Does regularization penalize models that are simpler than needed? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Different regularization parameter per parameterWhy does not ridge regression perform feature selection although it makes use of regularization?Do discriminative models overfit more than generative models?Regularization for ARIMA modelsWhy do smaller weights result in simpler models in regularization?Why regularize all parameters in the same way?What are Regularities and Regularization?Are there empirical models that predict variance?Does regularization leads to stucking in local minima?Is there a theoretical reason why simple models perform better than complex models on time series forecasting tasks?
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Does regularization penalize models that are simpler than needed?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Different regularization parameter per parameterWhy does not ridge regression perform feature selection although it makes use of regularization?Do discriminative models overfit more than generative models?Regularization for ARIMA modelsWhy do smaller weights result in simpler models in regularization?Why regularize all parameters in the same way?What are Regularities and Regularization?Are there empirical models that predict variance?Does regularization leads to stucking in local minima?Is there a theoretical reason why simple models perform better than complex models on time series forecasting tasks?
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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
$endgroup$
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
$endgroup$
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 |
$begingroup$
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
1
active
oldest
votes
$begingroup$
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
$$frac1Nsum_i=1^N left|y_i - f_theta(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
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 |
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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
$$frac1Nsum_i=1^N left|y_i - f_theta(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
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
$$frac1Nsum_i=1^N left|y_i - f_theta(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
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
$$frac1Nsum_i=1^N left|y_i - f_theta(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
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
$$frac1Nsum_i=1^N left|y_i - f_theta(x_i)right|_2^2 + lambdaleft|thetaright|_2^2,$$
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