lambda_0(|Ax-y|_2^2-epsilon)right].
$$If we have $lambda_0 = 0$, then the minimizer should be $x=0$ and $|y|^2_2leqepsilon.$ Let us exclude this trivial case and assume that $lambda_0>0$. By complementary slackness, we should have $|Ax_0-y|_2^2 =epsilon$ where $x_0$ is the solution of the first problem.
In the same manner, the second problem can be equivalently described as
$$
min_{x'}max_{muge 0}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]=max_{muge 0}min_{x'}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]quadcdots(*),
$$ where $t = |x_0|_1$. If we look at the formulation of the second problem carefully, we can see that the solution $(mu_0, x_0')$ should be $(frac{1}{lambda_0},x_0)$. To see this, one can plug $mu=lambda_0^{-1}$ and $x_0'=x_0$ into the $(*)$ to check if equality holds. In fact, equality holds since given $mu=lambda_0^{-1}$, $x=x_0$ is the minimizer and conversely, given $x=x_0$, $mu =lambda_0^{-1}$ is the maximizer. This establishes that Lagrangian dual problems in this case (in fact, generally) are equivalent.
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
add a comment |
$begingroup$
The first problem
$$min_x |x|_1 text { subject to } |Ax-y|^2_2 le epsilon$$
can be described by the equivalent saddle point problem:
$$
min_xmax_{lambdage 0}left[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right],
$$ by Lagrange's multiplier theorem. Let $lambda_0ge 0$ be the solution of the above problem such that
$$
max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=min_xleft[ |x|_1 +lambda_0(|Ax-y|_2^2-epsilon)right].
$$If we have $lambda_0 = 0$, then the minimizer should be $x=0$ and $|y|^2_2leqepsilon.$ Let us exclude this trivial case and assume that $lambda_0>0$. By complementary slackness, we should have $|Ax_0-y|_2^2 =epsilon$ where $x_0$ is the solution of the first problem.
In the same manner, the second problem can be equivalently described as
$$
min_{x'}max_{muge 0}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]=max_{muge 0}min_{x'}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]quadcdots(*),
$$ where $t = |x_0|_1$. If we look at the formulation of the second problem carefully, we can see that the solution $(mu_0, x_0')$ should be $(frac{1}{lambda_0},x_0)$. To see this, one can plug $mu=lambda_0^{-1}$ and $x_0'=x_0$ into the $(*)$ to check if equality holds. In fact, equality holds since given $mu=lambda_0^{-1}$, $x=x_0$ is the minimizer and conversely, given $x=x_0$, $mu =lambda_0^{-1}$ is the maximizer. This establishes that Lagrangian dual problems in this case (in fact, generally) are equivalent.
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
$endgroup$
The first problem
$$min_x |x|_1 text { subject to } |Ax-y|^2_2 le epsilon$$
can be described by the equivalent saddle point problem:
$$
min_xmax_{lambdage 0}left[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right],
$$ by Lagrange's multiplier theorem. Let $lambda_0ge 0$ be the solution of the above problem such that
$$
max_{lambdage 0}min_xleft[ |x|_1 +lambda(|Ax-y|_2^2-epsilon)right]=min_xleft[ |x|_1 +lambda_0(|Ax-y|_2^2-epsilon)right].
$$If we have $lambda_0 = 0$, then the minimizer should be $x=0$ and $|y|^2_2leqepsilon.$ Let us exclude this trivial case and assume that $lambda_0>0$. By complementary slackness, we should have $|Ax_0-y|_2^2 =epsilon$ where $x_0$ is the solution of the first problem.
In the same manner, the second problem can be equivalently described as
$$
min_{x'}max_{muge 0}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]=max_{muge 0}min_{x'}left[ |Ax'-y|_2^2 +mu (|x'|_1-t)right]quadcdots(*),
$$ where $t = |x_0|_1$. If we look at the formulation of the second problem carefully, we can see that the solution $(mu_0, x_0')$ should be $(frac{1}{lambda_0},x_0)$. To see this, one can plug $mu=lambda_0^{-1}$ and $x_0'=x_0$ into the $(*)$ to check if equality holds. In fact, equality holds since given $mu=lambda_0^{-1}$, $x=x_0$ is the minimizer and conversely, given $x=x_0$, $mu =lambda_0^{-1}$ is the maximizer. This establishes that Lagrangian dual problems in this case (in fact, generally) are equivalent.
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
answered Dec 20 '18 at 10:38
SongSong
15.4k1736
15.4k1736
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
add a comment |
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
$begingroup$
Thanks a lot. Not sure I fully understand your answer for now but all that I get seems very logical and meaningfull. I will look further into it.
$endgroup$
– nicomezi
Dec 20 '18 at 10:50
add a comment |
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