Does regularization penalize models that are simpler than needed?
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Yes, regularization penalizes models that are more complex than needed. But does it also penalize models that are simpler than needed?
machine-learning predictive-models modeling regularization
asked Mar 31 at 11:44
alienflowalienflow
275
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Yes, regularization penalizes models that are more complex than needed. But does it also penalize models that are simpler than needed?
machine-learning predictive-models modeling regularization
asked Mar 31 at 11:44
alienflowalienflow
275
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1
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Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
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– usεr11852
Mar 31 at 12:04
add a comment |
$begingroup$
Yes, regularization penalizes models that are more complex than needed. But does it also penalize models that are simpler than needed?
machine-learning predictive-models modeling regularization
asked Mar 31 at 11:44
alienflowalienflow
275
$endgroup$
Yes, regularization penalizes models that are more complex than needed. But does it also penalize models that are simpler than needed?
machine-learning predictive-models modeling regularization
machine-learning predictive-models modeling regularization
asked Mar 31 at 11:44
alienflowalienflow
275
asked Mar 31 at 11:44
alienflowalienflow
275
asked Mar 31 at 11:44
alienflowalienflow
275
asked Mar 31 at 11:44
alienflowalienflow
275
asked Mar 31 at 11:44
alienflowalienflow
275
275
1
$begingroup$
Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
$endgroup$
– usεr11852
Mar 31 at 12:04
add a comment |
1
$begingroup$
Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
$endgroup$
– usεr11852
Mar 31 at 12:04
1
1
$begingroup$
Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
$endgroup$
– usεr11852
Mar 31 at 12:04
$begingroup$
Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
$endgroup$
– usεr11852
Mar 31 at 12:04
add a comment |
1 Answer
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For regularization terms similar to $left|thetaright|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.
Error terms such as $sum_i left|y_i - f_{theta}(x_i)right|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $theta = 0$, leads to a high error.
We balance these two forces by using a regularization parameter ($lambda$) in a summation like
$$frac{1}{N}sum_{i=1}^{N} left|y_i - f_{theta}(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
where higher $lambda$ forces the model toward more simplicity.
answered Mar 31 at 12:00
EsmailianEsmailian
42615
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So, regularization like L2, L1 correspond to the first case, right?
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– alienflow
Mar 31 at 12:05
1
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@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For regularization terms similar to $left|thetaright|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.
Error terms such as $sum_i left|y_i - f_{theta}(x_i)right|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $theta = 0$, leads to a high error.
We balance these two forces by using a regularization parameter ($lambda$) in a summation like
$$frac{1}{N}sum_{i=1}^{N} left|y_i - f_{theta}(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
where higher $lambda$ forces the model toward more simplicity.
answered Mar 31 at 12:00
EsmailianEsmailian
42615
$endgroup$
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
1
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
add a comment |
$begingroup$
For regularization terms similar to $left|thetaright|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.
Error terms such as $sum_i left|y_i - f_{theta}(x_i)right|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $theta = 0$, leads to a high error.
We balance these two forces by using a regularization parameter ($lambda$) in a summation like
$$frac{1}{N}sum_{i=1}^{N} left|y_i - f_{theta}(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
where higher $lambda$ forces the model toward more simplicity.
answered Mar 31 at 12:00
EsmailianEsmailian
42615
$endgroup$
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
1
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
add a comment |
$begingroup$
For regularization terms similar to $left|thetaright|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.
Error terms such as $sum_i left|y_i - f_{theta}(x_i)right|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $theta = 0$, leads to a high error.
We balance these two forces by using a regularization parameter ($lambda$) in a summation like
$$frac{1}{N}sum_{i=1}^{N} left|y_i - f_{theta}(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
where higher $lambda$ forces the model toward more simplicity.
answered Mar 31 at 12:00
EsmailianEsmailian
42615
$endgroup$
For regularization terms similar to $left|thetaright|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.
Error terms such as $sum_i left|y_i - f_{theta}(x_i)right|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $theta = 0$, leads to a high error.
We balance these two forces by using a regularization parameter ($lambda$) in a summation like
$$frac{1}{N}sum_{i=1}^{N} left|y_i - f_{theta}(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
where higher $lambda$ forces the model toward more simplicity.
answered Mar 31 at 12:00
EsmailianEsmailian
42615
edited Mar 31 at 12:10
answered Mar 31 at 12:00
EsmailianEsmailian
42615
answered Mar 31 at 12:00
EsmailianEsmailian
42615
answered Mar 31 at 12:00
EsmailianEsmailian
42615
42615
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
1
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
add a comment |
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
1
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
$begingroup$
So, regularization like L2, L1 correspond to the first case, right?
$endgroup$
– alienflow
Mar 31 at 12:05
1
1
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
$begingroup$
@alienflow yes they all force toward zero (most simple).
$endgroup$
– Esmailian
Mar 31 at 12:06
add a comment |
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Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1)
$endgroup$
– usεr11852
Mar 31 at 12:04